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CA Foundation Maths Solutions For Chapter 4 Time Value of Money

July 5, 2024July 5, 2024 by Marksparks

Time Value of Money Introduction:

People can’t make money or spend it on housing. food, clothing, education, entertainment, etc. Sometimes extra expenditures have also to be met. For example, there might be a marriage in the family; one may want to line house, one may want to set up his or her business, one may want to have a car, and so on.

Some people can manage to put aside some money for such expected and unexpected expenditures. But most people have to borrow money for such contingencies.
From where they can borrow money? Money can be borrowed from friends money lenders or Banks.
If you can arrange a loan from your friend it might be interest-free but if you borrow money from lenders or Banks you will have to pay some charge periodically for using money from money lenders or Banks. This charge is called interest.
Interest can be defined as the price paid by a borrower for the use of a lender’s money.

Why is Interest Paid?

Time value of money: The time value of money means that the value of the unity of money is different in different periods. The sum of money received in the future is less valuable than it is today. In other words, the present worth of money received after some time will be less than the money received today. Since money received today has more value rational investors would prefer current receipts to future receipts. If they postpone their receipts, they will certainly charge some money i.e., interest.
Opportunity Cost: The lender has a choice between using his money in different investments. If he chooses one, he forgoes the return from all others. In other words, lending incurs an opportunity cost due to the possible alternative uses of the lent money.
Inflation: Most economies generally exhibit inflation. Inflation is a fall in the purchasing power of money. Due to inflation a given amount of money buys fewer goods in the future than it will now. The borrower needs to compensate the lender for this.
Liquidity Preference: People prefer to have their resources available in a form that can immediately be converted into cash rather than a form that takes time or money to realize.
Risk Factor: There is always a risk that the borrower will go bankrupt or otherwise default on the loan. Risk is a determinable factor in fixing the rate of interest

Read and Learn More CA Foundation Maths Solutions

Simple Interest

If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called Simple Interest.
Simple interest is the interest computed on the principal for the entire period of borrowing.
It is calculated on the outstanding principal balance and not on interest previously earned. It means no interest is paid on interest earned during the term of the loan.

Simple Interest – Important Facts and Formulae:

Principal: The money borrowed or lent out for a certain period is called the principal or the sum.

Interest: Extra money paid for using other’s money is called interest.

Simple Interest (S.I.): If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.), and Time = T years. Them

⇒ $\text { S.I. }=\left(\frac{P \times R \times T}{100}\right)$

⇒ $\mathrm{P}=\left(\frac{100 \times 5 . I}{R \times T}\right) $

⇒ $\mathrm{R}=\left(\frac{100 \times 5 . I}{P \times T}\right)$

⇒ $\mathrm{T}=\left(\frac{100 \times 5 . I}{P \times R}\right)$

A = Accumulated amount

[Final value of investment]

P = Principal. [Initial value of an investment]

r= Rate of interest

t = time (years.]

I = Amount interest

A = P + I

I = A – P

⇒ $\mathrm{A}=\mathrm{P}\left(1+\frac{1 t}{100}\right)$

Solved Sums with Solution:

Question 1. Find the simple interest on Rs. 651000 at $16 \frac{2}{3} \%$per annum for 9 months.
Solution:

⇒ $\mathrm{P}=\mathrm{Rs} 68000, \mathrm{R}=\frac{50}{3} \% \text { p.a and } \mathrm{T}=\frac{9}{12} \text { Years }=\frac{3}{4} \text { years. }$

⇒ $\mathrm{S} .1 .=\left(\frac{P \times R \times S}{100}\right)=R s .\left(68000 \times \frac{50}{3} \times \frac{3}{4} \times \frac{1}{100}\right)=R s .8500 $

Question 2. Find the simple interest on Rs. 3000 at 6ÿ% per annum for the period from 4″’ Fell., 2005 to 18 th April, 2005.
Solution:

⇒ $\text { Time }=(24+31+18) \text { days= } 73 \text { days }=\frac{73}{365} \text { year }=\frac{1}{5} \text { year. }$

⇒ $P=\text { Rs. } 3000 \text { and } R=6 \frac{1}{4} \% \text { p.a. }=\frac{25}{4} \% \text { p.a. }$

⇒ $\text { S.I. }=\text { Rs. Rs. }\left(3000 \times \frac{25}{4} \times \frac{1}{5} \times \frac{1}{100}\right)=\text { Rs. } 37.50 $

Remark: the day on which money is deposited is not counted while the day on which money is withdrawn is counted.

Question 3. A sum at simple interest at 131/2% per annum amounts to Rs. 2502.50 after 4 years, find the sum.
Solution:

Let the sum be Rs. X then, S.l. = Rs$\left(x \times \frac{27}{2} \times 4 \times \frac{1}{100}\right)={Rs} . \frac{27 x}{50}$

⇒ $\frac{77 x}{50}=250250 \Leftrightarrow x=\frac{250250 \times 50}{77}=1625 .$

Question 4. A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to how much?
Solution:

S.l. =Rs. (920 – 800) = Rs. 120; P=Rs. 800, T= 3 yrs.

⇒ $\mathrm{R}=\left(\frac{100 \times 120}{800 \times 3}\right) \%=5 \% $

⇒ $\text { New rate }=(5+3) \%=8 \%$

⇒ $\text { New S.I. }=\text { Rs. }\left(\frac{800 \times 8 \times 3}{100}\right)=\text { Rs. } 192$

New amount – Rs. (800+192) = Rs. 992.

Question 5. Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond five years, If he pays a total interest of Rs. 11,400 at the end of nine years, how much money did he borrow?
Solution:

Let the sum borrowed be x. then,

⇒ $\left(\frac{x \times 6 \times 2}{100}\right)+\left(\frac{x \times 9 \times 3}{100}\right)+\left(\frac{x \times 14 \times 4}{100}\right)=11400$

⇒ $\Leftrightarrow\left(\frac{3 x}{25}+\frac{27 x}{100}+\frac{14 x}{25}\right)=11400 \Leftrightarrow \frac{95 x}{100}=11400 \Leftrightarrow\left(\frac{11400 \times 100}{95}\right)=12000$

Hence, sum borrowed = Rs. 12,000

Question 6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs. 1164 in 31/2 years find the sum and the rate of interest.
Solution:

⇒ $\text { S.I. for } 1 \frac{1}{2} \text { years }=\text { Rs. }(1164-1008)=\text { Rs. } 156 .$

⇒ $\text { S.I. for } 2 \text { years }=\text { Rs. }\left(156 \times \frac{2}{3} \times 2\right)=\text { Rs. } 208 \text {. }$

⇒ $\text { Principal= Rs. }(1008-208)=\text { Rs. } 800 .$

⇒ $\text { Now, } P=800, T=2 \text { and S.I. }=208$

⇒ $\text { Rate }=\left(\frac{100 \times 208}{800 \times 2}\right) \%=13 \% .$

Question 7. At what rate present per annum will a sum of money double in 16 years?
Solution:

Let principals P. then, S.l. = P and T = 16yrs.

⇒ $\text { Rate }=\left(\frac{100 \times P}{P \times 16}\right) \%=6 \frac{1}{4} \% p . a .$

Question 8. The simple interest on a sum of money is $\frac{4}{9}$ of the principal, find the rate present and time, if both are numerically equal.
Solution:

Let sum =Rs. x. then, S.L =Rs.$\frac{4 x}{9}$

Let rate = R % and time = R years.

Then$\left(\frac{x \times R \times R}{100}\right)=\frac{4 x}{9} \text { or } R^2=\frac{400}{9} \text { or } R=\frac{20}{3}=6 \frac{2}{3}$

⇒ $\text { Rate }=6 \frac{2}{3} \% \text { and time }=6 \frac{2}{3} y \text { rs }=6 \text { yrs } 8 \text { Months. }$

Question 9. The simple interest on a certain sum of money for 2ÿ years at 12% per annum is Rs.40 less than the simple interest on the same sum for 3 years at 10% per annum. Find the sum.
Solution:

Let the sum be Rs. X. then,$\left(\frac{x \times 10 \times 7}{100 \times 2}\right)-\left(\frac{x \times 12 \times 5}{100 \times 2}\right)=40$.

⇒ $\frac{7 x}{20}-\frac{3 x}{10}=40 \Leftrightarrow x=(40 \times 20)=800 \text {. }$

Hence, the sum is Rs.800.

Question 10. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.
Solution:

Let sum= P and original rate= R. then,$\left[\frac{P \times(R+2) \times 3}{100}\right]-\left[\frac{P \times R \times 3}{100}\right]=360$

3PR + 6P – 3PR =36000« 6P =36000o P =6000

Hence, sum =Rs. 6000.

Exercise – 1

Question 1. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest; He immediately lends it to another person at $6 \frac{1}{4} \%$ p.a. for 2 years. Find his gain in the transaction per year.

Rs. 112.50
Rs. 125
Rs. 150
Rs. 29

Solution:

⇒ $ \text { Gain in } 2 \text { yrs. }=\text { Rs. }\left[\left(5000 \times \frac{25}{1} \times \frac{2}{100}\right)-\left(\frac{5000 \times 4 \times 2}{100}\right)\right]=\text { Rs. }(625.400)=\text { Rs. } 225$

⇒ $\text { Gain in } 1 \text { year }=\text { Rs. }\left(\frac{225}{2}\right)=\text { Rs. } 112.50 .$

Question 2. How much time will it take for an amount of Rs. 450 to yield Rs.81 as interest at 4.5% per annum of simple interest?

3.5 years
4 years
4.5 years
5 years

Solution:

⇒ $\text { Time }=\left(\frac{100 \times 81}{150 \times 4.5}\right) \text { years }=4 \text { years. }$

Question 3. A sum of Rs. 12,500 amounts to Rs.15, 500 in 4 years at the rate of simple interest what is the rate of interest?

5%
$6 \frac{1}{4} \%$
$6 \frac{1}{4} \%$
$6 \frac{3}{4} \%$

Solution:

S.l.= Rs. (15500-12500] =Rs.3000.

⇒ $\text { Rate }=\left(\frac{100 \times 3000}{12500 \times 4}\right) \%=6 \% \% .$

Question 4. Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

3.6
6
Cannot be determined
None of these

Solution:

Let rate = R% and time = R years. Then,

⇒ $\left(\frac{1200 \times R \times R}{100}\right)=432 \Leftrightarrow \quad 12 R^2=432 \quad \Leftrightarrow \quad R^2=\quad 36 \Leftrightarrow \mathrm{R}=6$

Question 5. A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs.5400 interest only for the period. The principal amount borrowed by him was:

Rs 2000
Rs 10,000
Rs 15,000
Rs. 20,000

Solution:

⇒ $\text { Principal }=\text { Rs. }\left(\frac{100 \times 5100}{12 \times 3}\right)=\text { Rs. } 15000 .$

Question 6. What is the present worth of Rs.132 due in 2 years at 5% simple interest per annum?

Rs. 112
Rs. 118.80
Rs. 15,000
Rs. 122

Solution:

Let the present worth be Rs. x, then, S.l. =Rs. (132-x).

⇒ $ \left(\frac{x \times 5 \times 2}{100}\right) 132-x \Leftrightarrow 10 x=13200-100 x \Leftrightarrow 110 x=13200 \Leftrightarrow x=120 $

Question 7. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p. c. p.a. in 5 years. What is the sum?

Rs. 4462.50
Rs. 8032.50
Rs. 8900
Rs. 8925
None of these

Solution:

⇒ $\text { principal }=\text { Rs. }\left(\frac{100 \times 4016.25}{9 \times 5}\right)={Rs} .\left(\frac{401625}{45}\right)={Rs} .8925 .$

Question 8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest, if the rate ofinterest is increased by 4%, what amount will Rs. 800 become in 3 years?

Rs. 1020.80
Rs 1025
Rs. 1052
Data inadequate
None of these

Solution:

S.l. = Rs. (956-800) = Rs. 1 56.

⇒ $\text { Rate }=\left(\frac{100 \times 156}{800 \times 3}\right) \%=6 \frac{1}{2} \%$

⇒ $\text { New rate }=\left(6 \frac{1}{2}+4\right) \%=10 \frac{1}{2} \%$

⇒ $\text { New S.I. }=\text { Rs. }\left(800 \times \frac{21}{2} \times \frac{3}{100}\right)=\text { Rs. } 252$

New amount =Rs.(800+252)= Rs.1052

Question 9. A certain amount earns a simple interest of Rs.1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?

Rs. 35
Rs. 245
Rs. 350
Cannot be determined
None of these

Solution:

We need to know the S.I. principal and time to find the rate since the principal is not given, so data is inadequate.

Question 10. In how many years, Rs. 150willproduce the same interest @ 8%as Rs.800 produce in 3 years $\text { @ } 4 \frac{1}{2} \% \text { ? }$

Solution:

P= Rs 800, R=4ÿ% = %, T = 3 years, then,

$\text { S.I. }=\text { Rs. }\left(800 \times \frac{9}{2} \times \frac{3}{100}\right)=R s .108 .$

Now, P =Rs. 150, S.l. = Rs. 108, R= 8%

⇒ $\text { Time }=\left(\frac{100 \times 108}{150 \times 8}\right) \text { years }=9 \text { years. }$

Question 11. A sum invested at 5% simple interest per annum grows to Rs. 504 in 4 years. The same amount at 10% simple interest per annum in 2ÿyears will grow to:

Rs. 420
Rs. 450
Rs. 525
Rs. 550

Solution:

let the sum be Rs. X. then, S.l. = Rs.(504-x).

⇒ $\left(\frac{x \times 5 \times 4}{100}\right)=504 \cdot \mathrm{x} \Leftrightarrow 20 \mathrm{x}=50400-100 \mathrm{x} \Leftrightarrow 120 \mathrm{x}=50400 \Leftrightarrow \mathrm{x}=420$

Now, P= Rs. 420, R= 10%. $\mathrm{T}=\frac{5}{2} \text { years. }$

⇒ ${\text { S.I. }=Rs} .\left(\frac{420 \times 10}{100} \times \frac{5}{2}\right)={Rs} .105$

Amount = Rs, (420+105] = Rs. 525.

Question 12. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 6 years and that for 9 years?

1 : 3
1: 4
2 : 3
Data inadequate
None Of these

Solution:

Let the principal boI’ ami rale ol interest be R%

Required ratio $=\left\lfloor\frac{\frac{P \times R \times 6}{100}}{\frac{P\times R\times 9}{100}}\right\rfloor=\frac{6 PR}{9 P R}=\frac{6}{9}=2: 3$

Question 13. Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years, and 13% p.a. for the period beyond eight years, If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow?

Rs. 8000
Rs. 10,000
12,000
Data inadequate

Solution:

let the sum to Its. X. Then,

⇒ $\left(\frac{x \times 6 \times 3}{100}\right)+\left(\frac{x \times 9 \times 5}{100}\right)+\left(\frac{x \times 13 \times 3}{100}\right)=8160$

18x+45x+39x=(8 1 60 x 100) <=> 102x = 816000 <=> x = 8000.

Question 14. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10% the effective rate of interest becomes:

10%
10. 25%
10.5%
None of these

Solution:

Let the sum be Rs. 100 then

S.I. for first 6 months Rs.$\text { Rs. }\left(\frac{100 \times 10 \times 1}{100 \times 2}\right)={Rs} .5$

S.l. for last 6 months= Rs.$\left(\frac{105 \times 10 \times 1}{100 \times 2}\right)=Rs .5 .25$

So, the amount at the end of 1 year is Rs. (100+5+5.25)= Rs.110.25.

Effective rate = (110.25-100)= 10.25 %

Question 15. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs.854 in 4 years. The sum is:

Rs. 650
Rs. 690
Rs. 698
Rs 700

Solution:

S.l. for 1 year= RS. (854-815)=Rs.39.

S.l. for 3 Years= RS.(39 x 3) = Rs.117.

Principal = Rs.( 815 -117) = Rs. 698.

Question 16. A sum of money lent out at simple interest amounts to Rs. 720 after 2 years and to Rs. 1020 after a further period of 5 years. The sum is:

Rs. 500
Rs. 600
Rs. 700
Rs. 710

Solution:

S.I. for 5 years RS. (1020-720)=Rs.300

S.I. for 2 Years=$\text { RS. }\left(\frac{300}{5} \times 2\right)={Rs} .120$

Principal = Rs.( 720 -120) = Rs. 600.

Question 17. A sum of money amorist to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest, the rate of interest per annum is:

Solution:

S.I. for 3 years RS. (12005-9800)=Rs.2205

⇒ $\text { S.I. for } 5 \text { Years }=\text { RS. }\left(\frac{2205}{3} \times 5\right)=\text { Rs. } 3675 \text {. }$

⇒ $\text { Principal }=\text { Rs. }(9800-3675)=\text { Rs. } 6125 \text {. }$

⇒ $\text { Hence, rate }=\left(\frac{100 \times 3675}{6125 \times 5}\right) \%=12 \%$

Question 18. At what rate present of simple interest will a sum of money double itself in 12 years?

$ 8 \frac{1}{4} \%$
$ 8 \frac{1}{3} \%$
$ 8 \frac{1}{2} \%$
$ 9 \frac{1}{2} \%$

Solution:

Letsum=x. then, S.l. = x

⇒ $\text { Rate }=\left(\frac{100 \times S . J}{P \times T}\right)=\left(\frac{100 \times x}{x \times 12}\right) \%=\frac{25}{3} \%=8 \frac{1}{3} \%$

Question 19. At what rate present per annum will the simple interest sum of money be $\frac{2}{5}$ of the amount in 10 years?

4%
$5 \frac{2}{3} \%$
6%
$6 \frac{2}{3} \%$

Solution:

Let sum= x. then, S.I. =$\frac{2 x}{5} \text { Time }=10 \text { Years. }$

⇒ $\text { Rate }=\left(\frac{100 \times 2 x}{x \times 5 \times 10}\right) \%=4 \%$

Question 20. In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?

$1 \frac{1}{4}years$
$1 \frac{3}{4}years$
$2 \frac{1}{4}years$
$2 \frac{3}{4}years$

Solution:

Let sum= x. then, S.l. = 0.125x=$\frac{1}{8} x, R=10 \%$

⇒ $\text { Time }=\left(\frac{100 \times x}{x \times 8 \times 10}\right) \text { Years }=\frac{5}{4} \text { Years }=1 \frac{1}{4} \text { Years }$

Question 21. A sum of money becomes of itself in 3 years at a certain rate of simple interest the rate annum is:

$5 \frac{5}{9} \%$
$ 6 \frac{5}{9} \%$
18%
25%

Solution:

Let sum= x. then, Amount=$\frac{7 x}{6}, \mathrm{~S} . \mathrm{I}=\left(\frac{7 x}{6}-x\right)=\frac{x}{6} ;$

Time = 3 Years

⇒ $\text { Rate }=\left(\frac{100 \times x}{x \times 6 \times 3}\right) \%=\frac{50}{9} \%=5 \frac{5}{9} \%$

Question 22. Simple Interest on a certain amount is $\frac{9}{16}$of the principal. If the numbers representing principal are lent out, is:

$5 \frac{1}{2}years$
$ 6 \frac{1}{2}years$
7 years
$7 \frac{1}{2}years$

Solution:

let sum = x. tlum, S.l. $=\frac{9}{16} x$

Let rate= R % and Time = R years

⇒ $\left(\frac{x \times R \times R}{100}\right)=\frac{9 x}{16} \Leftrightarrow R^2=\frac{900}{16} \Leftrightarrow R=\frac{30}{4}=7 \frac{1}{2}$

Hence, time = $7 \frac{1}{2} \text { Years. }$

Question 23. A lends Rs. 2500 to B and a certain sum to C at the same time at 7% p.a. simple Interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is:

Rs. 700
Rs. 1500
Rs. 4000
Rs. 6500

Solution:

Let the sum lent to C he Rs. X. then,$\left(\frac{2500 \times 7 \times 4}{100}\right)+\left(\frac{x \times 7 \times 4}{100}\right)=1120$

⇒ $ \Leftrightarrow \quad \frac{7}{25} x=(1120-700) \Rightarrow x=\left(\frac{420 \times 25}{7}\right)=1500$

Question 24. A lent Rs.5000 for 2 years and Rs. 300 for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:

5%
7%
71 %
10%

Solution:

Let the rate be R% p.a. then,$\left(\frac{5000 \times R \times 2}{100}\right)+\left(\frac{3000 \times R \times 4}{100}\right)=2200$

⇒ $\Leftrightarrow \quad 100 R+120 R=2200 \Leftrightarrow R=\left(\frac{2200}{220}\right)=10 \text { Rate }=10 \%$

Question 25. A sum of Rs 725 is lent at the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former at the end of the year, Rs.33.50 is earned as interest from both the loans. What was the original rate of interest?

3.6%
4.5%
5%
6%
None

Solution:

Let the original rate be R% then, the new rate = (2R) %.

⇒ $\left(\frac{725 \times R \times 1}{100}\right)+\left(\frac{36250 \times 2 R \times 1}{100 \times 3}\right)=33.50$

(2175+725) R= 33.50x 100 x 3=10050 $\Leftrightarrow \quad R=\frac{10050}{2900}=3.46$

Original rate = 3.46%

Question 26. The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50 the difference between their rates of interest is:

0.1%
0.2%
0.3%
0.4%
None

Solution:

⇒ $\left(\frac{1500 \times R_1 \times 3}{100}\right)-\left(\frac{1500 \times R_2 \times 3}{100}\right)=13.50$

4500 (R1 – R2) = 1350o (R1 – R2)=$=\frac{1350}{4500}=0.3 \%$

Question 27. Peter invested an amount of 12000 at the rate of 10 p.c.p.a. simple interest and another amount at the end of 20 p.c.p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14 p.c.p.a. Find the total amount invested.

Rs. 20,000
Rs. 22,000
Rs 24,000
Rs. 25,000
None

Solution:

Let the second amount be RS. X. Then,

⇒ $\left(\frac{12000 \times 10 \times 1}{100}\right)+\left(\frac{x \times 20 \times 1}{100}\right)=\left[\frac{(12000+x) \times 14 \times 1}{100}\right]$

12000+20x=1 68000+14x ⇔ 6x= 48000 ⇔ x = 8000

Total investment = Rs. (12000+8000) = Rs. 20000

Question 28. If the annual rate of simple interest increases from 10 % to $12 \frac{1}{2}$ % a man’s yearly income increases by Rs. 1250 His principal (in Rs.) is:

45,000
50,000
Rs. 24,600
Rs. 26,000

Solution:

Let the sum be Rs. X, then,$\left(x \times \frac{25}{2} \times \frac{1}{100}\right)-\left(\frac{x \times 10 \times 1}{100}\right)=1250$

⇔ 25x-20x = 250000 ⇔ 5x = 250000 o x = 50000

Question 29. A moneylender finds that due to a fall in the annual rate of interest from 8% to 7- %, his 4 yearly income diminishes by Rs. 61.50 His capital is:

Rs. 22,400
Rs. 23,800
Rs. 24,600
Rs. 26,000

Solution:

Let the capital be Rs. X, then,$\left(\frac{x \times 8 \times 1}{100}\right)-\left(x \times \frac{31}{4} \times \frac{1}{100}\right)=61.50$

32x- 31x = 6150x 4 o x = 24600

Question 30. The price of a T.V. set worth Rs. 20,000 is to be paid in 20 installments of Rs. 1000 each. If the rate of interest is 6 per annum, and the first installment is paid at the time of purchase, then the value of the last instalment covering the interest as well will be:

Rs. 1050
Rs. 2050
Rs. 3000
None of these

Solution:

Money paid in cash = Rs. 1000. Balance payment = Rs. (20000-1000) = Rs. 19000.

Question 31. Mr. Thomas invested an amount of Rs. 13900 divided into two different schemes A and B at the simple interest earned in 2 years Rs. 3508, what was the amount invested in scheme B?

Rs. 6400
Rs. 6500
7200
Rs. 7500
None

Solution:

Let the sum invested in scheme A be Rs. X and that in scheme B be Rs. (13900 – x).

⇒ $\text { Then, }\left(\frac{x \times 14 \times 2}{100}\right)+\left[\frac{13900-x) \times 11 \times 2}{100}\right]=3508$

28x- 22x = 350800 – (13900 x 22) cs 6x = 45000 ⇔ x = 7500.

Question 32. An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is, 9% then the amount invested in each share was:

Rs. 52,500; Rs. 47,500
Rs. 62,500; Rs. 37,500
Rs. 72,500; Rs. 27,500
Rs. 82,500; Rs. 17,500

Solution:

Let the sum invested at 9% be Rs. X and that invested at 11% be Rs. (100000 – x)

Then, $\left(\frac{x \times 9 \times 1}{100}\right)+\left[\frac{(100000-x) \times 11 \times 1}{100}\right]=\left(100000 \times \frac{39}{4} \times \frac{1}{100}\right)$

⇒ $\Leftrightarrow \frac{9 x+1100000-11 x}{100}=\frac{39000}{4}=9750$

2x = (1100000-975000) = 125000 ox = 62500.

The sum invested at 9%= Rs. 62500. The sum invested at 1 1% = Rs. (100000- 62500) = Rs. 37500.

Question 33. David invested a certain amount in three different schemes A, B, and C with the rate of interest of 10 % p.a. 12% p.a., and 15% P.a. Respectively. If the total interest. Accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme B.

Rs. 5000
Rs. 6500
Rs. 8000
Cannot be determined
None of these

Solution:

let x,y, and z be the amounts invested in schemes A, B, and C respectively. Then,

⇒ $\left(\frac{x \times 10 \times 1}{100}\right)+\left(\frac{y \times 12 \times 1}{100}\right)+\left(\frac{z \times 15 \times 1}{100}\right)=3200$

10x+12y+15z= 320000

Now, z=240% of y$=\frac{12}{5} y$

And,z= 150% of x$x=\frac{3}{2} x \Rightarrow x=\frac{2}{3} z=\left(\frac{2}{3} \times \frac{12}{5}\right) y=\frac{8}{5} y$

From (1), (2) and (3), we have

16y+ 12y + 36y= 320000 o 64y = 320000 ⇔y = 5000

Question 34. A person invested in all Rs. 2600 at 4%, 6%, and 8% per annum simple interest, and at the end of the year, he got the same interest in all three cases, the money invested at 4% is:

Rs. 200
Rs. 600
Rs. 800
Rs. 1200

Solution:

Let the parts be x, y, and [2600-(x + y)]. Then,

⇒ $\frac{x \times 4 \times 1}{100}=\frac{y \times 6 \times 1}{100}=\frac{[2600-(x+y)] \times 8 \times 1}{100}$

⇒ $\frac{y}{x}=\frac{4}{6}=\frac{2}{3} \text { or } y=\frac{2}{3} x$

⇒ $So, \frac{x \times 4 \times 1}{100}=\frac{\left(2600-\frac{5}{3} x\right) \times 8}{100}$

⇒ $\Rightarrow 4 x=\frac{(7800-5 x) \times 8}{3} \Leftrightarrow 52 x=(7800 \times 8) \Leftrightarrow x=\left(\frac{7800 \times 8}{52}\right)=1200 .$

Money invested at4%= Rs.1200

Question 35. Divide Rs. 2379 into 3 parts so that their amounts after 2, 3, and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is:

Rs. 759
Rs. 792
Rs. 818
Rs. 828

Solution:

Let the parts be x, y and [2379 – (x +y)]

⇒ $\mathrm{X}+\left(x \times 2 \times \frac{5}{100}\right)=y+\left(y+3 \times \frac{5}{100}\right)=z+\left(z \times 4 \times \frac{5}{100}\right)$

⇒ $\Rightarrow \frac{11 x}{10}=\frac{23 y}{20}=\frac{6 \mathrm{z}}{5}=\mathrm{k} \quad \Rightarrow \quad \mathrm{x}=\frac{10 \mathrm{k}}{11}, \mathrm{y}=\frac{20 \mathrm{k}}{23}, \mathrm{z}=\frac{5 \mathrm{k}}{6}$

⇒ $ text { But } \mathrm{x}+\mathrm{y}+\mathrm{z}=2379 $

⇒ $ \Rightarrow \frac{10 \mathrm{k}}{11}+\frac{20 \mathrm{k}}{23}+\frac{5 \mathrm{k}}{6}=2379 \quad \Rightarrow \quad 1380 \mathrm{k}+1320 \mathrm{k}+1265 \mathrm{k}=2379 \times 11 \times 23 \times 6 $

⇒ $ \mathrm{k}=\frac{2379 \times 11 \times 23 \times 6}{3965}=\frac{3 \times 11 \times 23 \times 6}{5} $

⇒ $\mathrm{x}=\left(\frac{10}{11} \times \frac{3 \times 11 \times 23 \times 6}{5}\right)=828$

Hence, the first part is Rs. 828

Question 36. Two equal sums of money were lent at simple interest at ll%p.a. for 3 years and 4 years respectively.If the difference in interest for two periods was ₹ 412.50, then each sum is:

₹3,250
₹3,500
₹ 3.750
₹4,350

Solution:

The difference in is due to the time

Rate of interest for the whole

Duration = [11×4.5 – 11×3.5) = 11%

⇒ $P=\frac{\text { Total S.I }}{\text { Interest on ₹1 }}=\frac{412.50}{0.11}=₹ 3750$

Question 37. In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?

$1 \frac{1}{4}$
$1 \frac{3}{4}$
$2\frac{1}{4}years$
$2 \frac{3}{4}years$

Solution:

⇒ $\mathrm{t}=\frac{1 / P}{r \%}=\frac{0.125}{0.10}=1.25 \mathrm{yrs}$

Details: S . I$ =\frac{p . r. t}{100} $

0r ;$ 0.125 \mathrm{P}=\frac{P, 10 \times t}{100}$

Or t = 0.125×10 = 1.25 yrs

Question 38. Find the number of years in which a sum doubles itself at the rate of 8% per annum.

$11 \frac{1}{2}$
$12 \frac{1}{2}$
$9 \frac{1}{2}$
$13 \frac{1}{2}$

Solution:

⇒ $text { Tricks: } \mathrm{t}=\frac{(x-1) \times 100}{r}$

⇒ $=\frac{(2-1) \times 100}{B}=12.5 \mathrm{yrs} $

Question 39. The time by which a sum of money is 8 times its self if it doubles itself in 15 years.

42 years
43 years
45 years
46 years

Solution:

It is compound interest Qts.

2t2 =815

Or 212 = 815 Or 212 = (23)15 : tz = 45 yrs

Question 40. What is the rate of simple interest if a sum of money amount ₹2,784 in 4 years and ₹2,688 in. 3 years?

l%p.a.
4% p.a.
5% p.a.
8% p.aa

Solution:

⇒ $\text { S.l. pa }=\frac{\text { difference in s.t }}{\text { difference in time }}$

⇒ $\frac{s t_2-s t_1}{t_2-t_1}=\frac{2784-2688}{4-3}=90$

Principal = 1(20(0) – 3 x 00) = 2400

⇒ $\mathrm{r}=\frac{t \times 100}{p \times t}=\frac{96 \times 100}{2400 \times 1}=4 \%$

Question 41. If a simple interest on a sum of money at 6% p.a. for 7 years is equal to twice of simple interest on another sum for 9 years at 5% p.a. the ratio will be:

2:15
7:15
15:7
1:7

Solution:

⇒ $P_1 \frac{6.7}{100}=2 \times \frac{P_{2 .} .9 .5}{100} $

⇒ $\text { Or } \frac{P_1}{P_2}=2 \times \frac{9 \times 5}{6 \times 7}=\frac{15}{7} \Rightarrow \frac{P_1}{P_2}=\frac{15}{7}$

Question 42. By mistake, a clerk, calculated the simple interest on the principal for 5 months at 6.5% p.a. instead of 6 months at 5.5% p.a. if the calculation error was ₹25.40. The original sum of principal was____.

₹60,690
₹60,960
₹90,660
₹90690

Solution:

⇒ $P=\frac{25.40}{\frac{5.5}{100} \times \frac{6}{12}-\frac{6.5}{100} \times \frac{5}{12}}$

⇒ $\frac{25.40 \times 1200}{5.5 \times 6-6.5 \times 5}=₹ 60,960 $

Question 43. If the simple interest on ₹1,400 for 3 years is less than the simple interest on ₹1,800 for the same period by ₹80, then the rate of interest is:

5.67%
6.67%
7.20%
5.00%

Solution:

⇒ $ r=\frac{80 \times 100}{(1800-1400) \times 3}$

= 6.67%

Question 44. The S.l on a sum of money is $ \frac{4}{9}$ of the principal and the No. of years is equal to the rate of interest per annum. Find the rate of interest per annum.

5%
20/3%
22/7%
6%

Solution:

⇒ $\text { S.I. }=\frac{p . r. r}{100} \Rightarrow \frac{1}{9} p .=p \cdot\left(\frac{r}{10}\right)^2 $

⇒ $\frac{r}{10}=\frac{2}{3}r=\frac{20}{3} \%$

Question 45. Simple interest on *2,000 for 5 months at 16% p.a. is_.

₹133.33
₹133.26
₹134.00
₹132.09

Solution:

⇒ $\text { S.I. }=2000 \times \frac{5}{12} \times \frac{16}{100}=₹ 133.33$

Question 46. How much investment is required to yield an Annual income of ₹420 at 7% p.a. simple interest.

₹6,000
₹6,420
₹5,580
₹5,000

Solution:

⇒ $\mathrm{P}=\frac{420 \times 100}{7 \times 1}=₹ 6000$

2 GBC: P= 420+7% button = 16000

Question 47. Mr. X invests ₹90,500 in the post office at 7.5% p.a. simple interest. While calculating the rate was wrongly taken as 5.7% p.a. the difference in amounts at maturity is ₹9,774. Find the period for which the sum was invested.

7 years
5.8years
6 years
8 years

Solution:

⇒ $\mathrm{t}=\frac{9774 \times 100}{90.500 \times(7.5-5.7)}=6 \mathrm{yrs}$

Question 48. If the sum of money when compounded annually becomes ₹1140 in 2 years and ₹1710 in 3 years at rate of interest

Solution:

Interest in 3rd yr= 1710- 1140 = 570

For 3rd yrs.; it will be like S.l

⇒ $\mathrm{r}=\frac{l \times 100}{P . t}=\frac{570 \times 100}{1140 \times 1}=50 \%$

For (c) A = 1140 + 50% (calculator) = 1710

Question 49. In what time will a sum of money double itself at 6.25% p.a at simple interest

5 yrs.
8 years
12 yrs.
16 yrs.

Solution:

⇒ $\mathrm{t}=\frac{(2-1) \times 100}{6.25}=16 \text { years }$

Question 50. What principal will amount to ₹370 in 6 years at 8% p.a. at simple interest

₹210
₹250
₹310
₹350

Solution:

⇒ $ P=\frac{370}{1+6 \times 0.08}=₹ 250$

Question 51. If a sum trip Per in 15 years at the simple rate of interest then the rate of interest per annum will be

13.0%
13.3%
13.5%
18%

Solution:

$\text { Tricks }=\frac{(3-1) \times 100}{1 \times 15}=13.3$

Calculator Tricks: GBC

(2) r = 15 x 13.333%= 200%

A = 1 + 200% (button) = 3

(2) is correct

Question 52. A certain sum of money was invested at a simple rate of interest for three years. If it was invested at 7% higher, the interest have been *882 more, then the sum has been invested at that rate was

₹12,600
₹6,800
₹4,200
₹2,800

Solution:

t=3 years

⇒ $\mathrm{P}=\frac{1 \times 100}{r t}=\frac{882 \times 100}{7 \times 3}=₹ 4200$

Question 53. The sum of money will be doubled in 8 years at S.I. In how many years the sum will be tripled?

20 years
12 years
16 years
None

Solution:

⇒ $ \frac{t_2}{8}=\frac{3-1}{2-1} \quad t_2=16 \text { yrs. }$

Question 54. A sum of 44,000 is divided into 3 parts such that the corresponding interest earned after 2 years, 3 years and 6 years may be equal at the rate of simple interest are 6% p.a. 8% p.a. & 6% p.a., respectively. Then the smallest part of the sum will be.

₹4,000
₹8,000
₹10,000
₹12,000

Solution:

⇒ $ P_1: P_2: P_3=\frac{1}{r_1 t_1}: \frac{1}{r_2 t_2}: \frac{1}{r_3 t_3}$

⇒ $\frac{1}{2 \times 6}: \frac{1}{8 \times 3}: \frac{1}{6 \times 6} $

⇒ $\left[\frac{1}{12}: \frac{1}{24}: \frac{1}{36}\right] \times 72 \text { LCM of denominators }$

=6: 3: 2

So, smallest principal $ =\frac{44000}{6+3+2} \times 2=₹ 8000$

Question 55. No. of years the sum of money becomes 4 times itself at 12% p.a. at simple interest:

Solution:

⇒ $\text { (3) Tricks: } \mathrm{t}=\frac{(4-1) \times 100}{1 \times 12}=25 \mathrm{yrs} \text {. }$

Question 56. No. years of a sum of money becomes 4 times itself at 12% p.a. at simple interest:

Solution:

(3) is correct

⇒ $\text { Tricks: } t=\frac{(4-1) \times 100}{1 \times 12}=25 \mathrm{yrs} \text {. }$

Question 57. If a person lends ₹6,000 for A years and 10,000 for 3 years at S.I. The total interest earned Is ₹2100 then the rate of interest is____.

Solution:

(1) is correct.

For (1): Total SI = 6000 x 4. x 5% + 8000 x 3 x 5% =R2400

So, (1) is correct.

Question 58. In simple interest, a certain sum becomes ₹97,920 in years, and ₹1,15,200 in 5 years, then the rate of interest is:

10%
11.2%
12%
13.6%

Solution:

(3)

⇒ $\text { Tricks: S.I p.a. }=\frac{1,15,200-97,920}{5-3}$=8640

Principal = 97,920 – 3 yrs interest = 97,920 – 3 x 8640 = 72,000

⇒ $r=\frac{8640 \times 100}{72000}=12 \%$

Amounts = 72000 + (12 x 3 = 36) %

Button = (True)

So, option (C) is correct.

Question 59. A person borrows Rs. 5,000 for 2 years at 4% per annual simple interest. He immediately lends to another person at $6 \frac{1}{4} \%$. Per annual for 2 years find his gain in the transaction.

Rs. 112.50
Rs. 225
Rs. 125
Rs. 107.50

Solution:

Interest Gain =$\left(6 \frac{1}{4}-4\right)=2 \frac{1}{4}=2.25 \%$

So, Interest Gain

⇒ $=\frac{5000 \times 2 \times 2.25}{100}=\text { Rs. } 225$

Question 60. A certain money doubles itself in 10 years when deposited on simple interest. It would triple itself in.

30 year
20 years
25 years
15 years

Solution:

⇒ $\frac{t_1}{t_1}=\frac{r_2-1}{r_1-1} \quad \Rightarrow \frac{t_2}{t_0}=\frac{3-1}{2-1} \Rightarrow t_2=20 \mathrm{yrs} .$

Question 61. A certain sum of money Q was deposited for 5 years and 4 months at 4.5% simple interest and amounted to ₹248, then the value of Q is

Solution:

t = 5 yrs 4 months = $5+\frac{1}{12}=\frac{16}{3} \mathrm{yrs}$

⇒ $A=Q\left(1+\frac{1 t}{100}\right)$

⇒ $\text { Or } 248=Q\left[1+\frac{45}{100} \times \frac{16}{3}\right]$

⇒ $ Q=\frac{248 \times 300}{372}={Rs} .200$

Rates for 5 yrs 4 Months = 5 ₹ 4.5% + one-third of 4.5% = 24% Note: 4 months means one-third of one year, so rate for 4 months = one-third of one-year interest rale.

Amounts = 200 + 24% = 248 (True) So, (b) is correct.

Question 62. The certain sum of money became Rs. 692 in 2 yrs. And Rs. 800 in 5 yrs. Then the principle amount is_____.

Rs. 520
Rs. 620
Rs. 720
Rs. 820

Solution:

If a certain sum of money becomes Ai in ti year and A.- in t2 years then

⇒ $\text { S.I. per annum }=\frac{A_2-A_1}{t_2-t_1}$

⇒ $\text { S.I. p. } a=\frac{800-692}{5-2}$

= Rs. 36.

Principal = A – Interest

= 692 – Interest of 2 yrs.

= 692 – 2 x 36 = Rs. 620

(2) is correct

Question 63. A sum of money amount of Rs. 6,200 in 2 years and Rs 7,400 in 3 years as per S.I. then the principal is

Rs. 3,000
Rs. 3,500
Rs. 3,800
None

Solution:

S.I. p.a $=\frac{7400-6200}{3-2}$

= Rs. 1200.

Principal = 6200 – 2 x 1200

= Rs. 3800.

Question 64. P = Rs. 5,000; R = 15% T $=4 \frac{1}{2} \text { using } I=\frac{P T K}{100} \text { then I will be } I=\frac{P r t}{100}$

Rs. 3,375
Rs. 3,300
Rs. 3,735
None

Solution:

⇒ $1=\frac{5000 \times 15 \times 4.5}{1000}=3375$

Question 65. In simple interest if the principal is Rs. 2,000 and the rate and time are the roots of the equation x2-2x- 30 = 0 then simple interest is

Rs. 500
Rs. 600
Rs. 700
Rs. 800

Solution:

x2 – 11 x + 3 0 = 0

Or x2 – 5x – 6x + 30 = 0

0r x(x-5) – 6(x-5) = 0

Or (x-5) (x-6) = 0

= 5 ; 6

If r = 5% then t = 6 yrs.

⇒ $S. I=\frac{{Prt}}{100}=\frac{2000 \times 5 \times 6}{100}$

= Rs. 600.

(2) is correct

Question 66. ₹ 8,000 becomes ₹10,000 in two years at simple interest. The amount that will become ₹6,875 in 3 years at the same rate of interest is:

₹ 4,850
₹ 5,000
₹ 5,500
₹ 5,275

Solution:

⇒ $\text { S.I. } / \text { year }=\frac{10000-8000}{2}=₹ 1000$

⇒ $\mathrm{r}=\frac{1000 \times 100}{8000}=12.5 \% $

⇒ $\mathrm{P}=\frac{A \mathrm{Amt}}{A m \mathrm{at} \ 1}=\frac{6875}{1+0.125 \times 3}$

=5,000

Question 67. The rate of simple: interest on a sum of money is 6% p.a. for the first 3 years, 8% p.a. for the next five years, and 10% years for the period beyond 8 years, if the simple interest accrued by the sum for the period for 10 years is ₹1,560. The sum is:

₹1,500
₹2,000
₹3,000
₹5,000

Solution:

(2) is correct

Single S.l for 1 year = (6×34-8×5+10×2)%

= 78%

⇒ $ \mathrm{P}=\frac{\text { Total } \mathrm{S} .1}{\text { S.I on Rsl }}=\frac{1560}{0.78}=₹ 2000$

Question 68. The rate of simple: interest on a sum of money is 6% p.a. for the first 3 years, 8% p.a. for the next five years, and 10% years for the period beyond 8 years, if the simple interest accrued by the sum for the period for 10 years is ₹ 1,560. The sum is:

₹1,500
₹2,000
₹3,000
₹5,000

Solution:

(2) is correct

Single S.l for 1 year = (6×34-8×5+10×2)%

= 78%

⇒ $\text { Tricks: } \mathrm{P}=\frac{\text { Total } \mathrm{S} .1}{\text { S.I on Rsl }}=\frac{1560}{0.78}=₹ 2000$

Compound Interest

Compound Interest: Sometimes so happens that the born and the lender agree to fix up a certain unit of time, say yearly half-yearly, or quarterly to settle the previous accounts.

In such cases, the amount after the first unit of tone becomes the principal for the second unit, the amount after the second unit becomes the principal for the third unit, and So on.
After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.l.) for that period.

Let Principal = P, Rate = R% per annum, Time = n years.

1. When interest is compound Annually:

Amount =$\mathrm{P}\left(1+\frac{R}{100}\right)^n$

2. When interest is compounded Half-yearly:

Amount = $P\left[1+\frac{(n / 2)}{100}\right]^{2 n}$

3. When interest is compounded Quarterly:

Amount =$\mathrm{P}\left[1+\frac{(R/ 4)}{100}\right]^{4 n}$

4. When interest is compounded Annually but time is in fraction, say $3 \frac{2}{5} \text { years. }$

Amount$=\mathrm{P}\left(1+\frac{R}{100}\right)^3 \times\left(1+\frac{\frac{2}{5} R}{100}\right)$

5. When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively.

Then, Amount = $\mathrm{P}\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$

6. Present worth of Rs. X due n years hence is given by:

Present Worth =$=\frac{x}{\left(1+\frac{R}{160}\right)^n} .$

Effective Rate of Interest

If interest is compounded more than once a year the effective interest rate for a year exceeds the per annum interest rate. Suppose you invest ₹1 0,000 for a year at the rate of 6% per annum compounded semi-annually. The effective interest rate for a year will be more than 6% per annum since interest is being compounded more than once a year. To compute the effective rate of interest first, we have to compute the interest. Let us compute the interest.

Interest for first six months = ₹10,000 ₹ 6/100 ₹ 6/12 = ₹300

Principal for calculation of interest for next six months

= Principal for first period one + Interest for first period

= ₹(10,000 + 300)

= ₹10,300

Interest for next six months = ₹ 10,300 x 6/100 x 6/12 =₹309

Total interest earned during the current year

= Interest for first six months + Interest for next six months

= ₹(300 + 309) = ₹ 609

The interest of ₹ 609 can also be computed directly from the formula or compound Interest, We can compute the effective rate of interest by following the formula

I = PEt

Where I = Amount of interest

E = Effective rate of interest in decimal

t = Period

P = Principal amount

Putting the values we have

609 = 10,000 x E x 1

⇒ $E=\frac{609}{10,000}$

= 0.0609 or

= 6.09%

Thus, if we compound the interest more than once a year effective interest rate for the year will be more than the actual interest rate per annum. But if interest is compounded annually effective interest rate for the year will be equal to the actual interest rate per annum.

So effective interest rate can be defined as the equivalent annual rate of interest compounded annually if interest is compounded more than once a year.

The effective interest rate can be computed directly by following the formula:

E = (1 + i)n – 1

Where E is the effective interest rate

i = actual interest rate in decimal

n = number of conversion period

Solved Examples

Question 1. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
Solution:

⇒ $\text { Amount = Rs. }\left[7500 \times\left(1+\frac{4}{100}\right)^2\right]=\text { Rs. }\left(7500 \times \frac{26}{25} \times \frac{26}{25}\right)=\text { Rs. } 8112$

C.I. = Rs. (8112 – 7500) = Rs. 612.

Question 2. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.
Solution:

time = 2 years 4 months $=2 \frac{4}{12} \text { years }=2 \frac{1}{3} \text { years. }$

⇒ $\text { Amount }=\text { Rs. }\left[8000 \times\left(1+\frac{15}{100}\right)^2 \times\left(1+\frac{\frac{1}{3} \times 15}{100}\right)\right]=\text { Rs. }\left(8000 \times \frac{23}{20} \times \frac{23}{20} \times \frac{21}{20}\right)$

= Rs. 11109.

C.I. = Rs. (11109- 8000) = Rs.3109.

Question 3. Find the compound interest on Rs. 10000 in 2 years at 4% per annum, the interest being compounded half-yearly.
Solution:

Principal = Rs. 10000; Rate= 2% per half- year; time = 2 years = 4 half- years,

Amount=$\text { Rs. }\left[10000 \times\left(1+\frac{2}{100}\right)^4\right]=\text { Rs. }\left(10000 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\right)$

Rs. 10824.32 C.I. = Rs. (10824-10000) = Rs. 824.32

Question 4. Find the compound Interest on It. 16000; it is 20% per annum for 0 months. Compounded quarterly.
Solution:

principal = Its. 16000; time = 9 months = 3 quarters;

Rate = 20 % per annum = 5% per quarter.

⇒ $\text { Amount = Rs. }\left[16000 \times\left(1+\frac{5}{100}\right)^3\right]=\text { Rs. }\left(16000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\right)=\text { Rs. } 18522 .$

C.I. = Rs. (18522-16000) = Rs. 2522

Question 5. If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.
Solution:

Clearly, Rate = 55 p.a., Time = 3 years, S.I. = Rs. 1200

So, principal = Rs.$\left(\frac{100 \times 1200}{3 \times 5}\right)=\text { Rs. } 8000 \text {. }$

⇒ $\text { Amount }=\text { Rs. }\left[8000\left(1 \times \frac{5}{100}\right)^3\right] \text { Rs. }\left(8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\right)=\text { Rs. } 9261 .$

C.I. = Rs. (9261-8000) – Rs. 1261

Question 6. At what time will Rs. 1000 become Rs. 1331 at 10% Per annum compounded annually?
Solution:

Principal = Rs. 1000; Amount = Rs.1331;

Rate= 10 %p.a.

Let the time be N years then,

⇒ $\left[1000 \times\left(1+\frac{10}{100}\right)^n\right]=1331 \text { or }\left(\frac{11}{10}\right)^n=\left(\frac{1331}{1000}\right)=\left(\frac{11}{10}\right)^3$

n =3 years.

Question 7. If Rs. 500 amounts to Rs. 583.20 in two years compounded annually find the rate of interest per annum.
Solution:

principal = Rs. 500; Amount =Rs. 583.20; Time = 2 years.;

Let the rate be R% per annum, then

⇒ ${\left[500 \times\left(1+\frac{R}{100}\right)^2\right]=583.20 \text { or }\left(1+\frac{R}{100}\right)^2=\left(\frac{5832}{5000}\right)=\left(\frac{11664}{10000}\right)}$

⇒ $\left(1+\frac{R}{100}\right)^2=\left(\frac{108}{100}\right)^2 \text { or } 1+\frac{R}{100}=\frac{108}{100} \text { or } R=8$

So, rate= 8% p.a.

Question 8. If the compound interest on a certain sum at 16% for 3 years is Rs. 1270 find the simple interest on the same sum at the same rate and for the same period.
Solution:

Let the sum be Rs. X. Then,

⇒ $\text { C.I. }=\left[x \times\left(1+\frac{50}{3 \times 100}\right)^3-x\right]=\left(\frac{343 x}{216}-x\right)=\frac{127 x}{216} $

⇒ $ \frac{127 x}{216}=1270 \text { or } \mathrm{x}=\frac{1270 \times 216}{127}=2160$

Thus, the sum is Rs 2160.

⇒ $\text { S.I. }=\text { Rs. }\left(2160 \times \frac{50}{3} \times 3 \times \frac{1}{100}\right)=\text { Rs. } 1080$

Question 9. The difference between compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.
Solution:

Let the sum be Rs. X. Then,

⇒ $\text { C.I. }=x\left(1+\frac{10}{100}\right)^2-x=\frac{21 . x}{100}, S . I .=\left(\frac{x \times 10 \times 2}{100}\right)=\frac{x}{5}$

⇒ $(\text { C. I. })-(\text { S. I. })=\left(\frac{21 \mathrm{x}}{100}-\frac{x}{5}\right)=\frac{\mathrm{x}}{100}$

⇒ $\frac{\mathrm{x}}{100}=631 \Leftrightarrow \mathrm{X}=63100 .$

Hence, the sum is Rs.63,100.

Question 10. Divide Rs. 1301 between A and X, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.
Solution:

Let the two parts be Rs. X and Rs. (1301- x)

C.I=$x\left(1+\frac{4}{100}\right)^7=(1301-x)\left(1+\frac{4}{100}\right)^9 \Leftrightarrow\left(\frac{26}{25} \times \frac{26}{25}\right)$

⇒ $\Leftrightarrow 625 x=676(1301-x) \Leftrightarrow 1301 x=676 \times 1301 \Leftrightarrow x=676$

So, the two parts are Rs. 676 and Rs. (1301-676) i.e. Rs. 676 and Rs. 625.

Question 11. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate present.
Solution:

S.I. on Rs. 7350 for 1 year Rs. (8575- 7350)= Rs. 1225.

⇒ $\text { Rate }=\left(\frac{100 \times 1225}{7350 \times 1}\right) \%=16 \frac{2}{3} \%$

Let the sum be Rs. x then,

⇒ $X\left(1+\frac{50}{3 \times 100}\right)^2=7350 \Leftrightarrow x \times \frac{7}{6} \times \frac{7}{6}=7350 \Leftrightarrow x=\left(7350 \times \frac{36}{49}\right)=5400$

⇒ $\text { sum }=\text { Rs. } 5400$

Question 12. A sum of money amounts to Rs. 6690 after 3 years and to RS. 10035 after 6 years on compound interest find the sum.
Solution:

Let the sum be Rs. P. then,

⇒ $\mathrm{P}\left(1+\frac{R}{100}\right)^3=6690 \quad \ldots \text { (1) } \quad \text { and } \mathrm{P}\left(1+\frac{R}{100}\right)^6=10035$

On dividing, we get$\left(1+\frac{R}{100}\right)^3=\frac{10035}{6690}=\frac{3}{2}$

Substituting this value in (i) we get :

⇒ $P \times \frac{3}{2}=6690 \text { or } P=\left(6690 \times \frac{2}{3}\right)=4460$

Hence, the sum is Rs. 4460

Question 13. A sum of money doubles itself at compound interest in 15 years, in how many years will it become eight times?
Solution:

⇒ $ \mathrm{P}\left(1+\frac{R}{100}\right)^{15}=2 P \quad \Rightarrow\left(1+\frac{R}{100}\right)^{15}=\frac{2 P}{P}=2$

Let$\mathrm{P}\left(1+\frac{R}{100}\right)^n=8 P \Rightarrow\left(1+\frac{R}{100}\right)^n=8=2^3=\left\{\left(1+\frac{R}{100}\right)^{15}\right\}^3[uning (i) ]$

⇒ $\Rightarrow\left(1+\frac{R}{100}\right)^n=\left(1+\frac{n}{100}\right)^{45} \Rightarrow n=45$

Tints, the required time = 45 years.

Question 14. What annual payment will discharge a debt of Rs. 7620 due in 3 years at $16 \frac{2}{3} \%$ per annum compound interest?
Solution:

let each installment be Rs. X then,

(P.W. of Rs. X due 1 year hence) + (P.W. of Rs. X due 2 years hence) + (P.W. of Rs. X due 3 years hence) = 7620.

⇒ $\frac{x}{\left(1+\frac{50}{3 \times 100}\right)}+\frac{x}{\left(1+\frac{50}{3 \times 100}\right)^2}+\frac{x}{\left(1+\frac{50}{3 \times 100}\right)^3}=7620$

⇒ $\Leftrightarrow \frac{6 x}{7}+\frac{36 x}{49}+\frac{216 x}{343}=7620 \Leftrightarrow 294 x+252 x+216 x=7620 \times 343$

⇒ $\Leftrightarrow X=\left(\frac{7620 \times 343}{762}\right)=3430 $

Amount of each instalment= Rs. 3430

Question 15.₹5,000 is invested in a Term Deposit Scheme that fetches interest of 6% per annum compounded quarterly. What will be the interest after one year? What is the effective rate of interest?
Solution:

We know that

I = P [(1 + i)n- 1]

Here P = ₹ 5,000

i = 6% p.a. = 0.06 p.a. or 0.015 per quarter

n = 4

and I = amount of compound interest

putting the values, we have

I = ₹ 5,000 [(1 + 0.015)4 — 1]

= ₹ 5,000 x 0.06136355

= ₹ 306.82

For the effective rate of interest using, I = PEt we find

306.82 = 5,000 x E x 1.

306.82

5000

= 0.0613 or 6.13%

Note: We may arrive at the same result by using

E = (1+i)4 – 1

E = (1 + 0.015)4-1

= 1.0613-1

= 0613 or6.13%

We may also note that the effective rate of interest is not related to the amount of principal. It is related to the interest rate and frequency of compounding the interest.

Question 16. Find the amount of compound interest and effective rate of interest if an amount of ₹20,000 is deposited in a bank for one year at the rate of 8% per annum compounded semi-annually.
Solution:

We know that

1 = P [(1 + i)n- 1]

Here P = ₹20,000

i = 8% p.n. = 8/2 % semi-annually = 0.04

n = 2

I = ₹20,000 IO + 0.04)2- 1 1

= ₹20,000 x 0.08 16

= ₹1,632

The effective rate of interest:

We know that

I = PEt

1,632 = 20,000 x E x 1

⇒ $E=\frac{1632}{20000}$ = =0.0816

= 8.16%

The effective rate of interest can also be computed by following the formula

E = (1 + i)n -1

= (1 + 0.04)2-1

= 0.0816 or 8.16%

Question 17. Which is a better investment 3% per year compounded monthly or 3.2% per year simple interest? Given that (1+0.0025) ‘2 =1.0304.
Solution:

i = 3/12 = 0.25% = 0.0025

= 12

= (1 + i)n -1

= (1 + 0.0025)12 -1

= 1.0304-1 = 0.0304

= 3.04%

With the effective rate of interest (E) being less than 3.2%, the simple interest of 3.2% per year is the better investment.

Exercise – 2

Question 1. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at a compound interest rate of 5 p.c.p.a How much amount will Albert get on maturity of the fixed deposit?

Rs. 8600
Rs. 8820
Rs. 8800
Rs. 8840
None of these

Solution:

(2) Amount = $\text { Rs. }\left[8000 \times\left(1+\frac{5}{100}\right)^2\right]=\text { Rs. }\left(8000 \times \frac{21}{20} \times \frac{21}{20}\right)=\text { Rs. } 8820$

Question 2. What will be the compound interest on a sum of Rs. 25000 after 3 years at the rate of 12 p.c.p.a.?

Rs. 9000.30
Rs. 9720
Rs. 10123.20
Rs. 10483.20
None of these

Solution:

(3) Amount = $\text { Rs. }\left[25000 \times\left(1+\frac{12}{100}\right)^3\right]=\text { Rs. }\left(25000 \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25}\right)=\text { Rs. } 35123.20 .$

C.I. = Rs. (35123.20-25000) = Rs. 10123.20

Question 3. A man saves Rs. 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years?

Rs. 565.25
635
Rs. 662.02
Rs. 666.50

Solution:

(3) Amount = $\text { Rs. }\left[200 \times\left(1+\frac{5}{100}\right)^3+200\left(1 \times \frac{5}{100}\right)^2+200\left(1+\frac{5}{100}\right)\right]$

⇒ $\text { Rs. }\left[200 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}+200 \times \frac{21}{20} \times \frac{21}{20}+200 \times \frac{21}{20}\right]$

⇒ $ =R s .\left[200 \times \frac{21}{20}\left(\frac{21}{20} \times \frac{21}{20}+\frac{21}{20}+1\right)\right]=\text { Rs. } 662.02 .$

Question 4. Sam invested Rs. 15000 @10% per annum for one year. If the interest is compounded halfyearly, then the amount received by Sam at the end of the year will be:

Rs. 16,500
Rs.525.50
Rs. 16,537.50
Rs. 18,150
None of these

Solution:

(3) P=15000; R=10%p.a.= 5% per half- year; T=1 year=2 half-years

Amount = ${Rs} .\left[15000 \times\left(1+\frac{5}{100}\right)^2\right]=\text { Rs. }\left(15000 \times \frac{21}{20} \times \frac{21}{20}\right)={Rs} .16537 .50$

Question 5. A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st of January and 1St July of a year. At the end of the year, the amount he would have gained by way of interest is

Rs. 120
Rs. 121
Rs. 122
Rs. 123

Solution:

⇒ $\text { Amount }=\text { Rs. }\left[1600 \times\left(1+\frac{5}{2 \times 100}\right)^2+1600 .\left(1+\frac{5}{2 \times 100}\right)\right]$

⇒ $\left[1600 \times \frac{41}{40} \times \frac{41}{40}+1600 \times \frac{41}{40}\right]$

⇒ $ \text { = Rs. }\left[1600 \times \frac{41}{40}\left(\frac{41}{40}+1\right)\right]=\text { Rs. }\left(\frac{1600 \times 41 \times 81}{40 \times 40}\right)=\text { Rs. } 3321 .$

C.I. = Rs. (3321-3200) = Rs. 121.

Question 6. What is the difference between the compound interests on RS.5000 for $1 \frac{1}{2}$ years at 4% per annum compounded yearly and half-yearly?

Rs. 2.04
Rs. 3.06
Rs. 4.80
Rs. 8.30

Solution:

(1) C.I. when interest is compounded yearly

⇒ $=\text { Rs. }\left[5000 \times\left(1+\frac{4}{100}\right) \times\left(1+\frac{\frac{1}{2} \times 4}{100}\right)\right]=\text { Rs. }\left(5000 \times \frac{26}{25} \times \frac{51}{50}\right)={Rs} .5304$

C.l. when interest is compounded half-yearly

⇒ $\text { Rs. }\left[5000 \times\left(1+\frac{2}{100}\right)^3\right]={Rs} .\left(5000 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\right)={Rs} .5306 .04$

Difference = Rs. (5306.04-5304) = Rs. 2.04.

Question 7. Find the compound interest on Rs. 15625 for 9 months at 16% per annum compounded quarterly.

Rs. 1851
Rs. 1941
Rs. 1951
Rs. 1961

Solution:

(3) P= RS.15625, n=9 months= 3 quarters, R= 1 6% p.a. = 4% per quarter.

Amount =Rs.$\left[15625 \times\left(1+\frac{4}{100}\right)^3\right]=R s .\left(15625 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\right)=\text { Rs. } 17576$

C.I. =Rs. (17576 – 15625) =Rs.l951.

Question 8. If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same sum at the same rate and for the same time?

Rs. 51.25
Rs. 52
Rs. 25
Rs. 60

Solution:

(1) Sum Rs.$\left(\frac{50 \times 100}{2 \times 6}\right)=R s .500$

Amount = $\text { Rs. }\left[500 \times\left(1+\frac{5}{100}\right)^2\right]={Rs} .\left(500 \times \frac{21}{20} \times \frac{21}{20}\right)={Rs} .551 .25$

C.l. =Rs. (551.25 -500) =Rs.51.25

Question 9. What will be the difference between simple and compound interest @ 10% per sum of Rs. 1000 after 4 years?

Rs. 31
Rs. 32.10
Rs. 40.40
Rs. 64.10
None of these

Solution:

S.I.= Rs. $\left(\frac{1000 \times 10 \times 4}{100}\right)=\text { Rs. } 400$

⇒ $\text { C.I. }=\text { Rs. }\left[1000 \times\left(1+\frac{10}{100}\right)^4-1000\right]=\text { Rs. } 464.10$

Difference = Rs. (464.10 – 400) = Rs.64.10.

Question 10. The difference between simple interest and compound interest on Rs. 1200 for one year at 10% per annum reckoned half-yearly is:

Rs. 2.50
Rs. 3
Rs. 4.75
Rs. 4
None

Solution:

⇒ $\text { Rs. }\left(\frac{1200-111-1}{100}\right)=Rs. 120$

⇒ $ Rs .\left|1200 \times\left(1+\frac{6}{100}\right)^2-1200\right|=Rs.123$

Difference = Rs. (123-120)= Rs.3

Question 11. The compound interest on Rs. 30000 at 7% per annum Is Rs. 4347. The period (in years) is:

2
$2 \frac{1}{2}$
3
4

Solution:

Amount = Rs. (30000 + 4347) Rs. 34347

let the time be n years then

⇒ $30000\left(1+\frac{7}{100}\right)^{11}=34347 \Leftrightarrow\left(\frac{1117}{1000}\right)^{11}=\frac{34347}{300010}=\frac{11449}{10000}=\left(\frac{107}{100}\right)^2n=2 \text { years. }$

Question 12. The principle that amounts to Rs. 4913 in 3 years at 6% per annum compound interest compounded annually, is:

Rs. 3096
Rs. 4076
Rs. 4085
Rs. 4096

Solution:

(4) $\text { (1) Principal }=\text { Rss. }\left|\frac{4913}{\left(11 \frac{25}{1.1000^3}\right)}\right|=\mathrm{Rs.}\left(4913 \times \frac{16}{17} \times \frac{16}{17} \times \frac{16}{17}\right)=\mathrm{Rs} .4096 \text {. }$

Question 13. In how many years will a sum of Rs. 800 at 10% per annum compounded semi-annually become Rs. 926.10?

$1 \frac{1}{3}$
$1 \frac{1}{2}$
$2 \frac{1}{3}$
$2 \frac{1}{2}$

Solution:

(2) Iet the time and years. Then,

⇒ $ 80\left(0 \times\left(1+\frac{5}{1001}\right)^{2 n}=926.100+\left(1+\frac{5}{100}\right)^{2 n}=\frac{926,1}{10001}\right.$

⇒ $\text { Or }\left(\frac{21}{20}\right)^{2 n}=\left(\frac{21}{20}\right)^3 \text { or } 2 n=3 \text { or } n=\frac{3}{2} n=1 \frac{1}{2} \text { years. }$

Question 14. If the compound interest on a sum for 2 years at $12 \frac{1}{2}$%per annum is Rs.510, the simple interest on the same sum at the same rate for the same period is:

Rs. 400
Rs. 450
Rs. 460
Rs. 480

Solution:

(4) Let the sum he Rs. P, then

⇒ $\left|P\left(\frac{25}{2 \times 100}\right)^2-P\right|=510 \text { or } P\left|\left(\frac{9}{11}\right)^2-1\right|=5100 \text { r } P=\left(\frac{510 \times 64}{17}\right)=1920 $

⇒ $\text { Sum }=\text { Rs. } 1920 .$

⇒ $\text { So, S.I. = Rs. }\left(\frac{1920 \times 25 \times 2}{2 \times 100}\right)=\text { Rs. } 480 .$

Question 15. The compound interest on a certain sum for 2 years at 10% per annum is 525. The simple interest on the same sum for double the time at half the rate present per annum is:

Rs. 400
Rs. 500
Rs. 600
Rs. 800

Solution:

(2) Let the sum be Rs. P, Then,

⇒ $\left|p\left(1+\frac{10}{1001}\right)^2-p\right|=525 \Leftrightarrow p\left|\left(\frac{11}{10}\right)^2-1\right|=525 \Leftrightarrow P=\left(\frac{525 \times 100}{21}\right)=2500$

⇒ $\text { Sum }=\text { Rs. } 25000 . $

⇒ $\text { So, S.I. }=\text { Rs. }\left(\frac{2500 \times 5 \times 1.1}{100}\right)=\text { Rs. } 500$

Question 16. The simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs.4000 for 2 years at 10% per annum the sum placed on simple interest is:

Rs. 1550
Rs. 1650
Rs. 1750
Rs. 2000

Solution:

(3) $\text { C.I. }=\text { Rs. }\left|4000 \times\left(1+\frac{10}{100}\right)^2-4000\right|=R s .\left(4000 \times \frac{11}{10} \times \frac{11}{10}-4000\right)=R s .840 . $

⇒ $ S u m=R s .\left(\frac{420 \times 100}{3 \times 8}\right)=\text { Rs. } 1750$

Question 17. There is a 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12000 after 3 years at the same rate?

2160
Rs. 3120
Rs. 3972
Rs. 6240
None of these

Solution:

Let P = Rs. 1 00 then, S.l. Rs. 60 anil T= 6 years

⇒ $R=\frac{100 \times 60}{100 \times 6}=10 \% / 0 \mathrm{p.a1} .$

Now, P= Rs. 1 2000, T= 3 years, and R = 1 0% p.a

⇒ $\text { C.I. }=\text { Rs. }\left[12000 \times\left\{\left(1+\frac{10}{100}\right)^3-1\right\}\right]=R s .\left(12000 \times \frac{331}{1000}\right)=R s .3972 .$

Question 18. The difference between compound interest and simple interest on an amount of Rs. 15000 for 2 years is Rs. 96. What is the rate of interest per annum?

8
10
12
8
None of these

Solution:

⇒ $\text { (a) }\left|15000 \times\left(1+\frac{R}{100}\right)^2-15000\right|-\left(\frac{15000 \times R \times 2}{100}\right)=96$

⇒ $\Leftrightarrow 15000\left[\left(1+\frac{R}{100}\right)^2-1-\frac{2 R}{100}\right]=96 \Leftrightarrow 15000\left[\frac{(100+R)^2-10000-200 R}{10000}\right]=96$

⇒ $\Leftrightarrow R^2=\frac{96 \times 2}{3}=64 \Leftrightarrow R=8$

Rate = 8%

Question 19. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:

Solution:

(1) Let the sum he Rs. X, Then

⇒ $\text { C.I. }=\left[x\left(1+\frac{4}{100}\right)^2-x\right]=\left(\frac{675}{625} x-x\right)=\frac{51}{625} x .$

⇒ $\text { S.I. }=\left(\frac{x \times 4 \times 2}{100}\right)=\frac{2 x}{25} $

⇒ $ \frac{51 x}{625}-\frac{2 x}{25}=1 \text { or } \mathrm{x}=625 $

Question 20. The difference between the simple interest on a certain sum at the rate of 10% per annum for 2 years and compound interest which is compounded every 6 months is Rs. 124.05. What is the principal sum?

Rs. 6000
Rs. 8000
Rs. 10,000
Rs. 12,000
None of these

Solution:

(2) Let the sum be Rs. P. then

⇒ $\mathrm{P}\left[\left(1+\frac{5}{100}\right)^4-1\right]-\frac{P \times 10 \times 2}{100}=124.05$

⇒ $\Rightarrow \mathrm{P}\left[\left(\frac{21}{20}\right)^4-1-\frac{1}{5}\right]=124.05 \Rightarrow P\left[\frac{194481}{160000}-\frac{6}{5}\right]=\frac{12405}{100} $

⇒ $\Rightarrow \mathrm{P}\left[\frac{194481-192000}{160000}\right]=\frac{12405}{100} \Rightarrow P=\left(\frac{12405}{100} \times \frac{160000}{2481}\right)=8000$

Question 21. On a sum of money, the simple interest for 2 years is Rs.660, while the compound interest is Rs. 696.30 the rate of interest being the same in both cases. The rate of interest is:

10%
10.5%
12%
None of these

Solution:

(4) Difference in C.I. and S.I for 2 years = Rs. (696.30- 660) = Rs. 36.30

S.I. for one year = Rs. 330

S.I. on Rs. 330 for1 year = Rs. 36.30

⇒ $\text { Rate }\left(\frac{100 \times 36.30}{330 \times 1}\right) \%=11 \%$

Question 22. The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:

6.06%
6.07%
6.08%
6.09%

Solution:

(4) Amount of Rs. 100 for a year when compounded half-yearly

⇒ $\text { Rs. }\left[100 \times\left(1+\frac{3}{100}\right)^2\right]={Rs} .106 .09$

Effective rate = (106.09- 100) % = 6.09 %

Question 23. Mr. Dua invested money in two schemes A and B offering compound interest @ 8 P.c.p.a. and 9 p.c.p.a. Respectively. If the total amount of interest accrued through two schemes together in two years was RS. 4818.30 And the total amount invested was Rs. 27000, what was the amount invested in scheme A?

Rs. 12,000
Rs. 13,500
Rs. 15,000
Cannot be determined
None Of these

Solution:

(1) Let the investment in the scheme be Rs. X.

Then, investment in scheme B = Rs. (27000-x)

⇒ $\left[x \times\left\{\left(1+\frac{8}{100}\right)^2-1\right\}+(27000-x)\left\{\left(1+\frac{9}{100}\right)^2-1\right\}\right]=4818.30$

⇒ $\Leftrightarrow\left(x \times \frac{104}{625}\right)+\frac{1881(27000-x)}{10000}=\frac{481830}{100}$

1664x+ 1881 (27000-x) = 48183000

o (1881x- 1664x) = (50787000- 48183000)

⇒ $217 x=2604000 \Leftrightarrow x=\frac{2604000}{217}=12000$

Question 24. A sum of money invested at compound interest amounts to Rs. 800 in 3 years and to Rs.840 in 4 years. The rate of interest per annum is:

$Rs. 2 \frac{1}{2} \%$
4%
5%
$6 \frac{2}{3} \%$

Solution:

(3) S.I. on Rs. 800 for1 year = Rs. (840-800) =Rs.40

⇒ $\text { Rate }=\left(\frac{40 \times 100}{800}\right) \%=5 \%$

Question 25. A sum of money placed at compound interest doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in:

7 years
15 years
24 years
36 years

Solution:

(2) $\mathrm{P}\left(1+\frac{R}{100}\right)^{\mathrm{g}}=2 \mathrm{P} \Rightarrow\left(1+\frac{R}{100}\right)^g=2$

⇒ $Let \mathrm{P}\left(1+\frac{R}{100}\right)^n=8 P \Rightarrow\left(1+\frac{R}{100}\right)^n=8=2^3=\left\{\left(1+\frac{R}{100}\right)^5\right\}^3
[using(i)]$

⇒ $\left(1+\frac{R}{100}\right)^n=\left(1+\frac{R}{100}\right)^{15} \Rightarrow \mathrm{n}=15$

Required time 15 Years.

Question 26. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

Solution:

(2)

⇒ $\mathrm{P}\left(1+\frac{20}{100}\right)^{\mathrm{n}}>2 \mathrm{P} \text { or }\left(\frac{6}{5}\right)^{\mathrm{n}}>2 \quad \text { Now, }\left(\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\right)>2 \text { so, } \mathrm{n}=4 \text { years. }$

Question 27. What annual payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest?

Rs. 550
Rs. 551.25
Rs. 560
Rs. 560.75

Solution:

(2)

bet each installment be Rs. X. Then,

⇒ $\frac{x}{\left(1+\frac{5}{100}\right)}+\frac{x}{\left(1+\frac{5}{100}\right)^2}=1025 \Leftrightarrow \frac{20 x}{21}+\frac{400 x}{441}=1025$

₹820x= 1025×441 ⇔ x = $\left(\frac{1025 \times 441}{120}\right)=551.25 .$. So, the value of each installment = Rs. 551.25

Question 28. A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% compound interest. The sum borrowed was:

Rs. 1620
Rs.1640
Rs. 1680
Rs. 1700

Solution:

(2) Principal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence

⇒ $=\left[\frac{882}{\left(1+\frac{5}{100}\right)}+\frac{882}{\left(1+\frac{5}{100}\right)^2}\right]=\left(\frac{882 \times 20}{21}+\frac{882 \times 400}{441}\right)={Rs} .1640 .$

Question 29. If the difference between simple interest and compound interest is ₹11 at the rate of 10% for two years, then find the sum:

₹1,200
₹1,100
₹ 1,000
None of these

Solution:

(2) is correct

⇒ $\mathrm{P}=\frac{\text { Difference } \times(100)^2}{(\text { rate })^2} \quad=\frac{11 \times(100)^2}{(10)^2}=₹ 1100$

Question 30. In how many years, a sum will become double at 5% p.a. compound interest?

14.0 years
14.1 years
14.2 years
14.3 Years

Solution:

⇒ $\mathrm{t}=\frac{\log \left(\frac{1}{p}\right)}{m e \log \left(1+\frac{r}{100 m}\right)}=\frac{\log 2}{\log (1.05)}=14.2 \text { yrs. (approx) }$

Question 31. A sum amount to ₹1331 at a principle of ₹1,000 at 10% compounded annually. Find the time.

3.31 years
4 years
3 years
2 years

Solution:

(3) is correct

For (C); A = 1000$\left(1+\frac{10}{100}\right)^3=₹ 1331$

So; t = 3 yrs

Question 32. The compound interest for a certain sum @ 5% p.a. for the first year is ₹25. The S-I for the same money @ 5% p.a. for 2 years will be.

₹40
₹50
₹60
₹70

Solution:

(2) S.I for 1st yrs. = C.l for 1st years. = 25

S.I. for 2 yrs. For same ‘p’ = 2×25 = 50

Question 33. At what % rate of compound interest corresponding (C.I) will the sum of money become 16 times in four years, if interest is being calculated compounding annually:

r=100%
r=10%
r=200%
r=20%

Solution:

(1) For (a) Let P = 1; A =1$\left(1+\frac{100}{100}\right)^4=(2)^4=16$

(1) is correct

Question 34. If the simple interest on a sum of money at 12% p.a. for two years is 3:3,600. The compound interest on the same sum for two years at the same rate is:

33,816
33,806
33,861
33,860

Solution:

(1) $ P=\frac{3600 \times 100}{12 \times 2}=₹ 15000$

⇒ $\text { C.I. }=15000\left(1+\frac{12}{100}\right)^2-15000=₹ 3816$

Question 35. The effective annual rate of interest corresponding to the nominal rate of 6% p.a. payable half yearly is

6.06%
6.07%
6.08%
6.09%

Solution:

⇒ $r_e=\left[\left(1+\frac{6}{200}\right)^2-1\right] \times 100=6.09 \%$

Question 36. The cost of machinery is 31,25,000/- if its useful life is estimated to be 20 years and the rate of depreciation of its cost is 10% p.a. then the scarp value of the Machinery is [given that (0.9)2=0.1215]

15,187
15,400
15,300
15,250

Solution:

(1) S (Scrap Value) = p$\left(1-\frac{d}{100}\right)^1$

Where P = Principle;

d=rate of depreciation

S = 1,25,000 $\left(1-\frac{10}{100}\right)^{20}=₹ 15,187.50$

Question 37. Mr. X invests the ‘P’ amount at a simple rate of 10% and Mr. Y invests the ‘Q’ amount at a compound interest rate of 5% compounded annually. At the end of two years, both get the same amount of interest, then the relation between two amounts P and Q is given by:

$P=\frac{41 Q}{80}$
$P=\frac{410}{40}$
$P=\frac{41 Q}{100}$
$P=\frac{41 Q}{200}$

Solution:

(1) is correct $\mathrm{S} .\mathrm{I}=\frac{P .10 \times 2}{100}=\frac{P}{5}$

⇒ $C. I=Q\left[\left(1+\frac{5}{100}\right)^2-1\right]=0.1025 . Q$

From Question

S.I = C.I

⇒ $\frac{11 Q}{80}=0.1025 Q$

Or P = 5×0.1025Q = 0.5125Q

⇒ $P=\frac{5125}{10000} Q=\frac{205 Q}{400}=\frac{41 Q}{80}$

⇒ $P=\frac{41 Q}{80}$

Question 38. If the difference between S.I and C. I am 372 at 12% for 2 years. Calculate the amount.

8,000
6,000
5,000
7,750

Solution:

(3)$\mathrm{P}=\frac{(C . I-S . I) \times(100)^2}{12 \times 12}=\frac{72 \times 100 \times 100}{12 \times 12}=₹ 5000$

Question 39. The nominal rate of interest is 9.9% p.a. if interest is compounded monthly. What will be the effective rate of interest?

⇒ $\left({Given}\left(\frac{4033}{4000}\right)^{12}=1.1036\right. \text { (approx)? }$

10.36%
9.36%
11.36%
9.9%

Solution:

(1) $\mathrm{r}_{\mathrm{e}}=\left[\left(1+\frac{9.9}{1200}\right)^{12}-1\right] \times 100 \quad=10.36 \%$

Question 40. The difference between Cl and SI on a certain sum of money for 2 years at 4% per annum is 1. The sum is

Solution:

(1) for 2 yrs Sum of Money =$\frac{\text { Diff.(100) }}{r^2}=\frac{1 \times(100)^2}{4^2}=₹ 625$

Question 41. If the sum of money when compounded annually becomes 1 140 in 2 years and 1710 in 3 years at a rate of interest

Solution:

(3) Interest in 3rd yr. = 511710 – 511140 = 51570

For 3rd yr; it will be like S.l r=$\frac{1 \times 100}{P t}=\frac{570 \times 100}{1140 \times 1}=50 \%$

For (3) A = 1140 + 50% (calculator) = ₹ 1710

Question 42. The difference between and C. I and S.I at 7% p.a. for 2 years is 329.4 then the principal is

35,000
35,5000
36,000
36,500

Solution:

(2) $\mathrm{P}=\frac{\text { Difference } \times(100)^2}{r^2}=\frac{29.4 \times(100)^2}{(7)^2}=₹ 6000 \text {. }$

Question 43. The partners A & B together lent 33903 at 4% p.a. interest compounded annually. After a span of 7 years. A gets the same amount as B gets after 9 years. The share of A in the sum of 33903/- would have been

31875
32280
32028
32820

Solution:

⇒ $A\left(1+\frac{4}{100}\right)^7=B\left(1+\frac{4}{100}\right)^9 \quad \text { Or } \frac{A}{B}=\left(1+\frac{4}{100}\right)^2=\left(\frac{26}{25}\right)^2 =\frac{676}{625}$

⇒ $A: B=676: 625 \quad A=\frac{676}{676+625} \times 3903=₹ 2028$

Question 44. A certain sum of money doubled itself in 4 years at C.I. In how many years it will become 32 times to itself

15 years
24 years
20 years
None

Solution:

2tz = 324

= 2t2 = (25)4 = 220

= t2= 20 yrs.

Question 45. On a certain sum rate of interest @ 10% p.a, S.I = 3 90 Term = 2 year, Find compound interest for the same:

₹544.5
₹94.5
₹ 450
₹ 18

Solution:

⇒ $\text { S.I p.a }=\frac{90}{2}=₹ 45$

Compound interest = 45 + (45+10%) = 94.5

Question 46. If an amount is kept at simple interest, it earns ₹600 in the first 2 years but when kept at compound interest it earns an interest of ₹660 for the same period: then the rate of interest and principal amount respectively are

20%; ₹1200
10%; ₹1200
20%; ₹1500
10%; ₹1500

Solution:

(2)$\text { S.I. }=\frac{1500 \times 2 \times 20}{100}=₹ 600 \text { (true) }$

⇒ $\text { C.I }=1500\left[\left(1+\frac{20}{100}\right)^2-1\right]=₹ 660 \text { (also true) }$

Question 47. Mr. X bought an electronic item for ₹1000. What would be the future value of the same item after two years,if the value is compounded semi-annually at the rate of 22% per annum?

₹1488.40
₹ 1518.07
₹2008.07
₹2200.00

Solution:

(b)

FV = P (1 + i)n

⇒ $1000\left(1+\frac{22}{200}\right)^{2 \times 2}$

= 1518.07 (approx.)

Question 48. The difference between simple interest and compound interest on a certain sum of money invested for 2 years at 5% P.a. is ₹ 30. Then the sum =

10,000
12,000
13,000
None

Solution:

(2)

P=30 + 5% / 5% button = 12,000

Question 49. A sum of money amounts to ₹7803 for one year at the rate of 4% compounded semi-annually then the sum invested is

7,000
7,500
7,750
8,000

Solution:

(2)

P= (4+200+1)+ = button 2 times x 7803 = button

= 7500

Question 50. The difference between simple and compound interest on a sum of ₹10000 for 4 years at the rate of interest 10 % per annum is____

Solution:

C.I-S.I

⇒ $=\left[10,000\left(1+\frac{10}{100}\right)^4-10,000\right]-{\left[\frac{10,000 \times 10 \times 4}{100}\right]}$

(3) is correct

= 4641 – 4000 = 641

Question 51. If the compound interest on a sum for two years at the rate of 5% p.a. is ₹512.50, then the principle is_____.

4,000
3,000
5,000
None of these

Solution:

(3)

Amount = 5000 + 5% + 5% button = 5512.50.

C.I= 5512.50 -5000 = 512.50

Question 52. Find the effective rate of interest corresponding to the nominal rate of interest 7% compounded monthly is_____.

7.68%
7.22%
8.1%
None of these

Solution:

(2)

⇒ $\mathrm{R}_{\mathrm{e}}=\left[\left(1+\frac{7}{1200}\right)^{12}-1\right] \times 100 \%$

= 7.229 % = 7.22%

Question 53. In compound interest, if the amount is 9 times its principles in two years then the rate of interest is?

100%
200%
150%
None of these

Solution:

(2), Given

⇒ $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^1 $

⇒ $\text { Or; } 9 \mathrm{P}=\mathrm{P}\left(1+\frac{r}{100}\right)^2$

⇒ $\text { Or; } 9=\left(1+\frac{r}{100}\right)^2$

⇒ $\text { Or } 3^2=\left(1+\frac{r}{100}\right)^2 \Rightarrow 3=1+\frac{r}{100}$

⇒ $\Rightarrow 2=\frac{r}{100} \Rightarrow r=200 \%$

1 + 200% + 200 % = 9

So, (2) is correct

Question 54. If the difference between compound interest and simple interest for 3 years is ₹ 912 at the rate of 4% p.a. the principle is

₹ 1,87,500
₹ 1,87,000
₹ 1,87,550
₹ 1,85,700

Solution:

(1)

P = 912 + 4% + 4% + (300 + 4)%

= ? 1,87,500

Question 55. If Rs. 1,000 is invested at an interest rate of 5% and the interest is added to the principle every 10 years, then the number of years in which it will amount to Rs. 2,000 is:

$16 \frac{2}{3}$
$6 \frac{1}{4}$
$16 years$
$6 \frac{2}{3}$

Solution:

(1)

Interest is added to the principal every 10 years.

So within 10 years; simple interest will apply.

So, the amount after 10 years

⇒ $=1000+1000 \times \frac{10 \times 5}{100}=R s .1500 \text {. }$

Total amount = Rs. 2000

Extra interest needed = 2000 – 1500 =Rs. 500

⇒ $\text { Time }=\frac{500 \times 100}{1500 \times 5}=\frac{20}{3}=6 \frac{2}{3} \mathrm{yrs} .$

⇒ $\text { So; Total time }=10+6 \frac{2}{3}=16 \frac{2}{3} \mathrm{yrs} .$

Question 56. If an amount is kept at S.I it earns an interest of Rs 600 in the first two years but when kept at the compound interest it earns an interest of Rs. 660 for the same period, then the rate of interest and principal amount respectively are:

20%, Rs. 1,200
20%, Rs. 1,500
10%, Rs. 1,200
10%, Rs. 1.500

Solution:

(2)

⇒ $\text { S.I }=\frac{1200 \times 2 \times 20}{100}=480 \neq 600$

So; (1) is false.

⇒ $S. I=\frac{1500 \times 2 \times 20}{100}=R s .600$

C.I = (-1500+20%+20%) (button)

= 660.

So; (b) is true.

Question 57. If 10,000 is invested at 8% per year compound quarterly, then the value of the investment after 2 years is (given (l+0.2)8= 1.171659)

R 10,716.59
11,716.59
17.1659
None of these

Solution:

(2)

⇒ $F V=100000\left(1+\frac{4}{800}\right)^{2 \times 4}=₹ 11,716.59$

Question 58. A bank pays a 10% rate of interest, interest is calculated half-yearly. A sum of is deposited in the bank. The amount at the end of 1 year will be

Solution:

⇒ $\text { (4) } F V=400\left(1+\frac{10}{200}\right)^2=441$

Question 59. A man deposited 8,000 in the bank for 3 years at 5% per annum compound interest, after 3 years he will get

9,000
8,800
₹9,200
9,261

Solution:

⇒ $\text { (4) } \quad F V=8000\left(1+\frac{5}{100}\right)^3=₹ 9261 \text {. }$

Question 60. If in two years, a principle amounts to when the interest at the rate of r% is compounded annually, then the value of r will be

14
10.5
15
10

Solution:

⇒ $121=100\left(1+\frac{r}{100}\right)^2 \Rightarrow \frac{121}{100}=\left(1+\frac{r}{100}\right)^2$

⇒ ${Or}\left(\frac{11}{10}\right)^2=\left(1+\frac{r}{100}\right)^2 \Rightarrow 1+\frac{r}{100}=\frac{11}{10}$

⇒ ${Or} \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10} $

r = 100%

For $\mathrm{FV}=100\left(1+\frac{10}{100}\right)^2=121 \text { (true) }$

Question 61. The effective rate of interest for one year deposit corresponding to a nominal 7% rate of interest per annum convertible quarterly is

7%
7.4%
7.5%
7.18%

Solution:

(4)

⇒ $r_e=\left[\left(1+\frac{7}{400}\right)^4-1\right] \times 100=7.18 \%$

Question 62. How much will 25,000 amount to in 2 years at compound interest if the rates for the successive years are 4% and 5% per year

27,000
27,300
27,500
27,900

Solution:

(2)

⇒ $F V=25000\left(1+\frac{4}{100}\right) \times\left(1+\frac{5}{100}\right)$

= ₹ 27,3007-

Question 63. at 10% per annum, interest compounded half yearly will become at the end of one year

8800
R 8,900
8820
9,600

Solution:

(3)

⇒ $FV=8000\left(1+\frac{10}{200}\right)^2= 8,820$

Question 64. The value of furniture depreciates by 10% a year, if the present value of the furniture in an office is calculate the value of furniture 3 years ago

30,000
% 40,000
35,000
R 50,000

Solution:

(1) 30000 – 10% – 10% – 10%

button = 2 1870.

Details Method

⇒ $21870 =P\left(1-\frac{10}{100}\right)^3 $

⇒ $P =\frac{21870}{(0.9)^3}=₹ 30,000$

Question 65. If compound interest on a sum for 2 years at 4% per annum is % 102, then the simple interest on the same period at the same rate will be

90
100
4101
93

Solution:

⇒ $\text { (b) } \mathrm{C} . \mathrm{I}=\mathrm{P}\left(1+\frac{4}{100}\right)^2-p=102$

Or P [(1.04)2- 1] = 102

Or Px 0.0816 =102

⇒ $ \text { Or } \mathrm{P}=\frac{102}{0.0816}=1250$

⇒ $\text { S.I }=\frac{p . r. t .}{100}=\frac{1250 \times 4 \times 2}{100}=₹ 100$

Question 66. If the difference between the compound interest compounded annually and simple interest on a certain amount at 10% per annum for two years is 372, then the principal amount is

37,000
37,200
37,500
None of the above

Solution:

(2) P= 372 v 10% + 10% = ₹ 37,200

Question 67. What is the net present value of a piece of property that would be valued at 2 lakh at the end of 2 years? (Annual rate of increase = 5%)

2.00 lakh
1.81 lakh
2.01 lakh
None of the above

Solution:

(2) NPV = $2\left(1+\frac{5}{100}\right)^{-2}=$ 1.81 lakh (approx).

Question 68. The difference between the simple and compound interest on a certain sum for 3 years at 5% p.a. is \228.75. The compound interest on the sum for 2 years at 5% p.a. is:

3,175
3,075
3,275
2,975

Solution:

⇒ $\mathrm{P}=\frac{\text { Difference } \times(100)^3}{r^2(300+r)}[For 3 years only]$

⇒ $=\frac{228.75 \times(100)^3}{5 \times 5(300+5)}=₹ 30,000$

A = 30000 + 5% + 5% buttons = 33075

Or P = = 45,317.33

C.l. = A – P = 33075 – 30000

= 7 3,075 (2) is correct

Question 69. In what time will amount to 8% per annum, when the interest is compounded semi-annually? [Given: (1.04)4 = 1.16986]

2 years
4 years
5 years
7 years

Solution:

⇒ $ A=P\left(1+\frac{r}{100 m}\right)^{m t}$

⇒ $\frac{4,56,976}{3,90,625}=\left(1+\frac{8}{200}\right)^{2 t}$

Or 1.16985856 = (1.04)2t

Or 1.6966 = (1.04)2t

Or (1.04)4 = (1.04)2t

2t = 4 t = 2 years

(1) is correct

Question 70. How long will take to amount to 5% p.a. converted quarterly? [Given: (1.0125)124 = 1.1666]

3 years
3.1 years
13.5 years
12.4 years

Solution:

⇒ $\text { (b) } \frac{A}{p}=\left(1+\frac{5}{400}\right)^{4 t}$

⇒ $ \text { Or } \frac{14000}{12000}=(1.0125)^{4 t}$

Or 1.6666……..(1.0125)4t

Or (1.0125)12.4 = (1.0125)4t

Or 4t = 12,4 t = 3.1 yrs.

(2) is correct

Question 71. If be invested at an interest rate of 5% and the interest is added to the principal every 10 years, then the number of years in which it will amount to is :

$16 \frac{2}{3} years$
$\frac{1}{10} years$
16 years
$6 \frac{2}{3} years$

Solution:

⇒ $Given P=\left\{1000 ; m=\frac{1}{10} ; n=m t=\frac{1}{10} \times t=\right.0.1 \mathrm{t}$

r = 5% p.a

⇒ $\mathrm{A}=\mathrm{P}\left[1+\frac{5}{0.1 \times 100}\right]^{0.1 t}$

⇒ $ \frac{2000}{1000}=(1.50)^{0.1 t}$

⇒ $\text { Or } 0.1 t=\frac{\log 2}{\log (1.5)}$

⇒ $\text { Or } 0.1 \mathrm{t}=1.709 \text { or } \mathrm{t}=\frac{1.709}{0.1}$

= 17.09 = 16$\frac{2}{3}$

(1) is correct

Question 72. A Person deposited ₹5,000 in a bank. The deposit was left to accumulate at 6% compounded quarterly for the first five years and at 8% compounded semiannually for the next eight years. The compound amount at the end of 13 years is:

12621.50
12613.10
13613.10
None.

Solution:

⇒ $A=5000\left(1+\frac{6}{400}\right)^{5 \times 4}\left(1+\frac{8}{200}\right)^{8 \times 2}$

₹12613.17 =₹ 12610.00 (approx.)

(2) is correct

Question 73. Anshul’s father wishes to have 75,000 in a bank account when his first college expenses begin. How much amount does his father deposit now at 6.5% compounded annually if Anshul is to start college 8 years hence from now?

45,320
₹46,360
55,360
1 48,360.

Solution:

(1)

⇒ $ A=P\left(1+\frac{r}{100 m}\right)^{m t}$

⇒ $\text { Or; } 75000=P\left(1+\frac{6.5}{100}\right)^{1 \times B}$

⇒ $\text { Or } P=\frac{75000}{(1.065)^8}=₹ 45,317.33$

(1) is correct

Question 74. The difference between compound interest and simple interest on a certain sum for 2 years @ 10% p.a. is Find the sum:

1,010
1,095
1,000
990

Solution:

(3)

⇒ $\mathrm{P}=\frac{\text { actual }(C . I-S . I)}{(C . I-S t .) a t R e .1 .}$

⇒ $\mathrm{P}=\frac{10}{\left|(1.10)^2-1\right|-0.2}=₹ 1000$

(3) is correct

Question 75. A machine worth Rs 4,90,740 is depreciated at 15% on its opening value each year when its value would reduce to 2,00,000:

5 years 6 months
5 years 7 months
5 years 5 months
None

Solution:

(1) is correct

⇒ $t=\frac{\log (2,00,00 / 4,90,740)}{\log (1-15 / 100)}$

= 5.5 years (approx..)

= 5 yrs. 6 months

Question 76. If the difference between simple interest and compound interest is 11 at the rate of10% for two years, then find the sum:

₹ 1,200
1,100
1,000
None of these

Solution:

(2) is correct

⇒ $\mathrm{P}=\frac{\text { Difference } \times(100)^2}{(\text { rate })^2}$

⇒ $=\frac{11 \times(100)^2}{(10)^2}=₹ 1100$

Annuity

Annuity Definition:

A sequence of payments, generally equal In size, made at equal intervals of time is called an annuity.

Monthly Rent; premiums of LIC; deposit into a recurring account in a bank; equal monthly payments got by a retired government servant as pension and loan installments to houses or automobiles etc.

Some Terms Related With Annuities

Periodic Payment: The size of each payment of an annuity is called the periodic payment of the annuity.
Annual Rent: The sum of all payments of an annuity made in one year is called its annual rent.
Payment Period or Interval: The duration between two successive payments of an annuity is called the payment period (or payment interval) of the annuity
Term: The total duration from the beginning time of the first payment period to the end of the last payment period is called the term of the annuity.
Present value of an Annuity: The sum of the present values of all the payments of an annuity is called the present value or capital value of the annuity.

Annuity regular and Annuity due or immediate

CA Foundation For Maths Chapter 4 Annuity

The annuity may be of two types:

Annuity regular: In an annuity, a regular first payment or receipt takes place at the end of the first period. Consider the following table:

CA Foundation For Maths Chapter 4 Annuity Regular

Annuity Due or Annuity Immediate: When the first receipt or payment is made today (at the beginning of the annuity) it is called annuity due or annuity immediate. Consider the following table:

CA Foundation For Maths Chapter 4 Annuity Due or Annuity Immediate

Annuity Regular Or Annuity Ordinary – Formulae

To Find Future Value Amount

⇒ $S=A\left[\frac{(1+i)^n-1}{r}\right] \times 100 \mathrm{~m} .$

Where S = Amount of an Annuity

A = Value of each installment

R = rate of interest

M = No. of conversion periods in a year

N = m.t = No. of in statements made in t yrs.

⇒ $\mathrm{i}=\frac{r}{100 \mathrm{~m}}$ = Rate of interest of one conversion Period

To find the Present Value for an Ordinary Annuity

PV = Present value $=\mathrm{A}\left[\frac{1-(1+i)^{-n}}{i}\right]$

Annuity Immediate Or Due – Formulae

To find the Future Value Amount

FV = Amount s = $\mathrm{A}\left[\left\{\frac{(1+i)^{n+1}-1}{r}\right\} \times 100 m-1\right]$

Present value of annuity due or annuity immediate Present value of annuity due or immediate for n years is the same as an annuity regular for (n – 1] years plus an initial receipt or payment at the beginning of the period.

Calculating the present value of the annuity due involves two steps.

Step 1: Compute the present value of the annuity as if it were an annuity regular for one period short.

Step 2: Add the initial cash payment or receipt to the Step 1 value.

Sinking Fund

It is the fund credited for a specified purpose by way of a sequence of periodic payments over some time at a specified interest rate. Interest is compounded at the end of every period. The size of the sinking fund deposit is computed from A = P.A(n, i) where A is the amount to be saved, P is the periodic payment, and n is the payment period.

Solved Example

Question 1. How much amount is required to be invested every year to accumulate ₹300000 at the end of 10 years if interest is compounded annually at 10%?
Solution:

Here A = 3,00,000

n = 10

i = 0.1

Since A = P.A (n, i)

300000 = P.A.(10, 0.1)

= P x 15.9374248

⇒ $\mathrm{P} \quad=\frac{3,00,000}{15.9374248}=₹ 18,823.62$

This value can also be calculated by the formula of the future value of annuity regular. We know that

⇒ $ A(n \text { i }) =A\left[\frac{(1+i)^n-1}{i}\right]$

⇒ $ 300000 =A\left[\frac{(1+0.1)^{10}-1}{0.1}\right] $

300000= A x 15.9374248

⇒ $\text { A } \quad=\frac{3.00,000}{15.9374248}=₹ 18,823.62$

Leasing

Leasing is a financial arrangement uniter which the owner of the asset (lessor) allows the user of the asset (lessee) to use the asset for a defined period (lease period) for a consideration (lease rental) payable over a given period, This is a kind of taking an asset on rent.

Solved Examples:

Question 1. ABC Ltd. wants to lease out an asset costing ₹3,60,000 for five years. It has fixed a rental of ₹1,05,000 per annum payable annually starting from the end of the first year, Suppose the rate of interest is 14% per annum compounded annually on which money can be invested by the company. Is this agreement favorable to the company?
Solution:

First, we have to compute the present value of the annuity of ₹1,05,000 for five years at the interest rate of 14% p.a. compounded annually.

The present value V of the annuity is given by

V = A.P (n, i)

= 1,05,000 xP(5, 0.14)

= 1,0,5000 x 3.43308 = ₹3,60,473.40

which is greater than the initial cost of the asset and consequently leasing is favourable to the lessor.

Question 2. A company is considering the proposal of purchasing a machine either by making a full payment of ₹4,000 or by leasing it for four years at an annual rate of ₹1,250. Which course of action is preferable if the company can borrow money at 14% compounded annually?
Solution:

The present value V of the annuity is given by

= A.P (n, i)

= 1,250 x p (4, 0.14)

= 1,250 x 2.91 371 = ₹3,642.11

which is less than the purchase price and consequently leasing is preferable

Capital Expenditure (investment decision)

Capital expenditure means purchasing an asset (which results in outflows of money) today in anticipation of benefits (cash inflow) that would flow across the life of the investment. For making investment decisions we compare the present value of cash outflow and the present value of cash inflows. If the present value of cash inflows is greater than the present value of cash outflows decision should be in favor of investment. Let us see how we make capital expenditure (investment) decisions.

Solved Examples:

Question 1. A machine can be purchased for ₹50000. The machine will contribute ₹12000 per year for the next five years. Assume borrowing cost is 10% per annum compounded annually. Determine whether the machine should be purchased or not.
Solution:

The present value of annual contribution

= A.P(n, i)

= 12,000 x P(5, 0.10)

= 12,000 x 3.79079

= ₹45,489.48

which is less than the initial cost of the machine. Therefore machine must not be purchased.

Question 2. A machine with a selling life of seven years costs f1 0,000 while another machine with a useful life of live years costs (111,001). The first machine saves labor expenses of ₹1,900 annually and the second one saves labor expenses of ₹2,200 annually. Determine the preferred course of action. Assume the cost of borrowing as 10% compounded per annum.
Solution:

The present value of annual cost savings for the first machine

= ₹1,900 KP (7, 0.10)

= ₹ 1,900 ⇔ 4.86842

= ₹9,249.99

= ₹9,250

The cost of the machine being ₹10,000 it costs more by ₹750 than it saves in terms of labour cost.

The present value of annual cost savings of the second machine

= ₹2,200 xP(5, 0.10)

= ₹2,200 x 3.79079

= ₹8,339.74

The cost of the second machine is ₹8,000 effective savings in labor cost is ₹339.74. Hence the second machine is preferable.

Valuation Of Bond

A bond is a debt security in which the issuer owes the holder a debt and is obliged to repay the principal and interest. Bonds are generally issued for a fixed term longer than one year.

Solved Example

Question 1. An investor intends to purchase a three-year ₹1,000 par value bond having a nominal interest rate of 10%. At what price the bond may be purchased now if it matures at par and the investor requires a rate of return of 14%?
Solution:

Present value of the bond

⇒ $=\frac{100}{(1+0.14)^1}+\frac{100}{(1+0.14)^2}+\frac{100}{(1+0.14)^3}+\frac{1,000}{(1+0.14)^3}$

= 100 x 0.87719 + 100 x 0.769467 + 100 x 0.674 972 + 1,000 x 0.674972

= 87.719+ 76.947+ 67.497+ 674.972 = ₹907.125

Thus, the purchase value of the bond is ₹907.125

Perpetuity

Perpetuity is an annuity in which the periodic payments or receipts begin on a fixed date and continue indefinitely or perpetually. Fixed coupon payments on permanently invested (irredeemable) sums of money are prime examples of perpetuities.

Solved Example

Question 1. Ramesh wants to retire and receive ₹3,000 a month. He wants to pass this monthly payment to future generations after his death. He can earn an interest of 8% compounded annually. How much will he need to set aside to achieve his perpetuity goal?
Solution:

R = ₹3,000

i = 0.08/12 or 0.00667

Substituting these values in the above formula, we get

⇒ PVA =$\frac{3,000}{0.00667}$

= ₹ 4,49,775

If he wanted the payments to start today, he must Increase the size of the funds to handle the first payment. This is achieved by depositing H52,775 of normal perpetuity received in the beginning = 4,49,775 + 3,000 which provides the immediate payment of ₹3,000 and leaves ₹4,49,775 in the fund to provide the future ₹3,000 payments.

Growing perpetuity

A stream of cash flows that grow at a constant rate forever is known as growing perpetuity.

The formula for determining the present value of growing perpetuity is as follows:

⇒ $\text { PVA }=\frac{R}{(1+i)^1}+\frac{R(1+g)}{(1+i)^2}+\frac{R(1+g)^2}{(1+i)^3}+\ldots \ldots \ldots . . .+\frac{R(1+g)^{\infty}}{(1+i)^{\infty}}$

⇒ $\sum_{n=1}^{\infty} \frac{R(1+g)^{n-1}}{(1+i)^n}$

⇒ $=\frac{R}{i-g}$

Solved Example:

Question 1. Assuming that the discount rate is 7% per annum, how much would you pay to receive ₹50, growing at 5%, annually, forever?
Solution:

⇒ $\text { PVA }=\frac{R}{i-g}=\frac{50}{0.07-0.05}=2,500$

Net Present Value Technique (NPV):

The net present value technique is a discounted cash flow method that considers the time value of money in evaluating capital investments. An investment has cash flows throughout its life, and it is assumed that a rupee of cash flow in the early years of an investment is worth more than a rupee of cash flow in a later year.

The net present value method uses a specified discount rate to bring all subsequent net cash inflows after the initial investment to their present values (the time of the initial investment is year 0).

Determining Discount Rate

Theoretically, the discount rate or desired rate of return on an investment is the rate of return the firm would have earned by investing the same funds in the best available alternative investment that has the same risk.

Many organizations choose to use the overall cost of capital or Weighted Average Cost of Capital (WACC) that an organization has incurred in raising funds or expects to incur in raising the funds needed for an investment.

The net present value of a project is the amount, in current value of rupees, the investment earns after paying the cost of capital in each period.

Net Present Value

Net present value = Present value of cash inflow – Present value of cash outflow

The steps to calculating net present value are:
Determine the net cash inflow in each year of the investment.
Select the desired rate of return discounting rate or Weighted Average Cost of capital.
Find the discount factor for each year based on the desired rate of return selected.
Determine the present values of the net cash flows by multiplying the cash flows by respective the discount factors of the respective period called the Present Value (PV) of Cash flows
Total the amounts of all PVs of Cash Flows

Decision Rule:

If NPV > 0 Accept the proposal.

lf NPV > 0 Reject the proposal

Solved Example:

Question 1. Compute the net present value for n pro|cct with n not Invoslment of 1,00,000 and cash flows year one Is ₹55,000; for year two Is ₹80,000 and for year three Is 15,000. Further, the company’s cost of capital Is 10%.

|PVIF @ 10% for three years are 0.909, 0.020 and 0.751 |

Solution:

CA Foundation For Maths Chapter 4 Net Present Value

Nominal Rate of Return:

The nominal rate is the stated interest rate. If a bank pays 5% annually on a savings account, then 5% is the nominal interest rate. So, if you deposit n00 for 1 year, you will receive in interest.

However, that will probably be worth less at the end of the year than it would have been at the beginning. This is because inflation lowers the value of money.

Nominal Rate of Return – Inflation = Real Rate of Return

Real Rate of Return:

The real interest rate is so named because it states the “real” rate that the lender or investor receives after inflation is factored in; that is, the interest rate that exceeds the inflation rate.

Nominal Interest Rate = Real Interest Rate + Inflation

Effective Rate:

It is the actual equivalent annual rate of interest at which an investment grows in value when interest is credited more often than once a year. If interest is paid m times in a year it can be found by calculating:

⇒ $\mathrm{E}_{\mathrm{i}}=\left(1+\frac{i}{m}\right)^m-1$

The chief advantage of knowing the difference between nominal, real, and effective rates is that it allows consumers to make better decisions about their loans and investments. A loan with frequent compounding periods will be more expensive than one that compounds annually.

Effective and nominal interest rates allow banks to use the number that looks most advantageous to the consumer. When banks are charging interest, they advertise the nominal rate, which is lower and does not reflect how much interest the consumer would owe on the balance after a full year of compounding. On the other hand, with deposit accounts where banks are paying interest, they generally advertise the effective rate because it is higher than the nominal rate.

Compound Annual Growth Rate (CAGR)

Compounded Aiuuml Growth Knlc (CAGR) is an IHISIIIORR mill Inverting specific terms for the smoothed in a mulled gain of mi Investnu it over; I gave time. Is not my accounting I r-rin, 1ml remains widely used, particularly in growth Industries or to compare the growth rates of two investments because CAGR dampens the effect or volatility of periodic returns that can render arithmetic means irrelevant. CAGR is often used to describe the growth over some time of some element of the business, for example, revenue, units delivered, registered users, etc.

⇒ ${CAGR}\left(t_0, t_{11}\right)=\left(\frac{V\left(t_n\right)}{V\left(t_0\right)}\right)^{\frac{1}{t_{01-}-t_0}}-1$

Where V(to) = Beginning Period ; V (tn) = End Period

Solved Example:

Question 1. Suppose the revenues of a company for four years, V(t) in the above formula, have been

CA Foundation For Maths Chapter 4 Compound Annual Growth Rate

Calculate Compound Annual Growth Rate.
Solution:

tn-to = 2016-2013 =3

The CAGR revenues over the three years from the end of 2013 to the end of 2016 is

CAGR (0, 3) = $\left(\frac{210}{100}\right)^{\frac{1}{3}}-1=1.2774-=27.74 \%$

Applications: These are some of the common CAGR applications:

Calculating average returns of investment funds.
Demonstrating and comparing the performance of investment advisors.
Comparing the historical returns of stocks with bonds or with a savings account.
Forecasting future values based on the CAGR of a data series.
Analyzing and communicating the behavior, over a series of years, of different business measures such as sales, market share, costs, customer satisfaction, and performance.

Solved Examples:

Question 1. Mr. X invests ₹310,000 every year starting today for the next 10 years suppose the interest rate is 8% per annum compounded annually. Calculate the future value of the annuity:

(given that (1+0.08)10 = 2.15892500)

3156454.88
3144865.625
3156554.88
None of these

Solution:

⇒ $A=F V=R\left[\left\{\frac{(1+i)^{n+1}-1}{r}\right\} \times 100 \mathrm{~m}-1\right]$

⇒ $=10,000\left[\frac{(1+0.08)^{10+1}-1}{8} \times 100-1\right]$

Question 2. The present value of an annuity of 33,000 for 15 years at 4.5% p.a. C.l. is: [Given that (1.045)15 = 1.935282]

323,809.67
332,218.67
332,908.67
None of these

Solution:

(2)

⇒ $ P V=R\left|\frac{1-(1+1)^{-11}}{1}\right|$

⇒ $ =3000\left\lceil\frac{1-(1.045)^{-13}}{0.045}\right\rceil$

Question 3. A machine can be purchased for ₹50,000. The machine will contribute Rs. 2,000 per year for five years. Assume the borrowing cost is 10% per annum. Determine whether the machine should be purchased or not:

Should be purchased
Should not be purchased
Can’t say about the purchase
None of the above

Solution:

(2)

⇒ $P V=R\left[\frac{1-(1+i)^{-n}}{i}\right]$

⇒ $P V=12000\left[\frac{1-(1.10)^{-5}}{0.10}\right]$

= ₹45,489.44

It should not be purchased

Question 4. How much amount is required to be invested every year so as to accumulate ₹3,00,000 at the end of 10 years? If interest is compounded annually at 10%? (Given (1.1)⇔=2.5937)

₹18,823.65
₹18,828.65
₹18,832.65
₹18,882.65

Solution:

(1)

⇒ $\mathrm{FV}=\mathrm{R}\left[\frac{(1+i)^n-1}{r} \times 100 \mathrm{~m}\right]$

⇒ $3,00,000=\mathrm{R}\left[\frac{\left(1+\frac{10}{100}\right)^{10}-1}{10} \times 100\right]$

⇒ $\mathrm{R}=\frac{3.00,000}{\left|\frac{(1.1)^{10}-1}{10} \times 100\right|}=₹ 18,823.65$

Question 5. A company is considering the proposal of purchasing a machine either by making a full payment of ₹4,000 or by leasing it for four years at an annual rate of ₹ 1,250. Which course of action is preferable, if the company can borrow money at 14% compounded annually? (Given: (1.14) = 1.68896)

Leasing is preferable
Should be purchased
No difference
None of these

Solution:

(1)

₹4000 = Present value

⇒ $PV=R\left[\frac{1-(1-i)^{-n}}{r} \times 100\right]$

⇒ $=1250\left[\frac{1-\left(1+\frac{14}{100}\right)^{-4}}{14} \times 100\right]=₹ 3642.14$

It is less than the real cost price.

Leasing is better

(1) is correct

Question 6. Vipul purchases a car for ₹5,50,000, nets, I loan of ₹5,00,000 at IS% p.a. from; i hank and balance ISO,000 ho pays as ₹ time of purchase, lie has to pay the whole amount of loan in 12 equal monthly Instalments with Interest starting from the end of the first month, The money he has to pay at the end of every month is:
(Given (1.0125)’₹’= 1.16075452)

₹45,130.43
₹45,230.43
₹45,330.43
None of these

Solution:

Loan value = ₹5,00,000 = PV

R = Instalment value =?

⇒ $P V=R\left[\frac{1-(1+i)^{-11}}{i}\right]$

⇒ $5,00,000=R\left[\frac{1-\left(1+\frac{15}{1200}\right)^{-12}}{i}\right]$

R = ₹45,130.43

Question 7. A company establishes a sinking fund to pay for ₹2,00,000 debt maturing in 20 years. Contributions to the fund are to be made at the end of every year. Find the amount of each annual deposit interest is 5% per annum:

₹6,142
6,049
₹6,052
6,159

Solution:

(2)

₹ 200,000

⇒ $200,000=R\left[\frac{(1+5 / 100)^{-20}-1}{5} \times 100\right] $

⇒ $ R=\frac{2,00,000 \times 5}{\left.\mid(1.05)^{20}-1\right) \mid \times 100}$

⇒ $=₹ 6049 \text { (Approx. })$

Question 8. A company may obtain a machine either by leasing it for 5 years (useful life) at an annual rent of ₹2,000 or by purchasing the machine for ₹8,100. If the company can borrow money at 18% per annum, which alternative is preferable?

Leasing
Purchasing
can’t say
None of these

Solution:

(1) PV = ₹8100

It is an ordinary annuity

= ₹ 6254.34

It is less than ₹8100. (1) Is correct

Question 9. A sinking fund is created for redeeming debentures worth ₹5 lacs at the end of 25 years. How much provision needs to be made out of profits each year provided sinking fund investments can earn interest at 4% p.a.?

12,006
12,040
12,039
12,035

Solution:

⇒ $₹ 5,00,000=R\left\lceil\frac{(1.04)^{25}-1}{0.04}\right]$

R = 12006.00 approx

Question 10. Future value of an ordinary annuity:

$(..) $A(n, 1)=A\left(\frac{(1+1)^n-1}{1}\right)$
$A(n, i)=A\left(\frac{(1+1)^n+1}{i}\right)$
$A(n, i)=A\left(\frac{1-(1+1)^n}{1}\right)$
$ A(n, i)=A\left(\frac{(1+i)^n-1}{\left((1+i)^n\right.}\right)$$

Solution:

(1) It is Formulae.

Question 11. Paul borrows ₹20,000 on condition that repay it with compound interest at 5% p.a. in an annual installment of ₹2,000 each. Find the number of years in which the debt would be paid off.

layers
12years
14years
15years

Solution:

(4)

⇒ $20,000=2000\left[\frac{1-\left(1+\frac{5}{10}\right)^{-1}}{5} \times 100\right]$

⇒ $\text { Or } 10=\left[\frac{1-(1.05)^{-t}}{5}\right] \times 100 $

⇒ $\text { Or } \frac{10 \times 5}{100}=1-(1.05)^{-t}$

Or 0.5-1= (1.05)-t

Or 0.5-1= -(1.05)-t

⇒ ${Or}(1.05)^t=\frac{1}{0.5}=2$

Or $ t =\frac{\log 2}{\log (1.05)}=15$ yrs. approx.$

Question 12. Find the value of an annuity of ₹1,000 payable at the end of each year for 10 years. If the rate of interest is 6% compounding per annum. (given (1.06)-111 = 0.558):

₹7,360
₹8,360
₹12,000
None of these

Solution:

(1) Is correct

⇒ $P V=1000\left[\frac{1-(1.06)^{10}}{0.06}\right] \quad=₹ 7360$

Question 13. The future value of an annuity of ₹ 5,000 is made annually for 8 years at an interest rate of 9% compounded annually (Given that (1.09)8 = 1.99256_____.

₹55,142.22
₹65,142.22
₹65,532.22
₹57,425.22

Solution:

(1)

⇒ $FV=5000\left[\frac{(1.09)^8-1}{0.09}\right]=₹ 55,142.22$

Question 14. How much amount is required to be invested every year to accumulate ₹ 6,00,000 at the end of 10, the year, if interest is compounded annually at a 10% rate of interest?

₹37,467
₹37,476
₹37,647
₹37,674

Solution:

(3)

Let the amount invested annually be = R

⇒ $R=\frac{6,00,000}{\left|\frac{\left(1+\frac{1}{180}\right)^{10}-1}{10} \times 100\right}=₹ 37,647 \text { (approx.) }$

Question 15. The future value of an annuity of ₹1,000 annually for 5 years at the rate of interest of 14% compound annually is

₹5610
₹6610
₹6160
₹5160

Solution:

(2)

⇒ $ \mathrm{FV}=1000\left[\frac{\left(1+\frac{14}{100}\right)^5-1}{14} \times 100\right]$

= ₹6610.104

Question 16. Suppose your mon decides to gift you ₹10,000 every year starting from today for the next sixteen years. You deposit this amount in a bank as and when you receive it and get an 8.5% per annum interest rate compounded annually. What is the present value of this money: (given that P (15,0.085) = 8.304236)

83,042
90,100
93,042
10,100

Solution:

(3)

⇒ $ PV=10,000\left[\frac{1-\left(1+\frac{8.5}{10}\right)^{(-16-1)}}{8.5} \times 100+1\right]$

= 10,000 (8.304236+1) = ₹93,042

Question 17. The future value of an annuity of ₹1500 made annually for 5 years with an interest rate of 10% compounded annually is_

9517.56
9157.65
9715.56
9175.65

Solution:

(2)

⇒ $F V=1500\left[\frac{\left(1+\frac{10}{100}\right)^5-1}{10} \times 100\right]=₹ 9157.65$

Exercise – 3

Question 1. The present value of an annuity of ₹3000 for 15 years at 4.5% p.a Cl is

₹23.809.41
₹32,214.60
₹3 2,908.41
None of these

Solution:

(2) Since no info is given, by default it is annuity regular

Present value =$=A\left[\frac{1-(1+i)^{-n}}{i}\right]$

A = Periodic Payment 1 = interest n = no. of yrs.

⇒ $\text { P.V. }=3000\left[\frac{1-(1.045)^{-15}}{0.045}\right]$ = ₹32,214.60 0.045

Question 2. The amount of an annuity is certain of V150 for 12 years at 3.5% p.a C. I am

₹2.190.28
₹290.28
₹2,180.28
None of these

Solution:

(1) v Amount of annuity certain will be the future value

⇒ $ \text { F.V. }=A\left[\frac{(1+i)^n-1}{i}\right]$

⇒ $150\left[\frac{(1.035)^{12}-1}{0.035}\right]=₹ 2190.28$

Question 3. A loan of ₹ 1 0,000 is to be paid back in 30 equal installments. The amount of each installment to cover the principal and at 4% p.a Cl is

₹587.87
₹587
₹578.30
None of these

Solution:

(3) loan amount is given i.e. P.V of annuity regular is given

⇒ $\text { P. V }=A\left[\frac{1-(1+i)^{-n}}{i}\right]$

⇒ $10,000=A\left[\frac{1-(1.04)^{-30}}{0.04}\right]$

Therefore, A=₹ 578.30

Question 4. A = ₹1,200 n = 12 yearsi = 0.08, V = ?

Using the formula$V=\frac{A}{i}\left[1-\frac{1}{(1+i)^n}\right]$ value of v will be

₹3,039
₹3,990
₹9930
₹9043.30

Solution:

(4)

⇒ $ \mathrm{V} =\frac{A}{i}\left[1 \frac{-1}{(1+i)^n}\right]$

⇒ $\mathrm{V} =\frac{A}{i}\left[1-(1+i)^{-n}\right]$

⇒ $\mathrm{V} =1200\left[\frac{1-(1.08)^{-12}}{0.08}\right]$

V = ₹9043.30

Question 5. a = ₹100 n = 10, /= 5% find the FV of annuity. Using the formula FV = a {(1 + i)” – l)/i, FV is equal to

₹1,258
₹2,581
₹1,528
None of these

Solution:

⇒ $ { (1) } F . V=a =100\left[\frac{(1.05)^{10}-1}{0.05}\right]$

⇒ $ =₹ 1258$

Question 6. If the amount of an annuity after 25 years at 5% p.a C.1 is ₹50,000 the annuity will be

₹1,406.90
₹1,047.62
₹1,146.90
None of these

Solution:

[2] F.V given = 50,000

⇒ $\text { F.V }=\mathrm{A}\left[\frac{(1+i)^n-1}{i}\right]$

⇒ $50,000 =\mathrm{A}\left[\frac{(1.05)^{25}-1}{0.05}\right]$

A = ₹1047.62

Question 7. An annuity of ₹100 amounts to ₹3137.12 at 4.5% p.a C. I. The number of years will be

25 years (appx.)
20 years (appx.)
22 years
None of these

Solution:

(2) F.V = ₹3137.12 A= 100 I = 0.045 n = ?

⇒ $ F \cdot V=A\left[\frac{(1+i)^n-1}{i}\right]$

⇒ $\frac{(F . V)}{A} \mathrm{i}+1=(1+\mathrm{i})^{\prime \prime}$

Question 8. A company borrows ₹10,000 on condition to repay it with compound interest at 5% p.a by annual installments of ₹1000 each. The number of years by which the debt will be clear is

14.2 years
10 years
12 years
None of these

Solution:

(1) P.V or annuity = loan amount = 10,000

⇒ $P \cdot V=A\left[\frac{1-(1+i)^{-n}}{i}\right]$

⇒ $ \frac{10.000}{1000}(0.05)=1-(1+i)^{-n}$

⇒ $(1+i)^n=\frac{1-10.000(0.05)}{1000}$

(1 + i)n =0.5

(l.05)n = 0.5

n = 14.2 yrs

Question 9. Mr. X borrowed ₹5,120 at 12 % % p.a C.I. At the end of 3 years, the money was repaid along with the interest accrued. The amount of interest paid by him is

₹2,100
₹2,170
₹2,000
None of these

Solution:

(2) Principal = 5120 Rate =12.5% Time: 3 yrs

Compound Interest = P [1 + i)”- 1]

= 5120 [(1+0.125)3 – 1]

= 5120 [(1.125P-1]

C.I = 2170

Interest paid by him is 2170

Question 10. Mr. Paul borrows ₹20,000 on condition that repay it with C.I. at 5% p.a in annual installments of ₹2000 each. The number of years for the debt to be paid off is

10 years
12 years
11 years
14.2 years

Solution:

10. (d) P.V of annuity = loan amount = 20,000

⇒ $ \text { P.V }=A\left[\frac{1-(1+i)^{-n}}{i}\right]$

⇒ $20,000=2000\left[\frac{1-(1.05)^{-n}}{0.05}\right]$

0.5=1-(1.05)-n

(1.05)n = 0.5

n = 14.2 yrs

Question 11. A person invests ₹500 at the end of each year with a bank that pays interest at 10% p.a C.I. annually. The amount standing to his credit one year after he has made his yearly investment for the 12th time is.

₹11,761.36
₹10,000
₹12,000
None of these

Solution:

(1) The future value of the annuity after 12 years will be

⇒ $ F .V =A\left[\frac{(1+i)^n-1}{i}\right]$

⇒ $=500\left[\frac{(1.1)^{12}-1}{0.1}\right]$

they are asking for one year after the 12th payment interest will be given at 10% but no 500 is deposited as the year has not ended.

Interest on @ 10% will be added

⇒ $ 10692.14188 \times \frac{10}{100}=1069.2141$

=₹ 11761.36

Question 12. The present value of an annuity of ₹5,000 per annum for 12 years at 4% p.a C.I. annually is

₹46,000
₹46,850
₹15,000
46925.40

Solution:

⇒ $ {P.V }=A\left[\frac{1-(1+i)^{-7}}{i}\right] $

⇒ $5000\left[\frac{1-(1.04)^{-12}}{0.04}\right]$

Question 13. A person desires to create a fund to be invested at 10% Cl per annum to provide for a prize of ₹300 every year. Using V = a/I find V and V will be

₹2,000
₹2,500
₹3,000
None of these

Solution:

(3) V=\frac{a}{1}=\frac{300}{0.1}=₹ 3000

Question 14. What sum should be invested at the end of every year to accumulate an amount of ’796570 at the end of 10 years at the rate of interest 10% compounded annually, (given that A(10;0.1) = 15.9374)

40,000
4,50,000
4,80,000
50,000

Solution:

(4) $R=\frac{796870}{\left|\frac{\left(1+\frac{10}{100}\right)^{10}-1}{10} \times 100\right|}=₹ 50,000$

Question 15. A person invests at the end of each month @ of interest 6% compounding monthly, find the amount of annuity after the 10th payment:

20,456
20,156
20,256
20,356

Solution:

(1) $F V=2000\left[\frac{\left(1+\frac{6}{1200}\right)^{10}-1}{6} \times 1200\right]=₹ 20,456$

Exercise 4- Mix Question And Answers

Question 1. A = ₹5,200. R = 5% p.a., T = 6 years, P will be

₹2,000
₹3.880
₹3,000
None of these

Solution: 1. ₹2,000

Question 2. If P = 1,000, n = 4 years., R = 5% p.a then C. I will be

₹215.50
₹210
₹220
None of these

Solution: 1. ₹215.50

Question 3. The time in which a sum of money will be doubled at 5% p.a C. I am

₹10 years
12 years
14.2 years
None of these

Solution: 3. 14.2 years

Question 4. ‘If A = ₹10,000, n = 18yrs., R = 4% p.a C.I, P will be

₹4,000
₹4,900
₹4,500
4936.30

Solution: 4. 4936.30

Question 5. The time by which a sum of money would treble itself at 8% p.a C. I am

14.28 years
14 years
12 years
None of these

Solution: 1. 14.28 years

Question 6. The present value of an annuity of ₹80 a year for 20 years at 5% p.a is

₹997 (appx.)
₹900
₹1,000
None of these

Solution: 1. ₹ 997 (appx.)

Question 7. A person bought a house paying ₹ 20,000 cash down and ₹4,000 at the end of each year for 25 yrs. at 5% p.a. C.l. The cash-down price is

₹75,000
₹76,000
₹76,375.80
None of these.

Solution: 3. ₹ 76,375.80

Question 8. A man purchased a house valued at ₹3,00,000. He paid ₹2,00,000 at the time of purchase and agreed to pay the balance with interest at 12% per annum compounded half-yearly in 20 equal half-yearly installments. If the first installment is paid after six months from the date of purchase then the amount of each installment is [Given log 10.6 = 1.0253 and log 31.19 = 1.494]

₹8,71 8.45
₹8,769.21
₹7,893.13
None of these.

Solution: 1. ₹8,71 8.45

Question 9. The difference between compound and simple interest at 5% per annum for 4 years on ₹20,000 is _____.

Solution: 4. 310

Question 10. The compound interest on half-yearly rests on ₹10,000 the rate for the first and second years being 6% and for the third year, 9% p.a. is _____.

2,200
2,287
2,285
None

Solution: 4. None

Question 11. The present value of ₹10,000 due in 2 years at 5% p.a. compound interest when the interest is paid every year is ______.

9,070
9,000
9,061
None

Solution: 1. 9,070

Question 12. The present value of ₹10,000 is due in 2 years at 5% p.a. compound interest when the interest is on a half-yearly basis _____.

9,070
9,069
9,061
None

Solution: 3. 9,061

Question 13. lohn son loft 1,00,000 with the direction that it should be divided in such a way that his minor sons Tom, Dick, and Harry aged 9, 12, and 15 years should each receive equally after attaining the age of 25 years. The rate of interest is 3.5%, how much does each son receive after getting 25 years old?

50,000
51,994
52,000
None

Solution: 4. None

Question 14. In how many years will a sum of money double at 5% p.a. compound interest?

15 years 3 months
14 years 2 months
14 years 3 months
15 years 2 months

Solution: 2. 14 years 2 months

Question 15. In how many years does a sum of money treble at 5% p.a. compound interest payable every year?

18 years 7 months
18 years 6 months
18 years 8 months
22 years 3 months

Solution: 15. 22 years 3 months

Question 16. A machine depreciates at 10% of its value at the beginning of a year. The cost and scrap value realized at the time of sale being and respectively. For how many years the machine was put to use?

7 years
8 years
9 years
10 years

Solution: 3. 9 years

Question 17. A machine’s worth is depreciated at 15% on its opening value each year. When its value would reduce to 2,00,000?

4 years 6 months
4 years 7 months
4 years 5 months
5 years 7 months approximately

Solution: 4. 5 years 7 months approximately

Question 18. A machine’s worth is depreciated at 15% of its opening value each year. When its value would reduce by 90%?

11 years 6 months
11 years 8 months
11 years 7 months
14 years 2 months approximately

Solution: 4. 14 years 2 months approximately

Question 19. Alibaba borrows lakhs Housing Loan at 6% repayable in 20 annual installments commencing at the end of the first year. How much annual payment is necessary?

52,420
52,419
52,310
52,320

Solution: 3. 52,310

Question 20. A sinking fund is created for redeeming debentures worth lakhs at the end of 25 years. How much provision needs to be made out of profits each year provided sinking fund investments can earn interest at 4% p.a.?

12,006
12,040
12,039
12,035

Solution: 1. 12,006

Question 21. A machine costs 5,20,000 with an estimated life of 25 years. A sinking fund is created to replace it with a new model at a 25% higher cost after 25 years with a scrap value realization of 25000. what amount should be set aside every year if the sinking fund investments accumulate at 3.5% compound interest p.a.?

16,000
16,500
16,050
16,005

Solution: 3. 16,050

Question 22. Raja aged 40 wishes his wife Rani to have ₹ 40 lakhs at his death. If his expectation of life is another 30 years and he starts making equal annual investments commencing now at 3% compound interest p.a. how much should he invest annually?

84,448
84,450
84,449
84,080

Solution: 4. 84,080

Question 23. Appu retires at 60 years receiving a pension of 14,400 a year paid in half-yearly installments for the rest of his life after reckoning his life expectation to be 13 years and that interest at 4% p.a. is payable half-yearly. What single sum is equivalent to his pension?

1,45,000
1,44,900
1,44,800
1,44,700

Solution: 2. 1,44,900

CA Foundation Maths Solutions For Chapter 1 Ratio And Proportion

July 5, 2024July 5, 2024 by Marksparks

Ratio And Proportion

Ratio Formulas

The ratio of two quantities a and b in the same units, is the fraction £ and we write it as a: b.
In the ratio a: b, we call “an as the first term or antecedent” and “b, the second term or consequent”.
1. Example: The ratio 5: 9 represents with antecedent = 5, consequent = 9.
The multiplication or division of each ratio term by the same non-zero number does not affect the ratio.
1. Example: 4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2 : 3 i.e. normally a ratio is expressed in simplest form.
The order of the terms in a ratio must be maintained because 3: 5 is not the same as 5 : 3.
Ratio exists only with quantities that have the same unit (kind).
Equality, Greater Equality, and Lesser Equality
- If x > y, then the ratio x: y is called of greater inequality.
- If x < y, then the ratio x: y is called of lesser inequality.
- If x = y, then the ratio a: b is called the ratio of Equal Equality.
Comparison of ratios:
1. We say that (a : b) > (c: d) $\frac{a}{b}>\frac{c}{d^{\prime}}$
2. Compounded Ratio: The compounded ratio of the ratios (a: b), (c : d), (e: f) is ace: pdf.
Duplicate ratio of (a: b)is (a2: b2).
- Sub-duplicate ratio of (a: b) is $(\sqrt{a}: \sqrt{b}).$
- The triplicate ratio of (a: b) is (a3: b3).
- The sub-triplicate ratio of (a: b) is (a1/3: b1/3).
- If $\frac{a}{b}=\frac{c}{d^{\prime}} \text { then } \frac{a+b}{b-b}=\frac{c+d}{c-d^{\prime}}$ (componendo and dividendo)
The inverse ratio of x y is y x.
Commensurable: If the terms of the ratio are integers, the ratio is called commensurable.
Answer: 3: 2
Incommensurable: If the terms of the ratio are not integers, the ratio is called
Incommensurable.
Answer:√3: √2 cannot be expressed in terms of integers. So, it is Incommensurable.

Read and Learn More CA Foundation Maths Solutions

Proportion:

1. The equality of two ratios is called Proportion.

If a: b = c: d, we write, a: b:: c: cl and say that a, b, c, d are in Proportion.

A and d are called extremes, while b and c are called mean terms.

2. Product of means = Product of extremes. This rule is also known as Cross – Product Rule

Thus, a: b: : c: d = (b x c) = (a x d).

3. (1). Fourth Proportional: If a b = c: d, then d is called the fourth proportional to a, b, c.

(2). Third Proportional: If a b = b; c, then c is called the third proportional to a and b.

(3). Mean Proportional: MThe mean proportional between a and b is fab.

4. Properties of Proportion

Cross – Product

If a:b::c:d. $\frac{a}{b}=\frac{c}{d}\mathrm{ad}=\mathrm{bc} .$

Invertendo

If a : b: : c: d.; Then its inverse

b : a: : d: c also becomes in proportion.

If $\frac{a}{b}=\frac{c}{d} \quad \text { Then, } \frac{b}{a}=\frac{a}{c} \text {. }$

Componendo

If a : b: : c: d.

Then a + b:b::c + d:d.

Proof: $\frac{a}{b}+1=\frac{c}{d}+1 \Rightarrow \frac{a+b}{b}=\frac{c+d}{d}$

Dividendo

If a : b = c: d.

Then a-b:b = c-d:d.

Proof:$\frac{a}{b}=\frac{c}{d} \Rightarrow \frac{a}{b}-1=\frac{c}{d}-1$ Or $\frac{a-b}{b}=\frac{c-d}{d }$

Componendo and Dividendo

If a : b : : c : d.; Dividing [3) by (4)

Then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$

Alternendo

Ifa : b: : c : d.

Then a : c:: b : d.

Therefore ratio of alternatives is also in proportion.

Addendum

If a:b = c:d = e:f= $\text { Then each ratio }=\frac{\text { Sum of antecedents of all ratios }}{\text { Sum of consequents of all ratios }}$

⇒ $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots \ldots \ldots \ldots . .=\frac{a+c+e+\cdots \ldots . ..}{b+d+f+\cdots \ldots .}$

Subtrahendo

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots \ldots \ldots \ldots \ldots$

Then each ratio

⇒ $=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots \ldots \ldots \ldots=\frac{a-c-e-\cdots}{b-d-f-…}$

Ratio Exercise – 1

Question 1. The inverse ratio of 11:15 Is

15: 11
$\sqrt{11}: \sqrt{15}$
121: 225
None of these

Answer: 1. 15:11

Ratio → A: B Inverse → B: A ->15:11

Question 2. The ratio of the two quantities is 3 4. If the antecedent is 15, the consequent Is

Answer: 4. 20

Ratio 3:1

3 is antecedent

4 is consequent

15 : x → x = 20

Question 3. The ratio of the quantities is 5: 7. If the consequent of its inverse ratio is 5, the antecedent Is

5
$\sqrt{5}$
7
None of these

Answer: 3. 7

Ratio 5: 7

7 is antecedent

5 consequence

Question 4. The ratio compounded of 2: 3, 9: 4, 5, 6 and 8 10 is

1: 1
1: 5
3 8
None of these

Answer: 1. 1: 1

⇒ $C R \rightarrow \frac{2 \times 4 \times 5 \times 8}{3 \times 4 \times 6 \times 10}=\frac{720}{720}=\frac{1}{1}=1 : 1$

Question 5. The duplicate ratio of 3: 4 is

$\sqrt{3}: 2$
4:3
9: 16
None of these

Answer: 3. 9:16

Duplicate Ratio of 3 : 4 = 32 : 42 = 9 : 16

Question 6. The sub-duplicate ratio of 25: 36 is

6: 5
36 25
50: 72
5: 6

Answer: 5: 6

Sub duplicate $\rightarrow \sqrt{25}: \sqrt{36}=5: 6$

Question 7. The triplicate ratio of 2 : 3 is

8: 27
6: 9
3: 2
None of these

Answer: 1. 8: 27

Triplicate a : b a³: b³ = 2³ : 3³ = 8:27

Question 8. The sub-triplicate ratio of 8: 27 is

27: 8
24: 81
2 : 3
None of these

Answer: 3. 2 : 3

Sub triplicate: $\sqrt[3]{a}: \sqrt[3]{b}=\sqrt[3]{8}: \sqrt[3]{27}=2: 3$

Question 9. The ratio compounded of4: 9 and the duplicate ratio of 3: 4 is

1 4
1 : 3
3 1
None of these

Answer: 1. 1: 4

CR = 4:9 and a duplicate of 34

= 4: 9 and 9 16

⇒ $=\frac{4}{9} \times \frac{9}{16}=\frac{1}{4}=1: 4$

Question 10. The ratio compounded of 4: 9, the duplicate ratio of 3 4, the triplicate ratio of 2: 3 and 9 7 is

2: 7
7 2
2: 21
None of these

Answer: 3. 2: 21

CR = 4: 9 and duplicate of 3: 4 and triplicate of 2 : 3 and 9 7

⇒ $\frac{4}{9} \times \frac{9}{16} \times \frac{8}{21} \times \frac{9}{7}=\frac{2}{21}$

Question 11. The fete compounded of the duplicate ratio of 4 5, triplicate ratio of 1: 3, sub duplicate ratio of81 256, and sub-triplicate ratio of125 512 is

4: 512
3 32
1 12
None of these

Answer: 4. None of these

⇒ $\mathrm{CR}=\frac{16}{25} \times \frac{1}{27} \times \frac{9}{16} \times \frac{5}{8} \rightarrow \frac{1}{120}=1: 120$

Question 12. If a : b = 3 : 4, the value of (2a+3b) : (3a+4b) is

54: 25
8: 25
17: 24
18: 25

Answer: 4. 18: 25

a : b = 3 : 4 → a = 3x, b = 4x

⇒ $\frac{2 a+3 b}{3 a+4 b}=\frac{6 x+12 x}{9 x+16 x}=\frac{18 x}{25 x}=18: 25$

Question 13. Two numbers are in the ratio 2 : 3. If 4 is subtracted from each, they are in the ratio 3:5. The numbers are

(16,24)
(4,6)
(2,30
None of these

Answer: 1. 16, 24

a : b = 2 : 3 → a = 2x, b = 3x

⇒ $\frac{2 x-4}{3 x-4}=\frac{3}{5}$

10x- 20 = 9x- 12

X = 8

a = 16, b = 24

Question 14. The angles of a triangle are in ratio 2:7:11. The angles are

(20°, 70°,90°)
(30°, 70°, 80°)
(18°, 63°, 99°)
None of these

Answer: 3. 18°, 63°, 99°

L in ratios 2:7:11

2x, 7x,1lx

2x + 7x + 11x = 180°

20x = 180 = x = 9

L→ 18, 63,99

Question 15. The division of 324 between X and Y is in the ratio 11: 7. X and Y would get Rupees

(204, 120)
(200, 124)
(180, 144)
None of these

Answer: 4. none of these

$324 x and y in ratio 11: 7

1x + 7x= 324

18x = 324

X= 18

X gets 11 (18) = 198 y gets 7(18) = 126

Question 16. Anand earns t 80 in 7 hours and Pramod in 12 hours. The ratio of their earnings is

32: 21
23: 12
8: 9
None of these

Answer: 1. (a) 32: 21

→ Anand earns in 7 hrs

⇒ $\text {i. e.} ₹ \frac{80}{7}$ in 1 hrs.

Pramodearns in 12 hrs

⇒ $\text {i. e. } ₹ \frac{90}{12}$in 1 hrs.

⇒ $\text { Ratio }=\frac{80}{7} \times \frac{12}{90}=\frac{96}{63}=\frac{32}{21}=32: 21$

Question 17. The ratio of the two numbers is 7: 10 and their difference is 105. The numbers are

(200,305)
(185,290)
(245,350)
None of these

Answer: 3. (245, 350)

Ratio → 7:10 difference = 105

7x – 10 x

10x- 7x = 105

3x = 105 → x= 35 ->245,350

Question 18. P, Q, and R are three cities. The ratio of average temperature between P and Q is 11 12 and that between P and R is 9: 8. The ratio between the average temperature of Q and R is

22: 27
27: 22
32: 33
None of these

Answer: 3. 32: 33

Avg temperature ratio between

P and Q = 11 : 12

P and R = 9 : 8

⇒ $\frac{P}{Q}=\frac{11}{12} \quad \frac{P}{R}=\frac{9}{8}$

⇒ $ \frac{P}{Q}=\frac{11}{12} \quad \frac{P}{R}=\frac{9}{8}$

⇒ $\frac{P}{Q} \times \frac{R}{P}=\frac{11}{12} \times \frac{9}{0}=\frac{R}{Q}=\frac{99}{96}$

Avg temp ratio between Q & R is

⇒ $\frac{Q}{R}=\frac{96}{99}=\frac{32}{33} \quad=32: 33$

Question 19. If x: y = 3 : 4, the value of x² y + xy² : x³ + y³ is

13: 12
12: 13
21: 31
None of these

Answer: 3. 12: 13

X : y = 3 : 4→ x = 3z

X²y + xy²: x³ + y³

= (3z)². 4z + 3z . (4z)² : (3z)³ + (4z)³

= 9z². 4z + 3z. 16z²: 27z³+ 64z³ + 84z³ : 91z³

= 84:91 = 12:13

= 84 : 91 = 12 : 13

Question 20. If p: q is the sub-duplicate ratio of p-x² q-x² then x2 is

$\frac{p}{p+q}$
$\frac{q}{p+q}$
$\frac{pq}{p+q}$
None of these

Answer: 3. $\frac{p q}{p+q}$

→ sub duplicate of = – x² = 9 – x²

⇒ $\frac{\sqrt{p-x^2}}{q-x^2}=\frac{p}{q}$

⇒ $\frac{p-x^2}{q-x^2}=\frac{p^2}{q^2}$

⇒ $q^2\left(p-x^2\right)=p^2\left(q-x^2\right)$

q²p-q² x²= p²q – p²x²

p² x² – q² x² = p²q – q²p

⇒ $x^2=\frac{p^2 q-q^2 p}{p^2-q^2}$

⇒ $=\frac{p q(p-q)}{(p-q)(p+q)}=\frac{p q}{p+q}$

Question 21. If 2s: 3t is the duplicate ratio of 2s – p : 3t – p then

p2 = 6st
p = 6st
2p = 3st
None of these

Answer: 2. p = 6st

⇒ $\frac{(2 s-p)^2}{(3 t-p)^2}=\frac{(2 s)}{(3 t)}$

⇒ $\frac{4 s^2-4 s p+p^2}{9 t^2-6 t p+p^2}=\frac{2 s}{3 t}$

12s² – 12spt + 3p2t = 18st² – 12stp + 2sp²

12s²t – 18st² = p²(2x – 3t) = p² = 6st

Question 22. If p : q = 2 : 3 and x : y = 4 : 5, then the value of 5px + 3qy : lOpx + 4qy is

71: 82
27: 28
17: 28
None of these

Answer: 3. 17: 28

5px + 3qy : 10px + 4qy

40rz + 45rz: 80rz + 60rz

85rz : 140rz

85: 140

17:28 = 17:28

Question 23. The number which when subtracted from each of the terms of the ratio 19:31 reducing it to 1: 4 is

15
5
1
None of these

Answer: 1. 15

Let,x be $\frac{19-x}{31-x}=\frac{1}{4}$

76- 4x. 31 – x

3x = 45

→ x= 15

Question 24. The daily earnings of two persons are in the ratio of 4:5 and their daily expenses are in the ratio of 7 9. If each saves1 50 per day, their daily earnings in 1 are

(40, 50)
(50, 40)
(400, 500)
None of these

Answer: (400.500)

Earning = 4:5

Expenses 7:9

4x, 5x

Savings = 50 each = 1:1

4x – 7y = 50 5x-9y = 50

20x- 35y = 250 (1) 20x – 36y

=200 (2)

Subtracts (2) from (1)

20x – 35y = 250

20x-36y= 200

Y = 50 x= 100

Daily earning = 4x, 5x

= 400,500

Question 25. The ratio between the speeds of two trains is 7: 8. If the second train runs 400 km. in 5 hours, the speed of the first train is

10 Km/hr
50 Km/hr
70 Km/hr
None of these

Answer: 70 km/hr

Ratio between speeds = 7:8

S1 = 7x S2 = 5x

The second train covers 400 km t = 5 hrs

We know

⇒ $\mathrm{S}_2=\frac{d}{t}=\mathrm{S}_2=\frac{400}{5}=80 \mathrm{~km} / \mathrm{hrs}$

S2 = 8x = 80 km/hrs = x = 10

S1 = 7x = 70km/hrs

Proportion Exercise – 2

Question 1. The fourth proportional to 4, 6, and 8 is

12
32
48
None of these

Answer: 1. 12

4,6,8.x

4 : 6 = 8: x

x= 12

Question 2. The third proportional to 12, 18 is

24
27
36
None of these

Answer: 2. 27

12,18,x

12 : 18 = 18 : x

X= 27

Question 3. The mean proportion between 25, and 81 is

40
50
45
None of these

Answer: 3. 45

25,x,81

⇒ $\frac{25}{x}=\frac{x}{81}=$ x2 = 2025 ⇒ x=45

Question 4. The number which has the same ratio to 26 that 6 has to 13 is

11
10
21
None of these

Answer: 4. None of these

⇒ $\frac{6}{13}=\frac{x}{26}=x=12 \text {. }$

Question 5. The fourth proportional to 2a, a², c is

ac/2
ac
2/ac
None of these

Answer: 2. ac/2

2a, a²,c, x

⇒ $\frac{2 a}{a^2}=\frac{c}{x}$

⇒ $\frac{2}{a}=\frac{c}{x}=2 x=a c$

⇒ $\mathrm{X}=\frac{ac}{2}$

Question 6.If four numbers 1/2, 1/3, 1 /5, 1/x are proportional then x is

6/5
5/6
15/2
None of these

Answer: 3. 15/2

⇒ $\frac{1}{2}: \frac{1}{3}=\frac{1}{5}: \frac{1}{x}$

3:2 = x:5

⇒ $X=\frac{15}{2}$

Question 7. The mean proportion between 12x² and 27y² is

18xy
81xy
8xy
None of these

Answer: 1. 18xy

12x² a, 27²2

⇒ $\frac{12 x^2}{a}=\frac{a}{27 y^2}$

a²= 324x²y²

a = 18xy

Question 8. If A = B/2 = C/5, then A: B: C is

3: 5: 2
2: 5 : 3
1: 2: 5
None of these

Answer: 3. 1 : 2: 5.

A: B: C

⇒ $\frac{A}{1}=\frac{B}{2}=\frac{C}{5}=1: 2: 5$

⇒ $\text { Tricks: } \frac{x}{a}=\frac{y}{b}=\frac{z}{a} \text { is given }$

x,y,z are in ratio

a: b: c

Question 9. If a/3 = b/4 = c/7, then a + b + c/c is

1
3
2
None of these

Answer: 3. 2

⇒ $\frac{a+b+c}{c}$

⇒ $\frac{a}{c}+\frac{b}{c}+1$

a:b:c = 3:4:7

a = 3x, b = 4x, c = 7x

⇒ $\frac{3 x}{7 x}+\frac{4 x}{7 x}+1=\frac{7 x}{7 x}+1=2$

Question 10. If p/q = r/s = 2.5/1.5, the value of Ps: QR is

3/5
1:1
5/3
None of these

Answer: 2. 1:1

⇒ $\frac{P}{q}=\frac{2.5}{1.5} \rightarrow p=2.5 x, q=1.5 x$

⇒ $\frac{r}{s}=\frac{2.5}{1.5} \rightarrow x=2.5 y, s=1.5 y$

⇒ $\frac{p s}{q r}=\frac{3.75 x y}{3.75 y x}=\frac{1}{1}=1: 1$

Question 11. If x : y = z : w = 2.5 : 1.5, the value of (x + z)/(w+ w) is

1
3/5
5/3
None of these

Answer: 2. 3/5

= x = 2.5x y = 1.5x z = 2.5p, w = 1.5p

⇒ $\frac{x+z}{y+w}=\frac{2.5(r+p)}{1.5(r+p)}=\frac{2.5}{1.5}=\frac{5}{3}$

Question 12. If (5x – 3y)/(5y – 3x) = 3/4, the value of x : y is

2: 9
7: 2
7: 9
None of these

Answer: 4. none of these

4(5x-3y) = 3(5y- 3x)

20x – 12y = 15y- 9x

29x= 27y

⇒ $\frac{x}{y}=\frac{27}{29}$

Question 13. If A : B = 3: 2 and B: C = 3:5 then A:B:c is

9:6:10
6:9:10
10:9:6
None of these

Answer: 9: 6: 10.

A : B: C =?

A:B = 3:2 = 9:6

B:C = B:S=6:10

A : B : C = 9 : 6 : 10

Question 14. If x/2 = y/3 = z/7, then the value of (2x – 5y + 4z)/2y is

6/23
23/6
3/2
17/6

Answer: 4. 17/6

x:y:z= 2:3:7 x = 2x, y=3x, z=7x

⇒ $\frac{2 x-5 y+4 z}{2 y}=\frac{4 x-15 x+28 x}{6 x}$

⇒ $=\frac{17 x}{6 x}=\frac{17}{6}$

Question 15. If x : y = 2 : 3,y: z = 4 : 3 then x:y : z is

2 : 3: 4
4 : 3: 2
3: 2: 4
None of these

Answer: 4. None of these

x:y = 2:3 = 8:12

y:z = 4:3 = 12:9

x:y:z=8: 12:9 =8:12:9

Question 16. Division of into 3 parts in the ratio 4: 5: 6 is

(200,250,300)
(250, 250, 250)
(350, 250,150)
(8 12 9)

Answer: 1. 200, 250, 300.

4x + 5x + 6x= 750

15x = 750

x = 50

→ 750 will be divided as 200,250,300

Question 17. The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio of 7:8:9. Their present ages are

(45, 50, 55)
(40, 60. 50)
(35,45, 70)
None of these

Answer: 1. 45,50,55

7x, 8x, 9x

Present age

7x + 10, 8x+ 10 + 9x + 10 = 150

24x= 120

x = 5

Present age = 45,50,55

Question 18. The numbers 14, 16, 35, and $2 ‘are not in proportion. The fourth term for which they will be in proportion is

45
40
18
None of these

Answer: 3. 40

14,16,35,x

⇒ $\frac{14}{16}=\frac{35}{x}$

⇒ $\frac{7}{8}=\frac{35}{x} \rightarrow x=40$

Question 19. If x/y = z/w, implies y/x = w/z, then the process is called /

Dividend
Componendo
Alternendo
None of these

Answer: 4. None of these. This is inverted

Question 20. Ifp/q = r/s = p-r/q-s, the process is called

Subtrahendo
Addendum
InVert£hdo
None of these

Answer: 1. subtrabendo

Question 21. If a/b = c/d, implies (a + b)/(a – b) = (c + d)/(c- d), the process is called

Componendo
Dividendo
Componendo and Dividendo
None of these

Answer: 3. Componendo & dividendo

Question 22. Ifu/v = w/p, then (u – v)/(u + v) = (w – p)/(w + p). The process is cached

Invertendo
Alternendo
Addendum
None of these

Answer: 4. None of these

Question 23. 12, 16, * 20 are in proportion. Then * is

25
14
15
None of these

Answer: 3. 15

⇒ $\frac{12}{16}=\frac{x}{20}$

⇒ $\frac{3}{4}=\frac{x}{20} \rightarrow x=15$

Question 24. 4, 9, 13V2 are in proportion. Then is

6
8
9
None of these

Answer: 1. 6

⇒ $\begin{aligned}
& \frac{4}{x}=\frac{9}{13.5}=\frac{36}{9 x}=\frac{9}{13.5} \\
& x \rightarrow 6
\end{aligned}$

Question 25. The mean proportional between 1.4 gms and 5.6 gms is

28 gms
2.8 gms
3.2 gms
None of these

Answer: 2. 2.8gms

x2 =1.4 x 5.6 = 7.84 = x2.8 gms

Question 26. $\frac{a}{4}=\frac{b}{5}=\frac{c}{9} \text { then } \frac{a+b+c}{c} \text { is }$

4
2
7
None of these.

Answer: 2. 2

a : b: c = 4 : 5 : 9

a = 4x

b = 5x

c = 9x

⇒ $\frac{a+b+c}{c}=\frac{18 x}{9 x}=\frac{2}{1}=2 $

Question 27. Two numbers are in the ratio 3: 4; if 6 is added to each term of the ratio, then the new ratio will be 4 5, then the numbers are

14, 20
17, 19
18 and 24
None of these

Answer: 3. 18 and 24

⇒ $\frac{a+6}{b+6}=\frac{4}{5}$

5a + 30 = 4b + 24

5a -4b = -6

15x- 16x = -6

-x = -6

x = 6

a = 18 b=24

Question 28. $\frac{a}{4}=\frac{b}{5} \text { then }$

$\frac{a+4}{a-4}=\frac{b-5}{b+5}$
$\frac{a+4}{a-4}=\frac{b+5}{b-5}$
$\frac{a-4}{a+4}=\frac{b+5}{b-5}$
None of these

Answer: 3. Applying componendo And dividendo

⇒ $\frac{a+4}{a-4}=\frac{b+5}{b-5}$

Question 29. If a : b = 4 : 1 then $\sqrt\frac{a}{b}+\sqrt{\frac{b}{a}} \text { is }$

5/2
4
5
None of these

Answer: 1. 5/2

$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=
\sqrt{\frac{4 x}{x}}+\sqrt{\frac{x}{4 x}}
=2+\frac{1}{2}=\frac{5}{2}$

Question 30. If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c} \text { then }$ (b- c)x + (c – a)y + (a – b)z is

1
0
5
None of these

Answer: 2. 0

⇒ $\frac{(b-c) x}{(b-c)(b+c-a)}=\frac{(c-a) y}{(c-a)(c+a-b)}=\frac{(a-b) z}{(a-b)(a+b-c)}$

(b-c)x : (c-a)y: (a-b)z

= (b-c) (b+c-a) : (c-a) (c+a-b) : (a-b)(a+b-c)

= (b-c)x + (c-a)y + (a-b)z

= b² + be- ab -bc – c² + ac- ac + c² + ac – cb – a² + ab + a² + ab – ac- ab – b² + bx

= 0

Ratio And Proportion Exercise – 3

Question 1. If A : B : C = 2: 3 : 4, then $\frac{A}{B}: \frac{B}{C}: \frac{C}{A}$is equal to :

4: 9: 16
8: 9: 12
8: 9: 16
8: 9: 24

Solution:

Let = 2x, It = 3x and C 4x. Then $\frac{A}{n}=\frac{2 x}{3 x}=\frac{2}{i^{\prime}} \cdot \frac{n}{C}=\frac{3 x}{4 x}=\frac{3}{4} \text { and } \frac{C}{A}=\frac{4}{2 x}=\frac{2}{1}$

⇒ $\Rightarrow \frac{A}{B}: \frac{B}{c}: \frac{c}{A}=\frac{2}{3}: \frac{3}{1}: \frac{2}{1}=8: 9: 24 .$

Question 2. A : B = 2 : 3 , B : C = 4 : 5 and C : D = 6 : 7,then A:B:C:D is

16:22:30:35
16:24:15:35
16:22:30:35
18:24:30:35

Solution:

A: B = 2 : 3. B : C = 4 : 5 =$\left(4 \times \frac{3}{4}\right):\left(5 \times \frac{3}{4}\right)=3: \frac{15}{4}$

And C:D 6:7 =$=\left(6 \times \frac{15}{24}\right):\left(7 \times \frac{15}{24}\right)=\frac{15}{4}: \frac{35}{8}$

A : B : C: D = 2: 3 :$\frac{15}{4}: \frac{35}{8}=16: 24: 30: 35 \text {. }$

Question 3. If 0.75 : x: : 5 : 8, then x is equal of x is :

22: 29
11.20
1.25
1.30

Solution:

(x X 5) = (0.75 x 8)

⇒ $x=\frac{4.8}{4}=1.20$

Question 4. If x: y = 5 : 2, then (8x + 9y) : (8x + 2y) is :

22: 29
6: 61
9: 22
61: 26

Solution:

Let x = 5k and y = 2k. Then $\frac{8 x+9 y}{8 x+2 y}=\frac{(8 \times 5 k)+(9 \times 2 k)}{(8 \times 5 k)+(2 \times 2 k)}=\frac{58 k}{44 k}=\frac{29}{22}$

(8x + 9y) : (8x + 2y) = 29 : 22

Question 5. The salaries of A, B, and C are in the ratio 2 : 3: 5. If the increments of 15%, 10%, and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?

3 : 3: 10
10 : 11: 20
23: 33: 60
Cannot be determined

Solution:

Let A = 2k, B = 3k and C = 5k.

A’s new salary =$\left.=\frac{115}{100} \text { of } 2 \mathrm{k}=\left(\frac{115}{100} \times 2 k\right)\right)=\frac{23}{10} \mathrm{k}$

B’s new salary = $\left.=\frac{110}{100} \text { of } 3 k=\left(\frac{110}{100} \times 3 k\right)\right)=\frac{33}{10} k$

Question 6. If Rs. 782 is divided into three parts, proportional to then the first part is:

Rs. 182
Rs. 190
Rs. 196
Rs. 204

Solution:

Given ratio = $\frac{1}{2}: \frac{2}{3}: \frac{3}{4}=6: 8: 9$

⇒ $1^{\text {st }} \text { part }=\operatorname{Rs}\left(782 \times \frac{6}{23}\right)=\operatorname{Rs} .204 .$

Question 7. Two numbers are in the ratio 3: 5. If 9 is subtracted from each, the new number is in the ratio 12: 23. The small number is :

Solution:

Let the number be 3x and 5x. Then $\text { Then, } \frac{3 x-9}{5 x-9}=\frac{12}{23} \Leftrightarrow 23(3 x-9)=12(5 x-9) \Leftrightarrow 9 x=99 x=11 \text {. }$

The smaller number = (3 x 11) = 33

Question 8. Two numbers are in the ratio 1: 2. If 7 is added to both, their ratio changes to 3: 5. The greatest number is :

Solution:

Then $\frac{x+7}{2 x+7}=\frac{3}{5} \rightarrow 5 x+35=6 x+21 \rightarrow x=14$

Greatest number = 2x = 28

Question 9. In a bag, there are coins of 25, p, 10 p, and 5 p in the ratio of 1: 2 : 3. If there are Rs. 30 in all, how many 5 p coins are there?

Solution:

Let the number of 25 p, 10 p, and 5 p coins be x, 2x, and 3x respectively.

Then, sum of their values = $\operatorname{Rs.}\left(\frac{25 x}{100}+\frac{10 \times 2 x}{100}+\frac{5 \times 3 x}{100}\right)=\operatorname{Rs} \cdot \frac{60 x}{100}$

⇒ $\frac{60 . x}{100}=30 \Leftrightarrow \mathrm{x}=\frac{30 \times 100}{60}=50 .$

Hence, the number of 5 p coins = (3 x 50) = 150

Question 10. Salaries the new ratio of Ravibecomesand Sumit40:are57. in What the isratioSumit’s2 : 3. present the salary salary? of each is increased by Rs. 4000,

Rs. 17,000
Rs. 20,000
Rs. 25,500
None of these

Solution:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then $\text { 1. } \frac{2 x+4000}{3 x+4000}=\frac{40}{57} \Leftrightarrow 57(2 x+4000)=40(3 x+4000) \Leftrightarrow 6 x=68000 \Leftrightarrow 3 x=34000$

Sumit’s present salary = (3x + 4000) = Rs. (34000 + 4000) = Rs. 38,000.

Question 11. If Rs. 510 is divided among A, B, and C in such a way that A gets what B gets and B gets what C gets, then their shares are respectively:

Rs. 120, Rs. 240, Rs. 150
Rs. 60, Rs. 90, Rs. 360
Rs. 150, Rs. 300, Rs. 60
None of these

Solution:

⇒ $\left(A=\frac{2}{3} B \text { and } B=\frac{1}{4} C\right) \Leftrightarrow \quad \frac{A}{B}=\frac{2}{3} \text { and } \frac{B}{C}=\frac{1}{4}$

A : B = 2 : 3 and B:C=1:4 = 3:12=>A:B:C = 2:3:12.

⇒ $\text { A’s share }=\text { Rs. }\left(510 \times \frac{2}{17}\right)=\text { Rs. } 60 \text {; B’s share }=\text { Rs. }\left(510 \times \frac{3}{17}\right)=\text { Rs. } 90 \text {; }$

⇒ $\text { C’s share }=\text { Rs. }\left(510 \times \frac{12}{17}\right)=\text { Rs. } 360 \text {. }$

Question 12. The sum of the three numbers is 98. If the ratio of the first to the second is 2 : 3 and that of the second to the third is 5: 8, then the second number is:

Solution:

Let the three parts be A, B, and C Then,

A : B : 2 : 3 and B : C = 5 : 8 $\left(5 \times \frac{1}{5}\right):\left(8 \times \frac{1}{5}\right)=3: \frac{14}{5}$

A : B : C = 2 : 3 :$\frac{24}{5}=10: 15: 24 \Rightarrow B=\left(98 \times \frac{15}{19}\right)=30 .$

Question 13. A fraction which bears the same ratio to $\frac{1}{27} \text { that } \frac{3}{11}$ does to $\frac{5}{9}$ is equal to.

$\frac{1}{55}$
$\frac{1}{11}$
$\frac{3}{11}$
55

Solution:

Let $ x: \frac{1}{27}:: \frac{1}{11}: \frac{5}{9}$

Then $x \times \frac{5}{9}=\frac{1}{27} \times \frac{3}{11} \Leftrightarrow x=\left(\frac{1}{27} \times \frac{3}{11} \times \frac{9}{5}\right)=\frac{1}{55} .$

Question 14. A sum of Rs. 1300 is divided amongst P, Q, R and S such that Then, $\frac{p^{\prime} \text { share }}{Q^{\prime} s \text { share }}=\frac{p^{\prime} \text { share }}{\text { Rsshare }}=\frac{R^{\prime} \text { share }}{S^{\prime} \text { share }}=\frac{2}{3} \text {. }$ P’s share is :

Rs. 140
Rs. 160
Rs. 240
Rs. 320

Solution:

Let P = 2x and Q = 3x. Then $\frac{19}{1}=\frac{2}{3} \Rightarrow R=\frac{3}{2}Q=\left(\frac{3}{2} \times 3 x\right)=\frac{1 x}{2} .$

⇒ $Also, \frac{R}{S}=\frac{2}{3} \Rightarrow S=\frac{3}{2} R=\left(\frac{3}{2} \times \frac{9 x}{2}\right)=\frac{27 x}{1}$

Thus, $\mathrm{P}=2 \mathrm{x}, \mathrm{Q}=3 \mathrm{x}, \mathrm{R}=\frac{9 \mathrm{x}}{2}$ and $\mathrm{S}=\frac{27 \mathrm{x}}{4}$.

⇒ P+(Q+R+S= & 1300 $\Leftrightarrow\left(2 x+3 x+\frac{9 x}{2}+\frac{27 x}{3}\right)=1300 $

⇒ $\Leftrightarrow(8 x+12 x+18 x+27 x)=5200$

⇒ $\Leftrightarrow 65 x=5200 \Leftrightarrow x=\frac{5200}{65}=80 $

P’s share = Rs. (2 * 80) = Rs, 1 60

Question 15. A and B together have Rs. 1210. If $\frac{4}{15}$ of A’s amount is equal to $\frac{2}{5}$ of B’s amount, how much amount does B have?

Rs. 460
484
Rs. 550
664

Solution:

⇒ $\frac{1}{15} A=\frac{2}{5} B=A=\left(\frac{2}{5} \times \frac{15}{4}\right) B \Leftrightarrow A=\frac{3}{2} B \Leftrightarrow \frac{A}{B}=\frac{3}{2} \Leftrightarrow A: B=3: 2$

⇒ $\text { B’s share }=\text { Rs. }\left(1210 \times \frac{2}{4}\right)=\text { Rs. } 484 .$

Question 16. Two numbers are respectively 20% and 50% more than a third number The ratio of the two numbers is:

2: 5

3: 5

4: 5

6: 7

Solution:

Let the third number be x.

Then, first number = 120% of x $=\frac{120 x}{100}=\frac{6 x}{5} ;$

Second number = $150 \% \text { of } \mathrm{x}=\frac{150 x}{100}=\frac{3 x}{2} \text {. }$

Ratio of first two numbers$=\frac{6 x}{5}: \frac{3 x}{2}=12 x: 15 x=4: 5 .$

Question 17. Seats for Mathematics, Physics, and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40%, 50%, and 75% respectively. What will be the ratio of increased seats?

2 : 3: 4
6: 7: 8
6: 8: 9
None of these

Solution:

Originally, let the number of seats for Mathematics, Physics, and Biolog}’ be 5x, 7x, and 8x respectively.

Number ofincreased seats are (140% of5x), (50%. of 7x) and (175% of8x) $\left(\frac{140}{100} \times 5 x\right),\left(\frac{150}{100} \times 7 x\right) \text { and }\left(\frac{175}{100} \times 8 x\right) \text { i.e. } 7 \times \text {, } \frac{21 x}{2} \text { and } 14 \times \text {. }$

⇒ Required ratio = $7 \mathrm{x}: \frac{21 x}{2}: 14 \mathrm{x}=21 \mathrm{x}: 21 \mathrm{x}: 28 \mathrm{x}=2: 3: 4$

Question 18. The ratio of the number of boys and girls in a college is 7: 8. If the percentage increase in the number of boys and girls is 20% and 10% respectively, what will be the new ratio?

8: 9
17:18
21: 22
Cannot be determined

Solution:

Originally, numberless the of boys and girls in the college was 7x and 8x respectively. Their increased is number(120%of7x) and (110%8x)

⇒ $\left(\frac{120}{100} \times 7 x\right) \text { and }\left(\frac{110}{100} \times 8 x\right) \text { i.e. } \frac{42 x}{5} \text { and } \frac{44 x}{5} \text {. }$

Required ratio = $=\frac{42 x}{5}: \frac{44 x}{5}=21: 22 .$

Question 19. A sum of money is to be distributed among A, B, C, and D in the proportion of 5: 2: 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

Rs. 500
Rs. 1500
Rs. 2000
None of these

Solution:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x – 3x = 1000 ⇔ x = 1000.

B’s share = Rs. 2x = Rs. (2 * 1000] = Rs. 2000

Question 20. If 40% of a number is equal to two-thirds of another number, what is the ratio of the first number to the second number?

2: 5
3: 7
5 : 3
7 : 3

Solution:

Let 40% of A =$\frac{2}{3} B \text {. Then, } \frac{40 A}{100}=\frac{2 B}{3} \Leftrightarrow \frac{2 A}{5}=\frac{2 B}{3} \Leftrightarrow \frac{A}{B}=\left(\frac{2}{3} \times \frac{5}{2}\right)=\frac{5}{3}$

A: B = 5 : 3.

Question 21. The ratio of the earnings of A and B is 4: 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earnings becomes 8: 7. What are A’s earnings?

Rs. 21,000
Rs. 26,000
Rs. 28,000
Data inadequate

Solution:

Let the original earnings of A and B be Rs. 4x and Rs. 7x.

Net earnings of A = 150% or Rs. 4x = Rs$\left(\frac{150}{100} \times 4 x\right)=\mathrm{Rs.} .6 \mathrm{x}$

Net earnings of B = 75% of Rs. 7x = Rs.$\left(\frac{75}{100} \times 7 x\right)=\operatorname{Rs} . \frac{21 x}{4} .$

⇒ $6 x: \frac{21 x}{4}=8: 7 \Leftrightarrow \frac{6 x \times 4}{21 x}=\frac{8}{7} .$

This does not give x. So, the given data is inadequate.

Question 22. The fourth proportional to 5, 8, 15, is:

Solution:

Let the fourth proportional to 5, 8, 15 be x.

Then, 5 : 8 : : 15 : x <=> 5x = (8 x 15) <=> $x=\frac{(8 \times 15)}{5}=24$

Question 23. x varies inversely as the square of y. Given that y = 2 for x = 1. The value of x for y = 6 will be equal to :

3
9
$\frac{1}{3}$
$\frac{1}{9}$

Solution:

Given x = $\frac{k}{y^2}$ . where 1< is a constant.

Now, y = 2 and x = 1 gives 1< = 4, $x \frac{4}{y^2} \Rightarrow x=\frac{4}{6^2} \text {, when } y=6 \Rightarrow x=\frac{4}{36}=\frac{1}{9} \text {. }$

Question 24. If 10% of x = 20% off, then x : y is equal to :

1:2
2: 1
5: 1
10: 1

Solution:

10%ofx = 20% of y a $\Leftrightarrow \frac{10 x}{100}=\frac{20 y}{100} \Leftrightarrow \frac{x}{10}=\frac{y}{5} \Leftrightarrow \frac{x}{y}=\frac{10}{5}=\frac{2}{1} . \mathrm{x} ; \mathrm{y}=2: 1 .$

Question 25. What number should be added to each term in the ratio 19:43, so that it becomes equal to 2:3?

Solution:

Let x be the number to be added.

(19+x) : (43+x)=2:3 57+3x=86+2x x=29

29 must be added to each term in the ratio 19:43 so that it becomes equal to 2:3.

Question 26. A construction company is planning to invest in road and railway line construction in the ratio4:5. If the amount invested in the railway line construction is 6 million, then how much money did the company invest in the road construction?

14 million
10.8 million
4.8 million
2.6 million
7.5 million

Solution:

let the company’s investment in the road construction be x.

4:5=x:6 or4/5=x/6

x=(6x 4)/5 =4.8

The company invested Rs.4.8 million in road construction. Hence, option c

Question 27. If the incomes of A and B are in the ratio 3;4 and their expenditures are in the ratio 2:3, then find the ratio of their savings.

1:1
1:9
1:2
cannot be determined

Solution:

Let the incomes of A and B be 3x and 4x respectively. Let their expenditures be 2y and 3y respectively.

Savings = Income – Expenditure

A’s savings/B’s savings=(3x-2y)/4x-3y) the values of x and y are not known.

Hence, the ratio of savings cannot be determined. Hence, option d

Question 28. The total money collected for New Year celebrations in a certain building was Rs.20,500. The ratio of the amount contributed by the people of the A-wing to that contributed by the people of the B wing was 8:5 also, the ratio of the amount contributed by the people of the B wing to that contributed by the people of the C wing was 2:3 find the amount contributed by the people of the wing.

Rs.5000
Rs.2000
Rs.2500
Rs.3000
Rs.500

Solution:

Since the amount collected by B wing is common to both the ratios; it is to be used to compare the collections of all 3 wings.

Hence, find the LCM of 5 and LCM of 5 and 2=10

The ratio of the amounts contributed by the people of all the three wings- 16:10:15

The amount contributed by each wing is 16x, lOx, and 15x respectively.
16x+10x+15x= 20500 x=500 i.e. 10x= 5000

Hence, the amount collected by B wing is Rs.5000. Hence, Option a

Question 29. 78 is divided into two parts such that the ratio between those two parts in 7:6 finds the product of those two parts.

1215
2808
l512
3276
1014

Solution:

Let one of the parts be x.

The other part is (78-x)

The ratio between the two parts is 7:6

⇒ $\frac{x}{(78-x)}=\frac{7}{6} 6 $

∴ 6x=546-7x

∴ 13x=546

x=42 and (78-x) = 36. Product of 42 and 36 = 1512 Hence, option c.

Question 30. During the elections for the post of building society chairman, the ratio of the number of members with Mr . Shah to that with Mr.Raheja was 6ÿ5 but 24 members from Mr. Shah’s side defected and joined Mr, Raheja. Now the ratio of members with Mr, Shah to that with Mr. Raheja is 2:3 find the number of members siding with Mr. Shah initially.

Solution:

Let the initial number of members with Mr. Shah be 6k and the number of members with Mr.

Raheja be sk. 24 members went over from Mr. Shah’s side to Mr. Itaheja’s side.

Heme, the number of members now supporting Mr. Shah is 6k-24 while the number of members with Mr. Raheja is 5k+24.

This ratio is now 2;3

(6k-24): (5k+24) =2:3

18k-72=10k + 48

8k= 1 20

k= 15

The number of members with Mr. Shah initially =6k=90. Hence, option a

Question 31. Vessel 1 contains 38 litres of milk and vessel 2 contains 24 litres of water. 8 liters of milk is taken from vessel 1 and placed in vessel 2. Then, 201 liters of the mixture is taken from vessel 2 and placed in vessel 1. Find the ratio of milk in vessel 1 to water in vessel 2.

4:9
15:35
15:4
35:9
35:3

Solution:

Business Mathematics Ratio And Proportion chapter 1

Now, the ratio of milk to water in vessel 2 is 8:24 = 1:3 and the ratio of milk to the total solution in vessel 2 is 8: 32 = 1:4

Of the 20 liters, 1/4″1 (i.e. 5 liters) is milk, and 3/4″‘ (i.e. 1 5 liters) is water.

After the 2lul iteration, amount of milk in vessel 1=30 +5=35 litres and amount of water in vessel 2= 24- 15 = 9 litres

The ratio of milk in vessel 1 to water in vessel 2 is 35:9. Hence, Option d

Question 32. Does Aakash have coins of 50 paise, 25 paise, and Rs.1.50 in the ratio 1:2:3 (Aakash(e) stays in a country where all are valid currency coins. Also, in his country. 1 rupee equals 100 paise) how many coins of 25 paise does Aakash have if he has got Rs. 6600 in all? 32.

2000
2200
2400
2600
2800

Solution:

The ratio of the number of coins is 1;2:3 for the 50 paise, 25 paise, and Rs. 1.50 coins respectively.

in terms of monetary value, the ratio becomes (1x 0.5):(2x 0.25): (3 x 1.5)which equals o.5:0.5:4.5, i.e., 1:1:9

∴ (1/11)th of the total value comes from 25 paise coins, (1/2)x6600= Rs.600 in the form of 25 paise coins.

The total no. of 25 paise coins is 600/0.25=2400. Hence, option c

Question 33. The annual income of Mr.X and Mr. Y is in the ratio of 9:8 and their expenditures are in the ratio of 5:4 if both individually manage to save Rs. 5000, then B’s expenditure is:

Rs.1,250
Rs.5,000
Rs.6,250
Rs.l1,250
Rs.10,000

Solution:

Let the annual income of Mr. Y be rs9x and 8x respectively.

Also, let their expenditures be Rs. 5y and Rs. 4y respectively.

Both individually save Rs. 5000

Income -Expenditure = savings for Mr. X

9x – 5y = 5000 —-(1)

8x-4y=5000 (2), x=1250

for Y,equations =(1)(2)andy=1250

B’s Mr.expenditure 4y = Rs.5000

Hence, option b

Question 34. If 5x-13y= 3x-8y, find the value of (2×2 + 3y2): [2×2 – 3y2)

50:12
62:39
25:4
31:19

Solution:

5x-13y=3x-8y

2x=5y

X:y=5:2

x²:y² = 25:4

2x²: 3y² =50:12

Using components and Dividendo law, Hence, option d

Question 35. A group of children went to play a game of marbles. Amar had 9 marbles, Akbar had 6 marbles and the youngest Anthony had none. So they decided to share their marbles equally among themselves. In return, Anthony offered to give them his 15 Pokemon cards. He gave the cards in the same proportion in which he received the marbles. How many cards did Akbar get from Anthony?

Solution:

total number of marbles =15

The marbles are distributed in a way that each of them has an equal number of marbles, each of them should have 5 marbles.

Anthony gets 4 marbles from Amar and 1 marble from Akbar, i.e. in the ratio of 4:1

The 15 cards are distributed proportionally

The number of cards Akbar gets = (1/5)x15=3

Hence, option d.

Question 36. Based on their performance in a test, Professor Shetty distributed Rs. 798 among Vinay and Vinit such that 6 times Vinod’s share is equal to 10 times Vinay’s share or 5 times Vinit’s share. How much does Vinod get?

Solution:

Let Vinod’s share be x.

6x= 10X (Vinay’s share)

Vinay’s share=3x/5

Similarly, visit’s share =6x/5

x+ (3x/5) +6x/5) =14x/5 = 798

x= (798/14)x5=57×5=285

Hence, option e,

Question 37. Find the fourth proportional to 3,5 and 27.

45
16.2
135
55

Solution:

Let x be the fourth proportional.

3/5 = 27/x

x = (27×5) /3=45

Hence, option a

Question 38. If $\frac{x-y}{x^2-y^2}=\frac{x^2-y^2}{k}$ , them K =?

(x-y)(x²-y²)
(XY) (x²– y²)
(x+y) (x²– y²)
(x²– y²) (x+y)

Solution:

From the given equation,$\mathrm{k}=\frac{\left(x^2-y^2\right)^2}{x-y} \text { now, }\left(x^2-y^2\right) \text { can be written as }(x+y)(x-y) \text { so we have } \mathrm{k}=\frac{(x+y)^2 \times(x-y)^2}{x-y}$

=(x + y² X (x- y)

= (x+y)(x+y)(x-y)

=[x+y] (x²– y²)

Question 39. The cost of manufacturing a circular cast from the plate is directly proportional to the square root of its diameter. A 24 cm (diameter) plate casts Rs. 346. How much more or less will it cost to many facture 2 plates with diameters 1 8 cm and 8cm?

Rs.143 more
Rs.l53.41 more
Rs.282.31 more
Rs.282.3l less
Rs.l82.3 less

Solution:

Cost is proportional to the square root of the diameter.

Cost=k

xÿ/diameter, Where k is a constant of proportionality

The cost of a plate with 24 cm diameter=k√24 =2k √6

Similarly, the cost of the plate with 18 cm diameter=k√l8 = 3k√2

And the cost of a plate with 8 cm diameter = k V8 = 2k√2

The cost of an 18 cm and 8cm plate = 31<√2 + 2/c√2 =Rs.5k√2

Let the cost of the 18 cm and 8cm plates put together be x.

⇒ $346: \mathrm{x}:: 2 \mathrm{k} \sqrt{6}: 5 \mathrm{k} \sqrt{2}\mathrm{x}=\frac{(346 \times 5 \sqrt{2})}{2 \sqrt{6}}=\frac{346 \times 5}{2 \sqrt{3}}=\frac{173 \times 5}{\sqrt{3}} \approx \frac{173 \times 5}{1.73} \approx 500$

The additional cost of making the plates « 500- 346 = RS. 1 54. Hence, option 2

Question 40. If $\frac{x}{y+z-x}=\frac{y}{z+x-y}=\frac{z}{x+y-z}=r$then r cannot take any value except.

1
$-\frac{1}{2}$
$1 \text { or } \frac{1}{2}$
$1 \text { or } \frac{-1}{2}$

Solution:

⇒ $\text { If } \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k \text {, }$

⇒ $\text { Then } \frac{a+c+e}{b+d+f}=k$

Using this property, add the numerators and denominators to get,$r=\frac{x+y+z}{x+y+z}=1$

Similarly, if we subtract the denominators and numerators of the first two ratios, you get$r^{\prime}=\frac{x-y}{(y+z-x)-(z+x-y)}=\frac{x-y}{2(y-x)}=\frac{-1}{2}$ r can take the values 1 and -1/2. Hence, option d

Question 41. A precious stone is accidentally broken into 2 pieces whose weights are in the ratio 4;5 the value of the stone is directly proportional to the square of its weight. What is the ratio of the total value of the original (unbroken) stone to the total value of the broken pieces?

41:81
81:41
40:81
81:40
None of these

Solution:

Let the weights of the two pieces be 4x and 5x.

Therefore, the weight of the original stone was 9x.

The value of the stone is directly proportional to the square of its weight.

The values of the original stone and the two pieces. Are proportional to 81×2,16×2 and 25×2.

Hence, the required ratio is 81 : ( 16+25) = 81:41.

Hence, option b

Question 42. Two numbers are in the ratio of 2: 3 and the difference of their squares is 320. The numbers are:

12, 18
16, 24
14,21
None

Solution:

Tricks: (1) ; (2) & (3) all are In ratio

2:3 Rut for option

182 – 122 # 320

242- 162 = (24 + 16) (24 – 16)

= 40 X 8 = 320

Question 43. If p : q is the sub-duplicate ratio ofp – x2: q – x-‘, then x’ is :

$\frac{p}{p+q}$
$\frac{q}{p+q}$
$\frac{q p}{p-q}$
None

Solution:

Detail Method:

⇒ $\frac{\sqrt{p-x^2}}{q-x^2}=\frac{p}{q}$

Squaring on both side; wc get

⇒ $\frac{p-x^2}{q-x^2}=\frac{p^2}{q^2}$

Or pq² – q² x²= p² q – p² x²

Or p2 x²– q² x² = p²q – pq²

Or x² (p² – q²) = Pq (P – q)

Or x2 (p + q) (p-q) = pq (p-q)

⇒ $\text { Or } x^2=\frac{p q}{p+q}$

(4) is correct

Question 44. An alloy is to contain copper and zinc in the ratio 9: 4. The zinc required to melt with 24kg of copper is:

$10 \frac{2}{3} \mathrm{~kg}.$
$10 \frac{1}{3} \mathrm{~kg}$
$9 \frac{2}{3} \mathrm{~kg}$
9 Kg

Solution:

Let Zinc = x kg

⇒ $\frac{9}{4}=\frac{24}{x} \mathrm{x}=\frac{4 \times 24}{9}=\frac{32}{3}$

⇒ $=10 \frac{2}{3} \mathrm{~kg}$

(1) is correct

Question 45. Two numbers are in the ratio 7:8. If 3 is added to each of them, their ratio becomes 8: 9. The numbers are

14, 16
24, 27
21, 24
16, 18

Solution:

Let x be common in the ratio

et x is common in the ratio

Number are 7x and 8x

⇒ $\frac{7 x+3}{8 x+3}=\frac{8}{9}$

Or 64x + 24 = 63x + 27

Or 64x – 63x = 27- 24

Or x = 3

1st number = 7x = 7×3 = 21

2nd number = 8x = 8 x 3 = 24

(3) is correct

Question 46. A box contains ₹56 in the form of coins of one rupee, 50 paise, and 25 pages. The number of 50 paise coins is double the number of 25 paise coins and four times the number of one rupee coins. The numbers of 50paise coins in the box are:

Solution:

Let No. of 50 Paise coins = x

No. of 1 coins =$\frac{x}{4}$

No. of 25 paise coins= $=\frac{x}{2}$

Total value = $\frac{x}{4} \times 1 \times \times 0.50+\frac{x}{2} \times 0.25=56 \quad \text { Or } 0.25 x+0.50 x+0.125 x$

⇒ $\text { Or } 0.875 x=56 \quad \text { Or } x=\frac{56}{0.875}=64$

(1) is correct

Question 47. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increases to:

$1 / 9$
$1 / 8$
$1 / 7$
$7 / 8$

Solution:

Per Person share increase =$\frac{1}{7}$ of total share

Question 48. A bag contains 187 in the form of 1 rupee, 50 paise, and 10 paise coins in the ratio of 3:4:5. Find the number of each type of coin:

102, 136, 170
136, 102, 170
170, 102, 136
None

Solution:

let x Be common In the 2m ratio

No, of I Kopec; 50 Pulse ;KI 10 Raise

Coins are 3x ; 4x and 5x

3x x I + 4x x 0.50 + 5x x 0.10 = 187

Or 5.50x = 187

⇒ $x=\frac{1117}{5.50}=34$

No.of 1 rupee coins = 3x=3×34=102

No of 50 paise coins = 4x= 4×34=136

No. of 10 Paise coins = 5x = 5 x 34 =170

(1) is correct

Question 49. P, Q, and R are three cities. The ratio of average temperature between P and Q is 11: 12 and that between P and R is 9: 8. The ratio between the average temperature of Q and R is:

22: 27
27: 22
32: 33
None

Solution:

P : Q = 11 : 12 P = 12 : 11

⇒ $\frac{Q}{p} \times \frac{p}{R}=\frac{12}{11} \times \frac{9}{8}$

⇒ $\frac{Q}{R}=\frac{27}{22} \quad Q: R=27: 22$

(2) is Correct

Question 50. ₹407 are to be divided among A, B, and C so that their shares are in the ratio $\frac{1}{4}: \frac{1}{5}: \frac{1}{6}.$ The respective shares of A, B, and C are:

₹165, ₹132,
₹165, ₹110, ₹132
₹132, ₹165
₹110, ₹132, ₹165

Solution:

⇒ $A: B: C=\frac{1}{4}: \frac{1}{5}: \frac{1}{6} \times$ LCM of denominators = 60 = 15:12:10

A’s share =$=\frac{407}{15+12+10} \times 15=\text { Rs. } 165$

B’s share$=\frac{407}{37} \times 12=\text { Rs. } 132$

C’s share$=\frac{407}{37} \times 10=\text { Rs. } 110$

Question 51. The incomes of A and B are in the ratio 3: 2 and their expenditures in the ratio 5 : 3. If each saves ₹1,500, then, B’s income is :

₹6,000
₹4,500
₹3,000
₹7,500

Solution:

Let x be common in the ratio.

A’s income = 3x

B’s income = 2x

⇒ $\frac{3 x-1500}{2 x-1500}=\frac{5}{3}$

Or 10x- 7500 = 9x- 4500

Or 10x-9x = 7500 -4500

Or x = 3000

B’s income = 2x = 2 x 3000

= 6000.

(1) is correct

Question 52. In 40 liter mixture of glycerine and water, the ratio of glycerine and water is 3: 1. The quantity of water added in the mixture to make this ratio 2:1 is:

15 litres
10 litres
8 litres
5 litres

Solution:

Glycerine = $\frac{40}{3+1} \times 3=30 \text { litres. }$

Water =$=\frac{40}{4} \times 1=10 \text { litres }$

Let x liters of water be added to the mixture

⇒ $\text { Then } \frac{30}{10+x}=\frac{2}{1}$

Or, 2x + 20 = 30 or x = 5

(4) is Correct

Question 53. The third proportional to (a² – b²) and (a + b)² is:

$\frac{a+b}{a-b}$
$\frac{a-b}{a+b}$
$\frac{(a-b)^2}{a+b}$
$\frac{(a+b)^3}{a-b}$

Solution:

3rd Proportion$I=\frac{(\text { Mean Prop. })^3}{\text { 1st Proportional }}$

⇒ $=\frac{\left((a+b)^2\right)^2}{a^2-b^2}=\frac{(a+b)^4}{3(a+b)(a+b)}=\frac{(a+b)^3}{a-b}$

(4) is Correct

Question 54. The ages of the two persons are in the ratio of 5:7. Eighteen years ago their ages were in the ratio of 8: 13, and their present ages (in years] are:

50, 70
70, 50
40, 56
None

Solution:

(1) and (3) in the ratio 5: 7 not (2)

For (a) 18 years ago

⇒ $\frac{50-18}{70-18}=\frac{32}{32}=\frac{8}{13} \text { (True) }$

So. (1) is Correct

Question 55. If A, B, and C started a business by investing ₹1,26,000, ₹84,000, and ₹2,10,000. If at the end of the year, profit is ₹2,42,000 then the share of each is:

₹72,600; ₹48,400; ₹1,21,000
₹48,400; ₹1,21,000; ₹72,600
₹72,000; ₹49,000; ₹1,21,000
₹48,000; ₹1,21,400; ₹72,600

Solution:

⇒ $ \text { A’s share }=\frac{\{242,000}{3+2+5} \times 3=₹ 72,600 $

⇒ $\text { B’s share }=\frac{242,000}{20} \times 2=₹ 48,400$

⇒ $\text { C’s share }=\frac{242,000}{10} \times 5=₹ 1,21,000$

Investment ratio is A : B : C = 126,000 : 84,000 : 2,10,000 + 14,000

= 9: 6: 15 + 3

=3:2:5

Question 56. If $\frac{p}{q}=-\frac{2}{3} \text { then the value of } \frac{2 p+q}{2 p-q} \text { is: }$

1
$-\frac{1}{7} $
$ \frac{1}{7}$
7

Solution:

⇒ $\frac{p}{q}=\frac{-2}{3}$

⇒ $\text { Tricks }\frac{2 p+q}{20-q}=\frac{2(-2)+3}{2(-2)-3}=\frac{-4+3}{-4-3}=\frac{-1}{-7}=\frac{1}{7}$

(3) is correct

Question 57. Fourth proportional to x, 2x, (x+1) is:

x+2
x+2
(2x+2)
2x-2

Solution:

Let Fourth Proportional is K

$\frac{x}{2 x}=\frac{x+1}{K}$

Or k.x = 2x (x+1)

Or 1< = 2 (x+1) = 2x+ 2

(3) is correct

Question 58. Wlwt must be added to each term of the ratio 49: 68 so that it becomes 3: 47

Solution:

Let x be added to each term

Then $\frac{49+x}{68+x}=\frac{3}{4}$

Or 196 + 4x = 204 + 3x

Or 4x – 3x = 204 – 196

Or x = 8

(3) is Correct

Question 59. The students of the two classes are in the ratios 5: 7. if 10 students left from each class, the remaining students are in the ratio of 4: 6, then the number of students in each class was.

30, 40
25, 24
40, 60
50, 70

Solution:

(1) : (2) and (3) are not in the ratio 5: 7

(4) is Correct.

Question 60. If A : B = 2:5, then (10A + 3B) is equal to

7:4
7:3
6: 5
7: 9

Solution:

It A : B = 2 : 5 Then

⇒ $\frac{10 A+3 B}{5 A+2 B}=\frac{10 \times 2+3 \times 5}{5 \times 2+2 \times 5}=\frac{35}{20}=\frac{7}{4}$ = 7:4

(1) is Correct

Question 61. In a film shooting, A and B received money in a certain ratio and B and C also received the money in the same ratio. If A gets ₹1,60,000 and C gets ₹2,50,000. Find the amount received by B.

₹2,00,000
₹2,50,000
₹1,00,000
₹1,50,000

Solution:

A: B = B: C

So, B2 = AC;

So, B = AC – y1,60,000 x 2,50,000

= 400 x 500 = 2,00,000

Question 62. The ratio compounded of 4: 5 and sub-duplicate of”a”: 9 is 8: 15. The value of”a” is

Solution:

⇒ $(c) \frac{1}{5} \times \sqrt{\frac{a}{9}}=\frac{8}{15} \quad$ Or \frac{1}{5} \times \sqrt{\frac{a}{3}}=\frac{8}{15}$

⇒ $\sqrt{a}=2 \Rightarrow a=4$

Question 63. Which of the numbers are not in proportion?

6,8,5,7
7,14,6,12
18,27,12,18
8,6,12,9

Solution:

For a $\frac{6}{8}=\frac{3}{4} \neq \frac{5}{7}$

(1) is not in proportion

Question 64. Find two numbers such that the mean proportional between them is 18 and the third proportional to them is 144

9 ; 36
8 ; 32
7 ; 28
6 ; 14

Solution:

(1) is correct

For (a) Mean Proportional of 9 and 36

= √9 x 36 = 18

It satisfies 1st condition.

If144 is its 3rd condition.

362 = 9 x 144

It also satisfies the 2nd Condition.

Question 65. The triplicate ratio of 4: 5 is

125: 64
16: 25
64: 125
120:46

Solution:

(3) Triplicate ratio of 4: 5

= 43 : 5:1= 64 : 125

Question 66. The mean proportion between 24 and 54 is _______.

Solution:

(4) Mean – Proportion = V24 x 54 = 36

Question 67. The ratio of numbers is 1: 2 : 3 and the sum of their squares is 504 then the numbers are

6,12,18
3,6,9
4,8,12
5,10,15

Solution: (1) is correct

Question 68. If P is 25% less than Q and R is 20% higher than Q the Ratio of R and P

5: 8
8: 5
5 : 3
3: 5

Solution:

(2) is correct

Let Q = 100, So, P = 100-025 = 75 and R = 100 + 20 = 120

⇒ $\frac{R}{P}=\frac{120}{75}=\frac{8}{5}$

Question 69. A person has assets worth ₹1, 48, 200. He wishes to divide it among his wife, son, and daughter in the ratio of 3:2:1 respectively. From these assets, the share of his son will be

₹74,100
₹37,050
₹49,400
₹24,700

Solution:

⇒ $=\frac{2}{3+2+1} \times 1,48,200=₹ 49,400$

(3) is correct Share of son

Question 70. If x : y = 2 : 3 then (5x + 2y) : (3x – y) =

19: 3
l6 : 3
7: 2
7 : 3

Solution:

(2) is correct $\frac{5 x+2 y}{3 x-y}=\frac{5 \times 2+2 \times 3}{3 \times 2-3}=\frac{16}{3}$

Question 71. The first, second, and third-month salaries of a person are in the ratio 2:4:5. The difference between the product of the salaries of the first 2 months & last 2 months is ₹4,80,00,000. Find the salary for the second month

₹4,000
₹6,000
₹12,000
₹8,000

Solution:

Let x be common in the ratio.

1st, 2nd and 3rd-month salaries of a person

= 2x; 4x; 5x

From Qts.

4x x 5x – 2x x

4x = 4,80,00,000.

Or, 12x² = 4.80,00,000.

Or, x² = 4,00,000

X = 2000.

2nd month salary = 4x = 4 x 2000

Question 72. 15 (2p2-q2)= 7 pq, where,q are positive then p: q

Solution:

(1) is correct

I5(2p- q ’) = 7

For (a) pul p = 5; q = 6 we get

15|2 X 52- 62| = 3*5*6

Or 15 x 14 = 210

Or 210 = 210

Question 73. Kind the ratio of the third proportional of 1 2; 30 and the mean proportional of 9; 25 :

7: 2
5: 1
9: 4
None of these

Solution:

3″1 proportional $=\frac{30^2}{12}=$ = 75

Mean Proportional = V9 x 25 = 15

Ratio = $=\frac{75}{15}=$ 5 : 1 (b) is correct

Question 74. What must be added to each of the numbers 10, 18, 22, and 38 to make them proportional:

Solution:

(b) is correct

Let x be added.

⇒ $\frac{10+x}{18+x}=\frac{22+x}{38+x}$

x = 2 satisfies it

Question 75. x, y, z together start a business, if x invests 3 times as much as y invests and y investwo-thirdsird of what z invests, then the ratio of capitals of x, y, z is______.

3: 9: 2
6 : 3: 2
3: 6: 2
6: 2 : 3

Solution:

6 = 3×2 and 2 = 3 x $\frac{2}{3}$

Question 76. A bag contains 23 numbers of coins in the form of 1 rupee, 2 rupee, and 5 rupee coins. The total sum of the coins is The ratio between 1 rupee and 2 rupees coins is 3: 2. Then the number of1 rupee coins is

Solution:

Let option (a) be correct.

Let x be common in the ratio.

So, coins = 3x = 12; So, x = 4

No. of ₹ 2 coins=2×4=8

Hence no. of coins of ?5 coins = 23 – 12-8 = 3

Satisfied, So (a) is correct

Question 77. If a : b = 2 : 3, b: c = 4: 5, c: d = 6 : 7 then a : d is

24: 35
8: 15
16: 35
7: 15

Solution:

Option (3) is correct.

Multiply all ratios.

⇒ $=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7}=\frac{16}{35}$

Question 78. The ratio of the number of five rupee coins to the number of ten rupee coins is 8: 15. If the total value of five rupee coins is 360, then the no. of ten rupee coins is

Solution:

Option (4) is correct.

Total No. of? 5 coins = 360/5 = 72

Let x be common in the ratio.

So, coins = 8x = 72 ; So, x = 9

No. ₹ 10 coins = 15 x 9= 135

Question 79. If $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{x}$ are in proportion then x =

$\frac{15}{2}$
$\frac{3}{15}$
$\frac{2}{15}$
$\frac{1}{15}$

Solution:

Option (1) is correct

Product of middle two terms

= Product of extremes $\text { So, } \frac{1}{2 x}=\frac{1}{15} ; x=15 / 2$

Question 80. If (a+b) : (b+c) : (c+a) = 7:8:9 and a + b + c=18 then a : b : c =

5: 4 : 3
3 : 4: 5
4 : 3: 5
4: 5 : 3

Solution:

(3)

Leta : h : c = 4: 3: 5

It is in ratio. So, it should satisfy the given ratio (a+h) : (b+c): (c+a) = 7: H: 9 i.c (4+3): (3+5): (5+4) = 7: 8: 9 (true) Avoid 2ml condition.

In detail, it will take too much time.

Question 81. The mean proportional between 24 and 54

Solution:

Formula Mean Proportion of a and b = √ab

(4) = √24 x 54 = 36

Question 82. $\frac{3 x-2}{5 x+6} $is the duplicate ratio of| then find the value of x:

Solution:

1. Given$\frac{3 x-2}{5 x+6}=\left(\frac{2}{3}\right)^2=\frac{4}{9}$

Tricks: Go by choices

For option (a) putting x = 6 in LHS; we get

⇒ $\frac{3 \times 2-2}{5 \times 2+6}=\frac{4}{9} \text { (R.H.S) }$

Question 83. x:6y : z = 7:4: 110000 then $\frac{x+y+z}{z}$ is:

Solution:

⇒ $\frac{x+y+z}{z}=\frac{7+4+11}{11}=2$

Question 84. If the ratio of two numbers is 7: 11. If 7 is added to each number then the new ratio will be 2 : 3 then the numbers are.

49,77
42,45
43,42
39,40

Solution:

⇒ $\left[\begin{array}{l}49 \div 7=777 \div 11=7\end{array}\right]$ both must be equal. Here it is correct.

now $\frac{49+7}{77+7}=\frac{56}{84}=\frac{2}{3}$

Divide 56 by numerator (2) and 84 by

Denominator (3) we get the same value “28”

Indices

Concepts And Tricks To Remember

1. Laws Of Indices:

a^m x an= a^m+n
$\frac{a^m}{a^n}=a^m-n$
(a^m)ⁿ= a^mn
(ab)ⁿ = aⁿbⁿ
$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
a° = 1

2. Surds: Let a be a rational number and n be a positive integer such that $a^{\frac{1}{n}}=\sqrt[n]{a}$ is irrational. Then, $\sqrt[n]{a}$ is called a surd of order n.

3. Laws Of Surds:

$\sqrt[n]{a}=a^{\frac{1}{n}}$
$\sqrt[n]{a b}=\sqrt[n]{a} \times \sqrt[n]{b}$
$\sqrt[n]{\frac{a}{b}+\sqrt[n]{a}} \sqrt[n]{b}$
$(\sqrt[n]{a})^n=a$
$\sqrt[m]{\sqrt[n]{a}}=\sqrt[m n]{a}$
$(\sqrt[n]{a})^m \cdot \sqrt[n]{a^m}.$

Some Related Formulae.

a^m = a x a x a x to m times.
a^e = 1 where a ≠ 0; or ∞
$a^{-1}=\frac{1}{a}$
$a^{-m}=\frac{1}{a^m}$
a^m x aⁿ = a^m+n
- amx a^m x aⁿ x a^r…..=a^m+n+r+
$\frac{a^m}{a^n}=a^{m-n}$
- $\frac{a^m}{a^n}=\frac{1}{a^{n-m}}$
(a^m)ⁿ = a^mn
- a^mn= a^m-n
If a^m = b^m then a=b
- if a^m = aⁿ then m = n
$\sqrt[m]{a^n}=a^{\frac{n}{m}}$
- $\sqrt{a}=a^{\frac{1}{2}}$
- $\sqrt[3]{a}=a^{\frac{1}{3}}$
$\text { if } a^m=k \Rightarrow a=k^{1 / m}$
- $\text { if } a^m=k^n \Rightarrow a=k^{m / m}$
- $\text { If } a^{1 / m}=k \Rightarrow a=k^m $
$\text { if } a^{1 / m}=k^n \Rightarrow a=k^{m / m}$
- $\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}$
$(a b \ldots)^m=a^m \cdot b^m$
- $\sqrt[m]{a b \ldots \ldots \ldots \ldots}=\sqrt[m]{a} \cdot \sqrt[m]{b}\qquad$
$\sqrt{a b}=\sqrt{a} \cdot \sqrt{b}.$
$\left(\frac{a}{b}\right)^m=\left(\frac{b}{a}\right)^{-m}$
If a^b = b² then

Either a = b

Or if a = 2

Then b=4

or if a = 4

Then b = 2

If a > 1 and x < y Then ax < ay

Exercise -1 – Indices

Question 1. $Value of \left(a^{1 / 8}+a^{\cdot 1 / 14}\right)\left(a^{1 / 8}-a^{-1 / 8}\right) \quad\left(a^{1 / 4}+a^{-1 / 4}\right)\left(a^{1 / 2}+a^{-1 / 2}\right) is:$

$a+\frac{1}{a}$
$\mathrm{a} \cdot \frac{1}{a}$
$a^2+\frac{1}{a^2}$
$a^2+\frac{1}{a^2}$

Solution:

⇒ $\left[a^{1 / 8}+a^{\cdot 1 / 8}\right]\left[a^{1 / 8}-a^{1 / 8}\right]\left[a^{1 / 4}+a^{1 / 4}\right]\left[a^{1 / 2}+a^{1 / 2}\right]$

⇒ ${\left[\text { Use formula }(a+b)(a-b)=a^2-b^2\right]}$

⇒ $\left[\left(a^{1 / 8}\right)^2-\left(a^{1 / 8}\right)^2\right]\left[a^{1 / 4}+a^{-1 / 4}\right]\left[a^{1 / 2}+a^{-1 / 2}\right]$

⇒ $\left(a^{1 / 4}-a^{-1 / 4}\right)\left(a^{1 / 4}+a^{-1 / 4}\right)\left(a^{1 / 2}+a^{-1 / 2}\right) $

⇒ $\left[\left(a^{1 / 4}\right)^2-\left(a^{-1 / 4}\right)^2\right]\left[a^{1 / 2}+a^{-1 / 2}\right] $

⇒ $\left(a^{1 / 2}-a^{-1 / 2}\right)\left(a^{1 / 2}+a^{-1 / 2}\right)=\left(a^{1 / 2}\right)^2-\left(a^{-\frac{1}{2}}\right)^2=a-a^{-1}=a-\frac{1}{a}$

(2) is correct

Question 2. Simplification of $\frac{x^{m+3 n} \cdot x^{4 m-9 n}}{x^{6 m-6 n}} \text { is: }$

x^m
x^-m
xⁿ
x^-n

Solution:

⇒ $\frac{x^{m+3 n} \cdot x^{4 m-9 n}}{x^{6 m-6 n}}=x^{m+3 n+4 m-9 n-6 m+6 n}=x^{-m}=x^{-m}$

(2) is correct

Question 3. On simplification$\frac{1}{1+z^{a-b}+z^{a-c}}+\frac{1}{1+z^{b-c}+z^{b-a}}+\frac{1}{1+z^{c-a}+z^{c-b}}$ reduce to:

$\frac{1}{z^{2(a+b+c)}}$
$\frac{1}{z^{2(a+b+c)}}$
1
0

Solution:

⇒ $\frac{1}{1+z^{a-b}+z^{n-c}}+\frac{1}{1+z^{b-c}+z^{b-n}}+\frac{1}{1+z^{c-a}+z^{c-b}}=1 \text { [it is in cyclic order] }$

Question 4. If 4X = 5y = 207- then z is equal to:

xy
$\frac{x+y}{x y}$
$\frac{1}{x y}$
$\frac{x y}{x+y}$

Solution:

(4)

Let$4^x=5^y=20^z \text { then } z \text { is equal to: }$

⇒ $ 4=\mathrm{k}^{1 / x} ; 5=\mathrm{k}^{1 / y} ; 20=\mathrm{k}^{1 / 2}$ 20 = 4×5

⇒ $\mathrm{k}^{1 / 2}=\mathrm{k}^{1 / x} \cdot \mathrm{k}^{\frac{1}{y}} \quad \mathrm{~K}^{1 / 2}=k^{\frac{1}{x}+\frac{1}{y}} \frac{1}{z}=\frac{1}{x}+\frac{1}{y} \Rightarrow \frac{1}{z}=\frac{y+x}{x y}$

⇒ $\mathrm{Z}=\frac{x y}{x+y}$

(4) is correct

Question 5. $\left(\frac{\sqrt{3}}{9}\right)^{5 /2}\left(\frac{9}{3 \sqrt{3}}\right)^{7 / 2} \times 9 \text { is equal to: }$

1
√3
3√3
$\frac{3}{9 \sqrt{3}}$

Solution:

⇒ $\left(\frac{\sqrt{3}}{9}\right)^{5 / 2}\left(\frac{9}{3 \sqrt{3}}\right)^{7 / 2} \times 9 $

⇒ $\left[\left(\frac{3^{1 / 2}}{3^2}\right)^5\left(\frac{3^2}{3 \cdot 3^{1 / 2}}\right)^7\right]^{1 / 2} \times 3^2=\left[3^{\left(\frac{1}{2}-2\right) 5} \times 3^{\left(2-1-\frac{1}{2}\right) 7}\right]^{1 / 2} \cdot 3^2$

⇒ $\left[3^{\frac{-15}{2}} \cdot 3^{\frac{7}{2}}\right]^{\frac{1}{2}} \times 3^2=\left(3^{\frac{-15}{2}+\frac{7}{2}}\right)^{\frac{1}{2}} \cdot 3^2=\left(3^{-4}\right)^{1 / 2} \cdot 3^2=3^{-2} \cdot 3^2=3^{-2+2}=3^0=1$

(1) is correct

Question 6. If 2x – 2x-1,=4, then the value of x x is:

Solution:

2y -2y =4

Or 2x-1(2-1) = 4

Or 2x-1 x 1 = 22 Or 2x-1 = 22

x – 1 = 2

x = 3 xx = 33 = 27

(4) is correct

Question 7. If x = ya, y=zb and z = xc then ABC is:

Solution:

It x = yd; y=zb and z=xc

Then abc=1

it is in cyclic order. :- (2) is correct

Question 8. Ifx = 31/3 + 3 -1/3then find value of 3×3 – 9x

Solution:

Detail Method

It x = 31/3 + 3-1/3______(1)

Cubing on both sides; we get x3:$\left(3^{1 / 3}\right)^3+\left(3^{-1 / 3}\right)^3+3 \cdot 3^{\frac{1}{3}} \cdot 3^{-\frac{1}{3}}\left(3^{\frac{1}{3}}+3^{-1 / 3}\right)$

= 3 + 3-1 + 3xlxx (from (I)

Question 9. Find the value of:$\left[1-\left\{1-\left(1-x^2\right)^{-1}\right\}^{-1}\right]^{-\frac{1}{2}} \text { is }$

1/x
x
1
None of these

Solution:

⇒ $\left[1-\left\{1-\left(1-x^2\right)^{-1}\right]^{-\frac{1}{2}}\left[1-\left\{1-\frac{1}{1-x^2}\right\}^{-1}\right]^{-1 / 2}\right.$

⇒ $
\left[1-\left\{\frac{1-x^2-1}{1-x^2}\right\}^{-1}\right]^{-1 / 2}=\left[1-\left\{\frac{-x^2}{1-x^2}\right\}^{-1}\right]^{-1 / 2}$

⇒ $\left[1-\frac{1-x^2}{-x^2}\right]^{-\frac{1}{2}}=\left[1+\frac{1-x^2}{x^2}\right]^{-\frac{1}{2}}=\left[\frac{x^2+1-x^2}{x^2}\right]^{-1 / 2}$

⇒ $\left(\frac{1}{x^2}\right)^{-1 / 2}=\left(x^{-2}\right)^{-\frac{1}{2}}=x (\mathrm{b})$

(2) is correct

Question 10.$\frac{2^n+2^{n-1}}{2^{n+1}-2^n}$

Solution:

⇒ $\frac{2^n+2^{n-1}}{2^{n+1}-2^n}=\frac{2^n\left(2^0+2^{-1}\right.}{2^n\left(2^1-2^0\right)}=\frac{1+\frac{1}{2}}{2-1}=\frac{3}{2}$

[2] is correct

Put n =1

Question 11. If 2X x3y x 52 = 360. Then what is the value of x,y,z.?

3,2,1
1,2,3
2,3,1
1,3,2

Solution:

If 2x x3y x 5z = 360

x= 3; y = 2; z = 1;

(1) is correct

Question 12. The recurring decimal 2.7777……can be expressed as

24/9
22/9
26/9
25/9

Solution:

⇒ $\frac{24}{9}=2.666 \ldots \ldots \ldots . \neq 2.777 \ldots \ldots$

⇒ $\frac{22}{9}=2.444 \ldots \ldots \ldots . \neq 2.777 \ldots \ldots$

⇒ $\frac{26}{9}=2.888 \ldots \ldots \ldots \neq 2.77 \ldots \ldots$

⇒ $\frac{25}{9}=2.777 \ldots \ldots \ldots . . .2 .777 \ldots \ldots …$

(4) is correct

Question 13. The value of$\frac{\left(3^{n+1}+3^n\right)}{\left(3^{n+3}-3^{n+1}\right)} \text { is equal to }$

Solution:

(2) Put n = 0 $\frac{3+3^0}{3^3-3}=\frac{3+1}{27-3}=\frac{4}{24}=\frac{1}{6}$

Detail Method$\frac{3^{n+1}+3^n}{3^{n+3}-3^{n+1}}=\frac{3^n(3+1)}{3^n\left(3^3-3\right)}=\frac{4}{24}=\frac{1}{6}$

Question 14. Find the value of x, if x.(x)1/3 = (x1/3)x

Solution:

⇒ $(b)x \cdot x^{\frac{1}{3}}=x^{\frac{x}{3}}
or x^{1+\frac{1}{3}}=x^{x / 3}1+\frac{1}{3}=\frac{x}{3}$

⇒ $\frac{4}{3}=\frac{x}{3}\mathrm{x}=4$

Question 15. $If\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{c}=0; then find the value of \left|\frac{a+b+c}{3}\right|^3=$

9abc
$\frac{1}{9 a b c}$
abc
$\frac{1}{a b c}$

Solution:

(3) is correct

Let a = -1; b = -1 and c = 8, because $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{-1}+\sqrt[3]{-1}+\sqrt[3]{8}$

$=-1-1+2=0 \text { (R.H.S) } \left[\frac{a+b+c}{3}\right]^3=\left[\frac{-1-1+8}{3}\right]^3=(2)^3=8$

(-l). (-l).(8) = abc

(3) is correct

Question 16. The value of$\left(\frac{y^{\prime}}{y^b}\right)^{a^2+a b+b^2}\left(\frac{y^b}{y^a}\right)^{b^2+b c+c^2}\left(\frac{y^c}{y^a}\right)^{c^2+c a+a^2}$

y
-1
1
6

Solution:

(3) is correct.

Question 17. Ifpx = q,qy = r,rz = p6, then the value of XYZ is…

Solution:

qy = r => (px)y = r => pxy

Now r2 = p6 => (pxy)z =p6=> pxyz = Pb :- xyz = 6

Question 18. The value of $\frac{x^2-(y-z)^2}{(x+z)^2-y^2}+\frac{y^2-(x-z)^2}{(x+y)^2-z^2}+\frac{z^2-(x-y)^2}{(y+z)^2-x^2}=$

0
1
-1
∞

Solution:

(2) is correct

⇒ $\frac{(x+y-z)(x-y+z)}{(x+y+z)(x-y+z)}+\frac{(y+x-z)(y-x+z)}{(x+y+z)(x+y-z)}+\frac{(x-y+z)(z-x+y)}{(x+y+z)(y+z-x)}$

⇒ $=\frac{x+y-z+y-x+z+x-y+z}{x+y+z}=\frac{x+y+z}{x+y+z}=1$

Question 19. If 3X = 5y = (75)z then

$\frac{1}{x}+\frac{2}{y}=\frac{1}{z}$
$\frac{2}{x}+\frac{1}{y}=\frac{1}{z}$
$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
None

Solution:

⇒ $3^x=5^y=(75)^z \ldots \ldots \text { (1) }$

⇒ $3^1 \times 5^2=75^1 \ldots \ldots .(2)$

Power of(2) + Power of(1)

And put the + sign in the place of”x”.

We get$\frac{1}{x}+\frac{2}{y}=\frac{1}{z}$

So (1) is correct

Question 20. If ABC = 2, then the value of$\frac{1}{1+a+2 b^{-1}}+\frac{1}{1+\frac{b}{2}+c^{-1}}+\frac{1}{1+a^{-1}+c}=$

1
2
$\text { (d) } \frac{1}{2}$
$\text { (d) } \frac{3}{4}$

Solution:

“Puta = 1, b = 2 & c = 1. So abc = 2” in the given question. We get

⇒ $\frac{1}{1+1+\frac{2}{2}}+\frac{1}{1+\frac{2}{2}+1^{-1}}+\frac{1}{1+1^{-1}+1}=1$

Option (1) is correct.

Question 21. If a $=\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6-\sqrt{5}}}, b=\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6+\sqrt{5}}} \text { then the value of } \frac{1}{a^2}+\frac{1}{b^2} \text { is }$

Solution:

⇒ $\frac{1}{a}+\frac{1}{b}= $

⇒ $\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}=22$

⇒ $\frac{1}{a^2}+\frac{1}{b^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2-2\left(\frac{1}{a} \cdot \frac{1}{b}\right)=22^2-2=482
$

(1) is correct.

Question 22. If u5x = v5y = w5z and u2 = vw then xy + 7.x- 2yz =_______.

a
1
2
None of these

Solution:

U5X=VSy=W5z => u2 – vy = wz

see quicker BMLRS chapter: Indices

⇒ $\mathrm{u}^2=\mathrm{VW} ; \frac{2}{x}=\frac{1}{y}+\frac{1}{z}=\frac{y+z}{y z}$

Or; xy + zx = 2yz

Or; xy + zx -2yz = 0

Question 23. $\left(x^{\sum_{n=1}^{\infty} a p^{n-1}}\right)^{1-p}\left(x^{\sum_{n=1}^{\infty} b q^{n-1}}\right)^{1-q}\left(x^{\sum_{n=1}^{\infty} c r^{n-1}}\right)^{1-r}$

$x-(a p+b(c)+c r)$
$x^{a+b+c}$
$x^{(a p+b \varphi+c r)}$
$X^{\text {abe }}$

Solution:

(2)

⇒ $\sum_{n=1}^{\infty} a p^{n-1}=a+a p+a p^2+\cdots=\frac{a}{1-p}$

⇒ $\left(x^{\sum_{n=1}^{\infty} a p^{n-1}}\right)^{1-p}=\left(x^{\frac{a}{1-p}}\right)^{1-p}=x^a $

Similarly doing as above; we get

⇒ $\left(x^{\sum_{n=1}^{\infty} a p^{n-1}}\right)^{1-p}\left(x^{\sum_{n=1}^{\infty} b q^{n-1}}\right)^{1-q}\left(x^{\sum_{n=1}^{\infty} c r^{n-1}}\right)^{1-r}=x^a \cdot \mathrm{x}^b \cdot \mathrm{x}^c \quad=\mathrm{x}^{\cdot+a+b+c}$

(2) is correct

Question 24. $\frac{2^n+2^{n-1}}{2^{n+1}-2^n}$

$\frac{1}{2}$
$\frac{3}{2}$
$\frac{2}{3}$
$\frac{1}{3}$

Solution:

(2)

Put minimum powers-l=0+1 or n=l in the question.

⇒ $ \frac{2^n+2^{n-1}}{2^{n+1}-2^n}=\frac{2^1+2^{1-1}}{2^{1+1}-2^1}=\frac{2+1}{4-2}=\frac{3}{2}$

Question 25. $\frac{2^{m+1} \times 3^{2 m-n+3} \times 5^{n+m+4} \times 6^{2 n+m}}{6^{2 m+n} \times 10^{n+1} \times 15^{m+3}}$

3 2m+2n
3 2n-2m
1
None.

Solution:

Tricks Put m = n = 0 in this equation.$\frac{2^{m+1} \times 3^{2 m-n+3} \times 5^{n+m+4} \times 6^{2 n+m}}{6^{2 m+n} \times 10^{n+1} \times 15^{m+3}}=1$

Question 26. If 2×2 = 2yZ = 12z2 then

$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$
$\frac{1}{x^2}+\frac{2}{y^2}=\frac{1}{z^2}$
$\frac{2}{x^2}+\frac{2}{y^2}=\frac{1}{z^2}$
None

Solution:

2×2 = 2y2 = 12z2 …(1) (given)

Factorize 12 in terms of 2&3. We get 22 x 31 = 121….(2)

Always write as a power of base of (2) +ÿ Power on the same base of 1; put the “+” sign in the place of the “x” sign.

So $\frac{2}{x^2}+\frac{1}{y^2}=\frac{1}{z^2} \quad \text { So (3) is correct. }$

Question 27. The value of $\frac{1}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{1}{(32)^{-\frac{1}{5}}} \text { is }$

Solution:

⇒ $\frac{1}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{1}{4}}}+\frac{1}{(32)^{-\frac{1}{5}}}=\frac{1}{\left(6^3\right)^{-\frac{2}{3}}}+\frac{1}{\left(4^4\right)^{\left(-\frac{1}{4}\right)}}+\frac{1}{\left(2^5\right)^{-\frac{1}{5}}}$

⇒ $\frac{1}{6^{3 \times \frac{(-2)}{3}}}+\frac{1}{4^{4 \times \frac{(-3)}{4}}}+\frac{1}{2^{5 \times \frac{(-1)}{5}}}=\frac{1}{6^{-2}}+\frac{1}{4^{-3}}+\frac{1}{2^{-1}} $

⇒ $\left(6^2+4^3+2^1\right)=(36+64+2)=102 $

Question 28. The value of[(10)150 + (10) 146] is

1000
10000
100000
106

Solution:

⇒ $(10)^{150} \div(10)^{146}=\frac{(10)^{150}}{(10)^{146}}=(10)^{(150-146)}=10^4=10000$

Question 29. (1000)7+ 1018 = ?

10
100
1000
10000

Solution:

⇒ $(1000)^7 \div 10^{18}=\frac{(1000)^7}{10^{18}}=\frac{\left(10^3\right)^7}{10^{18}}=\frac{10^{(3 \times 7)}}{10^{18}}=\frac{10^{21}}{10^{18}}=(10)^{(21-18)}=10^3=1000$

Question 30.(256)0.16 x (256)0.09 = ?

4
16
64
256.25

Solution:

⇒ $(256)^{0.16} \times(256)^{0.09} =(256)^{(0.16+0.09)}=(256)^{0.25}=(256)^{\left(\frac{25}{100}\right)}$

⇒ $ (256)^{\frac{1}{4}}=\left(4^4\right)^{\frac{1}{4}}=4^{\left(4 \times \frac{1}{4}\right)}=4^1=4$

Question 31. (0.04)-1.5 =?

Solution:

⇒ $(0.04)^{-1.5}=\left(\frac{4}{100}\right)^{-1.5}=\left(\frac{1}{25}\right)^{-\frac{3}{2}}=(25)^{\frac{3}{2}}=\left(5^2\right)^{\frac{3}{2}}=5^{\left(2 \times \frac{3}{2}\right)}=5^3=125$

Question 32. (17)3.5 x (17)? = 178

2.29
2.75
4.25
4.5

Solution:

Let (17)3.5 X (17)x = 178. Then, (17)35+x = (17)8.

3.5 + x = 8 <=> x = (8 – 3.5) » x = 4.5

Question 33. $(64)^{-\frac{1}{2}}-(-32)^{-\frac{4}{5}}=?$

$frac{1}{8}$
$\frac{3}{8}$
$\frac{1}{16}$
$\frac{3}{16}$
None of these

Solution:

⇒ $ (64)^{-\frac{1}{2}}-(-32)^{-\frac{4}{5}}=\left(8^2\right)^{-\frac{1}{2}}-\left\{(-2)^5\right\}^{-\frac{4}{5}}=8^{2 \times \frac{(-1)}{2}}-(-2)^{5 \times \frac{(-4)}{5}}=8^{-1}-(-2)^{-4}$

⇒ $ \frac{1}{8}-\frac{1}{(-2)^4}=\left(\frac{1}{8}-\frac{1}{16}\right)=\frac{1}{16}$

Question 34. (18)3.5 + (27)3.5 x 63.5 = 2? :

3.5
4.5
6
7
None of these

Solution:

⇒ $(18)^{3.5} \div(27)^{3.5} \times 6^{3.5}=2^x $

⇒ $\Leftrightarrow(18)^{3.5} \times \frac{1}{(27)^{3.5}} \times 6^{3.5}=2^x \Leftrightarrow\left(3^2 \times 2\right)^{3.5} \times \frac{1}{\left(3^2\right)^{3.5}} \times(2 \times 3)^{3.5}=2^x$

⇒ $\Leftrightarrow 3^{(2 \times 3.5)} \times 2^{3.5} \times \frac{1}{3^{(3 \times 3.5)}} \times 2^{3.5} \times 3^{3.5}=2^x $

⇒ $3^7 \times 2^{3.5} \times \frac{1}{3^{10.5}} \times 2^{3.5} \times 3^{3.5}=2^x \Leftrightarrow 2^x=2^x \Leftrightarrow \mathrm{x}=7 $

Question 35. (25)7.5 x (5)2.5 + (125)1.5 = 5?

8.5
13
16
17.5
None of these

Solution:

⇒ $\text { Let }(25)^{7.5} \times(5)^{2.5} \div(125)^{1.5}=5^x \text {. Then } \frac{\left(5^2\right)^{7.5} \times(5)^{2.5}}{\left(5^3\right)^{1.5}}=5^x \Leftrightarrow \frac{5^{(2 \times 7.5)} \times 5^{2.5}}{5^{(3 \times 1.5)}}=5^x $

⇒ $\Leftrightarrow \frac{5^{15} \times 5^{2.5}}{5^{4.5}}=5^x \Leftrightarrow 5^x=5^{(15+2.5-4.5)}=5^{13} \Leftrightarrow x=13$

36. The value of$\frac{(243)^{0.13} \times(243)^{0.07}}{(7)^{0.25} \times(49)^{0.075} \times(343)^{0.2}} \text { is : }$

$\frac{3}{7}$
$\frac{7}{3}$
$1 \frac{3}{7}$
$2 \frac{2}{7}$

Solution:

⇒ $ \frac{(243)^{0.13} \times(243)^{0.07}}{7^{0.25} \times(49)^{0.075} \times(343)^{0.2}}=\frac{(243)^{(0.13+0.07)}}{7^{0.25} \times\left(7^2\right)^{0.075} \times\left(7^3\right)^{0.2}}$

⇒ $\frac{(243)^{0.2}}{7^{0.25} \times 7^{(2 \times 0.075)} \times 7^{(3 \times 0.2)}}=\frac{\left(3^5\right)^{0.2}}{7^{0.25} \times 7^{0.15} \times 7^{0.6}}$

⇒ $\frac{3^{(5 \times 0.2)}}{7^{(0.25+0.15+0.6)}}=\frac{3^1}{7^1}=\frac{3}{7^{.}}$

Question 37. $\text { If }\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3} \text {, then the value of } \mathrm{x} \text { is: }$

$\frac{1}{2}$
1
2
$\frac{7}{2}$

Solution:

⇒ $\text { 7. }\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3} \Leftrightarrow\left(\frac{a}{b}\right)^{x-1}=\left(\frac{a}{b}\right)^{-(x-3)}=\left(\frac{a}{b}\right)^{(3-x)}$

⇔ x – 1 = 3 – x 2x = 4 <=> x = 2.

Question 38. If x = 3 + 2V2, then the value of $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right) \text { is: }$

1
2
2√2
3√2

Solution:

⇒ $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}-2=\left(3+2 \sqrt{2)}+\frac{1}{(3+2 \sqrt{2})}-2\right.$

⇒ $=(3+2 \sqrt{2})+\frac{1}{(3+2 \sqrt{2})} \times \frac{(3-2 \sqrt{2})}{(3-2 \sqrt{2})}-2=(3+2 \sqrt{2})+(3-2 \sqrt{2})-2=4 .$

⇒ $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)=2$

39. Given that 100 48 = x, IO070 = y and xz = y2, then the value of z is close to :

1.45
1.88
2.9
3.7

Solution:

⇒ $x^z=y^2 \Leftrightarrow\left(10^{0.48}\right)^z=\left(10^{0.70}\right)^2 \Leftrightarrow 10^{(0.48 z)}=10^{(2 \times 0.70)}=10^{1.40}$

⇒ $\Leftrightarrow 0.48 z=1.40 \Leftrightarrow z=\frac{140}{48}=\frac{35}{12}=2.9 \text { (approx). }$

Question 40. If m and n are whole numbers such that”m” = 121, then the value of (m – 1)n+1 is :

1
10
121
1000

Solution:

We know that ll2 = 121. Putting m = 11 and n = 2, we get :

(m- 1)n+1 = (11 – 1)(2+1) = 103 = 1000

Question 41. $\frac{(243)^n \times 3^{2 n+1}}{9^n \times 3^{n-1}}=\text { ? }$

Solution:

Given Expression =$\frac{(243)^{\frac{n}{5}} \times 3^{2 n+1}}{9^n \times 3^{n-1}}=\frac{\left(3^5\right)^{\frac{n}{5}} \times 3^{2 n+1}}{\left(3^2\right)^n \times 3^{n-1}}=\frac{3^{\left(5 \times \frac{n}{5}\right)} \times 3^{2 n+1}}{3^{2 n} \times 3^{n-1}}$

⇒ $\frac{3^n \times 3^{2 n+1}}{3^{2 n} \times 3^{n-1}}=\frac{3^{(n+2 n+1)}}{3^{(2 n+n-1)}}=\frac{3^{3 n+1}}{3^{3 n-1}}=3^{(3 n+1-3 n+1)}=3^2=9 .$

Question 42. $\frac{(24,3)^{\frac{n}{5}} \times 3^{2 n+1}}{9^n \times 3^{n-1}}=?$

0
1/2
1
a m+n

Solution:

⇒ $ \frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}}=\frac{1}{\left(1+\frac{a^n}{a^m}\right)}+\frac{1}{\left(1+\frac{a^m}{a^n}\right)}$

⇒ $\frac{a^m}{a^m+a^n}+\frac{a^n}{a^m+a^n}=\frac{a^m+a^n}{a^m+a^n}=1 .$

Question 43. $\frac{1}{1+x^{(b-a)}+x^{(c-a)}}+\frac{1}{1+x^{(n-b)}+x^{(c-b)}}+\frac{1}{1+x^{(b-c)}+x^{(a-c)}}=?$

0
1
x a-b-c-a
None of these

Solution:

Given Expression: $ \frac{1}{1+\frac{x^b}{x^a}+\frac{x^c}{x^a}}+\frac{1}{1+\frac{x^a}{x^b}+\frac{x^c}{x^b}}+\frac{1}{1+\frac{x^b}{x^c}+\frac{x^a}{x^c}}$

⇒ $=\frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^a+x^b+x^c}+\frac{x^c}{x^a+x^b+x^c}=\frac{x^a+x^b+x^c}{x^a+x^b+x^c}=1 .$

Question 44. $\left(\frac{x^b}{x^c}\right)^{(b+c-a)}\left(\frac{x^c}{x^a}\right)^{(c+a-b)}\left(\frac{x^a}{x^b}\right)^{(a+b-c)}=?$

x abc
1
xab+bc+ca
xa+b+c-a

Solution:

Given Exp. = x(b-c)(b+c-a)x.(c-a)(c+n-b)i x(a-b)(a+b-c)

=x(b~c)(b+c)-a(b-c) x(c-a)(c+a)-b(c-a) x(a-b)(a+b)-c(a-b)

=x(b²-c²+c²-a²+a²-b²) x-a(b-c)-b(c-a)-c(a-b) = (X° x X°) = (1x 1) = 1

Question 45. If 3(x-y) = 27 and 39(x+y) = 243, then x is equal to:

Solution:

3x~y = 27 = 33 ⇔ x-y = x-ÿ=3…..(1)

3X + y = 243 = 35 ⇔ x + y = 5…..(2)

On solving (1) and (2), we get x = 4.

Question 46. 4x-1/4 is expressed as

-4x 1/4
X-1
4/x1/4
none of these

Solution:

(4)

⇒ $4 / x^{1 / 4}$

⇒ $ 4 x^{1 / 4}=\frac{4}{x^{\frac{1}{4}}}$

Question 47. The value of 8 1/3 is

√2
4
2
none of these

Solution:

⇒ $8^{\frac{1}{3}}=\sqrt[3]{8}=\sqrt[3]{2^3}=2$

Question 48. The value of 2 x (32)1/5 is

2
10
4
None of these

Solution:

⇒ $=2 \times(32)^{\frac{1}{5}}=2 \times(32)^{\frac{1}{5}}=2 \times\left(2^5\right)^{\frac{1}{5}}=2 \times 2=4$

Question 49. The value of 4/(32) 1/5 is

8
2
4
None of these

Solution:

⇒ $(2)=\frac{4}{\left(2^5\right)^{\frac{1}{5}}}=\frac{4}{2}=2$

Question 50. The value of (8/27)1/3 is

2/3
3/2
2/9
none of these

Solution:

⇒ $(1)=\frac{\left(2^3\right)^{\frac{1}{3}}}{\left(3^3\right)^{\frac{2}{3}}}=\frac{2}{3}$

Question 51. The value of 2(256)-1/8 is

1
2
1/2
None of these

Solution:

⇒ $\text { (a) }=\frac{2}{(256)^{\frac{1}{8}}}=\frac{2}{\left(2^8\right)^{\frac{1}{8}}}=\frac{2}{2}=1$

Question 52. 21/2. 43/4 is equal to

a fraction
a positive integer
a negative integer
none of these

Solution:

(2) a positive Integer=$=2^{\frac{1}{2}} \cdot 4^{\frac{3}{4}} \quad=2^{\frac{1}{2}} \cdot\left(2^2\right)^{\frac{3}{4}}=2^{\frac{1}{2}} \cdot 2^{\frac{6}{4}}$

⇒ $=2^{\frac{1}{2}+\frac{6}{4}}=2^{\frac{8}{4}}=2^2=4 .$

Question 53.$\left(\frac{81 x^4}{y^{-8}}\right)^{\frac{1}{4}}$

xy2
x2y
9xy2
None Of these

Solution:

(4) None of these

⇒ $\left(81 x^4 \cdot y^8\right)^{\frac{1}{4}}$

⇒ $ \left(3^4 x^4 y^8\right)^{\frac{1}{4}}=3 \cdot x \cdot y^2 $

Question 54. Xa-b x x b-c is equal to

x
1
0
None of these

Solution:

(4)1

⇒ $=X^{a-b} \times X^{b-c} \times x^{c-a}$

⇒ $^x \cdot a^y=a^{x \cdot y}$

⇒ $ x^{a-b+b-c+c-a}$

⇒ $ =x^0=1$

Question 55. The value of$\left(\frac{2 p^2 q^3}{3 x y}\right)^0 \text { where } p, q, x, y \neq 0 \text { is equal to }$

0
2/3
1
None of these

Solution:

(3)l

= anything raised to 0 is 1.

Question 56. $\text { 56. } \quad\left\{\left(3^3\right)^2 \times\left(4^2\right)^1 \times\left(5^3\right)^2\right\} /\left\{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3\right\} \text { is }$

Solution:

(4) 0

⇒ $=\frac{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2}{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3}=1$

Question 57. Which is true?

$2^{\circ}>(1 / 2)^o$
$2^0<(1 / 2)^0$
$2^0=(1 / 2)^{\circ}$
None of these

Solution:

(3) $20=\left(\frac{1}{2}\right) $

⇒ $2^0=1 \quad\left(\frac{1}{2}\right)^0=1$

⇒ $2^0=\left(\frac{1}{2}\right)^0$

Question 58. If $\text { If } x^{1 / p}=y^{1 / 4}=z^{1 / r} \text { and } x y z=1 \text {, then the value of } p+q+r \text { is }$

1
0
1/2
None of these

Solution:

(2) 0

⇒ $x^{\frac{1}{p}}=y^{\frac{1}{q}}=z^{\frac{1}{r}}=\mathrm{k} $

⇒ $\mathrm{X} k^p \cdot y=k^q \mathrm{z}=k^r$

⇒ $\mathrm{Xyz}=1 $

⇒ $k^p \cdot k^{q} \cdot k^r=1$

⇒ $k^{p+q+r}=k^0=\mathrm{p}+\mathrm{q}+\mathrm{r}=0 .$

Question 59. The value of y a-b x y b-c x ya-b is

y a+b
y
1
1/ya+b-a

Solution:

(4) l/ya+b

⇒ $y^{a-b} \times y^{(1-c} \times y^{(-\cdot a)} \times y^{-a \cdot l)}=y-a-b=y^{-(a+b)}$

⇒ $ \frac{1}{y^{a+b}}$

Question 60. The True option is

$x^{2 / 3}=\sqrt[3]{x^2}$
$^{2 / \beta}=\sqrt{x^3}$
$x^{2 / 3}>\sqrt[3]{x^2}$
$ x^{2 / 3}<\sqrt[3]{x^2}$

Solution:

⇒ $\text { (a) } x^{2 / 3}=\sqrt[3]{x^2}$

⇒ $x^{\frac{m}{n}}=\sqrt[m]{x^m}$

⇒ $x^{\frac{2}{3}}=\sqrt[3]{x^2}$

Question 61. The simplified value of 16x3y2 x 8>x3y-2 is

2xy
xy/2
2
None of these

Solution:

(3) 2

⇒ $=\left(\frac{27}{8}\right)^{\frac{1}{3}} \cdot\left(\frac{213}{32}\right)^{\frac{1}{5}}$

⇒ $=\frac{16 y^2}{x^3} \cdot \frac{x^3}{8 y^2}=2$

Question 62. The value of (8/27)-i/3 x (32/243)-1/5 is

9/4
4/9
2/3
none of these

Solution:

(1) $\frac{9}{4}$

⇒ $=\left(\frac{27}{8}\right)^{\frac{1}{3}} \cdot\left(\frac{213}{32}\right)^{\frac{1}{5}}$

⇒ $\left(\frac{3}{2}\right)^{3 \cdot \frac{1}{3}} \cdot\left(\frac{3}{2}\right)^{5 \cdot \frac{1}{5}}$

⇒ $\left(\frac{3}{2}\right) \cdot\left(\frac{3}{2}\right)=\frac{9}{4}$

Question 63. The value of $\left\{(x+y)^{2 / 3}(x-y)^{3 / 2} / \sqrt{(x+y)} \times \sqrt{(x-y)^3}\right\}^6 \text { is }$

(x+y)2
(x-y)
x + y
None of these

Solution:

c=x+y

⇒ $\left\{\frac{(x+y)^{\frac{2}{3}}(x-y)^{\frac{3}{2}}}{(x+y)^{\frac{1}{2}} \cdot(x-y)^{\frac{3}{2}}}\right\}^6$

⇒ $\left\{(x+y)^{\frac{2}{3}-\frac{1}{2}}\right\}^6 $

⇒ $ (x+y)^{\frac{1}{6}}=x+y$

Question 64. Simplified value of $f(125)^{2 / 3} \times \sqrt{25} \times \sqrt[3]{5^3} \times 5^{1 / 2} \text { is }$

5
1/5
1
none of these

Solution:

(4) none of these

(5)3)2/3 x 5 x 5 x 51/2

= 52 x 5 x 5 x 51/2

5 2+1+1+1/2 = 59/2

Question 65. $\left[\left\{(2)^{1 / 2} \cdot(4)^{3 / 4} \cdot(8)^{5 / 6} \cdot(16)^{7 / 8} \cdot(32)^{9 / 110}\right\}^4\right]^{3 / 25} \text { is }$

A fraction
an integer
l
none of these

Solution:

(2) an integer

⇒ ${\left[\left\{(2)^{1 / 2} \cdot 2^{6 / 4} \cdot 2^{15 / 6} \cdot 2^{28 / 8} \cdot 2^{15 / 10}\right\}^4\right]^{3 / 25}}$

⇒ ${\left[\left\{2^{1 / 2+6 / 4+15 / 6+28 / 8+15 / 10}\right\}^1\right]^{3 / 25}}$

⇒ ${\left[\left\{2^{1 / 2+3 / 2+5 / 2+7 / 2+9 / 2}\right\}^4\right]^{3 / 25}}$

⇒ ${\left[\left\{2^{25 / 2}\right\}^4\right]^{3 / 25}}$

⇒ $=\left[\left\{2^{\frac{25}{2} \times 4}\right\}\right]^{\frac{3}{25}}=\left[2^{\frac{25 \times 2.3}{25}}\right]$

= 26 = 64

Question 66. [1-{1-(1-X2)-1}-1]-1/2 is equal to

x
1/x
1
none of these

Solution:

(1) x

⇒ $\left[1-\left\{1-\left(1-x^2\right)^{-1}\right\}^{-1}\right]^{\frac{-1}{2}}$

⇒ $=\left[1-\left\{1-\left(\frac{1}{1-x^2}\right)\right\}^{-1}\right]^{\frac{-1}{2}}$

⇒ $\left[1-\left\{1-\left(\frac{1-x^2-1}{1-x^2}\right)\right\}^{-1}\right]^{\frac{-1}{2}}$

⇒ $\left[1-\left\{\left(\frac{-x^2}{1-x^2}\right)\right\}^{-1}\right]^{\frac{-1}{2}}$

⇒ $\left[1-\left\{\left(\frac{-x^2}{-x^2}\right)\right\}\right]^{\frac{-1}{2}}=\left[\frac{-x^2-1+x^2}{-x^2}\right]^{\frac{-1}{2}}$

⇒ $ \left[\frac{-1}{-x^2}\right]^{\frac{-1}{2}}=\left[x^2\right]^{\frac{1}{2}}=\mathrm{x}$

Question 67.$\left[\left(x^n\right)^{n-\frac{1}{n}}\right]^{\frac{1}{n+1}}$

xn
xn-1
xn-1
None of these

Solution:

(3) xn-1

⇒ $\left[\left(x^n\right) \frac{n^2-1}{n}\right]^{\frac{1}{n+1}}$

⇒ $ \left[x^{n^2-1}\right]^{\frac{1}{n+1}}$

⇒ $ \left[x^{(n-1)(n+1)}\right]^{\frac{1}{n+1}}=x^{n-1}$

Question 68. If a3- b3 = (a-b) (a2 + ab + b2), then the simplified form of $\left[\frac{x^l}{x^m}\right]^{l^2+l m+m^2} \times\left[\frac{x^m}{x^n}\right]^{m^2+m n+n^2} \times\left[\frac{x^n}{x^l}\right]^{l^2+\ln +n^2}$

0
1
x
None of these

Solution:

(2)

⇒ ${\left[\frac{x^l}{x^m}\right]^{l^2+l m+m^2} \times\left[\frac{x^m}{x^n}\right]^{m^2+m n+n^2} \times\left[\frac{x^n}{x^l}\right]^{l^2+l n+n^2}}$

⇒ ${\left[x^{(l-m)\left(l^2+l m+m^2\right)}\right] \times\left[x^{(m-n)\left(m^2+m n+n^2\right)}\right] \times\left[x^{(n-l)\left(n^2+n l+l^2\right)}\right]}$

⇒ $x^{l^3-m^3} \times x^{m^3-n^3} \times x^{n^3-l^3}$

⇒ $x^{l^3-m^3+m^3-n^3+n^3-l^3}=x^0=1 $

Question 69. Using (a-b)3 = a3-b3-3ab(a-b) tick the correct of these when x = p 1/3 – p-1/3

$x^3+3 x=p+1 / p$
$x^3+3 x=p-1 / p$
$x^3+3 x=p+1$$
none of these

Solution:

(2)$x^3+3 x=p \frac{-1}{p} $

⇒ $ x=p^{1 / 3}-p^{1 / 3} $

⇒ $x^3=\left(p^{1 / 3}-p^{-1 / 3}\right)^3 =p-\frac{1}{p}-3 p^{1 / 3} \cdot p^{-1 / 3}\left(p^{1 / 3}-p^{-1 / 3}\right)$

⇒ $x^3=p-\frac{1}{p}-3 x x^3+3 x=p \frac{-1}{p}$

Question 70. On simplification, $1 /\left(1+a^{m-n}+a^{m \cdot p}\right)+1 /\left(1+a^{n \cdot m}+a^{n-p}\right)+1 /\left(1+a^{p-m}+a^{p-n}\right)$ is equal to

Solution:

(3)

⇒ $=\frac{1}{1+\frac{a^m}{a^n}+\frac{a^m}{a^p}}+\frac{1}{1+\frac{a^n}{a^m}+\frac{a^n}{a^p}}+\frac{1}{1+\frac{a^p}{a^m}+\frac{a^p}{a^n}}$

⇒ $=\frac{a^{n+p}}{a^{n+p}+a^{m+p}+a^{m+n}}+\frac{a^{m+p}}{a^{m+p}+a^{n+p}+a^{n+m}}+\frac{a^{m+n}}{a^{m+n}+a^{p+n}+a^{p+m}}=\frac{a^{n+P}+a^{m+P}+a^{m+n}+}{a^{n+P}+a^{m+P}+a^{m+n}+}=1$

Question 71. The value of$\left(\frac{x^a}{x^b}\right)^{a+b} \times\left(\frac{x^b}{x^c}\right)^{b+c} \times\left(\frac{x^c}{x^a}\right)^{c+a}$

1
0
2
none of these

Solution:

(1) 1

=x(a-b)(a+b).x(b-c)(b+c).x(c-a)(c+a)

⇒ $x^{a^2-b^2} \cdot x^{b^2-c^2} \cdot x^{c^2-a^2} \Rightarrow x^{a^2-b^2+b^2-c^2+c^2-a^2}=x^0=1$

Question 72. If x =$3^{\frac{1}{3}}+3^{\frac{1}{3}} \text {, then } 3 x^3-9 x \text { is }$

15
10
12
None of these

Solution:

(2) 10

⇒ $x^3=3^{\frac{1}{3}}+3^{-\frac{1}{3}}$

⇒ $ x^3=3+\frac{1}{3}+3.3^{\frac{-1}{3}}\left(3^{\frac{1}{3}}+3^{-\frac{1}{3}}\right) x^3=\frac{10}{3}+\frac{9 x}{3}=3 x^3-9 x=10$

Question 73. If ax = b, by = c, c2 = a, then XYZ is

1
2
3
none of these

Solution:

(1)

⇒ $ a^x=b, b^y=c$

⇒ $\left(a^x\right)^y=c \quad a^{x y^{\prime}}=c, c^z=a \quad\left(c^z\right)^{x y}=c$

⇒ $c^{x y z}=c^1 \quad x y z=1$

Question 74. The value of $\left(\frac{x^a}{x^b}\right)^{\left(a^2+a b+b^2\right)} \times\left(\frac{x^b}{x^c}\right)^{\left(b^2+b c+c^2\right)} \times\left(\frac{x^c}{x^a}\right)^{\left(c^2+c a+a^2\right)}$

1
0
-1
None of these

Solution:

(1) 1

⇒ $x^{(a-b)\left(a^2+a b+b^2\right)} \cdot x^{(b-c)\left(b^2+b c+c^2\right)} \cdot x^{(c-a)\left(c^2+c a+a^2\right)}$

⇒ $x^{a^3-b^3} \cdot x^{b^3-c^3} \cdot x^{c^3-a^3} \cdot=x^{a^3-b^3+b^3-c^3+c^3-a^3} \quad \Rightarrow x^0=1$

Question 75. $\text { If } 2^x=3^y=6 \cdot z, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \text { is }$

1
0
2
none of these

Solution:

⇒ $2^x=3^y=6^{-z}=k$

⇒ $2=k^{\frac{1}{x}} \cdot 3=k^{\frac{1}{y}} 4=k^{-\frac{1}{z}} $

⇒ $2 \times 3=6 $

⇒ $k^{\frac{1}{x}} \cdot k^{\frac{1}{y}}=k^{-\frac{1}{z}}$

⇒ $ k^{\frac{1}{x}}+k^{\frac{1}{y}}=k^{-\frac{1}{z}}$

⇒ $ \frac{1}{x}+\frac{1}{y}=\frac{-1}{z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Logarithms

Concepts and Tricks to Remember

1. Logarithm: If a is a positive real number, other than 1 and a'” = x, then we write: m = loga x and say that the value of log x to the base a is m.

Example:

103 = 1000⇔log₁₀ 1000 = 3.

34 = 81 ⇔log₃ 81 = 4

2-3=$\frac{1}{8}$ « log2 =$\frac{1}{8}=-3$

(1)2 = .01 ⇔ log₍₁₀₎0.1 = 2.

2. Properties of Logarithms:

Loga (xy] = log_a X + log_a y

loga$\left(\frac{x}{y}\right)$ = logan-gay

log_a x — 1

Log_a 1 = 0

log_a (xp) = p (log_a x)

log_a X = $\frac{1}{\log _x a}$

log_a X = $\frac{\log _a x}{\log _b a}=\frac{\log _a x}{\log _a a}$

3. Common Logarithms: Logarithms to the base 10 are known as common logarithms. When the base is not mentioned, it is taken as 10.

4. The logarithm of a number contains two parts, namely characteristic and mantissa.

Characteristic: The integral part of the logarithm of a number is called its characteristic.

Case I: When the number is greater than 1.

In this case, the characteristic is one less than the number of digits on the left of the decimal point in the given number.

Case 2: When the number is less than 1.

In this case, the characteristic is one more than the number of zeros between the decimal point and, which is negative.

Instead of -1,-2, etc. We write 1(one bar), 2(two bar)etc.

Example:

Business Mathematics Ratio And Proportion chapter 2 The First Signification Digit Of The Number

Mantissa: The decimal part of the logarithm of a number is known as its mantissa.

For Mantissa, we look through the log table.

Some more tricks and formulae:

1. If a^b= c = log_a c=b; where a = 1.

2. a^loga^b = b^x

3. log_aⁿ=1

4. log_b _a =log_b x logxa =log_xalog_bx

5. log_ba=log_xa.log_yx.log_zy….log_bk

6. log_ba=log_bx.log_xy.log_yz….log_k

If log_ba=x

Then

⇒ $\log _{\frac{1}{5}} a=-x$

⇒ $\log _b \frac{1}{a}=-x$

⇒ $\log _{\frac{1}{b}} \frac{1}{a}=+x$

⇒ $\log _a b\left(m^n\right)=\frac{n}{b} \log _{a^n}$

⇒ $\log _{a^{\left(m m^n\right)}}=n \log _{a^{m i}}$

⇒ $\log _{\frac{1}{b}} m=\frac{1}{b} \log _{a^m}$

If log_am = log_am ⇒ a = b.

If log_am = log_an ⇒ m = n.

Logarithms Exercise – 1

Question 1. The value of log2 16is:

$\frac{1}{\mathrm{8}}$
4
8
16

Solution:

Let log1 16 = n. Then, 2″ = 16 = 24 => n = 4

log₂ 16 = n.

Question 2. If log_x4=0.4then the value of x is

Solution:

logs 4 = 0.4

⇒ $\Leftrightarrow \log _{\mathrm{x}} 4=\frac{4}{10}=\frac{2}{5} \Leftrightarrow x^{2 / 5}=4 \Leftrightarrow \mathrm{x}=4^{5 / 2}=\left(2^2\right)^{5 / 2}$

$\Leftrightarrow x=2^{\left(2 \times \frac{5}{2}\right)}=2^5 \Leftrightarrow x=32$

Question 3. If log10 y = 100 log₂X= 10, then the value of:

210
2100
2100
210000

Solution:

log₂ X = 10 ⇒ x = 2¹⁰

log_x y = 100 ⇒ y = x¹⁰⁰ = (2¹⁰)¹⁰⁰ = y= 2¹⁰⁰⁰

Question 4. $\frac{\log \sqrt{8}}{\log 8}$

$\frac{1}{\sqrt{8}}$
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{8}$

Solution:

⇒ $\frac{\log \sqrt{B}}{\log 8}=\frac{\log (B)^{1 / 2}}{\log 8}=\frac{\frac{1}{2} \log 8}{\log 8}=\frac{1}{2} .$

Question 5. Which of the following statements is not correct?

log₁₀ 10=1
log (2 + 3) = log (2 x 3)= log (2×3)
log₁₀ 1 = 0
log (1 + 2 + 3) = log ₁ + log₂+ log₃

Solution:

Since logaa = 1, so log₁₀ 10 = 1.

log (2 + 3)= 5 and log (2 x 3) = log 6 = log 2 + log 3

log (2 + 3) = log (2 + 3).

Since log_a 1 = 0, so log₁₀ 1 = 0.

log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log₁ + log ₂ + log ₃.

So, (2) is incorrect.

Question 6. If ax – by, then:

$\log \frac{a}{b}=\frac{x}{y}$
$\frac{\log a}{\log b}=\frac{x}{y}$
$\frac{\log n}{\log b}=\frac{y}{x}$
None

ax = by⇒ log a^x= log b^y⇒ x log a = y log b $\Rightarrow \frac{\log a}{\log b}=\frac{y}{x}$

Question 7. 2 log10 5+ log10 8 – $\frac{1}{2} \log _{10} 4=?$

2
4
2+2log10 2
4-4 log10 2

Solution:

2 log₁₀ 5 + login 8 login 4 = log₁₀ (52) + login 8 – log₁₀$\left(4^{\frac{1}{2}}\right)$

= log₁₀ 25 + log₁₀ 8 – log₁₀ 2 = log₁₀ $\left(\frac{25 \times 8}{2}\right)=\log _{11} 100=2 \text {. }$

Question 8. If loga (ab) = x, then logi, (ab) is:

$\frac{1}{x}$
$\frac{x}{x+1}\$
$\frac{x}{1-x}$
$\frac{x}{x-1}$

Solution:

Loga (ab) = x $\Leftrightarrow \quad \frac{\log a b}{\log a}=\mathrm{x} \quad \Leftrightarrow \quad \frac{\log a+\log b}{\log a}=\mathrm{x}$

⇒ $\Leftrightarrow \quad 1+\frac{\log b}{\log a}=x \quad \Leftrightarrow \quad \frac{\log b}{\log a}=x-1$

⇒ $\Leftrightarrow \quad \frac{\log a}{\log b}=\frac{1}{x-1} \quad \Leftrightarrow \quad 1+\frac{\log a}{\log b}=1+\frac{1}{x-1}$

⇒ $\Leftrightarrow \quad \frac{\log b}{\log a}+\frac{\log a}{\log b}=\frac{x}{x-1} \quad \Leftrightarrow \quad \frac{\log b+\log a}{\log b}=\frac{x}{x-1}$

⇒ $\Leftrightarrow \quad \frac{\log (a b)}{\log b}=\frac{x}{(x-1)} \quad \Leftrightarrow \quad \log _b(a b)=\frac{x}{x-1} .$

Question 9. log2 = x, log₃ = y and log 7 = z, then the value of log (4. √63) is:

$2 x+\frac{2}{3} y=\frac{1}{3} z$
$\text 2 x+\frac{2}{3} y+\frac{1}{3} z$
$2 x-\frac{2}{3} y+\frac{1}{3} z$
$-2 x+\frac{2}{3} y+\frac{1}{3} z$

Solution:

⇒ $\log (4 \cdot \sqrt[3]{63})=\log 4+\log (\sqrt[3]{63})=\log 4+\log (63)^{1 / 3}=\log \left(2^2\right) \log \left(7 \times 3^2\right)^{1 / 3}$

⇒ $=2 \log 2+\frac{1}{3} \log 7+\frac{2}{3} \log 3=2 x+\frac{1}{3} z+\frac{2}{3} y .$

Question 10. If log₁₂27=a,then log₆ 6 is:

$\frac{3-a}{4(3+a)}$
$\frac{3+a}{4(3-a)}$
$\frac{4(3+a)}{(3-a)}$
$\frac{4(3-a)}{(3+a)}$

Solution:

Logi2 27 = a $\frac{\log 27}{\log 12}=a \quad \Rightarrow \quad \frac{\log 3^3}{\log \left(3 \times 2^2\right)}=a$

⇒ $\frac{3 \log 3}{\log 3+2 \log 2}=a \quad \Rightarrow \quad \frac{\log 3+2 \log 2}{3 \log 3}=\frac{1}{a}$

⇒ $\frac{\log 3}{3 \log 3}+\frac{2 \log 2}{3 \log 3}=\frac{1}{a} \quad \Rightarrow \quad \frac{2}{3} \frac{\log 2}{\log 3}=\frac{1}{a}-\frac{1}{3}=\left(\frac{3-a}{3 a}\right)$

⇒ $\frac{\log 2}{\log 3}=\left(\frac{3-a}{2 a}\right) \quad \Rightarrow \quad \log 3=\left(\frac{2 a}{3-a}\right) \log 2$

⇒ $\log _6 16=\frac{\log 16}{\log 6}=\frac{\log 2^4}{\log (2 \times 3)}=\frac{4 \log 2}{\log 2+\log 3}=\frac{4 \log 2}{\log 2\left[1+\left(\frac{2 a}{3-a}\right)\right]}=\frac{4}{\left(\frac{3+\pi}{3-a}\right)}=\frac{4(3-a)}{(3+a)}$

Question 11. If log₁₀5+log₁₀(x+5)+1,then x is equal to:

Solution:

⇒ $\log _{10} 5+\log _{10}(5 x+1)=\log _{10}(x+5)+1$

⇒ $\Rightarrow \log _{10} 5+\log _{10}(5 x+1)=\log _{10}(x+5)+\log _{10} 10$

⇒ $\left.\Rightarrow \log _{10}[5(5 x+1)]=\log _{10} \mid 10(x+5)\right] \Rightarrow 5(5 x+1)=10(x+5)$

⇒ $\Rightarrow 5 x+1=2 x+10 \Rightarrow 3 x=9 \Rightarrow x=3$

Question 12. If log10 7=a then log10$\left(\frac{1}{70}\right)$ is equal to:

-(1+a)
(1+a)-1
$\frac{a}{10}$
$\frac{1}{10 a}$

Solution:

⇒ $\log _{10}\left(\frac{1}{70}\right)=\log _{10} 1-\log _{10} 70=-\log _{10}(7 \times 10)=-\left(\log _{10} 7+\log _{10} 10\right)=-(a+1) .$

Question 13. If log 27 = 1.431, then the value of log 9 is:

0.934
0.945
0.954
0.958

Solution:

log₂₇=1.431=log(3)3=1.431=3log₃=1.431

log₃=0.477 log₉=log(3)2=2log₃=(2×0.477)=0.954

Question 14. If log₁₀2 = 0.3010, then log₂ 10 is equal to:

$\frac{699}{301}$
$\frac{1000}{301}$
0.3010
0.6990

Solution:

⇒ $\log _2 10=\frac{1}{\log _{10} 2}=\frac{1}{0.3010}=\frac{10000}{3010}=\frac{1000}{301} .$

Question 15. If log₁₀2 = 0.3010, then value of log₁₀ 5 is:

0.3241
0.6911
0.6990
0.7525

Solution:

⇒ $\log _{10} 5=\log _{10}\left(\frac{10}{2}\right)=\log _{10} 10-\log _{10} 2=1-\log _{10} 2=(1-0.3010)=0.6990$

Question 16.If log₃=0.477 and (1000)x=3 then x equals

0.0159
0.0477
0.159
10

Solution:

(1000)x = 3⇒ log [(1000)x] = log₃ ⇒ x log 1000 = log₃

⇒ xlog (10) = log 3 ⇒ 3x log₁₀ = log₃

⇒ 3x = log 3 $x=\frac{0.477}{3}=0.159$

Question 17. If log3 = 0.3010 and log 3 = 0.4771, then the value of log 512 is:

2.870
2.967
3.876
3.912

Solution:

⇒ $\log _5 512=\frac{\log 512}{\log 5}=\frac{\log 2^9}{\log \left(\frac{10}{2}\right)}=\frac{9 \log 2}{\log 10-\log 2} \quad=\frac{(9 \times 0.3010)}{1-0.3010}=\frac{2.709}{0.699}=\frac{2709}{699}=3.876 .$

Question 18. If log₁₀ 2 = 0.3010 and log₁₀ 7 = 0.8451, then the value of login 2.8 is:

0.4471
1.4471
2.4471
None of these

Solution:

= login (7 x 22) -1 = logm7 + 2 login2-1

= 0.8451 + 2 x 0.3010 – 1 = 0.8451 + 0.602 -1 = 0.4471

Question 19. If log (0.57) = 1.756, then the value of log 57 + log (0.57)3 + log √0.57 is:

0.902
2.146
1.902
1.146

Solution:

Log (0.57) = 1.756 => log 57 = 1.756 [-. mantissa will remain the same]

log 57 + log (0.57)3 + log V0.57

⇒ $\log 57+3 \log \left(\frac{57}{100}\right)+\log \left(\frac{57}{100}\right)^{1 / 2}$

⇒ $\log 57+3 \log 57-3 \log 100+\frac{1}{2} \log 57-\frac{1}{2} \log 100$

⇒ $\frac{9}{2} \log 57-\frac{7}{2} \log 100=\frac{9}{2} \times 1.756-\frac{7}{2} \times 2=7.902-7=0.902$

Question 20. If log₁₀ 2 = 0.30103, then number of digits in 264 is:

Solution:

Log characteristics(264) = 64 x log₁₀ 19.2 =Hence,(64 x

0.30103)the number 19.26592.of digits in 264 is 20.

Question 21. log₆ + log₅ is expressed as

log 11
log 30
log 5/6
none of these

Solution:

(2) log 30

Log₆ + log₅

We know logab + loga C = log_a b

log₆ + log₅ = log_6×5 = log₃₀

Question 22. log₁₀ 8 is equal to

2
8
3
none of these

Solution:

(3) 3

Log2 8 = log223 = 3 log₂ 2 = 3 x 1 = 3

Question 24. log 32/4 is equal to

log 32/log 4
log 32 – log 4
2
none of these

Solution:

log 32 -log 4

We know log$\frac{b}{c}$ = loga b – loga c

⇒ $\log \frac{32}{4}=\log 32-\log 4$

Question 24. log (1 x 2 x 3) is equal to

log1 + log 2 + log 3
log 3
log 2
none of these

Solution:

(a) log t + log 2 + log 3

We know Ion., be x = Ion,, b + log,, c +

loga x

log (1x2x3) =log₁+ log₂ +log₃

Question 25. The value of log 0.0001to the base 0.1 is

-4
4
1/4
None of these

Solution:

(2) 4

log_0.1 0.0001 = log_0.1 (0.1)⁴ = 4×1 = 4

Question 26. lf 2log_x = 41og₃,thexisequalto

3
9
2
none of these

Solution:

(2)

log x² = log 3⁴ = x² = 3⁴

x²– 81 = x = 9

Question 27. log¹² 64 is equal to

12
6
1
none of these

Solution:

(1) 12

log √2 (√2)12 = 12

Question 28. log² 1728 is equal to

2x/3
2
6
none of these

Solution:

(3)

log2√326.√36

log2√3(2√3)6 =6

Question 29. log (1/81) to the base 9 is equal to

2
1/2
-2
none of these

Solution:

(3) -2

log9 1 – log9 81

0 – 2 =- 2

Question 30. log 0.0625 to the base 2 Is equal to

4
5
l
none of these

Solution:

(4)None of the above

as 0.0625 is divisible only by 2

Question 31. Given log2 = 0.3010 and log3 = 0.4771 the value of log 6 is

0.9030
0.9542
0.778 1
none of these

Solution:

(3) 0.7781

as log 6 = log 2 + log 3

= 0.3010 + 0.4771 =0.7781

Question 32. The value of log2 log2 loga 16

0
2
1
none of these

Solution:

log₂ log₂ log₂ 16 = log₂ log₂ 4

= log₂ 2 =1

Question 33. The value of log $\frac{1}{3}$ to the base 9 is

$\frac{1}{2}$
$\frac{1}{2}$
1
None of these

Solution:

(1)$\frac{-1}{2}$

log₉ $\frac{1}{3}=$ log₉ 1 – log₉ 3

= 0 -log₉ (9)√2 =$\frac{-1}{2}$

Question 34. If log x + log y = log (x+y), y can be expressed as

x-1
x-1
x/x-1
None of these

Solution:

(3) x / x – 1

Log xy = log (x + y)

Xy = x + y

Xy-y = x

Y(x-l) = x

⇒ $Y=\frac{x}{x-1}$

Question 35. The value of log₂[log₂ {logs (log₃₂73)}] is equal to

1
4
0
None of these

Solution:

(2) 0

=log₂[log₂{log₃9}]

=log₂[log₂2]=log₂1=0

Question 36. If log₂ x + log₁₀x + log₁₆ x = 21/4, these x is equal to

8
4
16
None of these

Solution:

⇒ $=\frac{\log _{10} x}{\log _{10} 2}+\frac{\log _{10} x}{\log _{10} 4}+\frac{\log _{10} x}{\log _{10} 16}$

⇒ $ \frac{\log _{10} x}{\log _{10} 2}+\frac{\log _{10} x}{2 \log _{10} 2}+\frac{\log _{10} x}{4 \log _{10} 2}$

⇒ $\frac{4 \log _{10} x+2 \log _{10} x+\log _{10} x}{4 \log _{10} 2} $

⇒ $\frac{\log _{10} x^4+\log _{10} x^2+\log _{10} x}{4 \log _{10} 2}$

⇒ $\frac{\log _{10}\left(x^7\right)}{4 \log _{10}^2}=\frac{21}{4}=\frac{\log _{10} x^7}{\log _{10}{ }^2}=21 $

⇒ $\log _2 X^7=21=\log _2$

Question 37. Given that, log₁₀2 = x and log₃ = y, the value of login 60 is expressed as

x- y + 1
x + y + 1
x — y — 1
None of these

Solution:

(2) x + y+ 1.

Log₁₀ 60 = log₁₀6x 10

= log₁₀ 6 + log₁₀ 10

= log₁₀2 + log₁₀ 3 + 1

= x + y +1

Question 38. Given that log₁₀ 2 = x, log₁₀ 3 = y, then log₁₀ 12 is expressed in terms of x and y as

x + 2y-1
x + y-1
2x +y-l
None of these

Solution:

(3)2x + y-l

= log₁₀ 1-2

= log₁₀6 + log₁₀ 0.2

= log₁₀ (2×3) + log₁₀ $\frac{2}{10}$

= log₁₀ 2 + log₁₀3 + log₁₀ 2 – log₁₀10.

= x + y + x-l=2x + y-l

Question 39. Given that log x = m + n and log y = m -n, the value of log lOx / y2 is expressed in terms of m and n as

1 – m + 3n
m -1 + 3n
m + 3n + 1
None of these

Solution:

(1) 1 – m + 3n.

Log$\frac{10 x}{y^2}$ = log10 + logs- logy2

= log₁₀ + logx- 2logy

=l + m + n- 2 (m-n)

= 1 + 3n – m = 1 – m + 3n

Question 40. The simplified value of 2 log₁₀ 5 + log₁₀ 8 – y2 log₁₀ 4 is

1/2
4
2
None of these

Solution:

(3) 2.

= 21og₁₀ 5 + log₁₀ 8 – log₁₀2

= log₁₀25 + log₁₀8-log₁₀2

= log₁₀25 + log₁₀2 + log₁₀4 – log₁₀2

= log₁₀25 + Iog₁₀4 = log₁₀lOO = 2.

Question 41. log [1 – {1 – (1 – x2)’1}1]1/2 can be written as

log x2
log x
log1/x
None of these

Solution:

(2) log x

⇒ $=\log \left|1-\left(1 \cdot\left(\frac{1}{1-1^{\prime}}\right)\right)+1\right|^{1 / 2} \quad=\log \left|1-\left(\frac{1-x^{-}-1}{1-x^{\prime}}\right)\right|^{1 / / 2}$

⇒ $=\log \left|1-\left(\frac{1-x^1}{-x^2}\right)\right|^{1 / 2}=\log \left|\frac{-x^2-\left(1-x^2\right)}{-x^2}\right|^{-1 / 2}$

Question 42. The simplified value of log $\sqrt[4]{729 \cdot \sqrt[3]{9^{-1} \cdot 27^{-4 / 3}}} \text { is }/$

log 3
log 2
log 1/2
None of these

Solution:

(1) log₃

⇒ $=\log \sqrt[1]{729 \sqrt[2]{9^{-1}\left(3^3\right)^{-4 / 3}}} \quad=\log \sqrt[4]{3^6 \sqrt[3]{3^{-2} \cdot 3^{-4}}}$

⇒ $=\log \sqrt[1]{3^6 \cdot 3^{-6 / 3}} \quad=\log \sqrt[4]{3^4}=\log 3$

Question 43. The value of (log₁₀a x log₁₀b x log₁₀c)3 is equal to

3
0
1
None of these

Solution:

(3) 1

⇒ $\left(\frac{\log _{a^a}}{\log _{a^b}} \times \frac{\log _{a^a}}{\log _{a^c}} \times \log _{a^c}\right)^3=(1)^3=1$

Question 44. The logarithm of 64 to the base 2\2 is

2
√2
1/2
None of these

Solution:

(4) None of these

⇒ $=\log _{\sqrt[2]{2}} 2^6=\log _{\sqrt[2]{2}} 2^4 \sqrt{2}^4=\log _{\sqrt[2]{2}}(\sqrt[2]{2})^4=4$

Question 45. The value of Ingn 25 given log₁₀ 2 = 0.3010 Is

1
2
1.5482
None of these

Solution:

(3) 1.5482

⇒ $ \log _8 5^2=2 \log _8 5=\frac{2 \log _{10} 5}{3 \log _{10^2}} \quad=\frac{2 \log _{10} \frac{1}{2}}{3 \log _{10} 2} \quad=\frac{2}{3} \frac{\left(\log _{10} 10-\log _{10} 2\right)}{\log _{10^2}}$

⇒ $\frac{2}{3} \frac{(1-0.3010)}{0.3010}=1.54817=1.5482 $

Question 46. The value of$\frac{\log _1 8}{\log _3 16 . \log _6 10}$

3log₁₀2
7log₁₀3
3log₆z
None

Solution:

⇒ $\frac{\log _3 B}{\log _7 16 \log _4 10}=\frac{\log _3 2^3}{\log _3 2^4 \times \log _2 210} \quad=\frac{3 \log _3 2}{\frac{4}{2} \log _3 2 \times \frac{1}{2} \log _2 10}=\frac{3}{\log _2 10}=3 \log _{10} 2(\text { a) is correct }$

Question 47. log 144 is equal to:

2log₄+2log₂
4log₂+2log₃
3log₂+4log₃
3log₂-4log₃

Solution:

Log 144 = log (16×9] = log 16 + log 9= log 2⁴ + log 3² = 4 log 2 + 2 log 3

(2) is correct.

Question 48. If log $\left(\frac{a+b}{4}\right)=\frac{1}{2}(\log a+\log b)$ then$: \frac{a}{b}+\frac{b}{a}$.

Solution:

⇒ $\log \left(\frac{a+b}{4}\right)=\frac{1}{2}(\log a+\log \mathrm{b})$

⇒ $\text { Or } \log \left(\frac{a+b}{4}\right)=\log (a b)^{1 / 2} \quad \text { Or } \frac{a+b}{4}=\sqrt{a b} \quad \text { Or } \mathrm{a}+\mathrm{b}=\sqrt[4]{a b}$

Squaring on both sides; we get (a + b)² = 16ab

Or a² + b² + 2ab= 16ab. Or a² + b² = 14ab

⇒ ${Or} \frac{a^2}{a b}+\frac{b^2}{a b}=\frac{14 a b}{a b} [Dividing by ab o both sides]$

⇒ ${Or} \frac{a}{b}+\frac{b}{a}=14$

(2) is correct

Question 49. If log²g x = 3$\frac{1}{3}$ find the value of x

Solution:

(1)

⇒ $\log _{\sqrt{8}^x}=\frac{10}{3} \Leftrightarrow x=(\sqrt{8})^{10 / 3}=\left(2^{2 / 3}\right)^{10 / 3}=2^{\left(\frac{2}{2} \times \frac{10}{3}\right)}=2^5=32 .$

Question 50. Evaluate :

log₅3 x log₂₇ 25
log₉ 27 – log₂₇ 9
1. 2,2
2. $\frac{4}{2}, \frac{1}{6}$
3. $\frac{2}{3}, \frac{5}{6}$
4. None of these

Solution:

(3)

⇒ $(1) \log _5 3 \times \log _{27} 25=\frac{\log 3}{\log 5} \times \frac{\log 25}{\log 27}=\frac{\log 3}{\log 5} \times \frac{\log \left(5^2\right)}{\log \left(3^2\right)}=\frac{\log 3}{\log 5} \times \frac{2 \log 5}{3 \log 3}=\frac{2}{3}$

$Let \log _y 27=n.$

Then 9n = 27 – 32n = 33 ⇔ 2n = 3 »

⇒ $\mathrm{n}=\frac{2}{3}$

Again, let log₂₂ 9 =m.

Then, 27^m= 9 ⇔ 3^3m = 32 ⇔ 3m = 2 ⇔ m = $\mathrm{n}=\frac{2}{3}$

log₉ 27 – log ₂₇ 9=(n-m)=$\left(\frac{3}{2}-\frac{2}{3}\right)=\frac{5}{6}$

Question 51.Simplify:$\left(\log \frac{75}{16}-2 \log \frac{5}{7}+\log \frac{32}{243}\right)$

log 3
log2
2
None of these

Solution:

⇒ $\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log \frac{75}{16}-\log \left(\frac{5}{9}\right)^2+\log \frac{32}{243}=\log \frac{75}{16}-\log \frac{25}{81}+\log \frac{32}{243}$

⇒ $=\log \left(\frac{75}{16} \times \frac{32}{243} \times \frac{81}{25}\right)=\log 2$

Question 52. If log x = a-b; log y = a + b then log$\left(\frac{10 x}{y^2}\right)$

l-a+3b
a-l+3b
l+3b-l
l-b+3a

Solution:

(1) is correct

log x = a + b; log y = a-b

$\log \left(\frac{10 x}{y^2}\right)=\log _{10}+\log x-\log y^2$ =1 + a + b- 21ogy=l + a + b- 2 (a-b)

=l + a + b-2a + 2b =l- a + 3b

Question 53. If log x = m + n ; logy = m – n then log$\left(\frac{10 x}{y^2}\right)=$

1-m + 3n
m-1+3n
m+3n+1
None

Solution:

(a) If log x = m + n; log y = m-n

Then $\log \left(\frac{10 x}{y^2}\right)$ =log!0 + logx-logy2= 1 + logx- 2 logy = 1 + (m+n) – 2(m-n)

Question 54. The integral part of a logarithm is called______, and the decimal part of a logarithm is called______.

Mantissa, Characteristic
Characteristic, Mantissa
Whole, Decimal
None of these

Solution:

(2) is correct.

Question 55. The value of$\frac{1}{\log _3 60}+\frac{1}{\log _4 60}+\frac{1}{\log _5 60}=$

Solution:

(2) is correct. log₆₀ 30 + log₆₀ 4 + log₆₀ 5 = log₆₀ (3x4x5)+ log₆₀60 =1

Question 56.If log10 (Xÿ + x) – log (x + 1) = 2 then the value of x is

Solution:

(3) is correct.

⇒ $\log _4 \frac{\left(x^2+x\right)}{(x+1)}=2 \quad \text { Or } \log _4\left\{\frac{x(x+1)}{(x+1)}\right\}=2 \quad \text { Or } \log _4 \mathrm{X}=2 \Rightarrow \mathrm{x}=4^2=16$

Question 57. log (l3 + 23 + 3s + + n) = ______.

2 log n + 2 log (n+1) – 2 log 2
log n + 2 log (n + 1) – 2 log
2 log n + log (n + 1) – 2 log 2
None

Solution:

log (l3 + 23 + 33+ + n3)

⇒ $=\log \left(\frac{n(n+1)}{2}\right)^2=2 \log \frac{n(n+1)}{2} \quad=2[\log n+\log (n+1)-\log 2]$

= 2 log n + 2 log(n+l) – 2 log 2 So, (1) is correct.

Question 58. $7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+31 \log \left(\frac{81}{80}\right)$

0
1
log2
log3

Solution:

Tricks:- $7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right)$

⇒ $\log \left(\frac{16}{15}\right)^7+\log \left(\frac{25}{24}\right)^5+\log \left(\frac{81}{80}\right)^3=\log \left[\left(\frac{16}{15}\right)^7 \cdot\left(\frac{25}{24}\right)^5 \cdot\left(\frac{81}{80}\right)^3\right]$

We get; it is approx. 2 (3) is correct

Question 59. The value of the expression alog_ab.log_bC.log_cd.log_dt.

t
abcdt
(a+b+c+d+t)
None.

Solution:

alog_ab.log_bc.log_cd.log_dt

=a¹^loga^t=t¹=t

∴ (1) is correct

Question 60. If login on x= then x is given by:

1/100
1/10
1/20
None of these

Solution:

log100000x=$\frac{-1}{4}$

⇒ $ (10000)^{-\frac{1}{4}}=x$

⇒ $ \left(10^4\right)^{-\frac{1}{4}}=x$

⇒ $\mathrm{x}=10^{-1}=\frac{1}{10}$

(2) is correct.

Question 61. Find the value of x which satisfies the relation log10 3 + log10 (4x + 1) = log10 (x + 1) + 1

$\frac{3}{2}$
$\frac{5}{2}$
4
$\frac{7}{2}$

Solution:

log ₁₀ 3=log10(4x+1)=log₁₀(x+1)+1
$\log _{10} 3+\log _{10}(4 x+1)=\log _{10}(x+1)+\log _{10} 10 \Leftrightarrow \log _{10}[3(4 x+1)]=\log _{10}[10(x+1)]$
$\Leftrightarrow 3(4 x+1)=10(x+1) \Leftrightarrow 12 x+3=10 x+10 \Leftrightarrow 2 x=7 \Leftrightarrow x=\frac{7}{2}$

Question 62. Simplify:$\left[\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}\right]$

0
1
2
None of these

Solution:

(3)

Given expression = logxyz (xy) + \ogxyz (yz) + logxyz ) (zx)

= logxyz (xy x yz x zx) = logxyz (xyz)2

= 2logxyz(XYZ)=2×1=2

Question 63. If log₁₀ 2 = 0.30103, find the value of log₁₀ 50.
Solution:

log₁₀ 50 = log $\left[\frac{100}{2}\right]$ = log₁₀(, 100- log₁₀ 2 = 2- 0.30103 = 1.69897.

Question 64. Iflog (2a – 3b) = log a – log b, then a = ?

$\frac{3 b^2}{2 b-1}$
$\frac{3 b}{2 b-1}$
$\frac{b^2}{2 b+1}$
$\frac{3 b^2}{2 b+1}$

Solution:

Log(2a-3b)=log$\frac{a}{b}$

Or 2a – 3b =$\frac{a}{b}$

Or 2ab-3b2 Or a(2b- 1) = 3b2

Or $a=\frac{3 b^2}{2 b-1}$

(1) is Correct

Question 65. If log 2 = 0.3010 and log 3 = 0.4771, find the values of:

Log 25
Log 4.5
1. 1.398, 0.6532
2. 1.389, 0.649
3. 1.19, 0.697
4. None of these

Solution:

Log 25 = log$\left[\frac{100}{4}\right]$ = log 100 – log4 = 2-2 log2 = (2 – 2 x 0.3010) = 1.398.

Log 4.5 = log$\left[\frac{9}{2}\right]=$ log 9 – log2 = 21og 3 -log 2 = (2 x 0.4771 – 0.3010) = 0.6532

Question 66. If log 2 = 0.30103, find the number of digits in 256

Solution:

Log (256) = 56 log 2 = (56 x 0.30103) = 16.85768.

Its characteristic is 16 hence, the number of digits in 2s6 is 17.

Question 67. $\frac{1}{\log _{a b}(a b c)}+\frac{1}{\log _{b c}(a b c)}+\frac{1}{\log _{(a b c)}} \text { is equal to: }$

0
1
2
-1

Solution:

⇒ $\frac{1}{\log _{a b}(a b c)}+\frac{1}{\log _{b c}(a b c)}+\frac{1}{\log _{c a}(a b c)}$ = Logabc (ab.bc.ca) = logabc (abc)2

= 2 logabc (abc) = 2×1 = 2

Question 68. Log4 (x2 + x) — log4(x + 1) = 2. Find x

16
0
-1
None of these

Solution:

⇒ $\log _4\left(x^2+x\right)-\log _4(x+1)=2$

Log₁₀(x2 + x)- logA(x + 1) = 2

⇒ $\log _4 \frac{x(x+1)}{x+1}=4^2$

Or x = 14

(1) is correct

Question 69. If logab + loga c = 0 then

b=c
b = -c
b=c=l
b and c are reciprocals.

Solution:

Loga b + Loga c = 0 or Loga (bc) = Loga1 bc =1 b =$\frac{1}{c}$

(4) is correct

Question 70. If log 2 x + log.(X = 6), then the value of x is

Solution:

(1)

Detail method: log₂ x + log , x = 6 (Or )log₂ x + log-22 x = 6 Or log₂ x + log₂x = 6

⇒ $\left(1+\frac{1}{2}\right) \log _2 x=6 \quad \text { Or } \log _2 x=\frac{6 \times 2}{3}=4 x=2^4=16$

Question 71. If log₁₀ Y = 100 and log₂ x = 10, then the value of’Y’is’:

210
2100
21000
210000

Solution:

Log x x = 100

∴ x = 2¹⁰ Now logx y = 100 y = x1000 y = (2¹⁰)¹⁰⁰ = 2¹⁰⁰⁰

Question 72. Which of the following is true?$\text { If } \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{1}{a b c}$

log (ab+bc+ca) = abc
$\log \left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=a b c$
Log (abc) = 0
Log (a+b+c) = 0

Solution:

(4) $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{1}{a b c}$

Multiplying both sides by a b c $ \frac{a b c}{a b}+\frac{a b c}{b c}+\frac{a b c}{c a}=\frac{a b c}{a b c}$

Or; c + a + b = 1 Or; a + b + c =1

Taking logs on both sides; we get. Log (a+b+c) = log¹⁰ =

Question 73. Find the value of log49.1og32 =

Solution:

(4) log4 9.1og:< 2 = log (22) (32).log3 2 $=\frac{2}{2} \log _2 3 \cdot \log _3 2$=1×1=1

Question 74. If x = log24 12; y=log3G 24; z = log48 36 then xyz +1 =?

Solution:

(c)

XYZ + 1 = log₂₄ 12; y=log36

if 24; z = log₄₈ 36 +1

= log₄₈ 12+log₄₈ 48

= login (12×48) = log₄₈ (12 x 2)2 =2 log₄₈ 24 = 2log3t,24.log₄₈ 36 = 2yz

Question 75. If x2 + y2 = 7xy then log$\frac{1}{3}$(x + y) =

logs + logy
$\frac{1}{2}(\log x+\log y) $
$\frac{1}{3}(\log x+\log y)$
$\frac{1}{3}(\log x \cdot \log y)$

Solution:

(2) is correct $\log _3 \frac{1}{3}(x+y)=\frac{1}{2} 2 \log \left\{\frac{1}{3}(x+y)\right\} $

⇒ $=\frac{1}{2} \log\left\{\frac{1}{3}(x+y)\right\}^2$

⇒ $=\frac{1}{2} \log \left(\frac{x^2+y^2+2 x y}{9}\right)$

⇒ $=\frac{1}{2} \log \left(\frac{7 x y+2 x y}{9}\right)$

⇒ $=\frac{1}{2} \log (x y)=\frac{1}{2}(\log x+\log y)$

CA Foundation Maths Solutions For Chapter 2 Equations

July 5, 2024July 5, 2024 by Marksparks

CA Foundation Maths Solutions For Chapter 4 Equations

Introduction and Meaning:

The equation is defined to be a mathematical statement of equality.

Determination of the value of the variable that satisfies an equation is called a solution of the equation or root of the equation.
An equation in which the highest power of the variable is 1 is called a Linear (or simple) equation.
This is also called the equation of degree 1.
Two or more linear equations involving two or more variables are called Simultaneous Linear Equations.
An equation of degree 2 (Highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation.

Simple Equation

A simple equation is one unknown x in the form of a x + b = 0.

where a, and b are known constants and a1 0

Note: A simple equation has only one root.

Read and Learn More CA Foundation Maths Solutions

Simultaneous Linear Equations in Two Unknowns

Two such equations a₁x+b₁y +c₁= 0 and a₂ x+b₂y + c₂ = 0 form a pair of simultaneous equations in x and y.
A value for each unknown that satisfies simultaneously both equations will give the roots of the equations.

Method of solution

1. Elimination Method: In this method, two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknown and then solving for the other unknown

Example 1: Solve, 2x+5y = 9 and 3x- y =5.
Solution:

2x + 5y = 9 …..(1)

3x – y = 5 …..(2)

By making (1) x 1, 2x+5y = 9

And by making (2) x 5, 15x- 5y = 25

Adding 17x- 34 or x- 2. Substituting this values of x in (1) i.e. 5y = 9 – 2x we find;

5y = 9 – 4 =5

y = 1

x = 2,

y =1.

2. Cross Multiplication Method:

Let two equations be:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

We write the coefficients of x, y, and constant terms and two more columns by repeating the coefficients of x and y as follows:

1 2 3 4

b₁ c₁ a₁ b₁

b₂ c₂ a₂ b₂

⇒ and the result is given by$y: \frac{x}{b 1 c 2-b 2 c 1}=\frac{y}{c 1 a 2-c 2 a 1}=\frac{x}{a 1 b 2-a 2 b 1}$

⇒ so the solution is: x = $=\frac{b 1 c 2-b 2 c 1}{a 1 b 2-a 2 b 1} \quad y=\frac{c 1 a 2-c 2 a 1}{a 1 b 2-a 2 b 1}$

Example 2: Solve 3x+2y+17 =0, 5x-6y-9=0

Solution: 3x +2y 17= 0…..(1)

5x – 6y -9 = 0…..(3)

Method of elimination:

By (1) x 3 we get 9x + 6y + 51 = 0 …

Adding (2) and (3) we get 14x+42=0

Or x=$x=\frac{42}{14}=-3$

Putting x = -3 in (1) we get 3(-3) + 17 = 0

Or, 2y + 8 $=0 \text { or, } y=-\frac{8}{2}-4$

So,x = -3 andy = -4

Method of cross-multiplication:

3x + 2y + 17 = 0

5x – 6y- 9 = 0

⇒ $ \frac{x}{2(-9)-17(-6)}=\frac{y}{17(5)-3(-9)}=\frac{1}{3(-6)-5(2)}$

Or, $ \quad \frac{x}{84}=\frac{y}{112}=\frac{1}{-28}$

Or $\frac{x}{3}=\frac{y}{4}=\frac{1}{-28}$

Or x = -3 , y = -4

Method of Solving Simultaneous Linear Equation with Three Variables

Example 1: Solve for x, y and z:

2x-y + z = 3, x + 3y-2z=ll, 3x-2y + 4z=l

Solution:

(1) Method Of Elimination

2x – y + z = 3…..(1)

x + 3y — 2z =11…..(2)

3x – 2y + 4z = 1…..(3)

By (1) x 2 we get

4x – 2y + 2z = 6…..(4)

By (2) + (4),5x + y= 17…..(5)

By (2) x 2, 2x + 6y- 4z = 22…..(6)

By (3) + (6), 5x + 4y = 23…..(4)

By (5) – (7), – 3y = -6 ory = 2

Putting y = 2 in (5) 5x + 2 = 17, or 5x = 15 or, x=3

Putting = 3andy = 2 in (1)

2×3-2+ z =3

Or 6 – 2 + z = 3

Or 4 + z = 3

Or z=-l

So x- 3, y- 2, Z = -1 is the required solution

(2) Method Of Cross-Multiplication

We write the equations as follows:

2x-y + (z – 3) = 0

x + 3y + (-2z-ll) = 0

By cross multiplication

⇒ $ \frac{x}{-1(-2 z-11)-3(z-3)}=\frac{y}{(z-3)-2(-2 z-11)}=\frac{1}{23-1(-1)}$

⇒ $\frac{x}{20-z}=\frac{y}{5 z+19}=\frac{1}{7}$

⇒ $X=\frac{20-z}{7}, y=\frac{5 z+19}{7}$

Substituting above values for x and y in equation (iii) i.e. 3x- 2y + yz = 1, we have

⇒ $3{20-z}{7}-2{5 z-19}{7}+4 z=1$

Or 60 – 3z = lOz- 38 + 28 z = 7

Or 15z = 7-22 orl5z = -15 or Z=l

⇒ $x=\frac{20-(-1)}{7}=\frac{21}{7}=3, \quad y=\frac{5(-1)+19}{7}=\frac{14)}{7}=2$

Thus x = 3,y= 2, Z= -1.

An equation of the form ax2 + fax + c = 0 where x is a variable and a, b, c are constants with a ≠ 0 is called a quadratic equation or equation of the second degree.

When b = 0 the equation is called a pure quadratic equation; when b * 0 the equation is called an affected quadratic.

Examples:

1. 2x² + 3x + 5 = 0

2. x²-x = 0

3. 5x² – 6x – 3 = 0

The value of the variable say x is called the root of the equation. A quadratic equation has two roots.

How to find out the roots of a Quadratic Equation:

ax² +bx + c = 0 (a ≠ 0)

⇒ $\text { Or } x^2+\frac{b}{a} x+\frac{c}{a}=0$

Or $x^2+2 \frac{b}{2 a} x+\frac{b^2}{4 a^2}=\frac{b^2}{4 a^2}-\frac{c}{a}$

Or$\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}$

Or$\mathrm{x}+\frac{b}{2 a}=\frac{ \pm \sqrt{b^2-4 a c}}{2 a}$

Or$\mathrm{x}=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

Sum and Product of the Roots:

Let one root be a and the other root be p

Now $ \alpha+\beta & =\frac{-b+\sqrt{b^2-4 a c}}{2 a}+\frac{-b-\sqrt{b^2-4 a c}}{2 a}=\frac{-b+\sqrt{b^2-4 a c}-b-\sqrt{b^2-4 a c}}{2 a}$

⇒ $\frac{-2 b}{2 a}=\frac{-b}{a}$

Thus sum of roots = $-\frac{b}{a}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}$

⇒ $x \beta=\left(\frac{-b+\sqrt{b^2-4 a c}}{2 a}\right)\left(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\right)=\frac{c}{a}$

So the product of the roots =$\frac{\mathrm{e}}{a}=\frac{\text { Constant term }}{\text { coefficient of } x^2}$

How to Construct a Quadratic Equation

For the equation ax² + bx + c = 0 we have

⇒ $\text { Or } x^2+\frac{b}{a} x+\frac{c}{a}=0$

⇒ $\text { Or } x^2-\left(-\frac{b}{a}\right) x+\frac{c}{a}=0$

Or A-2 – (Sum of the roots) x + Product of the roots = 0

Nature of the Roots

⇒ $X=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

If b² — 4ac = 0 the roots are real and equal;

If b²– 4ac > 0 then the roots are real and unequal (or distinct);

If b²– 4ac < 0 then the roots are imaginary;

If b²– 4ac is a perfect square (≠0) the roots are real, rational, and unequal (distinct);

If b² – 4ac > 0 but not a perfect square the roots are real, irrational, and unequal.

Since b²– 4ac discriminates the roots b²– 4ac is called the discriminant in the equation ax² + bx + c = 0 as it discriminates between the roots.

Note:

Irrational roots occur in conjugate pairs that is if (m + Vn) is a root then (m – Vn) is the other of the same equation.
If one root is reciprocal to the other root then their product is 1 and so $\frac{c}{a}=1 \text { i.e. } c=\mathrm{a}$
If one root is equal to another root but opposite in sign then.
- Their sum = 0 and so $\frac{b}{a}=1 \text { i.e. } c=\mathrm{a}$

Summary, Tips, Tricks And Formulae

Equations

An equation is defined as a mathematical statement of equality.

Types of equations

Linear equation in one variable.

Linear simultaneous equations in 2 or 3 variables.
Quadratic equations.
Cubic equations.
Bi-quadratic equations.
Exponential equations.

Quadratic Equations

A quadratic equation is defined as the polynomial equation of degree 2.
A quadratic equation can be expressed in the following general form: ax² + ba + x = 0, (a ≠ 0)
A quadratic equation can also be expressed in the factor form as follows: a(x-α){x-β)
1. Here, a and /3 are the roots or solutions of quadratic equations;
2. The general solution of the quadratic equation can be obtained as follows:
$\alpha=\frac{-b+\sqrt{b^2-4}}{2 a} \text { and } \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
- Sum of roots =$a+\beta=-\frac{b}{a}$
Product of roots =$\alpha\beta=\frac{c}{a}$

Structure of Quadratic Equations

If the sum (S) (α + β) and Product (P) (αβ) of the roots are known, then the quadratic is x² – Sx + P = 0

Sign of Roots of a Quadratic Equation

When c = 0, one root of the equation must be 0,
When h and c are 0, then both the roots must be 0.
If a,b, and c all are 0, both roots are negative.
If a and c are of some sign, opposite to that of, then both the roots will be positive.
If a and c are of opposite signs, one root is positive, and another root is negative.

Nature of Roots

The Expression “b² – 4ac” is called the “Discriminant (D)” of the quadratic equation.

When D> 0, Roots are real and distinct.
When D = 0, Roots are real and equal.
When D < 0, the Roots are imaginary.
When D > 0, Roots are real.
When D is a perfect square, Roots are real, rational, and unequal.
When D is not a perfect square, Roots are real, rational, and unequal.
If roots are equal use b² = 4ac.
If roots are reciprocal of each other, use a = c
If roots are equal but of opposite sign, use b = 0
If roots are reciprocal but opposite in sign, use c = -0

Note

1. Irrational roots will always appear in conjugate pairs.

it = (a- √b) and (1= (a + √b)

2. Imaginary roots will always appear in conjugate pairs.

u = (a- ib) and β = (a + ib)

Cubic Equations

1. A cubic equation is a polynomial equation of degree 3, and the general form is represented as follows:

ax³+ bx² + cx + d = 0; (a ≠ 0)

2. The factor form of the cubic equation is given as follows:

a(x- α)(x- β)(x- γ) = 0

Here, α,β, and γ are the roots or solutions of the cubic equation.

3. Sum of roots = α+β+γ= -b/a

4. Product of the roots = αβγ = -d/a

Bi-Quadratic Equations

A bi-quadratic equation is a polynomial of degree 4, and the general form is represented as follows:

ax⁴+ bx³ + cx² + dx + e = 0; (a * 0)

The factor form of a cubic equation is given as follows:

a(x- α)(x- β)(x-γ)(x-δ) = 0

Here, α, β,andγ are the roots or solutions of the bi-quadratic equation.

The sum of roots = α + β +γ + δ = -b/a

Product of the roots = αβγδ = e/a

Exercise – 1

Simple Or Linear Equation in one Variable

Question 1. The equation -7x +1 = 5 – 3 x will be satisfied for x equal to:

2
-1
1
None of these

Solution:

(2)

7x + 1 = 5 – 3x

-4 = 4x

X = -1

Question 2. The root of the equation$1 \frac{x+4}{4}+\frac{x-5}{3}=11$ is

20
10
2
None of these

Solution:

(1)

⇒ $\frac{x+4}{4}+\frac{x-5}{3}=11$

3x+ 12 + 4x – 20 = 132

7x = 140 → x = 20

Question 3. Pick up the correct value of x for$\frac{x}{30}=\frac{2}{45}$

x = 5
x = 7
$x=1 \frac{1}{3}$
None Of these

Solution:

(3)

⇒ $\frac{x}{30}=\frac{2}{45} \rightarrow \frac{x}{2}=\frac{2}{3} \rightarrow \mathrm{x}=\frac{1}{{ }^3} $

Question 4. The solution of the equation $\frac{x+24}{5}=4+\frac{x}{4}$

6
10
16
None of these

Solution:

(3)

⇒ $\frac{x+24}{5}=\frac{4+x}{4}$

⇒ $\frac{4 x+96-5 x}{20}=4$

⇒ $-\mathrm{x}=-16 \rightarrow \mathrm{x}=16$

Question 5. 8 is the solution of the equation

$\frac{x+4}{4}+\frac{x-5}{3}=11$
$\frac{x+4}{2}+\frac{x+10}{9}=8$
$\frac{x+24}{5}=4+\frac{x}{4}$
$\frac{x-15}{10}+\frac{x+5}{5}=4$

Solution:

(2)

⇒ $\frac{x+4}{2}+\frac{x+10}{9}=8 \Rightarrow \frac{8+4}{2} \rightarrow \frac{8+10}{9}=6+2=8$

Question 6. The value of that satisfies the equation$\frac{y+11}{6}-\frac{y+1}{9}=\frac{y+7}{4} \text { is }$

-1
7
1
$-\frac{1}{7}$

Solution:

(3)

⇒ $\frac{3 y+33-2 y+2}{18}=\frac{y+7}{4}$

⇒ $\frac{y+35}{18}=\frac{y+7}{4} \rightarrow \frac{2 y+70}{36}=\frac{9 y+63}{36}$

⇒ 7y = 7 ⇒ y = 1

Question 7. The solution of the equation (p+2) (p-3) + (p+3) (p-4) = p(2p-5) is

6
7
5
None of these

Solution: (1)

Question 8. The equation $\frac{12 x+1}{4}=\frac{15 x-1}{5}+\frac{2 x-5}{3 x-1}$ is true for.

A = 1
x = 2
A = 5
x = 7

Solution:

(4)

⇒ $\frac{12(7)+1}{1}=\frac{85}{4}$

⇒ $\frac{15(7)-1}{1}+\frac{2(7)-5}{3(7)-1}=\frac{425}{20}=\frac{85}{4}$

Question 9. Pick up the correct value x for which$\frac{x}{0.5}-\frac{1}{0.05}+\frac{x}{0.005}-\frac{1}{0.0005}=0$

x = 0
x = 1
x = 10
None of these

Solution:

(3)

⇒ $\frac{10 x}{5}-\frac{100}{5}+\frac{1000 x}{5}-\frac{10000}{5}=0$

⇒ $\frac{1010 x}{5}=\frac{10000+100}{5}$

10l0x = 10100

X = 10

Question 10. The sum of two numbers is 52 and their difference is 2. The numbers are

17 and 15
12 and 10
27 and 25
None of these

Solution:

(3)

Let x lie bigger no, y be a smaller number

27and25

27 + 25 = 52 and 27 – 25 = 2

Question 11. The diagonal of a rectangle is 5 cm and one of at sides is 4 cm. Its area is

20 sq. cm.
l2 sq. cm.
10 sq. cm.
None of these

Solution:

(2)

Diagonal = 5cm

Sides = 4cm x x

42 + x2 = 52 by Pythagoras theorem

x2 = 9

x = 3

Area = 4×3 = I2sq cm

Question 12. Divide 56 into two parts such that three times the first part exceeds.one-third of the second by 48. The parts are.

(20, 36)
(25.31)
(24, 32)
None of these

Solution:

(1)

3x = $\frac{1}{3}$y + 48

⇒ $\frac{1}{3}$

⇒ $\frac{y}{3}$

9x – 144 = y

56 = x + 9x – 144

200 = lOx ⇒ x = 20, y = 36

Question 13. The sum of the digits of a two-digit number is 10. If 18 is subtracted from it the digits in the resulting number will be equal. The number is

37
73
75
None of these

Solution:

(2)

7 + 3 = 10

73-18 = 55

Question 14. The fourth part of a number exceeds the sixth part by 4. The number is

84
44
48
None Of these

Solution:

(3)

⇒ $\frac{x}{4}=\frac{x}{6}+4$

⇒ $\frac{x}{4}-4=\frac{x}{6} \rightarrow 3 x-48=2$

x = 48

Question l5. Ten years ago the age of a father was four times of his son. Ten years hence the age of the father will be twice that of ofIlls son. The present ages of the father and the son are.

(50. 20)
60, 20)
(55,25)
None of these

Solution:

(1)

4 (y – 10] = x – 1o

4y – 40 = x – 10

4y = 3(30)- y = $\frac{x+30}{4}$

(x+ 10] = 2 (y+ 10)

X+ 10 = 2y + 20

⇒ $\frac{x-10}{2}=\frac{x-10}{2}$

X + 30 = 2x – 20 => x = 50, y = 20

Question 16. The product of two numbers is 3200 and the quotient when the larger number is divided by the smaller is 2. The numbers are

(16. 200)
(160, 20)
(60,30)
(80, 40)

Solution:

(4)

xy = 3200 x = 2; y= x= 2y

⇒ $\frac{x}{2}=y$

x2 = 6400 x = 80 y = 40

Question 17. The denominator of a fraction exceeds the numerator by 2. If 5 is added to the numerator the fraction increases by unity. The fraction is.

$\frac{5}{7}$
$\frac{1}{3}$
$\frac{7}{9}$
$\frac{3}{5}$

Solution:

(4)

⇒ $\frac{x}{x+2}$

⇒ $\frac{1+5}{x+2}=\frac{x}{x+2}+1$

x+5=2 x+2

3=x

function $ =\frac{3}{5}$

Question 18. Three persons Mr. Roy, Mr. Paul, and Mr. Singh together have Mr. Paul has less than Mr. Roy and Mr. Singh has got less than Mr. Roy. They have the money.

(₹20,₹16,₹15)
(₹15,₹20,₹16)
(₹25,₹11,₹5)
None of these

Solution:

(1)

20- 16 = 4,

2a- 15 = 5

Question 19. A number consists of two digits. The digits in the ten’s place are 3 times the digits in the unit’s place. If 54 is subtracted from the number the digits are reversed. The number is

Solution:

(3)

9 = 3 (3)

93-54= 39

Question 20. One student is asked to divide half of a number by 6 and the other half by 4 and then to add the two quantities. Instead of doing so, the student divides the given number by 5. If the answer is 4 short of the correct answer then the number was

320
400
480
None of these.

Solution:

(3)

Let number be x $\frac{x}{12}+\frac{x}{8}=4+\frac{x}{5}$

⇒ $\frac{2 x+3 x}{24}=\frac{4+x}{5} \Rightarrow \frac{5 x}{24}-\frac{x}{5}=4$

⇒ $\frac{25 x-24 x}{24 \times 5}=4$

x = 4 x 24 x 5

x = 480

Question 21. If several which the half is greater than $\frac{1}{5}$th of the number by 15 then the number is

50
40
80
None of these

Solution:

(1)

let number be x according to question

⇒ $\frac{x}{2}=\frac{x}{5}+15$

⇒ $\frac{x}{2}-\frac{x}{5}=15$

3x = 150 ⇒ x = 50

Question 22. $\text { If } \frac{x-b c}{b+c}+\frac{x-c a}{c+a}+\frac{x-a b}{a+b}=$ a + b + c the value of x is

a2+b2+c2
a(a+b+c)
(a+b)(b+c)
ab+bc+ca

Solution:

(4)

ab + be + ca

⇒ $\rightarrow \frac{a b+c a}{b+c}+\frac{a b+b c}{a+c}+\frac{b c+c a}{a+b}$

= a+b+c

Question 23. A man rowing at the rate of 5 km in an hour in still water takes thrice as much time going 40 km up the river as in going 40 km down. Find the rate at which the river flows:

4.5km/hr
7.5 km/hr
2.5 km/hr
None

Solution:

Let the speed of the river be x km/hr

Speed of man in still water = 5 km/hr

Speed while upward rowing will be 5 – x

Speed while downward rowing will be 5 + x

We know speed = distance or time

Let the time taken for upward rowing be T1 and the time taken for downward rowing be T2.

According to question TI=3T2

⇒ $\frac{40}{5-x}=3 \frac{40}{5+x}$

5+x = 15 – 3x

x= 2.5 km/hr

(3) is correct

Question 24. If the length of a rectangle is 5 cm more than the breadth and if the perimeter of the rectangle is 40 cm, then the length and breadth of the rectangle will be:

7.5 cm, 2.5 cm
10cm, 5cm
12.5cm, 7.5cm
15.5cm, 10.5cm

Solution:

(3)

1st condition = length is 5cm more than breadth

All options satisfy this condition

II perimeter = 2(l+b) = 40 of rectangle

Only option (c) satisfies it

Detail Method

Let breadth = x; :-length = x + 5

Perimeter = 40

2(x+5+x) = 40

Or 2x + 5 = 20

Or x = $\frac{15}{2}$= 7.5cm

Length = x + 5 = 7.5 + 5 = 12.5

Breadth = x = 7.5cm

(3) is correct

Question 25. For all 2 e R, the line (2+ 2)x + (3- A) y + 5 = 0 Passing through a fixed point, then the fixed point is_____.

(-1,-1)
0,0
2,2
1. 1

Solution:

(2)

For option (b) Point (-1; -1) satisfies the equation

LHS = =(2+ A)x + (3-/1) y + 5

Or (2 + A)(-l]+(3- A) (-l) + 5

-2-A-3+A + 5 = 0 = RHS.

(2) is correct

Question 26. If kx – 4 = (k – l).x which of the following is true

x = -5
x = -4
x = -3
x = 4

Solution:

(4) is correct

Kx- 4 = (k-l)x

Orkx-4 = kx-x

Or -4 = -x

∴ x = 4

Question 27. The age of a person is 8 years more than thrice the age of the sum of his two grandsons who were twins. After 8 years his age will be 10 years more than twice the sum of the ages of his grandsons. Then the age of the person when the twins were born is_

86 yrs
73 yrs
68 yrs
63 yrs

Solution:

(2) let the age of 1st grandson = x

Person’s age = P = 3(x+x) + 8

P = 6x + 8

After 8 years

P + 8 = 2 [x+8+x+8] + 10

= 2(2x+16) + 10

Or 6x + 8 + 8 = 4x + 32 + 10

Or 2x = 42 -16 = 26

x = 13

Age of person when grandsons

were born

= 6x + 8 – x

= 6 x 13 + 8-13 = 73

(2) is correct

Question 28. In a school number of students in each section is 36. If 12 new students are added, then the number of sections is increased by 4 and the number of students in each section becomes 30. The original number of sections at first is

Solution:

(4) let original No. of sections = x

Total students = 36x

36x+ 12 = (x+4) 30

Or 36x + 12 = 30x + 120

Or 6x = 108 ⇒ x = 18

Question 29. A person on a tour has for his expenses. But the tour was extended for another 16 days, so he must cut his daily expenses by 20. The original duration of the tour had been?

48 days
64 days
80days
96 days

Solution:

Let No. of tour days = x

Expenses per day =$=\frac{9600}{x}$

Now Expense per day =$=\frac{9600}{x+16}$

⇒ $\text { From } \frac{9600}{x}-\frac{9600}{x+16}=20$

From here we get

For (3) LHS

⇒ $\frac{9600}{80}-\frac{9600}{80+16}=20 \text { RHS. }$

(3) is correct

Question 30. The particular company produces some articles in a day. The cost of production per article is more than thrice the number of articles and the total cost of production is on a day then the number of articles is:

Solution:

(1) is correct.

Let (A) be correct.

So cost, per unit = 800/16 =

It is 2 more than 3 times of 16. (as given in Qts.)

Question 31. The sides of an equilateral triangle are shortened by 3 units, 4 units, and 5 units respectively, and then a right-angle triangle is formed. The side of the equilateral triangle was

Solution:

For option [3]

1st side of right-angled A = 8 – 3 = 5

2nd side = 8-4 = 4

And 3rd side = 8-5 = 3

Here; 5; 4 and 3 are making a right-angled

triangle.

So, 52 = 42 + 32

Hence, option (c) is correct

Question 32. A number consists of two digits such that the digit in one’s place is thrice the digit in ten’s place. If 36 is added then the digits are reversed. Find the number

62
26
39
None

Solution:

(2)

[1] 62 → 2 3 x 6 [False]

And 62 + 36 = 98* 26 (false]

(2) 26 clearly 6 = 3×2 (True]

And 26 + 36 = 62 (Orders of digits reversed)

So; (2) is correct.

Exercise – 2

Simultaneous Or Linear Equation In Two Or Three Variable

Question 1. The solution of the set of equations 3x + 4y = 7, 4x-y = 3 is

(1.-1)
(1. 1)
(2.1)
(1.-2)

Solution:

(2)

ax + by = c

px + qy = r

(aq – bp) x = ac

→(-3 -16) x = -7-12

-19 x = – 19

X = 1

3 + 4y = 7 → y = 1

Question 2. The values of X and y satisfying the equations $\frac{x}{2}+\frac{y}{3}=$= 2,x + 2y = 8 are given by the pair.

(3.2)
(-2, -3)
(2.3)
None of these

Solution:

(3)

3x + 2y = 12

x + 2y = 8

2x = 4 ⇒ x = 2

⇒ $y=\frac{8-x}{2} \quad \Rightarrow \frac{8-2}{2}=3$

Question 3.$\frac{x}{p}+\frac{y}{q}=2,$x + y = p + q are satisfied by the values given by the pair.

(x=p,y=q)
(x=q,y=p)
(x=l, y=l)
None of these

Solution:

(1)

GBC

x = p, y = q

Question 4. The solution for the pair of equations$\frac{1}{16 x}+\frac{1}{15 y}=\frac{9}{20}, \frac{1}{20 x}-\frac{1}{27 y}=\frac{4}{45}$ is given by

$\left(\frac{1}{4}, \frac{1}{3}\right)$
$\left(\frac{1}{3}, \frac{1}{4}\right)$
(3,4)
(4,3)

Solution:

(1)

Let 1/x = a, 1/y=b

⇒ $\text { 1) }\frac{a}{16}+\frac{b}{15}=\frac{9}{20}=15 a+16 b$

⇒ $\frac{9}{20} \times 16 \times 15 \Rightarrow 108$

2) $\frac{a}{20}-\frac{b}{27}=\frac{4}{45}$

⇒ $27 a-20 b=\frac{4 \times 20 \times 27}{45}=48$

15a + 16b = 108

27a -20b = 48

By trick refer to question 1

(-300 -432) a = (-2160-768)

a = 4

15 (4) + 16b = 108

16 b = 48

B = 3

⇒ $a=4, b=3 \Rightarrow x=\frac{1}{a}, y=\frac{1}{b} \Rightarrow x=\frac{1}{4} ,y=\frac{1}{3}$

Question 5. Solve for x and y: $\frac{4}{x}-\frac{5}{y}=\frac{x+y}{x y}+\frac{3}{10} \text { and } 3 x y=10(y-x) \text {. }$

(5,2)
(-2,-5)
(2,-5)
(2,5)

Solution:

⇒ $\frac{4}{x}-\frac{5}{y}=\frac{x+y}{x y}+\frac{3}{10}$

⇒ $3 \mathrm{xy}=10(\mathrm{y}-\mathrm{x})$

⇒ $\frac{4}{x}-\frac{5}{y}=\frac{1}{y}+\frac{1}{x}+\frac{3}{10}$

⇒ $3 \mathrm{xy}=10 \mathrm{y}-10 \mathrm{x} \rightarrow 3=\frac{10}{x}-\frac{10}{y}$

⇒ $\frac{3}{x}-\frac{6}{y}=\frac{3}{10} \quad \frac{10}{x}-\frac{10}{y}=3$

⇒ $\text { Let } \frac{1}{y}=a \frac{1}{y}=b$

3a -6b = 3 ….(1)

10a – 10b = 3 → 30a – 30b – 9….. (2)

30a- 60b = 3

⇒ $\frac{30 a-30 b=9}{30 b=6}=\mathrm{b}=\frac{6}{30}$

30a -60b = 3

⇒ $ 30 a-60 \frac{(6)}{30}=3$

⇒ $\rightarrow 30 a-12=3$

⇒ $a=\frac{15}{30}$

⇒ $x=\frac{1}{a}=\frac{30}{15}=2 .$

⇒ $ y=\frac{1}{b}=\frac{30}{6}=5 .$

(4) (2,5)

Question 6. The pair satisfying the equations x + 5y = 36,ÿ = is given by Qpfrbft 1 ” Y>

(l6,4)
(4, 16)
(4, 8)
None of these

Solution:

x + 5y = 36 …….(1)

⇒ $\frac{x+y}{x-y}=\frac{5}{3} \ldots \ldots \ldots \ldots(2)$

3x + 3y = 5x- 5y

2x + lOy = 72….(1)

2x-8y = 0 (2)

2x+10y

Subtract $\frac{2 x-8 y}{18 y}$= 72 —> y = 4.

x + 20 = 36 → x — 16.

(a) (16,4)

Question 7. Solve for x and y : x-3y = 0, x+2y = 20CM 3 Opt >

x = 4, y = 12
x= 12, y = 4
x = 5, y = 4
None of these.

Solution:

x – 3y = 0 x + 2y = 20

x = 3y = 0

x + 2y = 20

by trick (Ql)

(2 + 3)x= (0-(-60))

x = +12

x- 3y = 0 —> 12 -3y = 0 —> 12 = 3y

y = 4

x = 12, y = 4

(2) is correct

Question 8. The simultaneous equations 7x-3y = 31, 9x-5y = 41 have solutions given by

(-4,-1)
(-1.4)
(4,-1)
(3,7)

Solution:

7x-3y = 31, 9x – 5y = 41

By tricks

7x- 3y = 31

9x-5y = 41

(-35 + 27)x = (-155 + 123)

-8x = -32

x = 4

7(4) – 3y = 31

28 – 3y = 31

-3y = 3→ y = -l

(3) (4 .-1)

Question 9. 1.5x + 2.4 y = 1.8, 2.5(x+l) = 7y have solutions as

(0.5, 0.4)
(0.4, 0.5)
$\left(\frac{1}{2}, \frac{2}{5}\right)$
(2,5)

Solution:

1.5x + 2.4y= 1.8 …(1)

2.5x + 2.5 = 7y

15x+ 24y = 18

25x- 70y = -25

15x+ 24y = 18

25x+ 70 = -25

By trick

(-1050 -600)x = (-1260 + 600) – 1650x = -660

x = 0.4

6 + 24y = 18

⇒ $ y=\frac{12}{24}=\frac{1}{2}=0.5$

⇒ $x=0.4, y=0.5$$

Question 10. The values of x and y satisfying the equations$\frac{3}{x+y}+\frac{2}{x-y}=3, \frac{2}{x+y}+\frac{3}{x-y}=3 \frac{2}{3}$ are given by

(1,2)
(-l.-2)
$\left(1, \frac{1}{2}\right)$
(2, 1)

Solution:

(1)

GBC

Question 11. 1.5x + 3.6y = 2.1, 2.5 (x+1) = 6y ‘/’JfY,1

(0-2, 0.5)
(0.5, 0.2)
(2,5)
(-2,-5)

Solution:

By trick

[(-900) -900]x= (-1260 + 900) – 1800x = -360

$x=\frac{360}{1800}=0.2$

3 + 36y = 21

36y= 18 → y = 0.5

Question 12. $\frac{x}{5}+\frac{y}{6}+1=\frac{x}{6}+\frac{y}{5}=28$

(6,9)
(9, 6)
(60,90)
(90, 60)

Solution:

(3)

6x + 5y = 810

5x + 6y = 840

By tricks

6x + 5y = 810

5x + 6y = 840

(36 -25)x = 4860 -4200

llx= 660 x = 60, y = 90.

Question 13. $\frac{x}{4}=\frac{y}{3}=\frac{z}{2} ; 7 x+8 y+5 z=62$

(4,3,2)
(2,3,4)
(3,4,2)
(4,2,3)

Solution:

(1)

7x + 8y + 5z = 62

7(4) + 8(3) + 5(2) = 62

28 + 24 + 10 = 62

⇒ $=\frac{x}{4}=\frac{y}{3}=\frac{z}{2} \rightarrow(4,3,2)$

Question 14. $\frac{x y}{x+y}=20, \frac{y z}{y+z}=40, \frac{z x}{z+x}=24$

(120,60,30)
(60,30, 120)
(30,120,60)
(30,60,120)

Solution:

(4)

⇒ $\frac{x+y}{x y}=\frac{1}{20} \frac{y+z}{y z}=\frac{1}{40} \frac{z+x}{2 x}=\frac{1}{24}$

⇒ $\frac{1}{y}+\frac{1}{x}=\frac{1}{20} \ldots(1) \frac{1}{z}+\frac{1}{y}=\frac{1}{40} \ldots(2) \frac{1}{x}+\frac{1}{z}=\frac{1}{24} \ldots .(3)$

⇒ $\frac{1}{y}=\frac{1}{20}-\frac{1}{x} \frac{1}{24}-\frac{1}{x}+\frac{1}{y}=\frac{1}{40} \frac{1}{z}=\frac{1}{24}-\frac{1}{x}$

⇒ $\frac{1}{24}-\frac{1}{x}+\frac{1}{20}-\frac{1}{x}=\frac{1}{40} $

⇒ $\frac{1}{x}+\frac{1}{x}=\frac{1}{24}+\frac{1}{20}-\frac{1}{40} =\frac{2}{x}=\frac{8}{120} \quad=\frac{x}{2}=\frac{120}{8} \rightarrow x=30 \quad 1 / y=\frac{1}{20}=\frac{1}{3}=10 / 600$

⇒ $=\frac{2}{x}=\frac{5+6-3}{120} x=\frac{1}{z}=\frac{1}{24}-\frac{1}{30}=\frac{6}{720} z=120$

(d) (30, 60, 120)

Question 15. 2x + 3y + 4z = 0,x + 2y – 5z = 0, lOx + 16y- 6z = 0

(0,0,0)
(1,-1, 1)
(3, 2,-1)
(1,0,2)

Solution:

(1)

GBC:(0,0,0)

Question 16. $\frac{1}{3}(x+y)+2z=21,3 x-\frac{1}{2}(y+z)=65, x+\frac{1}{2}(x+y-z)=38$

(4, 9. 5)
(2,9,5)
(24, 9, 5)
(5, 24,9)

Solution:

(3)

⇒ $\frac{1}{3}(24+9)+2(5)=213(24) \cdot \frac{1}{2}(9+5)=65$

⇒ $24+\frac{1}{2}(24+9-5)=38$

Question 17. $\frac{x}{0.01}+\frac{y+0.03}{0.05}=\frac{y}{0.02}+\frac{x+0.03}{0.04}=2$

(1,2)
(0.1, 0.2)
(0.01, 0.02)
(0.02,0.01)

Solution:

⇒ $\frac{x}{0.01}+\frac{y+0.03}{0.05}=\frac{y}{0.02}+\frac{x+0.03}{0.04}=2$

⇒ $100 x+\frac{100 y+3}{5}=2 ; \quad 50 y+\frac{100 x+3}{4}=2$

500x + 100y + 3 = 10; 200y +100x + 3 = 8

500x + lOOy = 7 100x + 200y = 5

Multiply by 2

Subtract

$\begin{gathered}
1000 x+200 y=14 \\
100 x+200 y=5 \\
\hline 900 x=9
\end{gathered}$

x = 0.01 y = 0.02

Question 18. $\frac{x y}{y-x}=110, \frac{y z}{z-y}=132, \frac{z x}{z+x}=\frac{60}{11}$

(12, 11, 10)
(10, 11, 12)
(11,10,12)
(12, 10, 11)

Solution:

(2)

⇒ $ \frac{y-x}{x y}=1 / 110 \frac{z-y}{y z}=1 / 132 \frac{z+x}{z x}=\frac{11}{60}$

⇒ $\frac{1}{x}-\frac{1}{y}=\frac{1}{110} \frac{1}{y}-\frac{1}{z}=\frac{1}{132} \frac{1}{x}+\frac{1}{z}=\frac{11}{60}$

GBC = 10,11,12

(2) is correct

Question 19. 3x-4y+70z = 0, 2x+3y-10z = 0, x+2y+3z = 13

(1,3, 7)
(1,7, 3)
(2, 4, 3)
(-10, 10, 1)

Solution:

(4)

3x-4y+70z = 0, 2x+3y-10z = 0, x+2y+3z = 1 3

For three variables GBC x = -10,y = 10, z=l

Question 20. The monthly incomes of two persons are in the ratio 4: 5 and their monthly expenses are in the ratio of 7: 9. If each saves 350 per month find their monthly incomes.

(500, 400)
(400, 500)
(300, 600)
(350, 550)

Solution:

(2)

Monthly Income = 4 : 5 i.e., 4x, 5x

Monthly expenses = 7 : 9 i.e. 7y,9y

Saving = Income – Expenses = 50

4x-7y = 50

5x-9y = 50

By trick (-36 + 35)x-450 + 350

-x = -100 = x = 100

Monthly Income = 4x, 5x = (400, 500)

Question 21. Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are increased by 12.

Solution:

(1)

Let Fraction be$\frac{x}{y}$

⇒ $ \frac{x+2}{y+2}=\frac{1}{2} \quad, \frac{x+12}{y+12}=\frac{3}{4}$

GBC

⇒ $\frac{x}{y}=\frac{3}{8}$

Question 22. The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age.

60 years
52 years
51 years
50 years

Solution:

(4)

Let the sum of ages be x, Man’s age = x

Acc. To quest, x = 2(y)…….(1)

x – 5 = 3(y- 10)

x – 5 = 3y- 30

Put (1) in (2) 2y-3y = -25,

y=25

x = 50

Question 23. A number between 10 and 100 is five times the sum of its digits. If 9 is added to it the digits are reversed find the number.

Solution:

(3)

Let the number be (x is one digit, y another)

Number = lOx + y

= 10x + y + 9 = lOy + x = 9x-9y = -9

Also

10x + y = 5(x + y)

10x + y = 5x + 5y

5x = 4y

x = $\frac{4}{5} y \quad=9\left(\frac{4}{5}\right) y-9 y=-9$

= 36y-45y = -45

-9=-45

y=5, x =$\frac{4}{5} \cdot 5$ = 4

Numbers = lOx + y = 45

Question 24. The wages of 8 men and 6 boys amount to 333. If 4 men earn 34.50 more than 5 boys determine the wages of each man and boy.

(31.50,33)
(33,31.50)
(32.50, 32)
(32,32.50)

Solution:

(2)

Let each man’s wage be x & boy’s be y

8x + 6y = 33 …….. (1)

4x- 4.50 = 5y …….(2)

4x = 5y + 4.50

8x= 10y+ 9 ……. (3) (3) Put (3) in (1)

lOy + 9 + 6y = 33

16y = 24 → y = 1.5.v = 3

Question 25. A number consisting of two digits is four times the sum of its digits and if 27 is added to it the digits are reversed. The number is:

Solution:

(3)

Let number be lOx + y

Acc to quest.

lOx + y = 4(x+y)

1 Ox – 4x = 4y- y

6x = 3y

2x = y

Also

10x + y + 27= lOy + x 5y + y + 24= 10y +$\frac{y}{2}$

10y+ 2y + 54 = 20y + y

12y + 54 = 21y

9y = 64-4y = 6 x = 3

Number =36

Question 26. Of two numbers, l/5lh of the greater is equal to 1 /3rd of the smaller, and their sum is 16. The numbers are:

(6,10)
(9,7)
(12,4)
(11,5)

Solution:

(1)

Let two numbers be x,y such that x > y

Acc to question

$ \frac{x}{5}=\frac{y}{3}, \quad x+y=16$ $3 x=5 y \rightarrow x=\frac{5}{3} y$ $\frac{5}{3} y+y=16=5 y+3 y=48 \rightarrow y=6, x=10$

Two numbers are 6, 10

Question 27. y is older than A- by 7 years 15 years back x’s age was 3/4 of/s age. Their present ages are

(x-36,y=43)
(x=50,y=43)
(x=43,y=50)
(x=40,y=47)

Solution:

(1)

$Y=x+7 \text { also, }(y-15) \cdot \frac{3}{4}=x=15$ $y-7=x \quad \frac{3}{4} y-\frac{45}{4}=x-15$

$\frac{3}{4} y-\frac{45}{4}=y-7-15=3 y-45=$4y- 88

= y = 43 y = 43 x = 36

Present ages are 36, 43

Question 28. The sum of the digits in a three-digit number is 12. If the digits are reversed the number is increased by 495 but reversing only of the ten’s and unit digits increases the number by 36. The number is

Solution:

(c)

Three digit no. be lOOx + lOy + z

Reversing all digits

x + y + z = 12,

100z+ lOy + x = lOOx + lOy + z + 495

100z + x = lOOx + z + 495

99z-99x = 495

z-x = 5…(1) z = 5 + x

Reversing ten’s T unit digit no. increases by 36

100x + lOz + y = 100x + lOy + z + 36

9z-9y = 36

z-y = 4

z = 4 + y…..[2]

x + y + z = 12

from (1) and [2]

3z-9 = 12

3z = 21

z-5+z-4+z=12

z = 7

x = 2, y = 3, z = 7

Numbers = 237

Question 29. Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and l/3rd of the smaller and 1/5th of the greater number are together 21. The numbers are:

(36, 45)
(45, 36)
(50, 41)
(55, 46)

Solution:

(2)

Let two numbers be x and y. such that x>y

2x = 2y + 18 And $\frac{1}{3} y+\frac{1}{5} x=21$

2x-2y=18

x-y = 8 5y+3x = 315

x = 9 + y 5y + 27 + 3y = 315

8y = 288 y = 36

x = 45

Question 30. The demand and supply equations for a certain commodity are 4q + 7p = 17 and p = $\frac{q}{3}+\frac{7}{4}$ respectively where p is the market price and q is the quantity than the equilibrium price and quantity are:

$2, \frac{3}{4}$
$3, \frac{1}{2}$
$5, \frac{3}{5}$
None of these

Solution:

⇒ $4 q+7 p=17 \rightarrow p=\frac{17-4 q}{7}$

⇒ $P=\frac{q}{3}+\frac{7}{4}$

For equilibrium

⇒ $\frac{q}{3}+\frac{7}{4}=\frac{17}{7}-\frac{4}{7} q$

⇒ $ \frac{7 q+12 q}{21}=\frac{68-49}{28}=\frac{19 q}{21}=\frac{19}{28}=q=\frac{3}{4}$

⇒ $P=\frac{17-4 q}{7}=\frac{17-3}{7}=2 .$

Question 31. Solving 6x+5y-16=0 and 3x-y-l=0 we get values of andy as

1,1
1,2
-1,2
0,2

Solution:

(2)

6x + 5y= 16

3x-y= 1

By trick

(-6-15)x = -16 -5

-21x = -21 = x = 1

6 + 5y= 16 → y = 2

Question 32. Solving 9x+3y-4z=3 ,x+y-z=0 and 2x-5y-4z=-20 following roots are obtained

2, 3, 4
1,3,4
l, 2,3
None

Solution:

(3)

9x + 3y- 4x = 3, x+y-z=0

2x- 5y- 4x = -2

For three variables

Question 33. A man went to the Reserve Bank of India with 1,000. He asked the cashier to give him 5 and % 10 notes only in return. The man got 175 notes in all. Find how many notes of K 5 and 10 he receives.

(25,150)
(40,110)
(150,25)
None

Solution:

For (1) 25 x 5 + 150 x 10$\neq$ 1000

(2) 40 x 5 + no x 10$\neq$ RS. 1000

(3)150 x5 + 25 x 10 = Rs. 1000

Question 34. The Point of intersection of the lines 2x- 5y = 6 and x+y=3 is

(0,3)
(3,0)
(3,3)
(0,0)

Solution:

(2)

Intersecting Point lies on both straight lines. It will satisfy both ends.

For 00 (0,3) Point 2×0 – 5×3

it is incorrect 6

0Ption (2) (3; 0) satisfies both eqns.

(2) is correct

Question 35. If the equations kx + 2y = 5, 3x + y = 1 has no solution then the value of k is

Solution:

Kx + 2y = 5

3x + y=l

They have no soln. (given)$\frac{k}{3}=\frac{2}{1} \neq \frac{5}{1} ; \Rightarrow$-k = 6

Question 36. The equation x + 5y = 33;$\frac{x+y}{x-y}=\frac{13}{3}$ Has the solution (x,y) as:

(4,8)
(8,5)
(4,16)
(16,4)

Solution:

(2)

For LHS = x + 5y = 8 + 5×5 = 33

⇒ $\frac{x+y}{x-y}+\frac{8+5}{8-5}=\frac{13}{5}$

Clearly (b) satisfies both equations.

Question 37. If 2 x+y = 2 2x-ÿ = v8 then the respective value of and y are_

$1, \frac{1}{2}$
$\frac{1}{2}, 1$
$\frac{1}{2}, \frac{1}{2}$
None of these

Solution:

(a) is correct

+y _ 22x-y = V8 = V21 = 23/2

x+y =$\frac{3}{2}$ (1)

2x-y=$\frac{3}{2}$ (2)

(a)satisfies (1] and (2) both

Question 38. Let Ei,thenE2 are E1, and E2 two E2 are(2,-1)linear_.isequationsa solution in software equation variables E1 only d and y,(-2,-1)(0,1) is a Solution of for equation both the E2.only,equations

x= 0,y = 1;

2x-y = -l, 4x + y = 1

x + y = l,x-y = -1

x + 2y = 2, x + y =1

Solution:

(0; 1) satisfies E1 and E2 both

(2, -l) satisfies 1 Eqn. 2-1 = 1 (true)

But -2; -13 also satisfies E2

i.e -2 –13 = -1 (true)

Question 39. $\text { If } \frac{3}{x+y}+\frac{2}{x-y}=-1 \text { and } \frac{1}{x+y}-\frac{1}{x-y}=\frac{4}{3} \text { then }(x, y) \text { is }$

(2.1)
(1.2)
(-1,2)
(-24)

Solution:

⇒ $\frac{3}{1+2}+\frac{2}{1-2}=1-2=-1 \text { (True) }$

⇒ $\frac{1}{1+2}-\frac{1}{1-2}=\frac{1}{3}+1=\frac{4}{3} \text { (True) }$

So;(2) is correct.

Question 40. The line 3x + 2y = 6 intersects the line 3x – y = 12 in_ quadrant:

Solution:

(4), Eqn. (1) – Eqn. (2); we get

3x + 2y = 6

3x- y = 12

– + –

3y = -6 ⇒ y = -2

From (1); 3x = 6-2y=6-2 (-2) = 10

$ x=\frac{10}{3}$

Coordinate the point of intersection

= (x;y) = $\left(\frac{10}{3} ;-2\right)$

It is in the 4th Quadrant.

Exercise – 3

Quadratic Equations

Question 1. If the roots of the equation 2×2 + 8x – m’ = 0 are equal then value of m is

-3
– 1
l
-2

Solution:

(4) 2x² 8x- m3 = 0 for equal b2- 4ac = 0

→ 64 + 4.2. m³ = 0

8m³ = – 64

m3 = – 8

m= – 2

Question 2. If 2^x+3 – 32. 2x + 1 = 0 then values of x are

0,1
1,2
0,3
0,-3

Solution:

(4) 2^x+3– 32. 2x + 1 = 0

2^x+3– 9.2x +1 = 0

Question 3. The values of 4+_____1_____

4+_____1_____
4+_____1_____
4+…….CO
1. 1± √2
2. 2+ √5
3. 2± √5
4. None of these

Solution:

4x+1=x²

x² — 4x —1 = 0

x²– 4x- 1

⇒ $X=\frac{4 \pm \sqrt{16+4}}{2}$

⇒ $=2 \pm \sqrt{5}$

Question 4. If be the roots of the equation 2×2 – 4,v -3 = 0 the value of a2 + 32 is

5
7
3
-4

Solution:

(2) 2x²– 4x- 3 = 0

ap is root

→ x² – 2x- $\frac{3}{2}$

α²+ β² = (α + β)² – 2 α β

⇒ $(2)^2+2 \cdot \frac{3}{2}$=7

Question 5. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals then$\frac{a^2}{a c}+\frac{b c}{a^2}$ is equal to

2
-2
1
-1

Solution: (1) 2

Question 6. The equation x² -(p+4)x + 2p + 5 = 0 has equal roots the values of p will be.

± 1
2
± 2
-2

Solution:

(3) x² – (p + 4) x + 2p + 5 = 0

For equal roots

b² – 4ac = 0

(p + 4)² – 4 (2p + 5) = 0

p² – 4 = 0

p² = 4

P = ± 2

Question 7. The roots of the equation x2 + (2p-l)x + p2 = 0 are real if.

p>l
p<4
p>1/4
p<1/4

Solution:

(4) x² + (2p-1) x + pz = 0

For real roots

b² – 4ac > 0

(2p-1)²-4(p²)>0

4p² + 1 – 4p – 4p² > 0

1 – 4p > 0

1 > 4p

⇒ $\frac{1}{4} \geq \mathrm{p}$

Question 8. If p and q are the roots of x2 + 2x + 1 = 0 then the values of* + q:i becomes

2
-2
4
-4

Solution:

(2) p and q roots of x² +2x + 1 = 0

p + q = -2

pq = 1

p³+ q³=(p + q) (p²-pq + q²)

= -2 (p² + q²– 1)

= -2 (p+q)²– 2pq – 1)

= -2 (4 – 2 – 1) = -2

Question 9. If L + M + N = 0 and L, M, N are rationales the roots of the equation
(M+N-L)x2+(N+L-M)x+(L+M-N) = 0 are

Real and irrational
Real and rational
Imaginary and equal
Real and equal

Solution:

(2) Real and Rational

Question 10. If one root of 5x² + 13x + p = 0 is reciprocal of the other then the value is

-5
5
1/5
-1/5

Solution:

(2) 5×2 + 13x + p = 0

⇒ $\text { roots } \rightarrow \alpha \cdot \frac{1}{\alpha}$

⇒ ${sum}=\alpha+\frac{1}{\alpha} \quad \text { product }=\frac{p}{5}$

⇒ $\frac{p}{5}=1 \quad \text { p }=5$

Question 11. A solution of the quadratic equation (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0 is

x = 1
x = -1
x= 2
x = – 2

Solution:

(2) (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0

For x = 1

a + b-2c + 2a-b-c + c + a-2b≠0

for x = -1

a + b -2c -2a + b + c + c + a -2b = 0

∴ x = – 1

Question 12. If the root of the equation x2-8x+m = 0 exceeds the other by 4 then the value of m is

m = 10
m= 11
m = 9
m = 12

Solution:

(4) x² – 8x + m = 0

roots α, α + 4

sum = 8

α+α+4=82 α=4

Product = m

α (α + 4) = m

α² + 4α = m

4 + 8 = m m = 12

Question 13. The values of x in the equation 7(x+2p)² + 5p² = 35xp + 1 17p² are

(4p,-3p)
(4p,3p)
(-4p, 3p)
(-4p,-3p)

Solution:

(1) 7(x + 2p)² + 5p² = 35 xp + 117p²

(4p, -3p)

Question 14. H. The solutions of the equation $\frac{6 x}{x+1}+\frac{6(x+1)}{x}=13 \text { are }$

(2, 3)
(3,-2)
(-2,-3)
(2,-3)

Solution:

⇒ $\frac{6 x}{x+1}+\frac{6(x+1)}{x}=13$

(2,-3)

Question 15. The satisfying values of.v for the equation$\frac{1}{x+p+q}=\frac{1}{x}+\frac{1}{p}+\frac{1}{q} \text { are }$

(p. q)
(-p. -q)
(p, -p)
(-P, q)

Solution:

⇒ $\text { (2) } \frac{1}{x+p+q}=\frac{1}{x}+\frac{1}{p}+\frac{1}{q}$

(-p. -q)

Question 16. The values of for the equation x² + 9x + 18 = 6 – 4x are

(1,12)
(-1,-12)
(1,-12)
(-1,12)

Solution:

(2) x² + 9x + 18 = 6 – 4x

⇒ x² + 13x + 12 = 0

⇒ x² + 12x + x + 12 = 0

⇒ x(x+12) + 1 (x+ 12) = 0

x = -1,-12

Question 17. The values of x satisfying the equation $\sqrt{\left(2 x^2+5 x-2\right)}-\sqrt{\left(2 x^2+5 x-9\right)}=1 \text { are }$

(2, -9/2)
(4,-9)
(2, 9/2)
(-2, 9/2)

Solution:

⇒ $\text { (1) } \sqrt{2 x^2+5 x-2}-\sqrt{2 x^2+5 x-9}=1$

(2. -9/2)

Question 18. The solution of the equation 3x²-17x + 24 = 0 are

(2,3)
$(b) \left(2,3 \frac{2}{3}\right)$
$\left(3,2 \frac{2}{3}\right)$
$\left(3, \frac{2}{3}\right)$

Solution:

(3) 3x² – 17x + 24 = 0

3x² – 8x -9x + 24 = 0

3x (x- 3) – 8 (x- 3) = 0

(3x – 8) (x- 3) = 0

⇒ $x=\frac{8}{3}, x=3$

⇒ $x=3,2 \frac{2}{3}$

Question 19.The equation $\frac{3\left(3 x^2+15\right)}{6}+2 x^2+9=\frac{2 x^2+96}{7}+6$ has got the solution as

(1, 1)
(1/2, -1)
(1. -1)
(2,-1)

Solution:

⇒ $\text { (c) } \frac{3\left(3 x^2+15\right)}{6}+2 x^2+9=\frac{2 x^2+96}{7}+6$

⇒ $\frac{9 x^2+45+12 x^2+54}{6}=\frac{2 x^2+96+42}{7}$

⇒ $\frac{21 x^2+99}{6}=\frac{2 x^2+138}{7}$

Multiply 42

⇒ 147x² + 693 = 12x² + 828

⇒ 135x² = 135

x² =1 → x ±1

Question 20. The sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are

(7, 10)
(4,4)
(3, 5)
(2,6)

Solution:

3 + 5 = 8 ⇒ 3² + 5² = 34

(3,5)

Question 21. The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are

(7,4)
(5, 8)
(3, 6)
(2,5)

Solution:

(2) let two positive integers be a, 3

⇒ α-β = 3

⇒ α² +β² = 89

(5,8)

Question 22. Five times of a positive whole number is 3 less than twice the square of the number. The number is

3
4
-3
2

Solution:

(1) let the number be x

⇒ 5x = 2x² -3

⇒ 2x² – 5x – 3 = 0

⇒ 2x² – 6x + lx – 3 = 0

⇒ 2x(x-3) + 1 (x-3) = 0

⇒ $x=\frac{-1}{2}, x=3$

x is positive

x = 3

Question 23. The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solving it to find the length and breadth of the field. The length and breadth are

(205m, 80m)
(50m, 40m)
(60m, 50m)
None

Solution:

(2) Perimeter = 180m

2(1 + b) = 180

L + b = 90

We know I = x

b = 90 – x

Area = 2000 m

∴ x (90 – x) = 2000

⇒ 90x- X² = 2000

⇒ x²-90x + 2000 = 0

⇒ x² – 5Ox – 40x + 2000 = 0

⇒ x (x-50) – 40 (x- 50) = 0

x = 40, 50

Question 24. Two squares have sides p cm and (p + 5) cms. The sum of their squares is 625 sq. cm. The sides of the squares are

(10 cm, 30 cm)
(12 cm, 25 cm)
(15 cm, 20 cm)
None of these

Solution:

(3) Two squares with sides p, p + s cm

⇒ p² + (p+5)² = 625

⇒ p² + p² + 25 + lOp = 625

⇒ 2p²+10p- 600 = 0

⇒ p² + 5p – 300 = 0

⇒ p2 + 20p-15p- 300 = 0

⇒ p (p + 20) – 15 (p+20) = 0

⇒ p = 15,p = -20

⇒ p≠ -20

⇒ p = 15

⇒ p + 5 = 20

Question 25. Divide 50 into two parts such that the sum of their reciprocals is 1/12. The numbers are

(24, 26)
(28, 22)
(27,23)
(20,30)

Solution:

(4) let the first part be x,

second = 50 – x

⇒ $\frac{1}{x}+\frac{1}{50-x}=\frac{1}{2}$

⇒ $\frac{50-x+x}{x(50-x)}=\frac{1}{12}$

⇒ 600 = x (50 – x)

⇒ 600 = 50x – x²

⇒ x²-50x + 600 = 0

⇒ x²– 30x- 20x + 600 = 0

⇒ x (x – 30) – 20 (x- 30) = 0

∴ x = 20, 30

Question 26. There are two consecutive numbers such that the difference of their reciprocals is 1/240. The numbers are

(15, 16)
(17, 18)
(13, 14)
(12, 13)

Solution:

(1) Two numbers be x, x + 1

⇒ $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{240}$

⇒ $\frac{x-x+1}{x(x+1)}=\frac{1}{240} \rightarrow+240=x(x+1)$

⇒ x² + x – 240 = 0

⇒ x² + x- 240 = 0

⇒ x²+ 16x- 15x- 240 = 0

⇒ x (x + 16) – 15 (x +16) = 0

⇒ x = 15, x = -16

∴ x= 15

(15,16)

Question 27. The hypotenuse of a right-angled triangle is 20cm. The difference between its other two sides is 4cm. The sides are

(11cm, 15cm)
(12cm, 16cm)
(20cm, 24cm)
None of these

Solution:

(2) hypotenuse = 20 cm

The difference between the other two sides a and, b is 4

⇒ a – b = 4

⇒ a = 4 + b

⇒ a² + b² = c²

⇒ a² + (a-4)²= c²

⇒ a² + a²+ 16 – 8a = 20²

⇒ 2a² + 8a = 384

⇒ 2a²+ 8a – 384 = 0

⇒ a² – 4a – 192 – 0

⇒ a²– 16a + 12a – 192 = 0

⇒ a (a- 16) + 12 (a—16) = 0

⇒ a = 16

∴ b = a-4 = 12

Question 28. The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are

(15,30)
(32, 13)
(36,9)
(25,20)

Solution:

(3) → 4 (1 = 45)

⇒ $\frac{a}{x}=\frac{x}{\beta} \rightarrow \alpha \beta=x^2$

αβ= 18²

(36.9)

Question 29. The sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right-angle triangle is formed. The side of the equilateral triangle is

17 units
16 units
15 units
18 units

Solution:

(1) let the side be x

New sides

x- 12, x- 13, x- 14

new A is night-angled

(x – 14)²+ (x- 13)² = (x – 12)²

Question 30. A distributor of Apple Juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D (in number of bottles) is given by D = -2000p2 + 2000p + 17000. The price per bottle that will result in zero inventory is

3
5
2
none of these.

Solution:

(1) D = -2000p²+ 2000p + 17000

For zero inventory D = supply

5000 = – 2000p² + 2000p + 17000

2000p² = -2000p – 12000 = 0

p² – p – 6 = 0

p²– 3p + 2p- 6 = 0

P (P – 3) + 2 (p- 3) = 0

P = 3

Question 31. The sum of two irrational numbers multiplied by the larger one is 70 and their difference multiplied by the smaller one is 12; the two numbers are

3√2, 2√3
5√2, 3√5
2√2, 5√2
none of these.

Solution:

⇒ $\alpha(\alpha+\beta)=70$

⇒ $\alpha^2+\alpha \beta=70$

⇒ $\alpha \beta=70-\alpha^2$

⇒ $\beta(\alpha-\beta)=12$

⇒ $\alpha \beta-\beta^2=12$

⇒ $\alpha \beta=12+\beta^2$

⇒ $70-\alpha^2=12+\beta^2$

⇒ $ \alpha^2+\beta^2=58$

⇒ $2 \sqrt{2}, 5 \sqrt{2}$

Question 32. Solving equation x² – (a+b) x + ab = 0 are, value(s) of

a, b
a
b
None

Solution:

(1)

x² – (a + b) x + ab = 0

a, b are values of x

a, b

Question 33. Solving equation x² – 24x + 135 = 0 are, value(s) of x

9, 6
9,15
15,6
None

Solution:

(2) 135-0

x² – 15x -9×4 135 = 0

x- 15,9

Question 34. If $\frac{x}{b}+\frac{b}{x}=\frac{a}{b}+\frac{b}{a}$ the roots of the equation are

a, b² / a
a² , b/a²
a² , b² /a
a, b²

Solution:

⇒ $\text { (a) } \frac{x}{b}+\frac{b}{x}=\frac{a}{b}+\frac{b}{a}$

⇒ $\frac{x^2+b^2}{x b}=\frac{a^2+b^2}{a b}=a x^2+a b^2=a^2 x+b^2 x$

⇒ $a x^2-a^2 x-b^2 x+a b^2=0$

⇒ $a x^2-\left(a^2+b^2\right) x+a b^2=0$

⇒ $x^2-\left(\frac{a^2+b^2}{a}\right) x+b^2=0 $

⇒ $ x^2-\frac{\left(a+b^2\right)}{a} x+b^2=0$

⇒ $\text { Iwo roots } \rightarrow \mathrm{a}, \frac{b^2}{a}$

Question 35. Solving equation $\frac{6 x+2}{4}+\frac{2 x^2-1}{2 x^2+2}=\frac{10 x-1}{4 x}$We get roots as

±l
+l
-1
0

Solution:

⇒ $\frac{6 x+2}{4}+\frac{2 x^2-1}{2 x^2+2}=\frac{10 x-1}{4 x}$

⇒ $x= \pm 1 \rightarrow 2+\frac{1}{4}=\frac{9}{4}$

⇒ $\text { for } x=-1 \rightarrow-1+\frac{1}{4}=\frac{11}{4}$

⇒ $x=1 \rightarrow 2+\frac{1}{4}=\frac{9}{4}$

Question 36. Solving equation 3x² – 14x +16 = 0 we get roots as

±l
2 and$\frac{8}{3}$
0
None

Solution:

(2) 3x² – 8x – 6x 416 = 0

3x² – 6x – 8 x 416 = 0

3x (x – 2) – 8 (x- 2) = 0

⇒ $x=\frac{8}{3},2$

Question 37. Solving equation 3x² – 14x + 8 = 0 we get roots as

±4
+2
$4,\frac{2}{3}$
None

Solution:

(3) 3x² -14x + 8 = 0

3x² – 12x -2x+9 = 0

3x (x – 4) -2 (x – 4) = 0

⇒ $x=\frac{2}{3}, x=4$

Question 38. Solving equation (b-c) x² + (c-a)x + (a-b) = 0 following roots are obtained

$\frac{a-b}{b-c}, 1$
(a-b) (a-c),1
$\frac{b-c}{a-b}, 1$
None

Solution:

(a) (b- c) x2 4 (c – a) x 4 (a – b) = 0

Sum of roots=$\frac{-(c-a)}{(b-c)}=\frac{a-c}{b-c}$

Product =$\frac{a-b}{b-c}$

⇒ $\frac{a-b}{b-c}, 1$

Question 39. Solving equation 7 $\sqrt{\frac{x}{x-1}}+8 \sqrt{\frac{x-1}{x}}=15$ Following roots are obtained.

$\frac{64}{113}, \frac{1}{2}$
$\frac{1}{50}, \frac{1}{65}$
$\frac{49}{101} \frac{1}{65}$
$\frac{1}{50}, \frac{64}{65}$

Solution:

⇒ $\text { (a) } \sqrt[7]{\frac{x}{x-1}}+\sqrt[8]{\frac{x-1}{x}}=15$

Question 40. Solving equalion 6 $\sqrt{\frac{x}{x-1}}+8 \sqrt{\frac{x-1}{x}}=13$ Following roots are obtained.

$\frac{4}{13}, \frac{9}{13}\frac{-4}{13}, \frac{-4}{19}$
$frac{4}{13}, \frac{5}{13}$
$\frac{6}{13},$
$\frac{7}{13}$

Solution:

⇒ $6\left[\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\right]=13$

⇒ $\frac{4}{13}, \frac{9}{13}$

Question 41. Solving equation z² -6z + 9 = 4$\sqrt{z^2-6 z+6}$ following roots are obtained.

$\text 3+2 \sqrt{3}, 3-2 \sqrt{3}$
5, 1
all the above
None

Solution:

(3 $z^2-6 z+9=4 \sqrt{z^2-6 z+6}$

All of the above

Question 42. Solving equation$\frac{x 1 \sqrt{12 p-x}}{x-\sqrt{12 p-x}}=\frac{\sqrt{p+1}}{\sqrt{p-1}}$ following roots are obtained.

3p
both 3p and – 4p
Only -4p
-3p4p

Solution:

(1) 3p

Question 43. Solving equation $z+\sqrt{z}=\frac{6}{25}$

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{1}{25}$
$\frac{2}{25}$

Solution:

(3) let √z = a

⇒ $ a 2+a=\frac{6}{25}$

⇒ $25 a 2+25 a-6=0$

⇒ $a=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

⇒ $\frac{-25 \pm \sqrt{625-4(25)(-6)}}{50}$

⇒ $\frac{-25 \pm \sqrt{625+600}}{50}$

⇒ $\frac{-25 \pm \sqrt{1225}}{50}$

⇒ $\frac{-25 \pm 35}{50}=\frac{10}{50} \text { or } \frac{-60}{50}=\frac{1}{5} \text { or } \frac{-6}{5}$

⇒ $ a=\sqrt{z} \rightarrow a 2=z $

⇒ $z=\frac{1}{25} \text { or } \frac{36}{25}$

Question 44. Solving equation$z^{10}-33z^5+32=0$the following values of x are obtained

1,2
2,3
2, 1
1,2,3

Solution:

(1) z¹⁰-33z⁵ +32=0

Let z⁵ = a

a² -33a + 32 = 0

a² -32a -1a + 32 = 0

a (a – 32) – 1 (a – 32) = 0

a = 1,32

z⁵= 1,32

= 1,2

Question 45. When $\sqrt{2z+1}+\sqrt{3z+4}=$ 7 the value of z is given by

Solution:

(4) 4

Question 46. Solving equation 6x⁴ + 11x³– 9x²– 11 x + 6 = 0 following roots are obtained

$\frac{1}{2}, -2 \frac{-11 \sqrt{37}}{6}$
$-\frac{1}{2}, 2 \frac{-11 \sqrt{17}}{6}$
$\frac{1}{2}, 2 \frac{5}{6} \frac{-7}{6}$
None

Solution:

(1) 6x⁴ + 11x³– 9x²– llx + 6 = 0

⇒ $\text { For } \frac{1}{2}$

⇒ $6\left(\frac{1}{2}\right)^4+11\left(\frac{1}{2}\right)^3-9\left(\frac{1}{2}\right)^2-11\left(\frac{1}{2}\right)+6$

⇒ $6\left(\frac{1}{16}\right)+11\left(\frac{1}{8}\right)-9\left(\frac{1}{4}\right)-\frac{11}{2}+6$

⇒ $\frac{6}{16}+\frac{22}{16}-\frac{36}{16}-\frac{88}{16}+\frac{96}{16}=0$

For-2

6(—2)⁴+l 1(—2)³ -9 ( —2)2—11 (-2)+6

= 96-88-36 + 22 + 6 = 0

⇒ $\frac{1}{2},-2, \frac{-1 \pm \sqrt{37}}{6}$

Question 47. $\text { If } \frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{x-1}{x+3}-\frac{x+3}{x-3}$ then the values of x are

$0, \pm \sqrt{6}$
$0, \pm \sqrt{3}$
$0, \pm 2 \sqrt{3}$
None of these

Solution:

$\text { For } 0 \frac{2}{-2}+\frac{2}{2}=\frac{-1}{3}+\frac{3}{-3}$ $-1+1 \neq \frac{-1}{3}-1$

0 is not a solution but all options have 0

None of these

Question 48. If $\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}$ then the values of x are

0,(a+b),(a-b)
$0,(a+b), \frac{a^2+b^2}{a+b}$
$0,(a-b) \cdot \frac{a^2+b^2}{a+b}$
$\frac{a^2+b^2}{a+b}$

Solution:

⇒ $0,(a-b) \cdot \frac{a^2+b^2}{a+b}$

Question 49. Solving equation $\left(x-\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+12=0$ we get roots as follows

0
1
-l
None

Solution:

⇒ $\text { (2) }\left(x-\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+12=0$

For 1 (O)2 – 6 (2) + 12 = 0

Question 50. Solving equation $\left(x-\frac{1}{x}\right)^2-10\left(x-\frac{1}{x}\right)+24=0$ we get roots as follows

0
1
-l
$(2 \pm \sqrt{5}),(3 \pm \sqrt{10})$

Solution:

⇒ $\text { (4) }\left(x-\frac{1}{x}\right)^2-10\left(x-\frac{1}{x}\right)+24=0$

(2±√5), (3±√I0)

Question 51. Solving equation $ 2\left(x-\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}+2\right)+18=0$ we get roots as under

0
1
-1
-2± √3

Solution:

⇒ $\text { (2) } 2\left(x-\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}+2\right)+18=0$

a, b, c do not satisfy

-2±√3

Question 52. If αβ the roots of equation x² – 5x + 6=0 and α > β then the equation with roots (α + β)and (α – β) is

x² -6x+5=0
2x² -6x+5=0
2x² -5x+6=0
x² -5x+6=0

Solution:

(1) α+ β = 5 αβ=6

x2- 5x + 6

x2 – 3x- 2x + 6

x (x- 3) – 2 (x – 3)

x = 2 or x = 3

α = 3, β = 2 α = 9 β2 = 4

α – β = 1

new equation→

x² – (6) x + 5 = 0

x² – 6x + 5 = 0

Question 53. If αβ are the roots of equation x²-5x+6=0 α > β then the equation with roots(αβ+γ + β) and (α + β²) is

x²-9x+99=0
x²-18x+90=0
x²-18x+77=0
None

Solution:

x²-18x+77=0

Question 54. If alpha beta are the roots of equation x²-5x+6=0 and γ > β then the equation with roots (γβ +γ+β)and )γβ – γ-β)

x²-12x+11=0
2x² -6x+12=0
x²-12x+12=0
None

Solution:

(1) x² -5x+6=0

x²– 3x- 2x + 6 = 0

x = 3, x = 2

α= 3, β = 2

αβ= 6

αβ + α + β= 6 + 3 + 2 =11

αβ – α – β= 6- 3- 2=1

new equation→

x² – (sum) x + product = 0

x²-12x +ll = 0

Question 55. The condition that one or ax2+bx+c=0 the roots of is twice the other is

b²=4ca
2b²=9(c+a)
2b²=9ca
2b²=9(c-a)

Solution:

(3) let two roots be a, 2a

⇒ $3 \alpha=\frac{-b}{a}, 2 \alpha^2=\frac{c}{a}$

⇒ $\alpha=\frac{-b}{3 a} \quad \alpha^2=\frac{c}{2 a}$

⇒ $a^2=\frac{b^2}{9 a^2}\frac{b^2}{9 a^2}=\frac{c}{2 a}$

⇒ $\rightarrow b^2=\frac{c .9 a^2}{2 a}$

⇒ $\rightarrow 2 b^2=c a 9 \quad \rightarrow 2 b^2=9 c a$

Question 56. The condition that one of ax2+bx+c=0 the roots of is thrice the other is

3b²=16ca
b²=9ca
3b²=-16ca
b²=-9ca

Solution:

(1) ax²+bx+c=0

roots → α, 3α

α + 3α =$\frac{-b}{a}$

⇒ $4 \alpha=\frac{-b}{a} \rightarrow \alpha$

⇒ $\frac{-b}{4 a} \rightarrow \alpha^2$

⇒ $\frac{b^2}{16 a^2}$

⇒ $a(3 a)=3 t^2=\frac{c}{a}$

⇒ $a^2=\frac{c}{3 a}=\frac{b^2}{16 a^2}$

⇒ $b^2=\frac{16 a^2 c}{3 a}$

⇒ $3 b^2=16 \mathrm{ac}$

Question 57. If the roots of ax²+bx+c=0 are in the ratic$\frac{p}{q}$ then the value of$\frac{b^2}{(c a)} \text { is }$

$\frac{(p+q)^2}{(p q)}$
$\frac{(p+q)}{(p q)}$
$\frac{(p-q)^2}{(p q)}$
$\frac{(p-q)}{(p q)}$

Solution:

⇒ $\frac{-b}{a}=p y+q y \frac{c}{a}=p q y^2$

⇒ $-b=\mathrm{a}(\mathrm{py}+\mathrm{qy}) \mathrm{c}=a p q y^2$

⇒ $\mathrm{~b}^2=\mathrm{a}^2(\mathrm{py}+\mathrm{qy})^2$

⇒ $\frac{b^2}{c a}=\frac{a^2(p y+q y)^2}{a p q y^2 a}=\frac{(p+q)^2}{p q}$

Question 58. Solving equation$\sqrt{x^2-9 x+18}+\sqrt{x^2+2 x-15}=\sqrt{x^2-4 x+3}$ following roots are obtained

$3, \frac{2 \pm \sqrt{94}}{3}$
$\frac{2 \pm \sqrt{94}}{3}$
$4,-\frac{8}{3}$
$3,4-\frac{8}{3}$

Solution:

⇒ $\sqrt{x^2-9 x+18}+\sqrt{x^2+2 x-15}=\sqrt{x^2-4 x+3}$

⇒ $\sqrt{x^2-6 x-3 x+18}+\sqrt{x^2+5 x-3 x-15}$

⇒ $\sqrt{(x)(x-6)-3}$

⇒ $\sqrt{(x)(x-6)-3(x-6)}+\sqrt{x(x+5)-3(x+5)}$

⇒$\sqrt{x(x-3)-1(x-3)}$

⇒$\sqrt{(x-3)(x-6)}+\sqrt{(x-3)(x+5)}$

⇒ $\sqrt{(x-1)(x-3)}$

⇒ $3, \frac{2 \pm \sqrt{94}}{3}$

Question 59. Solving equation $\sqrt{y^2+4 y-21}+\sqrt{y^2-y-6}=\sqrt{6 y^2-5 y-39}$ following roots are obtained

2, 3, 5/3
2, 3, -5/3
-2, -3, 5/3
-2, -3, -5/3

Solution:

⇒$\sqrt{y^2+4 y-21}+\sqrt{y^2-y-6}=\sqrt{6 y^2-5 y-39}$

⇒$\sqrt{y^2+7 y-3 y-21}+\sqrt{y^2-3 y+2 y-6}$

⇒$\sqrt{6 y^2 18 y+13 y-39}$

⇒$\sqrt{y(y+7)-3(y+7)}+\sqrt{y(y-3)+2(y-3)}$

⇒$\sqrt{6 y(y-3)+13(y-3)}$

⇒$\sqrt{(y-3)(y+7)}+\sqrt{(y+2)(y-3)}$

⇒$\sqrt{(y-3)(6 y+13)}$

⇒$2,3, \frac{-5}{3}$

Question 60. Find the positive value of k for which the equations: x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots:

Solution:

(2) for real roots

D = b2 – 4ac> 0

For eqn (1]; k² ≥ 4.1.64 ⇒ k > 16……(1)

For eqn (2) (-8)^2≥ 4.1k⇒ k < 16……(2)

∴ (2) is correct.

Question 61. If one root of an equation is 2 + √5. Then the quadric equation is:

x² + 4x -1 = 0
x² – 4x – 1 = 0
x² + 4x +1 = 0
x² – 4x + 1 = 0

Solution:

(2) if one root = 2 + √5 (Irrational conjugate)

Eqn is

X² – (sum of roots) x + product of roots = 0

Or x²– (2 + √5 + 2 -√5)x+ (2 + √5) (2 – √5) = 0

Or x²– 4x + (4- 5) = 0

0r x²-4x-1 = 0

(2) is correct.

Question 62. A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was K 1,500 after 4 years of service and 1,800 after 10 years of service, what was his starting salary and what is the annual increment in rupees?

1,300, 50
1,00, 50
1,500, 30
None

Solution:

For (1) salary after 4 years 1300 + 4 x 50 =1500……..(1)

Salary after 10 years = 1300 + 10 x 50 =1800……….(2)

Detail Method

Let fixed salary = a

Increment per yr = b

Salary after t yrs = s

s = a + bt

a + b x 4 = 1500 = a + 4b = 1500…

a + bx 10= 1800 = a + 10b = 1800.

Eqn. (2) – eqn. (1); we get

a + bx 10= 1800

a + 4b =1500

6b = 300 b =$b=\frac{300}{6} ₹ 50$

From (1) a + 4 x 50 = 1500

Or a = 1500 – 200 = 1300

(1) is correct

Question 63. The sides of an equilateral triangle are shortened by 1 2 units, 13 units, and 14 units respectively, and a right-angled triangle is formed. The side of the equilateral triangle is:

17 units
16 units
15 units
18 units

Solution:

For (a] let each side = 17 units.

∴ sides are 17 – 12; 17 – 13 and 17 – 14

= 5; 4 and 3 units respectively

They make a right-angled triangle because the sum of squares of two sides is equal to
square of the largest side.

i.e., 42 + 32 = 52

(1) is correct.

Detail method

Let the length of each side = x

Side of the right angle triangle are x – 12 ; x – 13 and x – 14

by Pythagoras theorem

(x- 12)² = (x- 13)² + (X- 14)²

Or x² – 24x + 144 = x² – 26x + 167 + x² – 28x + 196

Or 0 = x² – 30x + 221

Or x²-17x-13x + 221 = 0

Orx(x-17)-13(x-17) =0

Or (x-17) (x-13) = 0

x = 17; 13

No side = x-13 = 0

x = 17

(a) is correct.

Question 64.The value of $\sqrt{6+\sqrt{6+\sqrt{6+\cdots \ldots to \infty}}} \text { is: }$

Solution:

⇒ $\text { (3) } \sqrt{6+\sqrt{6+\sqrt{6+\cdots \ldots \text { to } \infty}}}=+3$

Find two factors of 6 such that their difference becomes 1.

6 = 3×2

(1) Give + sign for = greater factor

(2) – sign for smaller factor

(3) is correct

Question 65. The area of a rectangular garden is 8000 square meters. The ratio in length and breadth is 5: 4 A path of uniform width, runs all-round the inside of the garden. If the part occupies 3200 m2, what is its width?

Solution:

(3) let x be common in the ratio

length = 5x anil breadth = 4x

Area = 5x x 4x = 8000

Or 20x² = 8000

Or x² = 400 = 20²

∴ x = 20

Length = 5x = 5 x 20 = 100m

Breadth = 4x = 4 x 20 = 80m

Let Width of path = ym

∴ Area of rest part = 8000 – 3200

Or (100 – 2y) (80- 2y) = 4800

Tricks: go by choices.

∴ (3) y = 10 m satisfies it

Width = 10m is correct

Question 66. If (2 +V3) is a root of a quadratic equation x² + px + q = 0 then find the value of p and q.

(4,-1)
(4, 1)
(-4, 1)
(2,3)

Solution:

One root = 2 + V3

Another root = 2- V3

Eqn is

x² — (2 + √3 + 2 – √3)x + √2+√3) (2 – √3) = 0

Or x² – 4x + (4- 3) = 0

Comparing it with x + px + q = 0

p = -4 and q =1

(3) is correct

Question 67. If the area and perimeter of a rectangle are 6000 cm² and 340cm respectively, then the length of a rectangle is:

Solution:

(2)

let length =1

and breadth = b

lb = 6000 _

2(l + b) = 340…….(1)

Or 1 +b = 170…….(2)

Then Go by choices

For (a) if 1 = 140 then b = 170 – 140 = 30

lb = 140 x 30 = 4200* 6000

(a) is not the answer

For (2) if 1 = 120

Then 120 + b = 170

b= 170 -120 = 50

Area = lb = 120 x 50 = 6000

Question 68. A straight line passes through the point (3,2). Find the equations of the straight line.

x + y =1
x + y = 3
x + y = 5
x + y = 2

Solution:

(3) it cannot be solved properly because only one condition is given for the answer

(3) Point (3 ; 2) satisfies x + y = 5

(3) is correct

Question 69. One root of the equation: x² – 2(5+m)x+3(7+m>0 is reciprocal of the other. Find the value of m.

-20/3
7
1/7
117

Solution:

(1)

Let one root = or, then other root =$=\frac{1}{\alpha}$

⇒ $\alpha \times \frac{1}{a}=\frac{3(7+m)}{1}$

1 = 21 + 3m

⇒ $3 m=-2m=\frac{-20}{3}$

(1) is correct

Question 70. Roots of the equation 3x² – 14x – k = 0 will be reciprocal of each other if:

k = -3
k = 0
k = 3
k = 14

Solution:

(3) is correct.

Let one root = a and another root =$=\frac{1}{\alpha}$ (given)

Product of roots = $=\frac{c}{a}$

⇒ $\alpha \times \frac{1}{\alpha}=-\frac{k}{3}$

⇒ $-1=\frac{k}{3} \text { So, K }=-3$

(1) is correct

Question 71. Positive value of for which the roots at equation 12x² + kx + 5 = 0 are in ratio 3 : 2, is

5/12
12/5
$\frac{5 \sqrt{10}}{2}$
5√10

Solution:

(4)

Let a be common in the ratio.

Roots are 3 a and 2 a

Sum of roots =$-\frac{b}{a}$
.
3a + 2a = 5a =$-\frac{k}{12}$

⇒ $\alpha=-\frac{k}{60}$

Product of roots = $3 \alpha+2 \alpha=\frac{c}{a}$

⇒ $6 x^2=\frac{5}{12}$

⇒ $\text { 6. }\left(-\frac{k}{60}\right)^2=\frac{5}{12}, \text { so, } \mathrm{k}^2=250$

k = 5√10

Question 72. If one root of the equation x²– 3x + k = 0 is 2, then value of k will be

Solution:

(3) is correct

2 is a root of given eqn.

2²– 3 x 2 + k = 0

Or-1 + k = 0

∴ k = 2

Question 73. It roots of equation x² + x + r = 0 are ‘a’ and ‘α’ and β = -6. Find the value of V.

$\frac{-5}{3}$
$\frac{7}{3}$
$\frac{-4}{3}$
1

Solution:

(1) is correct

⇒ $\alpha+\beta=\frac{-1}{1}=-1 \ and \alpha \beta=\frac{r}{1}=r$

α³ + β³ = -6

or (α + β)³– 3αβ (α + β) = -6

or (-1)3 — 3r(—1) = -6

Or- 1 + 3r = -6; or 3r = -5

r = -5/3

Question 74. If one root of the equation Px²+ qx + r = 0 is r then the other root of the equation will be

1/q
1/r
1/p
$\frac{1}{p+q}$

Solution:

(3) let a is another root.

⇒ $\mathrm{r} \alpha=\frac{r}{p} \alpha=\frac{1}{p}$

Question 75. If the ratio of the root of the equation 4x² – 6x + p = 0 is 1 : 2 then the value of is

1
2
-2
-l

Solution:

(2) let α be common in the ratio

⇒ $ \alpha+2 \pi=\frac{p}{4}p=8 \alpha^2=8 \cdot \frac{1}{4}=2$

Question 76. If p and q are the root of the equations x² – bx + c = 0, what is the equation whose roots are (pq + p + q) and (pq – p – q)?

x² – 2cx + c² – b² = 0
x² – 2cx + c² + b²= 0
8cx² – 2(a+c)x + c² = 0
cx² + 2bx – (c² – b²) = 0

Solution:

(1)

x² – (p + q)x + pq = 0

b = p + q ; C = pq

New roots are

pq + (p + q) = c + b

pq-(p + q) = c-b

Eqn is

x² – (c + b + c- b)x +(c + b)(c- b) = 0

or x² – 2cx + c² – b² = 0

Question 77. If the arithmetic mean between the roots of a quadratic equation is 8 and the geometric mean between them is 5, the equation is-

x²– 16x-25 = 0
x²– 16x + 25 =
x² – 16x + 5 = 0
None of these

Solution:

(2) let α and β be two roots.

(α+β)/2 = 8 and √αβ = 5

⇒ α+β = 16 and αβ = 25

Eqn. is x²-16x + 25 = 0

Question 78. The minimum value of the function x²– 6x + 10 is-

Solution:

(1) coeff. Of x2 = 1 > 0; function is minimum (formula)

Minimum value =$\frac{4 a c-b^2}{4 a}$

⇒ $=\frac{4.1 .10-(-b)^2}{4 \times 1}=\frac{4}{4}=1$

Question 79. If one of the roots of the equation x² + px + a is √3 + 2. Then the value of and ‘a’ is

-4,-1
4, -1
4,1
4,1

Solution:

(3) Roots are 2 + √3 and 2 —V3

(conjugate of 2+√3)

Eqn is

x² – (sum of roots) x + product of roots = 0

x²– 4x + (4- 3) = 0

x² + px + a = 0

∴ P = -4; a = 1

Question 80. The roots of equation 2x² + 3x + 7 = 0 are α and β. The value of a αβ^-1+βα^-1

2
3/7
7/2
-19/14

Solution:

⇒ $\text { (d) } \alpha+\beta=-\frac{3}{2} ; \alpha \beta=7 / 2$

⇒ $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-\alpha \beta}{\alpha \beta}$

⇒ $\frac{\frac{9}{4}-2 \cdot \frac{7}{2}}{\frac{7}{2}}=\frac{9-28}{4} \times \frac{2}{7}=\frac{-19}{14}$

Question 81. The quadratic equation x²– 2kx + 1 6 = 0 will have equal roots when the value of is –

Solution:

(4) let roots are or; or

⇒ $\alpha+\alpha=\frac{-(-2 k)}{1} \Rightarrow \alpha=k$

= α.α = k.k => k² = 16 => k = ±4

Question 82. If α, βare roots of x² + 7x+ 11 = 0 then the equation whose roots as (α+β)² And (α – β)²

x²-54x+ 245 = 0
x² — 14x + 49 = 0
x²-24x+ 144 = 0
x² – 50x + 49 = 0

Solution:

(1) is correct

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{7}{1}=-7$

⇒ $\alpha \beta=\frac{c}{a}=\frac{11}{1}=11$

(α-β)² = (α+β)²– 4ap

= (—7)² -4×11 = 5

Require eqn. is

x² (α+β)² + (α-β)² = 0

or x² – (49+5)x + 49 x 5 = 0

or x²-5x + 245 = 0

(1) is correct

Question 83. If b² – 4ac is a perfect square but not equal to zero then the roots of the equation ax2 + bx + c = 0 are

Real and equal
Real irrational and equal
Real rational and unequal
Imaginary

Solution:

(3) b²– 4ac > 0 and perfect square

Question 84. Divide 80 into two parts so that their products are maximum then the numbers are

15,65
25,55
35,45
40,40

Solution:

(2) is correct

Let 1st part = x

2ml part = 80 – x

Lety = x(80-x) = -x² + 80x

Here co. eff. of x²< 0

Numbers are (40; 40)

Question 85. If a,p is the roots of a quadratic equation if x² + 2x – 3 find the quadratic equation:

x² + 2x – 7 = 0
x² + 2x – 3 = 0
x² – 2x – 3 = 0
x² – 2x +7 = 0

Solution:

(2)

Quadratic eqn is

x2 – (a + p)x + ap = 0

x2 – (-2)x + (-3) = 0

x2 + 2x- 3 = 0

Question 86. Value of k for which roots are equal of given equation 4x² – 12x + k = 0:

144
9
5
None of these

Solution:

(2)

4×2- 12x + k = 0

D = b2 – 4.ac = 0

(-12)2 = 4 x 4.k

Or. 144= 16k

k = 9

Question 87. If difference between the roots of the equation x² – kx + 8 = 0 is 4 then the value of K is

0
±4
±8√3
±4√3

Solution:

(4)

Let a; /? are roots of x² – kx + 8 = 0

⇒ $\alpha+\beta=b / a=-\frac{(-k)}{1}=k \& \alpha \cdot \beta=c / a=8 / 1 $

⇒ $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=4^2$

⇒ $\Rightarrow k^2-4 \times 8=16$

⇒ $\text { Or } k^2=48 \Rightarrow k= \pm \sqrt{16 \times 6} \Rightarrow k= \pm 4 \sqrt{3}$

Question 88. If a: /? be the roots of x²+x+5 = 0 then$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=$

$\frac{16}{5}$
2
3
$\frac{14}{5}$

Solution:

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{1}{1}=-1$

⇒ $\alpha \beta=\frac{c}{a}=\frac{5}{1}=5$

⇒ $\frac{a}{\beta}+\frac{\beta^2}{u}=\frac{\alpha^3+\beta^3}{\alpha \beta}$

⇒ $\frac{(\alpha+\beta)^3-3 a \beta(\alpha+\beta)}{\alpha \beta}=\frac{(-1)^3-3.5 \cdot(-1)}{5}$

⇒ $\frac{-1+15}{5}=\frac{14}{5}$

Question 89. If the sum of two numbers is 13 and the sum of their squares is 85 then the numbers

6,7
4,9
10,3
5,8

Solution:

(1)

6 + 7= 13 (true)

62+ 72 = 36 + 49 = 85 (True)

Question 90. The difference between the roots of the equation x² – 7x- 9 = 0 is

7
V85
9
2x/85

Solution:

(2)

⇒ $\text { Let } \alpha+\beta=-\frac{b}{a}=-\frac{-7}{1}=7$

⇒ $\propto \beta=\frac{c}{a}=\frac{-9}{1}=-9 $

⇒ $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \propto \beta=7^2-4(-9)=85 $

⇒ $\alpha-\beta=\sqrt{85}$

Question 91. If the roots of the equation kx² — 3x —1 = 0 are the reciprocal of the roots of the equation x² + 3x – 4 = 0 then k=

4
-4
3
-3

Solution:

(1) x² + 3x- 4 = 0

Or x² – 4x + x – 4 =0

Orx(x- 4) +1 (x- 4) = 0

Or; (x- 4) (x + 1) = 0

x = 4; -1

Eqn. having rots $\frac{1}{4} \& \frac{1}{-1}=\frac{1}{4} \&-1 \text { is. }$

⇒ $x^2-\left(\frac{1}{4}-1\right) x+\frac{1}{4}(-1)=0$

⇒ $\text { or } x^2+\frac{-1}{4} x-\frac{1}{4}=0$

Multiplying by 4; we get

4x² + 3c- 1 = 0

Comparing it with le tx² + 3x-1 = 0

We get K = 4

Eqn. having roots the reciprocal of the roots of ax²+ bx + c = 0 is ax² + bx + a = 0 i.e. 1st and last term interchanges

Question 92. If cr + ft = -2 mid aft = -3 where u and \ I are the roots of the equation, which is

x² – 2x -3 = 0
x² + 2x -3 = 0
x² + 2x + 3 = 0
x² – 2x + 3 = 0

Solution:

(2)

Quadratic Eqn. having roots a and |3 is

x² – (a + (3) x + a(3 = 0

or; x² – (-2)x + (-3) = 0

or; x² + 2x- 3 = 0

(2) is correct

Question 93. If a and (1 are the roots of the equation x² + x + 5 = 0 the$\frac{a^2}{\beta}+\frac{\beta^2}{\alpha}$

$\frac{16}{5}$
3
$\frac{14}{5}$
2

Solution:

(3)

Given Eqn. is x² + x + 5 = 0

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{1}{1}=-1$

⇒ $\alpha \beta=\frac{c}{a}=\frac{5}{1}=5 $

⇒ $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{a \beta}$

⇒ $\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}$

⇒ $\frac{(-1)^3-3 \cdot 5 \cdot(-1)}{5}=\frac{-1+15}{5}=\frac{14}{5}$

Question 94. When two roots of quadratic equation are a,$\alpha, \frac{1}{\alpha}$ then what will be the quadratic equation:

ux² – (cr² + 1)x + cr = 0
ax² – ax² + 1 = 0
ax² – (cr ² + l)x +1=0
None of these

Solution:

(1)

For $\alpha \cdot \frac{1}{a}=1=\frac{c}{a}=\frac{a}{a}=1 \text { (True) }$

Question 95. Let cr and ft be the roots of x2 + 7x + 12 = 0, The the value of$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} \text { will be }$

$\frac{19}{141}+\frac{141}{49}$
$\frac{7}{12}+\frac{12}{7}$
$-\frac{91}{12}$
None of these

Solution:

x² + 7x+ 12 = 0

or x²+ 4x + 3x + 12 = 0

or x(x + 4) + 3 (x + 4) = 0

or (x + 4) (x + 3) = 0

x = -3; -4

Letα = —3 ; β = -4

⇒ $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{9}{-4}+\frac{16}{-3}$

⇒ $-\left[\frac{9}{4}+\frac{16}{3}\right]=-\frac{91}{12}$

2 and Method

⇒ $alpha+\beta=\frac{-7}{1}=-7$

⇒ $\alpha \beta=\frac{c}{a}=\frac{12}{1}=12$

⇒ $ \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta}=\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}$

⇒ $\frac{(-7)^3-3 \times 12(-7)}{12}=-\frac{91}{12}$

Question 96. Find the condition that one root is double the of ax² + bx + c = 0

2b² = 3ac
b²= 3ac
2b² = 9ac
None

Solution:

Let 1st root =1

Then 2nJ root = 2

The Eqn. is

x2-(l + 2)x+lx2 = 0

or x² – 3x + 2 = 0

Comparing it with ax² + bx + c = 0

We get;

a = 1; b = 3; c = 2

Go by choices (GBC)

(1) 2b² = 9ac

2.(-3)2 = 9.1.2

⇒ 18 = 18 (True)

Hence, (3) is (true)

Question 97. Find the quadratic equation in x whose roots are 2 and 3.

x² + 5x + 6 = 0
x² — 5x — 6 = 0
x² – 5x + 6 = 0
x² + 5x- 6 = 0

Solution:

The quadratic equation is (x-2) (x-3) = 0

Or x² — 2x — 3x + 6 = 0 or x²– 5x + 6 = 0

Question 90. If the sum of the roots of a quadratic equation in x is 4 and the constant term of the equation is 15, then find the quadratic equation.

x²– 4x + 1 5 = 0
x² – 4x + 1 5 = 0
x² + 4x+ 15 = 0
x² + 4x+ 15 = 0

Solution:

we know that if s is the sum and p is the product of the roots of a quadratic equation then the equation is x²– 5x + p = 0

Note that the constant term in the equation is nothing but the product of the roots.

Given s=4, p=1 5

The quadratic equation is x² – 4x + 15 = 0

Question 99. Find the roots of the quadratic equation x²– 8Y + 9 = 0

1,9
-1,-9
4+√7, 4-√7
3+√6, 3-√6

Solution:

⇒ $x^2-8 x+9=0 \text { the roots are } \frac{-\Delta \pm \sqrt{D^2-4 a c}}{2 a}$

⇒ $x=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(0)}}{2(1)}$

⇒ $\frac{8 \pm \sqrt{s t-3 a)}}{2}=\frac{8 \pm \sqrt{2 s}}{2}=\frac{8 \pm 2 \sqrt{7}}{2}=4 \pm \sqrt{7}$

⇒ $\text { roots are } 4+\sqrt{7} \text { and } 4-\sqrt{7}$

Question 100. Find the sum of the roots of the equation, 9x² — 36x + 35 = 0.

36
-36
4
-4

Solution:

he sum of the roots of equation ax² +bx+ c = 0 is -b/a

the sum of the roots of 9x² — 36x+ 35 = 0 is $\frac{-(-36)}{9}=4$

Question 101. If the sum of the roots of a quadratic equation is 11 and the product of the roots is 24, find the roots of the equation.

6,4
12,2
8,3
1,24

Solution:

let a, and p be the roots of the equation. Given, α + β = 11 and αβ = 24

(α- β)² = (α + β)²– 4αβ = ll2- 4(24) = 121- 96 = 25

α- β = 5

(1) ÷ (2)⇒ 22=16⇒α = 8

β= 11-8 = 3

the roots are 8,3.

Question 102. If the sum of the roots of a quadratic equation is 7 and the product of the roots is 12, then find the equation.

x² + 7x + 12 = 0
x² + 7x — 12 = 0
x²– 7x + 12 = 0
x²– 7x — 12 = 0

Solution:

Given the sum of the roots is 7 and the product of the roots is 12.

the equation is

x² -(sum of the roots)x+ (product of the roots) = 0 i. e.x² — 7x + 12 = 0

Question 103. Find the discriminant of the equation 2x² + 3,Y + 4 = 0

Solution:

discriminant of the quadratic equation ax² + bx + c = 0 is b²– 4ac

the discriminant of 2.v2 + 3.v 4-4 = 0 Is

32 — 4(2)(4) = 9 — 32 = —23

Question 104. Construct a quadratic equation whose roots are 4 more than the roots orx2 + lx + 16 = 0

x2 + 15x + 60 = 0
x2 + 7x + 60 = 0
x2- x + 4 = 0
x2 + 15x + 16 = 0

Solution:

to find a new quadratic equation, whose roots are 4 more than the roots of x²+ 7x + 16 = 0

we have to replace x by (x-4)

The equation is (x- 4)² +7(x- 4) + 16 = 0

⇒ x²– 8. Y 4- 16 + 7x – 28 + 16 = 0

⇒ x²– x 4- 4 = 0

Question 105. Construct a quadratic equation whose roots are reciprocals of the roots of the equation 3x² + 5x + 7 = 0

7x² + 5x + 3 = 0
5x² + 7x + 3 = 0
5x² + 3x + 7 = 0
7x² + 3x + 5 = 0

Solution:

to find a new quadratic equation, whose roots are reciprocals of the roots of 3x²+ 5x +7 = 0 we have replaced x by1/x

The equation is$3\left(\frac{1}{x}\right)^2+5\left(\frac{1}{x}\right)+7=0$

7x² + 5A + 3 = 0

Question 106. Construct a quadratic equation whose roots are half roots of the equation x²+5x+9=0

4x² + lOx + 9 = 0
x² + lOx + 36 = 0
4x² + 5x + 36 = 0
4x² + 5x + 9 = 0

Solution:

To find a new quadratic equation, whose roots are half of the roots of the equation,

x² + 5x + 9 = 0 we have dot replace x by 2x the equation is (2,y)2 + 5(2x) + 9=0

⇒ (4×2) + 10.Y + 9 = 0

Question 107. How many distinct roots does the quadratic equation (x- 4)² = 0 have?

Solution:

Here the equation (x- 4)² = 0 is given.

Number of distinct roots = 1

Question 108. Solve for x x⁴– 61x² + 900 = 0

±4. ±5
±5, ±6
±3,±5
±4, ±6

Solution:

x⁴ – 61x² + 900 = 0=>(x²– 36)(x² – 25) = 0

⇒ x=±5,±6

Question 109. Solve for x 3^2x+1 _ 39(3x+2) + 8748 = 0

4
log336
Either (1) or (2)
6

Solution:

3(3^2x)- (39)(9)(3^x) + 8748 = 0

Dividing by 3 and setting y=3^x

y² – 117y + 2916 = 0

⇒ 3^x = 36 or 3X = 81

⇒ x= log336 or x = 4

Question 110. Solve for x$\sqrt{2 x-5}+\sqrt{3 x+4}=8$

8
615
7
Either (2) or [3]

Solution:

⇒ $ \sqrt{2 x-5}+\sqrt{3 x+4}=8 \rightarrow(\mathrm{A})$

By squaring on both sides.

⇒ $5 x-1+2 \sqrt{6 x^2-7 x-20}=64$

⇒ $ 2 \sqrt{6 x^2-7 x-20=-5(x-3)}$

By squaring on both sides

⇒ 4(6x² – 7x- 20) = 25(x² – 26x + 169)

⇒ x²– 622x + 4305 = 0

⇒ (x — 61 5)(x — 7) = 0

Substituting x =7 in (A) we see that V9 + V25 = 8 but for x=615

⇒ $=\sqrt{2(615)-5}+\sqrt{3(615)+4}=\sqrt{1225}+\sqrt{1849}$

35+43=78

To get 8, we have to take the negative root for 1225 The expression V1225 normally stands for the positive square root 35 we would say that x=615 does not satisfy [A)

Hence x =7

Question 111. Which of the following is the root [s] of $\sqrt{3 x+34}+\sqrt{3-2 x}=\sqrt{67+x}$

3
-3
-41/6
1. A or B
2. B or C
3. C or A
4. Only B

Solution:

⇒ $ \sqrt{3 x+34}+\sqrt{3-2 x}=\sqrt{67+x} \rightarrow(A)$

By squaring on both sides,

⇒ $x+37+2 \sqrt{-6 x^2-59 x+102}=67+x $

⇒ $2 \sqrt{-6 x^2-59 x+102}=30$

By squaring on both sides,

⇒ -6x²– 59x + 102=225

⇒ 6x² + 59x+123=0

⇒ (6x+41) (x+3)=0

⇒ x= -41/6 or -3

Substituting -3 in (A) the left-hand side (LIIS)

And if we substitute -41/6 the LHS of (A

⇒ $=\sqrt{3(-3)+34}+\sqrt{3-(-6)}=5+3=8 R H S$

⇒ $\sqrt{-\frac{41}{2}+\frac{68}{2}}+\sqrt{\frac{9}{3}+\frac{41}{3}}=\sqrt{\frac{27}{2}}+\sqrt{\frac{50}{3}}$

⇒ $\frac{9}{\sqrt{6}}+\frac{10}{\sqrt{6}}=\frac{19}{\sqrt{6}} \text { and the RHS }$

⇒ $\sqrt{\frac{402}{6}-\frac{41}{6}}=\sqrt{\frac{361}{6}}=\frac{19}{\sqrt{6}}$

⇒ $\text { Both }-3 \text { and }-41 / 6 \text { satisfy (A) }$

Question 112. Find the value of if one of the roots of x² – 9x + 2p = 0 is 3 more than the other.

Solution:

Let the roots of a and a + 3.

2α + 3 = 9

⇒ α = 3

The other root is 6 and 2p = 18 or p=9

Question 113. Four friends have some coins. Pavan has 2 less than Samir, who has 2 less than Tarun,ifPra has 2 less than Pavan and the product of the number of coins that each of them has is 5760, how many coins in all do they have among themselves?

Solution:

Let the number of coins with Pranav, Pavan, and Samir Tarun be (a-3), (a-1), and (a+1) respectively.

(a² + 3²)(a²– l²)(a + 3)= 5760

Let a2 = k(k- 9)(/c- 1) = 5760

k² – 10k- 5751 = (k- 81)(fc + 71) = 0

⇒k = 81 or- 71

As k is positive, a² = 81⇒a = 9

⇒ (a- 3) + (a- 1) + (a + 1) + (a + 3) = 36

Question 114. If the roots of ax² + bx+c-0 are α and β and the roots of px² + qx + r = 0 are α- k and β-k, then

r= ak² – bk + c
r= ak²– bk- ck²
r= ak + bk + ck²
r= ak² + bk + c

Solution:

(d)

As α— k and β — n are the roots ax² + bx + c = 0

Now, the equation with α — k and β- kas roots are

f(x + k) i.e.a(x-k)² + b(x + k) + c = 0

ax2 + (fa + 2ak)x + (a/c2 + fa/c + c) = 0

By comparing this equation with px² + qx +r = 0, r=ak² + bk + c

Question 115. if the roots of x² + 7x + 10 = 0 are a and /?, find the equation whose roots are $\left(\frac{1-\alpha}{\beta}\right)$ and $\left(\frac{1-\beta}{\alpha}\right)$

x² + 360x + 180 = 0

x²+ 0.36x + 0.18 = 0

x² +36x + 18 = 0

x² + 3.6x + 1.8 = 0

Solution:

(4)

The roots of x² +7x + 10 = 0 area and (3)

α + β = -7 and αβ = 10

⇒ $\text { If } \mathrm{p}=\frac{1-\alpha}{\beta} \text { and } \mathrm{q}=\frac{1-\beta}{\alpha}$

⇒ $P+q=\frac{(\alpha+\beta)-\left(\alpha^2+\beta^2\right)}{\alpha \beta}$

⇒ $=\frac{(\alpha+\beta)-(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}$

⇒ $=\frac{(-7)-\left(7^2-2(10)\right)}{10}=-3.6$

⇒ $\text { and } \mathrm{pq}=-\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}$

⇒ $=\frac{1-(-7)+10}{10}=1.8$

The equation where roots are p,q is x² + 3.6x + 1.8 = 0

Question 116. A total of Rs.3, 300 is raised by collecting equal amounts from a certain number of people. If there were 22 people more, each person would have to contribute Rs. 200 less to raise the same amount. How many people contributed?

Solution:

Let the number of people and the contribution of each person be n and a respectively,

(n + 22)(a- 200) = na

⇒ 22a – 200n = 4400 → (1)

And na= 3300→ (2)

Substituting the value of from (2) in (1)

22(3300/n)-200n = 4400

⇒ ll(33)/n- n = 22⇒ n² + 22n- 11(33) = 0

⇒ (n + 33)(n- 11) = 0 =>n = 11

Question 117. The sum of two positive numbers is 32. The square of the largest number is 223 more than twice the square of the smaller number. Find the smaller number.

Solution:

(1)

Let the smaller and larger numbers be x and yx + y=32→ (1)

2A2 + 23 = y2 → (2)

⇒ x² + 64x- 1001 = 0⇒(x + 77)(x- 13) = 0

⇒ x = -77 or x = 13

As x cannot be negative x = 13

Question 118. Akash and Badri, together buy 45 notebooks. Each of them buys a different number of books and both of them spend the same amount if Akash had bought his notebooks at the price at which Badri had bought them and Badri had bought his notebooks at the price where Akash had bought them, they would havespentRs. 160 and Rs. 250 respectively. How many notebooks did Badri buy?

Solution:

(1) Akash Badri

No. of notebooks mn

Price a b

We have m + n=45→ (1)

Ma = nb→ (2)

Mb = 160 → (3)

Na = 250 → (4)

From (2) ⇒ m/n = b/a

(3) +(4)⇒ (m/n)2 = (16/25) = (4/5)2

m: n = 4:5

And n=5/9 (45) =25

Question 119. If the roots α and β of x2- 4x- c = 0 satisfy the condition 2α + β = 1, then which of the following is true?

α = -3, β = 7
c=-21
α = -15, β = 3.5
α = 3, β = -7

Solution:

(1)

The α,β are the roots of x²– 4x- c = 0

α+β = 4 and 2a + p = 1 (given)

⇒ a = -3 and P = 7

Question 120. How many of the roots of x² – A + 2 are also the roots of x² + 4x² + 5x + 2 = 0?

Solution:

(2)

The roots of x²– x + 2 are x = 2 and x = -1

of these two roots, only, one of the roots satisfies the cubic equation, i.e. (2)³ + 4(2)² + 5(2) + 2≠0 but (-1)3+ 4(—l)² + 5(-l) + 2 = 0

has real and distinct roots.

Question 121. If one of the roots of x² – ax + b – 0 is 2+√3 and b=a+2+3√3, find the other root.

2-√3
√3- 2
4
-4

Solution:

(4)

the roots of 3x² + 9x + 5 = 0 are p and q p + q=9/3=3 and p q=5/3

⇒ $\frac{1}{p^2}+\frac{1}{q^2}=\frac{p^2+q^2}{p^2 q^2}=\frac{(p+q)^2-2 p q}{p^2 q^2}$

⇒ $\frac{3^2-(10 / 3)}{25 / 9}=\frac{51}{25} \text { and } \sqrt{3 p q}=\sqrt{5}$

⇒ $ p^2+q^2=\sqrt{3 p q}=\frac{51}{25}+\sqrt{5}$

Question 122. The roots of 35x² + x = 12 are

Rational and equal.
Rational and distinct.
Irrational and equal.
Irrational and unequal

Solution:

(2)

for the equation 35x² + x- 12 = 0 the determinant

s=1² + 4(35)(12) = If,81 = 41² = 18681=41²

the roots are rational (because A is a perfect square)and distinct (because Δ≠0)

Question 123. How many real roots does the equation x⁴ + 2x³ – 10x² – 2x + 1 = 0 have?

Zero
One
Two
Four

Solution:

For equations of the form Ax⁴ + Bx³ + Cx² ± Bx + A = 0, the method illustrated below should be used.

Given the equation, x⁴ + 2x³– 10x²– 2x +1 = 0 …. (1)

Divide (1) by x2 we get $x^2+2 x-10-\frac{2}{x}+\frac{1}{x^2}=0$

⇒ $\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2\left(x-\frac{1}{x}\right)-10=0$

⇒ $\text { Put } \mathrm{x}-\frac{1}{x}=y, \text { then } x^2+\frac{1}{x^2}=y^2+2$

⇒ y² + 2 + 2y- 10 = 0

⇒ (y+4)(y-2) =0

⇒ y= -4 or 2

⇒ $x-\frac{1}{x}=-4 \text { or } x-\frac{1}{x}=2$

⇒ x² + 4x-1 = 0 or x²– 2x-1 = 0 as the discriminant of both the quadratic equations is positive, hence equation (1) has 4 real roots.

Question 124. If the roots of 3 (x² + l)=kx area and 1/α(both real), which of the following could be the value of k?

10
5
-3
3

Solution:

(1)

The roots of 3x² + kx + 3 = 0 are a and 1/α both real quantize.

k²-4(3) (3) ≥ O or k² ≥ 36

only k = 10 satisfies this condition.

Question 125. If the roots of 3x² + (2k + 4)x = -12 are equal, then the value of k is

-5
5
-8
4

Solution:

(4)

As the roots of 3x²+ (2k + 4)x + 12 = 0 are equal A= 0 i.e. (k + 2)² = 3(12) = 36

⇒ k = 4 or- 8 =>k = 4 or- 8

Question 126. For which of the following equations with roots, αβ is α² + β²= 25 And αβ= -12?

x(x – 1) = 12
X(x + 9) = 12
x(x + 1) = 12
Either (1) or (3)

Solution:

(4)

⇒ $a^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

and $ \frac{1}{\alpha^2}-\frac{1}{\beta^2}=\frac{\beta^2-\alpha^2}{\alpha^2 \beta^2}=\frac{(a+\beta)(\beta-\alpha)}{\alpha^2 \beta^2}$

for the required equation with roots a, [3

α² + β² = 25 and a/? = -12

α² + β² + 2αβ = 25 + (2)(—12) = l⇒α + β = 1 or —1

The equation could be x² – x — 12 = 0 or x² + x- 12 = 0

Question 127. If the speed of a vehicle decreases by 10 km, it takes 2 hours more than what it usually takes, to cover a distance of 1 800 km, find the time it usually takes,

24 hours
30 hours
36 hours
18 hours

Solution:

(4)

the data is tabulated below

Speed time Distance

ut 1800⇒ut=1800

u-10 t + 2 1800

ut = (u – 10)(t + 2) = ut + 2u- lOt- 20

⇒ $2 u-10\left(\frac{18 00}{u}\right)=20 \Rightarrow u^2-10 u-9000=0$

(u – 100)(u + 90) = 0 ⇒ u=100

and t = 1800/100=18

Question 128. If one root of x2- 9x + 14 = 0 is the same as one root of x2 – 5x + k = 0, then k =

±6
±14
6,-14
-6,14

Solution:

(3)

the roots of x² – 9x + 14 = 0 are 7 and 2 if 7 is a root of the second equation 7²– 5(7) + k = 0 ⇒ K = 6

Question 129. If the sum of the roots of x²– 5x⁴+ k = 0 and x + k = 0 is 80 then what is/ are the possible real value (s) of k?

±5
±2
±4
±10

Solution:

(2)

the given equation is x²– 5x⁴ + k = 0

the sum of the roots is 80

5k4 = 80⇒k⁴ = 16

±2 are the real values of.

Question 130. How many equations of the form x² + 6x + p = 0 where p is an integer and 0< p < 15, have real roots?

Solution:

(1)

The given equation is x² + 6x + p = 0

This has real roots if and only 6² – 4p≥ 0 i.e., p≤9

10 integral values satisfy this condition.

Question 131. What is the minimum value of the Squares of the roots of the equation x²-(α – 3)x+(α-8)=0 where α is a positive number?

Solution:

(4)

x² – (a- 3) + (a- 8) = 0

The sum of the roots =a-3

The product of the roots= a-8

The sum of the squares of the roots=(a- 3)²– 2(a- 8) = a² — 6a + 9- 2a + 16 = (a² –
8a + 25)

the minimum value of a²– 8a + 25

⇒ $=\frac{4(1)(25)-(-8)^2}{4(1)}=\frac{100-64}{4}=9$

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.2

June 3, 2024September 29, 2023 by Marksparks

Question. Write a multi-step equations to solve real-life problems.

There are a lot of ways where you can apply multi-step equations to solve real-life problems.

One of them is in bookkeeping which we call balancing.

You will use equations and use multiple steps to balance the account.

It is also like taking one step at a time.

Sometimes when you encounter a problem, you resolve it by thinking about the possible ways to solve it.

Then, prepare yourself in making a move.

Taking action, and waiting for the result.

Multi-step equations can be used to solve real-life problems that include complex calculations.

$Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.2$

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Question. Evaluate the angles and number of sides and find the value of x.

Given: Sum of angles S and number of sides as n.

To find the value of x

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 1

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( Substitute number of terms)

S=180(3−2)

S=180(1)

S=180

Add the angle measurements and equate them to the obtained sum.

(30+x)+9x+30 =180 ( Combine like terms)

10x+60=180

10x+60−60=180−60 ( Subtract 60 on both sides)

10x=120

10x/10=120/10 ( Divide by 10 on both sides)

x=12.

To get the actual angle measures, substitute x to the original equation of each angle measure.

1.=(30+x)

=30+12

=42°

2.=9x

9(12)=108°

3.x=30

Checking:

42+108+30=180

180=180°

The three angles are 42°,108°, and 30°.

The three angles obtained are 40°,108°, and 30°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 2

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 3

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures as

S=180(n−2) ( substitute the number of sides )

S=180(3−2)

S=180(1) ( combine like terms )

S=180.

Add the angle measurements and equate it to the obtained sum.

(x+10)+(x+20)+50=180 (combine like terms)

2x+80=180

2x+80−80=180−80 (subtract 80 on both sides)

2x=100

2x/2=100/2 (divide both sides by 2)

x=50

To get the actual angle measures, substitute x to the original equation of each angle measure.

=(x+10)

=(x+20)

=50°

(x+10)=50+10=60°

(x+20)=50+20=70°

Checking: 60+70+50=180

180=180

The three angles are 60°,70° & 50°.

The obtained values of angles are 60°,70°, and 50°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 4

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( substitute the number of sides )

S=180(4−2)

S=180(2) (combine like terms )

S=360.

Add the angle measurements and equate it to the obtained sum.

50+(2x+30)+(2x+20)+x =360 ( combine like terms)

5x+100 = 360

5x+100−100 &=360−100 ( subtract 100 on both sides)

5x=260

5x=260/5

x=52. (divide both sides by 5)

To get the actual angle measures substitute x to the original equation of each angle measure.

(2x+30)=2(52)+30

=134°

(2x+20)=2(52)+20

=124°

x=50° ,52°

Checking :

134+124+52+50=360

360=360

The angles are134,124° ,52° & 50°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 5

The angles obtained are134,124°,52°and 50°.

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 6

Count the number of sides and n of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) (substitute the number of sides )

S=180(4−2)

S=180(2) (Combine like terms)

S=360.

Add the angle measurements and equate it to the obtained sum.

(x−17)+(x+35)+(x+42)+x =360 ( combine like terms )

4x+60 =360

4x+60−60=360−60 ( subtract 60 on both sides)

4x=300

4x/4=300/4 (divide both sides by 4)

x=75.

To get the actual angle measures, substitute x to the original equation of each angle measure.

1.(x−17)=75−17

=58°.

2.(x+35)=75+35

=110° .

3. (x+42)=75+42

=117°.

4. x=75°

Checking:

58+110+117+75=360

360=360

The angles are 58°,110°,117° and 75°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 7

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 8

Count the number of sides,n, of the given polygon. Then solve for the sum S of the angle measures using n.

S=180(n−2)

S=180(5−2) ( Substitute number of terms.)

S=180(3) (Combine like terms)

S=540°.

Add the angle measurements and equate it to the obtained sum.

(5x+2)+(3x+5)+(8x+8)+(4x+15)+(5x+10)=540

25x+40=540 (combine like terms )

25x+40−40=540−40 ( subtract 40 on both sides)

25x=500

25x/25=500/25 (Divide both sides by 25)

x=20.

To get the actual angle measures substitutes to the original equation of each angle measure.

1. (5x+2)=5(20)+2

=102°

2. (3x+5)=3(20)+5

=65°

3. (8x+8) =8(20)+8

=168°

4.(4x+15) =4(20)+15

=95°

5.(5x+10)=5(20)+10

=110°

Checking:

102+65+168+95+110=540

540=540

The angles are 102° ,65°,168°,95° &110°.

The obtained angles are 102,65,168 and 110. angles

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 9

Given: Sum of angles S=180(n−2) and number of sides asn

To find the Value of x.

Evaluate to get an answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 10

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( substitute the number of sides )

S=180(6−2)

S=180(4) ( combine like terms )

S=720.

Add the angle measurements and equate it to the obtained sum.

(3x+16)+(3x+16)+(2x+8)+(4x−18)+(2x+25)+(3x−7)=720

17x+40=720 (combine like terms )

17x+40−40=720−40 (subtract 40 on both sides )

17x=680

17x/17=680/17 (divide both sides by 17)

x=40.

To get the actual angle measures, substitute x to the original equation of each angle measure.

1.(3x+16)=3(40)+16

=136°.

2.(2x+8)=2(40)+8

=88°.

3.(4x−18)=4(40)−18

=142°.

4.(2x+25)=2(40)+25

=105°

5.(3x−7)=3(40)−7

= 113°.

Checking:

136+136+88+142+105+113=720

The angles are 136°,136°,88°,142°,105°&113°.

The obtained angles are136° ,136°,88°,142°,105°, and 113°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 11

Question. Evaluate the irregular polygon and find the measure of angles.

The irregular polygon can be drawn as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 1

The obtained irregular polygon is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 2

Given: An irregular polygon.

To find The measure of angles.

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 3

The measure of angles are evaluated to be

m∠A=100°

m∠B=68°

m∠C=99°

m∠D=93°

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 4

The measure of angles are 100 °,68°,99°and 93°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 5

Given: An irregular polygon.

To find The measure of angles in terms of x.

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 6

Let us consider the value for, x=20.

We will use the backward method to create equations.

m∠A=5x=5(20)=100°

m∠B=3x+8=3(20)+8=68°

m∠C=4x+19=4(20)+19=99°

m∠D=4x+13=4(20)+13=93°

The equation representations are 5x,4x+19,4x+13, and 3x+8.

Given: To trade polygons with a partner.

To find the polygon obtained after trading with a partner.

Evaluate to get the answer.

After trading with the partner, the obtained polygon is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 7

The obtained polygon is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 8

Given: Sum of angles S=180(n−2) and number of sides as n.

To find value of x.

Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 9

Count the number of sides and n of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) (substitute the number of sides)

S=180(n−2)

S=180(4−2) (Substitute number of sides)

S=180(2) ( Combine like terms)

S=360

Add the angle measurements and equate it to the obtained sum.

(x−17)+(x+35)+(x+42)+x=360 (Combine like terms)

4x+60=360

4x+60−60=360−60 ( subtract 60 on both sides)

4x/4=300/4 (Divide by 4 on both sides)

x=75.

To get the actual angle measures substitutes to the original equation of each angle measure.

1.(x−17)=75−17

=58°.

2.(x+35)=75+35

=110°.

3.(x+42)=75+42

=117°.

4.x=75°.

On checking

58+110+117+75=360

360=360

The angles obtained are 58°,110°,117° and75°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .2 Solving Multi step Equations image 10

Question. Write a multi-step equation to solve real-life problems.

There are a lot of ways where you can apply multi-step equations to solve real-life problems.

One of them is in bookkeeping which we call balancing.

You will use equations and use multiple steps to balance the account.

It is also like taking one step at a time.

Sometimes when you encounter a problem, you resolve it by thinking about possible solutions.

Then, prepare yourself to make a move.

We are taking action, and waiting for the result.

Multi-step equations can be used to solve real-life problems that include complex calculations.

Question. Evaluate the formula for sum S and number of sides. Find the reason why the formula works.

Given: A formula for sum S and number of sides.

To find The reason why the formula works.

Evaluate to get the answer.

We know that the sum of interior and exterior angles should be equal to at each vertex.

And that sum of all exterior angles should be 360∘.

Let n be the number of vertices,

Let be the sum of the exterior angles,360,

S be the sum of the angle measures of a polygon, then:

n(180)=S+e (substitute the value of e )

n(180)=S+360

n(180)−360=S+360−360 (subtract 360 on both sides)

180(n−2)=S (Factor out of 180).

It can be concluded that the sum of interior and exterior angles should equal 180 ° at each vertex. And that sum of ALL exterior angles should be 360∘.

Question. Evaluate the Sum of angles and find the number of sides of a polygon.

Given: Sum of angles S=1080°.

To find The number of sides of a polygon.

Evaluate to get the answer.

Use the backward method for the formula of the sum of the angle measures of a polygon with n sides.

Since n is unknown and S is known then,

S=180(n−2) ( substitute the value of S)

1080=180(n−2)

1080/180=180(n−2)/180 (divide both sides by 180).

6=n−2

6+2=n−2+2 (add 2 on both sides )

n=8.

The polygon has 8 sides.

The number of sides of a polygon is n=8.

Question. Evaluate the equation 3x + 4 = 19 and find the value of x.

Given: An equation 3x+4=19.

To find the Value of x.

Evaluate to get the answer.

Let us consider and solve the equation for x.

3x+4=19

3x+4−4=19−4 (subtract 4on both sides)

3x/3=15/3 (divide both sides by 3).

x=5

Substitute the obtained value of x to verify if it’s true.

3x+4=19

3(5)+4=19

15+4=19

19=19

LHS=RHS.

The obtained value after evaluating is x=5.

Question. Evaluate the equation 17 = z-(-9). And find the value of z.

Given: An equation17=z−(−9).

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

17=z−(−9) ( use keep-change-change )

17=z+9

17−9=z+9−9 & ( subtract 9 on both sides )

z=8.

Substitute the obtained value of z to verify if it’s true.

17=8−(−9)

17=8+9

17=17

LHS=RHS.

The obtained value for the equation is z=8.

Question. Evaluate the equation 15 = 2 + 4 – d. And find the value of d.

Given: An equation15=2+4−d.

To find the Value of d.

Evaluate to get the answer.

Let us consider and solve the equation ford.

15=2+4−d

15−2=2+4−d−2 (Subtract 2 on both sides )

15−2−4=4−d−4 (Subtract 4 on both sides )

9=−d (combine like terms )

−1×9=−d×−1 ( multiply −1 on both sides )

d=−9.

Substitute the obtained value of d to verify if it’s true.

15=2+4−d

15=2+4−(−9)

15=6+9

15=15

LHS=RHS.

The obtained value after evaluation is d=−9.

Question. Evaluate the equation f/4 – 5 = -9. Find the value of f.

Given: An equation f/ 4−5=−9.

To find the Value of f.

Evaluate to get the answer.

Let us consider and solve the equation for f.

f/4−5=−9

f/4−5+5=−9+5 (add 5 on both sides )

4×f/4=−4×4 ( multiply both sides by 4)

f−16

Substitute the obtained value to verify if it’s true.

4−5=−9

−4−5=−9

−9=−9

LHS = RHS.

The obtained value after evaluation is f=−16.

Question. Evaluate the equation q + (-5)/3 = 8. Find the value of q.

Given: An equation q+(−5)/3=8.

To find the Value of q.

Evaluate to get the answer.

Let us consider and solve the equation for q.

q+(−5)/3=8

3×q+(−5)/3=8×3 (multiply both sides by 3).

q+(−5)+5=24+5 (add 5 on both sides )

q=29.

Substitute the obtained value of q to verify if it’s true.

q+(−5)/3=8

29+(−5)/3=8

24/3=8

8=8

LHS=RHS.

Question. Evaluate the equation 5x + 3x = 28. Find the value of x.

Given: An equation 5x+3x=28.

To find the Value of x.

Evaluate to get the answer.

Let us consider and solve the equation for x.

5x+3x=28 (combine like terms)

8x=28

8x/8 =28/8 (divide both sides by 8)

x=28/8 ( lowest term )

x=7/2

Substitute the obtained value of x to verify if it’s true.

5(7/2)+3(7/2)=28

35/2+21/2=28

56/2=28

28=28

LHS= RHS.

The value obtained after evaluating is x=7/2.

Question. Evaluate the equation 3(z+7) = 21. Find the value of z.

Given: An equation 3(z+7)=21.

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

3(z+7) =21 (divide both sides by 3).

3(z+7)=21/3

z+7−7=7−7 (subtract 7 on both sides )

z=0.

Substitute the obtained value of z to verify if it’s true.

3(z+7)=21

3(0+7)=21

3(7)=21

21=21

LHS=RHS.

The obtained value after evaluating is z=0.

Question. Evaluate the equation -4(z-12) = 42. Find the value of z.

Given: An equation−4(z−12)=42.

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

−4(z−12)=42

−4(z−12)/−4=42/−4 (divide both sides by 4)

z−12+12=21/−2+12 (add 12 on both sides)

z=3/2

Substitute the obtained value of z to verify if it’s true.

−4(z−12)=42

−4(3/2−12)=42

−4(−21/2)=42

42=42

LHS=RHS.

The obtained value after evaluating is z=3/2.

Question. Evaluate the equation 33 = 12r – 3(9-r). Find the value of r.

Given: An equation33=12r−3(9−r).

To find Value of r.

Evaluate to get the answer.

Let us consider and solve the equation for r.

33=12r−3(9−r) (distribute 3)

33=12r−27+3r

33=15r−27 (combine like terms)

33+27=15r−27+27 (add 27 on both sides)

60/15=15r/15 (divide both sides by 15)

4=r

Substitute the obtained value of r to verify if it’s true.

33=12r−3(9−r)

33=12(4)−3(9−4)

33=48−3(5)

33=48−15

33=33

LHS = RHS.

The obtained value after evaluating is r=4.

Question. Evaluate the expression for odd integers 2n + 1. Find the consecutive odd integers summing up to 63.

Given: Expression for odd integers 2n+1.

To find The 3 consecutive odd integers summing up to 63.

Evaluate to get the answer.

Since odd numbers can be represented by the expression 2n+1, then, adding 2 to the expression will make us obtain the next odd integer and so on.

Let 2n+1 be the first odd integer,

2n+3 is the second odd integer,

2n+5 be the third odd integer, then:

Solve for the three consecutive off integers that have a sum of 63.

(2n+1)+(2n+3)+(2n+5)=63

6n+9−9=63−9 (subtract 9 on both sides )

6n/6=54/6 (divide both sides by 6)

n=9

Substitute the value of n.

2n+1=2(9)+1=19

2n+3=2(9)+3=21

2n+5=2(9)+5=23

The three consecutive numbers obtained are 19,21,23.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.1

May 23, 2024September 27, 2023 by Marksparks

Read the table and find the conjecture and three quadrilaterals that are different.

Given: The table

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations table 1

To find The conjecture and three quadrilaterals that are different.

Evaluate to get the required result.

$Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.1$

Conjecture: Using the table from part Exploration 1, we can say that the sum of the angle measures of a quadrilateral is equal to 360°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations table 2

Upon drawing three other quadrilaterals and measuring their angles, we can conclude that the conjecture we have is TRUE

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 1

Conjecture: Using the table from part Exploration 1, we can say that the sum of the angle measures of a quadrilateral is equal to 360°.
Upon drawing three other quadrilaterals and measuring their angles, we can conclude that the conjecture we have is TRUE.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 2

Evaluate to get the required result and find the value of x.

Given: The angle

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 1

To find: The value of x

Evaluate to get the required result.

Conjecture: The sum of the angle measures of a quadrilateral is equal to 360°.

Add the angle measurements for each quadrilateral and equate it to 360 to obtain the value of x.

85+100+80+x=360

265+x=360

Combine like terms,

265+x−265=360−265

Subtract 265 on both sides,

x=95°

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 2

The value of x is 95°

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 3

Given: The angle

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 4

To find: The value of x

Evaluate to get the required result.

Add the angle measurements for each quadrilateral and equate it to 360 to obtain the value of x.

x+78+72+60=360

x+210=360

Combine like terms

x+210−210=360−210

Subtract 210 on both sides

x=150°

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 5

The value of x is x=150°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 6

Given: The angle

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 7

To find: The value of x

Evaluate to get the required result.

Add the angle measurements for each quadrilateral and equate it to 360 to obtain the value of x

90+30+x+90=360

x+210=360

Combine like terms

x+210−210=360−210

Subtract 210 on both sides

x=150°

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 8

The value of x is x=150°.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 Solving Linear Equations image 9

One real-life example of using simple equations is when you are using your money.

When you buy something from a store, you will give your money to pay for it.

If the price is less than your payment, you will have a change.

You can compute your change by subtracting your money from the price of the item.

If you buy an item for a certain quantity, for example, 3 packs of biscuits, the price will be multiplied by the quantity.

The use of simple equations to solve real-life problems is stated.

Tear off the four corners of the quadrilateral and rearrange them to affirm the conjecture Evaluate to get the required result. Find the conjecture.

Given: Tear off the four corners of the quadrilateral and rearrange them to affirm the conjecture

To find The conjecture.

Evaluate to get the required result.

Consider that the sum of the angle measures of a quadrilateral is equal to 360 degrees.

The same goes for a circle.

Upon cutting off the corners of the quadrilaterals and rearranging them, it will form a circle.

This proves that the angles will still sum up to 360, proving the conjecture.

$Big Ideas Math Algebra 1Student Journal1st EditionChapter 1 Solving Linear image 1$

The angles will still sum up to 360, proving the conjecture.

$Big Ideas Math Algebra 1Student Journal1st Edition Chapter 1 Solving Linear image 2$

Evaluate the equation w + 4 = 16 and find the equation is true or not.

Given: The equation w+4=16

To find: If the equation is true or not.

Evaluate to get the required result.

Let us solve for w.

w+4=16

w+4−4=16−4 (Subtract 4 on both sides)

w=12

Substitute the obtained value back to the original equation to verify the solution.

w+4=16

12+4=16

16=16

The equation is true w=12.

Evaluate the equation -15 + w = 6. Find the value of the variable.

Given: An equation−15+w=6

To find Value of the variable.

Evaluate to get the answer.

Using the equation, we solve for w

−15+w=6

−15+w+15=6+15 (Add 15 on both sides)

w=21

Substitute the obtained value back to the original equation to verify the solution.

−15+w=6

−15+21

LHS=RHS.

The obtained value for the equation is w=21.

Evaluate the equation -2 = y – 9. Find the value of the variable.

Given: An equation−2=y−9

To find Value of the variable.

Evaluate to get the answer.

Using the equation, we solve for y.

−2=y−9

−2+9=y−9+9 (Add 9 on both sides)

y=7.

Substitute the obtained value back to the original equation to verify the solution.

−2=y−9

−2=7−9

−2=−2

LHS= RHS.

The obtained value for the equation is y=7.

Evaluate the equation 3 = q/11. Find the value of variable.

Given: An equation 3=q/11.

To find Value of the variable.

Evaluate to get the answer.

Using the equation, we solve for.

3=q/11

3.11=q/11.11 (Multiply 11 on both sides)

33=q.

Substitute the obtained value back to the original equation to verify the solution.

3=q/11

3=33/11

3=3

LHS=RHS.

The obtained value for the equation is q=33.

Evaluate the equation n/-2 = -15. Find the value of the variable.

Given: An equation n/−2 =−15.

To find Value of the variable.

Evaluate to get the answer.

Using the equation, we solve for n.

n/−2=−15

n/−2.(−2)=−15.(−2) (Multiply -2 on both sides)

n= 30.

Substitute the obtained value back to the original equation to verify the solution.

n/−2=−15

30/−2=−15

−15=−15

LHS=RHS.

The obtained value for the equation is n=30.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1

May 23, 2024September 23, 2023 by Marksparks

Question. Evaluate the value -1+(-3). Find the Addition or Subtraction.

Given: The value −1+(−3)

To Find Addition or subtraction.

Evaluate to get the required result.

Since both numbers have the same sign, we can simply add the numbers and keep the negative sign.

−1+(−3)

−4

We get the value −4 by adding both numbers.

$Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1$

Question. Evaluate the value 5-(-2). Find the Addition or subtraction.

Given: The value 5−(−2)

To Find Addition or subtraction.

Evaluate to get the required result.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

When subtracting numbers, you can do the method add the opposite or the keep-change-change method.

Keep the sign of the first number,

5−(−2)

Change the sign of the expression,

5+(−2)

Change the sign of the second number,

5+2

The answer is 7.

Question. Evaluate the value -4-7. Find the Addition or Subtraction.

Given: The value −4−7

To Find Addition or subtraction.

Evaluate to get the required result.

When subtracting numbers, you can do the method add the opposite or the keep-change-change method.

Keep the sign of the first number,

−4−7

Change the sign of the expression,

−4+7

Change the sign of the second number and add them,

−4+(−7)

We get

-11

The answer is −11.

Question. The sum of two integers is -6 find two pairs of integers whose sum is -6.

Given: The sum of two integers is −6

To Find Two pairs of integers whose sum is −6

By using the basic math operations, two such pairs can be concluded.

The first pair is −3,−3 whose sum is −6 and it is verified below:

−3+(−3)=−6

−6=−6

The second pair is −4,−2 whose sum is −6 and it is verified below:

−4+(−2)=−6

−6=−6

The two pairs of integers whose sum is −6 are −3,−3, and −4,−2.

Question. In a city, the record monthly high temperature for March is 56°F and record monthly for low temperature is -4°F. Find the range of temperatures for the month of March. Subtract the highest and lowest temperatures to get the range.

Given: In a city, the record monthly high temperature for March is 56°F and record monthly for low temperature is −4°F.

To Find The range of temperatures for the month of March.

Subtract the highest and lowest temperatures to get the range.

The range of the temperatures for the month of March is calculated below:

56−(−4)=56+4

⇒ 56+4=60

The range of temperature for the month of the March is 60°F.

Question. Find the evaluated result of the -8.(-5) expression. Multiply the provided integers and evaluate the result.

Given: An expression, −8. (−5)

To Find The evaluated result of the above expression.

Multiply the provided integers and evaluate the result.

The integers are to be simply multiplied to get their product.

⇒ −8. (−5)=40

The evaluated result of the expression −8. (−5) is 40.

Question. Find the evaluated result of the -14÷2 expression. Divide the integers to obtain the required result.

Given: An expression, −14÷2

To Find The evaluated result of the above expression.

Divide the integers to obtain the required result.

Division simply means breaking a number into parts. In this case, we have to divide 14 in 2 parts.

−14÷2=−14/2

=−7

The evaluated result of the expression −14÷2 is −7.

Question. Find the evaluated result of the -30÷(-3) expression. Divide the integers to obtain the required result.

Given: An expression, −30÷(−3)

To Find The evaluated result of the above expression.

Divide the integers to obtain the required result.

Division simply means breaking a number into parts. In this case, we have to divide −30 in −3 equal parts.

⇒ −30÷(−3)=−30/−3

⇒ −30÷(−3)=10

The evaluated result of the expression −30÷(−3) is 10.

Question. Find the two pairs using basic math operations. Two pairs of integers whose product evaluates to -20.

Given: A product of two integers is −20

To Find Two pairs of integers whose product evaluates to −20

Find the two pairs using basic math operations.

The first pair is −5,4 whose product is −20 and it is verified below:

−5×4=−20

−20=−20

The second pair is −2,10 whose product is −20 and it is verified below:

−2×10=−20

−20=−20

The two pairs whose product evaluates to −20 are −4,5 and −2,10.

Question. A football team loses 3 yards in 3 consecutive plays. Find total yardage gained.

Given: A football team loses 3 yards in 3 consecutive plays.

To Find Total yardage gained.

Evaluate the result by multiplying the losses to plays.

Multiply the number of yards gained (which will become a negative number signifying loss) by the number of plays to get the total yardage gained.

−3×3=−9

The total yardage gained is −9.

Go Math! The Practice of Statistics Chapter 6 Random Variables Solutions Introduction Answer Key

April 30, 2024September 23, 2023 by Marksparks

Go Math! The Practice of Statistics for the AP Exam 5th Edition Chapter 6 Random Variables Answer Key

Introduction Chapter 6: Random Variables Solutions

In this chapter, students will study discrete and continuous random variables, transforming and combining random variables, and binomial and geometric random variables. It also consists of a chapter review, exercise review, and records exercise test. By the end of the section, students will have the ability to:

1. Compute changes in the usage of the chance distribution of a discrete random variable.

2. Calculate and interpret a discrete random variable’s suggestion (anticipated value).

3. Compute and interpret the standard deviation of a discrete random variable.

4. Compute changes in the usage of the chance distribution of positive non-stop random variables.

5. Describe the consequences of remodeling a random variable by including or subtracting a steady and multiplying or dividing through a steady.

6. Find the suggested and general deviation of the sum or distinction of unbiased random variables.

7. Determine whether or not the situations for using a binomial random variable are met.

8. Compute and interpret chances regarding binomial distributions.

9. Calculate the suggested and general deviation of a binomial random variable. Interpret those values in context.

10. Find chances regarding geometric random variables.

11. When applicable, use the normal approximation to the binomial distribution to calculate chances.

Learn about calculating Discrete and Continuous random variables.

Learn about solving Transforming and combining random variables.

Learn about computing Binomial and Geometric random variables.

The Introduction exercise section provides students with an overview of the concepts and topics covered throughout the chapter. It prepares them to kickstart their learning process.

From simple to pretty complicated, statistical calculations never depend on mere guessing. Instead, it involves knowledge of acquiring and operating data. For example, if people are asked to comment on babies’ health at birth, they will give wrong deductions if they opt for guesswork. Hence, coming to a correct conclusion involves knowing probability and variables, handling numerical data sets, etc. In this chapter, students will learn the concepts of random variables, probability distributions, and types of random variables.

It constitutes several compelling examples that help learners connect random variables with real-life situations. Moreover, it assists design practical sessions for learning random variables and probabilities. Each section ends with a summary and an exercise. It helps in quick memorization and evaluation of acquired knowledge.

Understand the concept of discrete variables.

Gain knowledge of the expected value of a discrete random variable.

Learn the concept of continuous random variables.

Students will learn to calculate discrete and continuous random variables in the exercises section. Furthermore, it will equip them with the knowledge of effective formulas and terms. To determine the calculation of standard deviation and variance of the discrete random variable, let x be the discrete random variable have a probability distribution. Value sets constitute x1,x2,x3 and so on with the probability of having p1,p2,p3 an

$Go Math The Practice of Statistics for the AP Exam 5th Edition Chapter 6 Random Variables Answer Key 8$

Learn about probability distribution.

Understand how to operate calculations of expected value.

Gain concepts of mathematical operations related to discrete and random variables.

Context of the probability distribution, standard deviation, and expected value In this exercise section, mathematical problems related to standard deviation and random variables use practical contexts like a game, housing, life insurance, and pregnancy length—these aid in connecting real-life situations with statistical calculations. Students also gain concepts of mathematical operations related to discrete and random variables under this exercise.

Identify the different linear transformations.

Learn the concepts of combining random variables.

Understand the combination of a normal random variable.

This section focuses on the combination of random variables that constitutes many statistical operations that require the handling of more than one random variable. These exercises involve the sum and difference of random variables. Then, students will learn about the transformation of random variables including subtracting/adding and multiplying/dividing. This transformation produces no change in the shape of a probability distribution.

Finally, they will understand that a combination of normal random variables includes sums and differences from normal random variables. The sum of random variables involves adding independent normal random variables. In addition, they will learn how histograms are generated by adding and subtracting random variables that can be studied to understand the shape, center, and spread of distributions.

Learn to calculate the mean and standard deviation.

Operate statistical problems using knowledge of variance.

Apply the knowledge on the combination/transformation of random variables.

From buying stocks to finding the distribution of resistance in electronic circuits, many practical situations in life require statistical calculations. These involve calculating insurers’ income, finding the possible values of stock, solving resistance distribution in electric circuits, and calculating stock returns. This exercises section incorporates working knowledge of probability distribution, random variables, standard deviation, and variance.

Understand binomial random variables.

Acquaint knowledge of the mean and standard deviation of the binomial distribution.

Understand geometric random variables.

Gain knowledge of binomial distributions in statistical sampling.

The exercises in this section discuss binomial distribution and offer a useful approximation if the sample does not exceed 10% of the population. Hence, the 10% condition in binomial distributions. The expected number of trials of successes and failures is a minimum of 10. There are plenty of examples of geometric settings. Games provide good examples, such as rolling a pair of dice for getting doubles or trying to score three-point shots in basketball until making one.

Apply knowledge of variables to real-life situations.

Learn to calculate the standard deviation and mean.

Understand the concepts of the random selection process.

In this section, students will be applying statistical knowledge to the law courtroom process. The Case Closed exercises section provides an exciting backdrop to mentioned conditions. It gives the question of assessing the distribution of random variables. Also, it gives two different conditions associated with the same event and shows additional conceptual questions for the two. It helps students apply knowledge of variables to real-life situations, memorize to calculate the standard deviation and mean and comprehend the concepts of the random-selection process.

Understand what binomial distribution is.

Learn and review concepts of geometric settings.

Apply the knowledge of binomial and geometric variables.

Application of binomial and geometric variables to real-life conditions The exercises section contains many practical situations like playing games, airport security, and the impacts of smoking. Moreover, it constitutes an application of Benford’s law for giving probability operations. Several problems require the simultaneous application of binomial and geometric variables. It helps students learn and revise concepts of geometric settings and understand what binomial distribution is.

Apply the knowledge of probability and random selection.

Understand concepts of standard deviation.

Acquaint yourself working with normal distribution.

Students will review the probability, the mean, and the standard deviation of weight calculation. The Frappy! exercises section starts with the story of the production of homemade potato chips. Then, based on the information provided, students must calculate the probability of weight. Next, students are given calculations of mean weight and standard deviation based on changes in the values of the given information. It helps students apply the understanding of probability and random selection and comprehend concepts of standard deviation.

Re-examine every topic of the chapter.

Review learned methods associated with random variables.

Implement learned practices of statistical operations.

The Review Exercises section evaluates the depth of knowledge learned in previous sections. It gives problems related to roulette, games, balancing of coins, and keno. It helps in interpreting statistical outcomes and helps strengthen statistical knowledge along with reviewing learned methods associated with random variables and Implementing learned practices of statistical operations.

Learn about the concepts of random variables.

Get to know the concepts of probability, variables, and probability distribution.

Apply developed knowledge.

In this section, students will review and practice random variables, standard deviation, probability distribution, binomial distribution, and previously learned concepts. The exercises in the Statistics Practice Test section help them conduct extra practice. In addition, it strengthens their ideas of the binomial, geometric distribution, probability calculations, and interpretation of the standard

Go Math! The Practice of Statistics Chapter 7 Sampling Distributions Solutions Introduction Answer Key

April 30, 2024September 22, 2023 by Marksparks

Go Math! The Practice of Statistics for the AP Exam 5th Edition Chapter 7 Sampling Distributions Solutions Introduction Answer Key

Introduction Chapter 7: Sampling Distributions Solutions

In this chapter, students will find out about What Is a Sampling Distribution, Sample Proportions, and Sample Mean. It also consists of a Chapter review, exercise review, and records exercise test. By the end of the section, students will have the ability to:

1. First, distinguish between a parameter and a statistic.

2. Use the sampling distribution of a statistic to assess a declare approximately a parameter.

3. Distinguish the distribution of a populace, issuance of a pattern, and the sampling distribution of a statistic.

4. Specify whether or not now no longer a statistic is an independent estimator of a populace parameter.

5. Describe the connection between pattern length and the variety of a statistic.

6. Find the suggested and trendy deviation of the sampling distribution of a pattern percentage p^. Check the 10% circumstance earlier than calculating sp^.

7. Determine if the sampling distribution of p^ is about Normal.

8. If appropriate, use a Normal distribution to calculate chances concerning p^.

9. Find the suggested and trendy deviation of the sampling distribution of a pattern meant x. Check the 10% circumstance earlier than calculating sigma x

10. Explain how the sampling distribution of x is suffering from the form of the populace distribution and the pattern length.

11. If appropriate, use a Normal distribution to calculate chances concerning x.

The sampling distribution is a statistical process of interpreting the data produced by random sampling. As the random samples have an element of chances, the laws of probability are used to certify the amount of the chances conclusions drawn from sampling distributions will hold valid.

In this chapter, students shall learn the following:

The basic concept of sampling distributions

Description of sampling distributions

Sampling distribution of sample proportion

Sampling distribution of the sample means

Central Limit theorem for the sampling distribution of sample mean

Learn about the basic concept of sampling distributions.

Get to know the description of sampling distributions.

Understand parameters of statistics.

The sample variance is a measure of the spread of a sampling distribution.

The shape of a sampling distribution may be left-skewed, right-skewed, roughly symmetric, and even approximately normal.

This exercises section will teach will students about sampling distribution. They will begin the learning journey by describing sampling distributions. To trust a statistic for estimating a parameter, its sampling distribution must be analyzed for shape, center, and spread. Then, they will understand what the center of the sampling distribution is. If the mean of a sampling distribution is equal to the parameter’s estimated value, then such statistics is an unbiased estimator. Otherwise, it is a biased estimator. Lastly, they will observe the spread of the sampling distribution in this section.

$Go Math The Practice of Statistics for the AP Exam 5th Edition Chapter 6 Random Variables Answer Key 10$

SLearn to solve problems based on the sample proportion.

Understand sampling distribution of sample proportion.

Perform activities related to the topics above.

The exercises section will focus on understanding the sample proportion of successes that are closely related to x The sampling distribution and mean of sample proportion measure the proportion of success. The sampling distribution of proportion follows the binomial probability law if the random sample of ‘n’ is obtained with replacement.

Learn about the sampling distribution of the sample mean.

Understand the center and spread of the sample mean.

Solve problems based on the sample mean.

In the section, students will begin learning about the sample mean, which is an average of a data set. The sample mean can be used to calculate the central tendency, standard deviation, and variance. Furthermore, students can apply the sample mean to many uses, including calculating population averages. The following results are obtained from the exercises of the sampling distribution of the sample mean.

Lastly, they will understand that the sample mean has a sampling distribution that is (approximately) normal with a mean equal to the population mean for X and a standard deviation equal to a standard deviation of X divided by the square root of the sample size.

Implement the concept of sampling distribution using a case study.

Solve problems based on the case study.

Understand the concept of the sampling distribution.

Details of the case study

The chapter-opening Case Study described a sampling distribution problem in “Building Better Batteries”. Here, a randomly selected battery sample is drained to find the hours till it works and the data is provided. The data mean and standard deviation will give the center and spread of the distribution. It is an unbiased estimator if the sample mean is near the population (17 givens). If the standard deviation value is small, its spread is small and will follow a normal distribution.

Implement the concept of sampling distribution using a case study.

Gain insight into case study-based questions.

Solve questions based on acquired knowledge.

The sample variance is a measure of the spread of a sampling distribution.

The shape of a sampling distribution may be left-skewed, right-skewed, roughly symmetric, and even approximately normal.

Learn about the sampling distribution of sample proportion.

Study the shape, center, and spread of sample proportion.

Solve problems on the above topics.

The exercises of the sampling distribution of sample proportion are studied like this. The following results can be observed from the sampling distribution p^ by shape, which is roughly symmetric and somewhat bell-shaped (Approximate normal distribution). It seems to depend on n and, the center, the mean of p^ is quite close to p (Unbiased estimator), and the spread, standard deviation of p^ is very small. It depends on n and p.

Learn to solve problems based on the sample proportion.

Understand sampling distribution of sample proportion.

Perform activities related to the topics above.

Learn about the sampling distribution of the sample mean.

Understand the center and spread of the sample mean.

Solve problems based on the sample mean.

Implement the concept of sampling distribution using a case study.

Solve problems based on the case study.

Understand the concept of the sampling distribution.

Details of the case study
The chapter-opening Case Study described a sampling distribution problem in “Building Better Batteries”. Here, a randomly selected battery sample is drained to find the hours till it works and the data is provided. The data mean and standard deviation will give the center and spread of the distribution. It is an unbiased estimator if the sample mean is near the population (17 givens). If the standard deviation value is small, its spread is small and will follow a normal distribution.

Implement the concept of sampling distribution using a case study.

Gain insight into case study-based questions.

Solve questions based on acquired knowledge.

In the exercises section, students will solve problems based on sampling distributions with the help of reliable case studies and understand the background of each problem encountered within the said case study. The exercises are centered around the distribution samples, sample means, and the formula needed to solve them efficiently. It is further explained through the existing knowledge about distributions and samples from earlier chapters.

Learn to implement the concept of sampling distribution in a situation.

Solve problems using the spread, center, and shape of the distribution.

Understand random sample.

Implementation of experimental study
This Frappy section is modeled after actual AP ® Statistics exam free-response questions. It contains inquiries related to the sampling distribution. In the question, a population of 2500 students is given for which population mean, and population standard deviation are known. Then, a random sample of 50 students is selected for which we have to draw the sampling distribution. It can be done after determining the mean and standard deviation of the sample, which will provide information about the center and spread, respectively. After which, everything can be solved with the method discussed above.

Practice on some detailed problems to brush the concepts you have learned through the journey of this chapter.

Solve problems on key topics of the lesson.

Under in-depth the concept of population and sample.

Some of the brief concepts studied under Review Exercises are where the parameter defines the character of the population, and statistics determines the characteristics of the sample. Students must solve problems on how to draw a sample distribution if its center, spread, and shape are known, which is dealt with while analyzing a problem. They will also revise how sampling distribution of mean and sample proportion can be drawn similarly.

Practice on some objective and numerical problems to brush the concepts you have learned through the journey of this chapter.

Understand the sampling distribution of the sample mean.

Solve problems on the above concept.

This extra AP® Statistics Practice Test exercises section contains multiple-choice questions and numerical questions associated with the basic concept of sampling distributions, the description of sampling distributions, the sampling distribution of sample proportion, the sampling distribution of the sample mean, and the Central Limit theorem for the sampling distribution of the sample mean that are already discussed above

Practice on some objective and detailed problems to brush the concepts you have learned through the journey of three chapters.Solve problems on discrete and continuous random variables. Understand the Central Limit theorem for the sampling distribution of the sample mean.

The Cumulative AP® Practice Test 2 section contains multiple-choice questions and numerical questions associated with randomness, probability, and simulation, probability rules, conditional probability, and independence, discrete and continuous random variables, transforming and combining random variables, binomial and geometric random variables, the basic concept of sampling distributions, description of sampling distributions, sampling distribution of sample proportion, sampling distribution of the sample mean, and central limit theorem for the sampling distribution of the sample mean that are already discussed in the chapters altogether.

Go Math! Practice Fluency Workbook Grade 6 Chapter 6 Representing Ratios and Rates Exercise 6.2 Answer Key

May 25, 2024September 6, 2023 by Marksparks

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6: Representing Ratios and Rates

Question. David drove 135 miles in 3 hours. Find the unit rate.

Given: David drove 135 miles in 3 hours.

To find: We have to find the unit rate.

Divide the distance by the time to get the unit rate.

David drove 135 miles in 3 hours.

Unit rate = 135/3

Unit rate = 45 miles/hour.

The unit rate of David is 45 miles per hour.

Question. Three medium apples have about 285 calories. Find the unit rate.

Given: Three medium apples have about 285 calories.

To find: We have to find the unit rate.

Divide the calories by the number of apples to get the unit rate.

Three medium apples have about 285 calories,

Unit rate = 285/3

Unit rate = 95 calories/apple.

The unit rate is 95 calories/apple.

$Go Math! Practice Fluency Workbook Grade 6 Chapter 6 Representing Ratios and Rates Exercise 6.2 Answer Key$

Question. A 13-ounce package of pistachios costs $5.99. Find the unit rate.

Given: A 13-ounce package of pistachios costs $5.99.

To find: We have to find the unit rate.

Divide the number of ounces of packages by the total cost of pistachios to get the unit rate.

A 13-ounce package of pistachios costs $5.99.

Unit rate = 5.99/13

Unit rate = 0.4608 $/ ounce of package.

The unit rate is $0.4608 per ounce of a package of pistachios.

Question. Morgan’s favorite spaghetti sauce is available in two sizes pint and quart. Each size and its price are shown in the table. Find the unit rate to the nearest cent per ounce for pint size.

Given: Morgan’s favorite spaghetti sauce is available in two sizes: pint and quart. Each size and its price are shown in the table.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 59$

To find: We have to find the unit rate to the nearest cent per ounce for pint size.

Divide the price of the spaghetti sauce by its quantity to get the unit rate.

The “pint” size has 16 quantities and its cost is $3.98

The unit rate of pint-size =3.98/16

The unit rate of pint-size =0.25 dollars/ounce.

The unit rate to the nearest cent per ounce for pint-size is $ 0.25 /ounce.

Question. Morgan’s favorite spaghetti sauce is available in two sizes pint and quart. Each size and its price are shown in the table. If a coupon offers $1.00 off the 16 ounce size. Which size is the better buy then?

Given: Morgan’s favorite spaghetti sauce is available in two sizes: pint and quart. Each size and its price are shown in the table.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 59$

To find: If a coupon offers $1.00 off the 16-ounce size. Which size is the better buy then?

Just subtract the coupon amount from the pint-size and then recalculate the unit rate.

After that compare both unit rates and decide on the better one.

Now the initial price for the pint size is $3.98, then if a coupon is applied to it, then its cost will be reduced by $1.00.

The new price will be =3.98−1.00 = 2.98

So the unit rate of pint size will be = 2.98/16 = 0.18 and the unit rate of quart size was $0.18/ounce.

As we can see here, after applying the coupon their unit rate became equal so we can say that both are the better choices as there is no difference between the unit price of both sizes.

After applying the coupon their unit rate became equal so we can say that both are the better choices as there is no difference between the unit price of both sizes.

Question. A 24-ounce box of cornflakes costs 4.59. Find the unit rate to the nearest cent per ounce.

Given: A 24-ounce box of cornflakes costs 4.59.

To find: We have to find the unit rate to the nearest cent per ounce

Divide the cost of the box of cornflakes by the ounces to get the unit rate per ounce.

An A24-ounce box of cornflakes costs $4.59

Dividing to find the unit rate, we get

Unit rate = 4.59/24

Unit rate = $0.191/ounce.

For part (a), the unit rate to the nearest cent per ounce is $0.191/ounce.

Question. Karyn proof reads 15 pages in 2 hours for $40. Find her proofreading rate in pages per hour.

Given: Karyn proofreads 15 pages in 2 hours for $40

To find: We have to find her proofreading rate in pages per hour.

Just divide the number of pages by the number of hours given.

Karyn proofreads 15 pages in 2 hours for $40.

Her proofreading rate will be = 15/2 =7.5

Her proofreading rate is 7.5 pages per hour.

Karyn proofreads 7.5 pages per hour.

Question. Jack shells 315 peanuts in 15 minutes. Find the unit rate.

Given: Jack shells 315 peanuts in 15 minutes.

To find: We have to find the unit rate.

Just divide the number of peanuts shelled by Jack in a given number of minutes.

Jack shells 315 peanuts in 15 minutes.

Now, let’s divide the number of peanuts by the number of minutes, we get

Unit rate = 315/15

Jack shells 315 peanuts in 15 minutes.

Now, let’s divide the number of peanuts by the number of minutes, we get

= \(\frac{315}{15} \times \frac{15}{15}[latex]

= [latex]\frac{21}{1}\)

Unit rate = 21/1

This means the unit rate = is 21 peanuts per minute.

Jack shells 21 peanuts per minute.

Question. Sharmila received 81 texts in 9 minutes. Find the unit rate.

Given: Sharmila received 81 texts in 9 minutes.

To Find: We have to find the unit rate.

Just divide the number of texts received by the number of minutes.

Sharmila received 81 texts in 9 minutes.

Now, let’s divide the number of texts by the number of minutes in order to find out the unit rate.

Unit rate = 81/9

Make the denominator “1”, by multiplying and dividing the above expression by 9, we get

Unit rate =81/9 =9/1

This means the unit rate = 9 texts per minute.

Sharmila receives 9 texts per minute.

Question. Karim reads 56 pages in 2 hours. Find the unit rate.

Given: Karim reads56 pages in 2 hours

To find: Unit rate

We want to know the pages Karim reads per hour so we set up a ratio with hours in the denominator.

The total pages go in the numerator.

So the fraction is 56/2

To make denominator 1 divide both the numerator and denominator with 2.

To find pages read by Karim in 1 hour we will divide the total pages read by him by the hour.

So fraction is

→56pages/2hours =56/2

Divide numerator and denominator by 2 ,

$\frac{56}{2} \div \frac{2}{2}=\frac{28}{1}$

Which means 28pages/hour =28 pages per hour

Karim read 28 pages per hour.

Question. The weight of whole wheat bread is 16 oz for $2.24. The weight of Pita bread is 20 oz for $3.60 and the weight of 7-grain bread is 16 oz for $2.56. Find the best buy.

Given: The weight of whole wheat bread is 16oz for $2.24. The weight of Pita bread is 20oz for $3.60 and the weight of 7-grain bread is 16oz for $2.56.

To find: best buy

To find the best buy find a unit rate for each type of bread per oz that is divided the cost by weight. The unit rate which has the lowest price is the best buy of bread.

For whole wheat, the bread weight is 16oz and the cost is $2.24.

The unit rate of whole wheat bread is 2.24/16 =0.14

The unit rate of whole wheat bread is 0.14 $ per oz.

For Pita, the bread weight is 20oz and the cost is $3.60.

The unit rate of Pita bread is 3.60/20 =0.18

The unit rate of Pita bread is 0.18 $ per oz.

For 7-grain bread, the weight is 16oz and the cost is $2.56.

The unit rate of 7-grain bread is 2.56/16 =0.16

The unit rate of 7-grain bread is 0.16 $ per oz.

By comparing unit rates we can say that 0.14 $ per oz is the lowest unit rate which is whole wheat bread.

The best buy is whole wheat bread which weighs 16oz and cost $2.24.

Go Math Answer Key

Go Math! Practice Fluency Workbook Grade 6 Chapter 6 Representing Ratios and Rates Exercise 6.1 Answer Key

May 25, 2024September 1, 2023 by Marksparks

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6: Representing Ratios and Rates

Question. Read the table and find the ratio for lion and elephant by using the method of ratio and proportion.

Given table is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 1$

To find the ratio for lion and elephant.

Using the method of ratio and proportion.

The number of lions in the zoo is 9 and the number of elephants in the zoo is 12

now to find the ratio of a lion to elephant,

ratio of lion to elephant = lion / elephant,

Now substitute all the values in the above formula,

= $\frac{9}{12}$

$Go Math! Practice Fluency Workbook Grade 6 Chapter 6 Representing Ratios and Rates Exercise 6.1 Answer Key$

simplify,

= $\frac{3}{4}$

The ratio of a lion to elephant is 3:4

Question. Read the table and find the ratio of giraffes and otters by using the method of ratio and proportion.

Given table is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 3$

To find the ratio of giraffes and otters.

Using the method of ratio and proportion.

The number of giraffes in the zoo is 8 and the number of otters in the zoo is 16

now to find the ratio of giraffes to otters,

the ratio of giraffes to otters = giraffes/otters

Now substitute all the values in the above formula,

= $\frac{8}{16}$

simplify,

= $\frac{1}{2}$

The ratio of giraffes to otters is 1:2

Question. Read the table find the ratio of lions and seals by using the method of ratio and proportion.

Given table is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 5$

To find the ratio of lions and seals.

Using the method of ratio and proportion.

The number of lions in the zoo is 9 and the number of seals in the zoo is 10

now to find the ratio,

the ratio of the lions to seals = lion/seals

Now substitute all the values in the above formula,

= $\frac{9}{10}$

The ratio of lions to seals is 9:10

Question. Read the table find the ratio of seals to elephants by using the method of ratio and proportion.

Given table is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 8$

To find the ratio of seals to elephants.

Using the method of ratio and proportion.

The number of seals in the zoo is 10 and the number of elephants in the zoo is 12

now to find the ratio,

the ratio of the seals to elephants = seals/elephants

Now substitute all the values in the above formula,

= $\frac{10}{12}$

simplify,

= $\frac{5}{6}$

The ratio of the seals to elephants is 5:6

Question. Read the table find the ratio of elephants to lions by using the method of ratio and proportion.

Given table is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 6 Representing Ratios and Rates 11$

To find the ratio of elephants to lions.

Using the method of ratio and proportion.

The number of elephants in the zoo is 12 and the number of lions in the zoo is 9 to find the ratio,
the ratio of the elephant to lion = elephant/lion

Now substitute all the values in the above formula,

= $\frac{12}{9}$

simplify,

= $\frac{4}{3}$

The ratio of elephants to lions is 4:3

Question. Find the three equivalent ratios for the 4/3 expression using the method of ratio.

Given table is

The given expression is 4/3

To find the three equivalent ratios for the given expression.

Using the method of ratio.

To find the three equivalent ratios for the given expression, 4/3

This ratio is in reduced form, the equivalent ratio can be found by multiplying both parts of the ratios by the same constants.

$\frac{4 \times 2}{3 \times 2}=\frac{8}{6}$

Likewise, do the same procedure for the next two ratios,

$\begin{aligned}
& \frac{4 \times 3}{3 \times 3}=\frac{12}{9} \\
& \frac{4 \times 4}{3 \times 4}=\frac{16}{12}
\end{aligned}$

The three equivalent ratios for the given ratio of 4/3

are 8:3 or 12:9 or 16:12

Question. Find the three equivalent ratios for the 12/14 expression. Using the method of ratio.

The given expression is 12/14

To find the three equivalent ratios for the given expression.

Using the method of ratio.

The given ratio is 12/14

to write the reduced form of this ratio,

$\frac{6 \times 2}{7 \times 2}=\frac{12}{14}$

so its reduced form is 6/7

multiply and divide the ratio by the same constant,

$\frac{6 \times 3}{7 \times 3}=\frac{18}{21}$

Now do the same procedure for the next two ratios,

$\begin{aligned}
& \frac{6 \times 4}{7 \times 4}=\frac{24}{28} \\
& \frac{6 \times 5}{7 \times 5}=\frac{30}{35}
\end{aligned}$

The three equivalent ratios for the given ratio12 by 14

is 18:21 or 24:28 or 30:35

Question. Find the three equivalent ratios for the 6/9 expression using the method of ratio.

The given expression is 6/9

To find the three equivalent ratios for the given expression.

Using the method of ratio.

The given ratio is 6/9

to find its reduced form,

$\frac{6}{9}=\frac{2 \times 3}{3 \times 3}$

so its reduced form is 2/3

now multiply and divide the ratio by the same constant,

$\frac{2 \times 2}{3 \times 2}=\frac{4}{6}$

Now do the procedure to find the next two ratios,

$\begin{aligned}
& \frac{2 \times 4}{3 \times 4}=\frac{8}{12} \\
& \frac{2 \times 5}{3 \times 5}=\frac{10}{15}
\end{aligned}$

The three equivalent ratios for the given ratio 6/9 are

$\frac{4}{6}$ or 4:6

\(\frac{8}{12}[\latex] or 8:12

[latex]\frac{10}{15}\) or 10:15

Question. Find the three ratios equivalent to the ratio of cats to dogs in a park is 3 to 4 using the method of ratio.

Given the ratio of cats to dogs in a park is 3 to 4

To find the three ratios equivalent to the given ratio.

Using the method of ratio.

The given ratio of cats to dogs is 3:4

now multiply and divide the ratio by the same constant,

= $\frac{3 \times 2}{4 \times 2}$

= $\frac{6}{8}$

Now do the same procedure to find the next two ratios,

$\begin{aligned}
& \frac{3 \times 3}{4 \times 3}=\frac{9}{12} \\
& \frac{3 \times 4}{4 \times 4}=\frac{12}{16}
\end{aligned}$

The three ratios of cats to dogs in a park are

$\frac{6}{8}$ or 6 : 8

$\frac{9}{12}$ or 9 : 12

$\frac{12{16}$ or 12 : 16

Question. Find the three ratios of rainy days to sunny days is 5/7 using the method of ratios.

Given the ratio of rainy days to sunny days is 5/7

To find the three ratios of rainy days to sunny days.

Using the method of ratios.

The ratio of rainy days to sunny days is 5/7

now multiply and divide the ratio by the same constant,

= $\frac{5 \times 2}{7 \times 2}$

simplify,

= $\frac{10}{14}$

Now repeat the same procedure to find the ratios of the next two,

$\begin{aligned}
& \frac{5 \times 3}{7 \times 3}=\frac{15}{21} \\
& \frac{5 \times 4}{7 \times 4}=\frac{20}{28}
\end{aligned}$

The three equivalent ratios of rainy days to sunny days are

$\frac{10}{14}$ or 10 : 14

$\frac{15}{21}$ or 15 : 21

$\frac{20}{28}$ or 20 : 28

Question. Find the three equivalent ratios for the protein to fiber in a granola bar is 9/2 using the method of ratio.

Given the ratio of protein to fiber in a granola bar is 9/2

To find the three equivalent ratios for the protein to fiber in a granola bar.

Using the method of ratio.

The given ratio of the protein to fiber in a granola bar is 9/2

now to multiply and divide the ratio by the same constant,

= $\frac{9 \times 2}{2 \times 2}$

= $\frac{18}{4}$

Now repeat the same procedure to find the next two ratios,

$\begin{aligned}
& \frac{9 \times 3}{2 \times 3}=\frac{27}{6} \\
& \frac{9 \times 4}{2 \times 4}=\frac{36}{8}
\end{aligned}$

The three equivalent ratios of protein to fiber in a granola bar are

$\frac{18}{4}$ or 18 : 4

$\frac{27}{6}$ or 27 : 6

$\frac{36}{8}$ or 36 : 8

Question. Find how many angelfishes are there in the pet shop. Using the method of ratio.

Given the ratio of the clownfish to angelfish is 5:4

and the angelfish to golden fish is 4:3

To find how many angelfishes are there in the pet shop. Using the method of ratio.

The given ratio of clownfish to angelfish is 5:4 and the ratio of angelfish to goldfish is 4:3 there are 60 fishes are there in the pet shop.

from this, it is clear that clownfish=5

angelfish=4

goldfish=3

Using the above values now find the total number of angelfish in the pet shop,

= $\frac{60}{5}$

= 12

multiply by 4

= 12 x 4

= 48

There are 48 angelfishes are there in the pet shop.

Question. Find the ratio between the days of May month to days in a particular year.

Given: Days in May to days in a year.

To find: We have to find the ratio between the days of May month to days in a particular year.

Just find out the number of days present in May month.

Then find the number of days in a year.

Then divide the days of May month by days in a year to get the ratio.

The May month consists of 31 days.

and in a year, there are usually 365 days.

Now if we divide these days, we get

Ratio = $\frac{31}{365}$

Ratio = 31 : 365

The ratio between the days of May month to days in a particular year is 31:365.

Question. Find the ratio between the numbers of sides of a triangle to the number of sides a square.

Given: Sides of a triangle to sides of a square.

To find: We have to find the ratio between the number of sides of a triangle to the number of sides a square has.

Just find out the number of sides of a triangle.

Then find the number of sides of a square.

Then divide them to find the ratio.

As we know, any triangle has “3” sides.

and any square always has “4” sides.

Now let’s take the division of the number of sides of a triangle and the number of sides of a square, we get the ratio as,

Ratio = $\frac{3}{4}$

Ratio = 3 : 4

The ratio of the sides of a triangle to the sides of a square is 3:4.

Question. Find the ratio between 8 triangles to 12 circles.

Given: There are 8 triangles and 12 circles are given.

To find: We have to find the ratio between 8 triangles to 12 circles.

Take division of them.

Then simplify them to obtain different ratios.

We have given 8 triangles and 12 circles.

The ratio for the above statement is 8/12 i.e. 8:12.

Dividing the ratio 8:12 i.e. 8 by 12 by 4 in both numerator and denominator, we get,

$\frac{\frac{8}{4}}{\frac{12}{4}}=\frac{2}{3}$

i.e. the ratio is 2:3.

Dividing the ratio 8:12 i.e. 8 by12 by 2 in both numerator and denominator, we get

$\frac{\frac{8}{2}}{\frac{12}{2}}=\frac{4}{6}$

i.e. the ratio is 4:6.

Multiplying the ratio 8:12 i.e. 8 by12 by 2 in both numerator and denominator, we get

$\frac{8}{12} \times \frac{2}{2}$,

= $\frac{16}{24}$

i.e. the ratio is 16:24.

The three equivalent ratios of 8 triangles and 12 circles are,

2 : 3

4 : 6

16 : 24

Question. Find the three equivalent ratios between 20 pencils and 25 erasers.

Given: There are 20 pencils and 25 erasers are given.

To find: We have to find the three equivalent ratios between 20 pencils and 25 erasers.

Take division of them.

Then simplify them to obtain different ratios.

We have been given 20 pencils and 25 erasers.

The ratio for the above statement is 20:25.

If we multiply and divide the numerator and denominator of the ratio 20/25 by “2”, we get

Ratio =20

= $\frac{20}{25} \times \frac{2}{2}$

= $\frac{40}{50}$

= 40 : 50

Dividing the ratio 40/50, by 10 in both the numerator and denominator, we get

$\frac{\frac{40}{10}}{\frac{50}{10}}=\frac{4}{5}$

i.e. the ratio is 4:5.

Dividing the ratio 40/50, by 5 in both the numerator and denominator, we get

$\frac{\frac{40}{5}}{\frac{50}{5}}=\frac{8}{10}$

i.e. the ratio is 8:10.

Three equivalent ratios of 20 pencils and 25 erasers are

40 : 50

4 : 5

8 : 10

Question. Find the three equivalent ratios between 5 girls and 6 boys.

To find: There are 5 girls and 6 boys are given.

To find: We have to find the three equivalent ratios between 5 girls and 6 boys.

Take division of them.

Then simplify them to obtain different ratios.

We have given 5 girls and 6 boys.

Now divide them to get the different ratios.

Now let’s divide the number of girls and the number of boys, we get

= $\frac{5}{6}$

Ratio = 5 : 6

Now if we multiply and divide this ratio by “2”, we get

$\frac{5}{6} \times \frac{2}{2}$

= $\frac{10}{12}$

Ratio = 10 : 12

Now if we multiply and divide this ratio by “3”, we get

$\frac{15}{18}$

Ratio = 15 : 18

Three equivalent ratios of 5 girls and 6 boys are,

⇒ 5 : 6
⇒ 10 : 12
⇒ 15 : 18

Question. Find the three equivalent ratios between 10 pants and 14 shirts.

Given: There are 10 pairs of pants and 14 shirts given.

To find: We have to find the three equivalent ratios between 10 pants and 14 shirts.

Take division of them.

Then simplify them to obtain different ratios.

We have given 10 pairs of pants and 14 shirts.

The ratio is expressed as 10:14.

If we multiply the numerator and denominator by “2”, we get

= $\frac{10}{14} \times \frac{2}{2}$

= $\frac{20}{28}$

Ratio = 20 : 28

The ratio obtained after the simplification of 20:28 is determined by dividing the numerator and denominator of 20/28 by 4, thus, we obtain

$\frac{20}{28}$

= $\frac{5}{7}$

i.e. the ratio is 5:7

The ratio 10:14, i.e. 10 by 14 is multiplied by 3 in both the numerator and denominator.

We obtain

$\frac{10}{14} \times \frac{3}{3},$

= $\frac{30}{42}$

i.e. the ratio is 30:42.

Three equivalent ratios of10 pants and14 shirts are,

20 : 28

5 : 7

30 : 42

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