CA Foundation Maths Solutions For Chapter 1 Ratio And Proportion

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.

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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\} $

⇒ $=\frac{1}{2} \log\left\{\frac{1}{3}(x+y)\right\}^2$

⇒ $=\frac{1}{2} \log \left(\frac{x^2+y^2+2 x y}{9}\right)$

⇒ $=\frac{1}{2} \log \left(\frac{7 x y+2 x y}{9}\right)$

⇒ $=\frac{1}{2} \log (x y)=\frac{1}{2}(\log x+\log y)$

Ratio And Proportion

Ratio Exercise – 1

Proportion Exercise – 2

Ratio And Proportion Exercise – 3

Indices

Concepts And Tricks To Remember

Exercise -1 – Indices

Logarithms

Logarithms Exercise – 1

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