CA Foundation Maths Solutions For Chapter 2 Equations

CA Foundation Maths Solutions For Chapter 4 Equations

Introduction and Meaning:

The equation is defined to be a mathematical statement of equality.

Determination of the value of the variable that satisfies an equation is called a solution of the equation or root of the equation.
An equation in which the highest power of the variable is 1 is called a Linear (or simple) equation.
This is also called the equation of degree 1.
Two or more linear equations involving two or more variables are called Simultaneous Linear Equations.
An equation of degree 2 (Highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation.

Simple Equation

A simple equation is one unknown x in the form of a x + b = 0.

where a, and b are known constants and a1 0

Note: A simple equation has only one root.

Read and Learn More CA Foundation Maths Solutions

Simultaneous Linear Equations in Two Unknowns

Two such equations a₁x+b₁y +c₁= 0 and a₂ x+b₂y + c₂ = 0 form a pair of simultaneous equations in x and y.
A value for each unknown that satisfies simultaneously both equations will give the roots of the equations.

Method of solution

1. Elimination Method: In this method, two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknown and then solving for the other unknown

Example 1: Solve, 2x+5y = 9 and 3x- y =5.
Solution:

2x + 5y = 9 …..(1)

3x – y = 5 …..(2)

By making (1) x 1, 2x+5y = 9

And by making (2) x 5, 15x- 5y = 25

Adding 17x- 34 or x- 2. Substituting this values of x in (1) i.e. 5y = 9 – 2x we find;

5y = 9 – 4 =5

y = 1

x = 2,

y =1.

2. Cross Multiplication Method:

Let two equations be:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

We write the coefficients of x, y, and constant terms and two more columns by repeating the coefficients of x and y as follows:

1 2 3 4

b₁ c₁ a₁ b₁

b₂ c₂ a₂ b₂

⇒ and the result is given by$y: \frac{x}{b 1 c 2-b 2 c 1}=\frac{y}{c 1 a 2-c 2 a 1}=\frac{x}{a 1 b 2-a 2 b 1}$

⇒ so the solution is: x = $=\frac{b 1 c 2-b 2 c 1}{a 1 b 2-a 2 b 1} \quad y=\frac{c 1 a 2-c 2 a 1}{a 1 b 2-a 2 b 1}$

Example 2: Solve 3x+2y+17 =0, 5x-6y-9=0

Solution: 3x +2y 17= 0…..(1)

5x – 6y -9 = 0…..(3)

Method of elimination:

By (1) x 3 we get 9x + 6y + 51 = 0 …

Adding (2) and (3) we get 14x+42=0

Or x=$x=\frac{42}{14}=-3$

Putting x = -3 in (1) we get 3(-3) + 17 = 0

Or, 2y + 8 $=0 \text { or, } y=-\frac{8}{2}-4$

So,x = -3 andy = -4

Method of cross-multiplication:

3x + 2y + 17 = 0

5x – 6y- 9 = 0

⇒ $ \frac{x}{2(-9)-17(-6)}=\frac{y}{17(5)-3(-9)}=\frac{1}{3(-6)-5(2)}$

Or, $ \quad \frac{x}{84}=\frac{y}{112}=\frac{1}{-28}$

Or $\frac{x}{3}=\frac{y}{4}=\frac{1}{-28}$

Or x = -3 , y = -4

Method of Solving Simultaneous Linear Equation with Three Variables

Example 1: Solve for x, y and z:

2x-y + z = 3, x + 3y-2z=ll, 3x-2y + 4z=l

Solution:

(1) Method Of Elimination

2x – y + z = 3…..(1)

x + 3y — 2z =11…..(2)

3x – 2y + 4z = 1…..(3)

By (1) x 2 we get

4x – 2y + 2z = 6…..(4)

By (2) + (4),5x + y= 17…..(5)

By (2) x 2, 2x + 6y- 4z = 22…..(6)

By (3) + (6), 5x + 4y = 23…..(4)

By (5) – (7), – 3y = -6 ory = 2

Putting y = 2 in (5) 5x + 2 = 17, or 5x = 15 or, x=3

Putting = 3andy = 2 in (1)

2×3-2+ z =3

Or 6 – 2 + z = 3

Or 4 + z = 3

Or z=-l

So x- 3, y- 2, Z = -1 is the required solution

(2) Method Of Cross-Multiplication

We write the equations as follows:

2x-y + (z – 3) = 0

x + 3y + (-2z-ll) = 0

By cross multiplication

⇒ $ \frac{x}{-1(-2 z-11)-3(z-3)}=\frac{y}{(z-3)-2(-2 z-11)}=\frac{1}{23-1(-1)}$

⇒ $\frac{x}{20-z}=\frac{y}{5 z+19}=\frac{1}{7}$

⇒ $X=\frac{20-z}{7}, y=\frac{5 z+19}{7}$

Substituting above values for x and y in equation (iii) i.e. 3x- 2y + yz = 1, we have

⇒ $3{20-z}{7}-2{5 z-19}{7}+4 z=1$

Or 60 – 3z = lOz- 38 + 28 z = 7

Or 15z = 7-22 orl5z = -15 or Z=l

⇒ $x=\frac{20-(-1)}{7}=\frac{21}{7}=3, \quad y=\frac{5(-1)+19}{7}=\frac{14)}{7}=2$

Thus x = 3,y= 2, Z= -1.

An equation of the form ax2 + fax + c = 0 where x is a variable and a, b, c are constants with a ≠ 0 is called a quadratic equation or equation of the second degree.

When b = 0 the equation is called a pure quadratic equation; when b * 0 the equation is called an affected quadratic.

Examples:

1. 2x² + 3x + 5 = 0

2. x²-x = 0

3. 5x² – 6x – 3 = 0

The value of the variable say x is called the root of the equation. A quadratic equation has two roots.

How to find out the roots of a Quadratic Equation:

ax² +bx + c = 0 (a ≠ 0)

⇒ $\text { Or } x^2+\frac{b}{a} x+\frac{c}{a}=0$

Or $x^2+2 \frac{b}{2 a} x+\frac{b^2}{4 a^2}=\frac{b^2}{4 a^2}-\frac{c}{a}$

Or$\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}$

Or$\mathrm{x}+\frac{b}{2 a}=\frac{ \pm \sqrt{b^2-4 a c}}{2 a}$

Or$\mathrm{x}=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

Sum and Product of the Roots:

Let one root be a and the other root be p

Now $ \alpha+\beta & =\frac{-b+\sqrt{b^2-4 a c}}{2 a}+\frac{-b-\sqrt{b^2-4 a c}}{2 a}=\frac{-b+\sqrt{b^2-4 a c}-b-\sqrt{b^2-4 a c}}{2 a}$

⇒ $\frac{-2 b}{2 a}=\frac{-b}{a}$

Thus sum of roots = $-\frac{b}{a}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}$

⇒ $x \beta=\left(\frac{-b+\sqrt{b^2-4 a c}}{2 a}\right)\left(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\right)=\frac{c}{a}$

So the product of the roots =$\frac{\mathrm{e}}{a}=\frac{\text { Constant term }}{\text { coefficient of } x^2}$

How to Construct a Quadratic Equation

For the equation ax² + bx + c = 0 we have

⇒ $\text { Or } x^2+\frac{b}{a} x+\frac{c}{a}=0$

⇒ $\text { Or } x^2-\left(-\frac{b}{a}\right) x+\frac{c}{a}=0$

Or A-2 – (Sum of the roots) x + Product of the roots = 0

Nature of the Roots

⇒ $X=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

If b² — 4ac = 0 the roots are real and equal;

If b²– 4ac > 0 then the roots are real and unequal (or distinct);

If b²– 4ac < 0 then the roots are imaginary;

If b²– 4ac is a perfect square (≠0) the roots are real, rational, and unequal (distinct);

If b² – 4ac > 0 but not a perfect square the roots are real, irrational, and unequal.

Since b²– 4ac discriminates the roots b²– 4ac is called the discriminant in the equation ax² + bx + c = 0 as it discriminates between the roots.

Note:

Irrational roots occur in conjugate pairs that is if (m + Vn) is a root then (m – Vn) is the other of the same equation.
If one root is reciprocal to the other root then their product is 1 and so $\frac{c}{a}=1 \text { i.e. } c=\mathrm{a}$
If one root is equal to another root but opposite in sign then.
- Their sum = 0 and so $\frac{b}{a}=1 \text { i.e. } c=\mathrm{a}$

Summary, Tips, Tricks And Formulae

Equations

An equation is defined as a mathematical statement of equality.

Types of equations

Linear equation in one variable.

Linear simultaneous equations in 2 or 3 variables.
Quadratic equations.
Cubic equations.
Bi-quadratic equations.
Exponential equations.

Quadratic Equations

A quadratic equation is defined as the polynomial equation of degree 2.
A quadratic equation can be expressed in the following general form: ax² + ba + x = 0, (a ≠ 0)
A quadratic equation can also be expressed in the factor form as follows: a(x-α){x-β)
1. Here, a and /3 are the roots or solutions of quadratic equations;
2. The general solution of the quadratic equation can be obtained as follows:
$\alpha=\frac{-b+\sqrt{b^2-4}}{2 a} \text { and } \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
- Sum of roots =$a+\beta=-\frac{b}{a}$
Product of roots =$\alpha\beta=\frac{c}{a}$

Structure of Quadratic Equations

If the sum (S) (α + β) and Product (P) (αβ) of the roots are known, then the quadratic is x² – Sx + P = 0

Sign of Roots of a Quadratic Equation

When c = 0, one root of the equation must be 0,
When h and c are 0, then both the roots must be 0.
If a,b, and c all are 0, both roots are negative.
If a and c are of some sign, opposite to that of, then both the roots will be positive.
If a and c are of opposite signs, one root is positive, and another root is negative.

Nature of Roots

The Expression “b² – 4ac” is called the “Discriminant (D)” of the quadratic equation.

When D> 0, Roots are real and distinct.
When D = 0, Roots are real and equal.
When D < 0, the Roots are imaginary.
When D > 0, Roots are real.
When D is a perfect square, Roots are real, rational, and unequal.
When D is not a perfect square, Roots are real, rational, and unequal.
If roots are equal use b² = 4ac.
If roots are reciprocal of each other, use a = c
If roots are equal but of opposite sign, use b = 0
If roots are reciprocal but opposite in sign, use c = -0

Note

1. Irrational roots will always appear in conjugate pairs.

it = (a- √b) and (1= (a + √b)

2. Imaginary roots will always appear in conjugate pairs.

u = (a- ib) and β = (a + ib)

Cubic Equations

1. A cubic equation is a polynomial equation of degree 3, and the general form is represented as follows:

ax³+ bx² + cx + d = 0; (a ≠ 0)

2. The factor form of the cubic equation is given as follows:

a(x- α)(x- β)(x- γ) = 0

Here, α,β, and γ are the roots or solutions of the cubic equation.

3. Sum of roots = α+β+γ= -b/a

4. Product of the roots = αβγ = -d/a

Bi-Quadratic Equations

A bi-quadratic equation is a polynomial of degree 4, and the general form is represented as follows:

ax⁴+ bx³ + cx² + dx + e = 0; (a * 0)

The factor form of a cubic equation is given as follows:

a(x- α)(x- β)(x-γ)(x-δ) = 0

Here, α, β,andγ are the roots or solutions of the bi-quadratic equation.

The sum of roots = α + β +γ + δ = -b/a

Product of the roots = αβγδ = e/a

Exercise – 1

Simple Or Linear Equation in one Variable

Question 1. The equation -7x +1 = 5 – 3 x will be satisfied for x equal to:

2
-1
1
None of these

Solution:

(2)

7x + 1 = 5 – 3x

-4 = 4x

X = -1

Question 2. The root of the equation$1 \frac{x+4}{4}+\frac{x-5}{3}=11$ is

20
10
2
None of these

Solution:

(1)

⇒ $\frac{x+4}{4}+\frac{x-5}{3}=11$

3x+ 12 + 4x – 20 = 132

7x = 140 → x = 20

Question 3. Pick up the correct value of x for$\frac{x}{30}=\frac{2}{45}$

x = 5
x = 7
$x=1 \frac{1}{3}$
None Of these

Solution:

(3)

⇒ $\frac{x}{30}=\frac{2}{45} \rightarrow \frac{x}{2}=\frac{2}{3} \rightarrow \mathrm{x}=\frac{1}{{ }^3} $

Question 4. The solution of the equation $\frac{x+24}{5}=4+\frac{x}{4}$

6
10
16
None of these

Solution:

(3)

⇒ $\frac{x+24}{5}=\frac{4+x}{4}$

⇒ $\frac{4 x+96-5 x}{20}=4$

⇒ $-\mathrm{x}=-16 \rightarrow \mathrm{x}=16$

Question 5. 8 is the solution of the equation

$\frac{x+4}{4}+\frac{x-5}{3}=11$
$\frac{x+4}{2}+\frac{x+10}{9}=8$
$\frac{x+24}{5}=4+\frac{x}{4}$
$\frac{x-15}{10}+\frac{x+5}{5}=4$

Solution:

(2)

⇒ $\frac{x+4}{2}+\frac{x+10}{9}=8 \Rightarrow \frac{8+4}{2} \rightarrow \frac{8+10}{9}=6+2=8$

Question 6. The value of that satisfies the equation$\frac{y+11}{6}-\frac{y+1}{9}=\frac{y+7}{4} \text { is }$

-1
7
1
$-\frac{1}{7}$

Solution:

(3)

⇒ $\frac{3 y+33-2 y+2}{18}=\frac{y+7}{4}$

⇒ $\frac{y+35}{18}=\frac{y+7}{4} \rightarrow \frac{2 y+70}{36}=\frac{9 y+63}{36}$

⇒ 7y = 7 ⇒ y = 1

Question 7. The solution of the equation (p+2) (p-3) + (p+3) (p-4) = p(2p-5) is

6
7
5
None of these

Solution: (1)

Question 8. The equation $\frac{12 x+1}{4}=\frac{15 x-1}{5}+\frac{2 x-5}{3 x-1}$ is true for.

A = 1
x = 2
A = 5
x = 7

Solution:

(4)

⇒ $\frac{12(7)+1}{1}=\frac{85}{4}$

⇒ $\frac{15(7)-1}{1}+\frac{2(7)-5}{3(7)-1}=\frac{425}{20}=\frac{85}{4}$

Question 9. Pick up the correct value x for which$\frac{x}{0.5}-\frac{1}{0.05}+\frac{x}{0.005}-\frac{1}{0.0005}=0$

x = 0
x = 1
x = 10
None of these

Solution:

(3)

⇒ $\frac{10 x}{5}-\frac{100}{5}+\frac{1000 x}{5}-\frac{10000}{5}=0$

⇒ $\frac{1010 x}{5}=\frac{10000+100}{5}$

10l0x = 10100

X = 10

Question 10. The sum of two numbers is 52 and their difference is 2. The numbers are

17 and 15
12 and 10
27 and 25
None of these

Solution:

(3)

Let x lie bigger no, y be a smaller number

27and25

27 + 25 = 52 and 27 – 25 = 2

Question 11. The diagonal of a rectangle is 5 cm and one of at sides is 4 cm. Its area is

20 sq. cm.
l2 sq. cm.
10 sq. cm.
None of these

Solution:

(2)

Diagonal = 5cm

Sides = 4cm x x

42 + x2 = 52 by Pythagoras theorem

x2 = 9

x = 3

Area = 4×3 = I2sq cm

Question 12. Divide 56 into two parts such that three times the first part exceeds.one-third of the second by 48. The parts are.

(20, 36)
(25.31)
(24, 32)
None of these

Solution:

(1)

3x = $\frac{1}{3}$y + 48

⇒ $\frac{1}{3}$

⇒ $\frac{y}{3}$

9x – 144 = y

56 = x + 9x – 144

200 = lOx ⇒ x = 20, y = 36

Question 13. The sum of the digits of a two-digit number is 10. If 18 is subtracted from it the digits in the resulting number will be equal. The number is

37
73
75
None of these

Solution:

(2)

7 + 3 = 10

73-18 = 55

Question 14. The fourth part of a number exceeds the sixth part by 4. The number is

84
44
48
None Of these

Solution:

(3)

⇒ $\frac{x}{4}=\frac{x}{6}+4$

⇒ $\frac{x}{4}-4=\frac{x}{6} \rightarrow 3 x-48=2$

x = 48

Question l5. Ten years ago the age of a father was four times of his son. Ten years hence the age of the father will be twice that of ofIlls son. The present ages of the father and the son are.

(50. 20)
60, 20)
(55,25)
None of these

Solution:

(1)

4 (y – 10] = x – 1o

4y – 40 = x – 10

4y = 3(30)- y = $\frac{x+30}{4}$

(x+ 10] = 2 (y+ 10)

X+ 10 = 2y + 20

⇒ $\frac{x-10}{2}=\frac{x-10}{2}$

X + 30 = 2x – 20 => x = 50, y = 20

Question 16. The product of two numbers is 3200 and the quotient when the larger number is divided by the smaller is 2. The numbers are

(16. 200)
(160, 20)
(60,30)
(80, 40)

Solution:

(4)

xy = 3200 x = 2; y= x= 2y

⇒ $\frac{x}{2}=y$

x2 = 6400 x = 80 y = 40

Question 17. The denominator of a fraction exceeds the numerator by 2. If 5 is added to the numerator the fraction increases by unity. The fraction is.

$\frac{5}{7}$
$\frac{1}{3}$
$\frac{7}{9}$
$\frac{3}{5}$

Solution:

(4)

⇒ $\frac{x}{x+2}$

⇒ $\frac{1+5}{x+2}=\frac{x}{x+2}+1$

x+5=2 x+2

3=x

function $ =\frac{3}{5}$

Question 18. Three persons Mr. Roy, Mr. Paul, and Mr. Singh together have Mr. Paul has less than Mr. Roy and Mr. Singh has got less than Mr. Roy. They have the money.

(₹20,₹16,₹15)
(₹15,₹20,₹16)
(₹25,₹11,₹5)
None of these

Solution:

(1)

20- 16 = 4,

2a- 15 = 5

Question 19. A number consists of two digits. The digits in the ten’s place are 3 times the digits in the unit’s place. If 54 is subtracted from the number the digits are reversed. The number is

Solution:

(3)

9 = 3 (3)

93-54= 39

Question 20. One student is asked to divide half of a number by 6 and the other half by 4 and then to add the two quantities. Instead of doing so, the student divides the given number by 5. If the answer is 4 short of the correct answer then the number was

320
400
480
None of these.

Solution:

(3)

Let number be x $\frac{x}{12}+\frac{x}{8}=4+\frac{x}{5}$

⇒ $\frac{2 x+3 x}{24}=\frac{4+x}{5} \Rightarrow \frac{5 x}{24}-\frac{x}{5}=4$

⇒ $\frac{25 x-24 x}{24 \times 5}=4$

x = 4 x 24 x 5

x = 480

Question 21. If several which the half is greater than $\frac{1}{5}$th of the number by 15 then the number is

50
40
80
None of these

Solution:

(1)

let number be x according to question

⇒ $\frac{x}{2}=\frac{x}{5}+15$

⇒ $\frac{x}{2}-\frac{x}{5}=15$

3x = 150 ⇒ x = 50

Question 22. $\text { If } \frac{x-b c}{b+c}+\frac{x-c a}{c+a}+\frac{x-a b}{a+b}=$ a + b + c the value of x is

a2+b2+c2
a(a+b+c)
(a+b)(b+c)
ab+bc+ca

Solution:

(4)

ab + be + ca

⇒ $\rightarrow \frac{a b+c a}{b+c}+\frac{a b+b c}{a+c}+\frac{b c+c a}{a+b}$

= a+b+c

Question 23. A man rowing at the rate of 5 km in an hour in still water takes thrice as much time going 40 km up the river as in going 40 km down. Find the rate at which the river flows:

4.5km/hr
7.5 km/hr
2.5 km/hr
None

Solution:

Let the speed of the river be x km/hr

Speed of man in still water = 5 km/hr

Speed while upward rowing will be 5 – x

Speed while downward rowing will be 5 + x

We know speed = distance or time

Let the time taken for upward rowing be T1 and the time taken for downward rowing be T2.

According to question TI=3T2

⇒ $\frac{40}{5-x}=3 \frac{40}{5+x}$

5+x = 15 – 3x

x= 2.5 km/hr

(3) is correct

Question 24. If the length of a rectangle is 5 cm more than the breadth and if the perimeter of the rectangle is 40 cm, then the length and breadth of the rectangle will be:

7.5 cm, 2.5 cm
10cm, 5cm
12.5cm, 7.5cm
15.5cm, 10.5cm

Solution:

(3)

1st condition = length is 5cm more than breadth

All options satisfy this condition

II perimeter = 2(l+b) = 40 of rectangle

Only option (c) satisfies it

Detail Method

Let breadth = x; :-length = x + 5

Perimeter = 40

2(x+5+x) = 40

Or 2x + 5 = 20

Or x = $\frac{15}{2}$= 7.5cm

Length = x + 5 = 7.5 + 5 = 12.5

Breadth = x = 7.5cm

(3) is correct

Question 25. For all 2 e R, the line (2+ 2)x + (3- A) y + 5 = 0 Passing through a fixed point, then the fixed point is_____.

(-1,-1)
0,0
2,2
1. 1

Solution:

(2)

For option (b) Point (-1; -1) satisfies the equation

LHS = =(2+ A)x + (3-/1) y + 5

Or (2 + A)(-l]+(3- A) (-l) + 5

-2-A-3+A + 5 = 0 = RHS.

(2) is correct

Question 26. If kx – 4 = (k – l).x which of the following is true

x = -5
x = -4
x = -3
x = 4

Solution:

(4) is correct

Kx- 4 = (k-l)x

Orkx-4 = kx-x

Or -4 = -x

∴ x = 4

Question 27. The age of a person is 8 years more than thrice the age of the sum of his two grandsons who were twins. After 8 years his age will be 10 years more than twice the sum of the ages of his grandsons. Then the age of the person when the twins were born is_

86 yrs
73 yrs
68 yrs
63 yrs

Solution:

(2) let the age of 1st grandson = x

Person’s age = P = 3(x+x) + 8

P = 6x + 8

After 8 years

P + 8 = 2 [x+8+x+8] + 10

= 2(2x+16) + 10

Or 6x + 8 + 8 = 4x + 32 + 10

Or 2x = 42 -16 = 26

x = 13

Age of person when grandsons

were born

= 6x + 8 – x

= 6 x 13 + 8-13 = 73

(2) is correct

Question 28. In a school number of students in each section is 36. If 12 new students are added, then the number of sections is increased by 4 and the number of students in each section becomes 30. The original number of sections at first is

Solution:

(4) let original No. of sections = x

Total students = 36x

36x+ 12 = (x+4) 30

Or 36x + 12 = 30x + 120

Or 6x = 108 ⇒ x = 18

Question 29. A person on a tour has for his expenses. But the tour was extended for another 16 days, so he must cut his daily expenses by 20. The original duration of the tour had been?

48 days
64 days
80days
96 days

Solution:

Let No. of tour days = x

Expenses per day =$=\frac{9600}{x}$

Now Expense per day =$=\frac{9600}{x+16}$

⇒ $\text { From } \frac{9600}{x}-\frac{9600}{x+16}=20$

From here we get

For (3) LHS

⇒ $\frac{9600}{80}-\frac{9600}{80+16}=20 \text { RHS. }$

(3) is correct

Question 30. The particular company produces some articles in a day. The cost of production per article is more than thrice the number of articles and the total cost of production is on a day then the number of articles is:

Solution:

(1) is correct.

Let (A) be correct.

So cost, per unit = 800/16 =

It is 2 more than 3 times of 16. (as given in Qts.)

Question 31. The sides of an equilateral triangle are shortened by 3 units, 4 units, and 5 units respectively, and then a right-angle triangle is formed. The side of the equilateral triangle was

Solution:

For option [3]

1st side of right-angled A = 8 – 3 = 5

2nd side = 8-4 = 4

And 3rd side = 8-5 = 3

Here; 5; 4 and 3 are making a right-angled

triangle.

So, 52 = 42 + 32

Hence, option (c) is correct

Question 32. A number consists of two digits such that the digit in one’s place is thrice the digit in ten’s place. If 36 is added then the digits are reversed. Find the number

62
26
39
None

Solution:

(2)

[1] 62 → 2 3 x 6 [False]

And 62 + 36 = 98* 26 (false]

(2) 26 clearly 6 = 3×2 (True]

And 26 + 36 = 62 (Orders of digits reversed)

So; (2) is correct.

Exercise – 2

Simultaneous Or Linear Equation In Two Or Three Variable

Question 1. The solution of the set of equations 3x + 4y = 7, 4x-y = 3 is

(1.-1)
(1. 1)
(2.1)
(1.-2)

Solution:

(2)

ax + by = c

px + qy = r

(aq – bp) x = ac

→(-3 -16) x = -7-12

-19 x = – 19

X = 1

3 + 4y = 7 → y = 1

Question 2. The values of X and y satisfying the equations $\frac{x}{2}+\frac{y}{3}=$= 2,x + 2y = 8 are given by the pair.

(3.2)
(-2, -3)
(2.3)
None of these

Solution:

(3)

3x + 2y = 12

x + 2y = 8

2x = 4 ⇒ x = 2

⇒ $y=\frac{8-x}{2} \quad \Rightarrow \frac{8-2}{2}=3$

Question 3.$\frac{x}{p}+\frac{y}{q}=2,$x + y = p + q are satisfied by the values given by the pair.

(x=p,y=q)
(x=q,y=p)
(x=l, y=l)
None of these

Solution:

(1)

GBC

x = p, y = q

Question 4. The solution for the pair of equations$\frac{1}{16 x}+\frac{1}{15 y}=\frac{9}{20}, \frac{1}{20 x}-\frac{1}{27 y}=\frac{4}{45}$ is given by

$\left(\frac{1}{4}, \frac{1}{3}\right)$
$\left(\frac{1}{3}, \frac{1}{4}\right)$
(3,4)
(4,3)

Solution:

(1)

Let 1/x = a, 1/y=b

⇒ $\text { 1) }\frac{a}{16}+\frac{b}{15}=\frac{9}{20}=15 a+16 b$

⇒ $\frac{9}{20} \times 16 \times 15 \Rightarrow 108$

2) $\frac{a}{20}-\frac{b}{27}=\frac{4}{45}$

⇒ $27 a-20 b=\frac{4 \times 20 \times 27}{45}=48$

15a + 16b = 108

27a -20b = 48

By trick refer to question 1

(-300 -432) a = (-2160-768)

a = 4

15 (4) + 16b = 108

16 b = 48

B = 3

⇒ $a=4, b=3 \Rightarrow x=\frac{1}{a}, y=\frac{1}{b} \Rightarrow x=\frac{1}{4} ,y=\frac{1}{3}$

Question 5. Solve for x and y: $\frac{4}{x}-\frac{5}{y}=\frac{x+y}{x y}+\frac{3}{10} \text { and } 3 x y=10(y-x) \text {. }$

(5,2)
(-2,-5)
(2,-5)
(2,5)

Solution:

⇒ $\frac{4}{x}-\frac{5}{y}=\frac{x+y}{x y}+\frac{3}{10}$

⇒ $3 \mathrm{xy}=10(\mathrm{y}-\mathrm{x})$

⇒ $\frac{4}{x}-\frac{5}{y}=\frac{1}{y}+\frac{1}{x}+\frac{3}{10}$

⇒ $3 \mathrm{xy}=10 \mathrm{y}-10 \mathrm{x} \rightarrow 3=\frac{10}{x}-\frac{10}{y}$

⇒ $\frac{3}{x}-\frac{6}{y}=\frac{3}{10} \quad \frac{10}{x}-\frac{10}{y}=3$

⇒ $\text { Let } \frac{1}{y}=a \frac{1}{y}=b$

3a -6b = 3 ….(1)

10a – 10b = 3 → 30a – 30b – 9….. (2)

30a- 60b = 3

⇒ $\frac{30 a-30 b=9}{30 b=6}=\mathrm{b}=\frac{6}{30}$

30a -60b = 3

⇒ $ 30 a-60 \frac{(6)}{30}=3$

⇒ $\rightarrow 30 a-12=3$

⇒ $a=\frac{15}{30}$

⇒ $x=\frac{1}{a}=\frac{30}{15}=2 .$

⇒ $ y=\frac{1}{b}=\frac{30}{6}=5 .$

(4) (2,5)

Question 6. The pair satisfying the equations x + 5y = 36,ÿ = is given by Qpfrbft 1 ” Y>

(l6,4)
(4, 16)
(4, 8)
None of these

Solution:

x + 5y = 36 …….(1)

⇒ $\frac{x+y}{x-y}=\frac{5}{3} \ldots \ldots \ldots \ldots(2)$

3x + 3y = 5x- 5y

2x + lOy = 72….(1)

2x-8y = 0 (2)

2x+10y

Subtract $\frac{2 x-8 y}{18 y}$= 72 —> y = 4.

x + 20 = 36 → x — 16.

(a) (16,4)

Question 7. Solve for x and y : x-3y = 0, x+2y = 20CM 3 Opt >

x = 4, y = 12
x= 12, y = 4
x = 5, y = 4
None of these.

Solution:

x – 3y = 0 x + 2y = 20

x = 3y = 0

x + 2y = 20

by trick (Ql)

(2 + 3)x= (0-(-60))

x = +12

x- 3y = 0 —> 12 -3y = 0 —> 12 = 3y

y = 4

x = 12, y = 4

(2) is correct

Question 8. The simultaneous equations 7x-3y = 31, 9x-5y = 41 have solutions given by

(-4,-1)
(-1.4)
(4,-1)
(3,7)

Solution:

7x-3y = 31, 9x – 5y = 41

By tricks

7x- 3y = 31

9x-5y = 41

(-35 + 27)x = (-155 + 123)

-8x = -32

x = 4

7(4) – 3y = 31

28 – 3y = 31

-3y = 3→ y = -l

(3) (4 .-1)

Question 9. 1.5x + 2.4 y = 1.8, 2.5(x+l) = 7y have solutions as

(0.5, 0.4)
(0.4, 0.5)
$\left(\frac{1}{2}, \frac{2}{5}\right)$
(2,5)

Solution:

1.5x + 2.4y= 1.8 …(1)

2.5x + 2.5 = 7y

15x+ 24y = 18

25x- 70y = -25

15x+ 24y = 18

25x+ 70 = -25

By trick

(-1050 -600)x = (-1260 + 600) – 1650x = -660

x = 0.4

6 + 24y = 18

⇒ $ y=\frac{12}{24}=\frac{1}{2}=0.5$

⇒ $x=0.4, y=0.5$$

Question 10. The values of x and y satisfying the equations$\frac{3}{x+y}+\frac{2}{x-y}=3, \frac{2}{x+y}+\frac{3}{x-y}=3 \frac{2}{3}$ are given by

(1,2)
(-l.-2)
$\left(1, \frac{1}{2}\right)$
(2, 1)

Solution:

(1)

GBC

Question 11. 1.5x + 3.6y = 2.1, 2.5 (x+1) = 6y ‘/’JfY,1

(0-2, 0.5)
(0.5, 0.2)
(2,5)
(-2,-5)

Solution:

By trick

[(-900) -900]x= (-1260 + 900) – 1800x = -360

$x=\frac{360}{1800}=0.2$

3 + 36y = 21

36y= 18 → y = 0.5

Question 12. $\frac{x}{5}+\frac{y}{6}+1=\frac{x}{6}+\frac{y}{5}=28$

(6,9)
(9, 6)
(60,90)
(90, 60)

Solution:

(3)

6x + 5y = 810

5x + 6y = 840

By tricks

6x + 5y = 810

5x + 6y = 840

(36 -25)x = 4860 -4200

llx= 660 x = 60, y = 90.

Question 13. $\frac{x}{4}=\frac{y}{3}=\frac{z}{2} ; 7 x+8 y+5 z=62$

(4,3,2)
(2,3,4)
(3,4,2)
(4,2,3)

Solution:

(1)

7x + 8y + 5z = 62

7(4) + 8(3) + 5(2) = 62

28 + 24 + 10 = 62

⇒ $=\frac{x}{4}=\frac{y}{3}=\frac{z}{2} \rightarrow(4,3,2)$

Question 14. $\frac{x y}{x+y}=20, \frac{y z}{y+z}=40, \frac{z x}{z+x}=24$

(120,60,30)
(60,30, 120)
(30,120,60)
(30,60,120)

Solution:

(4)

⇒ $\frac{x+y}{x y}=\frac{1}{20} \frac{y+z}{y z}=\frac{1}{40} \frac{z+x}{2 x}=\frac{1}{24}$

⇒ $\frac{1}{y}+\frac{1}{x}=\frac{1}{20} \ldots(1) \frac{1}{z}+\frac{1}{y}=\frac{1}{40} \ldots(2) \frac{1}{x}+\frac{1}{z}=\frac{1}{24} \ldots .(3)$

⇒ $\frac{1}{y}=\frac{1}{20}-\frac{1}{x} \frac{1}{24}-\frac{1}{x}+\frac{1}{y}=\frac{1}{40} \frac{1}{z}=\frac{1}{24}-\frac{1}{x}$

⇒ $\frac{1}{24}-\frac{1}{x}+\frac{1}{20}-\frac{1}{x}=\frac{1}{40} $

⇒ $\frac{1}{x}+\frac{1}{x}=\frac{1}{24}+\frac{1}{20}-\frac{1}{40} =\frac{2}{x}=\frac{8}{120} \quad=\frac{x}{2}=\frac{120}{8} \rightarrow x=30 \quad 1 / y=\frac{1}{20}=\frac{1}{3}=10 / 600$

⇒ $=\frac{2}{x}=\frac{5+6-3}{120} x=\frac{1}{z}=\frac{1}{24}-\frac{1}{30}=\frac{6}{720} z=120$

(d) (30, 60, 120)

Question 15. 2x + 3y + 4z = 0,x + 2y – 5z = 0, lOx + 16y- 6z = 0

(0,0,0)
(1,-1, 1)
(3, 2,-1)
(1,0,2)

Solution:

(1)

GBC:(0,0,0)

Question 16. $\frac{1}{3}(x+y)+2z=21,3 x-\frac{1}{2}(y+z)=65, x+\frac{1}{2}(x+y-z)=38$

(4, 9. 5)
(2,9,5)
(24, 9, 5)
(5, 24,9)

Solution:

(3)

⇒ $\frac{1}{3}(24+9)+2(5)=213(24) \cdot \frac{1}{2}(9+5)=65$

⇒ $24+\frac{1}{2}(24+9-5)=38$

Question 17. $\frac{x}{0.01}+\frac{y+0.03}{0.05}=\frac{y}{0.02}+\frac{x+0.03}{0.04}=2$

(1,2)
(0.1, 0.2)
(0.01, 0.02)
(0.02,0.01)

Solution:

⇒ $\frac{x}{0.01}+\frac{y+0.03}{0.05}=\frac{y}{0.02}+\frac{x+0.03}{0.04}=2$

⇒ $100 x+\frac{100 y+3}{5}=2 ; \quad 50 y+\frac{100 x+3}{4}=2$

500x + 100y + 3 = 10; 200y +100x + 3 = 8

500x + lOOy = 7 100x + 200y = 5

Multiply by 2

Subtract

$\begin{gathered}
1000 x+200 y=14 \\
100 x+200 y=5 \\
\hline 900 x=9
\end{gathered}$

x = 0.01 y = 0.02

Question 18. $\frac{x y}{y-x}=110, \frac{y z}{z-y}=132, \frac{z x}{z+x}=\frac{60}{11}$

(12, 11, 10)
(10, 11, 12)
(11,10,12)
(12, 10, 11)

Solution:

(2)

⇒ $ \frac{y-x}{x y}=1 / 110 \frac{z-y}{y z}=1 / 132 \frac{z+x}{z x}=\frac{11}{60}$

⇒ $\frac{1}{x}-\frac{1}{y}=\frac{1}{110} \frac{1}{y}-\frac{1}{z}=\frac{1}{132} \frac{1}{x}+\frac{1}{z}=\frac{11}{60}$

GBC = 10,11,12

(2) is correct

Question 19. 3x-4y+70z = 0, 2x+3y-10z = 0, x+2y+3z = 13

(1,3, 7)
(1,7, 3)
(2, 4, 3)
(-10, 10, 1)

Solution:

(4)

3x-4y+70z = 0, 2x+3y-10z = 0, x+2y+3z = 1 3

For three variables GBC x = -10,y = 10, z=l

Question 20. The monthly incomes of two persons are in the ratio 4: 5 and their monthly expenses are in the ratio of 7: 9. If each saves 350 per month find their monthly incomes.

(500, 400)
(400, 500)
(300, 600)
(350, 550)

Solution:

(2)

Monthly Income = 4 : 5 i.e., 4x, 5x

Monthly expenses = 7 : 9 i.e. 7y,9y

Saving = Income – Expenses = 50

4x-7y = 50

5x-9y = 50

By trick (-36 + 35)x-450 + 350

-x = -100 = x = 100

Monthly Income = 4x, 5x = (400, 500)

Question 21. Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are increased by 12.

Solution:

(1)

Let Fraction be$\frac{x}{y}$

⇒ $ \frac{x+2}{y+2}=\frac{1}{2} \quad, \frac{x+12}{y+12}=\frac{3}{4}$

GBC

⇒ $\frac{x}{y}=\frac{3}{8}$

Question 22. The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age.

60 years
52 years
51 years
50 years

Solution:

(4)

Let the sum of ages be x, Man’s age = x

Acc. To quest, x = 2(y)…….(1)

x – 5 = 3(y- 10)

x – 5 = 3y- 30

Put (1) in (2) 2y-3y = -25,

y=25

x = 50

Question 23. A number between 10 and 100 is five times the sum of its digits. If 9 is added to it the digits are reversed find the number.

Solution:

(3)

Let the number be (x is one digit, y another)

Number = lOx + y

= 10x + y + 9 = lOy + x = 9x-9y = -9

Also

10x + y = 5(x + y)

10x + y = 5x + 5y

5x = 4y

x = $\frac{4}{5} y \quad=9\left(\frac{4}{5}\right) y-9 y=-9$

= 36y-45y = -45

-9=-45

y=5, x =$\frac{4}{5} \cdot 5$ = 4

Numbers = lOx + y = 45

Question 24. The wages of 8 men and 6 boys amount to 333. If 4 men earn 34.50 more than 5 boys determine the wages of each man and boy.

(31.50,33)
(33,31.50)
(32.50, 32)
(32,32.50)

Solution:

(2)

Let each man’s wage be x & boy’s be y

8x + 6y = 33 …….. (1)

4x- 4.50 = 5y …….(2)

4x = 5y + 4.50

8x= 10y+ 9 ……. (3) (3) Put (3) in (1)

lOy + 9 + 6y = 33

16y = 24 → y = 1.5.v = 3

Question 25. A number consisting of two digits is four times the sum of its digits and if 27 is added to it the digits are reversed. The number is:

Solution:

(3)

Let number be lOx + y

Acc to quest.

lOx + y = 4(x+y)

1 Ox – 4x = 4y- y

6x = 3y

2x = y

Also

10x + y + 27= lOy + x 5y + y + 24= 10y +$\frac{y}{2}$

10y+ 2y + 54 = 20y + y

12y + 54 = 21y

9y = 64-4y = 6 x = 3

Number =36

Question 26. Of two numbers, l/5lh of the greater is equal to 1 /3rd of the smaller, and their sum is 16. The numbers are:

(6,10)
(9,7)
(12,4)
(11,5)

Solution:

(1)

Let two numbers be x,y such that x > y

Acc to question

$ \frac{x}{5}=\frac{y}{3}, \quad x+y=16$ $3 x=5 y \rightarrow x=\frac{5}{3} y$ $\frac{5}{3} y+y=16=5 y+3 y=48 \rightarrow y=6, x=10$

Two numbers are 6, 10

Question 27. y is older than A- by 7 years 15 years back x’s age was 3/4 of/s age. Their present ages are

(x-36,y=43)
(x=50,y=43)
(x=43,y=50)
(x=40,y=47)

Solution:

(1)

$Y=x+7 \text { also, }(y-15) \cdot \frac{3}{4}=x=15$ $y-7=x \quad \frac{3}{4} y-\frac{45}{4}=x-15$

$\frac{3}{4} y-\frac{45}{4}=y-7-15=3 y-45=$4y- 88

= y = 43 y = 43 x = 36

Present ages are 36, 43

Question 28. The sum of the digits in a three-digit number is 12. If the digits are reversed the number is increased by 495 but reversing only of the ten’s and unit digits increases the number by 36. The number is

Solution:

(c)

Three digit no. be lOOx + lOy + z

Reversing all digits

x + y + z = 12,

100z+ lOy + x = lOOx + lOy + z + 495

100z + x = lOOx + z + 495

99z-99x = 495

z-x = 5…(1) z = 5 + x

Reversing ten’s T unit digit no. increases by 36

100x + lOz + y = 100x + lOy + z + 36

9z-9y = 36

z-y = 4

z = 4 + y…..[2]

x + y + z = 12

from (1) and [2]

3z-9 = 12

3z = 21

z-5+z-4+z=12

z = 7

x = 2, y = 3, z = 7

Numbers = 237

Question 29. Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and l/3rd of the smaller and 1/5th of the greater number are together 21. The numbers are:

(36, 45)
(45, 36)
(50, 41)
(55, 46)

Solution:

(2)

Let two numbers be x and y. such that x>y

2x = 2y + 18 And $\frac{1}{3} y+\frac{1}{5} x=21$

2x-2y=18

x-y = 8 5y+3x = 315

x = 9 + y 5y + 27 + 3y = 315

8y = 288 y = 36

x = 45

Question 30. The demand and supply equations for a certain commodity are 4q + 7p = 17 and p = $\frac{q}{3}+\frac{7}{4}$ respectively where p is the market price and q is the quantity than the equilibrium price and quantity are:

$2, \frac{3}{4}$
$3, \frac{1}{2}$
$5, \frac{3}{5}$
None of these

Solution:

⇒ $4 q+7 p=17 \rightarrow p=\frac{17-4 q}{7}$

⇒ $P=\frac{q}{3}+\frac{7}{4}$

For equilibrium

⇒ $\frac{q}{3}+\frac{7}{4}=\frac{17}{7}-\frac{4}{7} q$

⇒ $ \frac{7 q+12 q}{21}=\frac{68-49}{28}=\frac{19 q}{21}=\frac{19}{28}=q=\frac{3}{4}$

⇒ $P=\frac{17-4 q}{7}=\frac{17-3}{7}=2 .$

Question 31. Solving 6x+5y-16=0 and 3x-y-l=0 we get values of andy as

1,1
1,2
-1,2
0,2

Solution:

(2)

6x + 5y= 16

3x-y= 1

By trick

(-6-15)x = -16 -5

-21x = -21 = x = 1

6 + 5y= 16 → y = 2

Question 32. Solving 9x+3y-4z=3 ,x+y-z=0 and 2x-5y-4z=-20 following roots are obtained

2, 3, 4
1,3,4
l, 2,3
None

Solution:

(3)

9x + 3y- 4x = 3, x+y-z=0

2x- 5y- 4x = -2

For three variables

Question 33. A man went to the Reserve Bank of India with 1,000. He asked the cashier to give him 5 and % 10 notes only in return. The man got 175 notes in all. Find how many notes of K 5 and 10 he receives.

(25,150)
(40,110)
(150,25)
None

Solution:

For (1) 25 x 5 + 150 x 10$\neq$ 1000

(2) 40 x 5 + no x 10$\neq$ RS. 1000

(3)150 x5 + 25 x 10 = Rs. 1000

Question 34. The Point of intersection of the lines 2x- 5y = 6 and x+y=3 is

(0,3)
(3,0)
(3,3)
(0,0)

Solution:

(2)

Intersecting Point lies on both straight lines. It will satisfy both ends.

For 00 (0,3) Point 2×0 – 5×3

it is incorrect 6

0Ption (2) (3; 0) satisfies both eqns.

(2) is correct

Question 35. If the equations kx + 2y = 5, 3x + y = 1 has no solution then the value of k is

Solution:

Kx + 2y = 5

3x + y=l

They have no soln. (given)$\frac{k}{3}=\frac{2}{1} \neq \frac{5}{1} ; \Rightarrow$-k = 6

Question 36. The equation x + 5y = 33;$\frac{x+y}{x-y}=\frac{13}{3}$ Has the solution (x,y) as:

(4,8)
(8,5)
(4,16)
(16,4)

Solution:

(2)

For LHS = x + 5y = 8 + 5×5 = 33

⇒ $\frac{x+y}{x-y}+\frac{8+5}{8-5}=\frac{13}{5}$

Clearly (b) satisfies both equations.

Question 37. If 2 x+y = 2 2x-ÿ = v8 then the respective value of and y are_

$1, \frac{1}{2}$
$\frac{1}{2}, 1$
$\frac{1}{2}, \frac{1}{2}$
None of these

Solution:

(a) is correct

+y _ 22x-y = V8 = V21 = 23/2

x+y =$\frac{3}{2}$ (1)

2x-y=$\frac{3}{2}$ (2)

(a)satisfies (1] and (2) both

Question 38. Let Ei,thenE2 are E1, and E2 two E2 are(2,-1)linear_.isequationsa solution in software equation variables E1 only d and y,(-2,-1)(0,1) is a Solution of for equation both the E2.only,equations

x= 0,y = 1;

2x-y = -l, 4x + y = 1

x + y = l,x-y = -1

x + 2y = 2, x + y =1

Solution:

(0; 1) satisfies E1 and E2 both

(2, -l) satisfies 1 Eqn. 2-1 = 1 (true)

But -2; -13 also satisfies E2

i.e -2 –13 = -1 (true)

Question 39. $\text { If } \frac{3}{x+y}+\frac{2}{x-y}=-1 \text { and } \frac{1}{x+y}-\frac{1}{x-y}=\frac{4}{3} \text { then }(x, y) \text { is }$

(2.1)
(1.2)
(-1,2)
(-24)

Solution:

⇒ $\frac{3}{1+2}+\frac{2}{1-2}=1-2=-1 \text { (True) }$

⇒ $\frac{1}{1+2}-\frac{1}{1-2}=\frac{1}{3}+1=\frac{4}{3} \text { (True) }$

So;(2) is correct.

Question 40. The line 3x + 2y = 6 intersects the line 3x – y = 12 in_ quadrant:

Solution:

(4), Eqn. (1) – Eqn. (2); we get

3x + 2y = 6

3x- y = 12

– + –

3y = -6 ⇒ y = -2

From (1); 3x = 6-2y=6-2 (-2) = 10

$ x=\frac{10}{3}$

Coordinate the point of intersection

= (x;y) = $\left(\frac{10}{3} ;-2\right)$

It is in the 4th Quadrant.

Exercise – 3

Quadratic Equations

Question 1. If the roots of the equation 2×2 + 8x – m’ = 0 are equal then value of m is

-3
– 1
l
-2

Solution:

(4) 2x² 8x- m3 = 0 for equal b2- 4ac = 0

→ 64 + 4.2. m³ = 0

8m³ = – 64

m3 = – 8

m= – 2

Question 2. If 2^x+3 – 32. 2x + 1 = 0 then values of x are

0,1
1,2
0,3
0,-3

Solution:

(4) 2^x+3– 32. 2x + 1 = 0

2^x+3– 9.2x +1 = 0

Question 3. The values of 4+_____1_____

4+_____1_____
4+_____1_____
4+…….CO
1. 1± √2
2. 2+ √5
3. 2± √5
4. None of these

Solution:

4x+1=x²

x² — 4x —1 = 0

x²– 4x- 1

⇒ $X=\frac{4 \pm \sqrt{16+4}}{2}$

⇒ $=2 \pm \sqrt{5}$

Question 4. If be the roots of the equation 2×2 – 4,v -3 = 0 the value of a2 + 32 is

5
7
3
-4

Solution:

(2) 2x²– 4x- 3 = 0

ap is root

→ x² – 2x- $\frac{3}{2}$

α²+ β² = (α + β)² – 2 α β

⇒ $(2)^2+2 \cdot \frac{3}{2}$=7

Question 5. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals then$\frac{a^2}{a c}+\frac{b c}{a^2}$ is equal to

2
-2
1
-1

Solution: (1) 2

Question 6. The equation x² -(p+4)x + 2p + 5 = 0 has equal roots the values of p will be.

± 1
2
± 2
-2

Solution:

(3) x² – (p + 4) x + 2p + 5 = 0

For equal roots

b² – 4ac = 0

(p + 4)² – 4 (2p + 5) = 0

p² – 4 = 0

p² = 4

P = ± 2

Question 7. The roots of the equation x2 + (2p-l)x + p2 = 0 are real if.

p>l
p<4
p>1/4
p<1/4

Solution:

(4) x² + (2p-1) x + pz = 0

For real roots

b² – 4ac > 0

(2p-1)²-4(p²)>0

4p² + 1 – 4p – 4p² > 0

1 – 4p > 0

1 > 4p

⇒ $\frac{1}{4} \geq \mathrm{p}$

Question 8. If p and q are the roots of x2 + 2x + 1 = 0 then the values of* + q:i becomes

2
-2
4
-4

Solution:

(2) p and q roots of x² +2x + 1 = 0

p + q = -2

pq = 1

p³+ q³=(p + q) (p²-pq + q²)

= -2 (p² + q²– 1)

= -2 (p+q)²– 2pq – 1)

= -2 (4 – 2 – 1) = -2

Question 9. If L + M + N = 0 and L, M, N are rationales the roots of the equation
(M+N-L)x2+(N+L-M)x+(L+M-N) = 0 are

Real and irrational
Real and rational
Imaginary and equal
Real and equal

Solution:

(2) Real and Rational

Question 10. If one root of 5x² + 13x + p = 0 is reciprocal of the other then the value is

-5
5
1/5
-1/5

Solution:

(2) 5×2 + 13x + p = 0

⇒ $\text { roots } \rightarrow \alpha \cdot \frac{1}{\alpha}$

⇒ ${sum}=\alpha+\frac{1}{\alpha} \quad \text { product }=\frac{p}{5}$

⇒ $\frac{p}{5}=1 \quad \text { p }=5$

Question 11. A solution of the quadratic equation (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0 is

x = 1
x = -1
x= 2
x = – 2

Solution:

(2) (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0

For x = 1

a + b-2c + 2a-b-c + c + a-2b≠0

for x = -1

a + b -2c -2a + b + c + c + a -2b = 0

∴ x = – 1

Question 12. If the root of the equation x2-8x+m = 0 exceeds the other by 4 then the value of m is

m = 10
m= 11
m = 9
m = 12

Solution:

(4) x² – 8x + m = 0

roots α, α + 4

sum = 8

α+α+4=82 α=4

Product = m

α (α + 4) = m

α² + 4α = m

4 + 8 = m m = 12

Question 13. The values of x in the equation 7(x+2p)² + 5p² = 35xp + 1 17p² are

(4p,-3p)
(4p,3p)
(-4p, 3p)
(-4p,-3p)

Solution:

(1) 7(x + 2p)² + 5p² = 35 xp + 117p²

(4p, -3p)

Question 14. H. The solutions of the equation $\frac{6 x}{x+1}+\frac{6(x+1)}{x}=13 \text { are }$

(2, 3)
(3,-2)
(-2,-3)
(2,-3)

Solution:

⇒ $\frac{6 x}{x+1}+\frac{6(x+1)}{x}=13$

(2,-3)

Question 15. The satisfying values of.v for the equation$\frac{1}{x+p+q}=\frac{1}{x}+\frac{1}{p}+\frac{1}{q} \text { are }$

(p. q)
(-p. -q)
(p, -p)
(-P, q)

Solution:

⇒ $\text { (2) } \frac{1}{x+p+q}=\frac{1}{x}+\frac{1}{p}+\frac{1}{q}$

(-p. -q)

Question 16. The values of for the equation x² + 9x + 18 = 6 – 4x are

(1,12)
(-1,-12)
(1,-12)
(-1,12)

Solution:

(2) x² + 9x + 18 = 6 – 4x

⇒ x² + 13x + 12 = 0

⇒ x² + 12x + x + 12 = 0

⇒ x(x+12) + 1 (x+ 12) = 0

x = -1,-12

Question 17. The values of x satisfying the equation $\sqrt{\left(2 x^2+5 x-2\right)}-\sqrt{\left(2 x^2+5 x-9\right)}=1 \text { are }$

(2, -9/2)
(4,-9)
(2, 9/2)
(-2, 9/2)

Solution:

⇒ $\text { (1) } \sqrt{2 x^2+5 x-2}-\sqrt{2 x^2+5 x-9}=1$

(2. -9/2)

Question 18. The solution of the equation 3x²-17x + 24 = 0 are

(2,3)
$(b) \left(2,3 \frac{2}{3}\right)$
$\left(3,2 \frac{2}{3}\right)$
$\left(3, \frac{2}{3}\right)$

Solution:

(3) 3x² – 17x + 24 = 0

3x² – 8x -9x + 24 = 0

3x (x- 3) – 8 (x- 3) = 0

(3x – 8) (x- 3) = 0

⇒ $x=\frac{8}{3}, x=3$

⇒ $x=3,2 \frac{2}{3}$

Question 19.The equation $\frac{3\left(3 x^2+15\right)}{6}+2 x^2+9=\frac{2 x^2+96}{7}+6$ has got the solution as

(1, 1)
(1/2, -1)
(1. -1)
(2,-1)

Solution:

⇒ $\text { (c) } \frac{3\left(3 x^2+15\right)}{6}+2 x^2+9=\frac{2 x^2+96}{7}+6$

⇒ $\frac{9 x^2+45+12 x^2+54}{6}=\frac{2 x^2+96+42}{7}$

⇒ $\frac{21 x^2+99}{6}=\frac{2 x^2+138}{7}$

Multiply 42

⇒ 147x² + 693 = 12x² + 828

⇒ 135x² = 135

x² =1 → x ±1

Question 20. The sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are

(7, 10)
(4,4)
(3, 5)
(2,6)

Solution:

3 + 5 = 8 ⇒ 3² + 5² = 34

(3,5)

Question 21. The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are

(7,4)
(5, 8)
(3, 6)
(2,5)

Solution:

(2) let two positive integers be a, 3

⇒ α-β = 3

⇒ α² +β² = 89

(5,8)

Question 22. Five times of a positive whole number is 3 less than twice the square of the number. The number is

3
4
-3
2

Solution:

(1) let the number be x

⇒ 5x = 2x² -3

⇒ 2x² – 5x – 3 = 0

⇒ 2x² – 6x + lx – 3 = 0

⇒ 2x(x-3) + 1 (x-3) = 0

⇒ $x=\frac{-1}{2}, x=3$

x is positive

x = 3

Question 23. The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solving it to find the length and breadth of the field. The length and breadth are

(205m, 80m)
(50m, 40m)
(60m, 50m)
None

Solution:

(2) Perimeter = 180m

2(1 + b) = 180

L + b = 90

We know I = x

b = 90 – x

Area = 2000 m

∴ x (90 – x) = 2000

⇒ 90x- X² = 2000

⇒ x²-90x + 2000 = 0

⇒ x² – 5Ox – 40x + 2000 = 0

⇒ x (x-50) – 40 (x- 50) = 0

x = 40, 50

Question 24. Two squares have sides p cm and (p + 5) cms. The sum of their squares is 625 sq. cm. The sides of the squares are

(10 cm, 30 cm)
(12 cm, 25 cm)
(15 cm, 20 cm)
None of these

Solution:

(3) Two squares with sides p, p + s cm

⇒ p² + (p+5)² = 625

⇒ p² + p² + 25 + lOp = 625

⇒ 2p²+10p- 600 = 0

⇒ p² + 5p – 300 = 0

⇒ p2 + 20p-15p- 300 = 0

⇒ p (p + 20) – 15 (p+20) = 0

⇒ p = 15,p = -20

⇒ p≠ -20

⇒ p = 15

⇒ p + 5 = 20

Question 25. Divide 50 into two parts such that the sum of their reciprocals is 1/12. The numbers are

(24, 26)
(28, 22)
(27,23)
(20,30)

Solution:

(4) let the first part be x,

second = 50 – x

⇒ $\frac{1}{x}+\frac{1}{50-x}=\frac{1}{2}$

⇒ $\frac{50-x+x}{x(50-x)}=\frac{1}{12}$

⇒ 600 = x (50 – x)

⇒ 600 = 50x – x²

⇒ x²-50x + 600 = 0

⇒ x²– 30x- 20x + 600 = 0

⇒ x (x – 30) – 20 (x- 30) = 0

∴ x = 20, 30

Question 26. There are two consecutive numbers such that the difference of their reciprocals is 1/240. The numbers are

(15, 16)
(17, 18)
(13, 14)
(12, 13)

Solution:

(1) Two numbers be x, x + 1

⇒ $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{240}$

⇒ $\frac{x-x+1}{x(x+1)}=\frac{1}{240} \rightarrow+240=x(x+1)$

⇒ x² + x – 240 = 0

⇒ x² + x- 240 = 0

⇒ x²+ 16x- 15x- 240 = 0

⇒ x (x + 16) – 15 (x +16) = 0

⇒ x = 15, x = -16

∴ x= 15

(15,16)

Question 27. The hypotenuse of a right-angled triangle is 20cm. The difference between its other two sides is 4cm. The sides are

(11cm, 15cm)
(12cm, 16cm)
(20cm, 24cm)
None of these

Solution:

(2) hypotenuse = 20 cm

The difference between the other two sides a and, b is 4

⇒ a – b = 4

⇒ a = 4 + b

⇒ a² + b² = c²

⇒ a² + (a-4)²= c²

⇒ a² + a²+ 16 – 8a = 20²

⇒ 2a² + 8a = 384

⇒ 2a²+ 8a – 384 = 0

⇒ a² – 4a – 192 – 0

⇒ a²– 16a + 12a – 192 = 0

⇒ a (a- 16) + 12 (a—16) = 0

⇒ a = 16

∴ b = a-4 = 12

Question 28. The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are

(15,30)
(32, 13)
(36,9)
(25,20)

Solution:

(3) → 4 (1 = 45)

⇒ $\frac{a}{x}=\frac{x}{\beta} \rightarrow \alpha \beta=x^2$

αβ= 18²

(36.9)

Question 29. The sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right-angle triangle is formed. The side of the equilateral triangle is

17 units
16 units
15 units
18 units

Solution:

(1) let the side be x

New sides

x- 12, x- 13, x- 14

new A is night-angled

(x – 14)²+ (x- 13)² = (x – 12)²

Question 30. A distributor of Apple Juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D (in number of bottles) is given by D = -2000p2 + 2000p + 17000. The price per bottle that will result in zero inventory is

3
5
2
none of these.

Solution:

(1) D = -2000p²+ 2000p + 17000

For zero inventory D = supply

5000 = – 2000p² + 2000p + 17000

2000p² = -2000p – 12000 = 0

p² – p – 6 = 0

p²– 3p + 2p- 6 = 0

P (P – 3) + 2 (p- 3) = 0

P = 3

Question 31. The sum of two irrational numbers multiplied by the larger one is 70 and their difference multiplied by the smaller one is 12; the two numbers are

3√2, 2√3
5√2, 3√5
2√2, 5√2
none of these.

Solution:

⇒ $\alpha(\alpha+\beta)=70$

⇒ $\alpha^2+\alpha \beta=70$

⇒ $\alpha \beta=70-\alpha^2$

⇒ $\beta(\alpha-\beta)=12$

⇒ $\alpha \beta-\beta^2=12$

⇒ $\alpha \beta=12+\beta^2$

⇒ $70-\alpha^2=12+\beta^2$

⇒ $ \alpha^2+\beta^2=58$

⇒ $2 \sqrt{2}, 5 \sqrt{2}$

Question 32. Solving equation x² – (a+b) x + ab = 0 are, value(s) of

a, b
a
b
None

Solution:

(1)

x² – (a + b) x + ab = 0

a, b are values of x

a, b

Question 33. Solving equation x² – 24x + 135 = 0 are, value(s) of x

9, 6
9,15
15,6
None

Solution:

(2) 135-0

x² – 15x -9×4 135 = 0

x- 15,9

Question 34. If $\frac{x}{b}+\frac{b}{x}=\frac{a}{b}+\frac{b}{a}$ the roots of the equation are

a, b² / a
a² , b/a²
a² , b² /a
a, b²

Solution:

⇒ $\text { (a) } \frac{x}{b}+\frac{b}{x}=\frac{a}{b}+\frac{b}{a}$

⇒ $\frac{x^2+b^2}{x b}=\frac{a^2+b^2}{a b}=a x^2+a b^2=a^2 x+b^2 x$

⇒ $a x^2-a^2 x-b^2 x+a b^2=0$

⇒ $a x^2-\left(a^2+b^2\right) x+a b^2=0$

⇒ $x^2-\left(\frac{a^2+b^2}{a}\right) x+b^2=0 $

⇒ $ x^2-\frac{\left(a+b^2\right)}{a} x+b^2=0$

⇒ $\text { Iwo roots } \rightarrow \mathrm{a}, \frac{b^2}{a}$

Question 35. Solving equation $\frac{6 x+2}{4}+\frac{2 x^2-1}{2 x^2+2}=\frac{10 x-1}{4 x}$We get roots as

±l
+l
-1
0

Solution:

⇒ $\frac{6 x+2}{4}+\frac{2 x^2-1}{2 x^2+2}=\frac{10 x-1}{4 x}$

⇒ $x= \pm 1 \rightarrow 2+\frac{1}{4}=\frac{9}{4}$

⇒ $\text { for } x=-1 \rightarrow-1+\frac{1}{4}=\frac{11}{4}$

⇒ $x=1 \rightarrow 2+\frac{1}{4}=\frac{9}{4}$

Question 36. Solving equation 3x² – 14x +16 = 0 we get roots as

±l
2 and$\frac{8}{3}$
0
None

Solution:

(2) 3x² – 8x – 6x 416 = 0

3x² – 6x – 8 x 416 = 0

3x (x – 2) – 8 (x- 2) = 0

⇒ $x=\frac{8}{3},2$

Question 37. Solving equation 3x² – 14x + 8 = 0 we get roots as

±4
+2
$4,\frac{2}{3}$
None

Solution:

(3) 3x² -14x + 8 = 0

3x² – 12x -2x+9 = 0

3x (x – 4) -2 (x – 4) = 0

⇒ $x=\frac{2}{3}, x=4$

Question 38. Solving equation (b-c) x² + (c-a)x + (a-b) = 0 following roots are obtained

$\frac{a-b}{b-c}, 1$
(a-b) (a-c),1
$\frac{b-c}{a-b}, 1$
None

Solution:

(a) (b- c) x2 4 (c – a) x 4 (a – b) = 0

Sum of roots=$\frac{-(c-a)}{(b-c)}=\frac{a-c}{b-c}$

Product =$\frac{a-b}{b-c}$

⇒ $\frac{a-b}{b-c}, 1$

Question 39. Solving equation 7 $\sqrt{\frac{x}{x-1}}+8 \sqrt{\frac{x-1}{x}}=15$ Following roots are obtained.

$\frac{64}{113}, \frac{1}{2}$
$\frac{1}{50}, \frac{1}{65}$
$\frac{49}{101} \frac{1}{65}$
$\frac{1}{50}, \frac{64}{65}$

Solution:

⇒ $\text { (a) } \sqrt[7]{\frac{x}{x-1}}+\sqrt[8]{\frac{x-1}{x}}=15$

Question 40. Solving equalion 6 $\sqrt{\frac{x}{x-1}}+8 \sqrt{\frac{x-1}{x}}=13$ Following roots are obtained.

$\frac{4}{13}, \frac{9}{13}\frac{-4}{13}, \frac{-4}{19}$
$frac{4}{13}, \frac{5}{13}$
$\frac{6}{13},$
$\frac{7}{13}$

Solution:

⇒ $6\left[\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\right]=13$

⇒ $\frac{4}{13}, \frac{9}{13}$

Question 41. Solving equation z² -6z + 9 = 4$\sqrt{z^2-6 z+6}$ following roots are obtained.

$\text 3+2 \sqrt{3}, 3-2 \sqrt{3}$
5, 1
all the above
None

Solution:

(3 $z^2-6 z+9=4 \sqrt{z^2-6 z+6}$

All of the above

Question 42. Solving equation$\frac{x 1 \sqrt{12 p-x}}{x-\sqrt{12 p-x}}=\frac{\sqrt{p+1}}{\sqrt{p-1}}$ following roots are obtained.

3p
both 3p and – 4p
Only -4p
-3p4p

Solution:

(1) 3p

Question 43. Solving equation $z+\sqrt{z}=\frac{6}{25}$

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{1}{25}$
$\frac{2}{25}$

Solution:

(3) let √z = a

⇒ $ a 2+a=\frac{6}{25}$

⇒ $25 a 2+25 a-6=0$

⇒ $a=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

⇒ $\frac{-25 \pm \sqrt{625-4(25)(-6)}}{50}$

⇒ $\frac{-25 \pm \sqrt{625+600}}{50}$

⇒ $\frac{-25 \pm \sqrt{1225}}{50}$

⇒ $\frac{-25 \pm 35}{50}=\frac{10}{50} \text { or } \frac{-60}{50}=\frac{1}{5} \text { or } \frac{-6}{5}$

⇒ $ a=\sqrt{z} \rightarrow a 2=z $

⇒ $z=\frac{1}{25} \text { or } \frac{36}{25}$

Question 44. Solving equation$z^{10}-33z^5+32=0$the following values of x are obtained

1,2
2,3
2, 1
1,2,3

Solution:

(1) z¹⁰-33z⁵ +32=0

Let z⁵ = a

a² -33a + 32 = 0

a² -32a -1a + 32 = 0

a (a – 32) – 1 (a – 32) = 0

a = 1,32

z⁵= 1,32

= 1,2

Question 45. When $\sqrt{2z+1}+\sqrt{3z+4}=$ 7 the value of z is given by

Solution:

(4) 4

Question 46. Solving equation 6x⁴ + 11x³– 9x²– 11 x + 6 = 0 following roots are obtained

$\frac{1}{2}, -2 \frac{-11 \sqrt{37}}{6}$
$-\frac{1}{2}, 2 \frac{-11 \sqrt{17}}{6}$
$\frac{1}{2}, 2 \frac{5}{6} \frac{-7}{6}$
None

Solution:

(1) 6x⁴ + 11x³– 9x²– llx + 6 = 0

⇒ $\text { For } \frac{1}{2}$

⇒ $6\left(\frac{1}{2}\right)^4+11\left(\frac{1}{2}\right)^3-9\left(\frac{1}{2}\right)^2-11\left(\frac{1}{2}\right)+6$

⇒ $6\left(\frac{1}{16}\right)+11\left(\frac{1}{8}\right)-9\left(\frac{1}{4}\right)-\frac{11}{2}+6$

⇒ $\frac{6}{16}+\frac{22}{16}-\frac{36}{16}-\frac{88}{16}+\frac{96}{16}=0$

For-2

6(—2)⁴+l 1(—2)³ -9 ( —2)2—11 (-2)+6

= 96-88-36 + 22 + 6 = 0

⇒ $\frac{1}{2},-2, \frac{-1 \pm \sqrt{37}}{6}$

Question 47. $\text { If } \frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{x-1}{x+3}-\frac{x+3}{x-3}$ then the values of x are

$0, \pm \sqrt{6}$
$0, \pm \sqrt{3}$
$0, \pm 2 \sqrt{3}$
None of these

Solution:

$\text { For } 0 \frac{2}{-2}+\frac{2}{2}=\frac{-1}{3}+\frac{3}{-3}$ $-1+1 \neq \frac{-1}{3}-1$

0 is not a solution but all options have 0

None of these

Question 48. If $\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}$ then the values of x are

0,(a+b),(a-b)
$0,(a+b), \frac{a^2+b^2}{a+b}$
$0,(a-b) \cdot \frac{a^2+b^2}{a+b}$
$\frac{a^2+b^2}{a+b}$

Solution:

⇒ $0,(a-b) \cdot \frac{a^2+b^2}{a+b}$

Question 49. Solving equation $\left(x-\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+12=0$ we get roots as follows

0
1
-l
None

Solution:

⇒ $\text { (2) }\left(x-\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+12=0$

For 1 (O)2 – 6 (2) + 12 = 0

Question 50. Solving equation $\left(x-\frac{1}{x}\right)^2-10\left(x-\frac{1}{x}\right)+24=0$ we get roots as follows

0
1
-l
$(2 \pm \sqrt{5}),(3 \pm \sqrt{10})$

Solution:

⇒ $\text { (4) }\left(x-\frac{1}{x}\right)^2-10\left(x-\frac{1}{x}\right)+24=0$

(2±√5), (3±√I0)

Question 51. Solving equation $ 2\left(x-\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}+2\right)+18=0$ we get roots as under

0
1
-1
-2± √3

Solution:

⇒ $\text { (2) } 2\left(x-\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}+2\right)+18=0$

a, b, c do not satisfy

-2±√3

Question 52. If αβ the roots of equation x² – 5x + 6=0 and α > β then the equation with roots (α + β)and (α – β) is

x² -6x+5=0
2x² -6x+5=0
2x² -5x+6=0
x² -5x+6=0

Solution:

(1) α+ β = 5 αβ=6

x2- 5x + 6

x2 – 3x- 2x + 6

x (x- 3) – 2 (x – 3)

x = 2 or x = 3

α = 3, β = 2 α = 9 β2 = 4

α – β = 1

new equation→

x² – (6) x + 5 = 0

x² – 6x + 5 = 0

Question 53. If αβ are the roots of equation x²-5x+6=0 α > β then the equation with roots(αβ+γ + β) and (α + β²) is

x²-9x+99=0
x²-18x+90=0
x²-18x+77=0
None

Solution:

x²-18x+77=0

Question 54. If alpha beta are the roots of equation x²-5x+6=0 and γ > β then the equation with roots (γβ +γ+β)and )γβ – γ-β)

x²-12x+11=0
2x² -6x+12=0
x²-12x+12=0
None

Solution:

(1) x² -5x+6=0

x²– 3x- 2x + 6 = 0

x = 3, x = 2

α= 3, β = 2

αβ= 6

αβ + α + β= 6 + 3 + 2 =11

αβ – α – β= 6- 3- 2=1

new equation→

x² – (sum) x + product = 0

x²-12x +ll = 0

Question 55. The condition that one or ax2+bx+c=0 the roots of is twice the other is

b²=4ca
2b²=9(c+a)
2b²=9ca
2b²=9(c-a)

Solution:

(3) let two roots be a, 2a

⇒ $3 \alpha=\frac{-b}{a}, 2 \alpha^2=\frac{c}{a}$

⇒ $\alpha=\frac{-b}{3 a} \quad \alpha^2=\frac{c}{2 a}$

⇒ $a^2=\frac{b^2}{9 a^2}\frac{b^2}{9 a^2}=\frac{c}{2 a}$

⇒ $\rightarrow b^2=\frac{c .9 a^2}{2 a}$

⇒ $\rightarrow 2 b^2=c a 9 \quad \rightarrow 2 b^2=9 c a$

Question 56. The condition that one of ax2+bx+c=0 the roots of is thrice the other is

3b²=16ca
b²=9ca
3b²=-16ca
b²=-9ca

Solution:

(1) ax²+bx+c=0

roots → α, 3α

α + 3α =$\frac{-b}{a}$

⇒ $4 \alpha=\frac{-b}{a} \rightarrow \alpha$

⇒ $\frac{-b}{4 a} \rightarrow \alpha^2$

⇒ $\frac{b^2}{16 a^2}$

⇒ $a(3 a)=3 t^2=\frac{c}{a}$

⇒ $a^2=\frac{c}{3 a}=\frac{b^2}{16 a^2}$

⇒ $b^2=\frac{16 a^2 c}{3 a}$

⇒ $3 b^2=16 \mathrm{ac}$

Question 57. If the roots of ax²+bx+c=0 are in the ratic$\frac{p}{q}$ then the value of$\frac{b^2}{(c a)} \text { is }$

$\frac{(p+q)^2}{(p q)}$
$\frac{(p+q)}{(p q)}$
$\frac{(p-q)^2}{(p q)}$
$\frac{(p-q)}{(p q)}$

Solution:

⇒ $\frac{-b}{a}=p y+q y \frac{c}{a}=p q y^2$

⇒ $-b=\mathrm{a}(\mathrm{py}+\mathrm{qy}) \mathrm{c}=a p q y^2$

⇒ $\mathrm{~b}^2=\mathrm{a}^2(\mathrm{py}+\mathrm{qy})^2$

⇒ $\frac{b^2}{c a}=\frac{a^2(p y+q y)^2}{a p q y^2 a}=\frac{(p+q)^2}{p q}$

Question 58. Solving equation$\sqrt{x^2-9 x+18}+\sqrt{x^2+2 x-15}=\sqrt{x^2-4 x+3}$ following roots are obtained

$3, \frac{2 \pm \sqrt{94}}{3}$
$\frac{2 \pm \sqrt{94}}{3}$
$4,-\frac{8}{3}$
$3,4-\frac{8}{3}$

Solution:

⇒ $\sqrt{x^2-9 x+18}+\sqrt{x^2+2 x-15}=\sqrt{x^2-4 x+3}$

⇒ $\sqrt{x^2-6 x-3 x+18}+\sqrt{x^2+5 x-3 x-15}$

⇒ $\sqrt{(x)(x-6)-3}$

⇒ $\sqrt{(x)(x-6)-3(x-6)}+\sqrt{x(x+5)-3(x+5)}$

⇒$\sqrt{x(x-3)-1(x-3)}$

⇒$\sqrt{(x-3)(x-6)}+\sqrt{(x-3)(x+5)}$

⇒ $\sqrt{(x-1)(x-3)}$

⇒ $3, \frac{2 \pm \sqrt{94}}{3}$

Question 59. Solving equation $\sqrt{y^2+4 y-21}+\sqrt{y^2-y-6}=\sqrt{6 y^2-5 y-39}$ following roots are obtained

2, 3, 5/3
2, 3, -5/3
-2, -3, 5/3
-2, -3, -5/3

Solution:

⇒$\sqrt{y^2+4 y-21}+\sqrt{y^2-y-6}=\sqrt{6 y^2-5 y-39}$

⇒$\sqrt{y^2+7 y-3 y-21}+\sqrt{y^2-3 y+2 y-6}$

⇒$\sqrt{6 y^2 18 y+13 y-39}$

⇒$\sqrt{y(y+7)-3(y+7)}+\sqrt{y(y-3)+2(y-3)}$

⇒$\sqrt{6 y(y-3)+13(y-3)}$

⇒$\sqrt{(y-3)(y+7)}+\sqrt{(y+2)(y-3)}$

⇒$\sqrt{(y-3)(6 y+13)}$

⇒$2,3, \frac{-5}{3}$

Question 60. Find the positive value of k for which the equations: x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots:

Solution:

(2) for real roots

D = b2 – 4ac> 0

For eqn (1]; k² ≥ 4.1.64 ⇒ k > 16……(1)

For eqn (2) (-8)^2≥ 4.1k⇒ k < 16……(2)

∴ (2) is correct.

Question 61. If one root of an equation is 2 + √5. Then the quadric equation is:

x² + 4x -1 = 0
x² – 4x – 1 = 0
x² + 4x +1 = 0
x² – 4x + 1 = 0

Solution:

(2) if one root = 2 + √5 (Irrational conjugate)

Eqn is

X² – (sum of roots) x + product of roots = 0

Or x²– (2 + √5 + 2 -√5)x+ (2 + √5) (2 – √5) = 0

Or x²– 4x + (4- 5) = 0

0r x²-4x-1 = 0

(2) is correct.

Question 62. A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was K 1,500 after 4 years of service and 1,800 after 10 years of service, what was his starting salary and what is the annual increment in rupees?

1,300, 50
1,00, 50
1,500, 30
None

Solution:

For (1) salary after 4 years 1300 + 4 x 50 =1500……..(1)

Salary after 10 years = 1300 + 10 x 50 =1800……….(2)

Detail Method

Let fixed salary = a

Increment per yr = b

Salary after t yrs = s

s = a + bt

a + b x 4 = 1500 = a + 4b = 1500…

a + bx 10= 1800 = a + 10b = 1800.

Eqn. (2) – eqn. (1); we get

a + bx 10= 1800

a + 4b =1500

6b = 300 b =$b=\frac{300}{6} ₹ 50$

From (1) a + 4 x 50 = 1500

Or a = 1500 – 200 = 1300

(1) is correct

Question 63. The sides of an equilateral triangle are shortened by 1 2 units, 13 units, and 14 units respectively, and a right-angled triangle is formed. The side of the equilateral triangle is:

17 units
16 units
15 units
18 units

Solution:

For (a] let each side = 17 units.

∴ sides are 17 – 12; 17 – 13 and 17 – 14

= 5; 4 and 3 units respectively

They make a right-angled triangle because the sum of squares of two sides is equal to
square of the largest side.

i.e., 42 + 32 = 52

(1) is correct.

Detail method

Let the length of each side = x

Side of the right angle triangle are x – 12 ; x – 13 and x – 14

by Pythagoras theorem

(x- 12)² = (x- 13)² + (X- 14)²

Or x² – 24x + 144 = x² – 26x + 167 + x² – 28x + 196

Or 0 = x² – 30x + 221

Or x²-17x-13x + 221 = 0

Orx(x-17)-13(x-17) =0

Or (x-17) (x-13) = 0

x = 17; 13

No side = x-13 = 0

x = 17

(a) is correct.

Question 64.The value of $\sqrt{6+\sqrt{6+\sqrt{6+\cdots \ldots to \infty}}} \text { is: }$

Solution:

⇒ $\text { (3) } \sqrt{6+\sqrt{6+\sqrt{6+\cdots \ldots \text { to } \infty}}}=+3$

Find two factors of 6 such that their difference becomes 1.

6 = 3×2

(1) Give + sign for = greater factor

(2) – sign for smaller factor

(3) is correct

Question 65. The area of a rectangular garden is 8000 square meters. The ratio in length and breadth is 5: 4 A path of uniform width, runs all-round the inside of the garden. If the part occupies 3200 m2, what is its width?

Solution:

(3) let x be common in the ratio

length = 5x anil breadth = 4x

Area = 5x x 4x = 8000

Or 20x² = 8000

Or x² = 400 = 20²

∴ x = 20

Length = 5x = 5 x 20 = 100m

Breadth = 4x = 4 x 20 = 80m

Let Width of path = ym

∴ Area of rest part = 8000 – 3200

Or (100 – 2y) (80- 2y) = 4800

Tricks: go by choices.

∴ (3) y = 10 m satisfies it

Width = 10m is correct

Question 66. If (2 +V3) is a root of a quadratic equation x² + px + q = 0 then find the value of p and q.

(4,-1)
(4, 1)
(-4, 1)
(2,3)

Solution:

One root = 2 + V3

Another root = 2- V3

Eqn is

x² — (2 + √3 + 2 – √3)x + √2+√3) (2 – √3) = 0

Or x² – 4x + (4- 3) = 0

Comparing it with x + px + q = 0

p = -4 and q =1

(3) is correct

Question 67. If the area and perimeter of a rectangle are 6000 cm² and 340cm respectively, then the length of a rectangle is:

Solution:

(2)

let length =1

and breadth = b

lb = 6000 _

2(l + b) = 340…….(1)

Or 1 +b = 170…….(2)

Then Go by choices

For (a) if 1 = 140 then b = 170 – 140 = 30

lb = 140 x 30 = 4200* 6000

(a) is not the answer

For (2) if 1 = 120

Then 120 + b = 170

b= 170 -120 = 50

Area = lb = 120 x 50 = 6000

Question 68. A straight line passes through the point (3,2). Find the equations of the straight line.

x + y =1
x + y = 3
x + y = 5
x + y = 2

Solution:

(3) it cannot be solved properly because only one condition is given for the answer

(3) Point (3 ; 2) satisfies x + y = 5

(3) is correct

Question 69. One root of the equation: x² – 2(5+m)x+3(7+m>0 is reciprocal of the other. Find the value of m.

-20/3
7
1/7
117

Solution:

(1)

Let one root = or, then other root =$=\frac{1}{\alpha}$

⇒ $\alpha \times \frac{1}{a}=\frac{3(7+m)}{1}$

1 = 21 + 3m

⇒ $3 m=-2m=\frac{-20}{3}$

(1) is correct

Question 70. Roots of the equation 3x² – 14x – k = 0 will be reciprocal of each other if:

k = -3
k = 0
k = 3
k = 14

Solution:

(3) is correct.

Let one root = a and another root =$=\frac{1}{\alpha}$ (given)

Product of roots = $=\frac{c}{a}$

⇒ $\alpha \times \frac{1}{\alpha}=-\frac{k}{3}$

⇒ $-1=\frac{k}{3} \text { So, K }=-3$

(1) is correct

Question 71. Positive value of for which the roots at equation 12x² + kx + 5 = 0 are in ratio 3 : 2, is

5/12
12/5
$\frac{5 \sqrt{10}}{2}$
5√10

Solution:

(4)

Let a be common in the ratio.

Roots are 3 a and 2 a

Sum of roots =$-\frac{b}{a}$
.
3a + 2a = 5a =$-\frac{k}{12}$

⇒ $\alpha=-\frac{k}{60}$

Product of roots = $3 \alpha+2 \alpha=\frac{c}{a}$

⇒ $6 x^2=\frac{5}{12}$

⇒ $\text { 6. }\left(-\frac{k}{60}\right)^2=\frac{5}{12}, \text { so, } \mathrm{k}^2=250$

k = 5√10

Question 72. If one root of the equation x²– 3x + k = 0 is 2, then value of k will be

Solution:

(3) is correct

2 is a root of given eqn.

2²– 3 x 2 + k = 0

Or-1 + k = 0

∴ k = 2

Question 73. It roots of equation x² + x + r = 0 are ‘a’ and ‘α’ and β = -6. Find the value of V.

$\frac{-5}{3}$
$\frac{7}{3}$
$\frac{-4}{3}$
1

Solution:

(1) is correct

⇒ $\alpha+\beta=\frac{-1}{1}=-1 \ and \alpha \beta=\frac{r}{1}=r$

α³ + β³ = -6

or (α + β)³– 3αβ (α + β) = -6

or (-1)3 — 3r(—1) = -6

Or- 1 + 3r = -6; or 3r = -5

r = -5/3

Question 74. If one root of the equation Px²+ qx + r = 0 is r then the other root of the equation will be

1/q
1/r
1/p
$\frac{1}{p+q}$

Solution:

(3) let a is another root.

⇒ $\mathrm{r} \alpha=\frac{r}{p} \alpha=\frac{1}{p}$

Question 75. If the ratio of the root of the equation 4x² – 6x + p = 0 is 1 : 2 then the value of is

1
2
-2
-l

Solution:

(2) let α be common in the ratio

⇒ $ \alpha+2 \pi=\frac{p}{4}p=8 \alpha^2=8 \cdot \frac{1}{4}=2$

Question 76. If p and q are the root of the equations x² – bx + c = 0, what is the equation whose roots are (pq + p + q) and (pq – p – q)?

x² – 2cx + c² – b² = 0
x² – 2cx + c² + b²= 0
8cx² – 2(a+c)x + c² = 0
cx² + 2bx – (c² – b²) = 0

Solution:

(1)

x² – (p + q)x + pq = 0

b = p + q ; C = pq

New roots are

pq + (p + q) = c + b

pq-(p + q) = c-b

Eqn is

x² – (c + b + c- b)x +(c + b)(c- b) = 0

or x² – 2cx + c² – b² = 0

Question 77. If the arithmetic mean between the roots of a quadratic equation is 8 and the geometric mean between them is 5, the equation is-

x²– 16x-25 = 0
x²– 16x + 25 =
x² – 16x + 5 = 0
None of these

Solution:

(2) let α and β be two roots.

(α+β)/2 = 8 and √αβ = 5

⇒ α+β = 16 and αβ = 25

Eqn. is x²-16x + 25 = 0

Question 78. The minimum value of the function x²– 6x + 10 is-

Solution:

(1) coeff. Of x2 = 1 > 0; function is minimum (formula)

Minimum value =$\frac{4 a c-b^2}{4 a}$

⇒ $=\frac{4.1 .10-(-b)^2}{4 \times 1}=\frac{4}{4}=1$

Question 79. If one of the roots of the equation x² + px + a is √3 + 2. Then the value of and ‘a’ is

-4,-1
4, -1
4,1
4,1

Solution:

(3) Roots are 2 + √3 and 2 —V3

(conjugate of 2+√3)

Eqn is

x² – (sum of roots) x + product of roots = 0

x²– 4x + (4- 3) = 0

x² + px + a = 0

∴ P = -4; a = 1

Question 80. The roots of equation 2x² + 3x + 7 = 0 are α and β. The value of a αβ^-1+βα^-1

2
3/7
7/2
-19/14

Solution:

⇒ $\text { (d) } \alpha+\beta=-\frac{3}{2} ; \alpha \beta=7 / 2$

⇒ $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-\alpha \beta}{\alpha \beta}$

⇒ $\frac{\frac{9}{4}-2 \cdot \frac{7}{2}}{\frac{7}{2}}=\frac{9-28}{4} \times \frac{2}{7}=\frac{-19}{14}$

Question 81. The quadratic equation x²– 2kx + 1 6 = 0 will have equal roots when the value of is –

Solution:

(4) let roots are or; or

⇒ $\alpha+\alpha=\frac{-(-2 k)}{1} \Rightarrow \alpha=k$

= α.α = k.k => k² = 16 => k = ±4

Question 82. If α, βare roots of x² + 7x+ 11 = 0 then the equation whose roots as (α+β)² And (α – β)²

x²-54x+ 245 = 0
x² — 14x + 49 = 0
x²-24x+ 144 = 0
x² – 50x + 49 = 0

Solution:

(1) is correct

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{7}{1}=-7$

⇒ $\alpha \beta=\frac{c}{a}=\frac{11}{1}=11$

(α-β)² = (α+β)²– 4ap

= (—7)² -4×11 = 5

Require eqn. is

x² (α+β)² + (α-β)² = 0

or x² – (49+5)x + 49 x 5 = 0

or x²-5x + 245 = 0

(1) is correct

Question 83. If b² – 4ac is a perfect square but not equal to zero then the roots of the equation ax2 + bx + c = 0 are

Real and equal
Real irrational and equal
Real rational and unequal
Imaginary

Solution:

(3) b²– 4ac > 0 and perfect square

Question 84. Divide 80 into two parts so that their products are maximum then the numbers are

15,65
25,55
35,45
40,40

Solution:

(2) is correct

Let 1st part = x

2ml part = 80 – x

Lety = x(80-x) = -x² + 80x

Here co. eff. of x²< 0

Numbers are (40; 40)

Question 85. If a,p is the roots of a quadratic equation if x² + 2x – 3 find the quadratic equation:

x² + 2x – 7 = 0
x² + 2x – 3 = 0
x² – 2x – 3 = 0
x² – 2x +7 = 0

Solution:

(2)

Quadratic eqn is

x2 – (a + p)x + ap = 0

x2 – (-2)x + (-3) = 0

x2 + 2x- 3 = 0

Question 86. Value of k for which roots are equal of given equation 4x² – 12x + k = 0:

144
9
5
None of these

Solution:

(2)

4×2- 12x + k = 0

D = b2 – 4.ac = 0

(-12)2 = 4 x 4.k

Or. 144= 16k

k = 9

Question 87. If difference between the roots of the equation x² – kx + 8 = 0 is 4 then the value of K is

0
±4
±8√3
±4√3

Solution:

(4)

Let a; /? are roots of x² – kx + 8 = 0

⇒ $\alpha+\beta=b / a=-\frac{(-k)}{1}=k \& \alpha \cdot \beta=c / a=8 / 1 $

⇒ $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=4^2$

⇒ $\Rightarrow k^2-4 \times 8=16$

⇒ $\text { Or } k^2=48 \Rightarrow k= \pm \sqrt{16 \times 6} \Rightarrow k= \pm 4 \sqrt{3}$

Question 88. If a: /? be the roots of x²+x+5 = 0 then$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=$

$\frac{16}{5}$
2
3
$\frac{14}{5}$

Solution:

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{1}{1}=-1$

⇒ $\alpha \beta=\frac{c}{a}=\frac{5}{1}=5$

⇒ $\frac{a}{\beta}+\frac{\beta^2}{u}=\frac{\alpha^3+\beta^3}{\alpha \beta}$

⇒ $\frac{(\alpha+\beta)^3-3 a \beta(\alpha+\beta)}{\alpha \beta}=\frac{(-1)^3-3.5 \cdot(-1)}{5}$

⇒ $\frac{-1+15}{5}=\frac{14}{5}$

Question 89. If the sum of two numbers is 13 and the sum of their squares is 85 then the numbers

6,7
4,9
10,3
5,8

Solution:

(1)

6 + 7= 13 (true)

62+ 72 = 36 + 49 = 85 (True)

Question 90. The difference between the roots of the equation x² – 7x- 9 = 0 is

7
V85
9
2x/85

Solution:

(2)

⇒ $\text { Let } \alpha+\beta=-\frac{b}{a}=-\frac{-7}{1}=7$

⇒ $\propto \beta=\frac{c}{a}=\frac{-9}{1}=-9 $

⇒ $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \propto \beta=7^2-4(-9)=85 $

⇒ $\alpha-\beta=\sqrt{85}$

Question 91. If the roots of the equation kx² — 3x —1 = 0 are the reciprocal of the roots of the equation x² + 3x – 4 = 0 then k=

4
-4
3
-3

Solution:

(1) x² + 3x- 4 = 0

Or x² – 4x + x – 4 =0

Orx(x- 4) +1 (x- 4) = 0

Or; (x- 4) (x + 1) = 0

x = 4; -1

Eqn. having rots $\frac{1}{4} \& \frac{1}{-1}=\frac{1}{4} \&-1 \text { is. }$

⇒ $x^2-\left(\frac{1}{4}-1\right) x+\frac{1}{4}(-1)=0$

⇒ $\text { or } x^2+\frac{-1}{4} x-\frac{1}{4}=0$

Multiplying by 4; we get

4x² + 3c- 1 = 0

Comparing it with le tx² + 3x-1 = 0

We get K = 4

Eqn. having roots the reciprocal of the roots of ax²+ bx + c = 0 is ax² + bx + a = 0 i.e. 1st and last term interchanges

Question 92. If cr + ft = -2 mid aft = -3 where u and \ I are the roots of the equation, which is

x² – 2x -3 = 0
x² + 2x -3 = 0
x² + 2x + 3 = 0
x² – 2x + 3 = 0

Solution:

(2)

Quadratic Eqn. having roots a and |3 is

x² – (a + (3) x + a(3 = 0

or; x² – (-2)x + (-3) = 0

or; x² + 2x- 3 = 0

(2) is correct

Question 93. If a and (1 are the roots of the equation x² + x + 5 = 0 the$\frac{a^2}{\beta}+\frac{\beta^2}{\alpha}$

$\frac{16}{5}$
3
$\frac{14}{5}$
2

Solution:

(3)

Given Eqn. is x² + x + 5 = 0

⇒ $\alpha+\beta=-\frac{b}{a}=-\frac{1}{1}=-1$

⇒ $\alpha \beta=\frac{c}{a}=\frac{5}{1}=5 $

⇒ $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{a \beta}$

⇒ $\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}$

⇒ $\frac{(-1)^3-3 \cdot 5 \cdot(-1)}{5}=\frac{-1+15}{5}=\frac{14}{5}$

Question 94. When two roots of quadratic equation are a,$\alpha, \frac{1}{\alpha}$ then what will be the quadratic equation:

ux² – (cr² + 1)x + cr = 0
ax² – ax² + 1 = 0
ax² – (cr ² + l)x +1=0
None of these

Solution:

(1)

For $\alpha \cdot \frac{1}{a}=1=\frac{c}{a}=\frac{a}{a}=1 \text { (True) }$

Question 95. Let cr and ft be the roots of x2 + 7x + 12 = 0, The the value of$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} \text { will be }$

$\frac{19}{141}+\frac{141}{49}$
$\frac{7}{12}+\frac{12}{7}$
$-\frac{91}{12}$
None of these

Solution:

x² + 7x+ 12 = 0

or x²+ 4x + 3x + 12 = 0

or x(x + 4) + 3 (x + 4) = 0

or (x + 4) (x + 3) = 0

x = -3; -4

Letα = —3 ; β = -4

⇒ $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{9}{-4}+\frac{16}{-3}$

⇒ $-\left[\frac{9}{4}+\frac{16}{3}\right]=-\frac{91}{12}$

2 and Method

⇒ $alpha+\beta=\frac{-7}{1}=-7$

⇒ $\alpha \beta=\frac{c}{a}=\frac{12}{1}=12$

⇒ $ \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta}=\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}$

⇒ $\frac{(-7)^3-3 \times 12(-7)}{12}=-\frac{91}{12}$

Question 96. Find the condition that one root is double the of ax² + bx + c = 0

2b² = 3ac
b²= 3ac
2b² = 9ac
None

Solution:

Let 1st root =1

Then 2nJ root = 2

The Eqn. is

x2-(l + 2)x+lx2 = 0

or x² – 3x + 2 = 0

Comparing it with ax² + bx + c = 0

We get;

a = 1; b = 3; c = 2

Go by choices (GBC)

(1) 2b² = 9ac

2.(-3)2 = 9.1.2

⇒ 18 = 18 (True)

Hence, (3) is (true)

Question 97. Find the quadratic equation in x whose roots are 2 and 3.

x² + 5x + 6 = 0
x² — 5x — 6 = 0
x² – 5x + 6 = 0
x² + 5x- 6 = 0

Solution:

The quadratic equation is (x-2) (x-3) = 0

Or x² — 2x — 3x + 6 = 0 or x²– 5x + 6 = 0

Question 90. If the sum of the roots of a quadratic equation in x is 4 and the constant term of the equation is 15, then find the quadratic equation.

x²– 4x + 1 5 = 0
x² – 4x + 1 5 = 0
x² + 4x+ 15 = 0
x² + 4x+ 15 = 0

Solution:

we know that if s is the sum and p is the product of the roots of a quadratic equation then the equation is x²– 5x + p = 0

Note that the constant term in the equation is nothing but the product of the roots.

Given s=4, p=1 5

The quadratic equation is x² – 4x + 15 = 0

Question 99. Find the roots of the quadratic equation x²– 8Y + 9 = 0

1,9
-1,-9
4+√7, 4-√7
3+√6, 3-√6

Solution:

⇒ $x^2-8 x+9=0 \text { the roots are } \frac{-\Delta \pm \sqrt{D^2-4 a c}}{2 a}$

⇒ $x=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(0)}}{2(1)}$

⇒ $\frac{8 \pm \sqrt{s t-3 a)}}{2}=\frac{8 \pm \sqrt{2 s}}{2}=\frac{8 \pm 2 \sqrt{7}}{2}=4 \pm \sqrt{7}$

⇒ $\text { roots are } 4+\sqrt{7} \text { and } 4-\sqrt{7}$

Question 100. Find the sum of the roots of the equation, 9x² — 36x + 35 = 0.

36
-36
4
-4

Solution:

he sum of the roots of equation ax² +bx+ c = 0 is -b/a

the sum of the roots of 9x² — 36x+ 35 = 0 is $\frac{-(-36)}{9}=4$

Question 101. If the sum of the roots of a quadratic equation is 11 and the product of the roots is 24, find the roots of the equation.

6,4
12,2
8,3
1,24

Solution:

let a, and p be the roots of the equation. Given, α + β = 11 and αβ = 24

(α- β)² = (α + β)²– 4αβ = ll2- 4(24) = 121- 96 = 25

α- β = 5

(1) ÷ (2)⇒ 22=16⇒α = 8

β= 11-8 = 3

the roots are 8,3.

Question 102. If the sum of the roots of a quadratic equation is 7 and the product of the roots is 12, then find the equation.

x² + 7x + 12 = 0
x² + 7x — 12 = 0
x²– 7x + 12 = 0
x²– 7x — 12 = 0

Solution:

Given the sum of the roots is 7 and the product of the roots is 12.

the equation is

x² -(sum of the roots)x+ (product of the roots) = 0 i. e.x² — 7x + 12 = 0

Question 103. Find the discriminant of the equation 2x² + 3,Y + 4 = 0

Solution:

discriminant of the quadratic equation ax² + bx + c = 0 is b²– 4ac

the discriminant of 2.v2 + 3.v 4-4 = 0 Is

32 — 4(2)(4) = 9 — 32 = —23

Question 104. Construct a quadratic equation whose roots are 4 more than the roots orx2 + lx + 16 = 0

x2 + 15x + 60 = 0
x2 + 7x + 60 = 0
x2- x + 4 = 0
x2 + 15x + 16 = 0

Solution:

to find a new quadratic equation, whose roots are 4 more than the roots of x²+ 7x + 16 = 0

we have to replace x by (x-4)

The equation is (x- 4)² +7(x- 4) + 16 = 0

⇒ x²– 8. Y 4- 16 + 7x – 28 + 16 = 0

⇒ x²– x 4- 4 = 0

Question 105. Construct a quadratic equation whose roots are reciprocals of the roots of the equation 3x² + 5x + 7 = 0

7x² + 5x + 3 = 0
5x² + 7x + 3 = 0
5x² + 3x + 7 = 0
7x² + 3x + 5 = 0

Solution:

to find a new quadratic equation, whose roots are reciprocals of the roots of 3x²+ 5x +7 = 0 we have replaced x by1/x

The equation is$3\left(\frac{1}{x}\right)^2+5\left(\frac{1}{x}\right)+7=0$

7x² + 5A + 3 = 0

Question 106. Construct a quadratic equation whose roots are half roots of the equation x²+5x+9=0

4x² + lOx + 9 = 0
x² + lOx + 36 = 0
4x² + 5x + 36 = 0
4x² + 5x + 9 = 0

Solution:

To find a new quadratic equation, whose roots are half of the roots of the equation,

x² + 5x + 9 = 0 we have dot replace x by 2x the equation is (2,y)2 + 5(2x) + 9=0

⇒ (4×2) + 10.Y + 9 = 0

Question 107. How many distinct roots does the quadratic equation (x- 4)² = 0 have?

Solution:

Here the equation (x- 4)² = 0 is given.

Number of distinct roots = 1

Question 108. Solve for x x⁴– 61x² + 900 = 0

±4. ±5
±5, ±6
±3,±5
±4, ±6

Solution:

x⁴ – 61x² + 900 = 0=>(x²– 36)(x² – 25) = 0

⇒ x=±5,±6

Question 109. Solve for x 3^2x+1 _ 39(3x+2) + 8748 = 0

4
log336
Either (1) or (2)
6

Solution:

3(3^2x)- (39)(9)(3^x) + 8748 = 0

Dividing by 3 and setting y=3^x

y² – 117y + 2916 = 0

⇒ 3^x = 36 or 3X = 81

⇒ x= log336 or x = 4

Question 110. Solve for x$\sqrt{2 x-5}+\sqrt{3 x+4}=8$

8
615
7
Either (2) or [3]

Solution:

⇒ $ \sqrt{2 x-5}+\sqrt{3 x+4}=8 \rightarrow(\mathrm{A})$

By squaring on both sides.

⇒ $5 x-1+2 \sqrt{6 x^2-7 x-20}=64$

⇒ $ 2 \sqrt{6 x^2-7 x-20=-5(x-3)}$

By squaring on both sides

⇒ 4(6x² – 7x- 20) = 25(x² – 26x + 169)

⇒ x²– 622x + 4305 = 0

⇒ (x — 61 5)(x — 7) = 0

Substituting x =7 in (A) we see that V9 + V25 = 8 but for x=615

⇒ $=\sqrt{2(615)-5}+\sqrt{3(615)+4}=\sqrt{1225}+\sqrt{1849}$

35+43=78

To get 8, we have to take the negative root for 1225 The expression V1225 normally stands for the positive square root 35 we would say that x=615 does not satisfy [A)

Hence x =7

Question 111. Which of the following is the root [s] of $\sqrt{3 x+34}+\sqrt{3-2 x}=\sqrt{67+x}$

3
-3
-41/6
1. A or B
2. B or C
3. C or A
4. Only B

Solution:

⇒ $ \sqrt{3 x+34}+\sqrt{3-2 x}=\sqrt{67+x} \rightarrow(A)$

By squaring on both sides,

⇒ $x+37+2 \sqrt{-6 x^2-59 x+102}=67+x $

⇒ $2 \sqrt{-6 x^2-59 x+102}=30$

By squaring on both sides,

⇒ -6x²– 59x + 102=225

⇒ 6x² + 59x+123=0

⇒ (6x+41) (x+3)=0

⇒ x= -41/6 or -3

Substituting -3 in (A) the left-hand side (LIIS)

And if we substitute -41/6 the LHS of (A

⇒ $=\sqrt{3(-3)+34}+\sqrt{3-(-6)}=5+3=8 R H S$

⇒ $\sqrt{-\frac{41}{2}+\frac{68}{2}}+\sqrt{\frac{9}{3}+\frac{41}{3}}=\sqrt{\frac{27}{2}}+\sqrt{\frac{50}{3}}$

⇒ $\frac{9}{\sqrt{6}}+\frac{10}{\sqrt{6}}=\frac{19}{\sqrt{6}} \text { and the RHS }$

⇒ $\sqrt{\frac{402}{6}-\frac{41}{6}}=\sqrt{\frac{361}{6}}=\frac{19}{\sqrt{6}}$

⇒ $\text { Both }-3 \text { and }-41 / 6 \text { satisfy (A) }$

Question 112. Find the value of if one of the roots of x² – 9x + 2p = 0 is 3 more than the other.

Solution:

Let the roots of a and a + 3.

2α + 3 = 9

⇒ α = 3

The other root is 6 and 2p = 18 or p=9

Question 113. Four friends have some coins. Pavan has 2 less than Samir, who has 2 less than Tarun,ifPra has 2 less than Pavan and the product of the number of coins that each of them has is 5760, how many coins in all do they have among themselves?

Solution:

Let the number of coins with Pranav, Pavan, and Samir Tarun be (a-3), (a-1), and (a+1) respectively.

(a² + 3²)(a²– l²)(a + 3)= 5760

Let a2 = k(k- 9)(/c- 1) = 5760

k² – 10k- 5751 = (k- 81)(fc + 71) = 0

⇒k = 81 or- 71

As k is positive, a² = 81⇒a = 9

⇒ (a- 3) + (a- 1) + (a + 1) + (a + 3) = 36

Question 114. If the roots of ax² + bx+c-0 are α and β and the roots of px² + qx + r = 0 are α- k and β-k, then

r= ak² – bk + c
r= ak²– bk- ck²
r= ak + bk + ck²
r= ak² + bk + c

Solution:

(d)

As α— k and β — n are the roots ax² + bx + c = 0

Now, the equation with α — k and β- kas roots are

f(x + k) i.e.a(x-k)² + b(x + k) + c = 0

ax2 + (fa + 2ak)x + (a/c2 + fa/c + c) = 0

By comparing this equation with px² + qx +r = 0, r=ak² + bk + c

Question 115. if the roots of x² + 7x + 10 = 0 are a and /?, find the equation whose roots are $\left(\frac{1-\alpha}{\beta}\right)$ and $\left(\frac{1-\beta}{\alpha}\right)$

x² + 360x + 180 = 0

x²+ 0.36x + 0.18 = 0

x² +36x + 18 = 0

x² + 3.6x + 1.8 = 0

Solution:

(4)

The roots of x² +7x + 10 = 0 area and (3)

α + β = -7 and αβ = 10

⇒ $\text { If } \mathrm{p}=\frac{1-\alpha}{\beta} \text { and } \mathrm{q}=\frac{1-\beta}{\alpha}$

⇒ $P+q=\frac{(\alpha+\beta)-\left(\alpha^2+\beta^2\right)}{\alpha \beta}$

⇒ $=\frac{(\alpha+\beta)-(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}$

⇒ $=\frac{(-7)-\left(7^2-2(10)\right)}{10}=-3.6$

⇒ $\text { and } \mathrm{pq}=-\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}$

⇒ $=\frac{1-(-7)+10}{10}=1.8$

The equation where roots are p,q is x² + 3.6x + 1.8 = 0

Question 116. A total of Rs.3, 300 is raised by collecting equal amounts from a certain number of people. If there were 22 people more, each person would have to contribute Rs. 200 less to raise the same amount. How many people contributed?

Solution:

Let the number of people and the contribution of each person be n and a respectively,

(n + 22)(a- 200) = na

⇒ 22a – 200n = 4400 → (1)

And na= 3300→ (2)

Substituting the value of from (2) in (1)

22(3300/n)-200n = 4400

⇒ ll(33)/n- n = 22⇒ n² + 22n- 11(33) = 0

⇒ (n + 33)(n- 11) = 0 =>n = 11

Question 117. The sum of two positive numbers is 32. The square of the largest number is 223 more than twice the square of the smaller number. Find the smaller number.

Solution:

(1)

Let the smaller and larger numbers be x and yx + y=32→ (1)

2A2 + 23 = y2 → (2)

⇒ x² + 64x- 1001 = 0⇒(x + 77)(x- 13) = 0

⇒ x = -77 or x = 13

As x cannot be negative x = 13

Question 118. Akash and Badri, together buy 45 notebooks. Each of them buys a different number of books and both of them spend the same amount if Akash had bought his notebooks at the price at which Badri had bought them and Badri had bought his notebooks at the price where Akash had bought them, they would havespentRs. 160 and Rs. 250 respectively. How many notebooks did Badri buy?

Solution:

(1) Akash Badri

No. of notebooks mn

Price a b

We have m + n=45→ (1)

Ma = nb→ (2)

Mb = 160 → (3)

Na = 250 → (4)

From (2) ⇒ m/n = b/a

(3) +(4)⇒ (m/n)2 = (16/25) = (4/5)2

m: n = 4:5

And n=5/9 (45) =25

Question 119. If the roots α and β of x2- 4x- c = 0 satisfy the condition 2α + β = 1, then which of the following is true?

α = -3, β = 7
c=-21
α = -15, β = 3.5
α = 3, β = -7

Solution:

(1)

The α,β are the roots of x²– 4x- c = 0

α+β = 4 and 2a + p = 1 (given)

⇒ a = -3 and P = 7

Question 120. How many of the roots of x² – A + 2 are also the roots of x² + 4x² + 5x + 2 = 0?

Solution:

(2)

The roots of x²– x + 2 are x = 2 and x = -1

of these two roots, only, one of the roots satisfies the cubic equation, i.e. (2)³ + 4(2)² + 5(2) + 2≠0 but (-1)3+ 4(—l)² + 5(-l) + 2 = 0

has real and distinct roots.

Question 121. If one of the roots of x² – ax + b – 0 is 2+√3 and b=a+2+3√3, find the other root.

2-√3
√3- 2
4
-4

Solution:

(4)

the roots of 3x² + 9x + 5 = 0 are p and q p + q=9/3=3 and p q=5/3

⇒ $\frac{1}{p^2}+\frac{1}{q^2}=\frac{p^2+q^2}{p^2 q^2}=\frac{(p+q)^2-2 p q}{p^2 q^2}$

⇒ $\frac{3^2-(10 / 3)}{25 / 9}=\frac{51}{25} \text { and } \sqrt{3 p q}=\sqrt{5}$

⇒ $ p^2+q^2=\sqrt{3 p q}=\frac{51}{25}+\sqrt{5}$

Question 122. The roots of 35x² + x = 12 are

Rational and equal.
Rational and distinct.
Irrational and equal.
Irrational and unequal

Solution:

(2)

for the equation 35x² + x- 12 = 0 the determinant

s=1² + 4(35)(12) = If,81 = 41² = 18681=41²

the roots are rational (because A is a perfect square)and distinct (because Δ≠0)

Question 123. How many real roots does the equation x⁴ + 2x³ – 10x² – 2x + 1 = 0 have?

Zero
One
Two
Four

Solution:

For equations of the form Ax⁴ + Bx³ + Cx² ± Bx + A = 0, the method illustrated below should be used.

Given the equation, x⁴ + 2x³– 10x²– 2x +1 = 0 …. (1)

Divide (1) by x2 we get $x^2+2 x-10-\frac{2}{x}+\frac{1}{x^2}=0$

⇒ $\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2\left(x-\frac{1}{x}\right)-10=0$

⇒ $\text { Put } \mathrm{x}-\frac{1}{x}=y, \text { then } x^2+\frac{1}{x^2}=y^2+2$

⇒ y² + 2 + 2y- 10 = 0

⇒ (y+4)(y-2) =0

⇒ y= -4 or 2

⇒ $x-\frac{1}{x}=-4 \text { or } x-\frac{1}{x}=2$

⇒ x² + 4x-1 = 0 or x²– 2x-1 = 0 as the discriminant of both the quadratic equations is positive, hence equation (1) has 4 real roots.

Question 124. If the roots of 3 (x² + l)=kx area and 1/α(both real), which of the following could be the value of k?

10
5
-3
3

Solution:

(1)

The roots of 3x² + kx + 3 = 0 are a and 1/α both real quantize.

k²-4(3) (3) ≥ O or k² ≥ 36

only k = 10 satisfies this condition.

Question 125. If the roots of 3x² + (2k + 4)x = -12 are equal, then the value of k is

-5
5
-8
4

Solution:

(4)

As the roots of 3x²+ (2k + 4)x + 12 = 0 are equal A= 0 i.e. (k + 2)² = 3(12) = 36

⇒ k = 4 or- 8 =>k = 4 or- 8

Question 126. For which of the following equations with roots, αβ is α² + β²= 25 And αβ= -12?

x(x – 1) = 12
X(x + 9) = 12
x(x + 1) = 12
Either (1) or (3)

Solution:

(4)

⇒ $a^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

and $ \frac{1}{\alpha^2}-\frac{1}{\beta^2}=\frac{\beta^2-\alpha^2}{\alpha^2 \beta^2}=\frac{(a+\beta)(\beta-\alpha)}{\alpha^2 \beta^2}$

for the required equation with roots a, [3

α² + β² = 25 and a/? = -12

α² + β² + 2αβ = 25 + (2)(—12) = l⇒α + β = 1 or —1

The equation could be x² – x — 12 = 0 or x² + x- 12 = 0

Question 127. If the speed of a vehicle decreases by 10 km, it takes 2 hours more than what it usually takes, to cover a distance of 1 800 km, find the time it usually takes,

24 hours
30 hours
36 hours
18 hours

Solution:

(4)

the data is tabulated below

Speed time Distance

ut 1800⇒ut=1800

u-10 t + 2 1800

ut = (u – 10)(t + 2) = ut + 2u- lOt- 20

⇒ $2 u-10\left(\frac{18 00}{u}\right)=20 \Rightarrow u^2-10 u-9000=0$

(u – 100)(u + 90) = 0 ⇒ u=100

and t = 1800/100=18

Question 128. If one root of x2- 9x + 14 = 0 is the same as one root of x2 – 5x + k = 0, then k =

±6
±14
6,-14
-6,14

Solution:

(3)

the roots of x² – 9x + 14 = 0 are 7 and 2 if 7 is a root of the second equation 7²– 5(7) + k = 0 ⇒ K = 6

Question 129. If the sum of the roots of x²– 5x⁴+ k = 0 and x + k = 0 is 80 then what is/ are the possible real value (s) of k?

±5
±2
±4
±10

Solution:

(2)

the given equation is x²– 5x⁴ + k = 0

the sum of the roots is 80

5k4 = 80⇒k⁴ = 16

±2 are the real values of.

Question 130. How many equations of the form x² + 6x + p = 0 where p is an integer and 0< p < 15, have real roots?

Solution:

(1)

The given equation is x² + 6x + p = 0

This has real roots if and only 6² – 4p≥ 0 i.e., p≤9

10 integral values satisfy this condition.

Question 131. What is the minimum value of the Squares of the roots of the equation x²-(α – 3)x+(α-8)=0 where α is a positive number?

Solution:

(4)

x² – (a- 3) + (a- 8) = 0

The sum of the roots =a-3

The product of the roots= a-8

The sum of the squares of the roots=(a- 3)²– 2(a- 8) = a² — 6a + 9- 2a + 16 = (a² –
8a + 25)

the minimum value of a²– 8a + 25

⇒ $=\frac{4(1)(25)-(-8)^2}{4(1)}=\frac{100-64}{4}=9$

CA Foundation Maths Solutions For Chapter 4 Equations

Exercise – 1

Simple Or Linear Equation in one Variable

Exercise – 2

Simultaneous Or Linear Equation In Two Or Three Variable

Exercise – 3

Quadratic Equations

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