# Quadratic Equations MCQs for ICSE Class-10 Maths

Quadratic Equations MCQs for ICSE Class-10 Maths for Sem-1. These MCQ  / Objective Type Questions of Quadratic Equations  is based on latest reduced syllabus according 2021-22 session on bifurcated pattern. Main motto of MCQ Type Question is cracking the next upcoming exam of council. Visit official website CISCE for detail information about ICSE Board Class-10 Maths

## MCQs of Quadratic Equations for ICSE Class-10 Maths

 Board ICSE Class 10th ( x ) Subject Maths Chapter Quadratic Equations Syllabus on bifurcated syllabus (after reduction) bifurcated pattern Semester-1 Session 2021-22 Topic MCQ / Objective Type Question

### Multiple Choice Questions (MCQs) of Quadratic Equations for ICSE Class-10 Maths

Question :-1  Which of the following is not a quadratic equation

(a) x² + 3x – 5 = 0
(b) x² + x³ + 2 = 0
(c) 3 + x + x² = 0
(d) x² – 9 = 0

Answer: (b) x² + x³ + 2 = 0
Hint : Since it has degree 3.

Question :-2 The equation (x – 2)² + 1 = 2x – 3 is a
(a) linear equation
(c) cubic equation

hint  We have (x – 2)² + 1 = 2x – 3
⇒ x² + 4 – 2 × x × 2 + 1 = 2x – 3
⇒ x² – 4x + 5 – 2x + 3 = 0
∴ x² – 6x + 8 = 0, which is a quadratic equation.

Question :-3 The roots of the quadratic equation 6x² – x – 2 = 0 are

(a)    1/2, x = 2/ 3
(b) 1/2, x = – 2/ 3
(c)  1/2, x = –2/ 3
(d)   1/2, x = 2/ 3

Answer- (a)    1/2, x = 2/ 3

hint : We have 6×2 – x – 2 = 0
⇒ 6x² + 3x-4x-2 = 0
⇒ 3x(2x + 1) -2(2x + 1) = 0
⇒ (2x + 1) (3x – 2) = 0
⇒ 2x + 1 = 0 or 3x – 2 = 0
∴ x =1/2, x = 2/ 3

Question :-4  The quadratic equation whose one rational root is 3 + √2 is
(a) x² – 7x + 5 = 0
(b) x² + 7x + 6 = 0
(c) x² – 7x + 6 = 0
(d) x² – 6x + 7 = 0

Answer: (d) x² – 6x + 7 = 0
hint ∵ one root is 3 + √2
∴ other root is 3 – √2
∴ Sum of roots = 3 + √2 + 3 – √2 = 6
Product of roots = (3 + √2)(3 – √2) = (3)² – (√2)² = 9 – 2 = 7
∴ Required quadratic equation is x² – 6x + 7 = 0

Question :-5 . The equation 2x² + kx + 3 = 0 has two equal roots, then the value of k is
(a) ±√6
(b) ± 4
(c) ±3√2
(d) ±2√6

hint: Here a = 2, b = k, c = 3
Since the equation has two equal roots
∴ b² – 4AC = 0
⇒ (k)² – 4 × 2 × 3 = 0
⇒ k² = 24
⇒ k = ± √24
∴= ± 2√6

Question -6 The difference between the squares of two numbers is 45. The square of the smaller number is 4 times the larger number.  the numbers are

(a) 9, 6
(b) 3, 6
(c) 3, 9
(d) 9, 3

Let the larger number = x
then smaller number = y
Now according to the condition,

x2 – y2 = 45 … (i)

y2 = 4x … (ii)

substitute the value of y in equation (i),

x2 – 4x = 45

x2 – 4x – 45 = 0

on  factorize,

x2 – 9x + 5x – 45 = 0

x(x – 9) + 5 (x – 9) = 0

(x – 9) (x + 5) = 0

either (x – 9) = 0 or (x + 5) = 0

so x = 9 or x = -5

When x = 9, then

The larger number = x = 9

and Smaller number = y => y2 = 4x

y = 4x = 4(9) = 36 = 6

When x = -5, then

The larger number = x = -5

Smaller number = y => y2 = 4x

y = 4x = 4(-5) = -20 (which is not possible)

hence The value of x and y are 9, 6.

Question :-7  One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages are
(a) 7 years, 49 years
(b) 5 years, 25 years
(c) 1 years, 50 years
(d) 6 years, 49 years

Answer: (a) 7 years, 49 years

Question :-8 The sum of the squares of two consecutive natural numbers is 313. The numbers are
(a) 12, 13
(b) 13,14
(c) 11,12
(d) 14,15

Question :-9  A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

(A) 3

(B) 8

(C) 4

(D) 7

hint

Let the number be x

Then according question,

x + 12 = 160/x

x2 + 12x – 160 = 0

x2 + 20x – 8x – 160 = 0

(x + 20) (x – 8) = 0

x = -20, 8

Question :-10  The product of two successive integral multiples of 5 is 300. Then the numbers are:

(a) 25, 30

(b) 10, 15

(c) 30, 35

(d) 15, 20

hint

Let the consecutive integral multiple be 5n and 5(n + 1) where n is a positive integer.

According to the question:

5n × 5(n + 1) = 300

⇒ n2 + n – 12 = 0

⇒ (n – 3) (n + 4) = 0

⇒ n = 3 and n = – 4.

As n is a positive natural number so n = – 4 will be discarded.

Therefore the numbers are 15 and 20.

Question :-11  If p2x2 – q2 = 0, then x =?

(A) ± q/p

(B) ±p/q

(C) p

(D) q

hint:

p2x2 – q2 = 0

⇒p2x2 = q2

⇒x = ±p/q

Question :-12  Rohini had scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

(a) 14

(b) 16

(c) 15

(d) 18

hint

Let her actual marks be x

Therefore,

9 (x + 10) = x2

⇒x2 – 9x – 90 = 0

⇒x2 – 15x + 6x – 90 = 0

⇒x(x – 15) + 6 (x – 15) = 0

⇒(x + 6) (x – 15) = 0

Therefore  x = – 6 or x =15

Since x is the marks obtained, x ≠ – 6. Therefore, x = 15.

Question :-13  A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed?

(a) 42  Km/hr

(b) 40  Km/hr

(c) 44  Km/hr

(d) 42.5  Km/hr

Question :-14 The sum of two numbers is 27 and product is 182. The numbers are:

(a) 12 and 13

(b) 13 and 14

(c) 12 and 15

(d) 13 and 24

Hint: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x2 – 27x – 182 = 0

⇒ x2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

⇒ x = 13 or x = 14

Question :-15 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:

(a) Base=10cm and Altitude=5cm

(b) Base=12cm and Altitude=5cm

(c) Base=14cm and Altitude=10cm

(d) Base=12cm and Altitude=10cm

Hint Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base2 + Altitude2 = Hypotenuse2 (From Pythagoras theorem)

∴ x2 + (x – 7)2 = 132

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12-7 = 5cm

Question :-16  The roots of quadratic equation 2x2 + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Hint: 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

Dividing the equation by 2, we get

⇒ x2 + 1/2x = -2

⇒ x2 + 2 × x × 1/4 = -2

By adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2

⇒ (x + 1/4)2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16

Question :-17  The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:

(a) 7

(b) 10

(c) 5

(d) 6

Hint: Let, x is the present age of Rehman

Three years ago his age = x – 3

Five years later his age = x + 5

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x2 + 2x – 15

⇒ x2 – 4x – 21 = 0

⇒ x2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7

Question :-18  A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

(a) 30 km/hr

(b) 40 km/hr

(c) 50 km/hr

(d) 60 km/hr

Hint: Let x km/hr be the speed of train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/h

Question :-19 If one root of equation 4x2-2x+k-4=0 is reciprocal of the other. The value of k is:

(a) -8

(b) 8

(c) -4

(d) 4

Hint: If one root is reciprocal of others, then the product of roots will be:

α x 1/α = (k-4)/4

k-4=4

k=8

Question :-20  Which one of the following is not a quadratic equation?

(a) (x + 2)2 = 2(x + 3)

(b) x2 + 3x = (–1) (1 – 3x)2

(c) (x + 2) (x – 1) = x2 – 2x – 3

(d) x3 – x2 + 2x + 1 = (x + 1)3

Answer: (c) (x + 2) (x – 1) = x2 – 2x – 3

Hint

We know that the degree of a quadratic equation is 2.

By verifying the options,

(a) (x + 2)2 = 2(x + 3)

x2 + 4x + 4 = 2x + 6

x2 + 2x – 2 = 0

(b) x2 + 3x = (–1) (1 – 3x)2

x2 + 3x = -1(1 + 9x2 – 6x)

x2 + 3x + 1 + 9x2 – 6x = 0

10x2 – 3x + 1 = 0

(c) (x + 2) (x – 1) = x2 – 2x – 3

x2 + x – 2 = x2 – 2x – 3

x2 + x – 2 – x2 + 2x + 3 = 0

3x + 1 = 0

This is not a quadratic equation

Question :-21 The quadratic equation 2x2 – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than 2 real roots

Hint

Given,

2x2 – √5x + 1 = 0

Comparing with the standard form of a quadratic equation,

a = 2, b = -√5, c = 1

Now,

b2 – 4ac = (-√5)2 – 4(2)(1)

= 5 – 8

= -3 < 0

Therefore, the given equation has no real roots.

Question :-22 The equation (x + 1)2 – 2(x + 1) = 0 has

(a) two real roots

(b) no real roots

(c) one real root

(d) two equal roots

Hint

(x + 1)2 – 2(x + 1) = 0

x2 + 1 + 2x – 2x – 2 = 0

x2 – 1 = 0

x2 = 1

x = ± 1

Question :-23 The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by

(a) [-b ± √(b2-ac)]/2a

(b) [-b ± √(b2-2ac)]/a

(c) [-b ± √(b2-4ac)]/4a

(d) [-b ± √(b2-4ac)]/2a

Hint

The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by [-b ± √(b2-4ac)]/2a.

Question :-24 The quadratic equation x2 + 7x – 60 has

(a) two equal roots

(b) two real and unequal roots

(b) no real roots

(c) two equal complex roots

Answer: (b) two real and unequal roots

Hint

Given,

x2 + 7x – 60 = 0

Comparing with the standard form,

a = 1, b = 7, c = -60

b2 – 4ac = (7)2 – 4(1)(-60) = 49 + 240 = 289 > 0

Therefore, the given quadratic equation has two real and unequal roots

(b) b2 – 4ac = 0

(c) b2 – 4ac < 0

(d) b2 – ac < 0

Answer: (c) b2 – 4ac < 0

Hint A quadratic equation ax+ bx + c = 0 has no real roots, if b2 – 4ac < 0. That means, the quadratic equation contains imaginary roots

Question :-25  Which of the following is not a quadratic equation ?

(a) (x + 2)2 = 2(x + 3)
(b) x2 + 3x = ( – 1) (1 – 3x)
(c) (x + 2) (x – 1) = x2 – 2x – 3
(d) x3 – x2 + 2x + 1 = (x + 1)3

Answer -(d) x3 – x2 + 2x + 1 = (x + 1)3

Hint

(a) (x + 2)2 = 2(x + 3)
⇒ x2 + 4x + 4 = 2x + 6
⇒ x2 + 4x – 2x + 4 – 6 = 0
⇒ x2 + 2x – 2

(b) x2 + 3x = ( – 1) (1 – 3x)
⇒ x2 + 3x = –1 + 3x
⇒ x2 + 1 = 0

(c)(x + 2) (x – 1) = x2 – 2x – 3

x2 – x + 2x – 2 = x2 – 2x – 3
x2 – x2 + x + 2x – 2 + 3 = 0
⇒ 3x + 1 = 0
It is not a quadratic equation.

(d) x3 – x2 + 2x + 1 = (x + 1)3

= x3 + 3x2 + 3x + 1
x3 – x+ 2x + 1
3x2 + x– 2x – 1 + 3x + 1 = 0
⇒ 4x2 + x = 0

Question :-26  The roots of the equation x2 – 3x – 10 = 0 are

(a) 2,- 5
(b) – 2, 5
(c) 2, 5
(d) – 2, – 5

Hint

x2 – 3x – 10 = 0

either

= (3+7)/ 2

∴ x = 10/2 = 5
or
x = (3-7)/2
=-4/2

= –2

x = 5, – 2 or – 2, 5 (b)

Question :-27  The value(s) of k for which the quadratic equation 2x² – kx + k = 0 has equal roots is (are)

(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8

Hint

2x² – kx + k = 0
a = 2, b = -k, c = k
∴ b2 – 4ac
= (-k)2 – 4 x 2 x 4
= k2 – 8k
∴ Roots are equal.
∴ b2 – 4ac = 0
k2 – 8k = 0
⇒ k(k – 8) = 0
Either k = 0
or
k – 8 = 0,
then k = 8
k = 0, 8.
So option (d) is correct

Question :-28  If the equation 3x² – kx + 2k =0 roots, then the the value(s) of k is (are)

(a) 6
(b) 0 Only
(c) 24 only
(d) 0

Hint

3x² – kx + 2k = 0
Here, a = 3, b = -k, c = 2k

Question :-29  If the equation {k + 1)x² – 2(k – 1)x + 1 = 0 has equal roots, then the values of k are

(a) 1, 3
(b) 0, 3
(c) 0, 1
(d) 0, 1

Hint

(k + 1)x² – 2(k – 1)x + 1 = 0
Here, a = k + 1, b = -2(k – 1), c = 1
∴ b2 – 4ac
= [–2(k –- 1)]2 – 4(k + 1)(1)
= 4(k2 – 2k + 1) – 4k – 4
= 4k2 – 8k + 4 – 4k – 4
= 4k2 – 12k
∵ Roots are equal.
∴ b2 – 4ac = 0
∴ 4k2 – 12k = 0
4k(k – 3) = 0
⇒ 4k(k – 3) = 0
⇒ k(k – 3) = 0
Either k = 0
or
k – 3 = 0,
then k = 3
k = 0, 3

Option (b) is correct

Question :-30  The quadratic equation 2x² – √5x + 1 = 0 has

(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than two real roots

Hint

2x² – √5x + 1 = 0
Here, a = 2, b = -√5, c = 1
b2 – 4ac

= (-√5)² -4×2 x1
= 5 – 8
= -3
∵ b– 4ac < 0
∴ It has no real roots.

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