**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

(b) quadratic equation

(c) cubic equation

(d) bi-quadratic equation

Answer: (b) quadratic 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

Answer:(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

Answer- (a) 9, 6

Let the larger number = x

then smaller number = y

Now according to the condition,

x^{2} – y^{2} = 45 … (i)

y^{2} = 4x … (ii)

substitute the value of y in equation (i),

x^{2} – 4x = 45

x^{2} – 4x – 45 = 0

on factorize,

x^{2} – 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 => y^{2} = 4x

y = **√**4x = **√**4(9) = **√**36 = 6

When x = -5, then

The larger number = x = -5

Smaller number = y => y^{2} = 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

Answer: (a) 12, 13

**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

Answer: (B) 8

hint

Let the number be x

Then according question,

x + 12 = 160/x

x^{2} + 12x – 160 = 0

x^{2} + 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

Answer: (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

⇒ n^{2} + 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 p^{2}x^{2} – q^{2} = 0, then x =?

(A) ± q/p

(B) ±p/q

(C) p

(D) q

Answer: (A) ± q/p

hint:

p^{2}x^{2} – q^{2} = 0

⇒p^{2}x^{2} = q^{2}

⇒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

Answer: (c) 15

hint

Let her actual marks be x

Therefore,

9 (x + 10) = x^{2}

⇒x^{2} – 9x – 90 = 0

⇒x^{2} – 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

Answer: (a) 42 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

Answer: (b) 13 and 14

Hint: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x^{2} – 27x – 182 = 0

⇒ x^{2} – 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

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

Hint Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base^{2} + Altitude^{2} = Hypotenuse^{2} (From Pythagoras theorem)

∴ x^{2} + (x – 7)^{2} = 13^{2}

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 2x^{2} + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Answer: (d) No real roots

Hint: 2x^{2} + x + 4 = 0

⇒ 2x^{2} + x = -4

Dividing the equation by 2, we get

⇒ x^{2} + 1/2x = -2

⇒ x^{2} + 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

Answer: (a) 7

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 = x^{2} + 2x – 15

⇒ x^{2} – 4x – 21 = 0

⇒ x^{2} – 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

Answer: (b) 40 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

⇒ x^{2} + 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 4x^{2}-2x+k-4=0 is reciprocal of the other. The value of k is:

(a) -8

(b) 8

(c) -4

(d) 4

Answer: (b) 8

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) x^{2} + 3x = (–1) (1 – 3x)^{2}

(c) (x + 2) (x – 1) = x^{2} – 2x – 3

(d) x^{3} – x^{2} + 2x + 1 = (x + 1)^{3}

Answer: (c) (x + 2) (x – 1) = x^{2} – 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)

x^{2} + 4x + 4 = 2x + 6

x^{2} + 2x – 2 = 0

This is a quadratic equation.

(b) x^{2} + 3x = (–1) (1 – 3x)^{2}

x^{2} + 3x = -1(1 + 9x^{2} – 6x)

x^{2} + 3x + 1 + 9x^{2} – 6x = 0

10x^{2} – 3x + 1 = 0

This is a quadratic equation.

(c) (x + 2) (x – 1) = x^{2} – 2x – 3

x^{2} + x – 2 = x^{2} – 2x – 3

x^{2} + x – 2 – x^{2} + 2x + 3 = 0

3x + 1 = 0

This is not a quadratic equation

**Question :-21** The quadratic equation 2x^{2} – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than 2 real roots

Answer: (c) no real roots

Hint

Given,

2x^{2} – √5x + 1 = 0

Comparing with the standard form of a quadratic equation,

a = 2, b = -√5, c = 1

Now,

b^{2} – 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

Answer: (a) two real roots

Hint

(x + 1)^{2} – 2(x + 1) = 0

x^{2} + 1 + 2x – 2x – 2 = 0

x^{2} – 1 = 0

x^{2} = 1

x = ± 1

**Question :-23** The quadratic formula to find the roots of a quadratic equation ax^{2} + bx + c = 0 is given by

(a) [-b ± √(b^{2}-ac)]/2a

(b) [-b ± √(b^{2}-2ac)]/a

(c) [-b ± √(b^{2}-4ac)]/4a

(d) [-b ± √(b^{2}-4ac)]/2a

Answer: (d) [-b ± √(b^{2}-4ac)]/2a

Hint

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

**Question :-24 **The quadratic equation x^{2} + 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,

x^{2} + 7x – 60 = 0

Comparing with the standard form,

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

b^{2} – 4ac = (7)^{2} – 4(1)(-60) = 49 + 240 = 289 > 0

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

(b) b^{2} – 4ac = 0

(c) b^{2} – 4ac < 0

(d) b^{2} – ac < 0

Answer: (c) b^{2} – 4ac < 0

Hint A quadratic equation ax^{2 }+ bx + c = 0 has no real roots, if b^{2} – 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) x^{2} + 3x = ( – 1) (1 – 3x)

(c) (x + 2) (x – 1) = x^{2} – 2x – 3

(d) x^{3} – x^{2} + 2x + 1 = (x + 1)^{3}

Answer -(d) x^{3} – x^{2} + 2x + 1 = (x + 1)^{3}

Hint

(a) (x + 2)^{2} = 2(x + 3)

⇒ x^{2} + 4x + 4 = 2x + 6

⇒ x^{2} + 4x – 2x + 4 – 6 = 0

⇒ x^{2} + 2x – 2

It is a quadratic equation.

(b) x^{2} + 3x = ( – 1) (1 – 3x)

⇒ x^{2} + 3x = –1 + 3x

⇒ x^{2} + 1 = 0

it is also quadratic equation.

(c)(x + 2) (x – 1) = x^{2} – 2x – 3

x^{2} – x + 2x – 2 = x^{2} – 2x – 3

x^{2} – x^{2} + x + 2x – 2 + 3 = 0

⇒ 3x + 1 = 0

It is not a quadratic equation.

(d) x^{3} – x^{2} + 2x + 1 = (x + 1)^{3}

= x^{3} + 3x^{2} + 3x + 1

x^{3} – x^{2 }+ 2x + 1

3x^{2} + x^{2 }– 2x – 1 + 3x + 1 = 0

⇒ 4x^{2} + x = 0

It is a quadratic equation

**Question :-26 **The roots of the equation x^{2} – 3x – 10 = 0 are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Answer-(b) – 2, 5

Hint

x^{2} – 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

Answer-(d) 0, 8

Hint

2x² – kx + k = 0

a = 2, b = -k, c = k

∴ b^{2} – 4ac

= (-k)^{2} – 4 x 2 x 4

= k^{2} – 8k

∴ Roots are equal.

∴ b^{2} – 4ac = 0

k^{2} – 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

Answer (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

Answer -(b) 0, 3

Hint

(k + 1)x² – 2(k – 1)x + 1 = 0

Here, a = k + 1, b = -2(k – 1), c = 1

∴ b^{2} – 4ac

= [–2(k –- 1)]^{2} – 4(k + 1)(1)

= 4(k^{2} – 2k + 1) – 4k – 4

= 4k^{2} – 8k + 4 – 4k – 4

= 4k^{2} – 12k

∵ Roots are equal.

∴ b^{2} – 4ac = 0

∴ 4k^{2} – 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

Answer -(c) no real roots

Hint

2x² – √5x + 1 = 0

Here, a = 2, b = -√5, c = 1

b^{2} – 4ac

= (-√5)² -4×2 x1

= 5 – 8

= -3

∵ b^{2 }– 4ac < 0

∴ It has no real roots.

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