Class 10 Maths MCQs for Chapter 4 (Quadratic Equations) are available online here with answers. All these objective questions are prepared as per the latest CBSE syllabus and NCERT guidelines. Solving these multiple-choice questions will help students to score good marks in the board exams, which they can verify with the help of detailed explanations given here. To get chapter-wise MCQs, click here.

## Class 10 Maths MCQs for Quadratic Equations

Here, we have given multiple-choice questions for Chapter 4 quadratic equations, to help students to solve different types of questions, which could appear in the board exam. They can build their problem-solving capacity and boost their confidence level by practising the questions here. Get important questions for class 10 Maths here at BYJU’S.

#### Below are the MCQs for Quadratic Equations

**1.Equation of (x+1) ^{2}-x^{2}=0 has number of real roots equal to:**

(a)1

(b)2

(c)3

(d)4

Answer: **(a)**

Explanation: (x+1)^{2}-x^{2}=0

X^{2}+2x+1-x^{2} = 0

2x+1=0

x=-½

Hence, there is one real root.

**2.The roots of 100x ^{2} – 20x + 1 = 0 is:**

(a)1/20 and 1/20

(b)1/10 and 1/20

(c)1/10 and 1/10

(d)None of the above

Answer:** (c)**

Explanation: Given, 100x^{2} – 20x + 1=0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)^{2} = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

**3. 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)**

Explanation: 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

**4. If ½ is a root of the quadratic equation x ^{2}-mx-5/4=0, then value of m is:**

(a)2

(b)-2

(c)-3

(d)3

Answer: **(b)**

Explanation: Given x=½ as root of equation x^{2}-mx-5/4=0.

(½)^{2} – m(½) – 5/4 = 0

¼-m/2-5/4=0

m=-2

**5. 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 Altitiude=5cm

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

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

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

Answer: **(b)**

Explanation: 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

**6. 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)**

Explanation: 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

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

**7. The value of \(\sqrt{6+\sqrt{6+\sqrt{6} \ldots \ldots \ldots}}\) is:**

(a)4

(b)3

(c)3.5

(d)-3

Answer: **(b)**

Explanation: Let, \(\sqrt{6+\sqrt{6+\sqrt{6} \ldots \ldots \ldots}} = x\)

Hence, we cna write, √(6+x) = x

6+x=x^{2}

x^{2}-x-6=0

x^{2}-3x+2x-6=0

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

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

x=-2,3

Since, x cannot be negative, therefore, x=3

**8. 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**

Explanation: 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

⇒ 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.

**9. 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)**

Explanation: 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/hr.

**10. If one root of equation 4x ^{2}-2x+k-4=0 is reciprocal of other. The value of k is:**

(a)-8

(b)8

(c)-4

(d)4

Answer: **b**

Explanation: 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