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 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: **(a)**

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

Snce, the side of 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 other, then the product of roots will be:

α x 1/α = (k-4)/4

k-4=4

k=8