---
title: "NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2"
url: https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-2
dateModified: 2026-10-07T15:49:35+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Chapter 4: Quadratic Equations. Every question from Exercise 4.2, with full working and the final answer.

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## EXERCISE 4.2

### Question 1

*3 marks · Short answer*

Find the roots of the following quadratic equations by factorisation:
(i) $x^2 - 3x - 10 = 0$
(ii) $2x^2 + x - 6 = 0$
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
(iv) $2x^2 - x + \frac{1}{8} = 0$
(v) $100x^2 - 20x + 1 = 0$

**Part (i)**

1. $x^2 - 3x - 10 = 0$
2. $x^2 - 5x + 2x - 10 = 0$
3. $x(x - 5) + 2(x - 5) = 0$
4. $(x - 5)(x + 2) = 0$
5. $x = 5 \text{ or } x = -2$

Answer (i): $x = 5, -2$

**Part (ii)**

1. $2x^2 + x - 6 = 0$
2. $2x^2 + 4x - 3x - 6 = 0$
3. $2x(x + 2) - 3(x + 2) = 0$
4. $(2x - 3)(x + 2) = 0$
5. $x = \frac{3}{2} \text{ or } x = -2$

Answer (ii): $x = \frac{3}{2}, -2$

**Part (iii)**

1. $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
2. $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$
3. $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$
4. $(x + \sqrt{2})(\sqrt{2}x + 5) = 0$
5. $x = -\sqrt{2} \text{ or } x = -\frac{5}{\sqrt{2}}$

Answer (iii): $x = -\sqrt{2}, -\frac{5}{\sqrt{2}}$

**Part (iv)**

1. $2x^2 - x + \frac{1}{8} = 0$
2. $16x^2 - 8x + 1 = 0$
3. $(4x)^2 - 2(4x)(1) + (1)^2 = 0$
4. $(4x - 1)^2 = 0$
5. $x = \frac{1}{4}, \frac{1}{4}$

Answer (iv): $x = \frac{1}{4}, \frac{1}{4}$

**Part (v)**

1. $100x^2 - 20x + 1 = 0$
2. $(10x)^2 - 2(10x)(1) + (1)^2 = 0$
3. $(10x - 1)^2 = 0$
4. $(10x - 1)(10x - 1) = 0$
5. $x = \frac{1}{10}, \frac{1}{10}$

Answer (v): $x = \frac{1}{10}, \frac{1}{10}$

**Answer:** Roots of the given quadratic equations are found by factorisation.

> Common mistake: Sign errors while splitting the middle term.

### Question 2

*3 marks · Short answer*

Solve the problems given in Example 1.

**Part (i)**

1. The quadratic equation representing the problem is $x^2 - 45x + 324 = 0$.
2. $x^2 - 36x - 9x + 324 = 0$
3. $x(x - 36) - 9(x - 36) = 0$
4. $(x - 36)(x - 9) = 0$
5. Thus, $x = 36$ or $x = 9$. If John had 36 marbles, Jivanti had 9, and vice versa.

Answer (i): John and Jivanti had 36 and 9 marbles (or 9 and 36 marbles).

**Part (ii)**

1. The quadratic equation representing the problem is $x^2 - 55x + 750 = 0$.
2. $x^2 - 25x - 30x + 750 = 0$
3. $x(x - 25) - 30(x - 25) = 0$
4. $(x - 25)(x - 30) = 0$
5. Thus, $x = 25$ or $x = 30$.

Answer (ii): The number of toys produced on that day was 25 or 30.

**Answer:** Solutions to the problems given in Example 1.

> Common mistake: Not verifying if both values obtained satisfy the physical conditions of the problem.

### Question 3

*3 marks · Short answer*

Find two numbers whose sum is $27$ and product is $182$.

**Solution**

1. Let the first number be $x$.
2. Then the second number is $27 - x$.
3. Their product is $x(27 - x) = 182$.
4. $27x - x^2 = 182$, which gives $x^2 - 27x + 182 = 0$.
5. $x^2 - 13x - 14x + 182 = 0$, so $(x - 13)(x - 14) = 0$.
6. Thus, $x = 13$ or $x = 14$.

**Answer:** The two numbers are 13 and 14.

> Common mistake: Errors in finding the factors of 182 that add up to -27.

### Question 4

*3 marks · Short answer*

Find two consecutive positive integers, sum of whose squares is $365$.

**Solution**

1. Let the first consecutive positive integer be $x$.
2. Then the next consecutive positive integer is $x + 1$.
3. According to the given condition, $x^2 + (x + 1)^2 = 365$.
4. $x^2 + x^2 + 2x + 1 = 365$, which simplifies to $2x^2 + 2x - 364 = 0$ or $x^2 + x - 182 = 0$.
5. $x^2 + 14x - 13x - 182 = 0$, so $(x + 14)(x - 13) = 0$.
6. Since $x$ must be a positive integer, $x = 13$, and $x + 1 = 14$.

**Answer:** The two consecutive positive integers are 13 and 14.

> Common mistake: Ignoring the condition that the integers must be positive.

### Question 5

*3 marks · Short answer*

The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides.

**Solution**

1. Let the base of the right triangle be $x\text{ cm}$.
2. Then its altitude is $(x - 7)\text{ cm}$ and hypotenuse is given as $13\text{ cm}$.
3. By Pythagoras theorem, $x^2 + (x - 7)^2 = 13^2$.
4. $x^2 + x^2 - 14x + 49 = 169$, which simplifies to $2x^2 - 14x - 120 = 0$ or $x^2 - 7x - 60 = 0$.
5. $x^2 - 12x + 5x - 60 = 0$, so $(x - 12)(x + 5) = 0$.
6. Since length cannot be negative, $x = 12$, and the altitude is $12 - 7 = 5\text{ cm}$.

**Answer:** The other two sides are $5\text{ cm}$ and $12\text{ cm}$.

> Common mistake: Taking negative value for the base of the triangle.

### Question 6

*3 marks · Short answer*

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was $\text{`} 90$, find the number of articles produced and the cost of each article.

**Solution**

1. Let the number of articles produced on that day be $x$.
2. The cost of production of each article (in rupees) is $2x + 3$.
3. The total cost of production is given as $\text{`} 90$, so $x(2x + 3) = 90$.
4. Expanding and rearranging gives $2x^2 + 3x - 90 = 0$, which can be split as $2x^2 - 12x + 15x - 90 = 0$.
5. Factorising gives $(2x + 15)(x - 6) = 0$, so $x = 6$ (since the number of articles cannot be negative).
6. Thus, the number of articles produced is $6$ and the cost of each article is $2(6) + 3 = \text{`} 15$.

**Answer:** Number of articles produced = 6, Cost of each article = ₹15

> Common mistake: Taking the total cost as $x + 2x + 3$ instead of multiplying the number of articles by the cost per article.

## Related pages

- [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions)
- [Exercise 4.1](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-1)
- [Exercise 4.3](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-3)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student