---
title: "NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2"
url: https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-2
dateModified: 2026-10-07T15:50:31+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

Chapter 3: Pair of Linear Equations in Two Variables. Every question from Exercise 3.2, with full working and the final answer.

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## EXERCISE 3.2

### Question 1

*3 marks · Short answer*

Solve the following pair of linear equations by the substitution method.
(i) $x + y = 14$, $x - y = 4$
(ii) $s - t = 3$, $\frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x - y = 3$, $9x - 3y = 9$
(iv) $0.2x + 0.3y = 1.3$, $0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0$, $\sqrt{3}x - \sqrt{8}y = 0$
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$, $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

**Part (i)**

1. Given equations are $x + y = 14$ and $x - y = 4$.
2. From $x - y = 4$, we get $x = y + 4$.
3. Substituting $x$ in the first equation, $(y + 4) + y = 14$, which gives $y = 5$.
4. Putting $y = 5$ in $x = y + 4$, we get $x = 9$.

Answer (i): $x = 9, y = 5$

**Part (ii)**

1. Given equations are $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$.
2. From $s - t = 3$, we get $s = t + 3$.
3. Substituting $s$ in the second equation, $\frac{t + 3}{3} + \frac{t}{2} = 6$, giving $t = 6$.
4. Putting $t = 6$ in $s = t + 3$, we get $s = 9$.

Answer (ii): $s = 9, t = 6$

**Part (iii)**

1. Given equations are $3x - y = 3$ and $9x - 3y = 9$.
2. From $3x - y = 3$, we get $y = 3x - 3$.
3. Substituting $y$ in the second equation, $9x - 3(3x - 3) = 9$, which simplifies to $9 = 9$.
4. This is a true statement for all values of $x$, hence the pair has infinitely many solutions.

Answer (iii): Infinitely many solutions

**Part (iv)**

1. Given equations are $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$.
2. From the first equation, $x = \frac{1.3 - 0.3y}{0.2}$.
3. Substituting this into the second equation gives $y = 3$.
4. Putting $y = 3$ gives $x = 2$.

Answer (iv): $x = 2, y = 3$

**Part (v)**

1. Given equations are $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$.
2. From the first equation, $x = -\frac{\sqrt{3}}{\sqrt{2}}y$.
3. Substituting $x$ in the second equation gives $\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - 2\sqrt{2}y = 0$, which yields $y = 0$.
4. Putting $y = 0$ gives $x = 0$.

Answer (v): $x = 0, y = 0$

**Part (vi)**

1. Given equations are $\frac{3x}{2} - \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$.
2. Simplifying the equations gives $9x - 10y = -12$ and $2x + 3y = 13$.
3. Expressing $x$ from the second equation as $x = \frac{13 - 3y}{2}$ and substituting in the first gives $y = 3$.
4. Substituting $y = 3$ gives $x = 2$.

Answer (vi): $x = 2, y = 3$

**Answer:** Solutions obtained by substitution method.

> Common mistake: Errors in algebraic simplification of fractions and decimals.

### Question 2

*3 marks · Short answer*

Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of '$m$' for which $y = mx + 3$.

**Solution**

1. Given equations are $2x + 3y = 11$ and $2x - 4y = -24$.
2. Subtract the second equation from the first equation: $(2x - 2x) + (3y - (-4y)) = 11 - (-24)$, which gives $7y = 35$, so $y = 5$.
3. Substitute $y = 5$ into the first equation: $2x + 3(5) = 11$, leading to $2x + 15 = 11$, so $2x = -4$, and $x = -2$.
4. Substitute $x = -2$ and $y = 5$ into the relation $y = mx + 3$ to find $m$.
5. We get $5 = m(-2) + 3$, which simplifies to $2 = -2m$, so $m = -1$.

**Answer:** $m = -1$

> Common mistake: Substituting the values of x and y incorrectly into the linear relation for m.

### Question 3

*3 marks · Short answer*

Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is ₹ 105 and for a journey of $15\text{ km}$, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25\text{ km}$?
(v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

**Part (i)**

1. Let the two numbers be $x$ and $y$ such that $x > y$.
2. Given $x - y = 26$ and $x = 3y$.
3. Substituting $x = 3y$ into the first equation, $3y - y = 26$, giving $y = 13$.
4. Thus, $x = 3(13) = 39$.

Answer (i): Numbers are 39 and 13.

**Part (ii)**

1. Let the larger angle be $x$ and smaller angle be $y$.
2. Given $x + y = 180^\circ$ and $x - y = 18^\circ$.
3. From $x = y + 18^\circ$, substituting in the first equation gives $(y + 18^\circ) + y = 180^\circ$, so $y = 81^\circ$.
4. Thus, $x = 99^\circ$.

Answer (ii): Angles are $99^\circ$ and $81^\circ$.

**Part (iii)**

1. Let the cost of one bat be $x$ and one ball be $y$.
2. Form equations: $7x + 6y = 3800$ and $3x + 5y = 1750$.
3. Express $x$ from the second equation as $x = \frac{1750 - 5y}{3}$ and substitute into the first.
4. Solving gives $y = 50$ and $x = 500$.

Answer (iii): Cost of a bat is ₹ 500 and of a ball is ₹ 50.

**Part (iv)**

1. Let fixed charge be ₹ $x$ and charge per km be ₹ $y$.
2. Form equations: $x + 10y = 105$ and $x + 15y = 155$.
3. From the first, $x = 105 - 10y$; substituting in the second gives $y = 10$ and $x = 5$.
4. For $25\text{ km}$, total charge is $x + 25y = 5 + 25(10) = 255$.

Answer (iv): Fixed charge ₹ 5, charge per km ₹ 10, total for 25 km is ₹ 255.

**Part (v)**

1. Let the fraction be $\frac{x}{y}$.
2. Given $\frac{x + 2}{y + 2} = \frac{9}{11}$ and $\frac{x + 3}{y + 3} = \frac{5}{6}$.
3. Simplify to form $11x - 9y = -4$ and $6x - 5y = -3$.
4. Solving by substitution yields $x = 7$ and $y = 9$.

Answer (v): The fraction is $\frac{7}{9}$.

**Part (vi)**

1. Let Jacob's present age be $x$ and his son's age be $y$.
2. Given equations are $x + 5 = 3(y + 5)$ and $x - 5 = 7(y - 5)$
3. Simplifying gives $x - 3y = 10$ and $x - 7y = -30$.
4. Solving by substitution gives $y = 10$ and $x = 40$.

Answer (vi): Jacob's age is 40 years and son's age is 10 years.

**Answer:** Solutions to the word problems.

> Common mistake: Forming incorrect linear equations from the word statements.

## Related pages

- [All Chapter 3 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions)
- [Exercise 3.1](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-1)
- [Exercise 3.3](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-3)

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