NCERT Solutions for Class 10 Maths Exercise 3.1 Chapter 3 Linear Equations In Two Variables contains the solutions to all the questions provided on page number 44 in the textbook. Pair of Linear Variables is an important topic for the examinations. The students are advised to go through NCERT Solutions for better preparation.

The NCERT Solutions Class 10 Maths contain step-wise solutions to all the Maths problems. These Solutions of NCERT are the best guide for the students studying with detailed study material including the important topics. The experts have tried to include all that is important. The students can refer the solutions for a better understanding of the topic. The students appearing for the board examinations will find it helpful in scoring well.

### Download PDF for NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.1

### NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.1

**1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.**

**Solutions:**Let the present age of Aftab be ‘x’.

And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,

Age of Aftab = x-7

Age of his daughter = y-7

According to the question,

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42 ………………………(i)

Also, three years from now or after three years,

Age of Aftab will become = x+3.

Age of his daughter will become = y+3

According to the situation given,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6 …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y = 6+42

⇒4y = 48

⇒y = 12

The algebraic equation is represented by

x−7y = −42

x−3y = 6

For, x−7y = −42 or x = −42+7y

The solution table is

For, x−3y = 6 or x = 6+3y

The solution table is

The graphical representation is:

**2. The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.**

**Solutions: **Let us assume that the cost of a bat be ‘Rs x’

And, the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

3x+6y = 3900

And x+3y = 1300

For, 3x+6y = 3900

Or x = (3900-6y)/3

The solution table is

For, x+3y = 1300

Or x = 1300-3y

The solution table is

The graphical representation is as follows.

**3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.**

**Solutions:**Let the cost of 1 kg of apples be ‘Rs. x’

And, cost of 1 kg of grapes be ‘Rs. y’

According to the question, the algebraic representation is

2x+y = 160

And 4x+2y = 300

For, 2x+y = 160 or y = 160−2x, the solution table is;

For 4x+2y = 300 or y = (300-4x)/2, the solution table is;

The graphical representation is as follows;

This is the first exercise of the chapter and contains a total of 3 questions. The solutions to this exercise are provided by subject experts in Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1 of NCERT. The students can refer to this page for stepwise solutions to the questions.

In all the 3 questions the students are asked to represent the given form of equations graphically or algebraically.

### Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.1

- NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.1 are framed by subject experts.
- The answers provided here are accurate.
- The solutions will help the students score well in the examinations.
- All the solutions to the question provided on page number 44 are provided here.