NCERT Solutions for Class 10 Maths (Chapter 3, Exercise 3.1)

NCERT Solutions for Class 10 Maths Chapter 3 is provided in this section. The Chapter 3 of Class 10 Mathematics under the NCERT Syllabus is about the Pair of Linear Equations in Two Variables. As per the Textbook published by NCERT, the chapter discusses the concepts of Pair of Linear Equations in Two Variables, Graphical Method of Solution of a Pair of Linear Equations, Algebraic Methods of Solving a Pair of Linear Equations, Substitution Method, Elimination Method, Cross-Multiplication Method, and Equations Reducible to a Pair of Linear Equations in Two Variables.

This post contains the step-by-step solutions of Exercise 3.1, Chapter 3 of Class 10 Mathematics designed in the easiest possible way. There is a total of 3 questions in the Ex. 3.1 and all the questions are covered in this section.

Topics Covered in Chapter 3: Pair of Linear Equations in Two Variables

Serial No.	Section	Topic
1	3.1	Introduction
2	3.2	Pair of Linear Equations in Two Variables
3	3.2	Graphical Method of Solution of a Pair of Linear Equations
4	3.4	Algebraic Methods of Solving a Pair of Linear Equations
5	3.4.1	Substitution Method
6	3.4.2	Elimination Method
7	3.4.3	Cross-Multiplication Method
8	3.5	Equations Reducible to a Pair of Linear Equations in Two Variables
9	3.6	Summary

NCERT Solutions for Class 10 Maths, Chapter 3 – Pair of Linear Equations in Two Variables (Ex. 3.1)

NCERT Solutions for Class 10 Maths, Chapter 3 (Pair of Linear Equations in Two Variables), Exercise 3.1 are described below. The Exercise 3.1 is based on Pair of Linear Equations in Two Variables and Graphical Method of Solution of a Pair of Linear Equations and each and every question of Ex. 3.1 are solved (step-by-step) below.

Board	CBSE
Textbook (Council)	NCERT (National Council of Educational Research and Training)
Class	Class 10 or Class X
Subject	Mathematics
Chapter Number	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise Number	Exercise 3.1
Number of Questions	3 Questions

Exercise 3.1

Question 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.

Solution:

Let, the present age of Aftab be x years and the present age of his daughter be y years.

7 years ago, Aftab’s age was =(x-7) years and

Aftab’s daughter’s age was=(y-7) years

According to the question,

(x-7)=7(y-7)

\Rightarrow x-7=7y-49

\Rightarrow x-7y=-42 ......... (i)

Again, after three years Aftab’s age will be (x+3) years, and his daughter’s age will be (y+3) years.

According to the question,

(x+3)=3(y+3)

\Rightarrow x+3=3y+9

\Rightarrow x-3y=6 ......... (ii)

From Equation (i),

x-7y=-42

\Rightarrow y=\frac{x+42}{7}

x	0	7	14
=\frac{x+42}{7}	6	7	8

From Equation (ii),

x-3y=6

\Rightarrow y=\frac{x-6}{3}

x	6	12	18
y=\frac{x-6}{3}	0	2	4

Plotting the values of x and y from Table 1 and 2,

The above figure is the graphical representation of the given question.

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

Solution:

Let us consider the cost of one bat be ₹ x and the cost of one ball be ₹ y.

Therefore, the cost of 3 bats is 3x and the cost of 6 balls is 6y.

According to the question,

3x+6y=3900

x+2y=1300 ......... (i)

Again, the cost of one bat is x and 3 balls is 3y.

According to the question,

x+3y=1300 ......... (ii)

From Equation (i),

x+2y=1300

\Rightarrow 2y=1300-x

\Rightarrow y=\frac{1300-x}{2}

x	0	300	650
y=\frac{1300-x}{2}	650	500	325

From Equation (ii),

x+3y=1300

\Rightarrow 3y=1300-x

\Rightarrow y=\frac{1300-x}{3}

x	100	400	700
y=\frac{1300-x}{3}	400	300	200

Plotting the values of x and y from Table 3 and 4,

The above figure is the graphical representation of the given question.

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

Solution:

Let us consider the cost of 1 kg apple be ₹ x and 1 kg grapes be ₹ y.

According to the question, the cost of 2 kg of apples and 1 kg of grapes is ₹160.

\therefore \,\,\,\,\, 2x+y=160 ......... (i)

Again, after a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300.

\therefore \,\,\,\,\, 4x+2y=300 ......... (ii)

From Equation (i),

2x+y=160

\Rightarrow y=160-2x

x	0	5	50
y=160-2x	160	150	60

From Equation (ii),

\therefore \,\,\,\,\, 4x+2y=300

\Rightarrow 2y=300-4x

\Rightarrow y=\frac{300-4x}{2}

x	0	25	50
y=\frac{300-4x}{2}	150	100	50

Plotting the values of x and y from Table 5 and 6,

The above figure is the graphical representation of the given question.

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NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.1 PDF Download

The PDF of NCERT solutions for Class 10 Maths Chapter 3, Exercise 3.1 is provided below.

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List of Exercises in Pair of Linear Equations in Two Variables, Chapter 3, Class 10 Maths

• Exercise 3.1 (NCERT Solutions for Ex. 3.1 Class 10 Maths, Chapter 3)
• Exercise 3.2 (NCERT Solutions for Ex. 3.2 Class 10 Maths, Chapter 3)
• Exercise 3.3 (NCERT Solutions for Ex. 3.3 Class 10 Maths, Chapter 3)
• Exercise 3.4 (NCERT Solutions for Ex. 3.4 Class 10 Maths, Chapter 3)
• Exercise 3.5 (NCERT Solutions for Ex. 3.5 Class 10 Maths, Chapter 3)
• Exercise 3.6 (NCERT Solutions for Ex. 3.6 Class 10 Maths, Chapter 3)
• Exercise 3.7 (NCERT Solutions for Ex. 3.7 Class 10 Maths, Chapter 3)

I hope the NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.1 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear.

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