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

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(Last Updated On: September 24, 2021)

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

Are you looking for the best Ex 3.1 Class 10 OF Maths NCERT Solutions Chapter 3 ? Then, grab them from our page and speed up your preparation for CBSE Class 10 Exams.

Pair of Linear Equations in Two Variables Ex 3.1 , NCERT Solutions for Class 10 Maths Chapter 3 is part of NCERT Solutions for Class 10 Maths. Class 10 NCERT Solutions Ex 3.1 , Here we have given Chapter 3 Pairs of Linear Equations in Two Variables .

Simple step-by-step explanations are given of NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with . These solutions for pairing linear equations in two variables are extremely popular among class 1 0 students, math solutions for pairing linear equations in two variables are useful for completing your homework quickly and preparing for exams.

 Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 3 Chapter Name Pair of Linear Equations in Two Variables Exercise Ex 3.1 Number of Questions Solved 3 Category NCERT Solutions

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

Page No: 44

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 and his daughter be x and y respectively.
Seven years ago,
Age of Aftab = x – 7
Age of his daughter = y – 7
According to the given condition,

Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,

Thus, the given conditions can be algebraically represented as:
x – 7y = -42
x – 3y = 6
x – 7y = -42 ⇒ × = -42 + 7y
Three solutions of this equation can be written in a table as follows:

 x -7 0 7 y 5 6 7

× -3y = 6 ⇒ × = 6 + 3y
Three solutions of this equation can be written in a table as follows:

 x 6 3 0 y 0 -1 -2

The graphical representation is as follows:

Concept insight: To represent a given situation mathematically, let’s first look at what we need to find out in the problem. Here, the age will be represented by the variables x and y , the present age of Aftab and his daughter needs to be found. The problem speaks of his age seven years ago and three years from now. Here, the word ‘seven years ago’ means we need to subtract 7 from his current age, and ‘three years from now’ or ‘three years later’ means we need to add 3 to his current age. Remember that to represent algebraic equations graphically, the solution set of the equations should only be taken as whole numbers for accuracy. The graph of two linear equations will be represented by a straight line.

Question 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.

Solution:
Let the cost of a bat and a ball be Rs x and Rs y respectively.
The given conditions can be algebraically represented as:
3x + 6y = 3900
x + 2y = 1300

Three solutions of this equation can be written in a table as follows:

 x 300 100 -100 y 500 600 700

x + 2y = 1300 ⇒ x = 1300 – 2y
Three solutions of this equation can be written in a table as follows:

 x 300 100 -100 y 500 600 700

The graphical representation is as follows:

Concept insight: The cost of bat and balls needs to be found hence cost of ball and bat will be taken as variable. Applying the terms total cost of bats and balls will yield the algebraic equations. To avoid any computational errors , Then, to represent the obtained equations take at least three ordered pairs graphically on baoth equations .

Question 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.

Solution:
Let the cost of 1 kg of apples and 1 kg grapes be Rs. x and Rs. y.
The given condtions can be algebraically represented as:

Three solutions of this equation can be written in a table as follows:

 x 50 60 70 y 60 40 20

Three solutions of this equation can be written in a table as follows:

 x 70 80 75 y 10 -10 0

The graphical representation is as follows:

Concept insight: The cost of apples and grapes needs to be found so the cost of 1 kg of apples and 1 kg of grapes will be taken as variables. From the given terms of the collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, to represent the obtained equations graphically, just take the values ​​of the variables as whole numbers. Since these values ​​are large, take a suitable scale such as 1 cm = 20.

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