# NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.3

(Last Updated On: September 26, 2021)

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 are part of NCERT Solutions for Class 10 Maths. Here we have given Chapter 3 Pair of Linear Equations in Two Variable Class 10 NCERT Solutions Ex 3.3.

 Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 3 Chapter Name Pair of Linear Equations in Two Variables Exercise Ex 3.3 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Page No: 53

Question 1. Solve the following pair of linear equations by the substitution method.
(i) = 14 ; – = 4
(ii) – = 3 ; s/3 + t/2 = 6
(iii) 3x – y = 3 ; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0 ; √3x – √8y = 0
(vi) 3/2x – 5/3y = -2 ; x/3 + y/2 = 13/6

Solution:
(i) x + y = 14      … (i)
x – y = 4        … (ii)
From (i), we obtain:
x = 14 – y      … (iii)
Substituting this value in equation (ii), we obtain: Substituting the value of y in equation (iii), we obtain:
x = 9
∴ x = 9 , y = 5 From (i), we obtain:
s = t + 3                   …(iii) Substituting this value in equation (ii), we obtain:
Substituting the value of t in equation (iii), we obtain:
s = 9
∴ s = 9 , t = 6
(iii)  3x – y = 3        … (i)
9x – 3y = 9        … (ii)
From (i), we obtain
y = 3x – 3        … (iii)
Substituting this value in equation (ii), we obtain: 9 = 9
This is always true.
Thus, the given pair of equations has infinitely many solutions and the relation between these variables can be given by
y  = 3x – 3
So, one of the possible solutions can x = 3, y = 6.  Substituting the value of y in equation (iii), we obtain:
x = 0
∴ x = 0, y = 0 Concept Insight: To solve the given pair of equations, we need to replace the value of one of the variables with one of the equations. But be sure to substitute the value of the variable which simplifies the calculations. For example, it is most convenient to substitute the value of x in part (iv) from the first equation to the second, since division by 0.2 is easier than dividing by 0.3, 0.4, and 0.5.

Question 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m‘ for which y =mx + 3.

Solution: Thus, the value of m is -1. Concept Insight: First of all the solution of the given pair of linear equations can be found by substituting the value of a variable, say x, from one equation to the other. Then after finding the values ​​of x and y, substitute them in the equation y = m x + 3 to find the value of m.

Question 3. Form a 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 thrice the other number. Find them

Solution:
Let one number be x and the other number be y such that y > x.
According to the question: Substituting this in equation (1), we obtain
y = 39
Hence, the numbers are 13 and 39.

Concept Insight: In this problem two relations between two numbers are given. So, there are two numbers to be found here. So the two numbers will be represented by the variables x and y which clearly denote the larger variable.
From the given conditions , A pair of equations can be obtained. The pair of equations can be solved by suitable substitution.

(ii) The larger of two supplementary angles is 18 degrees more than the smaller angle. Find them

Solution:
Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always 180º.
According to the given information,
x + y = 180º — (1)
x – y = 180º — (2)
From (1), we obtain
x = 180º – y — (3)
Substituting this in equation (2), we obtain
180º – y – y = 18º
162º = 2y
81º = y —- (4)
Putting this in equation (3), we obtain
x = 180º – 81º
= 99º
Hence, the angles are 99º and 81º.

Concept Insight: This problem talks about the measure of two complementary angles. Hence, both the angles will be written as variables. The pair of equations can be constructed using the fact that the sum of two supplementary angles is 180° and using the condition given in the problem. The pair of equations can be solved by suitable substitution.

NCERT Solutions for Class 10 Maths Chapter 2
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution:
Let the cost of a bat and a ball be x and y respectively.
According to the given information, Hence, a bat costs Rs 500 and a ball costs Rs 50.

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. The pair of equations can be solved by suitable substitution.

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(iv) Taxi charges in a city include a fixed fee along with the fee for the distance covered. For a distance of 10 km, the fee paid is Rs 105 and for a distance of 15 km the fee paid is Rs 155. What are the fixed charges and charges per km? How much does a person have to pay to cover a distance of 25 km?

Solution: Concept Insight: In this problem, we have to find out the fixed charge and charge per km. Therefore, we will represent these two using different variables. Now, two linear equations can be written using the given conditions in the problem. The pair of equations can be solved by suitable substitution.

(v) A fraction becomes 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 5/6. Find the fraction.

Solution:
Let the fraction be x/y.
According to the given information, Concept Insight: This problem talks about a fraction which is not known to us. So numerator and denominator will be considered as variables x and y respectively and y will not be completely zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can be solved by suitable substitution.

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(vi) After five years, Jacob’s age will be three times that of his son. Five years ago, the age of Jacob was seven times that of his son. What is his present age?

Solution:
Let the age of Jacob be x and the age of his son be y.
According to the given information, Hence, the present age of Jacob is 40 years while the present age of his son is 10 years.

Concept Insight: Here, the present age of Jacob and his son is not known. So we will write these two as variables. The problem talks about their age five years before and five years after. Here, five years ago means we have to subtract 5 from his current age, and five years means we have to add 3 to his current age. Hence, using the given conditions, a pair of linear equations can be formed. The pair of equations can be solved by suitable substitution.

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