# Pair of Linear Equations in Two Variables (Chapter 3, Exercise 3.1) Solutions – Class 10 Mathematics

The Chapter 3 of Class 10 Mathematics under the NCERT Syllabus is about the Pair of Linear Equations in Two Variables. The Chapter discusses the Pair of Linear Equation in Two Variables, Graphical Method of Solving the Pair of Linear Equation in Two Variables, Algebraic Methods of Solving the Pair of Linear Equations in Two Variables including Substitution Method, Elimination Method, and Cross-Multiplication Method and Equations reducible to a pair of linear equations in two variables. This post contains the Solutions of Exercise 3.1 of Chapter 3 of Class 10 Mathematics under the NCERT Syllabus.

The solution of all the questions of Exercise 3.1 of Class 10 Mathematics is given below.

## 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 $(i)$,

$x-7y=-42$

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

From Equation $(ii)$,

$x-3y=6$

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

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

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

## 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 $(i)$,

$x+2y=1300$

$\Rightarrow 2y=1300-x$

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

From Equation $(ii)$,

$x+3y=1300$

$\Rightarrow 3y=1300-x$

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

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

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

## 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 $(i)$,

$2x+y=160$

$\Rightarrow y=160-2x$

From Equation $(ii)$,

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

$\Rightarrow 2y=300-4x$

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

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

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

I hope the solutions of Class 10, Exercise 3.1, Chapter 3 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 3.1, feel free to contact me at [email protected] or [email protected] or fill the form here.

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