NCERT Solutions for Class 10 Maths Exercise 3.1 Chapter 3 Pair of Linear Equations in Two Variables – FREE PDF Download
NCERT Class 10 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Pair of Linear Equations in Two Variables solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
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.
Seven years ago, Age of Aftab = (x – 7) years and Age of his daughter = (y – 7) years.
According to the given condition,
Again, Three years hence, Age of Aftab = x + 3 and Age of his daughter = y + 3
According to the given condition,
Thus, the given conditions can be algebraically represented as:
x – 7y = –42
x = –42 + 7y
Three solutions of this equation can be written in a table as follows:
And x – 3y = 6
x = 6 + 3y
Three solutions of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight: In order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
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 graphically.
According to given conditions, we have
3x + 6y = 3900 ⇒ x + 2y = 1300… (1)
And x + 3y = 1300… (2)
For equation x + 2y = 1300, we have following points which lie on the line
For equation x + 3y = 1300, we have following points which lie on the line
We plot the points for both of the equations and it is the graphical representation of the given situation.
It is clear that these lines intersect at B (1300,0).
3. The cost of 2 kg of apples and 1 kg 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.
According to given conditions, we have
2x + y = 160… (1)
4x + 2y = 300
⇒ 2x + y = 150… (2)
So, we have equations (1) and (2), 2x + y = 160 and 2x + y = 150 which represent given situation algebraically.
For equation 2x + y = 160, we have following points which lie on the line
For equation 2x + y = 150, we have following points which lie on the line
We plot the points for both of the equations and it is the graphical representation of the given situation.