# Chapter 3 Pair of Linear Equations in Two Variables

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### Exercise 1 ( Page No. : 44 )

• Q1 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.
Ans:

Let the present age of Aftab and his father be x years and y years respectively.

According to question,

7 years ago, we have

x – 7 = 7 (y - 7)

Or,                                          x – 7 = 7y – 49

Or,                                          x – 7y = - 42       …………… (1)

3 years from now, we have

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

Or,                                        x + 3 = 3y + 9

Or,                                        x – 3y = 6            ……………. (2)

Graphical Representation

From equation (1),                   x – 7y = -42

Table value of x and y

 x: 0 -42 -35 y: 6 0 1

From equation (2),     x – 3y = 6

Table value of x and y

 x: 0 9 6 y: -2 1 0

Plotting the tables on the graph:

Q2 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.
Ans:

Let the cost of one bat and one ball be x Rupees and y Rupees respectively.

According to first condition,

3x + 6y = â‚¨ 3900    …………….(1)

According to second condition,

x + 3y = â‚¨ 1300   …………….(2)

Graphical Representation

From equation (1),                  3x + 6y = 3900

Table value of x and y

 x: 100 300 700 y: 600 500 300

From equation (2),                   x + 3y = 1300

Table value of x and y

 x: 100 700 400 y: 400 200 300

1 Unit = 100

Q3 The cost of 2 kg of apples and 1kg 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.
Ans:

Let the cost of one kg apple be x â‚¨ and 1 kg grapes be y â‚¨.

According to first condition,

2x + y = 160 â‚¨ …………..(1)

According to second condition,

4x + 2y = 300

2x + y = 150 ……………….(2)

Graphical Representation

Table for equation (1),                    2x + y = 160

Table value of x and y

 x: 40 60 80 y: 60 40 0

Table for equation (2),                    2x + y = 150

Table value of x and y

 x: 40 60 20 y: 70 30 110

### Exercise 3 ( Page No. : 53 )

•  Q1 Solve the following pair of linear equations by the substitution method. Ans: (i)                              x + y = 14                 …………….(1)            x – y = 4                  …………….(2)                                  From the equation (1), we get                                    x = 14 - y                 …………….(3)                                  Putting the value of x in equation (2), we get                                     (14 - y) – y = 4                                                                                        14 – y – y = 4                                                                                             - 2y = - 10                                                                                                                                Putting the value of y in equation (3),                                        x = 14 – 5                                         x = 9                                         Hence,     x = 9 and y = 5 (iii)                             3x - y = 3                 …………….(1)              9x – 3y = 9…………….(2)                                                                  From the equation (1), we get                                                                                      Putting the value of y in equation (2), we get                                      9x – 3 (3x - 3) = 9                                       9x – 9x + 9 = 9                                       9 = 9, which is true.                                   Therefore, pair of linear equation has infinite many solutions. Q2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. Ans: Q3 Form the 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 three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (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. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes, , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes  Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Ans: (i)                  Let the numbers be x and y, such that x > y Therefore, according to question                               x - y = 26                 …………….(1) x = 3y…………….(2)                                    Putting the value of x from equation (2) to equation (1), we get                                                                                     3y – y = 26                                                                                     2y = 26                                                                                      y = 13                                                       Putting the value in equation (2), we get                                                                                      x = 3 x 13                                                                                      x = 39                                        Hence, the numbers are 39 and 13. (ii) Let one be xâ—¦ and other be yâ—¦ such that (xâ—¦ > yâ—¦) Therefore, according to question                                     xâ—¦ + yâ—¦ = 180â—¦…………….(1) (Supplementary angles) xâ—¦ = 18 + yâ—¦…………….(2) Putting the value of x from equation (2) to equation (1), we get                                              18 + yâ—¦ + yâ—¦ = 180â—¦ 2yâ—¦ = 162â—¦ Putting the value of y in equation (2), we get                       xâ—¦ = 18 + 81 x = 99 (iv)                 Let the fixed charge be = â‚¨ x Let the charge for 1 km distance be = â‚¨ y According to first condition,                                                                                    x + 10y = â‚¨ 105                                                                                       x = 105 – 10y                  …………….(1)                            According to second condition,                                                                                      x + 15y = 155                  ………………(2)                                      Putting the value of x in equation (2), we get                                                                                     105 – 10y + 15y = 155                                                                                     5y = 50                                                                                      y = 10                                      Putting the value of y in equation (2), we get                                                                                      x + 15 x 10 = 155                                                                                                                     x = 5                                     Hence, the fixed charge for taxi is â‚¨ 5 and, the charge for one km distance is â‚¨ 50.                                     Charge for 25 km distance                                                                                      = 25 x 10 + 5                                                                                      = â‚¨ 255 (vi)                 Let the age of Jacob be = x years Let the age of Jacob’s father be = y years After 5 years,                                                 Jacob’s age               x + 5 years                                                    Son’s age                  y + 5 years                                        According to question,                        x + 5 = 3 (y + 5)            x + 5 = 3y + 15 x = 3y + 10………………(1) Five years ago, (x- 5) = 7 (y - 5) x – 5 = 7y – 35 x – 7y = -30……………….(2)                                     Putting the value of x in equation (2), we get                                                                                     3y + 10 - 7y = -30                                                                                     -4y = -40                                                                                      y = 10                                      Putting the value of y in equation (1), we get                                                                                      x = 3 (10) + 10                                                                                                                       x = 40                                     Hence, the present age of Jacob is 40 years and the age of his son is 10 years.