# Class 10 Mathematics Pair of Linear Equations in Two Variables: NCERT Solutions for Question 3

This page focuses on the detailed Pair of Linear Equations in Two Variables question answers for Class 10 Mathematics Pair of Linear Equations in Two Variables, addressing the question: '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?'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
Question 3

## 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?

(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 x◦ and other be y◦ such that (x◦ > y◦)

Therefore, according to question

x◦ + y◦ = 180◦…………….(1) (Supplementary angles)

x◦ = 18 + y◦…………….(2)

Putting the value of x from equation (2) to equation (1), we get

18 + y◦ + y◦ = 180◦

2y◦ = 162◦

Putting the value of y in equation (2), we get

x◦ = 18 + 81

x = 99

(iv)                 Let the fixed charge be = ₨ x

Let the charge for 1 km distance be = ₨ y

According to first condition,

x + 10y = ₨ 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 ₨ 5 and, the charge for one km distance is ₨ 50.

Charge for 25 km distance

= 25 x 10 + 5

= ₨ 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.