Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.9

In Class 10 Mathematics, Chapter 3 focuses on Pair of Linear Equations in Two Variables. This chapter deals with the solving systems of the equations where each equation is linear and has two variables. The importance of this topic lies in its applications in various fields such as economics, engineering, and everyday problem-solving.

Pair of Linear Equations in Two Variables

A pair of linear equations in the two variables can be represented as:

a₁x+b₁y=c₁

​a₂x+b₂y=c₂

Here, x and y are the variables and a₁,b₁,c₁,a₂,b₂,c₂ are constants.

Question 1. A father is three times as old as his son. After twelve years, his age will be twice as that of his son and then. Find their present ages.

Solution:

Given, father's age is 3 times as his son's age.

Let the present age of father be 'a' and the present age of his son is 'b'

So, by the given condition, a = 3b   ......(1)

Also given, after twelve years, Father's age is twice as his son's age.

Therefore, (a + 12) = 2(b + 12)  

⇒ a + 12 = 2b + 24

⇒ a = 2b + 12   ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 3b = 2b + 12

⇒ b = 12

On putting b = 12 in eq(1), we get 

 a = 2(12) + 12 = 36

Hence, the present age of Father is 36 years and the present age of son is 12 years. 

Question 2. Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B?

Solution:

Let, the present age of A and B are x, y respectively.

Given, Ten years later, A will be twice as old as B.

⇒ x + 10 = 2 (y + 10)

⇒ x = 2y + 10      ......(1)

Also given, Five years ago, A was thrice as old as B.

⇒ x - 5 = 3 (y - 5)

⇒ x = 3y - 10       ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 2y + 10 = 3y - 10

⇒ y= 20.

On putting y = 20 in eq(1), we get

x = 2(20) + 10 = 50

Hence, the present ages of A is 50 years and B is 20 years. 

Question 3. A is elder to B by 2 years. A's father F is twice as old as A and B is twice as old as his sister S. If the ages of the father and sister differ by 40 years. Then Find the age of A.

Solution:

Let the present ages of A, B, A's father F and A's sister S are 'a', 'b', 'c', 'd' respectively.

Given, A is 2 years older than B.

⇒ a = b + 2         ......(1)

Also, A's father F is twice as old as A and B is twice as old as his sister,

⇒ c = 2a and b = 2d      ......(2)

Also given, the age of Father and sister differ by 40 years.

⇒ c - d = 40    ......(3)

On substituting d = b/2 and c = 2a from eq(2) and b = (a - 2) from eq(1) in eq(3), we get,

⇒ 2a - b/2 = 40

⇒ 2a - (a - 2)/2 = 40 

⇒ 

⇒ 3a + 2 = 80

⇒ a = 26

Hence, the age of A is 26 years.

Question 4. Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Solution:

Let the present age of man and his son be x, y respectively.

Given, six years hence man's age will be thrice as his son's age.

⇒ x + 6 = 3 (y + 6)

⇒ x = 3y + 12       ......(1)

Also, three years ago, he was nine times as old as his son.

⇒ x - 3 = 9 (y - 3)

⇒ x = 9y - 24       ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 3y + 12 = 9y - 24 

⇒ y = 6

On putting y = 6 in eq(1), we get

x = 3(6) + 12 = 30

Hence, the present age of Man is 30 years and his son is 6 years. 

Question 5. Ten years ago, a father was twelve times as old as his son, and ten years hence, he will be twice as old as his son will be then. Find their present ages.

Solution: 

Let the present age of father and his son are 'a' and 'b' respectively.

Given, Ten years ago, father's age is twelve times as his son's age.

⇒ a - 10 = 12(b - 10)

⇒ a = 12b - 110     ......(1)

Also, ten years hence, his age will be twice as his son's age.

⇒ a + 10 = 2 (b + 10)

⇒ a = 2b + 10         ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 12b - 110 = 2b + 10

⇒ b = 12

On putting, b = 12 in eq(1), we get

a = 2(12) + 10 = 34

Hence, the present ages of Father is 34 years and his son is 12 years.

Question 6. The present age of a father is three years more than three times the age of his son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.

Solution: 

Let the age of father and his son are 'a' and 'b' respectively.

Given, Present age of father is three years more than three times his son's age.

⇒ a = 3b + 3         ......(1)

Also, Three years hence, father's age will be 10 years more than twice his son's age.

⇒ a + 3 = 2 (b + 3) + 10

⇒ a = 2b + 13       ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 3b + 3 = 2b + 13

⇒ b = 10

On putting, b = 10 in eq(1), we get

a = 3(10) + 3 = 33

Hence, present age of Father is 33 years and his son's age is 10 years.

Question 7. A father is three times as old as his son. In 12 years time, he will be twice as old as his son. Find the present ages of father and the son.

Solution:

Let the present ages of father and his son be 'a' and 'b' respectively.

Given, Father is three times as old as his son.

⇒ a = 3b        ......(1)

Also, After 12 years, he will be twice as old as his son.

⇒ a + 12 = 2 (b + 12)

⇒ a = 2b + 12     ......(2)

On substituting, eq(1) in eq(2), we get,

⇒ 3b = 2b + 12

⇒ b = 12

On putting, b = 12 in eq(1), we get

a = 3(12) = 36

Hence, the present age of Father is 36 years and his son is 12 years.

Question 8. Father's age is three times the sum of ages of his two children. After 5 years, his age will be twice the sum of ages of two children. Find the age of father.

Solution: 

Let the present ages of father and sum of ages of his two children are 'a', 'b' respectively.

Given, Father's age is three times the sum of ages of his children.

⇒ a = 3b          ......(1)

Also, after 5 years, his age will be twice the sum of ages of his two children.

⇒ a + 5 = 2(b + 10)

⇒ a = 2b + 15        ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 3b = 2b + 15

⇒ b = 15

On putting b = 15 in eq(1), we get

a = 3(15) = 45

Hence, the age of Father is 45 years.

Question 9. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Solution:

Let the present ages of Father and his son are 'a' and 'b' respectively.

Given, two years before, father's age is five time the age of his son.

⇒ a - 2 = 5 (b - 2)

⇒ a = 5b - 8        ......(1)

Also, after two years, his age will be 8 more than thrice the age of the son.

⇒ a + 2 = 3 (b + 2) + 8

⇒ a = 3b + 12        ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 5b - 8 = 3b + 12

⇒ b = 10

On putting, b = 10 in eq(1), we get 

a = 3(10) + 12 = 42

Hence, the present age of Father is 42 years and his son is 10 years.

Question 10. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nurl and Sonu?

Solution:

Let the present ages of Nuri and Sonu are 'a' and 'b' respectively.

Given, five years ago, Nuri was thrice as old as Sonu.

⇒ a - 5 = 3 (b - 5)

⇒ a = 3b -10         ......(1)

Also, after ten years Nuri will be twice as old as Sonu.

⇒ a + 10 = 2 (b + 10) 

⇒ a = 2b + 10        ......(2)

On substituting eq(1) in eq(2), we get,

⇒ 3b - 10 = 2b + 10

⇒ b = 20.

On putting b = 20 in eq(1), we get 

a = 2(20) + 10 = 50

Hence, the present age of Nuri is 50 years and Sonu is 20 years.

Question 11. The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

Let the present ages of Ani, Biju, Cathy, and Dharam are 'a', 'b', 'c' and 'd' respectively.

Given, Ani and biju differ by 3 years.

⇒ a - b = 3         ......(1)

Also, Dharam is twice as old as Ani and Biju is twice as old as his Cathy.

⇒ d = 2a and b = 2c      ......(2)

Also given, the ages of Cathy and Dharam differ by 30 years.

⇒ d - c = 30         ......(3)

On substituting, d = 2a in eq(3) and a= b + 3 in eq(1), we get,

⇒ 2a - c = 3.

⇒ 2(b + 3) - b/2 = 30

⇒ 3b/2 = 24

⇒ b = 16

On substituting b in eq(1), we get

a = 3 + 16 = 19

Hence, the present age of Ani is 19 years and Biju is 16 years.

Question 12. Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Solution:

Let the present ages of Salim and his daughter are 'a' and 'b' respectively.

So, two years ago, Salim was thrice as old as his daughter 

(a - 2) = 3(b - 2)

a - 2 = 3b - 6

a - 3b + 4 = 0

a = 3b - 4        ......(1)

Also, after six years, Salim will be four years older than twice his daughter's age

(a + 6) = 2(b + 6) + 4

a + 6 = 2b + 12 + 4

a + 6 = 2b + 16

a - 2b - 10 = 0 

a = 2b + 10      ......(2)

On substituting eq(1) in eq(2), we get,

3b - 4 = 2b + 10

b = 14

On putting b = 14 in eq(1), we get 

a = 3(14) - 4

a = 38

Hence, the present age of Salim is 38 years and his daughter is 14 years.

Question 13. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Solution:

Let the ages of father and his two children are 'a' and 'b' respectively.

So, a = 2b    .....(1)

So, after 20 years, his age will be equal to the sum of the ages of his children

a + 20 = b + 40

a = b + 20   .....(1)

On substituting eq(1) in eq(2), we get,

2b = b + 20

b = 20

On putting b = 20 in eq(1), we get 

a = 2(20)

a = 40

Hence, the age of father is 40 years.

Read More:

• Pair of Linear Equations in Two Variables
• Linear Equation in Two Variables

Conclusion

Understanding how to solve pairs of linear equations in the two variables is a critical skill in algebra. It provides a foundation for the more advanced mathematical concepts and real-world problem-solving. Exercise 3.9 in RD Sharma's textbook helps students practice and master these techniques ensuring they are well-prepared for the more complex problems. Mastery of these methods also prepares students for the future studies in the algebra and related fields.