Chapter 3 - Pair Of Linear Equations In Two Variables, RD Sharma Solutions - (Part-21) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 3.89

Q.7. The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction.
Ans. Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and the denominator of the fraction is 18. Thus, we have
x + y = 18
⇒ x + y - 18 = 0
If the denominator is increased by 2, the fraction reduces to 1/3. Thus, we have

⇒ 3x = y + 2
⇒ 3x - y - 2 = 0
So, we have two equations
x + y - 18 = 0
3x - y - 2 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

Hence, the fraction is 5/13.

Q.8. If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3. Find the fraction.
Ans. Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y If 2 is added to the numerator of the fraction, it reduces to 1/2. Thus, we have

⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x - y + 4 = 0
If 1 is subtracted from the denominator, the fraction reduces to 1/3. Thus, we have

⇒ 3x = y - 1
⇒ 3x - y + 1 = 0
So, we have two equations
2x - y + 4 = 0
3x - y + 1 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

⇒ x = 3, y = 10
Hence, the fraction is 3/10

Q.9. A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.
Ans. Let the numerator of the fraction be x and the denominator be y.
According to the question,

and
Solving equation (i), we get
3(x − 2) = y
⇒3x − 6 = y      ...(iii)
Substituting the value of y in equation (ii), we get

⇒ 2 (x) = 3x − 6 − 1
⇒ 2x = 3x − 7
⇒ 2x − 3x = −7
⇒ −x = −7
⇒ x = 7      ...(iv)
From (iii) and (iv), we get
x = 7 and y = 15
Hence, the fraction is 7/15.

Q.10. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
Ans. Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have
x + y = 2x = 4
⇒ 2x + 4 - x - y = 0
⇒ x - y + 4 = 0
If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have
x + 3 : y + 3 = 2 : 3
⇒
⇒ 3(x + 3) = 2(y + 3)
⇒ 3x + 9 = 2y + 6
⇒ 3x - 2y + 3 = 0
So, we have two equations
x - y + 4 = 0
3x - 2y + 3 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

⇒ x = 5, y = 9
Hence, the fraction is 5/9.

Q.11. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Ans. Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. Thus, we have
x + y = 2y - 3
⇒ x + y - 2y + 3 = 0
⇒ x - y + 3 = 0
If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

⇒ 2(x - 1) = y - 1
⇒ 2x - 2 = y - 1
⇒ 2x - y - 1 = 0
So, we have two equations
x - y + 3 = 0
2x - y - 5 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

⇒ x = 4, y = 7
Hence, the fraction is 4/7.

Q.12. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Ans. Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y
The sum of the numerator and denominator of the fraction is 12. Thus, we have
x + y = 12
⇒ x + y - 12 = 0
If the denominator is increased by 3, the fraction becomes 1/2. Thus, we have

⇒ 2x = y + 3
⇒ 2x - y - 3 = 0
So, we have two equations
x + y - 12 = 0
2x - y - 3 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

⇒ x = 5, y = 7
Hence, the fraction is 5/7.

Page No 3.92

Q.1. A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find the their present ages.
Ans. Let the present age of father be x years and the present age of son be y years.
Father is three times as old as his son. Thus, we have
x = 3y
⇒ x - 3y = 0
After 12 years, father’s age will be (x + 12) years and son’s age will be (y + 12) years. Thus using the given information, we have
x + 12 = 2(y + 12)
⇒ x +12 = 2y + 24
⇒ x - 2y - 12 = 0
So, we have two equations
x - 3y = 0
x - 2y - 12 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

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

Q.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?
Ans. Let the present age of A be x years and the present age of B be y years.
After 10 years, A’s age will be (x + 10) years and B’s age will be (y + 10) years. Thus using the given information, we have
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y - 10 = 0
Before 5 years, the age of A was (x - 5) years and the age of B was (y - 5) years Thus using the given information, we have
x - 5 = 3(y - 5)
⇒ x - 5 = 3y - 15
⇒ x - 3y + 10 = 0
So, we have two equations
x - 2y - 10 = 0
x - 3y + 10 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

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

Q.3. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Ans. Let the present age of Nuri be x years and the present age of Sonu be y years.
After 10 years, Nuri’s age will be(x + 10) years and the age of Sonu will be(y + 10) years. Thus using the given information, we have
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y - 10 = 0
Before 5 years, the age of Nuri was(x – 5)years and the age of Sonu was(y – 5)years. Thus using the given information, we have
x - 5 = 3(y - 5)
⇒ x - 5 = 3y - 15
⇒ x - 3y + 10 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

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

Q.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.
Ans. Let the present age of the man be x years and the present age of his son be y years.
After 6 years, the man’s age will be (x + 6) years and son’s age will be (y + 6) years. Thus using the given information, we have
x + 6 = 3(y + 6) 
⇒ x + 6 = 3y + 18
⇒ x - 3y - 12 = 0
Before 3 years, the age of the man was (x - 3) years and the age of son’s was (y - 3) years. Thus using the given information, we have
x - 3 = 9(y - 3)
⇒ x - 3 = 9y - 27
⇒ x - 9y + 24 = 0
So, we have two equations
x - 3y - 12 = 0
x - 9y + 24 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

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

Q.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.
Ans. Let the present age of father be x years and the present age of his son be y years.
After 10 years, father’s age will be (x + 10) years and son’s age will be (y + 10) years. Thus using the given information, we have
x + 10 = 2(y + 10)
⇒  x + 10 = 2y + 20
⇒ x - 2y - 10 = 0
Before 10 years, the age of father was (x - 10) years and the age of son was (y - 10) years. Thus using the given information, we have
x - 10 = 12(y -10)
⇒ x - 10 = 12y - 120
⇒ x - 12y + 110 = 0
So, we have two equations
x - 2y - 10 = 0
x - 12y + 110 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

Hence, the present age of father is 34 years and the present age of son is 12 years.

Q.6. The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.
Ans. Let the present age of father be x years and the present age of his son be y years.
The present age of father is three years more than three times the age of the son. Thus, we have
x = 3y + 3
⇒ x - 3y - 3 = 0
After 3 years, father’s age will be (x + 3) years and son’s age will be (y + 3) years.
Thus using the given information, we have
x + 3 = 2(y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x - 2y - 13 = 0
So, we have two equations
x - 3y - 3 = 0
x - 2y - 13 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

Hence, the present age of father is 33 years and the present age of son is 10 years.

Q.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.
Ans. Let the present age of father be x years and the present age of his son be y years.
The present age of father is three times the age of the son. Thus, we have
x = 3y
⇒ x - 3y = 0
After 12 years, father’s age will be (x + 12) years and son’s age will be (y + 12) years. Thus using the given information, we have
x + 12 = 2(y + 12)
⇒ x + 12 = 2y + 24
⇒ x - 2y - 12 = 0
So, we have two equations
x - 3y = 0
x - 2y - 12 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have

Hence, the present age of father is 36 years and the present age of son is12 years.