# RD Sharma Class 10 Ex 3.8 Solutions Chapter 3 Pair of Linear Equations in Two Variables

In this chapter, we provide RD Sharma Class 10 Ex 3.8 Solutions Chapter 3 Pair of Linear Equations in Two Variables for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 3.8 Solutions Chapter 3 Pair of Linear Equations in Two Variables pdf, Now you will get step by step solution to each question.

# Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.8

### Question: 1

The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

### Solution:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y

The numerator of the fraction is 4 less the denominator. Thus, we have

x = y – 4

⇒ x – y = − 4

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have

y + 1 = 8(x – 2)

⇒ y + 1 = 8x – 16

⇒ 8x – y = 1 + 16

⇒ 8x – y = 17

So, we have two equations

x – y = – 4

8x – y = 17

Here x and y are unknowns. We have to solve the above equations for x and y. Subtracting the second equation from the first equation, we get

(x – y) – (8x – y) = – 4 – 17

⇒ x − y − 8x + y = −21

⇒ −7x = −21

⇒ −7x = −21

⇒ x = 21/7

⇒ x = 3

Substituting the value of x in the first equation, we have

3 – y = – 4

⇒ y = 3 + 4

⇒ y = 7

Hence the fraction is 3/7

### Question: 2

A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction

### Solution:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y

If 2 is added to both numerator and the denominator, the fraction becomes 9/11. Thus, we have

⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = 18 – 22 ⇒ 11x – 9y + 4 = 0

If 3 is added to both numerator and the denominator, the fraction becomes 5/6

⇒ 6(x + 3) = 5(y + 3)

⇒ 6x + 18 = 5y + 15

⇒ 6x – 5y = 15 –18

⇒ 6x – 5y + 3 = 0

So, we have two equations

11x – 9y + 4 = 0

6x – 5y + 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 = 7, y = 9

The fraction is 7/9

### Question: 3

A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. If 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction.

### Solution:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y. If 1 is subtracted from both numerator and the denominator, the fraction becomes 1/3. Thus, we have

⇒3(x – 1) = (y – 1)

⇒ 3x – 3 = y – 1

⇒ 3x – y – 2 = 0

If 1 is added to both numerator and the denominator, the fraction becomes 1/2. Thus, we have

⇒ (2x + 1) = (y + 1) ⇒ 2x + 2 = y + 1

⇒ 2x – y + 1 = 0

So, we have two equations

3x – y – 2 = 0

2x – 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 = 7

The fraction is 3/7

### Question: 4

If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?

### Solution:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y.

If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes 1. Thus, we have

⇒(x + 1) = (y–1)

⇒ x + 1– y + 1 = 0

⇒ x – y + 2 = 0

If 1 is added to the denominator, the fraction becomes 1/2. Thus, we have

⇒ 2x = (y + 1)

⇒ 2x – y – 1 = 0

So, we have two equations

x – y + 2 = 0

2x – 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 = 5

The fraction is 3/5

### Question: 5

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.

### Solution:

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

The fraction is 5/7

### Question: 6

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.

### Solution:

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 18. Thus, we have

x + y = 18

⇒ x + y – 18 = 0

If the denominator is increased by 2, the fraction becomes 1/3. Thus, we have

⇒ 3x = (y + 2)

⇒ 3x – y – 2 = 0

⇒ 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

⇒ x = 5, y = 13

The fraction is 5/13

### Question: 7

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.

### Solution:

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

The fraction is 3/10

### Question: 8

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.

### Solution:

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

The fraction is 5/9

### Question: 9

If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes 6/5. And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes 2/5. Find the fraction.

### Solution:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is x/y. If the numerator is multiplied by 2 and denominator is reduced by 5, the fraction becomes 6/5. Thus, we have

⇒ 10x = 6(y – 5)

⇒ 10x – 6y + 30 = 0

⇒ 2(5x – 3y + 15) = 0

⇒ 5x – 3y + 15 = 0

If the denominator is doubled and the numerator are increased by 8, the fraction becomes 2/5. Thus we have

⇒ 5(x + 8) = 4y

⇒ 5x + 40 = 4y

5x – 4y + 40 = 0

So, we have two equations

5x – 3y + 15 = 0

5x – 4y + 40 = 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 = 12, y = 25

The fraction is 12/25

### Question: 10

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

### Solution:

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 a 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 – 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 = 4, y = 7

The fraction is 4/7

All Chapter RD Sharma Solutions For Class10 Maths

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.