Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.8

In this section, we explore Chapter 3 of the Class 10 RD Sharma textbook, which focuses on Pairs of Linear Equations in Two Variables. Exercise 3.8 is designed to help students understand the methods for solving pairs of linear equations, enhancing their problem-solving skills in algebra.

Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.8

This section provides comprehensive solutions for Exercise 3.8 from Chapter 3 of the Class 10 RD Sharma textbook. These solutions are intended to assist students in mastering the techniques for solving linear equations in two variables, ensuring a strong foundation in algebra.

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

Solution:

Let's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x = y – 4,

x – y = − 4

and y + 1 = 8(x - 2)

y + 1 = 8x – 16

8x – y = 1 + 16

8x – y = 17

Therefore, we have two equations

x – y = -4     ----------------(i)

8x – y = 17         ------------------(ii)

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 = 3

Substituting the value of x in the (i) eqn, we have

3 – y = – 4

y = 3 + 4 = 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x+2 / y+2 = 9/11 

11(x+2) = 9(y+2)

11x + 22 = 9y + 18 

11x – 9y = 18 – 22  

11x – 9y + 4 = 0        ----------------(i)

and x+3 / y+3 = 5/6

6(x + 3) = 5(y + 3)

6x + 18 = 5y + 15

6x – 5y = 15 –18

6x – 5y + 3 = 0             ----------------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we have

x / (-9 x 3 -(-5) x 4) = -y / (11 x 3 - 6 x 4) = 1 / (11 x (-5) - 6 x (-9))

x / 7 = y / 9 = 1

x = 7 and y = 9

Hence, 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y. 

According to given condition's,

x-1 / y-1 = 1/3

3(x – 1) = (y – 1)

3x – 3 = y – 1

3x – y – 2 = 0           -----------------(i)

and x+1 / y+1 = 1/2

(2x + 1) = (y + 1) ⇒ 2x + 2 = y + 1

2x – y + 1 = 0           ------------------(ii)

We have to solve the above equations for x and y,

By using cross-multiplication, we got

x / -1-2 = -y / 3+4 = 1 / -3+2

x / -3 = -y / 7 = 1 / -1

x = 3 and y = 7

Hence, 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y.

According to given condition's,

x+1 / y-1 = 1

(x + 1) = (y – 1)

x + 1– y + 1 = 0

x – y + 2 = 0               --------------(i)

and x / y+1 = 1/2

2x = (y + 1)

2x – y – 1 = 0                     -------------(ii)

We have to solve the above equations for x and y,

By using cross-multiplication, we got

x / 1+2 = -y / -1-4 = 1 / -1+2

x/3 = y/5 = 1

x = 3 and y = 5

Hence, 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x + y = 12

x + y – 12 = 0              ---------------(i)

and x / y+3 = 1/2

2x = (y + 3)

2x – y – 3 = 0               ----------------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we got

x / (-3-12) = -y / (-3+24) = 1 / (-1-2)

x/15 = y/21 = 1/3

x = 5 and y = 7

Hence the fraction is 5/7.

Question 6. When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes 14. And, when 6 is added to numerator and the denominator is multiplied by 3, it becomes 23. Find the fraction.

Solution: 

Let's assume that the numerator of a fraction be x and denominator be y,

According to given condition,

x-2 / y+3 = 1/4 

4x - 8 = y + 3

4x - y = 11         --------------(i)

and x+6 / 3y = 2/3 

3x + 18 = 6y

x - 2y = -6     --------------(ii)

x = 2y - 6                  (from eqn. (ii))

substitute value of x in eqn. (i)

4(2y - 6) - y = 11

8y - 24 - y = 11

y = 5

x = 2 x 5 - 6 = 4

Hence, x / y = 4 / 5

Question 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.

Solution: 

Let's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x + y = 18

x + y – 18 = 0                ---------------(i)

and x / y+2 =1/3

3x = (y + 2)

3x – y – 2 = 0

3x – y – 2 = 0                 ----------------(ii)

We have to solve the above equations for x and y,

By using cross-multiplication, we got

x / (-2-18) = -y / (-2+54) = 1 / (-1-3)

x/-20 = -y/52 = 1/-4

x = 5 and y = 13

Hence, the fraction is 5/13

Question 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.

Solution: 

Let's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

2(x + 2) = y  

2x + 4 = y

2x – y + 4 = 0          -------------(i)

and x / y-1 = 1/3

3x = (y – 1)

3x – y + 1 = 0          ----------------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we got

x / (-1+4) = -y / (2-12) = 1 / (-2+3)

x / 3 = y / 10 = 1

x = 3 and y = 10

Hence, the fraction is 3/10.

Question 9. 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x + y = 2x + 4

2x + 4 – x – y = 0

x – y + 4 = 0              ----------------(i)

and x + 3 : y + 3 = 2 : 3

3(x + 3) = 2(y + 3)

3x + 9 = 2y + 6  

3x – 2y + 3 = 0                --------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we got

x / (-120+60) = y / (200-75) = 1 / (-20+15)

x / 60 = y / 125 = 1 / 5

x = 5 and y = 9

Hence, the fraction is 5/9.

Question 10. 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's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

2x / y-5 = 6/5

10x = 6(y – 5)

10x – 6y + 30 = 0

2(5x - 3y + 15) = 0

5x - 3y + 15 = 0      --------------(i)

and x+8 / 2y = 2/5

5(x + 8) = 4y  

5x + 40 = 4y

5x – 4y + 40 = 0          -----------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we got

x / (-120+60) = -y / (200-75) = 1 / (-20+15)

x / 60 = y / 125 = 1 / 5

x = 12 and y = 25

Hence, the fraction is 12/25.

Question 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

Solution:

Let's assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y

According to given condition's,

x-1 = 1/2 x (y-1)

x-1 / y-1 = 1/2

x + y = 2y – 3

x + y – 2y + 3 = 0

x – y + 3 = 0       ------------(i)

and 2(x - 1) = (y – 1)

2x – 2 = (y – 1)

2x – y – 1 = 0         ---------------(ii)

We have to solve the above equations for x and y.

By using cross-multiplication, we got

x / (1+3) = -y / (-1-6) = 1 / (-1+2)

x / 4 = y / 7 = 11

x = 4 and y = 7

Hence, the fraction is 4/7.

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• Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7
• Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.9