Class 10 RD Sharma Solutions – 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’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.