Ex-3.8 Pair Of Linear Equations In Two Variables, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q.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.

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

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

⇒ x = 21/7

⇒ 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

Q.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

Sol. 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

x + 2/y + 2 = 9/11

⇒ 11(x + 2) = 9(y + 2)

⇒ 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

x + 3/y + 3 = 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  =  1/−1

⇒ x/7 = y/9 = 1

x = 7, y = 9

The fraction is 7/9

Q.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.

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

If 1 is subtracted from both numerator and the denominator, the fraction becomes 1/3 . Thus, we have

x−1/y−1 = 1/3

⇒ 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

x + 1/y + 1 = 1/2

⇒ 2(x + 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  =  1/−1

⇒ x/3 = y/7 = 1

⇒ x = 3,y = 7

The fraction is 3/7

Q.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?

Sol.

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

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 = 1

⇒ (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

x/y + 1 = 1/2

⇒ 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  =  1/1

⇒ x/3 = y/5 = 1

⇒ x = 3,y = 5

The fraction is 3/5

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.

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

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

x/y + 3 = 1/2

⇒ 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/−15  =  −y/21  =  1/−3

⇒ x/15 = y/21 = 1/3

⇒ x = 15/3,y = 21/3

⇒ x = 5,y = 7

The fraction is 5/7

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.

Sol. 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

x/y + 2 = 1/3

⇒ 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

⇒ x/−20  =  −y/52  =  1/−4

⇒ x/20 = y/52 = 1/4

⇒ x = 20/4,y = 52/4

⇒ x = 5,y = 13

The fraction is 5/13

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.

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

If 2 is added to the numerator of the fraction, it reduces to 1/2. Thus we have

x + 2/y  =  1/2

⇒ 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

x/y−1 = 1/3

⇒ 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  =  1/1

⇒ x/3 = y/10 = 1

⇒ x = 3,y = 10

The fraction is 3/10

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.

Sol. 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

⇒ 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 = 1

⇒ x = 5,y = 9

The fraction is 5/9

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.

Sol. 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 25.. 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/−60  =  −y/125  =  1/−5

⇒ x/60  =  y/125  =  1/5

x = 60/5,y = 125/5

⇒ x = 12,y = 25

The fraction is 12/25

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

Sol. 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

x−1 = 1/2(y−1)

⇒ x−1/y−1 = 1/2

⇒ 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  =  1/1

⇒ x/4  =  y/7  =  1

⇒ x = 4,y = 7

The fraction is 4/7