# RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.8

## RD Sharma Solutions Class 10 Chapter 3 Exercise 3.8

### RD Sharma Class 10 Solutions Chapter 3 Ex 3.8 PDF Free Download

#### Exercise 3.8

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.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

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

x = y – 4

$\Rightarrow x – y= -4$

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)

$\Rightarrow y + 1 = 8x – 16$

$\Rightarrow 8x – y = 1 + 16$

$\Rightarrow 8x – y = 17$

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

$\Rightarrow x- y- 8x + y = -21$

$\Rightarrow -7x = -21$

$\Rightarrow x =\frac{21}{7}$

$\Rightarrow\frac{21}{7}$ x = 3

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

3 – y = – 4

$\Rightarrow y = 3+4$

$\Rightarrow y = 7$

Hence the fraction is $\frac{3}{7}$

Q.2) A fraction becomes $\frac{9}{11}$ if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

If 2 is added to both numerator and the denominator, the fraction becomes $\frac{9}{11}$ . Thus, we have

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x + 2) = 9(y + 2)$

$\Rightarrow 11x + 22 = 9y +18$

$\Rightarrow 11x – 9y = 18 – 22$

$\Rightarrow 11x – 9y + 4 = 0$

If 3 is added to both numerator and the denominator, the fraction becomes $\frac{5}{6}$

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x + 3) = 5(y + 3)$

$\Rightarrow 6x + 18 = 5y + 15$

$\Rightarrow 6x – 5y = 15 – 18$

$\Rightarrow 6x – 5y + 3 = 0$

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

$\frac{x}{-9*3-\left ( -5 \right )*4}$ = $\frac{-y}{11*3-6*4}$ = $\frac{1}{11*\left(-5 \right )-6*\left (-9 \right )}$

$\Rightarrow \frac{x}{-27+20}$ = $\Rightarrow \frac{-y}{33-24}$ = $\frac{1}{-55+54}$

$\Rightarrow \frac{x}{-7}$ = $\frac{-y}{9}$ = $\frac{1}{-1}$

$\Rightarrow \frac{x}{7}=\frac{y}{9}=1$

x = 7, y =9

The fraction is $\frac{7}{9}$

Q.3) A fraction becomes $\frac{1}{3}$ if 1 is subtracted from both its numerator and denominator. If 1 is added to both the numerator and denominator, it becomes $\frac{1}{2}$. Find the fraction.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

If 1 is subtracted from both numerator and the denominator, the fraction becomes $\frac{1}{3}$ . Thus, we have

$\frac{x-1}{y-1}=\frac{1}{3}$

$\Rightarrow 3(x – 1) = (y – 1)$

$\Rightarrow 3x – 3 = y – 1$

$\Rightarrow 3x – y – 2 = 0$

If 1 is added to both numerator and the denominator, the fraction becomes $\frac{1}{2}$. Thus, we have

$\frac{x+1}{y+1}=\frac{1}{2}$

$\Rightarrow 2(x + 1) = (y + 1)$

$\Rightarrow 2x + 2 = y + 1$

$\Rightarrow 2x – y + 1 = 0$

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

$\frac{x}{\left ( -1 \right )*1-\left ( -1 \right )*\left ( -2 \right )}$= $\frac{-y}{3*1-2*\left ( -2 \right )}$= $\frac{1}{3*\left ( -1 \right )-2*\left ( -1 \right )}$

$\Rightarrow \frac{x}{-1-2}$ = $\Rightarrow \frac{-y}{3+4}$ = $\frac{1}{-3+2}$

$\Rightarrow \frac{x}{-3}$ = $\frac{-y}{7}$ = $\frac{1}{-1}$

$\Rightarrow \frac{x}{3}=\frac{y}{7}=1$

$\Rightarrow x = 3, y = 7$

The fraction is $\frac{3}{7}$

Q.4) If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

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

$\frac{x+1}{y-1}= 1$

$\Rightarrow (x + 1) = (y – 1)$

$\Rightarrow x + 1 – y + 1 = 0$

$\Rightarrow x – y + 2 = 0$

If 1 is added to the denominator, the fraction becomes $\frac{1}{2}$. Thus, we have

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x = (y + 1)$

$\Rightarrow 2x – y – 1 = 0$

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

$\frac{x}{\left ( -1 \right )*\left ( -1 \right )-\left ( -1 \right )*2}$ = $\frac{-y}{1*\left ( -1 \right )-2*2}$ = $\frac{1}{1*\left ( -1 \right )-2*\left ( -1 \right )}$

$\Rightarrow \frac{x}{1+2}$ = $\Rightarrow \frac{-y}{-1-4}$ = $\frac{1}{-1+2}$

$\Rightarrow \frac{x}{3}$ = $\frac{-y}{-5}$ = $\frac{1}{1}$

$\Rightarrow \frac{x}{3}=\frac{y}{5}=1$

$\Rightarrow x = 3, y = 5$

The fraction is $\frac{3}{5}$

5) The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes $\frac{1}{2}$. Find the fraction.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

The sum of the numerator and denominator of the fraction is 12. Thus, we have

x + y = 12

$\Rightarrow x + y – 12 = 0$

If the denominator is increased by 3, the fraction becomes $\frac{1}{2}$. Thus, we have

$\frac{x}{y+3}=\frac{1}{2}$

$\Rightarrow 2x = (y + 3)$

$\Rightarrow 2x – y – 3 = 0$

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

$\frac{x}{\left ( 1 \right )*\left ( -3 \right )-\left ( -1 \right )*-12}$ = $\frac{-y}{1*\left ( -3 \right )-2*-12}$ = $\frac{1}{1*\left ( -1 \right )-2*\left ( 1 \right )}$

$\Rightarrow \frac{x}{-3-12}$ = $\Rightarrow \frac{-y}{-3+24}$ = $\frac{1}{-1-2}$

$\Rightarrow \frac{x}{-15}$ = $\frac{-y}{21}$ = $\frac{1}{-3}$

$\Rightarrow \frac{x}{15}=\frac{y}{21}=\frac{1}{3}$

$\Rightarrow x = \frac{15}{3}, y =\frac{21}{3}$

$\Rightarrow x = 5, y =7$

The fraction is $\frac{5}{7}$

7) The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to $\frac{1}{3}$. Find the fraction.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

The sum of the numerator and denominator of the fraction is 18. Thus, we have

x + y = 18

$\Rightarrow x + y – 18 = 0$

If the denominator is increased by 2, the fraction becomes $\frac{1}{3}$. Thus, we have

$\frac{x}{y+2}=\frac{1}{3}$

$\Rightarrow 3x = (y + 2)$

$\Rightarrow 3x – y – 2 = 0$

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

$\frac{x}{\left ( 1 \right )*\left ( -2 \right )-\left ( -1 \right )*-18}$ = $\frac{-y}{1*\left ( -2 \right )-3*-18}$ = $\frac{1}{1*\left ( -1 \right )-3*\left ( 1 \right )}$

$\Rightarrow \frac{x}{-2-18}$ = $\Rightarrow \frac{-y}{-2+54}$ = $\frac{1}{-1-3}$

$\Rightarrow \frac{x}{-20}$ = $\frac{-y}{52}$ = $\frac{1}{-4}$

$\Rightarrow \frac{x}{20}=\frac{y}{52}=\frac{1}{4}$

$\Rightarrow x = \frac{20}{4}, y =\frac{52}{4}$

$\Rightarrow x = 5, y =13$

The fraction is $\frac{5}{13}$

8) If 2 is added to the numerator of a fraction, it reduces to $\frac{1}{2}$ and if 1 is subtracted from the denominator, it reduces to $\frac{1}{3}$. Find the fraction.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

If 2 is added to the numerator of the fraction, it reduces to $\frac{1}{2}$. Thus we have

$\frac{x+2}{y}$ =  $\frac{1}{2}$

$\Rightarrow 2(x + 2) = y$

$\Rightarrow 2x + 4 = y$

$\Rightarrow 2x – y + 4 = 0$

If 1 is subtracted from the denominator, the fraction reduces to $\frac{1}{3}$. Thus, we have

$\frac{x}{y-1}=\frac{1}{3}$

$\Rightarrow 3x = (y – 1)$

$\Rightarrow 3x – y + 1 = 0$

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

$\frac{x}{\left ( -1 \right )*\left ( 1 \right )-\left ( -1 \right )*4}$ = $\frac{-y}{2*\left ( 1 \right )-3*4}$ = $\frac{1}{2*\left ( -1 \right )-3*\left ( -1 \right )}$

$\Rightarrow \frac{x}{-1+4}$ = $\Rightarrow \frac{-y}{2-12}$ = $\frac{1}{-2+3}$

$\Rightarrow \frac{x}{3}$ = $\frac{-y}{-10}$ = $\frac{1}{1}$

$\Rightarrow \frac{x}{3}=\frac{y}{10}=1$

$\Rightarrow x = 3, y =10$

The fraction is $\frac{3}{10}$

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.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have

x + y = 2x + 4

$\Rightarrow 2x + 4 –x –y = 0$

$\Rightarrow x – y + 4 = 0$

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

$\Rightarrow \frac{x+3}{y+3}=\frac{2}{3}$

$\Rightarrow 3(x+3) = 2(y + 3)$

$\Rightarrow 3x +9 = 2y + 6$

$\Rightarrow 3x – 2y + 3 = 0$

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

$\frac{x}{\left ( -1 \right )*\left ( 3 \right )-\left ( -2 \right )*4}$ = $\frac{-y}{1*\left ( 3 \right )-3*4}$ = $\frac{1}{1*\left ( -2 \right )-3*\left ( -1 \right )}$

$\Rightarrow \frac{x}{-3+8}$ = $\Rightarrow \frac{-y}{3-12}$ = $\frac{1}{-2+3}$

$\Rightarrow \frac{x}{5}$ = $\frac{-y}{-9}$ = $\frac{1}{-2+3}$

$\Rightarrow \frac{x}{5}=\frac{y}{9}= 1$

$\Rightarrow x = 5, y = 9$

The fraction is $\frac{5}{9}$

10) If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes $\frac{6}{5}$. And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes $\frac{2}{5}$. Find the fraction.

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

If the numerator is multiplied by 2 and denominator is reduced by 5, the fraction becomes $\frac{6}{5}$. Thus, we have

$\Rightarrow \frac{2x}{y-5}=\frac{6}{5}$

$\Rightarrow 10x = 6(y – 5)$

$\Rightarrow 10x – 6y + 30 = 0$

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

$\Rightarrow 5x-3y+15 = 0$

If the denominator is doubled and the numerator are increased by 8, the fraction becomes $\frac{2}{5}$.. Thus we have

$\Rightarrow \frac{x+8}{2y}=\frac{2}{5}$

$\Rightarrow 5(x+8) = 4y$

$\Rightarrow 5x +40 = 4y$

$\Rightarrow 5x – 4y + 40 = 0$

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

$\frac{x}{\left ( -3 \right )*\left ( 40 \right )-\left ( -4 \right )*15}$ = $\frac{-y}{5*\left ( 40 \right )-5*15}$ = $\frac{1}{5*\left ( -4 \right )-5*\left ( -3 \right )}$

$\Rightarrow \frac{x}{-120+60}$ = $\Rightarrow \frac{-y}{200-75}$ = $\frac{1}{-20+15}$

$\Rightarrow \frac{x}{-60}$ = $\frac{-y}{125}$ = $\frac{1}{-5}$

$\Rightarrow \frac{x}{60}$ = $\frac{y}{125}$ = $\frac{1}{5}$

$x=\frac{60}{5},y=\frac{125}{5}$

$\Rightarrow x = 12, y = 25$

The fraction is $\frac{12}{25}$

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

Soln:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{x}{y}$

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. Thus, we have

x + y = 2y – 3

$\Rightarrow x + y – 2y + 3 = 0$

$\Rightarrow x – y + 3 = 0$

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

$x-1=\frac{1}{2}\left ( y-1 \right )$

$\Rightarrow \frac{x-1}{y-1}=\frac{1}{2}$

$\Rightarrow 2(x-1) = (y – 1)$

$\Rightarrow 2x – 2 = (y – 1)$

$\Rightarrow 2x – y – 1 = 0$

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

$\frac{x}{\left ( -1 \right )*\left ( -1 \right )-\left ( -1 \right )*3}$ = $\frac{-y}{1*\left ( -1 \right )-2*3}$ = $\frac{1}{1*\left ( -1 \right )-2*\left ( -1 \right )}$

$\Rightarrow \frac{x}{1+3}$ = $\Rightarrow \frac{-y}{-1-6}$ = $\frac{1}{-1+2}$

$\Rightarrow \frac{x}{4}$ = $\frac{-y}{-7}$ = $\frac{1}{1}$

$\Rightarrow \frac{x}{4}$ = $\frac{y}{7}$ = $1$

$\Rightarrow x = 4, y = 7$

The fraction is $\frac{4}{7}$

#### Practise This Question

The number 98754256 is divisible by: