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

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

Exercise 3.8

 

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 of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

The numerator of the fraction is 4 less the denominator.

The equation becomes,

x = y – 4

\(\Rightarrow x – y= -4\)

According to the question,

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

Thus, the equation becomes,

y+1 = 8(x-2)

\(\Rightarrow y + 1 = 8x – 16\)

\(\Rightarrow 8x – y = 1 + 16\)

\(\Rightarrow 8x – y = 17\)

The two equations are:

x – y = -4

8x – y = 17

Subtracting the equation(ii) from (i), 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 =3 in the first equation, we get

3 – y = – 4

\(\Rightarrow y = 3+4\)

\(\Rightarrow y = 7\)

Therefore, the fraction is \(\frac{3}{7}\)

 

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

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

If 2 is added to both numerator and the denominator, the fraction becomes \(\frac{9}{11}\) .

Thus, we get the equation,

\(\frac{x+2}{y+2}=\frac{9}{11}\)

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

\(\Rightarrow 11x + 22 = 9y +18\)

\(\Rightarrow 11x – 9y = 18 – 22\)

\(\Rightarrow 11x – 9y + 4 = 0\)

According to the question,

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

We get,

\(\frac{x+3}{y+3}=\frac{5}{6}\)

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

\(\Rightarrow 6x + 18 = 5y + 15\)

\(\Rightarrow 6x – 5y = 15 – 18\)

\(\Rightarrow 6x – 5y + 3 = 0\)

The two equations are,

11x – 9y +4 = 0

6x – 5y + 3 = 0

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 \(\frac{7}{9}\)

 

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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

If 1 is subtracted from both numerator and the denominator, the fraction becomes \(\frac{1}{3}\) .

We get the equation,

\(\frac{x-1}{y-1}=\frac{1}{3}\)

\(\Rightarrow 3(x – 1) = (y – 1)\)

\(\Rightarrow 3x – 3 = y – 1\)

\(\Rightarrow 3x – y – 2 = 0\)

According to the question,

We get the equation,

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

\(\frac{x+1}{y+1}=\frac{1}{2}\)

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

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

\(\Rightarrow 2x – y + 1 = 0\)

The two equations are,

3x – y – 2 = 0

2x – y + 1 = 0

Using cross-multiplication, we have

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

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

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

\(\Rightarrow \frac{x}{3}=\frac{y}{7}=1\)

\(\Rightarrow x = 3, y = 7\)

Therefore, the fraction is \(\frac{3}{7}\)

 

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?

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

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

We get the equation,

\(\frac{x+1}{y-1}= 1\)

\(\Rightarrow (x + 1) = (y – 1)\)

\(\Rightarrow x + 1 – y + 1 = 0\)

\(\Rightarrow x – y + 2 = 0\)

According to the question,

If 1 is added to the denominator, the fraction becomes \(\frac{1}{2}\).

We get the equation,

\(\frac{x}{y+1}=\frac{1}{2}\)

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

\(\Rightarrow 2x – y – 1 = 0\)

The two equations are,

x – y + 2 = 0

2x – y – 1 = 0

Using cross-multiplication, we have

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

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

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

\(\Rightarrow \frac{x}{3}=\frac{y}{5}=1\)

\(\Rightarrow x = 3, y = 5\)

Therefore, 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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

The sum of the numerator and denominator of the fraction is 12.

We get the equation,

x + y = 12

\(\Rightarrow x + y – 12 = 0\)

According to the question,

If the denominator is increased by 3, the fraction becomes \(\frac{1}{2}\).

We get the equation,

\(\frac{x}{y+3}=\frac{1}{2}\)

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

\(\Rightarrow 2x – y – 3 = 0\)

The two equations are,

x + y – 12 = 0

2x – y – 3 = 0

Using cross-multiplication, we get

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

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

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

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

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

\(\Rightarrow x = 5, y =7\)

Therefore, 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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

The sum of the numerator and denominator of the fraction is 18.

We get the equation,

x + y = 18

\(\Rightarrow x + y – 18 = 0\)

According to the question,

If the denominator is increased by 2, the fraction becomes \(\frac{1}{3}\).

We get the equation,

\(\frac{x}{y+2}=\frac{1}{3}\)

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

\(\Rightarrow 3x – y – 2 = 0\)

The two equations are:

x + y – 18 = 0

3x – y – 2 = 0

Using cross-multiplication, we have

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

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

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

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

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

\(\Rightarrow x = 5, y =13\)

Therefore, 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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

If 2 is added to the numerator of the fraction, it reduces to \(\frac{1}{2}\).

We get the equation,

\(\frac{x+2}{y}\) =  \(\frac{1}{2}\)

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

\(\Rightarrow 2x + 4 = y\)

\(\Rightarrow 2x – y + 4 = 0\)

According to the question,

If 1 is subtracted from the denominator, the fraction reduces to \(\frac{1}{3}\).

We get the equation,

\(\frac{x}{y-1}=\frac{1}{3}\)

\(\Rightarrow 3x = (y – 1)\)

\(\Rightarrow 3x – y + 1 = 0\)

The two equations are:

2x – y + 4 = 0

3x – y + 1 = 0

Using cross-multiplication, we get

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

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

 

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

 

\(\Rightarrow \frac{x}{3}=\frac{y}{10}=1\)

 

\(\Rightarrow x = 3, y =10\)

Therefore, 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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

The sum of the numerator and denominator of the fraction is 4 more than twice the numerator.

We get the equation,

x + y = 2x + 4

\(\Rightarrow 2x + 4 –x –y = 0\)

 

\(\Rightarrow x – y + 4 = 0\)

According to the question,

If the numerator and denominator are increased by 3, they are in the ratio 2:3.

We get the equation,

x + 3:y + 3 = 2:3

\(\Rightarrow \frac{x+3}{y+3}=\frac{2}{3}\)

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

\(\Rightarrow 3x +9 = 2y + 6\)

\(\Rightarrow 3x – 2y + 3 = 0\)

The two equations are:

x – y + 4 = 0

3x – 2y + 3 = 0

Using cross-multiplication, we have

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

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

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

\(\Rightarrow \frac{x}{5}=\frac{y}{9}= 1\)

\(\Rightarrow x = 5, y = 9\)

Therefore, 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.

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

If the numerator is multiplied by 2 and denominator is reduced by 5, the fraction becomes \(\frac{6}{5}\). Thus, we get the equation,

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

\(\Rightarrow 10x = 6(y – 5)\)

\(\Rightarrow 10x – 6y + 30 = 0\)

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

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

According to the question,

If the denominator is doubled and the numerator are increased by 8, the fraction becomes \(\frac{2}{5}\). Thus, we get the equation

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

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

\(\Rightarrow 5x +40 = 4y\)

\(\Rightarrow 5x – 4y + 40 = 0\)

The two equations are:

5x – 3y + 15 = 0

5x – 4y + 40 = 0

Using cross-multiplication, we get

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

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

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

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

\(x=\frac{60}{5},y=\frac{125}{5}\)

\(\Rightarrow x = 12, y = 25\)

Therefore, 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

Solution:

Let the numerator of the fraction be x

Let the denominator of the fraction be y

We get the fraction as \(\frac{x}{y}\)

According to the question,

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator.

Thus, we get the equation

x + y = 2y – 3

\(\Rightarrow x + y – 2y + 3 = 0\)

\(\Rightarrow x – y + 3 = 0\)

According to the question,

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

Thus, we get the equation,

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

\(\Rightarrow \frac{x-1}{y-1}=\frac{1}{2}\)

\(\Rightarrow 2(x-1) = (y – 1)\)

\(\Rightarrow 2x – 2 = (y – 1)\)

\(\Rightarrow 2x – y – 1 = 0\)

The two equations are:

x – y + 3 = 0

2x – y – 1 = 0

Using cross-multiplication, we get

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

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

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

\(\Rightarrow \frac{x}{4}\) = \(\frac{y}{7}\) = \(1\)

\(\Rightarrow x = 4, y = 7\)

Therefore, the fraction is \(\frac{4}{7}\)

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