Question

11. The denominator of a fraction is 4 less than the numerator. If the sum of both numerator and denominator is 20, find the fraction.

half of the first

port

Answer 1

Answer:

The required fraction is 12/8.

Step-by-step explanation:

Given :

The denominator of a fraction is 4 less than the numerator. If the sum of both numerator and denominator is 20.

To find :

the fraction

Solution :

Let the numerator of the fraction be x and denominator be y.

According to the question,

y = x – 4

The sum of numerator and denominator is 20.

x + y = 20

x + x – 4 = 20

2x = 20 + 4

2x = 24

x = 24/2

x = 12

Value of y :

y = x – 4

y = 12 – 4

y = 8

The numerator is 12 and denominator is 8.

Therefore, the fraction is 12/8.

Answer 2

$\\ \textbf{ Let numerator \& denominator be x and y } \\ \\ \sf \: required \: fraction = \frac{x}{y} \\ \\ \underline{ {\pmb{ \textsf{ \maltese \: \: According to first condition} }}} \\ \\ \dashrightarrow \tt \: y = x - 4 \\ \\ \dashrightarrow \tt \: x - y = 4 \dots \dots {eq}^{n}( 1) \\ \\ \underline{ {\pmb{ \textsf{ \maltese \: \: According to second \: condition} }}} \\ \\\dashrightarrow \tt \: x + y = 20 \dots \dots {eq}^{n} (2) \\ \\ \\ \underline{ \pmb{ \sf{Adding \: \: eq{^n } \: \: (1) \: \: \& \: \: (2)}}} \\ \\ \\ \bf \: x - \cancel y = 4 \\ \underline{ \bf \: x + \cancel{y} = 20 }\\ \bf \: 2x = 24 \\ \bf \: x = \frac{ \cancel{24}}{ \cancel2} \\ \\ \underline {\boxed{ \bf \: x = 12}}\\ \\ \\ \\ \underline{ \pmb{ \sf{Substitute \: x = 12 \: \: in \: \: eq^{n} (1)}}} \\ \\ \dashrightarrow \tt \: x - y = 4 \\ \\ \dashrightarrow \tt12 - y = 4 \\ \\ \dashrightarrow \tt - y = 4 - 12 \\ \\ \dashrightarrow \tt \cancel- y = \cancel - 8 \\ \\ \dashrightarrow \tt \underline{ \boxed{ \sf{y = 8}}} \bigstar \\ \\ \\ \underline{ \boxed{ \textsf{The required fraction is} \: \sf\dfrac{x}{y} = \frac{12}{8} }}$