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.

Given:

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.

 To do:

We have to find the original fraction.

Solution:

Let the denominator of the original fraction be $x$.

This implies,

The numerator of the original fraction$=x-4$.

The original fraction$=\frac{x-4}{x}$.

The numerator is decreased by 2 and denominator is increased by 1.

This implies,

New fraction$=\frac{x-4-2}{x+1}=\frac{x-6}{x+1}$

According to the question,

The denominator of the new fraction is eight times the numerator.

$\Rightarrow x+1=8(x-6)$

$x+1=8x-48$

$8x-x=1+48$

$7x=49$

$x=\frac{49}{7}$

$x=7$

$\Rightarrow x-4=7-4=3$.

Therefore, the original fraction is $\frac{3}{7}$. 

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Updated on: 10-Oct-2022

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