Question

The numerator of a fraction is 3 less than its denominator. If 1 is added to both, it becomes 5/8. Find the original fraction

Answer 1

Given:

• The numerator of a fraction is 3 less than its denominator. If 1 is added to both, it becomes 5/8.

To Find:

• Find the original fraction?

Solution:

Let the denominator be x.

Numerator be (x-3).

So, Fraction is ${\sf{ \dfrac{x-3}{x} }}$ .

Again, adding 1 to both;

${\sf\red{ = \dfrac{x-3+1}{x+1} }}$

After putting the Values in the Equation;

$\colon\implies{\sf\red{ \dfrac{x-3+1}{x+1} = \dfrac{5}{8} }} \\ \\ \colon\implies{\sf{ \dfrac{x-2}{x+1} = \dfrac{5}{8} }} \\ \\ \colon\implies{\sf{ 8(x-2) = 5(x+1) }} \\ \\ \colon\implies{\sf{ 8x - 16 = 5x+5 }} \\ \\ \colon\implies{\sf{ 8x - 5x = 5 + 16 }} \\ \\ \colon\implies{\sf{ 3x = 21 }} \\ \\ \colon\implies{\sf{ x = \cancel{ \dfrac{21}{3} } }} \\ \\ \colon\implies{\sf\red{ x = 7 }}$

So, the Fraction is ${\sf\green{ \dfrac{7-3}{7} = \dfrac{4}{7} }}$

Hence,

The Original Fraction is ${\sf\bold{ \dfrac{4}{7} }}$.