Question

Numerator of a fraction is 3 less than the denominator. If 4 is added to both numerator and denominator, the value of the fraction increases by 1/8. Find the fraction?? Solve using quadratic

Answer 1

ANSWER :–

$\\ \to \: \: \: { \boxed { \bold{ Original \: \: fraction = \dfrac{5}{8} }}}$

EXPLANATION :–

GIVEN :–

▪︎ Numerator of a fraction is 3 less than the denominator.

▪︎ When 4 is added to both numerator and denominator, the value of the fraction increases by 1/8.

TO FIND :–

• Original fraction = ?

SOLUTION :–

▪︎ According to the question –

• Numerator of a fraction is 3 less than the denominator.

• Let the Denominator be 'x'.

• So the fraction be $\implies { \bold{ \dfrac{x - 3}{x} }}$

▪︎ When 4 is added to both numerator and denominator, the value of the fraction increases by 1/8.

$\\ \implies { \bold{ \frac{(x - 3) + 4}{x + 4} = \frac{x - 3}{x} + \frac{1}{8} }}$

$\\ \implies{ \bold{ \dfrac{x + 1}{x + 4} = \dfrac{8(x - 3) + x}{8x}}}$

$\\ \implies{ \bold{ \dfrac{x + 1}{x + 4} = \dfrac{9x - 24}{8x}}}$

$\\ \implies{ \bold{(x + 1)(8x) = (x + 4)(9x - 24)}}$

$\\ \implies{ \bold{8 {x}^{2} +8x= 9 {x}^{2} - 24x + 36x - 96 }}$

$\\ \implies{ \bold{ - {x}^{2} - 4x + 96= 0 }}$ $\\ \implies{ \bold{ {x}^{2} + 4x - 96= 0 }}$

$\\ \implies{ \bold{ {x}^{2} + 12x - 8x - 96= 0 }}$

$\\ \implies{ \bold{ x( x+ 12) - 8(x + 12)= 0 }}$

$\\ \implies{ \bold{ (x - 8)( x+ 12) = 0 }}$

$\\ \implies{ \bold{ x = 8 ,x = - 12}}$

(1) When (x = 8) :–

$\\ \implies \: \: \: { \boxed { \bold{ Original \: \: fraction = \dfrac{5}{8} }}}$

(2) When (x = -12) :–

$\\ \implies \: \: \: { \boxed { \bold{ Original \: \: fraction = \dfrac{5}{4} (Not \: \: satisfy \: \: the \: \: equation)}}}$

Hence, $\: \: \: \therefore \: \: \: { \boxed { \bold{ Original \: \: fraction = \dfrac{5}{8} }}}$