The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.

Given:

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. 

 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$=x-3$.

The original fraction$=\frac{x-3}{x}$

New fraction$=\frac{x-3+2}{x+2}=\frac{x-1}{x+2}$

According to the question,

$\frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{20}$

$\frac{(x-3)(x+2)+(x-1)x}{x(x+2)}=\frac{29}{20}$

$\frac{x^2-3x+2x-6+x^2-x}{x^2+2x}=\frac{29}{20}$

$\frac{2x^2-2x-6}{x^2+2x}=\frac{29}{20}$

$20(2x^2-2x-6)=29(x^2+2x)$    (O cross multiplication)

$40x^2-40x-120=29x^2+58x$

$(40-29)x^2-(40+58)x-120=0$

$11x^2-98x-120=0$

Solving for $x$ by factorization method, we get,

$11x^2-110x+12x-120=0$

$11x(x-10)+12(x-10)=0$

$(11x+12)(x-10)=0$

$11x+12=0$ or $x-10=0$

$11x=-12$ or $x=10$

$x=\frac{-12}{11}$ or $x=10$

The value of $x$ cannot be a fraction. Therefore, the value of $x$ is $10$.

$x-3=10-3=7$

The original fraction is $\frac{7}{10}$.

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Updated on: 10-Oct-2022

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