The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 115. The fraction is ____

# The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. The fraction is ____

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### Solution:

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. The fraction is $\frac{2}{5}$.
Given that the numerator of a fraction is 3 less than its denominator.
Let’s assume that the numerator and denominator are $n$ and $d$ respectively.
$n=d-3$            Then,
$\frac{n}{d}=\frac{d-3}{d}$                       … (1)
After adding 1 to the denominator the fraction decreased by $\frac{1}{15}$
$⟹\frac{d-3}{d+1}=\frac{1}{15}$                    … (2)
To find the fraction we will equate equation (1) and equation (2),

$⟹\frac{\left(d-3\right)\left(d+1\right)-d\left(d-3\right)}{d\left(d+1\right)}=\frac{1}{15}$
$⟹\frac{d\left(d+1\right)-3\left(d+1\right)-{d}^{2}+3d}{{d}^{2}+d}=\frac{1}{15}$
$⟹\frac{{d}^{2}+d-3d-3-{d}^{2}+3d}{{d}^{2}+d}=\frac{1}{15}$
$⟹\frac{d-3}{{d}^{2}+d}=\frac{1}{15}$
$⟹15\left(d-3\right)={d}^{2}+d$
$⟹15d-45={d}^{2}+d$
$⟹{d}^{2}-15d+45=0$
$⟹{d}^{2}-9d-5d+45=0$
$⟹d\left(d-9\right)-5\left(d-9\right)=0$
$⟹\left(d-5\right)\left(d-9\right)=0$

If $d=5$ then,
$⟹n=d-3=5-3=2$ Then the fraction is $\frac{2}{5}$.
If $d=9$ then,
$⟹n=d-3=9-3=6$ Then the fraction is $\frac{6}{9}=\frac{2}{3}$
∴ We will neglect this fraction because it is not satisfying the given condition.
Hence, the original fraction is $\frac{2}{5}$.

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