The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.

Given:

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. 

 To do:

We have to find the original fraction.

Solution:

Let the numerator and denominator of the original fraction be $x$ and $y$ respectively.

The original fraction$=\frac{x}{y}$

According to the question,

$x+y=2y-3$

$x+y-2y+3=0$

$x-y+3=0$

$y=x+3$.....(i)

When the numerator and denominator are decreased by 1, the numerator becomes half the denominator.

The new numerator $=x-1$

The new denominator $=y-1$

$\Rightarrow x-1=\frac{1}{2}\times(y-1)$

$2(x-1)=1(y-1)$     (On cross multiplication)

$2x-2=y-1$

$2x-y-2+1=0$

$2x-y-1=0$

$2x-(x+3)-1=0$    (From (i))

$2x-x-3-1=0$

$x-4=0$

$x=4$

$\Rightarrow y=x+3=4+3=7$.

Therefore, the original fraction is $\frac{4}{7}$.   

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Updated on: 10-Oct-2022

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