Question

The numerator of a fraction is 3 less than its denominator. If the numerator is increased by 1 and the denominator is increased by 3, the number obtained is 1/4. Find the original number

Answer 1

$n=d-3 \\ \\ \frac{n+1}{d+3} = \frac{1}{4} \\ \\ \frac{(d-3)+1}{d+3} = \frac{1}{4} \\ \\ 4[d-2]=[d+3] \\ \\ 3d=11 \\ \\ d=[11/3] \\ \\ n=[2/3]$

Answer 2

heya...

Here is your answer...

n=d−3d+3n+1​=41​d+3(d−3)+1​=41​4[d−2]=[d+3]3d=11d=[11/3]n=[2/3]​

It may help you...☺☺