Question

Find two numbers which differ by $4$, and such that one-half of the greater exceeds one-sixth of the lesser by $8$.

Answer 1

Let 2 nos. be x and x+4

According to the question, we have:

$\frac{x+4}{2}=\frac{x}{6}+8$

$\frac{x+4}{2}=\frac{x+48}{6}$

$6(x+4)=2(x+48)$

$6x+24=2x+96$

$4x=72orx=18$

Hence the nos. are 18 and 22