Question

One fourth of a number exceeds one-sixth of its succeeding number by $5$. Find the number.

Answer 1

Let the number be $a.$

Then,

$\begin{array}{cc}\frac{a}{4}-\frac{a+1}{6}& =5\\ \frac{6a-4\left(a+1\right)}{24}& =5\\ 6a-4a-4& =5\times 24\\ 2a& =120+4\\ 2a& =124\\ a& =62\end{array}$

Thus the required number is $62.$