Question

If $\frac{3}{5}$ of a number exceeds its $\frac{2}{7}$ by 44, find the number.

Answer 1

Let x be the required number.
We know that $\frac{3}{5}$ of the number exceeds its $\frac{2}{7}$ by 44.
That is,

$\frac{3}{5}\times x=\frac{2}{7}\times x+44$
$$\begin{gathered} \frac{3}{5}\times x-\frac{2}{7}\times x=44 \\ \left(\frac{3}{5}-\frac{2}{7}\right)\times x=44 \\ \left(\frac{3}{5}+\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{2}{7}\right)\times x=44 \\ \left(\frac{21-10}{35}\right)\times x=44 \end{gathered}$$
$\frac{11}{35}\times x=44$
$x=44÷\frac{11}{35}$

$$\begin{gathered} =44\times \frac{35}{11} \\ =\frac{44}{1}\times \frac{35}{11} \\ =\frac{44\times 35}{1\times 11} \\ =\frac{1540}{11} \\ =140 \end{gathered}$$

Hence, the number is 140.