Question

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{Let the two parts be}\mathit{\text{x}}\text{and}(184-x).\\ \text{Then, we have:}\\ \frac{\text{1}}{3}x=\frac{\text{1}}{7}(\text{184}-\text{x)+8}\\ \begin{array}{l}\text{⇒}\frac{\text{1}}{3}x-\frac{\text{1}}{7}(\text{184}-\text{x})=\text{8}\\ \text{⇒}\frac{\text{1}}{3}x-\frac{184}{7}+\frac{x}{7}=\text{8}\\ \text{⇒}\frac{1}{3}x+\frac{1}{7}x=\frac{184}{7}+8\\ \text{⇒}\frac{7x+3x}{21}=8+\frac{184}{7}\\ \text{⇒}\frac{10x}{21}=\frac{56+184}{7}\\ \text{⇒}\frac{10x}{21}=\frac{240}{7}\\ \text{⇒}x=\frac{240\times 21}{7\times 10}\\ =72\\ \mathrm{Now},\mathrm{ot}\text{her part =184}-72=112\end{array}\end{array} \\ \therefore \mathrm{The}\mathrm{two}\mathrm{parts}\mathrm{are}72\mathrm{and}112. \end{gathered}$$