Question

johnhad some pencils. 3/5 of them were red and the rest were green. he gave 2/3 of the red pencils and 1/4 of the greeenpencils to his sister. If he had 80 pencils left, how many pencils did he have at first?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{pencils}\mathrm{with}\mathrm{John}\mathrm{at}\mathrm{first}=\mathrm{x} \\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{pencils}=\frac{3\mathrm{x}}{5} \\ \mathrm{Number}\mathrm{of}\mathrm{green}\mathrm{pencils}=\mathrm{x}-\frac{3\mathrm{x}}{5}=\frac{2\mathrm{x}}{5} \\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{pencils}\mathrm{given}\mathrm{to}\mathrm{sister}=\frac{2}{3}\mathrm{of}\frac{3\mathrm{x}}{5}=\frac{2}{3}\times \frac{3\mathrm{x}}{5}=\frac{2\mathrm{x}}{5} \\ \mathrm{Number}\mathrm{of}\mathrm{green}\mathrm{pencils}\mathrm{given}\mathrm{to}\mathrm{sister}=\frac{1}{4}\mathrm{of}\frac{2\mathrm{x}}{5}=\frac{1}{4}\times \frac{2\mathrm{x}}{5}=\frac{\mathrm{x}}{10} \\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{pencils}\mathrm{left}\mathrm{with}\mathrm{John}=\frac{3\mathrm{x}}{5}-\frac{2\mathrm{x}}{5}=\frac{\mathrm{x}}{5} \\ \mathrm{Number}\mathrm{of}\mathrm{green}\mathrm{pencils}\mathrm{left}\mathrm{with}\mathrm{John}=\frac{2\mathrm{x}}{5}-\frac{\mathrm{x}}{10}=\frac{3\mathrm{x}}{10} \\ \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{pencils}\mathrm{left}\mathrm{with}\mathrm{John}=\frac{\mathrm{x}}{5}+\frac{3\mathrm{x}}{10}=\frac{5\mathrm{x}}{10}=\frac{\mathrm{x}}{2} \\ \mathrm{Now},\frac{\mathrm{x}}{2}=80 \\ \Rightarrow \mathrm{x}=160 \\ \mathrm{number}\mathrm{of}\mathrm{pencils}\mathrm{with}\mathrm{John}\mathrm{at}\mathrm{first}=\mathrm{x}=160 \end{gathered}$$