Question

The sum of the age of two brothers is 41, 5 years hence the elder brother will be twice as old as the younger brother. Find their present ages.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{younger}\mathrm{brother}=x\mathrm{yrs} \\ \mathrm{then}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{elder}\mathrm{brother}=\left(41-x\right)\mathrm{yrs} \\ \mathbf{Now}\mathbf{,}\mathbf{}\mathbf{After}\mathbf{}\mathbf{5}\mathbf{}\mathbf{yrs}\mathbf{}\mathbf{:} \\ \mathrm{Younger}\mathrm{brother}\text{'}\mathrm{s}\mathrm{age}=\left(x+5\right)\mathrm{yrs} \\ \mathrm{Elder}\mathrm{brother}\text{'}\mathrm{s}\mathrm{age}=\left(46-x\right)\mathrm{yrs} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}, \\ 46-x=2\left(x+5\right) \\ \Rightarrow 46-x=2x+10 \\ \Rightarrow 3x=36 \\ \Rightarrow \boxed{x=12} \\ \mathrm{Now},\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{younger}\mathrm{brother}=12\mathrm{yrs} \\ \mathrm{present}\mathrm{age}\mathrm{of}\mathrm{elder}\mathrm{brother}=41-12=29\mathrm{yrs} \end{gathered}$$