Question

fathers age 7 years ago was seven times his daughters age . three years hence he will be three times as old as his daughter . find their present ages ​

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{daughter}=\mathrm{x}\mathrm{yrs} \\ \mathrm{Let}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{Father}=\mathrm{y}\mathrm{yrs} \\ \underline{\mathbf{SEVEN}\mathbf{}\mathbf{YEARS}\mathbf{}\mathbf{AGO}\mathbf{}\mathbf{:}} \\ \mathrm{Daughter}\text{'}\mathrm{s}\mathrm{age}=\left(\mathrm{x}-7\right)\mathrm{yrs} \\ \mathrm{Father}\text{'}\mathrm{s}\mathrm{age}=\left(\mathrm{y}-7\right)\mathrm{yrs} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}: \\ \mathrm{y}-7=7\left(\mathrm{x}-7\right) \\ \Rightarrow \mathrm{y}-7=7\mathrm{x}-49 \\ \Rightarrow 7\mathrm{x}-\mathrm{y}=42.....\left(1\right) \\ \underline{\mathbf{AFTER}\mathbf{}\mathbf{3}\mathbf{}\mathbf{YEARS}\mathbf{}\mathbf{:}} \\ \mathrm{Daughter}\text{'}\mathrm{s}\mathrm{age}=\left(\mathrm{x}+3\right)\mathrm{yrs} \\ \mathrm{Father}\text{'}\mathrm{s}\mathrm{age}=\left(\mathrm{y}+3\right)\mathrm{yrs} \\ \mathrm{Now},\mathrm{y}+3=3\left(\mathrm{x}+3\right) \\ \Rightarrow \mathrm{y}+3=3\mathrm{x}+9 \\ \Rightarrow 3\mathrm{x}-\mathrm{y}=-6......\left(2\right) \\ \mathrm{Subtracting}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ 7\mathrm{x}-3\mathrm{x}=42+6 \\ \Rightarrow 4\mathrm{x}=48 \\ \Rightarrow \mathrm{x}=12 \\ \mathrm{Now},\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{y}=7\mathrm{x}-42=7\times 12-42=84-42=42 \\ \mathrm{So},\mathrm{father}\text{'}\mathrm{s}\mathrm{age}=42\mathrm{yrs} \\ \mathrm{Daughter}\text{'}\mathrm{s}\mathrm{age}=12\mathrm{yrs} \end{gathered}$$