Question

(ii) Ten years ago father was 12 times as old as his son at that time and 10 years hence he will be twice as old as his son. Find their present ages.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{age}\mathrm{of}\mathrm{son}\mathrm{be}x\mathrm{yrs}\mathrm{and}\mathrm{that}\mathrm{of}\mathrm{father}\mathrm{be}y\mathrm{yrs}. \\ \mathrm{Ten}\mathrm{years}\mathrm{ago}, \\ \mathrm{Age}\mathrm{of}\mathrm{son}=\left(x-10\right)\mathrm{yrs} \\ \mathrm{Age}\mathrm{of}\mathrm{father}=\left(y-10\right)\mathrm{yrs} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \left(y-10\right)=12\left(x-10\right) \\ \Rightarrow y-10=12x-120 \\ \Rightarrow y=12x-110...\left(i\right) \\ 10\mathrm{years}\mathrm{hence}, \\ \mathrm{Age}\mathrm{of}\mathrm{son}=\left(x+10\right)\mathrm{yrs} \\ \mathrm{Age}\mathrm{of}\mathrm{father}=\left(y+10\right)\mathrm{yrs} \\ y+10=2\left(x+10\right) \\ \Rightarrow y+10=2x+20 \\ \Rightarrow 2x-y=-10...\left(ii\right) \\ \mathrm{Substituting}\mathrm{eq}\left(\mathrm{i}\right)\mathrm{in}\mathrm{eq}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ 2x-\left(12x-110\right)=-10 \\ \Rightarrow -10x+110=-10 \\ \Rightarrow -10x=-120 \\ \Rightarrow x=12 \\ \mathrm{Substituting}\mathrm{x}=12\mathrm{in}\mathrm{eq}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ y=12\times 12-110 \\ \Rightarrow y=34\mathrm{yrs} \\ \therefore \mathrm{Age}\mathrm{of}\text{the}\mathrm{father}\mathrm{is}34\mathrm{yrs}\mathrm{and}\mathrm{age}\mathrm{of}\text{the}\mathrm{son}\mathrm{is}12\mathrm{yrs}. \end{gathered}$$