Question

A and B are friends. A is elder to B by 5 years. B's sister C is half the age of B while A's father D is 8 years older than twice the age of B. If the present age of D is 48 years, they find the present ages of A, B and C.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{B}\mathrm{be}x\mathrm{yrs}.\mathrm{Then}, \\ \mathrm{The}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{A}=\left(x+5\right)\mathrm{yrs}, \\ \mathrm{The}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{C}=\frac{x}{2}\mathrm{yrs}\mathrm{and} \\ \mathrm{The}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{D}=\left(2x+8\right)\mathrm{yrs} \\ \mathrm{But}\mathrm{the}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{D}=48\mathrm{yrs} \\ \Rightarrow 2x+8=48 \\ \Rightarrow 2x=48-8 \\ \Rightarrow 2x=40 \\ \Rightarrow x=\frac{40}{2} \\ \Rightarrow x=20 \\ \mathrm{So}, \\ \mathrm{The}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{A}=20+5=25\mathrm{years}, \\ \mathrm{The}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{B}=20\mathrm{years}\mathrm{and} \\ \mathrm{The}\mathrm{present}\mathrm{of}\mathrm{C}=\frac{20}{2}=10\mathrm{years} \end{gathered}$$