Question

The ages of A and B are in the ratio 4 : 3. Eight years ago, their ages were in the ratio 10 : 7. Find their present ages.

Answer 1

Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7

$$\begin{gathered} \Rightarrow \frac{4x-8}{3x-8}=\frac{10}{7} \\ \Rightarrow 28x-56=30x-80 \\ \Rightarrow 2x=24 \\ \Rightarrow x=12 \\ \mathrm{Hence},\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{A}=4\times 12\mathrm{yrs}=48\mathrm{yrs} \\ \mathrm{and}\mathrm{present}\mathrm{age}\mathrm{of}\mathrm{B}=3\times 12\mathrm{yrs}=36\mathrm{yrs} \end{gathered}$$