Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).
Four years ago Sonny was four times as old as Wale:
\[2w - 4 = 4(w - 4)\]
\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]
So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).
Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):
\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]
The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).