WAEC 2012 · Paper 2 · Q2

(a)

1. Let Wale's age now be $w$ years.
2. Sonny is twice as old: $2w$ years.
3. Four years ago they were $w - 4$ and $2w - 4$, and Sonny was four times as old as Wale: $2w - 4 = 4(w - 4)$.
4. Expand: $2w - 4 = 4w - 16$, so $2w = 12$ and $w = 6$.
5. Wale is 6 and Sonny is 12.
6. In $n$ years they will be $6 + n$ and $12 + n$, and their sum is 66: $(6 + n) + (12 + n) = 66$.
7. So $18 + 2n = 66$, $2n = 48$ and $n = 24$.
8. The sum of their ages will be 66 in 24 years' time.