WAEC 2013 · Paper 2 · Q3

(a)
The present ages of a father and his son are in the ratio $10 : 3$ . If the son is 15 years old now, in how many years will the ratio of their ages be $2 : 1$ ?
(b)
The arithmetic mean of $x$ , $y$ and $z$ is 6 while that of $x$ , $y$ , $z$ , $t$ , $u$ , $v$ and $w$ is 9. Calculate the arithmetic mean of $t$ , $u$ , $v$ and $w$ .

Worked solution (try it first)

The ratio is $10 : 3$ and the son is 15, so the father is $\frac{10}{3} \times 15 = 50$ years old.
In $n$ years they will be $50 + n$ and $15 + n$ , in the ratio $2 : 1$ : $\frac{50 + n}{15 + n} = 2$ .
Cross-multiply: $50 + n = 30 + 2n$ , so $n = 20$ .
The ratio will be $2 : 1$ in 20 years.