WAEC 2014 · Paper 2 · Q6

(a)
In a class of 52 students, 16 are Science students. If $\frac13$ of the boys and $\frac14$ of the girls are Science students, how many boys are in the class?
(b)
The sum of the first and third terms of a Geometric Progression (G.P.) is 40 while the fourth and sixth terms are in the ratio $1 : 4$ . Find the: (i) common ratio; (ii) fifth term.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let there be $x$ boys and $y$ girls.
The class has 52 students: $x + y = 52$ .
A third of the boys and a quarter of the girls do Science, making 16: $\frac x3 + \frac y4 = 16$ .
Multiply by 12: $4x + 3y = 192$ .
From the first equation $y = 52 - x$ , so $4x + 3(52 - x) = 192$ .
Then $4x + 156 - 3x = 192$ and $x = 36$ .
There are 36 boys.

(b)(i)

With first term $a$ and common ratio $r$ , the terms are $a, ar, ar^2, ar^3, \ldots$ The fourth and sixth terms are in the ratio $1 : 4$ : $\frac{ar^3}{ar^5} = \frac14$ , so $\frac1{r^2} = \frac14$ and $r^2 = 4$ .
So $r = 2$ or $r = -2$ .

(ii)

The first and third terms add up to 40: $a + ar^2 = 40$ , so $a + 4a = 40$ and $a = 8$ .
The fifth term is $ar^4 = 8 \times 16 = 128$ (the same for $r = 2$ or $r = -2$ ).

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