WAEC 2022 · Paper 2 · Q13

(a)
Find the sum of the first 87 multiples of 5.
(b)
Solve: $\dfrac{p - 5}{p} = \dfrac{5}{3p} - \dfrac15$ .
(c)
Find the equation of the line passing through the points $(-1, 4)$ and $(3, 6)$ .
Show the answer
$2y - x - 9 = 0$

Worked solution (try it first)

(a)

The first 87 multiples of 5 are $5, 10, 15, \ldots$ , an A.P. with $a = 5$ and $d = 5$ .
$S_n = \frac n2[2a + (n - 1)d]$ , so $S_{87} = \frac{87}{2}[10 + 86 \times 5]$
$= \frac{87}{2} \times 440$
$= 87 \times 220$
$= 19\,140$ .

(b)

Multiply every term by $15p$ , the LCM of the denominators: $15(p - 5) = 25 - 3p$ .
Expand: $15p - 75 = 25 - 3p$ .
So $18p = 100$ and $p = \frac{100}{18}$
$= \frac{50}{9}$
$= 5\frac59$ .

(c)

Gradient $= \frac{6 - 4}{3 - (-1)}$
$= \frac24$
$= \frac12$ .
Using the point $(-1, 4)$ : $y - 4 = \frac12(x + 1)$ .
Multiply by 2: $2y - 8 = x + 1$ .
So $2y - x - 9 = 0$ .

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