WAEC 2025 · Paper 2 · Q2✱✱

Variation

$y$ is partly constant and partly varies inversely as the square of $x$ .

(a)
Write down the relationship between $x$ and $y$ .
Show the answer
$y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants
(b)
When $x = 1$ , $y = 11$ , and when $x = 2$ , $y = 5$ . Find $y$ when $x = 4$ .

Worked solution (try it first)

(a)

Partly constant ( $a$ ) and partly varies inversely as the square of $x$ ( $\frac{b}{x^2}$ ): $y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants.

(b)

$x = 1$ , $y = 11$ : $a + b = 11$ .
$x = 2$ , $y = 5$ : $a + \frac{b}{4} = 5$ .
Subtract: $b - \frac{b}{4} = 6$ , so $\frac{3b}{4} = 6$ and $b = 8$ .
Then $a = 11 - 8 = 3$ , and $y = 3 + \dfrac{8}{x^2}$ .
When $x = 4$ : $y = 3 + \dfrac{8}{16} = 3\frac12$ .

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