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WAEC Mathematics 2025 Paper

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Question : 15

Total: 50

If y ∝

\frac{1}{x^2}

and x = 3, when y = 4, find y when x = 2

3
9
1
18

Solution:

Concept:

If

y

varies inversely as the square of

x

, then

y = \frac{k}{x^2}

where

k

is a constant.

Explanation:

Given

y \propto \frac{1}{x^2}

, write

y = \frac{k}{x^2}

.

When

x = 3

and

y = 4

, substitute into

y = \frac{k}{x^2}

:

4 = \frac{k}{3^2}

4 = \frac{k}{9}

Multiply both sides by 9:

k = 36

Now find

y

when

x = 2

:

y = \frac{36}{2^2}

y = \frac{36}{4}

y = 9

Answer:

y = 9

(Option B)

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