WAEC 2021 · Paper 2 · Q2

Expressions, formulae & change of subject

Given that $S = \sqrt{\dfrac{5d(L - d)}{7}}$ ,

(a)
make $L$ the subject of the relation;
(b)
find, correct to two decimal places, the value of $L$ when $d = 27$ and $S = 0.7$ .

Worked solution (try it first)

(a)

$L$ is inside a square root, so square both sides: $S^2 = \frac{5d(L - d)}{7}$ .
Multiply both sides by 7: $7S^2 = 5d(L - d) = 5dL - 5d^2$ .
Add $5d^2$ to both sides: $7S^2 + 5d^2 = 5dL$ .
Divide by $5d$ : $L = \frac{7S^2 + 5d^2}{5d}$ .

(b)

Substitute $d = 27$ and $S = 0.7$ : $L = \frac{7 \times 0.49 + 5 \times 729}{5 \times 27}$
$= \frac{3.43 + 3645}{135}$
$= \frac{3648.43}{135}$
$\approx 27.025$ .
Correct to two decimal places: $L = 27.03$ .

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