WAEC 2020 · Paper 2 · Q2

(a)
Given that $\cos60^\circ = \sin30^\circ = \frac12$ and $\cos30^\circ = \sin60^\circ = \frac{\sqrt3}{2}$ , evaluate $\dfrac{\tan60^\circ - 1}{1 - \tan30^\circ}$ .
(b)
In a class of 40 students, 30 read Chemistry and 20 read Physics. If all the students read at least one of the subjects, find the probability that a student selected at random from the class reads only Chemistry.

Worked solution (try it first)

(a)

$\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ}$
$= \sqrt3$ and $\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ}$
$= \frac{1}{\sqrt3}$ .
So the value is $\frac{\sqrt3 - 1}{1 - \frac{1}{\sqrt3}}$ .
Multiply the top and bottom by $\sqrt3$ : $\frac{\sqrt3(\sqrt3 - 1)}{\sqrt3 - 1} = \sqrt3$ .

(b)

Let $x$ students read both subjects.
Only Chemistry: $30 - x$ .
Only Physics: $20 - x$ .
Everyone reads at least one, so $(30 - x) + x + (20 - x) = 40$ , which gives $50 - x = 40$ and $x = 10$ .
Only Chemistry: $30 - 10 = 20$ students.
Probability $= \frac{20}{40} = \frac12$ .

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