WAEC 2019 · Paper 2 · Q3

(a)
Without using mathematical tables or calculators, simplify $3\sqrt{\frac32} - 4\sqrt{\frac23} - \sqrt{24}$ .
(b)
The probabilities of two candidates, $M$ and $N$ , passing an examination are $\frac23$ and $\frac45$ respectively. Find the probability that: (i) only one candidate will pass; (ii) at least one candidate will pass.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write each term as a multiple of $\sqrt6$ .
$3\sqrt{\frac32} = \frac{3\sqrt3}{\sqrt2}$
$= \frac{3\sqrt6}{2}$ .
$4\sqrt{\frac23} = \frac{4\sqrt2}{\sqrt3}$
$= \frac{4\sqrt6}{3}$ .
$\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt6$ .
Over the common denominator 6: $\frac{9\sqrt6}{6} - \frac{8\sqrt6}{6} - \frac{12\sqrt6}{6} = -\frac{11\sqrt6}{6}$ .

(b)

$P(M \text{ fails}) = \frac13$ and $P(N \text{ fails}) = \frac15$ .

(i)

Only one passes: $M$ passes and $N$ fails, $\frac23 \times \frac15 = \frac{2}{15}$ .
Or $M$ fails and $N$ passes, $\frac13 \times \frac45 = \frac{4}{15}$ .
Together: $\frac{6}{15} = \frac25$ .

(ii)

At least one passes is everything except "both fail": $1 - \frac13 \times \frac15 = 1 - \frac{1}{15}$
$= \frac{14}{15}$ .

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