WAEC 2019 · Paper 2 · Q1

(a)
Given that $110_x = 40_{\text{five}}$ , find the value of $x$ .
(b)
Simplify $\dfrac{15}{\sqrt{75}} + \sqrt{108} + \sqrt{432}$ , leaving the answer in the form $a\sqrt b$ , where $a$ and $b$ are positive integers.

Worked solution (try it first)

(a)

Change both to base ten: $110_x = x^2 + x$ and $40_{\text{five}} = 4 \times 5 = 20$ .
So $x^2 + x - 20 = 0$ , $(x + 5)(x - 4) = 0$ , and the base is $x = 4$ (a base can't be negative).

(b)

Rationalise and simplify each term: $\frac{15}{\sqrt{75}} = \frac{15}{5\sqrt3}$
$= \frac{3}{\sqrt3}$
$= \sqrt3$ .
$\sqrt{108} = 6\sqrt3$ .
$\sqrt{432} = \sqrt{144 \times 3}$
$= 12\sqrt3$ .
Total: $\sqrt3 + 6\sqrt3 + 12\sqrt3 = 19\sqrt3$ .

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