NECO 2023 · Paper 2 · Q6

(a)
If $135_k = 231_{\text{four}}$ , find the value of $k$ .
(b)
A sector of a circle of radius $21\text{ cm}$ has an angle of $120^\circ$ at the centre. Calculate its (i) perimeter; (ii) area. $\left[\pi = \frac{22}{7}\right]$
Separate values with commas, e.g. 3, −2
(c)
Simplify $\dfrac{3}{\sqrt5 + \sqrt2} - \dfrac{1}{\sqrt5 - \sqrt2}$ , leaving your answer in surd form.

Worked solution (try it first)

(a)

$231_{\text{four}} = 2 \times 16 + 3 \times 4 + 1$
$= 45$ .
In base $k$ , $135_k = k^2 + 3k + 5$ .
So $k^2 + 3k + 5 = 45$ , which gives $k^2 + 3k - 40 = 0$ and $(k + 8)(k - 5) = 0$ .
A base must be positive, so $k = 5$ .

(b)(i)

The sector is $\frac{120}{360} = \frac13$ of the circle.
Arc $= \frac13 \times 2 \times \frac{22}{7} \times 21$
$= 44$ cm.
Perimeter $= 44 + 21 + 21 = 86$ cm.

(ii)

Area $= \frac13 \times \frac{22}{7} \times 21^2$
$= 462\text{ cm}^2$ .

(c)

Rationalise each fraction.
$\frac{3}{\sqrt5 + \sqrt2} \times \frac{\sqrt5 - \sqrt2}{\sqrt5 - \sqrt2} = \frac{3(\sqrt5 - \sqrt2)}{3}$
$= \sqrt5 - \sqrt2$ , and $\frac{1}{\sqrt5 - \sqrt2} = \frac{\sqrt5 + \sqrt2}{3}$ .
So the expression is $\sqrt5 - \sqrt2 - \frac{\sqrt5 + \sqrt2}{3} = \frac{3\sqrt5 - 3\sqrt2 - \sqrt5 - \sqrt2}{3}$
$= \frac{2\sqrt5 - 4\sqrt2}{3}$ .

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