NECO 2022 · Paper 2 · Q8

(a)
If $x$ and $y$ are the surface area and volume of a sphere respectively and $\frac{x}{y} = \frac27$ , find the values of $x$ and $y$ . $\left(\text{Take } \pi = \frac{22}{7}\right)$
Separate values with commas, e.g. 3, −2
(b)
Find the equation of a line which is perpendicular to the line $3x + 3y = 5$ and passes through the point $(3, 2)$ . $y =$
(b)(x-intercept)
Find the $x$ -intercept of that line.

Worked solution (try it first)

(a)

For a sphere of radius $r$ : $x = 4\pi r^2$ and $y = \frac43 \pi r^3$ .
Divide: $\frac{x}{y} = \frac{4\pi r^2}{\frac43 \pi r^3}$
$= \frac{3}{r}$ .
Set $\frac{3}{r} = \frac27$ and cross-multiply: $2r = 21$ , so $r = 10.5$ .
Surface area: $x = 4 \times \frac{22}{7} \times 10.5^2$
$= 1386$ .
Volume: $y = \frac43 \times \frac{22}{7} \times 10.5^3$
$= 4851$ .
So $x = 1386$ (square units) and $y = 4851$ (cubic units).

(b)

Rearrange $3x + 3y = 5$ as $y = -x + \frac53$ : its gradient is $-1$ .
Perpendicular gradients multiply to $-1$ , so the new gradient is $1$ .
Through $(3, 2)$ : $y - 2 = 1(x - 3)$ , so $y = x - 1$ .
The $x$ -intercept is where $y = 0$ : $0 = x - 1$ , so $x = 1$ .
The line cuts the $x$ -axis at $(1, 0)$ .

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