Constructing direct/inverse proportional equations

3 exam-style questions. Answers and the mark for each step are on the last page.

Question 1Non-calculator · 3 marks
$y$ is directly proportional to $x^2$ . When $x = 3$ , $y = 18$ .
(a) Find a formula for $y$ in terms of $x$ . (2)
(b) Work out $y$ when $x = 5$ . (1)
Question 2Non-calculator · 1 mark
(a) y is inversely proportional to $x^{2}.$ Positive x increases by 20%. Which expression gives the new value of y in terms of its old value? Select one answer. (1)
1. (5/6) $\times$ old y
2. 1.44 $\times$ old y
3. (25/36) $\times$ old y
4. 0.8 $\times$ old y
Question 3Calculator · 5 marks
The brightness, $I$ , of a light is inversely proportional to the square of the distance, $d$ metres, from the light. When $d = 2$ , $I = 45$ .
(a) Work out $I$ when $d = 3$ . (2)
(b) The distance from the light is increased by $50\%$ . Work out the percentage decrease in the brightness. Give your answer to 1 decimal place. (3)

Answers and marks

(a) $y = 2x^2$

(b) $50$

(a) (25/36) $\times$ old y

(a) $20$

(b) $55.6\%$

P1 Recognising the multiplier for the distance is 1.5.
P1 Finding the brightness multiplier $\frac{1}{1.5^2} = 0.444\ldots$ (or using values: $d = 2$ to $d = 3$ gives $45 \to 20$ ).
A1 The correct answer, $55.6\%$ .

Constructing direct/inverse proportional equations

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