Direct & inverse proportion (algebraic) – IGCSE Maths exam questions with mark scheme

1Grade 4–5HigherCalculator3 marks

$y$ is directly proportional to $x$.

When $x = 8$, $y = 30$

Find the value of $y$ when $x = 14$ [3]

$y = kx$ (allow $y \propto x$ with $k$ introduced, or $30 = 8k$)M1
$k = \frac{30}{8} = 3.75$ (accept $\frac{15}{4}$); or unitary/scale method $30 \times \frac{14}{8}$M1
$52.5$ (accept $\frac{105}{2}$, $52\frac12$)A1
Answer: $y = 52.5$
Examiner tip: Write the proportion as an equation with a constant first: $y = kx$; a correct answer from a ratio method still scores full marks but an unsupported $k$ does not.

2Grade 6–7HigherCalculator6 marks

The cost, £$C$, of a circular rug is directly proportional to the square of its diameter, $d$ metres.

A rug of diameter 1.2 m costs £48

(a) Find a formula for $C$ in terms of $d$. [3]

(b) Work out the cost of a rug of diameter 1.8 m. [1]

(a) $C = kd^2$ (do not accept $C = d^2$ or $C = kd$)M1
(a) $48 = k \times 1.2^2$, i.e. $48 = 1.44k$M1
(a) $C = \frac{100}{3}d^2$ (accept $C = 33.3d^2$ or better; must be a formula, not just $k = \ldots$)A1
(b) $108$ (ft their $k \times 1.8^2$; allow 107.9 to 108 from rounded $k$)B1
(c) $75 = \frac{100}{3}d^2$ (ft their formula) or $d^2 = 2.25$M1
(c) $1.5$ (ignore $-1.5$ seen alongside $1.5$; do not accept $-1.5$ alone)A1
Answer: (a) $C = \frac{100}{3}d^2$ (b) £108 (c) 1.5 m
Examiner tip: Keep $k$ exact as $\frac{100}{3}$ (or use the calculator memory): using $k = 33$ loses accuracy in (b) and (c).

3Grade 6–7HigherCalculator4 marks

$F$ is inversely proportional to $r$.

$F = 15$ when $r = 4$

(a) Find the value of $F$ when $r = 10$ [3]

(b) Find the value of $r$ when $F = 24$ [1]

(a) $F = \frac{k}{r}$ (allow $Fr = k$)M1
(a) $15 = \frac{k}{4}$ or $k = 60$M1
(a) $6$A1
(b) $2.5$ (ft $\frac{\text{their } k}{24}$)B1
Answer: (a) $F = 6$ (b) $r = 2.5$
Examiner tip: Inverse proportion means $F = \frac{k}{r}$, so the product $Fr$ stays the same; students who use $F = kr$ get $k = 3.75$ and lose every mark.

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4Grade 6–7HigherCalculator5 marks

The intensity, $I$, of light from a lamp is inversely proportional to the square of the distance, $d$ metres, from the lamp.

When $d = 2$, $I = 180$

(a) Express $I$ in terms of $d$. [3]

(b) Find the distance from the lamp at which $I = 5$ [2]

(a) $I = \frac{k}{d^2}$ (do not accept $I = \frac{k}{d}$)M1
(a) $180 = \frac{k}{2^2}$ so $k = 720$M1
(a) $I = \frac{720}{d^2}$ (award M1M1A0 if $k = 720$ found but formula not stated)A1
(b) $5 = \frac{720}{d^2}$ (ft their $k$) or $d^2 = 144$M1
(b) $12$ (m) (ignore $\pm$ if $12$ is selected; do not accept $144$)A1
Answer: (a) $I = \frac{720}{d^2}$ (b) 12 m
Examiner tip: After finding $d^2 = 144$, remember the square root: $144$ is the most common wrong final answer.

5Grade 8–9HigherCalculator3 marks

$y$ is directly proportional to the cube of $x$.

The value of $x$ is increased by 20%.

Work out the percentage increase in the value of $y$. [3]

$y = kx^3$ with $x$ replaced by $1.2x$, i.e. $k(1.2x)^3$, or a correct numerical trial (e.g. $x = 10 \to 12$, $y$ from $1000k$ to $1728k$)M1
$1.2^3 = 1.728$ (or $\frac{1728}{1000}$)M1
$72.8$ (%) (do not accept $60\%$ or $1.728$ as final answer)A1
Answer: 72.8%
Examiner tip: A 20% increase cubed is not 60%: multipliers multiply, so use $1.2^3 = 1.728$ and then subtract 1.

6Grade 8–9HigherCalculator3 marks

$f$ is inversely proportional to the square root of $w$.

The value of $w$ is increased by 44%.

Work out the percentage decrease in the value of $f$.
Give your answer correct to 1 decimal place. [3]

$f = \frac{k}{\sqrt{w}}$ and new value $\frac{k}{\sqrt{1.44w}}$, or $\sqrt{1.44} = 1.2$ seen in a correct methodM1
new $f$ = $\frac{1}{1.2}f$ (= $0.833\ldots f$) or $1 - \frac{1}{1.2}$M1
$16.7$ (%) (accept $16\frac23$%; do not accept $20$% or $44$%)A1
Answer: 16.7%
Examiner tip: Dividing by 1.2 is not the same as a 20% decrease: $\frac{1}{1.2} = 0.833\ldots$, which is a 16.7% decrease.

7Grade 4–5HigherCalculator4 marks

The diagram shows four sketch graphs, A, B, C and D.

(a) Write down the letter of the graph that could show

(i) $y$ is directly proportional to $x$ [1]

(ii) $y$ is directly proportional to $x^2$ [1]

(iii) $y$ is inversely proportional to $x$ [1]

(b) Graph C is a straight line. Give a reason why graph C does not show $y$ directly proportional to $x$. [1]

(a)(i) DB1
(a)(ii) AB1
(a)(iii) BB1
(b) the line does not pass through the origin (accept: when $x = 0$, $y$ is not $0$; or it has a $y$-intercept; ignore 'not proportional' without reason)C1
Answer: (a)(i) D (ii) A (iii) B (b) It does not pass through the origin.
Examiner tip: A straight line only shows direct proportion if it goes through $(0, 0)$: check the intercept, not just the shape.

8Grade 6–7HigherCalculator4 marks

Here is a table of values for two variables $x$ and $y$.

$x$	2	3	5	$b$
$y$	12	27	$a$	147

(a) Leah says, “$y$ is directly proportional to $x$ because $y$ gets bigger when $x$ gets bigger.”
Use values from the table to show that Leah is wrong. [1]

$y$ is directly proportional to $x^2$, and $b$ is positive.

(b) Find the value of $a$ and the value of $b$. [3]

(a) e.g. $\frac{12}{2} = 6$ but $\frac{27}{3} = 9$, so $\frac{y}{x}$ is not constant (accept: $x$ multiplied by 1.5 but $y$ multiplied by 2.25; must use table values)C1
(b) $y = kx^2$ with $12 = k \times 2^2$ giving $k = 3$M1
(b) $a = 75$A1
(b) $b = 7$ (do not accept $b = \pm 7$ unless $7$ chosen)A1
Answer: (b) $a = 75$, $b = 7$
Examiner tip: 'Gets bigger together' is not proportion: show the ratio $\frac{y}{x}$ changes using actual numbers from the table.

9Grade 8–9HigherCalculator5 marks

$P$ is directly proportional to $Q^2$.
$Q$ is inversely proportional to $\sqrt{R}$.

When $R = 4$, $Q = 6$ and $P = 72$

(a) Find the value of $P$ when $R = 9$ [4]

(b) Express $P$ in terms of $R$. [1]

$Q = \frac{k}{\sqrt{R}}$ and $6 = \frac{k}{\sqrt 4}$ giving $k = 12$M1
$P = mQ^2$ and $72 = m \times 6^2$ giving $m = 2$M1
(a) when $R = 9$: $Q = \frac{12}{3} = 4$ (ft their $k$)M1
(a) $32$A1
(b) $P = \frac{288}{R}$ (ft $\frac{m k^2}{R}$ for their $m, k$; accept $P = 2\left(\frac{12}{\sqrt R}\right)^2$ only if simplified to $\frac{288}{R}$)B1
Answer: (a) $P = 32$ (b) $P = \frac{288}{R}$
Examiner tip: Use two different letters for the two constants; using $k$ for both is the commonest way to get a wrong answer here.

10Grade 6–7HigherCalculator6 marks

The time, $T$ seconds, for one swing of a pendulum is directly proportional to the square root of its length, $L$ cm.

When $L = 64$, $T = 1.6$

(a) Find a formula for $T$ in terms of $L$. [3]

(b) Work out the time for one swing when $L = 100$ [1]

(a) $T = k\sqrt{L}$ (do not accept $T = kL$)M1
(a) $1.6 = k \times \sqrt{64}$, i.e. $1.6 = 8k$M1
(a) $T = 0.2\sqrt{L}$ (accept $T = \frac{\sqrt L}{5}$)A1
(b) $2$ (s) (ft their $k \times 10$)B1
(c) $2.4 = 0.2\sqrt{L}$ (ft their formula) or $\sqrt{L} = 12$M1
(c) $144$ (cm) (do not accept $12$)A1
Answer: (a) $T = 0.2\sqrt{L}$ (b) 2 seconds (c) 144 cm
Examiner tip: In (c) you must square to undo the square root: stopping at $\sqrt L = 12$ loses the final mark.

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