5 pens cost £3.40.
Work out the cost of 8 of these pens. [2]
- $3.40 \div 5$ (= 0.68) or a complete equivalent methodM1
- £5.44A1
- Answer: £5.44
- Examiner tip: Find the cost of one pen first, then multiply – and write money to 2 decimal places.
6 workers can paint a fence in 10 hours.
(a) How long would it take 4 workers to paint the same fence? [2]
(b) Write down one assumption you have made. [1]
- (a) $6 \times 10 = 60$ (worker-hours) or $10 \times \frac{6}{4}$ or $10 \div 4 \times 6$M1
- (a) 15 hours (accept 900 minutes)A1
- (b) a valid assumption, e.g. all the workers work at the same rate; they keep working at the same speed (no breaks / do not get tired); the workers do not get in each other's way. Do not accept 'the fence is the same' (given in the question) or 'they work for 10 hours'.C1
- Answer: (a) 15 hours (b) all workers work at the same rate
- Examiner tip: Fewer workers means MORE time – this is inverse proportion, so don't scale 10 down to 6.67.
The table shows some values of $x$ and $y$.
| $x$ | 2 | 5 | 8 |
| $y$ | 7 | 17.5 | 28 |
(a) Show that $y$ is directly proportional to $x$. [2]
(b) Work out the value of $y$ when $x = 11$. [1]
- (a) works out $y \div x$ (or $x \div y$) for at least two pairs, e.g. $7 \div 2 = 3.5$, $17.5 \div 5 = 3.5$; or uses multipliers, e.g. $x \times 2.5$ and $y \times 2.5$ (2 to 5) or $x \times 4$ and $y \times 4$ (2 to 8)M1
- all three pairs shown consistent (e.g. $y \div x = 3.5$ each time, or $y = 3.5x$ checked for each pair, or both multipliers shown equal), with a conclusion that $y$ is directly proportional to $x$C1
- (b) 38.5B1
- Answer: (a) $y \div x = 3.5$ for every pair (b) $38.5$
- Examiner tip: Checking just one pair does not show proportion – you must show the multiplier is the same for every pair.
$y$ is directly proportional to $x^2$.
When $x = 4$, $y = 48$.
(a) Find the value of $y$ when $x = 7$. [3]
(b) Find the positive value of $x$ when $y = 27$. [1]
- (a) $y = kx^2$ (or $y \propto x^2$ with a constant)M1
- $48 = k \times 4^2$ leading to $k = 3$M1
- 147A1
- (b) 3B1
- Answer: (a) $147$ (b) $x = 3$
- Examiner tip: Use $x^2$ when finding $k$: $48 \div 4 = 12$ is the most common wrong value.
$d$ is inversely proportional to the square of $r$.
When $r = 3$, $d = 20$.
(a) Find a formula for $d$ in terms of $r$. [3]
(b) Find the value of $r$ when $d = 5$, given that $r > 0$. [2]
- (a) $d = \dfrac{k}{r^2}$M1
- $20 = \dfrac{k}{3^2}$ or $k = 180$M1
- $d = \dfrac{180}{r^2}$A1
- (b) $5 = \dfrac{180}{r^2}$ or $r^2 = 36$M1
- 6A1
- Answer: (a) $d = \dfrac{180}{r^2}$ (b) $r = 6$
- Examiner tip: The answer to (a) must be a full formula with the value of $k$ substituted – $d = k/r^2$ alone scores only 1 mark.
$f$ is inversely proportional to $w$.
Complete the table.
| $w$ | 2 | 5 | 40 | |
| $f$ | 50 | 4 |
[3]
- $f \times w = 100$ or $f = \dfrac{100}{w}$ seen or usedM1
- two of the three values correctA1
- all three correct: $f = 20$, $w = 25$, $f = 2.5$A1
- Answer: $w = 5 \Rightarrow f = 20$; $f = 4 \Rightarrow w = 25$; $w = 40 \Rightarrow f = 2.5$
- Examiner tip: For inverse proportion the product $f \times w$ stays the same – the values do not go up in equal steps.
$y$ is inversely proportional to $x^2$.
The value of $x$ is increased by 25%.
Work out the percentage decrease in the value of $y$. [4]
- $y = \dfrac{k}{x^2}$M1
- new $x = 1.25x$ (or uses a numerical example, e.g. $x = 4 \to 5$)M1
- new $y = \dfrac{k}{(1.25x)^2} = \dfrac{k}{1.5625x^2}$, or multiplier $0.64$ foundM1
- 36%A1
- Answer: 36% decrease
- Examiner tip: Increasing $x$ by 25% does not decrease $y$ by 25%: find the multiplier $1 \div 1.25^2 = 0.64$.
A printer prints 45 pages in 3 minutes.
It always prints at the same rate.
(a) Work out how many pages the printer prints in 7 minutes. [2]
(b) Work out how long it takes the printer to print 120 pages. [2]
- (a) $45 \div 3$ (= 15 pages per minute) or $45 \div 3 \times 7$ oeM1
- 105A1
- (b) $120 \div 15$ or $120 \div 45 \times 3$ oe (ft their pages per minute from (a))M1
- 8 minutes (accept 8 with no units)A1
- Answer: (a) 105 pages (b) 8 minutes
- Examiner tip: Find the rate for ONE minute first – multiplying 45 by 7 and forgetting to divide by 3 scores nothing.
Here are four sketch graphs.
(a) Write down the letter of the graph that shows $y$ is directly proportional to $x$. [1]
(b) Write down the letter of the graph that shows $y$ is inversely proportional to $x$. [1]
(c) Graph A passes through the origin.
Explain why graph A does not show that $y$ is directly proportional to $x$. [1]
- (a) CB1
- (b) BB1
- (c) it is a curve, not a straight line oe, e.g. 'the gradient is not constant' or 'when $x$ doubles, $y$ does not double'. Allow 'it shows $y$ is proportional to $x^2$'. Do not accept 'it does not go through the origin' or 'it gets steeper' on its own without reference to it not being a straight line.C1
- Answer: (a) C (b) B (c) it is a curve, not a straight line (direct proportion needs a straight line through the origin)
- Examiner tip: Direct proportion needs BOTH a straight line AND a line through the origin – passing through (0, 0) on its own is not enough.
The graph shows the mass, in grams, of different lengths of a copper wire.
(a) How does the graph show that the mass of the wire is directly proportional to its length? [1]
(b) Work out the mass of 26 metres of this wire. [2]
(c) A full reel of this wire has a mass of 1.2 kg.
Work out the length of wire on the reel. [2]
- (a) It is a straight line through the origin (both needed)C1
- (b) Correct reading used to find mass per metre, e.g. $120 \div 8$ or $150 \div 10$ (= 15), or a correct scaling such as $150 \times 2.6$ or $60 \times 6.5$M1
- 390 (g)A1
- (c) $1200 \div 15$ oe (must convert 1.2 kg to 1200 g; ft their mass per metre from (b))M1
- 80 (m)A1
- Answer: (a) straight line through the origin (b) 390 g (c) 80 m
- Examiner tip: In (c) change 1.2 kg into 1200 g before dividing – $1.2 \div 15 = 0.08$ is a very common wrong answer.
The time, $t$ hours, taken for a journey is inversely proportional to the average speed, $v$ km/h.
(a) Which of these statements means the same as '$t$ is inversely proportional to $v$'? [1]
$t \propto v$ $t \propto \dfrac{1}{v}$ $t \propto v^2$ $t \propto \sqrt{v}$
At an average speed of 60 km/h the journey takes 2.5 hours.
(b) Work out the time the journey takes at an average speed of 75 km/h. [2]
(c) Pip says, 'If the average speed is doubled, the time taken will also be doubled.'
Is Pip correct? Give a reason for your answer. [1]
- (a) $t \propto \dfrac{1}{v}$B1
- (b) $60 \times 2.5$ (= 150) or $t = \dfrac{150}{v}$ seenM1
- 2 (hours)A1
- (c) No, with reason, e.g. 'the time would be halved' or 'at 120 km/h it takes 1.25 hours, not 5 hours'C1
- Answer: (a) $t \propto \dfrac{1}{v}$ (b) 2 hours (c) No – doubling the speed halves the time
- Examiner tip: In inverse proportion the product $tv$ stays the same, so one quantity going up makes the other go down.
The distance, $d$ km, that a lookout can see to the horizon is directly proportional to the square root of their height, $h$ metres, above sea level.
When $h = 25$, $d = 18$.
(a) Find a formula for $d$ in terms of $h$. [3]
(b) Work out the value of $d$ when $h = 64$. [1]
(c) Work out the value of $h$ when $d = 43.2$ [2]
- (a) $d = k\sqrt{h}$ (allow $d \propto \sqrt{h}$ for this mark only if followed by use of a constant)M1
- $18 = k\sqrt{25}$ or $k = \dfrac{18}{5}$ oeM1
- $d = 3.6\sqrt{h}$ oeA1
- (b) 28.8 (ft $8 \times$ their $k$)B1
- (c) $\sqrt{h} = 43.2 \div 3.6$ (= 12) (ft their $k$)M1
- 144A1
- Answer: (a) $d = 3.6\sqrt{h}$ (b) 28.8 (c) 144
- Examiner tip: In (c) students often stop at 12 – that is $\sqrt{h}$, so it still has to be squared.
$T$ is inversely proportional to the square root of $L$.
When $L = 9$, $T = 20$.
(a) Work out the value of $T$ when $L = 50$.
Give your answer correct to 3 significant figures. [3]
(b) Work out the value of $L$ when $T = 8$ [2]
- (a) $T = \dfrac{k}{\sqrt{L}}$M1
- $k = 20 \times \sqrt{9}$ (= 60)M1
- 8.49 (accept 8.485…)A1
- (b) $8 = \dfrac{60}{\sqrt{L}}$ or $\sqrt{L} = 7.5$ (ft their $k$)M1
- 56.25 (accept $56\tfrac{1}{4}$)A1
- Answer: (a) 8.49 (b) 56.25
- Examiner tip: Using $T = k/L$ (forgetting the square root) gives $k = 180$ and loses every mark.
$a$ is directly proportional to the cube of $b$.
When $b = 6$, $a = 54$.
$b$ is inversely proportional to $c$.
When $c = 2$, $b = 6$.
(a) Find a formula for $a$ in terms of $c$. [4]
(b) Work out the value of $a$ when $c = 3$. [1]
- (a) $a = kb^3$ and $54 = k \times 6^3$, or $k = \dfrac{1}{4}$ oeM1
- $b = \dfrac{m}{c}$ and $m = 12$, or $b = \dfrac{12}{c}$M1
- $a = \dfrac{1}{4}\left(\dfrac{12}{c}\right)^3$ oe (substituting their $b$ into their $a$)M1
- $a = \dfrac{432}{c^3}$ oe, e.g. $a = \dfrac{1728}{4c^3}$A1
- (b) 16 (ft their formula from (a) if of the form $\dfrac{p}{c^3}$)B1
- Answer: (a) $a = \dfrac{432}{c^3}$ (b) 16
- Examiner tip: Use a different letter for each constant – using $k$ for both relationships is where most marks are lost.
The mass of a solid metal ball is directly proportional to the cube of its radius.
A ball of radius 3 cm has a mass of 810 g.
(a) Work out the mass of a ball of radius 5 cm made of the same metal. [3]
(b) Another ball of the same metal has a mass of 6.48 kg.
Work out its radius. [2]
(c) Fiona says, 'If the radius of a ball is doubled, its mass is also doubled.'
Explain why Fiona is wrong. [1]
- (a) $m = kr^3$M1
- $810 = k \times 3^3$ or $k = 30$M1
- 3750 (g) (accept 3.75 kg)A1
- (b) $r^3 = 6480 \div 30$ (= 216) (must convert to grams; ft their $k$)M1
- 6 (cm)A1
- (c) Mass is multiplied by $2^3 = 8$, e.g. $k(2r)^3 = 8kr^3$ or a correct numerical example such as 'radius 6 cm gives 6480 g, which is 8 times 810 g'C1
- Answer: (a) 3750 g (b) 6 cm (c) doubling the radius multiplies the mass by $2^3 = 8$
- Examiner tip: In (b), work in grams to match $k$ – using 6.48 gives $r^3 = 0.216$ and a radius of 0.6 cm.
Still losing marks on this topic?
Book a free introductory session and we'll find exactly which step is costing you.