Rates of change & proportion graphs – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

The graph can be used to convert between pounds (£) and euros (€).

(a) Use the graph to change £30 into euros. [1]

(b) Use the graph to change €150 into pounds. [2]

(a) €36 (accept 35–37)B1
(b) a correct reading used with a multiplier, e.g. €60 = £50 so $150 \div 60 \times 50$, or €30 = £25 and $\times 5$M1
£125 (accept 122–128)A1
Answer: (a) €36 (b) £125
Examiner tip: €150 is off the graph – read a value you can see (e.g. €30 or €60) and scale it up.

2Grade 4–5Non-calculator4 marks

Raj drives from home to a garden centre, stops there, and then drives home.

The distance–time graph shows his journey.

(a) Work out his speed for the first part of the journey. [1]

(b) For how many minutes did Raj stop at the garden centre? [1]

(a) 40 km/hB1
(b) 30 minutesB1
(c) $40 \div 2$M1
20 km/hA1
Answer: (a) 40 km/h (b) 30 minutes (c) 20 km/h
Examiner tip: Read the time for the return as 1.5 h to 3.5 h – a 2-hour journey, not 3.5 hours.

3Grade 4–5Non-calculator3 marks

Here are sketches of four graphs.

(a) Which graph shows that $y$ is directly proportional to $x$? [1]

(b) Which graph shows that $y$ is inversely proportional to $x$? [1]

(a) AB1
(b) CB1
(c) it does not pass through the origin (when $x = 0$, $y \neq 0$). Allow: it has a $y$-intercept that is not 0; it does not start at (0, 0) / at zero; it crosses the $y$-axis above the origin. Do not accept: "it is not steep enough", "it is a straight line", "the gradient is wrong".C1
Answer: (a) A (b) C (c) the line does not pass through the origin
Examiner tip: A straight line only shows direct proportion if it also passes through (0, 0).

4Grade 6–7HigherNon-calculator4 marks

The velocity–time graph shows the first 30 seconds of a cyclist's journey.

(a) Work out the acceleration of the cyclist during the first 10 seconds. [1]

(b) Work out the total distance travelled in the 30 seconds. [3]

(a) 1.2 m/s²B1
(b) one correct area, e.g. $\frac{1}{2} \times 10 \times 12 = 60$, $15 \times 12 = 180$ or $\frac{1}{2} \times 5 \times 12 = 30$M1
complete method for total area, e.g. $60 + 180 + 30$ or trapezium $\frac{1}{2}(15 + 30) \times 12$M1
270 mA1
Answer: (a) 1.2 m/s² (b) 270 m
Examiner tip: Distance is the AREA under a velocity–time graph; the GRADIENT gives acceleration.

5Grade 4–5Non-calculator4 marks

The graph shows the cost, in pounds, of a taxi journey for different distances.

(a) Work out the gradient of the line. [2]

(b) Explain what the gradient represents. [1]

(a) $\dfrac{\text{change in cost}}{\text{change in distance}}$ using two correct points, e.g. $\dfrac{18.5 - 2.5}{10}$M1
1.6 (accept £1.60 or 1.60; units not required)A1
(b) the cost per mile (e.g. each extra mile costs £1.60). Allow: the cost increases by £1.60 for every mile travelled. Do not accept "the rate", "the gradient", "how steep it is" or "speed" without the cost-per-mile context.C1
(c) the fixed (starting) charge of £2.50, paid before any distance is travelled. Allow: the cost of a 0-mile journey / the booking or call-out fee of £2.50. Do not accept "the $y$-intercept" or "where it starts" alone.C1
Answer: (a) 1.6 (b) cost per mile, £1.60 (c) a fixed charge of £2.50
Examiner tip: Interpretations must be in context – 'the rate' or 'the y-intercept' alone get no credit.

6Grade 6–7HigherCalculator5 marks

Water flows into a tank. The graph shows the volume of water, $V$ litres, in the tank $t$ minutes after the water is turned on.

(a) Work out the average rate of flow over the first 6 minutes. [2]

(b) By drawing a tangent, estimate the rate at which water is flowing into the tank when $t = 4$. [3]

(a) reads $V = 18$ at $t = 6$ (allow 17.5 to 18.5) and finds $18 \div 6$M1
3 litres per minute (accept 2.9 to 3.1 from a reading of $V$ in the range 17.5 to 18.5)A1
(b) a tangent drawn at $t = 4$M1
(dep on previous M1) $\dfrac{\text{difference in } V}{\text{difference in } t}$ using two points on their tangentM1
answer in the range 3.5 to 4.5 litres per minute (dep on a tangent drawn at $t = 4$)A1
Answer: (a) 3 litres per minute (b) approximately 4 litres per minute (3.5–4.5)
Examiner tip: Draw a long, straight tangent that touches the curve only at t = 4, then use a large triangle for the gradient.

7Grade 1–3FoundationNon-calculator5 marks

Mia unplugs her fully charged phone and uses it. The graph shows the battery charge, as a percentage, for the first 4 hours.

(a) Use the graph to find the battery charge after 2 hours. [1]

(b) Work out the rate at which the battery charge goes down. Give the units of your answer. [2]

(c) The battery charge keeps going down at the same rate.
Work out how long it takes, from when Mia unplugs the phone, for the battery charge to reach 0%.
Give your answer in hours and minutes. [2]

(a) 70(%)B1
(b) $\dfrac{\text{change in charge}}{\text{change in time}}$ from two correct points, e.g. $(100 - 40) \div 4$ or $30 \div 2$ or $60 \div 4$M1
15% per hour (oe, e.g. 15 per hour; accept −15% per hour). Units required for this markA1
(c) $100 \div 15$ or $40 \div 15$ (then added to 4), or extends the line to the time axis, or a build-up such as 85, 70, 55, … using their (b)M1
6 hours 40 minutes (accept $6\frac{2}{3}$ hours only if correctly converted; 6.6 or 6.67 hours alone, or 6 h 67 min, scores M1 A0)A1
Answer: (a) 70% (b) 15% per hour (c) 6 hours 40 minutes
Examiner tip: In (c), $6\frac{2}{3}$ hours is 6 hours 40 minutes, not 6 hours 67 minutes: multiply the decimal part by 60.

8Grade 4–5Non-calculator3 marks

The diagram shows the cross-sections of three containers, A, B and C. Each container is empty.

Water is poured into each container at the same constant rate until the container is full.

The graphs show how the depth of the water changes with time.

(a) Match each container to its graph. [2]

(b) Explain why the graph for container C has the shape it does. [1]

(a) at least one correct match (A and Graph 2, B and Graph 3, or C and Graph 1)B1
all three matches correct: A – Graph 2, B – Graph 3, C – Graph 1B1
(b) C is narrow at the bottom and wider at the top, so the depth rises quickly at first and then more and more slowly (the graph gets less steep). Accept: the same volume of water makes a smaller rise in depth where the container is wider. Do not accept just 'it is a curve'C1
Answer: (a) A – Graph 2, B – Graph 3, C – Graph 1 (b) C gets wider going up, so the depth rises more slowly over time
Examiner tip: Link the steepness of the graph to the width of the container: narrow means the depth rises fast (steep), wide means it rises slowly (shallow).

9Grade 1–3FoundationNon-calculator5 marks

The graph shows the volume of water used by two different showers, a power shower and an electric shower, when they are switched on.

(a) How does the graph show that the volume of water used by the power shower is directly proportional to time? [1]

(b) Work out the rate at which the electric shower uses water. [2]

(c) Sam has one 8-minute shower each day for a week.
He uses the power shower instead of the electric shower.
How many more litres of water does Sam use in the week? [2]

(a) it is a straight line that passes through the origin (0, 0). Both 'straight line' and 'through the origin' neededC1
(b) a correct reading divided by its time, e.g. $80 \div 10$ or $40 \div 5$M1
8 litres per minute (units not required for the mark)A1
(c) $96 - 64$ (= 32) or $(12 - 8) \times 8$, or $96 \times 7$ and $64 \times 7$ (672 and 448)M1
224 (litres)A1
Answer: (a) straight line through the origin (b) 8 litres per minute (c) 224 litres
Examiner tip: In (c) find the difference for one shower first, then multiply by 7 days; many students forget the 7.

10Grade 4–5Non-calculator5 marks

Amir walks from a village to a town 12 km away. Beth cycles along the same road from the village to the town.

The distance–time graph shows both of their journeys.

(a) Work out Amir's speed in km/h. [2]

(b) Write down the time at which Beth passes Amir. [1]

(a) $12 \div 3$ (distance ÷ time from a correct pair of readings, e.g. $4 \div 1$)M1
4 (km/h)A1
(b) 10:30 (accept 10.30 am or half past ten)B1
(c) Beth's speed found, e.g. $12 \div 1$ or $6 \div 0.5$ (= 12 km/h); or compares times: Beth takes 1 hour and Amir takes 3 hoursM1
yes, with correct supporting figures, e.g. $12 = 3 \times 4$ or $12 \div 4 = 3$ (ft their (a) if M1 awarded)C1
Answer: (a) 4 km/h (b) 10:30 (c) Yes: Beth's speed is 12 km/h, which is 3 × 4 km/h
Examiner tip: Convert the time to hours before dividing: Beth takes 60 minutes, which is 1 hour, so her speed is 12 km/h, not 12 ÷ 60.

11Grade 4–5Non-calculator5 marks

Ellie drives the same route every week. The time, $t$ hours, that the journey takes depends on her average speed, $s$ km/h.

The graph shows how $t$ changes with $s$.

(a) Use the graph to find how long the journey takes at an average speed of 40 km/h. [1]

(b) Use two points from the graph to show that $t$ is inversely proportional to $s$. [2]

(a) 3 (hours)B1
(b) finds $s \times t$ for at least two correct points, e.g. $20 \times 6$ and $40 \times 3$; or shows doubling $s$ halves $t$ using correct readings, e.g. 20 → 40 gives 6 → 3M1
states the products are the same (120) or 'when $s$ doubles, $t$ halves', so $t$ is inversely proportional to $s$C1
(c) $120 \div 1.25$ (oe, e.g. $120 \div \frac{5}{4}$ or $\frac{120}{75} \times 60$); allow $s \times 1.25 = 120$. $120 \div 1.15$ is M0M1
96 (km/h)A1
Answer: (a) 3 hours (b) e.g. $20 \times 6 = 120$ and $40 \times 3 = 120$, the product is constant (c) 96 km/h
Examiner tip: 1 hour 15 minutes is 1.25 hours, not 1.15 hours; convert the minutes before you divide.

12Grade 6–7HigherCalculator4 marks

The graph shows the value of a car, in pounds, from when it is new until it is 8 years old.

(a) Use the graph to estimate the value of the car when it is 3 years old. [1]

(b) Work out the average rate of decrease in the value of the car between 1 year old and 5 years old. Give the units of your answer. [2]

(a) answer in the range £9000 to £9400B1
(b) $\dfrac{V(1) - V(5)}{5 - 1}$ using readings in the ranges 14 200 to 14 600 and 5700 to 6100, e.g. $(14\,400 - 5900) \div 4$M1
answer in the range 2025 to 2225 with units £ per year (e.g. £2125 per year). Accept a negative rate, e.g. −£2125 per yearA1
(c) the graph is a curve, not a straight line: it gets less steep (the gradient decreases), so the car loses less value each year as it gets older. Accept figures read from the graph showing different losses, e.g. about £3600 in year 1 and about £1800 in year 4C1
Answer: (a) about £9200 (b) about £2125 per year (2025–2225) (c) the curve gets less steep, so the loss each year gets smaller
Examiner tip: An average rate of change is the gradient of the chord: (change in value) ÷ (change in time), and it needs units such as £ per year.

13Grade 6–7HigherCalculator6 marks

A cup of tea is left to cool. The graph shows the temperature of the tea, $T$ °C, $t$ minutes after it is poured.

(a) Work out the average rate of cooling of the tea over the first 10 minutes. [2]

(b) By drawing a tangent, estimate the rate of cooling of the tea when $t = 10$. [3]

(a) $(90 - T_{10}) \div 10$ with $T_{10}$ read in the range 43 to 46M1
answer in the range 4.4 to 4.7 (°C per minute). Accept a negative value, e.g. −4.56A1
(b) a tangent drawn at $t = 10$ (a straight line touching the curve at $t = 10$ and not crossing it there)M1
(dep) $\dfrac{\text{difference in } T}{\text{difference in } t}$ using two points on their tangentM1
answer in the range 2.2 to 3.0 (°C per minute) (accept negative values)A1
(c) the tea cools faster at the start: the curve is steeper between $t = 0$ and $t = 10$ than at $t = 10$, so the average over the first 10 minutes includes faster cooling. Accept: the rate of cooling decreases over time / the curve gets less steepC1
Answer: (a) about 4.6 °C per minute (4.4–4.7) (b) about 2.6 °C per minute (2.2–3.0) (c) the curve is steeper before $t = 10$, so the tea was cooling faster earlier
Examiner tip: A chord gives an average rate; a tangent gives the rate at one instant. Make the tangent long so your gradient triangle is large and accurate.

14Grade 8–9HigherNon-calculator5 marks

A ball is thrown upwards. Its height, $h$ metres, above the ground $t$ seconds after it is thrown is given by

$$h = 20t - 5t^2$$

(a) Work out the average rate of change of $h$ between $t = 0$ and $t = 1$. [2]

(b) Lena works out the average rate of change of $h$ between $t = 1$ and $t = 3$.
She says, “My answer shows that the ball was not moving between $t = 1$ and $t = 3$.”
Work out Lena's average rate of change and explain why her statement is wrong. [2]

(c) Is the rate of change of $h$ at the instant $t = 1$ greater than or less than your answer to part (a)?
Give a reason for your answer. [1]

(a) substitutes to find $h$ at $t = 1$: $20 - 5 = 15$, and uses $\dfrac{15 - 0}{1 - 0}$M1
15 (m/s)A1
(b) 0, from $h = 15$ at both $t = 1$ and $t = 3$B1
the ball goes up and then comes back down to the same height, so the overall change in height is zero even though the ball is moving (the chord is horizontal). Accept: the ball reaches its highest point at $t = 2$ in betweenC1
(c) less than, with a reason, e.g. the ball is slowing down as it rises / the graph of $h$ against $t$ gets less steep, so the tangent at $t = 1$ is less steep than the chord from $t = 0$ to $t = 1$. Accept a correct value of 10 m/s with comparison. 'Less' with no reason scores 0C1
Answer: (a) 15 m/s (b) 0 m/s; the ball rises and falls back to the same height, so it moved even though the net change is 0 (c) less than (it is 10 m/s), because the ball is slowing as it rises
Examiner tip: An average rate of change only compares the start and end values; it tells you nothing about what happens in between.

15Grade 8–9HigherNon-calculator4 marks

The graph shows the velocity of a car, $v$ m/s, during the first 10 seconds after it starts to move.

(a) Use 4 strips of equal width to estimate the distance the car travels in the first 8 seconds. [3]

(b) Is your answer to part (a) an underestimate or an overestimate of the actual distance?
Give a reason for your answer. [1]

(a) reads $v = 0, 7, 12, 15, 16$ at $t = 0, 2, 4, 6, 8$ and finds the area of at least one strip, e.g. $\frac{1}{2} \times 2 \times (0 + 7) = 7$ (allow a triangle for the first strip)M1
complete method for the sum of the 4 areas, e.g. $7 + 19 + 27 + 31$ or $\frac{1}{2} \times 2 \times [0 + 16 + 2(7 + 12 + 15)]$M1
84 (m) (accept 83 to 85 from slight misreadings with a correct method)A1
(b) underestimate, because the curve is above the straight tops of the trapezia (the trapezia miss some of the area under the curve). Accept: the graph curves outwards/is concave down so the straight lines lie below itC1
Answer: (a) 84 m (b) underestimate: the tops of the trapezia lie below the curve
Examiner tip: Strips of equal width means splitting 0 to 8 seconds into four 2-second strips; the first strip is a triangle because the car starts from rest.

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