Real-life graphs, gradient as rate, area under a graph – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

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

(a) Use the graph to change £20 into euros. [1]

(b) Gemma has €230. Use the graph to work out how much this is in pounds. [2]

(a) €23 (accept €22–€24)B1
(b) for a correct method using values read from the graph, e.g. €115 = £100 so €230 = $2 \times 100$, or €23 = £20 so $230 \div 23 \times 20$M1
(b) £200 (accept £195–£205)A1
Answer: (a) €23 (b) £200
Examiner tip: €230 is off the graph – read a value you CAN see (e.g. €115 → £100) and scale it up.

2Grade 4–5Non-calculator4 marks

Priya cycled from her home to a shop. She stayed at the shop and then cycled home.

The distance–time graph shows her journey.

(a) How far from home was Priya after 10 minutes? [1]

(b) How long did Priya stay at the shop? [1]

(a) 3 kmB1
(b) 15 minutesB1
(c) for $6 \div 30$ (km per minute) or $6 \div 0.5$, i.e. using 6 km and 30 minutesM1
(c) 12 km/hA1
Answer: (a) 3 km (b) 15 minutes (c) 12 km/h
Examiner tip: 30 minutes is 0.5 hours, not 0.3 – convert minutes to hours by dividing by 60.

3Grade 4–5Non-calculator3 marks

Water is drained from a tank at a constant rate. The graph shows the depth of water in the tank.

(a) Work out the rate at which the depth of the water decreases. Give your answer in cm per minute. [2]

(b) Explain what the point where the graph meets the vertical axis represents. [1]

(a) for a gradient method using two points on the line, e.g. $\dfrac{80 - 20}{40}$M1
(a) $1.5$ cm per minuteA1
(b) the depth of the water when draining started (at time 0), i.e. 80 cmC1
Answer: (a) 1.5 cm/min (b) the initial depth of water, 80 cm
Examiner tip: The gradient of a real-life graph is a RATE: (units up) per (unit across) – here cm per minute.

4Grade 6–7HigherNon-calculator5 marks

The velocity–time graph shows the first 40 seconds of a car's journey.

(a) Work out the acceleration of the car during the first 10 seconds. [2]

(b) Work out the total distance travelled by the car in the 40 seconds. [3]

(a) for $\dfrac{20}{10}$M1
(a) $2\text{ m/s}^2$A1
(b) for finding the area of one section, e.g. $\tfrac12 \times 10 \times 20 = 100$ or $20 \times 20 = 400$M1
(b) for a complete method for the total area, e.g. $\tfrac12(20 + 40) \times 20$ or $100 + 400 + 100$M1
(b) 600 mA1
Answer: (a) $2\text{ m/s}^2$ (b) 600 m
Examiner tip: On a velocity–time graph the AREA is the distance and the GRADIENT is the acceleration – don't mix them up.

5Grade 8–9HigherNon-calculator6 marks

The graph shows the velocity of a particle, in m/s, during the first 9 seconds of its motion.

(a) Work out an estimate for the distance travelled by the particle in the first 8 seconds. Use 4 strips of equal width. [3]

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

(a) for reading the velocities at $t = 0, 2, 4, 6, 8$: $0, 8, 12, 12, 8$M1
(a) for a correct method for the area of at least 3 of the 4 strips (trapezia, or a triangle and trapezia), e.g. $\tfrac12 \times 2 \times (8 + 12)$M1
(a) 72 mA1
(b) underestimate, as the trapezia (straight tops) lie below the curveC1
(c) for drawing a tangent to the curve at $t = 2$ and a method to find its gradientM1
(c) answer in the range $2.5$ to $3.5\text{ m/s}^2$A1
Answer: (a) 72 m (b) underestimate (c) about $3\text{ m/s}^2$
Examiner tip: For acceleration at an instant you must draw a TANGENT – a chord from the origin to $(2, 8)$ gives the average acceleration (4), not the answer.

6Grade 6–7HigherNon-calculator5 marks

A ball rolls down a slope. The distance–time graph shows the distance the ball has travelled.

(a) Work out the average speed of the ball between $t = 1$ and $t = 4$ seconds. [2]

(b) Work out an estimate of the speed of the ball at $t = 3$ seconds. [3]

(a) for $\dfrac{16 - 1}{4 - 1}$ (allow their readings of the distance at $t = 1$ and $t = 4$, each within $\pm 0.5$ m)M1
(a) 5 m/s (accept 4.8 to 5.2)A1
(b) for drawing a tangent to the curve at $t = 3$M1
(b) for a correct method to find the gradient of their tangent, e.g. $\dfrac{\text{change in distance}}{\text{change in time}}$ using two points on itM1
(b) answer in the range $5.5$ to $6.5$ m/sA1
Answer: (a) 5 m/s (b) about 6 m/s
Examiner tip: Average speed uses a chord between two points; speed at an instant uses a tangent that just touches the curve.

7Grade 1–3FoundationNon-calculator4 marks

A taxi company works out the cost of a journey using the graph.

(a) Use the graph to find the cost of a 7-mile journey. [1]

(b) The graph meets the vertical axis at £3. What does this £3 represent? [1]

(a) £17B1
(b) a fixed (starting / booking) charge paid before any miles are travelled, e.g. 'the cost for 0 miles' or 'the charge just for getting in the taxi'B1
(c) for a correct method using two points on the line, e.g. $(23-3)\div 10$ or $(17-3)\div 7$ (allow difference in cost ÷ difference in miles for any two correctly read points)M1
(c) £2 (per mile). Do not accept £2.30 from $23\div10$ or £2.43 from $17\div7$ (these ignore the fixed charge)A1
Answer: (a) £17 (b) the fixed charge (cost before any miles) (c) £2 per mile
Examiner tip: The cost of each extra mile is the gradient, not total cost ÷ miles: dividing £23 by 10 forgets the £3 fixed charge.

8Grade 1–3FoundationNon-calculator2 marks

Water is poured at a constant rate into three empty containers, P, Q and R.

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

Match each container to its graph. [2]

at least one correct match from P → Graph 2, Q → Graph 3, R → Graph 1B1
all three correct: P → Graph 2, Q → Graph 3, R → Graph 1B1
Answer: P → Graph 2, Q → Graph 3, R → Graph 1
Examiner tip: Narrow part of a container = depth rises quickly = steep graph; a container that gets wider as it fills gives a graph that gets less steep.

9Grade 4–5Non-calculator5 marks

Two plumbers, A and B, charge for their work in different ways.

The graph shows the cost, in pounds, of hiring each plumber for up to 8 hours.

(a) Write down the cost of hiring plumber A for 3 hours. [1]

(b) For how many hours do the two plumbers charge the same amount? [1]

(d) What does the gradient of the line for plumber A represent? [1]

(a) £100B1
(b) 4 (hours)B1
(c) reads both costs at 6 hours: £160 (A) and £180 (B)M1
(c) plumber A, cheaper by £20 (both plumber and amount needed)A1
(d) the amount plumber A charges per hour (£20 per hour); accept 'the hourly rate'. Do not accept just 'the cost'B1
Answer: (a) £100 (b) 4 hours (c) Plumber A, by £20 (d) the hourly rate of plumber A (£20 per hour)
Examiner tip: When a question asks 'which and by how much', you need both the name and the difference to get the final mark.

10Grade 4–5Non-calculator5 marks

Ali walks from Town A to Town B. Town B is 12 km from Town A.

Ben cycles along the same road from Town B to Town A.

The distance–time graph shows both journeys.

(a) Work out Ben's speed. Give your answer in km/h. [2]

(b) At what time do Ali and Ben pass each other? [1]

(d) Ali says, “Ben's speed is three times my speed.” Is Ali correct? You must show how you get your answer. [1]

(a) for $12 \div 1$ (12 km in 1 hour), or e.g. 6 km in 30 minutes scaled to an hourM1
(a) 12 (km/h)A1
(b) 10:30 (accept 10:30 am)B1
(c) 6 kmB1
(d) Yes, with Ali's speed shown as 4 km/h (e.g. $12\div3=4$) and $12 = 3\times 4$ or $12\div4=3$ (ft their (a) if 'Yes/No' is consistent)C1
Answer: (a) 12 km/h (b) 10:30 (c) 6 km (d) Yes: Ali's speed is 4 km/h and 12 = 3 × 4
Examiner tip: Read the time scale carefully: each small square is 10 minutes, so check the gradient uses hours, not squares.

11Grade 4–5Non-calculator3 marks

Here are four sketch graphs.

(a) The total cost, $y$ pounds, of $x$ litres of petrol is directly proportional to the number of litres. Which graph shows this? [1]

(b) A car travels 120 km. The time taken, $y$ hours, is inversely proportional to the average speed, $x$ km/h. Which graph could show this? [1]

(c) Kofi says, “Graph B shows that $y$ is directly proportional to $x$, because it is a straight line.” Explain why Kofi is wrong. [1]

(a) Graph AB1
(b) Graph CB1
(c) the line does not pass through the origin, e.g. 'when $x = 0$, $y$ is not 0' or 'doubling $x$ does not double $y$'C1
Answer: (a) Graph A (b) Graph C (c) Graph B does not go through the origin (0, 0), so y is not directly proportional to x
Examiner tip: Direct proportion needs a straight line AND it must go through the origin; inverse proportion is a curve that never touches either axis.

12Grade 6–7HigherNon-calculator6 marks

A train travels between two stations.

The speed–time graph shows its journey. The train accelerates uniformly for 20 seconds to a speed of $V$ m/s, travels at $V$ m/s for 60 seconds, then decelerates uniformly and stops 120 seconds after it set off.

The distance between the two stations is 1800 m.

(a) Work out the value of $V$. [3]

(b) Work out the deceleration of the train. [2]

(a) for an expression for the area under the graph in terms of $V$, e.g. $\tfrac12\times20\times V + 60V + \tfrac12\times40\times V$ or trapezium $\tfrac12(60+120)V$M1
(a) for equating to 1800, e.g. $90V = 1800$M1
(a) $V = 20$A1
(b) for $V \div 40$ (ft their $V$), e.g. $20\div40$M1
(b) $0.5\text{ m/s}^2$ (accept $-0.5\text{ m/s}^2$ given as an acceleration)A1
(c) 15 (m/s) from $1800\div120$ (units not required)B1
Answer: (a) $V = 20$ (b) $0.5\text{ m/s}^2$ (c) 15 m/s
Examiner tip: Distance is the area under a speed–time graph: write the area in terms of V before you set it equal to 1800.

13Grade 6–7HigherCalculator5 marks

A cup of tea is left to cool. The graph shows the temperature of the tea, in °C, during the first 20 minutes.

(a) Write down the temperature of the tea at the start. [1]

(b) Use the graph to estimate the time when the temperature of the tea is 40 °C. [1]

(d) Is the rate of decrease of the temperature at 15 minutes greater than or less than your answer to (c)? Give a reason for your answer. [1]

(a) 90 (°C)B1
(b) answer in the range 7.4 to 8 minutesB1
(c) for $(90 - T_{10}) \div 10$ where $T_{10}$ is read in the range 33 to 35, e.g. $(90-34)\div10$M1
(c) answer in the range 5.5 to 5.7 with units °C per minute (accept °C/min)A1
(d) less, with a reason, e.g. 'the graph is less steep at 15 minutes than the chord from 0 to 10 minutes' or 'the temperature changes very little after 15 minutes'C1
Answer: (a) 90 °C (b) about 7.7 minutes (7.4–8) (c) about 5.6 °C per minute (d) less, because the curve is less steep at 15 minutes
Examiner tip: Average rate of change over an interval is the gradient of the chord: change in temperature ÷ change in time, not the temperature at 10 minutes ÷ 10.

14Grade 8–9HigherCalculator5 marks

Water flows into a tank. The graph shows the rate of flow of the water, in litres per minute, during the first 6 minutes.

(a) Use 3 strips of equal width to work out an estimate for the volume of water that flows into the tank in the first 6 minutes. [3]

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

(a) reads values at 0, 2, 4, 6 minutes: 18, 16, 10, 0 (allow ±0.5 on each reading), or finds the area of one correct strip, e.g. $\tfrac12\times2\times(18+16)=34$M1
(a) for a complete method for the total area of the 3 trapezia (triangle for the last strip), e.g. $\tfrac12\times2\times(18+16) + \tfrac12\times2\times(16+10) + \tfrac12\times2\times10$M1
(a) 70 (litres); accept answers in the range 68 to 72 from readings within toleranceA1
(b) underestimate, because the curve lies above the tops of the strips (the straight lines are below the curve), so some area is missedC1
(c) their (a) $\div 6$, e.g. 11.7 litres per minute (accept 11.6 to 11.7; ft their (a))B1
Answer: (a) 70 litres (b) underestimate – the curve is above the tops of the trapezia (c) about 11.7 litres per minute
Examiner tip: Strips of equal width from 0 to 6 with 3 strips means width 2: read the graph at 0, 2, 4 and 6 and remember the last strip is a triangle.

15Grade 8–9HigherCalculator6 marks

A coach makes a journey of 180 km.

The graph shows the time taken, $t$ hours, for different average speeds, $s$ km/h.

(a) Use the graph to find the time taken when the average speed is 45 km/h. [1]

(b) By drawing a tangent, work out an estimate for the gradient of the graph when $s = 30$. [3]

(d) Using the graph, explain why increasing the average speed from 80 km/h to 90 km/h saves less time than increasing it from 30 km/h to 40 km/h. [1]

(a) 4 (hours)B1
(b) a tangent drawn to the curve at $s = 30$B1
(b) for vertical change ÷ horizontal change using two points on their tangentM1
(b) answer in the range $-0.24$ to $-0.16$ (must be negative)A1
(c) e.g. 'at 30 km/h, the journey time decreases by about 0.2 hours (12 minutes) for every 1 km/h increase in average speed' (ft their (b); must refer to time decreasing per km/h)C1
(d) the graph is less steep (gradient closer to 0) between 80 and 90 than between 30 and 40, so the same 10 km/h increase gives a smaller drop in time; or quotes readings, e.g. 6 − 4.5 = 1.5 hours compared with 2.25 − 2 = 0.25 hoursC1
Answer: (a) 4 hours (b) about −0.2 (c) at 30 km/h the time falls by about 0.2 hours for each extra 1 km/h (d) the curve is much flatter between 80 and 90 km/h, so the time falls by less (0.25 h compared with 1.5 h)
Examiner tip: A curve going down has a negative gradient – a positive answer in (b) loses the accuracy mark, and the units of the gradient here are hours per km/h.

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