IGCSE Maths 0580 Graphs in Practical Situations Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Algebra and Graphs · 2.9

Every graphs in practical situations question from 9 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.9. 9 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/31/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.

0580 Graphs in Practical Situations past paper questions: Core (Papers 1 and 3)

13 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q7

The travel graph shows the journey of a train from Central station to Parkway station (27 km away).

0580/31 May/June 2026 Q7 – travel graph of a train from Central to Parkway, 14 00 to 15 0014 0014 1014 2014 3014 4014 5015 00102030Parkway

(a) For how many minutes does the train stop between Central and Parkway?

(b) Write down the time the train arrives at Parkway.

(c) The train leaves Parkway at 14 33 and returns to Central at a constant speed. The return journey takes 21 minutes. Complete the graph to show this return journey.

Show solution
Part (a)
  1. A stop shows as a flat (horizontal) part of the graph, because the distance does not change.
  2. The flat part runs from 14 10 to 14 17: 7 minutes
Answer 7 minutes
Part (b)
  1. Find where the graph first reaches 27 km (Parkway) and read the time.
  2. The train arrives at 14 27
Answer 14 27
Part (c)
  1. Work out the arrival time: 14 33 + 21 minutes = 14 54.
  2. Draw a straight ruled line from Parkway at 14 33 down to Central at 14 54: (14 33, 27 km) to (14 54, 0 km)
0580/31 May/June 2026 Q7(c) solution – travel graph with the return journey drawn14 0014 1014 2014 3014 4014 5015 0010203014 33 → 14 54
Answer (14 33, 27 km) to (14 54, 0 km)
26 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q15

Tom leaves Alton at 08 00 and drives to Newtown, 50 km away. The travel graph shows Tom’s journey.

0580/33 May/June 2026 Q15 – travel graph of Tom's journey from Alton to Newtown, 08 00 to 11 0008 0008 3009 0009 3010 0010 3011 001020304050

(a) Find how long it takes Tom to reach Newtown. Give your answer in hours.

(b) Find Tom’s average speed, in km/h, from Alton to Newtown.

(c) Mei leaves Alton at 08 30 and drives to Newtown at a constant speed of 40 km/h. (i) Draw a line on the travel graph to show Mei’s journey. (ii) Find the distance between Tom and Mei at 09 00.

Show solution
Part (a)
  1. Read the time when the graph reaches 50 km: 10 54.
  2. From 08 00 to 10 54 is 2 hours 54 minutes.
  3. Change the minutes to hours: 54 ÷ 60 = 0.9. So the time is 2.9 hours
Answer 2.9 hours
Part (b)
  1. Average speed = distance ÷ time = 50 ÷ 2.9 = 17.24…
  2. 17.2 km/h
Answer 17.2 km/h
Part (c)(i)
  1. Time for Mei = distance ÷ speed = 50 ÷ 40 = 1.25 hours = 1 hour 15 minutes.
  2. She arrives at 08 30 + 1 h 15 min = 09 45.
  3. Rule a straight line from (08 30, 0 km) to (09 45, 50 km)
0580/33 May/June 2026 Q15(c) solution – travel graph with Mei's journey drawn08 0008 3009 0009 3010 0010 3011 001020304050MeiTom
Answer (08 30, 0 km) to (09 45, 50 km)
Part (c)(ii)
  1. At 09 00 Mei has driven for half an hour: 40 × 0.5 = 20 km.
  2. Read Tom’s distance at 09 00 from the graph: 27 km.
  3. Distance between them = 27 − 20 = 7 km
Answer 7 km
35 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q16

Bob travels from town A to town B, 625 km away. The travel graph shows his journey, starting at 08 00 and arriving at 14 15.

0580/31 October/November 2025 Q16 – travel graph of Bob's journey from 08 00 to 15 0008 0009 0010 0011 0012 0013 0014 0015 00100200300400500600

(a) Between which two times did Bob stop for a rest? Explain how you know.

(b) Calculate Bob’s average speed, in km/h, for the whole journey.

Show solution
Part (a)
  1. A rest shows as a flat (horizontal) part of the travel graph.
  2. The flat part runs from 10 30 and 11 15
  3. Reason: the graph is horizontal, so the distance from town A does not change
Answer 10 30 and 11 15 the graph is horizontal, so the distance from town A does not change
Part (b)
  1. Total time = 08 00 to 14 15 = 6 hours 15 minutes = 6.25 hours.
  2. Average speed = total distance ÷ total time = 625 ÷ 6.25 = 100 km/h
Answer 100 km/h
43 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q28

The travel graph shows Chi’s journey from home to the station, 5.4 km away, starting at 08 00 and arriving at 08 36. Calculate Chi’s average speed for the whole journey. Give your answer in kilometres per hour.

0580/33 October/November 2025 Q28 – travel graph of Chi's journey from home to the station08 0008 1008 2008 3008 40123456Station
Show solution
  1. The journey takes from 08 00 to 08 36, which is 36 minutes.
  2. Change to hours: 36 ÷ 60 = 0.6 hours.
  3. Average speed = distance ÷ time = 5.4 ÷ 0.6 = 9 km/h
Answer 9 km/h
55 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q10

Ky cycles from his office to a meeting and back again. The travel graph shows his time at the meeting and his journey back.

0580/11 May/June 2025 Q10 – travel graph of distance from office (km) against time from 09 00 to 12 0009 0010 0011 0012 00123456789101112131415Distance (km)Time

(a) How far is the meeting from his office?

(b) How long is Ky at the meeting for?

(c) Write down the time Ky arrives back at his office after the meeting.

(d) Ky cycles from his office to the meeting at a constant speed of 21 km/h. Complete the travel graph.

Show solution
Part (a)
  1. The horizontal part of the graph shows where Ky stays still at the meeting. Read its height on the distance axis.
  2. Distance = 14 km
Answer 14 km
Part (b)
  1. The horizontal part starts at 10 20 and ends at 11 00.
  2. Time at the meeting = 11 00 − 10 20 = 40 minutes
Answer 40 minutes
Part (c)
  1. Follow the return line down until the distance is 0 (back at the office). Read the time.
  2. He arrives back at 11 50
Answer 11 50
Part (d)
  1. Time = distance ÷ speed = 14 ÷ 21 = 2⁄3 hour.
  2. 2⁄3 of 60 minutes = 40 minutes.
  3. He reaches the meeting at 10 20, so he left 40 minutes earlier, at 09 40.
  4. Draw a straight ruled line from (09 40, 0) to (10 20, 14).
0580/11 May/June 2025 Q10 solution – completed travel graph with line from 09 40 to 10 2009 0010 0011 0012 00123456789101112131415Distance (km)Time09 40 → 10 20
Answer Draw a straight ruled line from (09 40, 0) to (10 20, 14).

0580 Graphs in Practical Situations past paper questions: Extended (Papers 2 and 4)

62 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q10(a)

Llew cycles to school. The speed–time graph shows the first 10 seconds of his journey. By drawing a suitable tangent, estimate the acceleration at t = 2.

0580/43 May/June 2026 Q10 – curved speed–time graph for the first 10 secondst0246810123
Show solution
  1. Rule a straight line (a tangent) that just touches the curve at t = 2 without crossing it.
  2. Pick two points far apart on the tangent and read their coordinates.
  3. Gradient = change in speed⁄change in time. This gradient is the acceleration.
0580/43 May/June 2026 Q10(a) solution – tangent drawn to the curve at t = 2t0246810123
  1. Acceleration ≈ 0.3 m/s²
Note: Any value from 0.15 to 0.4 is accepted, depending on your tangent.
Answer 0.3 m/s²
75 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q11

The speed–time graph shows 65 seconds of a journey. The speed increases steadily from 20 m/s to V m/s in the first 48 seconds, then decreases steadily to 0 at 65 seconds.

0580/42 February/March 2026 Q11 – speed–time graph from 20 m/s rising to V m/s at 48 s and falling to 0 at 65 s20V4865time (seconds)speed (m/s)NOT TOSCALE

(a) The acceleration during the first 48 seconds is 0.25 m/s². Show that V = 32 m/s.

(b) Work out the total distance travelled during the 65 seconds. Give your answer in kilometres.

Show solution
Part (a)
  1. Acceleration = change in speed⁄time, so V − 20⁄48 = 0.25.
  2. Multiply both sides by 48: V − 20 = 12.
  3. Add 20: V = 32 ✓
Answer V = 32
Part (b)
  1. Distance travelled = area under the speed–time graph. Split it into two shapes.
  2. 0 to 48 s is a trapezium with parallel sides 20 and 32: 1⁄2 × (20 + 32) × 48 = 26 × 48 = 1248 m.
  3. 48 to 65 s is a triangle with base 65 − 48 = 17 and height 32: 1⁄2 × 17 × 32 = 272 m.
  4. Total = 1248 + 272 = 1520 m.
  5. 1 km = 1000 m, so 1520 m = 1.52 km
Answer 1.52 km
84 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q18

The speed–time graph shows an athlete speeding up steadily from 0 to 8 m/s in the first 2 seconds, then running at 8 m/s up to 10 seconds.

0580/41 October/November 2025 Q18 – speed–time graph rising to 8 m/s at 2 s, then level to 10 s2108Time (seconds)Speed (m/s)NOT TOSCALE

(a) Calculate the distance the athlete runs in the first 10 seconds.

(b) The race is 100 m. After 10 seconds, the athlete continues at 8 m/s until the end. Calculate the total time taken.

Show solution
Part (a)
  1. Distance travelled = area under a speed–time graph.
  2. First 2 seconds (triangle): 1⁄2 × base × height = 1⁄2 × 2 × 8 = 8 m.
  3. From 2 s to 10 s (rectangle): 8 s × 8 m/s = 64 m.
  4. Total distance = 8 + 64 = 72 m
Answer 72 m
Part (b)
  1. Distance still to run = 100 − 72 = 28 m.
  2. Time = distance ÷ speed = 28 ÷ 8 = 3.5 s.
  3. Total time = 10 + 3.5 = 13.5 s
Answer 13.5 s
93 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q9

The speed–time graph shows part of a car journey: the speed changes steadily from 60 km/h to 80 km/h over the first 15 minutes, then falls steadily to 0 at 25 minutes. Find the total distance travelled in the 25 minutes.

0580/42 October/November 2025 Q9 – speed–time graph, 60 to 80 km/h over 15 minutes then down to 0 at 25 minutes60801525time (minutes)speed (km/h)NOT TOSCALE
Show solution
  1. Distance travelled = area under a speed–time graph.
  2. The speed is in km/h, so change the times into hours: 15 minutes = 0.25 h and 10 minutes = 10⁄60 h.
  3. 0 to 15 minutes is a trapezium: 1⁄2 × (60 + 80) × 0.25 = 70 × 0.25 = 17.5 km.
  4. 15 to 25 minutes is a triangle: 1⁄2 × 80 × 10⁄60 = 6.666… km.
  5. Total = 17.5 + 6.666… = 24.16…
  6. 24.2 km
Answer 24.2 km

Question papers used on this page

All graphs in practical situations questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q7
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q15
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q10(a)
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q11
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q16
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q28
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q18
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q9
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q10

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