GCSE Real-life graphs Practice Questions

Free GCSE Real-life graphs practice questions with full step-by-step worked solutions. Covers distance-time graph, gradient as speed, interpreting flat sections, fractions of an hour. Practise exam-style problems and check your method.

distance-time graphgradient as speedinterpreting flat sectionsfractions of an hourconversion graphreading a graph
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
On a distance–time graph for Ravi’s car journey, a straight line joins (0,0)(0, 0) to (2,100)(2, 100), where time is measured in hours and distance from home in km. Work out the speed of the car.
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Worked solution

  1. Sketch the graph from the description

    (0, 0)→(2, 100)(0,\ 0) \rightarrow (2,\ 100)

    The graph is one straight line from the origin to 100 km after 2 hours, so the speed is constant.

  2. Use gradient = speed

    speed=change in distancechange in time=100−02−0\text{speed} = \frac{\text{change in distance}}{\text{change in time}} = \frac{100 - 0}{2 - 0}

    On a distance–time graph the gradient tells you how much distance is covered per unit of time — that is the speed.

  3. State the answer with units

    1002=50 km/h\frac{100}{2} = 50\text{ km/h}

    Distance is in km and time is in hours, so the speed comes out in km/h.

Answer
50 km/h50\text{ km/h}
Question 2
1 markeasy
Water is poured at a constant rate into an empty cylinder (a container with the same width all the way up). Which statement describes the depth–time graph?
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Worked solution

  1. Think about equal time intervals

    same width⇒same volume per cm of depth\text{same width} \Rightarrow \text{same volume per cm of depth}

    Because the cylinder has the same cross-section all the way up, each extra litre raises the depth by the same amount.

  2. So the depth rises at a constant rate

    depth=(rate)×(time)\text{depth} = (\text{rate}) \times (\text{time})

    Equal amounts of water arrive in equal times, so the depth goes up in equal steps.

  3. Identify the shape of the graph

    straight line through the origin\text{straight line through the origin}

    Constant rate of change means a straight line; the cylinder starts empty so it passes through (0,0)(0, 0).

Answer
A straight line through the origin\text{A straight line through the origin}
Question 3
2 marksintermediate
A conversion graph between pounds (£) and US dollars is a straight line through (0,0)(0, 0) and (4,5)(4, 5), where xx is an amount in pounds and yy is the same amount in dollars. A jacket costs 8080 dollars. Use the graph to find its cost in pounds.
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Worked solution

  1. Sketch the conversion line

    (0, 0)→(4, 5)(0,\ 0) \rightarrow (4,\ 5)

    £4 is worth 5 dollars.

  2. Find the gradient (the exchange rate)

    gradient=5−04−0=1.25\text{gradient} = \frac{5 - 0}{4 - 0} = 1.25

    Every £1 is worth 1.25 dollars.

  3. Write the rule of the line

    y=1.25xy = 1.25x

    Dollars = 1.25 × pounds.

  4. Substitute the known dollar amount

    1.25x=801.25x = 80

    We are told the dollars (y=80y = 80) and want the pounds (xx).

  5. Divide by 1.25

    x=801.25x = \frac{80}{1.25}

    Going from dollars to pounds means dividing by the rate, not multiplying.

  6. State the answer

    801.25=64⇒£64\frac{80}{1.25} = 64 \Rightarrow \pounds 64

    Check: 64×1.25=8064 \times 1.25 = 80 dollars. ✓

Answer
£64\pounds 64
Question 4
3 markshard
A speed–time graph for a train is a straight line from (0,5)(0, 5) to (10,35)(10, 35), where time is in seconds and speed is in m/s. Work out the speed of the train after 44 seconds.
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Worked solution

  1. Sketch the graph

    (0, 5)→(10, 35)(0,\ 5) \rightarrow (10,\ 35)

    The train is already moving at 5 m/s when the timing starts.

  2. Find the gradient (the acceleration)

    acceleration=35−510−0=3010=3\text{acceleration} = \frac{35 - 5}{10 - 0} = \frac{30}{10} = 3

    The speed rises by 3 m/s every second.

  3. Write the rule of the line

    v=3t+5v = 3t + 5

    Gradient 3, intercept 5 — the starting speed.

  4. Substitute t=4t = 4

    v=3×4+5v = 3 \times 4 + 5

    Go up from 4 on the time axis to the line, then across to the speed axis.

  5. Multiply first

    3×4=12 m/s gained3 \times 4 = 12\text{ m/s gained}

    In 4 seconds the train gains 12 m/s of speed.

  6. Add the starting speed

    12+5=1712 + 5 = 17

    Forgetting the initial 5 m/s is the classic error.

  7. State the answer with units

    17 m/s17\text{ m/s}

    After 4 seconds the train is travelling at 17 m/s.

  8. Check the point lies on the line

    (4, 17) is between (0, 5) and (10, 35)(4,\ 17)\ \text{is between}\ (0,\ 5)\ \text{and}\ (10,\ 35)

    It sits below the final speed of 35 m/s and above the starting speed of 5 m/s. ✓

  9. Check with the acceleration

    5+3×10=35 ✓5 + 3 \times 10 = 35\ \checkmark

    The same rule reproduces the end point of the line.

  10. Interpret in context

    steady acceleration of 3 m/s2\text{steady acceleration of }3\text{ m/s}^2

    A straight speed–time graph always means a constant acceleration.

Answer
17 m/s17\text{ m/s}
Question 5
6 markschallenging
A distance–time graph for a sales rep joins (0,0)(0, 0), (1,60)(1, 60), (1.5,60)(1.5, 60), (2.5,120)(2.5, 120), (3,120)(3, 120) and (5,0)(5, 0), where time is in hours and distance from the office is in km. Work out her average speed while she is actually moving (that is, ignoring the time spent stopped).
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Worked solution

  1. Sketch the whole journey

    (0, 0)→(1, 60)→(1.5, 60)→(2.5, 120)→(3, 120)→(5, 0)(0,\ 0) \rightarrow (1,\ 60) \rightarrow (1.5,\ 60) \rightarrow (2.5,\ 120) \rightarrow (3,\ 120) \rightarrow (5,\ 0)

    Two driving legs away from the office, two stops, then a long drive back.

  2. Leg 1: distance and time

    60 km in 1 hour60\text{ km in }1\text{ hour}

    From (0,0)(0, 0) to (1,60)(1, 60) — she is moving.

  3. Leg 2 is a stop

    (1, 60)→(1.5, 60):gradient=0(1,\ 60) \rightarrow (1.5,\ 60): \text{gradient} = 0

    Flat line — she is stationary for 0.5 hours, and covers 0 km.

  4. Leg 3: distance and time

    120−60=60 km in 2.5−1.5=1 hour120 - 60 = 60\text{ km in } 2.5 - 1.5 = 1\text{ hour}

    From (1.5,60)(1.5, 60) to (2.5,120)(2.5, 120) — moving again.

  5. Leg 4 is another stop

    (2.5, 120)→(3, 120):gradient=0(2.5,\ 120) \rightarrow (3,\ 120): \text{gradient} = 0

    Another flat section, 0.5 hours long, 0 km covered.

  6. Leg 5: distance and time

    120−0=120 km in 5−3=2 hours120 - 0 = 120\text{ km in } 5 - 3 = 2\text{ hours}

    The drive home from 120 km back to the office.

  7. Total distance travelled

    60+0+60+0+12060 + 0 + 60 + 0 + 120

    The return leg counts as distance travelled, even though the distance FROM the office is falling.

  8. Work out the total distance

    60+60+120=240 km60 + 60 + 120 = 240\text{ km}

    240 km of driving in all.

  9. Total MOVING time

    1+1+2=4 hours1 + 1 + 2 = 4\text{ hours}

    Add only the three sloping legs; leave out the two flat sections.

  10. Check against the total time

    5 hours total−0.5−0.5=4 hours moving5\text{ hours total} - 0.5 - 0.5 = 4\text{ hours moving}

    The two half-hour stops take one hour out of the five. ✓

  11. Average speed while moving

    total distancemoving time=2404\frac{\text{total distance}}{\text{moving time}} = \frac{240}{4}

    This is what the question asks for.

  12. Work out the division

    2404=60\frac{240}{4} = 60

    Kilometres divided by hours.

  13. State the answer with units

    60 km/h60\text{ km/h}

    While she is actually driving, she averages 60 km/h.

  14. Compare with the overall average

    2405=48 km/h\frac{240}{5} = 48\text{ km/h}

    Including the stops, the average would only be 48 km/h — the stops drag it down.

  15. Check each leg speed is consistent

    60, 60, 1202=6060,\ 60,\ \frac{120}{2} = 60

    Every moving leg is at 60 km/h, so an average of exactly 60 km/h for the moving time is right. ✓

Answer
60 km/h60\text{ km/h}

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