Hard GCSE Real-life graphs Questions

Challenging, exam-style GCSE Real-life graphs questions with worked solutions. Stretch yourself on the hardest speed-time graph, area under graph, unit conversion, trapezium problems.

speed-time grapharea under graphunit conversiontrapeziumdecompositiondistance-time graph
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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).
Show worked solution

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}
Question 2
6 markschallenging
A distance–time graph for a car is a straight line from (0,0)(0, 0) to (2.5,100)(2.5, 100), where time is in hours and distance is in km. A conversion graph between miles and kilometres is a straight line through (0,0)(0, 0) and (5,8)(5, 8), where xx is a distance in miles and yy is the same distance in kilometres. Use both graphs to work out the speed of the car in miles per hour.
Show worked solution

Worked solution

  1. Sketch the distance–time graph

    (0, 0)→(2.5, 100)(0,\ 0) \rightarrow (2.5,\ 100)

    A single straight line, so the car travels at a constant speed.

  2. Find the gradient of the distance–time graph

    speed=100−02.5−0\text{speed} = \frac{100 - 0}{2.5 - 0}

    Change in distance ÷ change in time.

  3. Work out the division

    1002.5=40\frac{100}{2.5} = 40

    Multiply top and bottom by 2: 2005=40\frac{200}{5} = 40.

  4. State the speed in km/h

    40 km/h40\text{ km/h}

    Kilometres divided by hours. Now it must be changed into miles per hour.

  5. Sketch the conversion graph

    (0, 0)→(5, 8)(0,\ 0) \rightarrow (5,\ 8)

    5 miles is 8 km.

  6. Find the gradient of the conversion line

    gradient=8−05−0=1.6\text{gradient} = \frac{8 - 0}{5 - 0} = 1.6

    Each mile is 1.6 km, so y=1.6xy = 1.6x with xx in miles.

  7. Decide which way to convert

    km known⇒divide by 1.6\text{km known} \Rightarrow \text{divide by }1.6

    We have kilometres and want miles, so read ACROSS from the km axis and DOWN to the miles axis.

  8. Only the distance unit changes

    40 km in 1 hour→( ? ) miles in 1 hour40\text{ km in }1\text{ hour} \rightarrow (\ ?\ )\text{ miles in }1\text{ hour}

    The "per hour" is the same in both units, so leave the time alone.

  9. Form the equation

    1.6x=401.6x = 40

    xx is the number of miles that is the same as 40 km.

  10. Divide by 1.6

    x=401.6x = \frac{40}{1.6}

    Undo the multiplication.

  11. Clear the decimal

    401.6=40016\frac{40}{1.6} = \frac{400}{16}

    Multiply top and bottom by 10.

  12. Work out the division

    40016=25\frac{400}{16} = 25

    16×25=40016 \times 25 = 400. ✓

  13. State the answer with units

    25 mph25\text{ mph}

    The car is travelling at 25 miles per hour.

  14. Check by converting back

    25×1.6=40 km/h ✓25 \times 1.6 = 40\text{ km/h}\ \checkmark

    25 mph really is 40 km/h.

  15. Check the answer is sensible

    25<4025 < 40

    A mile is longer than a km, so the SAME speed is a SMALLER number of mph than of km/h. Multiplying by 1.6 (giving 64) would have been the classic error.

Answer
25 mph25\text{ mph}
Question 3
5 markschallenging
A distance–time graph joins (0,0)(0, 0), (2,80)(2, 80), (3,80)(3, 80) and (5,0)(5, 0), where time is in hours and distance from home is in km. A student says: “Between 22 hours and 33 hours the graph is horizontal at 8080, so the car is travelling at a steady 8080 km/h.” Which statement correctly explains the student’s mistake?
Show worked solution

Worked solution

  1. Sketch the journey

    (0, 0)→(2, 80)→(3, 80)→(5, 0)(0,\ 0) \rightarrow (2,\ 80) \rightarrow (3,\ 80) \rightarrow (5,\ 0)

    Look carefully at the horizontal middle section.

  2. What is on each axis?

    x:time (hours),y:distance from home (km)x: \text{time (hours)},\qquad y: \text{distance from home (km)}

    The vertical axis shows DISTANCE, not speed. This is the heart of the mistake.

  3. Read the two end points of the flat section

    (2, 80) and (3, 80)(2,\ 80) \text{ and } (3,\ 80)

    The distance is 80 km at both ends.

  4. Change in distance

    80−80=0 km80 - 80 = 0\text{ km}

    The car gets no further from home during that hour.

  5. Change in time

    3−2=1 hour3 - 2 = 1\text{ hour}

    A whole hour passes.

  6. Work out the gradient

    gradient=01=0\text{gradient} = \frac{0}{1} = 0

    A horizontal line has zero gradient.

  7. On a distance–time graph the gradient IS the speed

    speed=gradient=0 km/h\text{speed} = \text{gradient} = 0\text{ km/h}

    So the car is not moving at all.

  8. Say what really happens

    the car is parked 80 km from home for 1 hour\text{the car is parked }80\text{ km from home for }1\text{ hour}

    It is stationary — a break in the journey.

  9. Where has the 80 come from?

    80 is a DISTANCE, not a speed80\text{ is a DISTANCE, not a speed}

    The student has read the height of the line (80 km) and wrongly called it a speed.

  10. What would a steady 80 km/h look like?

    a straight line with gradient 80\text{a straight line with gradient }80

    It would be a SLOPING line, climbing 80 km for every hour — not a flat one.

  11. Compare with the first section

    80−02−0=40 km/h\frac{80 - 0}{2 - 0} = 40\text{ km/h}

    The car’s actual speed on the way out was 40 km/h.

  12. Compare with the last section

    0−805−3=−40⇒40 km/h\frac{0 - 80}{5 - 3} = -40 \Rightarrow 40\text{ km/h}

    On the way home it also travels at 40 km/h.

  13. Note where "flat = constant speed" IS true

    on a SPEED–time graph\text{on a SPEED–time graph}

    On a speed–time graph a horizontal line does mean constant speed — but this is a DISTANCE–time graph.

  14. Summarise the misconception

    horizontal on a d–t graph⇒stationary\text{horizontal on a }d\text{–}t\text{ graph} \Rightarrow \text{stationary}

    Flat means the distance is not changing, so the speed is zero.

  15. Choose the correct explanation

    the car is stationary, not moving at 80 km/h\text{the car is stationary, not moving at }80\text{ km/h}

    The 80 is how far from home it is parked, not how fast it is going.

Answer
The car is stationary; a flat d–t graph means zero speed\text{The car is stationary; a flat }d\text{–}t\text{ graph means zero speed}
Question 4
6 markschallenging
A van’s speed–time graph starts at (0,0)(0, 0), rises in a straight line to (6,24)(6, 24) and is then horizontal until it reaches the point (T,24)(T, 24), where time is in seconds and speed is in m/s. The van travels 336336 m in total during these TT seconds. Work out the value of TT.
Show worked solution

Worked solution

  1. Sketch the speed–time graph

    (0, 0)→(6, 24)→(T, 24)(0,\ 0) \rightarrow (6,\ 24) \rightarrow (T,\ 24)

    The van accelerates for 6 seconds and then holds 24 m/s for an unknown length of time.

  2. Total distance = total area

    areatriangle+arearectangle=336\text{area}_{\text{triangle}} + \text{area}_{\text{rectangle}} = 336

    Split the region at t=6t = 6.

  3. The first part is a triangle

    area=12×6×24\text{area} = \tfrac{1}{2} \times 6 \times 24

    Base 6 seconds, height 24 m/s.

  4. Work out the triangle

    12×6×24=72 m\tfrac{1}{2} \times 6 \times 24 = 72\text{ m}

    The van covers 72 m while speeding up.

  5. How much distance is left?

    336−72336 - 72

    Subtract the triangle from the total.

  6. Work out the remaining distance

    336−72=264 m336 - 72 = 264\text{ m}

    The constant-speed part must account for 264 m.

  7. The second part is a rectangle

    area=(T−6)×24\text{area} = (T - 6) \times 24

    Its width is the time spent at constant speed, T−6T - 6 seconds.

  8. Form an equation

    24(T−6)=26424(T - 6) = 264

    The rectangle’s area equals the remaining distance.

  9. Divide both sides by 24

    T−6=26424T - 6 = \frac{264}{24}

    Undo the multiplication.

  10. Work out the division

    26424=11\frac{264}{24} = 11

    24×11=26424 \times 11 = 264. ✓

  11. Solve for TT

    T=11+6T = 11 + 6

    Add the 6 seconds of acceleration back on — this step is easy to forget.

  12. Work out TT

    T=17T = 17

    The graph reaches (17,24)(17, 24).

  13. State the answer with units

    T=17 secondsT = 17\text{ seconds}

    The whole 336 m takes 17 seconds.

  14. Check the total area

    72+24×(17−6)=72+24×11=72+264=336 ✓72 + 24 \times (17 - 6) = 72 + 24 \times 11 = 72 + 264 = 336\ \checkmark

    The two areas add back to the given total distance.

  15. Check with a trapezium instead

    12(11+17)×24=12×28×24=336 ✓\tfrac{1}{2}(11 + 17) \times 24 = \tfrac{1}{2} \times 28 \times 24 = 336\ \checkmark

    Treating the whole region as one trapezium (parallel sides 11 s and 17 s, height 24 m/s) gives the same 336 m. ✓

Answer
T=17 secondsT = 17\text{ seconds}
Question 5
6 markschallenging
Ann and Ben travel along the same road away from home. On one distance–time graph, time is in hours and distance from home is in km. Ann walks: her graph is the straight line from (0,0)(0, 0) to (3,15)(3, 15). Ben sets off an hour later on his bike: his graph is the straight line from (1,0)(1, 0) to (3,30)(3, 30). How far from home is Ben when he catches up with Ann?
Show worked solution

Worked solution

  1. Sketch both journeys on one pair of axes

    Ann: (0, 0)→(3, 15),Ben: (1, 0)→(3, 30)\text{Ann: } (0,\ 0) \rightarrow (3,\ 15),\quad \text{Ben: } (1,\ 0) \rightarrow (3,\ 30)

    Ben’s line starts an hour later but is much steeper, so it will cross Ann’s line.

  2. He catches her where the lines cross

    same time, same distance from home\text{same time, same distance from home}

    The crossing point is the moment they are side by side.

  3. Ann’s speed

    15−03−0=5 km/h\frac{15 - 0}{3 - 0} = 5\text{ km/h}

    The gradient of Ann’s line — a steady walking pace.

  4. Ann’s rule

    d=5td = 5t

    She leaves home at t=0t = 0.

  5. Ben’s speed

    30−03−1=302=15 km/h\frac{30 - 0}{3 - 1} = \frac{30}{2} = 15\text{ km/h}

    The gradient of Ben’s line — three times Ann’s speed.

  6. Ben’s rule

    d=15(t−1)d = 15(t - 1)

    He is still at home (d=0d = 0) when t=1t = 1, so the time since he set off is t−1t - 1.

  7. Check Ben’s rule at t=3t = 3

    15(3−1)=30 ✓15(3 - 1) = 30\ \checkmark

    It reproduces his end point (3,30)(3, 30).

  8. Set the two rules equal

    5t=15(t−1)5t = 15(t - 1)

    At the crossing point the distances are equal.

  9. Expand the bracket

    5t=15t−155t = 15t - 15

    Multiply both terms inside the bracket by 15.

  10. Collect the tt terms

    15=15t−5t15 = 15t - 5t

    Add 15 to both sides and subtract 5t5t from both sides.

  11. Simplify

    15=10t15 = 10t

    Ten lots of tt make 15.

  12. Solve for tt

    t=1510=1.5 hourst = \frac{15}{10} = 1.5\text{ hours}

    Ben catches Ann 1.5 hours after Ann set off — that is, 30 minutes after HE set off.

  13. Substitute back into Ann’s rule

    d=5×1.5=7.5 kmd = 5 \times 1.5 = 7.5\text{ km}

    The question asks for the DISTANCE, not the time, so substitute back.

  14. Check with Ben’s rule

    15(1.5−1)=15×0.5=7.5 ✓15(1.5 - 1) = 15 \times 0.5 = 7.5\ \checkmark

    Both rules give 7.5 km, so the crossing point is (1.5,7.5)(1.5, 7.5).

  15. State the answer with units

    7.5 km from home7.5\text{ km from home}

    Ben catches Ann 7.5 km from home.

Answer
7.5 km7.5\text{ km}

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