GCSE Standard and compound units Practice Questions

Free GCSE Standard and compound units practice questions with full step-by-step worked solutions. Covers converting length, metric units, converting mass, converting capacity. Practise exam-style problems and check your method.

converting lengthmetric unitsconverting massconverting capacityconverting timeconverting money
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Convert 55 km into metres.
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Worked solution

  1. Recall the conversion

    1 km=1000 m1\text{ km} = 1000\text{ m}

    There are 1000 metres in every kilometre.

  2. Multiply

    5×1000=50005 \times 1000 = 5000

    Kilometres are larger than metres, so multiply by 1000.

  3. State the answer

    5000 m5000\text{ m}

    So 5 km is 5000 m.

Answer
5000 m5000\text{ m}
Question 2
2 markseasy
A runner runs 6060 m in 1212 s. Work out the runner's speed.
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Worked solution

  1. Write the formula

    speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}

    Speed is distance divided by time.

  2. Substitute the values

    speed=6012\text{speed} = \frac{60}{12}

    The distance is 60 m and the time is 12 s.

  3. Work out and add units

    =5 m/s= 5\text{ m/s}

    60 divided by 12 is 5, measured in metres per second.

Answer
5 m/s5\text{ m/s}
Question 3
2 marksintermediate
A car travels at a speed of 2020 m/s. How far does it travel in 1515 seconds?
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Worked solution

  1. Write the formula

    speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}

    Start from the speed formula.

  2. Rearrange for distance

    distance=speed×time\text{distance} = \text{speed} \times \text{time}

    Multiply both sides by the time.

  3. Note the values

    speed=20, time=15\text{speed} = 20,\ \text{time} = 15

    The speed is 20 m/s and the time is 15 s.

  4. Substitute

    distance=20×15\text{distance} = 20 \times 15

    Put the numbers into the rearranged formula.

  5. Work it out

    20×15=30020 \times 15 = 300

    20 times 15 is 300.

  6. State the answer with units

    300 m300\text{ m}

    The car travels 300 metres.

Answer
300 m300\text{ m}
Question 4
4 markshard
A driver goes 300300 km in 44 hours, then returns the same 300300 km in 66 hours. Work out the average speed for the whole trip.
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Worked solution

  1. Recall the rule

    average speed=total distancetotal time\text{average speed} = \frac{\text{total distance}}{\text{total time}}

    Average speed uses the totals, not the average of the two speeds.

  2. Find the total distance

    300+300=600 km300 + 300 = 600\text{ km}

    The trip is 300 km each way, so 600 km altogether.

  3. Find the total time

    4+6=10 h4 + 6 = 10\text{ h}

    4 hours out and 6 hours back is 10 hours.

  4. Substitute

    average=60010\text{average} = \frac{600}{10}

    Divide the total distance by the total time.

  5. Work it out

    600÷10=60600 \div 10 = 60

    600 divided by 10 is 60.

  6. Attach the units

    60 km/h60\text{ km/h}

    The average speed is 60 km/h.

  7. Check the outward speed

    300÷4=75 km/h300 \div 4 = 75\text{ km/h}

    The speed going was 75 km/h.

  8. Check the return speed

    300÷6=50 km/h300 \div 6 = 50\text{ km/h}

    The speed coming back was 50 km/h.

  9. Note the common error

    75+502=62.5≠60\frac{75 + 50}{2} = 62.5 \ne 60

    Averaging the two speeds gives the wrong answer of 62.5.

  10. State the final answer

    60 km/h60\text{ km/h}

    The correct average speed for the whole trip is 60 km/h.

Answer
60 km/h60\text{ km/h}
Question 5
6 markschallenging
A driver travels 9090 km to a meeting at an average speed of 6060 km/h. For the 9090 km return trip, the driver wants the average speed for the whole round trip to be 7272 km/h. Work out the required speed on the return trip.
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Worked solution

  1. Understand the goal

    whole trip average=72 km/h\text{whole trip average} = 72\text{ km/h}

    The 72 km/h applies to the entire there-and-back journey.

  2. Find the total distance

    90+90=180 km90 + 90 = 180\text{ km}

    The round trip is 90 km each way, so 180 km.

  3. Recall the rule

    average=total distancetotal time\text{average} = \frac{\text{total distance}}{\text{total time}}

    Average speed is total distance over total time.

  4. Rearrange for total time

    total time=total distanceaverage\text{total time} = \frac{\text{total distance}}{\text{average}}

    Divide the total distance by the target average speed.

  5. Substitute

    18072\frac{180}{72}

    Use 180 km and 72 km/h.

  6. Work out the total time

    180÷72=2.5180 \div 72 = 2.5

    The whole trip must take 2.5 hours.

  7. Attach units

    2.5 h2.5\text{ h}

    The total allowed time is 2.5 hours.

  8. Find the outward time

    9060\frac{90}{60}

    The way there is 90 km at 60 km/h.

  9. Work it out

    90÷60=1.590 \div 60 = 1.5

    The outward trip takes 1.5 hours.

  10. Find the return time

    2.5−1.52.5 - 1.5

    Subtract the outward time from the total time.

  11. Work it out

    2.5−1.5=12.5 - 1.5 = 1

    The return trip must take 1 hour.

  12. Find the return speed

    901\frac{90}{1}

    Return speed is 90 km in 1 hour.

  13. Work it out

    90÷1=9090 \div 1 = 90

    The required return speed is 90 km/h.

  14. Check the average

    1801.5+1=72\frac{180}{1.5 + 1} = 72

    180 km in 2.5 hours is 72 km/h, as required.

  15. State the final answer

    90 km/h90\text{ km/h}

    The driver must return at 90 km/h.

Answer
90 km/h90\text{ km/h}

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