Hard GCSE Standard and compound units Questions

Challenging, exam-style GCSE Standard and compound units questions with worked solutions. Stretch yourself on the hardest compound units, converting speed, speed, compound measures problems.

compound unitsconverting speedspeedcompound measuresconverting timedensity
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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}
Question 2
6 markschallenging
Rice is sold as: Shop A, 22 kg for £3.60\pounds 3.60; Shop B, 500500 g for £0.85\pounds 0.85; Shop C, 1.51.5 kg for £2.85\pounds 2.85. Which shop gives the best value per kilogram?
Show worked solution

Worked solution

  1. Decide the comparison

    compare price per kg\text{compare price per kg}

    Find the price of one kilogram at each shop.

  2. Shop A: set up

    £3.602 kg\frac{\pounds 3.60}{2\text{ kg}}

    £3.60 buys 2 kg at Shop A.

  3. Shop A: work out

    3.60÷2=1.803.60 \div 2 = 1.80

    Shop A costs £1.80 per kg.

  4. Shop B: convert the mass

    500 g=0.5 kg500\text{ g} = 0.5\text{ kg}

    500 grams is half a kilogram.

  5. Shop B: set up

    £0.850.5 kg\frac{\pounds 0.85}{0.5\text{ kg}}

    £0.85 buys 0.5 kg at Shop B.

  6. Shop B: work out

    0.85÷0.5=1.700.85 \div 0.5 = 1.70

    Shop B costs £1.70 per kg.

  7. Shop C: set up

    £2.851.5 kg\frac{\pounds 2.85}{1.5\text{ kg}}

    £2.85 buys 1.5 kg at Shop C.

  8. Shop C: work out

    2.85÷1.5=1.902.85 \div 1.5 = 1.90

    Shop C costs £1.90 per kg.

  9. List the rates

    A:1.80, B:1.70, C:1.90A: 1.80,\ B: 1.70,\ C: 1.90

    Now compare the three prices per kg.

  10. Find the cheapest

    1.70<1.80<1.901.70 < 1.80 < 1.90

    Shop B has the lowest price per kg.

  11. Identify the best value

    Shop B\text{Shop B}

    Shop B is the best value at £1.70 per kg.

  12. Check Shop B

    0.5×1.70=0.850.5 \times 1.70 = 0.85

    0.5 kg at £1.70 per kg is £0.85, as given.

  13. Check Shop A

    2×1.80=3.602 \times 1.80 = 3.60

    2 kg at £1.80 per kg is £3.60, as given.

  14. Note the biggest pack is not best

    Shop A≠best\text{Shop A} \ne \text{best}

    The largest pack (Shop A) is not the cheapest per kg.

  15. State the conclusion

    Shop B at £1.70 per kg\text{Shop B at } \pounds 1.70 \text{ per kg}

    Shop B gives the best value for money.

Answer
Shop B (£1.70 per kg)\text{Shop B (}\pounds 1.70\text{ per kg)}
Question 3
6 markschallenging
Water flows through a pipe at 0.50.5 m/s. The pipe has cross-sectional area 0.020.02 m2^2. Work out the volume of water passing through each minute, in litres.
Show worked solution

Worked solution

  1. Plan the method

    flow rate=area×speed\text{flow rate} = \text{area} \times \text{speed}

    The volume per second equals the cross-section area times the speed.

  2. Note the values

    area=0.02, speed=0.5\text{area} = 0.02,\ \text{speed} = 0.5

    The area is 0.02 m squared and the speed is 0.5 m/s.

  3. Substitute

    0.02×0.50.02 \times 0.5

    Multiply the area by the speed.

  4. Work out the flow rate

    0.02×0.5=0.010.02 \times 0.5 = 0.01

    0.02 times 0.5 is 0.01.

  5. Attach the units

    0.01 m3/s0.01\text{ m}^3\text{/s}

    So 0.01 cubic metres pass each second.

  6. Recall the minute fact

    1 minute=60 s1\text{ minute} = 60\text{ s}

    There are 60 seconds in every minute.

  7. Find volume per minute

    0.01×600.01 \times 60

    Multiply the per-second volume by 60.

  8. Work it out

    0.01×60=0.60.01 \times 60 = 0.6

    0.01 times 60 is 0.6.

  9. Attach the units

    0.6 m3 per minute0.6\text{ m}^3\text{ per minute}

    So 0.6 cubic metres pass each minute.

  10. Recall the litre fact

    1 m3=1000 litres1\text{ m}^3 = 1000\text{ litres}

    There are 1000 litres in every cubic metre.

  11. Convert to litres

    0.6×1000=6000.6 \times 1000 = 600

    0.6 cubic metres is 600 litres.

  12. Attach the units

    600 litres per minute600\text{ litres per minute}

    So 600 litres pass through each minute.

  13. Check per second

    600÷60=10 L/s600 \div 60 = 10\text{ L/s}

    That is 10 litres per second.

  14. Interpret

    a fast-flowing pipe\text{a fast-flowing pipe}

    10 litres a second is a strong flow.

  15. State the final answer

    600 litres600\text{ litres}

    600 litres pass through the pipe each minute.

Answer
600 litres600\text{ litres}
Question 4
5 markschallenging
A runner completes a 3636 km race in 22 hours 3030 minutes. Work out the average speed in metres per second.
Show worked solution

Worked solution

  1. Plan the method

    convert km to m and time to s, then s=dt\text{convert km to m and time to s, then } s = \tfrac{d}{t}

    Get both quantities into metres and seconds.

  2. Convert the distance

    36 km=36000 m36\text{ km} = 36000\text{ m}

    36 times 1000 is 36000 metres.

  3. Look at the time

    2 h 30 min2\text{ h } 30\text{ min}

    The time mixes hours and minutes.

  4. Write time as hours

    2.5 h2.5\text{ h}

    30 minutes is half an hour, so the time is 2.5 hours.

  5. Recall the seconds fact

    1 hour=3600 s1\text{ hour} = 3600\text{ s}

    There are 3600 seconds in every hour.

  6. Convert the time

    2.5×3600=90002.5 \times 3600 = 9000

    2.5 hours is 9000 seconds.

  7. Attach time units

    9000 s9000\text{ s}

    The total time is 9000 seconds.

  8. Write the speed formula

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

    Speed is distance divided by time.

  9. Substitute

    360009000\frac{36000}{9000}

    Use 36000 m and 9000 s.

  10. Work it out

    36000÷9000=436000 \div 9000 = 4

    36000 divided by 9000 is 4.

  11. Attach the units

    4 m/s4\text{ m/s}

    The average speed is 4 metres per second.

  12. Cross-check in km/h

    36÷2.5=14.4 km/h36 \div 2.5 = 14.4\text{ km/h}

    The speed is also 14.4 km/h.

  13. Convert that back

    14.4÷3.6=414.4 \div 3.6 = 4

    Dividing 14.4 km/h by 3.6 gives 4 m/s, which agrees.

  14. Interpret

    a steady running pace\text{a steady running pace}

    4 m/s is a realistic marathon-style pace.

  15. State the final answer

    4 m/s4\text{ m/s}

    The average speed is 4 m/s.

Answer
4 m/s4\text{ m/s}
Question 5
6 markschallenging
A gold bar is a cuboid measuring 44 cm by 33 cm by 22 cm. Gold has density 19.319.3 g/cm3^3 and is worth £50\pounds 50 per gram. Work out the value of the bar.
Show worked solution

Worked solution

  1. Plan the method

    volume→mass→value\text{volume} \to \text{mass} \to \text{value}

    Find the volume, then the mass, then the value.

  2. Write the volume formula

    V=l×w×hV = l \times w \times h

    The bar is a cuboid.

  3. Substitute

    4×3×24 \times 3 \times 2

    The sides are 4 cm, 3 cm and 2 cm.

  4. Work out the volume

    4×3×2=244 \times 3 \times 2 = 24

    The volume is 24 cubic centimetres.

  5. Attach volume units

    24 cm324\text{ cm}^3

    The volume of the bar is 24 cm cubed.

  6. Write the mass formula

    mass=density×volume\text{mass} = \text{density} \times \text{volume}

    Mass is density multiplied by volume.

  7. Substitute

    19.3×2419.3 \times 24

    Use density 19.3 and volume 24.

  8. Work out the mass

    19.3×24=463.219.3 \times 24 = 463.2

    19.3 times 24 is 463.2 grams.

  9. Attach mass units

    463.2 g463.2\text{ g}

    The bar has mass 463.2 g.

  10. Write the value formula

    value=mass×price per gram\text{value} = \text{mass} \times \text{price per gram}

    Multiply the mass by the price per gram.

  11. Substitute

    463.2×50463.2 \times 50

    Use 463.2 g at £50 per gram.

  12. Work out the value

    463.2×50=23160463.2 \times 50 = 23160

    463.2 times 50 is 23160.

  13. Attach money units

    £23160\pounds 23160

    The value is £23160.

  14. Sanity check the mass

    463.2 g≈0.46 kg463.2\text{ g} \approx 0.46\text{ kg}

    A 24 cm cubed gold bar weighing about 0.46 kg is reasonable.

  15. State the final answer

    £23160\pounds 23160

    The gold bar is worth £23160.

Answer
£23160\pounds 23160

Unlock 29 more Standard and compound units questions

Create a free account to work through every GCSE Standard and compound units question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Standard and compound units practice

Practice QuestionsWorked Solutions

Related Number topics

Upper and lower boundsError intervalsEstimating powers and rootsEstimation and checkingExact calculationFactors, multiples and primes