Hard GCSE Compound measures Questions

Challenging, exam-style GCSE Compound measures questions with worked solutions. Stretch yourself on the hardest average speed, multi-stage journey, total distance / total time, out and back problems.

average speedmulti-stage journeytotal distance / total timeout and backminutes to hoursdensity
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
Which of these is the greatest speed?
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Worked solution

  1. Decide on a common unit

    convert every speed into  km/h\text{convert every speed into }\,\mathrm{km/h}

    Speeds can only be compared once they are all written in the same unit.

  2. Convert 20 m/s20\,\mathrm{m/s}

    20×3.6=72 km/h20 \times 3.6 = 72\,\mathrm{km/h}

    Multiplying metres per second by 3.63.6 gives kilometres per hour, because there are 36003600 seconds in an hour and 10001000 metres in a kilometre.

  3. Leave 70 km/h70\,\mathrm{km/h} as it is

    70 km/h70\,\mathrm{km/h}

    This speed is already in the chosen unit.

  4. Convert 1.1 km1.1\,\mathrm{km} per minute

    1.1×60=66 km/h1.1 \times 60 = 66\,\mathrm{km/h}

    There are 6060 minutes in an hour, so multiply by 6060.

  5. Convert 19 m/s19\,\mathrm{m/s}

    19×3.6=68.4 km/h19 \times 3.6 = 68.4\,\mathrm{km/h}

    Nineteen metres per second is 68.468.4 kilometres per hour.

  6. Convert 11501150 metres per minute

    1150×601000=69 km/h\frac{1150 \times 60}{1000} = 69\,\mathrm{km/h}

    In an hour this covers 69 00069\,000 metres, which is 69 km69\,\mathrm{km}.

  7. Put the five speeds in order

    72>70>69>68.4>6672 > 70 > 69 > 68.4 > 66

    Once every speed is in km/h the comparison is straightforward.

  8. Identify the greatest

    20 m/s=72 km/h20\,\mathrm{m/s} = 72\,\mathrm{km/h}

    The greatest speed is 20 m/s20\,\mathrm{m/s}, which is 72 km/h72\,\mathrm{km/h}.

  9. Check the answer by working backwards

    72÷3.6=20 m/s72 \div 3.6 = 20\,\mathrm{m/s}

    Converting back gives 20 m/s20\,\mathrm{m/s}, so the conversion was right.

  10. Avoid the common error

    20<7020 < 70

    Comparing the bare numbers 2020 and 7070 without converting would wrongly pick 70 km/h70\,\mathrm{km/h}. The units must match first.

  11. Check the units

     m/s×3.6= km/h\,\mathrm{m/s} \times 3.6 =\,\mathrm{km/h}

    The factor 3.63.6 comes from multiplying by 36003600 seconds and dividing by 10001000 metres.

  12. Sanity-check the size of the answer

    20 m/s is about 45 mph20\,\mathrm{m/s}\text{ is about }45\,\mathrm{mph}

    Seventy-two km/h is roughly 45 mph45\,\mathrm{mph}, a fast road speed, which is a sensible answer for the greatest of the five.

  13. Note how the formula rearranges

    speed=distancetime,distance=speed×time,time=distancespeed\text{speed} = \frac{\text{distance}}{\text{time}}, \quad \text{distance} = \text{speed} \times \text{time}, \quad \text{time} = \frac{\text{distance}}{\text{speed}}

    No rearrangement is needed here; the work is entirely in the units.

  14. Reflect on the method

    always convert before comparing\text{always convert before comparing}

    Comparing compound measures always starts by writing them all in one unit.

  15. State the final answer

    20 m/s20\,\mathrm{m/s}

    The final answer is 20 m/s20\,\mathrm{m/s}, which is 72 km/h72\,\mathrm{km/h}.

Answer
20 m/s=72 km/h20\,\mathrm{m/s} = 72\,\mathrm{km/h}
Question 2
6 markschallenging
A concrete slab measures 2 m2\,\mathrm{m} by 1 m1\,\mathrm{m} by 0.1 m0.1\,\mathrm{m}. Concrete has density 2400 kg/m32400\,\mathrm{kg/m}^3. Each kilogram of mass exerts a force of 10 N10\,\mathrm{N}. The slab lies flat on the ground on its 2 m2\,\mathrm{m} by 1 m1\,\mathrm{m} face. Work out the pressure it exerts on the ground, in N/m2\mathrm{N/m}^2.
Show worked solution

Worked solution

  1. Work out the volume of the slab

    2×1×0.1=0.2 m32 \times 1 \times 0.1 = 0.2\,\mathrm{m}^3

    The slab is a cuboid, so its volume is length times width times thickness.

  2. Write down the mass formula

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

    Rearranging density = mass over volume gives the mass.

  3. Work out the mass

    2400×0.2=480 kg2400 \times 0.2 = 480\,\mathrm{kg}

    A fifth of a cubic metre of concrete has mass 480 kg480\,\mathrm{kg}.

  4. Work out the force on the ground

    480×10=4800 N480 \times 10 = 4800\,\mathrm{N}

    Each kilogram exerts 10 N10\,\mathrm{N}, so 480 kg480\,\mathrm{kg} exerts 4800 N4800\,\mathrm{N}.

  5. Work out the area in contact with the ground

    2×1=2 m22 \times 1 = 2\,\mathrm{m}^2

    The slab rests on its largest face, which measures 2 m2\,\mathrm{m} by 1 m1\,\mathrm{m}.

  6. Write down the pressure formula

    pressure=forcearea\text{pressure} = \frac{\text{force}}{\text{area}}

    Pressure is force divided by area.

  7. Substitute the values

    pressure=48002\text{pressure} = \frac{4800}{2}

    Newtons divided by square metres gives N\mathrm{N} per square metre.

  8. Work out the division

    48002=2400\frac{4800}{2} = 2400

    The slab presses on the ground with a pressure of 2400 N2400\,\mathrm{N} per square metre.

  9. Check the answer by working backwards

    2400×2=4800 N2400 \times 2 = 4800\,\mathrm{N}

    Pressure times area returns the 4800 N4800\,\mathrm{N} force, so the answer is consistent.

  10. Avoid the common error

    48000.2=24 000\frac{4800}{0.2} = 24\,000

    Dividing by the volume instead of the contact area gives 24 00024\,000, which is not a pressure at all.

  11. Check the units

     N m2= N/m2\frac{\,\mathrm{N}}{\,\mathrm{m}^2} =\,\mathrm{N/m}^2

    Newtons divided by square metres is the unit asked for.

  12. Sanity-check the size of the answer

    0.1 m thickness×2400×10=24000.1\,\mathrm{m}\text{ thickness} \times 2400 \times 10 = 2400

    For a flat slab the pressure is thickness times density times 1010, and 0.10.1 times 24002400 times 1010 is 24002400, which agrees.

  13. Note how the formula rearranges

    density=massvolume,mass=density×volume,volume=massdensity\text{density} = \frac{\text{mass}}{\text{volume}}, \quad \text{mass} = \text{density} \times \text{volume}, \quad \text{volume} = \frac{\text{mass}}{\text{density}}

    The density formula had to be rearranged for the mass before the pressure could be found.

  14. Reflect on the method

    chain the compound measures\text{chain the compound measures}

    Volume gives mass through density, mass gives force, and force over area gives pressure. Each link is one compound measure.

  15. State the final answer

    2400 N/m22400\,\mathrm{N/m}^2

    The final answer is 24002400 newtons per square metre.

Answer
2400 N/m22400\,\mathrm{N/m}^2
Question 3
6 markschallenging
A car has to cover 120 km120\,\mathrm{km} in total. It travels the first 40 km40\,\mathrm{km} at an average speed of 60 km/h60\,\mathrm{km/h}. Work out the speed needed for the remaining 80 km80\,\mathrm{km} so that the average speed for the whole 120 km120\,\mathrm{km} is 80 km/h80\,\mathrm{km/h}.
Show worked solution

Worked solution

  1. Write down the rule for average speed

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

    The required average of 80 km/h80\,\mathrm{km/h} fixes the total time allowed for the whole 120 km120\,\mathrm{km}.

  2. Work out the total time allowed

    total time=12080=1.5 h\text{total time} = \frac{120}{80} = 1.5\,\mathrm{h}

    To average 80 km/h80\,\mathrm{km/h} over 120 km120\,\mathrm{km} the whole journey must take one and a half hours.

  3. Work out the time already used

    time1=4060=23 h\text{time}_1 = \frac{40}{60} = \frac{2}{3}\,\mathrm{h}

    Forty km at 60 km/h60\,\mathrm{km/h} takes two thirds of an hour, which is 4040 minutes.

  4. Work out the time left

    1.5−23=32−23=9−46=56 h1.5 - \frac{2}{3} = \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}\,\mathrm{h}

    Using sixths, one and a half hours is nine sixths and two thirds is four sixths, leaving five sixths of an hour, which is 5050 minutes.

  5. Write down the formula for the speed needed

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

    The remaining 80 km80\,\mathrm{km} must be covered in the five sixths of an hour that is left.

  6. Substitute the values

    speed=8056\text{speed} = \frac{80}{\frac{5}{6}}

    Dividing by a fraction means multiplying by its reciprocal.

  7. Work out the division

    80×65=4805=9680 \times \frac{6}{5} = \frac{480}{5} = 96

    The car must average 96 km/h96\,\mathrm{km/h} over the remaining 80 km80\,\mathrm{km}.

  8. State the answer with its units

    96 km/h96\,\mathrm{km/h}

    The speed needed for the second part of the journey is 96 km/h96\,\mathrm{km/h}.

  9. Check the answer by working backwards

    8096=56 h,23+56=1.5 h,1201.5=80\frac{80}{96} = \frac{5}{6}\,\mathrm{h}, \quad \frac{2}{3} + \frac{5}{6} = 1.5\,\mathrm{h}, \quad \frac{120}{1.5} = 80

    Covering 80 km80\,\mathrm{km} at 96 km/h96\,\mathrm{km/h} takes five sixths of an hour, the total time is 1.51.5 hours, and 120 km120\,\mathrm{km} in 1.51.5 hours is exactly the required 80 km/h80\,\mathrm{km/h}.

  10. Avoid the common error

    2×80−60=100≠962 \times 80 - 60 = 100 \ne 96

    Treating the overall average as the mean of the two leg speeds would demand 100 km/h100\,\mathrm{km/h}. That is wrong, because the two legs cover different distances and take different times.

  11. Check the units

     kmhours= km/h\frac{\,\mathrm{km}}{\text{hours}} =\,\mathrm{km/h}

    Kilometres divided by hours gives the speed unit.

  12. Sanity-check the size of the answer

    96>8096 > 80

    The first leg was slower than the target average, so the second leg must be faster than 80 km/h80\,\mathrm{km/h} to make up the lost time.

  13. Note how the formula rearranges

    speed=distancetime,distance=speed×time,time=distancespeed\text{speed} = \frac{\text{distance}}{\text{time}}, \quad \text{distance} = \text{speed} \times \text{time}, \quad \text{time} = \frac{\text{distance}}{\text{speed}}

    This problem used all three arrangements: time from distance and speed, and then speed from distance and time.

  14. Reflect on the method

    work in times, not in speeds\text{work in times, not in speeds}

    Reverse average-speed problems are solved by budgeting the total time, never by averaging speeds.

  15. State the final answer

    96 km/h96\,\mathrm{km/h}

    The final answer is 96 km/h96\,\mathrm{km/h}.

Answer
96 km/h96\,\mathrm{km/h}
Question 4
5 markschallenging
A pressure of 5 N/cm25\,\mathrm{N/cm}^2 is applied to a surface. Which of these is the same pressure written in N/m2\mathrm{N/m}^2?
Show worked solution

Worked solution

  1. Write the pressure as a fraction

    5 N/cm2=5 N1 cm25\,\mathrm{N/cm}^2 = \frac{5\,\mathrm{N}}{1\,\mathrm{cm}^2}

    The force stays in newtons, so only the area unit changes.

  2. Convert the length

    1 m=100 cm1\,\mathrm{m} = 100\,\mathrm{cm}

    The conversion between the two area units comes from this single length fact.

  3. Square the length factor

    1 m2=1002=10 000 cm21\,\mathrm{m}^2 = 100^2 = 10\,000\,\mathrm{cm}^2

    An area conversion squares the length factor, so one square metre is ten thousand square centimetres.

  4. Work out the force on a whole square metre

    5×10 000=50 000 N5 \times 10\,000 = 50\,000\,\mathrm{N}

    If each of the ten thousand square centimetres carries 5 N5\,\mathrm{N}, the whole square metre carries 50 000 N50\,000\,\mathrm{N}.

  5. State the equivalent pressure

    5 N/cm2=50 000 N/m25\,\mathrm{N/cm}^2 = 50\,000\,\mathrm{N/m}^2

    The same pressure written in newtons per square metre is 50 00050\,000.

  6. Check by converting back

    50 00010 000=5 N/cm2\frac{50\,000}{10\,000} = 5\,\mathrm{N/cm}^2

    Dividing by ten thousand returns the original 5 N5\,\mathrm{N} per square centimetre.

  7. Note why the pressure looks bigger

    same pressure, larger area unit\text{same pressure, larger area unit}

    The physical pressure has not changed. The number is larger only because a square metre is a much larger area than a square centimetre.

  8. Test the rule on another value

    3 N/cm2=30 000 N/m23\,\mathrm{N/cm}^2 = 30\,000\,\mathrm{N/m}^2

    The factor of 10 00010\,000 converts any pressure from N\mathrm{N} per square centimetre to N\mathrm{N} per square metre.

  9. Check the answer by working backwards

    50 000÷10 000=550\,000 \div 10\,000 = 5

    The reverse conversion returns the original figure, so the answer is right.

  10. Avoid the common error

    5×100=500≠50 0005 \times 100 = 500 \ne 50\,000

    Multiplying by 100100 uses the length factor. Areas need the factor squared.

  11. Check the units

     N m2\frac{\,\mathrm{N}}{\,\mathrm{m}^2}

    The answer must be a force in newtons divided by an area in square metres.

  12. Sanity-check the size of the answer

    50 000 N/m2 is a large but ordinary pressure50\,000\,\mathrm{N/m}^2 \text{ is a large but ordinary pressure}

    Ordinary objects easily exert tens of thousands of newtons per square metre, so the size is believable.

  13. Note how the formula rearranges

    pressure=forcearea,force=pressure×area,area=forcepressure\text{pressure} = \frac{\text{force}}{\text{area}}, \quad \text{force} = \text{pressure} \times \text{area}, \quad \text{area} = \frac{\text{force}}{\text{pressure}}

    The formula triangle is unaffected by the change of unit.

  14. Reflect on the method

    convert numerator and denominator separately\text{convert numerator and denominator separately}

    Every compound-unit conversion is done by converting the top and the bottom of the fraction one at a time.

  15. State the final answer

    50 000 N/m250\,000\,\mathrm{N/m}^2

    The final answer is 50 00050\,000 newtons per square metre.

Answer
50 000 N/m250\,000\,\mathrm{N/m}^2
Question 5
6 markschallenging
Cyclist XX rides for 11 hour at 20 km/h20\,\mathrm{km/h} and then for 11 hour at 30 km/h30\,\mathrm{km/h}. Cyclist YY rides 20 km20\,\mathrm{km} at 20 km/h20\,\mathrm{km/h} and then 20 km20\,\mathrm{km} at 30 km/h30\,\mathrm{km/h}. Which statement is correct?
Show worked solution

Worked solution

  1. Write down the rule for average speed

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

    Both cyclists must be handled with total distance divided by total time.

  2. Work out the distances for cyclist XX

    20×1=20 km,30×1=30 km20 \times 1 = 20\,\mathrm{km}, \quad 30 \times 1 = 30\,\mathrm{km}

    Each hour is a whole hour, so the distances are just the speeds.

  3. Work out the average speed for cyclist XX

    20+301+1=502=25\frac{20 + 30}{1 + 1} = \frac{50}{2} = 25

    Cyclist XX covers 50 km50\,\mathrm{km} in 22 hours, so the average speed is 25 km/h25\,\mathrm{km/h}.

  4. Work out the times for cyclist YY

    2020=1 h,2030=23 h\frac{20}{20} = 1\,\mathrm{h}, \quad \frac{20}{30} = \frac{2}{3}\,\mathrm{h}

    The faster leg takes only two thirds of an hour, so the two legs take different times.

  5. Add the totals for cyclist YY

    distance=40 km,time=1+23=53 h\text{distance} = 40\,\mathrm{km}, \quad \text{time} = 1 + \frac{2}{3} = \frac{5}{3}\,\mathrm{h}

    Cyclist YY covers 40 km40\,\mathrm{km} in five thirds of an hour.

  6. Work out the average speed for cyclist YY

    4053=40×35=24\frac{40}{\frac{5}{3}} = 40 \times \frac{3}{5} = 24

    Cyclist YY averages 24 km/h24\,\mathrm{km/h}, which is less than 2525.

  7. Compare with the mean of the speeds

    20+302=25\frac{20 + 30}{2} = 25

    The mean of the two speeds is 2525. It happens to be right for XX and wrong for YY.

  8. Explain the difference

    X: equal timesY: equal distances\text{X: equal times} \quad \text{Y: equal distances}

    Averaging the speeds weights them equally, which only matches reality when equal times are spent at each speed. Cyclist YY spends longer on the slow leg, so the average is dragged below 2525.

  9. Check the answer by working backwards

    25×2=50 km,24×53=40 km25 \times 2 = 50\,\mathrm{km}, \quad 24 \times \frac{5}{3} = 40\,\mathrm{km}

    Each average speed multiplied by the matching total time returns that cyclist total distance, so both answers check out.

  10. Avoid the common error

    both average 25 km/h\text{both average }25\,\mathrm{km/h}

    Claiming both average 25 km/h25\,\mathrm{km/h} applies the mean of the speeds to cyclist YY, whose legs take different times.

  11. Check the units

     kmhours= km/h\frac{\,\mathrm{km}}{\text{hours}} =\,\mathrm{km/h}

    Kilometres divided by hours gives km/h for both cyclists.

  12. Sanity-check the size of the answer

    24<2524 < 25

    Cyclist YY spends more of the ride at the slower speed, so a slightly lower average is exactly what we should expect.

  13. Note how the formula rearranges

    speed=distancetime,distance=speed×time,time=distancespeed\text{speed} = \frac{\text{distance}}{\text{time}}, \quad \text{distance} = \text{speed} \times \text{time}, \quad \text{time} = \frac{\text{distance}}{\text{speed}}

    Cyclist XX needed distance = speed times time, and cyclist YY needed time = distance divided by speed.

  14. Reflect on the method

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

    The mean of the speeds is only ever safe when the times are equal. Total distance divided by total time is always safe.

  15. State the final answer

    X:25 km/h,Y:24 km/hX: 25\,\mathrm{km/h}, \quad Y: 24\,\mathrm{km/h}

    The final answer is 25 km/h25\,\mathrm{km/h} for XX and 24 km/h24\,\mathrm{km/h} for YY.

Answer
X averages 25 km/h and Y averages 24 km/hX\text{ averages }25\,\mathrm{km/h}\text{ and Y averages }24\,\mathrm{km/h}

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