Hard GCSE Unit conversion Questions

Challenging, exam-style GCSE Unit conversion questions with worked solutions. Stretch yourself on the hardest area conversion, two independent routes, mm2 to m2, volume conversion problems.

area conversiontwo independent routesmm2 to m2volume conversionmulti-stage chainm3 to ml
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
The mass, mm grams, of a length of wire is given by m=0.8Lm = 0.8L, where LL is the length of the wire in centimetres. A coil of this wire has mass 22 kg. Work out the length of the coil, in metres.
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Worked solution

  1. Read the formula carefully

    m=0.8Lm = 0.8L

    The formula only works if mm is in grams and LL is in centimetres.

  2. Check the units given

    m=2 kg, answer wanted in metresm = 2\text{ kg},\ \text{answer wanted in metres}

    The mass is in kilograms and the answer is wanted in metres — neither matches the formula.

  3. Convert the mass

    2 kg=2×1000=2000 g2\text{ kg} = 2 \times 1000 = 2000\text{ g}

    There are 1000 g in a kilogram, so the coil is 2000 g.

  4. Substitute into the formula

    2000=0.8L2000 = 0.8L

    Now the units on both sides agree.

  5. Rearrange for L

    L=20000.8L = \frac{2000}{0.8}

    Divide both sides by 0.8.

  6. Clear the decimal

    20000.8=200008\frac{2000}{0.8} = \frac{20000}{8}

    Multiplying top and bottom by 10 makes the division exact.

  7. Work out the length in cm

    200008=2500 cm\frac{20000}{8} = 2500\text{ cm}

    The wire is 2500 cm long.

  8. Convert to metres

    1 m=100 cm⇒2500÷100=251\text{ m} = 100\text{ cm} \Rightarrow 2500 \div 100 = 25

    The coil is 25 m long.

  9. Check by substituting back

    L=2500⇒m=0.8×2500=2000 gL = 2500 \Rightarrow m = 0.8 \times 2500 = 2000\text{ g}

    2000 g is 2 kg, so the answer satisfies the original formula.

  10. Note the first classic error

    m=2⇒L=2.5 cmm = 2 \Rightarrow L = 2.5\text{ cm}

    Putting the mass in as 2 (kilograms) gives a 2.5 cm coil — absurd for 2 kg of wire.

  11. Note the second classic error

    L=2500 m ?L = 2500\text{ m}\ ?

    Leaving the answer as 2500 without converting would claim a coil 2.5 km long.

  12. Rewrite the formula for metres

    L=100x⇒m=0.8×100x=80x gL = 100x \Rightarrow m = 0.8 \times 100x = 80x\text{ g}

    If the length is xx metres, the mass in grams is 80x80x — the wire weighs 80 g per metre.

  13. Convert that rule to kilograms

    m=80x1000=0.08x kgm = \frac{80x}{1000} = 0.08x\text{ kg}

    A wire xx metres long has mass 0.08x0.08x kg.

  14. Check the answer with the new rule

    0.08×25=2 kg0.08 \times 25 = 2\text{ kg}

    A 25 m coil has mass 2 kg — the rebuilt formula agrees.

  15. State the answer

    L=25 mL = 25\text{ m}

    The coil of wire is 25 metres long.

Answer
L=25 mL = 25\text{ m}
Question 2
6 markschallenging
A machine makes one part every 4545 seconds and runs for 33 hours 4545 minutes. Each part uses 250250 cm3^3 of plastic. Work out the volume of plastic used, in litres.
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Worked solution

  1. Write down what is given

    1 part per 45 s, 3 h 45 min, 250 cm3 per part1\text{ part per }45\text{ s},\ 3\text{ h }45\text{ min},\ 250\text{ cm}^3\text{ per part}

    The rate is in seconds but the running time is in hours and minutes.

  2. Convert the hours

    3×3600=10800 s3 \times 3600 = 10800\text{ s}

    There are 3600 seconds in an hour.

  3. Convert the minutes

    45×60=2700 s45 \times 60 = 2700\text{ s}

    There are 60 seconds in a minute.

  4. Add the two parts

    10800+2700=13500 s10800 + 2700 = 13500\text{ s}

    The machine runs for 13 500 seconds.

  5. Check the time another way

    3 h 45 min=3.75 h,3.75×3600=135003\text{ h }45\text{ min} = 3.75\text{ h},\quad 3.75 \times 3600 = 13500

    Writing the time as a decimal number of hours gives the same 13 500 s.

  6. Find the number of parts

    13500÷4513500 \div 45

    One part is made every 45 seconds.

  7. Work out the division

    13500÷45=30013500 \div 45 = 300

    The machine makes 300 parts.

  8. Check by multiplying back

    300×45=13500300 \times 45 = 13500

    300 parts really do take 13 500 seconds.

  9. Find the plastic used

    300×250=75000 cm3300 \times 250 = 75000\text{ cm}^3

    Each of the 300 parts uses 250 cm3^3.

  10. Recall the capacity fact

    1 litre=1000 cm31\text{ litre} = 1000\text{ cm}^3

    Now convert cubic centimetres to litres.

  11. Convert to litres

    75000÷1000=7575000 \div 1000 = 75

    The machine uses 75 litres of plastic.

  12. Check with a per-part conversion

    250 cm3=0.25 litres250\text{ cm}^3 = 0.25\text{ litres}

    Route 2: convert first, then multiply.

  13. Finish route 2

    300×0.25=75300 \times 0.25 = 75

    The second route gives 75 litres as well.

  14. Express it in cubic metres

    75 litres=0.075 m375\text{ litres} = 0.075\text{ m}^3

    Since 1 m3^3 = 1000 litres, the plastic occupies 0.075 m3^3.

  15. State the answer

    Volume of plastic=75 litres\text{Volume of plastic} = 75\text{ litres}

    The machine uses 75 litres of plastic in the shift.

Answer
75 litres75\text{ litres}
Question 3
5 markschallenging
Paint A: 55 litres covers 6060 m2^2 and costs £24\pounds 24. Paint B: 2.52.5 litres covers 250000250000 cm2^2 and costs £11\pounds 11. Which paint is better value for money?
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Worked solution

  1. Write down what is given

    A: 60 m2 for £24,B: 250000 cm2 for £11\text{A: }60\text{ m}^2\text{ for }\pounds 24,\quad \text{B: }250000\text{ cm}^2\text{ for }\pounds 11

    The coverages are in different area units, so they cannot be compared yet.

  2. Choose a common unit

    compare the cost per m2\text{compare the cost per m}^2

    Square metres are the natural unit for wall coverage.

  3. Write down the linear fact

    1 m=100 cm1\text{ m} = 100\text{ cm}

    Start every area conversion from the length fact.

  4. Square the linear factor

    1 m2=1002=10000 cm21\text{ m}^2 = 100^2 = 10000\text{ cm}^2

    A square metre holds 10 000 square centimetres.

  5. Convert Paint B's coverage

    250000÷10000=25 m2250000 \div 10000 = 25\text{ m}^2

    Paint B covers 25 m2^2, not the huge area the number 250 000 suggests.

  6. Check that conversion

    5 m×5 m=500 cm×500 cm=250000 cm25\text{ m} \times 5\text{ m} = 500\text{ cm} \times 500\text{ cm} = 250000\text{ cm}^2

    A 25 m2^2 square really does measure 250 000 cm2^2, so the factor 10 000 is right.

  7. Find the cost per m² for A

    24÷60=0.424 \div 60 = 0.4

    Paint A costs £0.40 per square metre.

  8. Find the cost per m² for B

    11÷25=0.4411 \div 25 = 0.44

    Paint B costs £0.44 per square metre.

  9. Compare the two

    0.40<0.440.40 < 0.44

    Paint A is cheaper for every square metre painted.

  10. Check the other way round

    60÷24=2.5,25÷11=2.27…60 \div 24 = 2.5,\quad 25 \div 11 = 2.27\ldots

    Square metres per pound: A gives 2.5 m2^2, B only 2.27 m2^2 — A again wins.

  11. Quantify the difference

    0.44−0.40=0.040.44 - 0.40 = 0.04

    Paint A saves 4p on every square metre.

  12. Note the classic error

    250000÷100=2500 "m2"250000 \div 100 = 2500\text{ "m}^2\text{"}

    Using the linear factor would claim Paint B covers 2500 m2^2 for £11, making it look 100 times better than it is.

  13. Sanity-check the coverage

    5 litres→60 m2,2.5 litres→25 m25\text{ litres} \rightarrow 60\text{ m}^2,\quad 2.5\text{ litres} \rightarrow 25\text{ m}^2

    Both paints cover roughly 10–12 m2^2 per litre, which is realistic.

  14. Answer the question asked

    cheaper per m2=better value\text{cheaper per m}^2 = \text{better value}

    Value for money is decided by the cost of painting the same area.

  15. State the answer

    Paint A is better value\text{Paint A is better value}

    Paint A costs £0.40 per m2^2 against £0.44 per m2^2 for Paint B.

Answer
Paint A\text{Paint A}
Question 4
6 markschallenging
A photograph measures 66 inches by 44 inches, where 11 inch =2.54= 2.54 cm. Work out the area of the photograph in cm2^2, giving your answer to 11 decimal place.
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Worked solution

  1. Write down what is given

    6 in×4 in, 1 in=2.54 cm6\text{ in} \times 4\text{ in},\ 1\text{ in} = 2.54\text{ cm}

    The sides are in inches but the area is wanted in square centimetres.

  2. Convert the length

    6×2.54=15.24 cm6 \times 2.54 = 15.24\text{ cm}

    Centimetres are shorter than inches, so the number gets bigger.

  3. Convert the width

    4×2.54=10.16 cm4 \times 2.54 = 10.16\text{ cm}

    Both sides are now in centimetres.

  4. Set up the area

    15.24×10.1615.24 \times 10.16

    Route 1: convert the lengths first, then multiply.

  5. Multiply without decimals

    1524×1016=15483841524 \times 1016 = 1548384

    Ignoring the decimal points for now keeps the multiplication whole-number.

  6. Replace the decimal point

    15.24×10.16=154.838415.24 \times 10.16 = 154.8384

    There are four decimal places altogether, so the answer is 154.8384 cm2^2.

  7. Start route 2: area in square inches

    6×4=24 in26 \times 4 = 24\text{ in}^2

    Route 2: find the area in the original unit first.

  8. Square the conversion factor

    1 in2=2.542=6.4516 cm21\text{ in}^2 = 2.54^2 = 6.4516\text{ cm}^2

    For an area the linear factor must be squared.

  9. Apply the squared factor

    24×6.4516=154.838424 \times 6.4516 = 154.8384

    Route 2 gives exactly the same area, so the conversion is sound.

  10. Note the classic error

    24×2.54=60.9624 \times 2.54 = 60.96

    Using the linear factor on an area gives 60.96 — far too small, and wrong by a factor of 2.54.

  11. Round to 1 decimal place

    154.8384≈154.8154.8384 \approx 154.8

    The digit after the tenths place is 3, so round down.

  12. Sanity-check the size

    15 cm×10 cm≈150 cm215\text{ cm} \times 10\text{ cm} \approx 150\text{ cm}^2

    A rough estimate gives about 150 cm2^2, which is close to 154.8 cm2^2.

  13. Convert to mm² as a check

    154.8384×100=15483.84 mm2154.8384 \times 100 = 15483.84\text{ mm}^2

    Since 1 cm2^2 = 100 mm2^2, this is another squared factor at work.

  14. Confirm the units

    cm×cm=cm2\text{cm} \times \text{cm} = \text{cm}^2

    Multiplying two lengths in centimetres always gives an area in square centimetres.

  15. State the answer

    Area=154.8 cm2\text{Area} = 154.8\text{ cm}^2

    To 1 decimal place, the photograph has an area of 154.8 cm2^2.

Answer
154.8 cm2154.8\text{ cm}^2
Question 5
5 markschallenging
An airline allows two cases with a combined mass of at most 4646 kg. Jon's case has mass 4444 lb and Amy's has mass 5555 lb, where 11 kg =2.2= 2.2 lb. Work out how many kilograms below the allowance their two cases are in total.
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Worked solution

  1. Write down what is given

    limit 46 kg, 44 lb, 55 lb, 1 kg=2.2 lb\text{limit }46\text{ kg},\ 44\text{ lb},\ 55\text{ lb},\ 1\text{ kg} = 2.2\text{ lb}

    The limit is in kilograms but the cases are weighed in pounds.

  2. Decide the operation

    lb→kg: ÷2.2\text{lb} \rightarrow \text{kg}:\ \div 2.2

    A kilogram is heavier than a pound, so the kilogram number must be smaller — divide.

  3. Convert Jon's case

    44÷2.2=20 kg44 \div 2.2 = 20\text{ kg}

    Since 2.2×20=442.2 \times 20 = 44, Jon's case is exactly 20 kg.

  4. Check that conversion

    20×2.2=4420 \times 2.2 = 44

    Converting back returns 44 lb.

  5. Convert Amy's case

    55÷2.2=25 kg55 \div 2.2 = 25\text{ kg}

    Since 2.2×25=552.2 \times 25 = 55, Amy's case is exactly 25 kg.

  6. Check that conversion

    25×2.2=5525 \times 2.2 = 55

    Converting back returns 55 lb.

  7. Add the two masses

    20+25=45 kg20 + 25 = 45\text{ kg}

    Together the cases weigh 45 kg.

  8. Compare with the allowance

    45<4645 < 46

    They are inside the 46 kg limit.

  9. Find the difference

    46−45=146 - 45 = 1

    The cases are 1 kg below the allowance.

  10. Start route 2: convert the limit

    46×2.2=101.2 lb46 \times 2.2 = 101.2\text{ lb}

    Instead, convert the allowance into pounds.

  11. Add the cases in pounds

    44+55=99 lb44 + 55 = 99\text{ lb}

    The two cases total 99 lb.

  12. Find the difference in pounds

    101.2−99=2.2 lb101.2 - 99 = 2.2\text{ lb}

    They are 2.2 lb under the limit.

  13. Convert that difference

    2.2÷2.2=1 kg2.2 \div 2.2 = 1\text{ kg}

    The two routes agree: 1 kg to spare.

  14. Note the classic error

    44×2.2=96.844 \times 2.2 = 96.8

    Multiplying instead of dividing would make Jon's case 96.8 kg — heavier than Jon.

  15. State the answer

    They are 1 kg below the allowance\text{They are }1\text{ kg below the allowance}

    The cases total 45 kg against an allowance of 46 kg.

Answer
1 kg below the allowance1\text{ kg below the allowance}

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