GCSE Ratio problem solving Practice Questions

Free GCSE Ratio problem solving practice questions with full step-by-step worked solutions. Covers recipe scaling, unitary method, scale factor, unit price. Practise exam-style problems and check your method.

recipe scalingunitary methodscale factorunit pricecurrency conversionmultiplying by a rate
GCSE Foundation70 questionsStep-by-step solutions
Question 1
2 markseasy
A recipe for 44 people uses 200200 g of flour. How much flour is needed for 66 people?
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Worked solution

  1. Find the amount for one person

    200÷4=50 g200 \div 4 = 50 \text{ g}

    Divide the flour by the number of people to get the amount per person.

  2. Scale up to six people

    50×6=300 g50 \times 6 = 300 \text{ g}

    Multiply the amount for one person by 66.

  3. State the answer

    300 g300 \text{ g}

    The recipe needs 300300 g of flour for 66 people.

Answer
300 g300 \text{ g}
Question 2
2 markseasy
A model car is made to a scale of 1:201:20. The real car is 44 m long. Work out the length of the model, in centimetres.
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Worked solution

  1. Convert the real length to centimetres

    4 m=400 cm4 \text{ m} = 400 \text{ cm}

    Work in one unit throughout; 11 m is 100100 cm.

  2. Divide by the scale

    400÷20=20 cm400 \div 20 = 20 \text{ cm}

    The model is 2020 times smaller than the real car.

  3. State the answer

    20 cm20 \text{ cm}

    The model is 2020 cm long.

Answer
20 cm20 \text{ cm}
Question 3
2 marksintermediate
A recipe for 1010 portions of soup uses 750750 ml of stock. Work out how much stock is needed for 44 portions.
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Worked solution

  1. Find the stock for one portion

    750÷10=75 ml750 \div 10 = 75 \text{ ml}

    Divide by 1010 to get the stock per portion.

  2. Scale down to four portions

    75×4=300 ml75 \times 4 = 300 \text{ ml}

    Multiply the amount for one portion by 44.

  3. Check with a scale factor

    4÷10=0.44 \div 10 = 0.4

    The new batch is 0.40.4 of the original.

  4. Apply the scale factor

    750×0.4=300 ml750 \times 0.4 = 300 \text{ ml}

    Both routes give 300300 ml.

  5. Check the answer is sensible

    300<750300 < 750

    Fewer portions need less stock, so the answer must be smaller than 750750 ml.

  6. State the answer

    300 ml300 \text{ ml}

    44 portions need 300300 ml of stock.

Answer
300 ml300 \text{ ml}
Question 4
4 markshard
In a company, the ratio of managers to staff is 1:91:9. The ratio of staff to trainees is 3:13:1. There are 2727 trainees. Work out how many managers there are.
Show worked solution

Worked solution

  1. Write both ratios down

    M:S=1:9,S:T=3:1M : S = 1 : 9, \quad S : T = 3 : 1

    Staff is the shared group, but it is 99 in one ratio and 33 in the other.

  2. Find the lowest common multiple of the staff parts

    LCM(9,3)=9\text{LCM}(9, 3) = 9

    Make staff equal to 99 in both ratios.

  3. Scale the second ratio

    S:T=3:1=9:3S : T = 3 : 1 = 9 : 3

    Multiply both parts by 33. The first ratio already has staff as 99.

  4. Join the ratios

    M:S:T=1:9:3M : S : T = 1 : 9 : 3

    Managers to staff to trainees is 1:9:31:9:3.

  5. Check the combined ratio

    1:9=1:9and9:3=3:11 : 9 = 1 : 9 \quad \text{and} \quad 9 : 3 = 3 : 1

    Both original ratios are recovered.

  6. Use the trainees to find one part

    3 parts=273 \text{ parts} = 27

    Trainees are 33 parts of the combined ratio.

  7. Work out one part

    27÷3=927 \div 3 = 9

    One part is 99 people.

  8. Find the managers

    M=1×9=9M = 1 \times 9 = 9

    Managers are 11 part.

  9. Check against both original ratios

    S=9×9=81,M:S=9:81=1:9 ✓S = 9 \times 9 = 81, \quad M : S = 9 : 81 = 1 : 9 \ \checkmark

    And S:T=81:27=3:1S : T = 81 : 27 = 3 : 1, so the answer is consistent.

  10. State the answer

    9 managers9 \text{ managers}

    There are 99 managers.

Answer
9 managers9 \text{ managers}
Question 5
6 markschallenging
In a biscuit recipe, flour and butter are in the ratio 5:25:2, and butter and sugar are in the ratio 4:34:3. A baker uses 800800 g of flour. Work out the mass of sugar he needs.
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Worked solution

  1. Write both ratios down

    F:B=5:2,B:S=4:3F : B = 5 : 2, \quad B : S = 4 : 3

    Butter is the shared ingredient, but it is 22 in one ratio and 44 in the other.

  2. Find the lowest common multiple of the butter parts

    LCM(2,4)=4\text{LCM}(2, 4) = 4

    Rewrite both ratios so that butter is 44.

  3. Scale the first ratio

    F:B=5:2=10:4F : B = 5 : 2 = 10 : 4

    Multiply both parts by 22.

  4. Leave the second ratio alone

    B:S=4:3B : S = 4 : 3

    Butter is already 44 in this ratio.

  5. Join the ratios

    F:B:S=10:4:3F : B : S = 10 : 4 : 3

    Butter is 44 in both, so the three-part ratio is complete.

  6. Check the combined ratio

    10:4=5:2and4:3=4:3 ✓10 : 4 = 5 : 2 \quad \text{and} \quad 4 : 3 = 4 : 3 \ \checkmark

    Both original ratios are recovered.

  7. Use the flour to find one part

    10 parts=800 g10 \text{ parts} = 800 \text{ g}

    Flour is 1010 parts of the combined ratio.

  8. Work out one part

    800÷10=80 g800 \div 10 = 80 \text{ g}

    One part is 8080 g.

  9. Find the butter

    B=4×80=320 gB = 4 \times 80 = 320 \text{ g}

    Butter is 44 parts.

  10. Find the sugar

    S=3×80=240 gS = 3 \times 80 = 240 \text{ g}

    Sugar is 33 parts.

  11. Check the flour-to-butter ratio

    800:320=5:2 ✓800 : 320 = 5 : 2 \ \checkmark

    Dividing both by 160160 gives 5:25:2.

  12. Check the butter-to-sugar ratio

    320:240=4:3 ✓320 : 240 = 4 : 3 \ \checkmark

    Dividing both by 8080 gives 4:34:3.

  13. Watch for the common error

    Using F:S=5:3 gives 480 g\text{Using } F : S = 5 : 3 \text{ gives } 480 \text{ g}

    Pairing the outer numbers without matching the butter is wrong.

  14. Note the total mass

    800+320+240=1360 g800 + 320 + 240 = 1360 \text{ g}

    The three ingredients weigh 13601360 g altogether.

  15. State the answer

    240 g240 \text{ g}

    The baker needs 240240 g of sugar.

Answer
240 g240 \text{ g}

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