Hard GCSE Ratio problem solving Questions

Challenging, exam-style GCSE Ratio problem solving questions with worked solutions. Stretch yourself on the hardest limiting ingredient, recipe scaling, rounding down, best buy problems.

limiting ingredientrecipe scalingrounding downbest buyunit pricecomparison
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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}
Question 2
6 markschallenging
A loaf of bread needs 500500 g of flour, 77 g of yeast and 300300 ml of water. Flour is sold in 1.51.5 kg bags costing £1.20 and yeast is sold in 100100 g tubs costing 9090p. A baker makes 99 loaves. Water is free. Work out the least amount he must pay for the flour and the yeast.
Show worked solution

Worked solution

  1. Work out the flour needed

    9×500=4500 g9 \times 500 = 4500 \text{ g}

    Nine loaves at 500500 g of flour each.

  2. Convert the flour needed to kilograms

    4500 g=4.5 kg4500 \text{ g} = 4.5 \text{ kg}

    The bags are measured in kilograms.

  3. Work out how many bags of flour are needed

    4.5÷1.5=34.5 \div 1.5 = 3

    Each bag holds 1.51.5 kg.

  4. Check three bags are exactly enough

    3×1.5=4.5 kg ✓3 \times 1.5 = 4.5 \text{ kg} \ \checkmark

    Three bags give exactly the 4.54.5 kg needed.

  5. Check two bags are not enough

    2×1.5=3 kg<4.5 kg2 \times 1.5 = 3 \text{ kg} < 4.5 \text{ kg}

    Two bags would leave the baker 1.51.5 kg short.

  6. Work out the cost of the flour

    3×1.20=3.603 \times 1.20 = 3.60

    Three bags at £1.20 each cost £3.60.

  7. Work out the yeast needed

    9×7=63 g9 \times 7 = 63 \text{ g}

    Nine loaves at 77 g of yeast each.

  8. Work out how many tubs of yeast are needed

    63÷100=0.6363 \div 100 = 0.63

    One tub holds 100100 g, which is more than enough.

  9. Round the tubs up

    ⌈0.63⌉=1\lceil 0.63 \rceil = 1

    You cannot buy 0.630.63 of a tub, so he must buy one whole tub.

  10. Check one tub is enough

    100≥63100 \ge 63

    One tub covers the 6363 g needed, with 3737 g left over.

  11. Work out the cost of the yeast

    1×0.90=0.901 \times 0.90 = 0.90

    One tub costs 9090p, which is £0.90.

  12. Add the two costs

    3.60+0.903.60 + 0.90

    Add the flour cost and the yeast cost.

  13. Work out the total

    3.60+0.90=4.503.60 + 0.90 = 4.50

    The total is £4.50.

  14. Write the money to two decimal places

    £4.50\pounds 4.50

    Money answers carry two decimal places.

  15. State the answer

    £4.50\pounds 4.50

    The baker must pay £4.50 for the flour and the yeast.

Answer
£4.50\pounds 4.50
Question 3
6 markschallenging
A tin holds 2020 litres of paint in which blue and yellow are in the ratio 2:32:3. How many litres of blue paint must be added to the tin so that blue and yellow are in the ratio 1:11:1?
Show worked solution

Worked solution

  1. Work out the parts in the tin

    2+3=5 parts2 + 3 = 5 \text{ parts}

    The 2020 litres is divided into 55 equal parts.

  2. Find the size of one part

    20÷5=4 litres20 \div 5 = 4 \text{ litres}

    One part is 44 litres.

  3. Find the blue paint in the tin

    2×4=8 litres2 \times 4 = 8 \text{ litres}

    The blue is 22 parts.

  4. Find the yellow paint in the tin

    3×4=12 litres3 \times 4 = 12 \text{ litres}

    The yellow is 33 parts.

  5. Check the two volumes

    8+12=20 ✓8 + 12 = 20 \ \checkmark

    The blue and yellow account for all 2020 litres.

  6. Note what stays the same

    yellow=12 litres throughout\text{yellow} = 12 \text{ litres throughout}

    Only blue paint is added, so the yellow never changes. This is the key idea.

  7. Write down the target ratio

    blue:yellow=1:1\text{blue} : \text{yellow} = 1 : 1

    Equal amounts of blue and yellow.

  8. Work out the blue needed at the end

    blue=12 litres\text{blue} = 12 \text{ litres}

    For a 1:11:1 ratio the blue must equal the yellow, which is 1212 litres.

  9. Work out how much blue to add

    12−8=4 litres12 - 8 = 4 \text{ litres}

    There are already 88 litres of blue in the tin.

  10. Check the new mixture

    blue=8+4=12,yellow=12\text{blue} = 8 + 4 = 12, \quad \text{yellow} = 12

    The tin now holds 1212 litres of each.

  11. Check the new ratio

    12:12=1:1 ✓12 : 12 = 1 : 1 \ \checkmark

    The mixture is now in the required ratio.

  12. Check the new total volume

    20+4=24 litres20 + 4 = 24 \text{ litres}

    The tin now holds 2424 litres of paint.

  13. Watch for the common error

    Adding 2 litres does not work\text{Adding } 2 \text{ litres does not work}

    Adding 22 litres would give 10:1210:12, which is 5:65:6, not 1:11:1.

  14. Watch for a second common error

    The answer is not 20÷2=10\text{The answer is not } 20 \div 2 = 10

    Splitting the new total equally is only correct once the yellow amount is taken into account.

  15. State the answer

    4 litres4 \text{ litres}

    44 litres of blue paint must be added.

Answer
4 litres4 \text{ litres}
Question 4
5 markschallenging
School A has 480480 students and the ratio of boys to girls is 5:75:7. School B has 350350 students and the ratio of boys to girls is 3:43:4. Which school has more boys, and how many more?
Show worked solution

Worked solution

  1. Work out the parts in School A

    5+7=12 parts5 + 7 = 12 \text{ parts}

    School A is divided into 1212 equal parts.

  2. Find one part in School A

    480÷12=40480 \div 12 = 40

    One part is 4040 students.

  3. Find the boys in School A

    5×40=2005 \times 40 = 200

    The boys are 55 parts.

  4. Find the girls in School A as a check

    7×40=280,200+280=480 ✓7 \times 40 = 280, \quad 200 + 280 = 480 \ \checkmark

    The boys and girls add back to the total.

  5. Work out the parts in School B

    3+4=7 parts3 + 4 = 7 \text{ parts}

    School B is divided into 77 equal parts.

  6. Find one part in School B

    350÷7=50350 \div 7 = 50

    One part is 5050 students.

  7. Find the boys in School B

    3×50=1503 \times 50 = 150

    The boys are 33 parts.

  8. Find the girls in School B as a check

    4×50=200,150+200=350 ✓4 \times 50 = 200, \quad 150 + 200 = 350 \ \checkmark

    The boys and girls add back to the total.

  9. Compare the two numbers of boys

    200>150200 > 150

    School A has more boys.

  10. Work out the difference

    200−150=50200 - 150 = 50

    School A has 5050 more boys than School B.

  11. Compare the fractions of boys

    512≈0.417,37≈0.429\frac{5}{12} \approx 0.417, \quad \frac{3}{7} \approx 0.429

    A slightly larger fraction of School B is boys.

  12. Explain why that does not decide it

    37>512 but 350<480\frac{3}{7} > \frac{5}{12} \text{ but } 350 < 480

    School B has the bigger share of boys, but School A is a much bigger school.

  13. Confirm with the fractions

    512×480=200,37×350=150 ✓\frac{5}{12} \times 480 = 200, \quad \frac{3}{7} \times 350 = 150 \ \checkmark

    Working directly with fractions gives the same two numbers of boys.

  14. Note the moral of the question

    A bigger share of a smaller whole can still be fewer\text{A bigger share of a smaller whole can still be fewer}

    Ratios cannot be compared across different totals without working out the actual amounts.

  15. State the answer

    School A, by 50 boys\text{School A, by } 50 \text{ boys}

    School A has 5050 more boys than School B.

Answer
School A, by 50 boys\text{School A, by } 50 \text{ boys}
Question 5
6 markschallenging
A camera costs 42 30042\,300 yen in Tokyo and £320 in London. The exchange rate is 11 pound =141= 141 yen. In which city is the camera cheaper, and by how much in pounds?
Show worked solution

Worked solution

  1. Decide on a common currency

    Convert the Tokyo price into pounds\text{Convert the Tokyo price into pounds}

    Prices can only be compared in one currency.

  2. Choose the operation

    yen÷141=pounds\text{yen} \div 141 = \text{pounds}

    Yen to pounds is the reverse of the rate, so divide.

  3. Set out the division

    42 300÷14142\,300 \div 141

    Divide the Tokyo price by the number of yen in a pound.

  4. Estimate first

    141×300=42 300141 \times 300 = 42\,300

    A sensible estimate: 141×3=423141 \times 3 = 423, so 141×300=42 300141 \times 300 = 42\,300 exactly.

  5. Read off the answer

    42 300÷141=30042\,300 \div 141 = 300

    The camera costs £300 in Tokyo.

  6. Check the conversion

    300×141=42 300 ✓300 \times 141 = 42\,300 \ \checkmark

    Converting back gives the Tokyo price.

  7. Write both prices in pounds

    Tokyo=£300,London=£320\text{Tokyo} = \pounds 300, \quad \text{London} = \pounds 320

    Now the comparison is fair.

  8. Compare

    300<320300 < 320

    The Tokyo price is lower.

  9. Work out the difference

    320−300=20320 - 300 = 20

    The camera is £20 cheaper in Tokyo.

  10. Check the difference in yen

    20×141=2820 yen20 \times 141 = 2820 \text{ yen}

    The London price is 320×141=45 120320 \times 141 = 45\,120 yen, and 45 120−42 300=282045\,120 - 42\,300 = 2820 yen.

  11. Confirm the two checks agree

    2820÷141=20 ✓2820 \div 141 = 20 \ \checkmark

    The difference is the same whichever currency it is measured in.

  12. Watch for the common error

    42 300×14142\,300 \times 141

    Multiplying instead of dividing gives an absurd price of nearly six million pounds.

  13. Sense-check the rate

    £1=141 yen⇒1 yen≈0.7p\pounds 1 = 141 \text{ yen} \Rightarrow 1 \text{ yen} \approx 0.7 \text{p}

    One yen is well under a penny, so a camera at 42 30042\,300 yen should be a few hundred pounds.

  14. State the currency of the answer

    The question asks for the difference in pounds\text{The question asks for the difference in pounds}

    So the answer is £20, not 28202820 yen.

  15. State the answer

    Tokyo, by £20\text{Tokyo, by } \pounds 20

    The camera is £20 cheaper in Tokyo.

Answer
Tokyo, by £20\text{Tokyo, by } \pounds 20

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