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Ratio, proportion and percentages

Ratio and scale, percentage change and reverse percentages, direct and inverse proportion, growth and decay, similarity of lengths, areas and volumes.

Spec M3.1-M3.11Section 1 Part 2: Higher GCSE mathematics15 questions2 real past questions
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1Ratio, proportion and percentages · spec M3.8 · Paper 1 style · ★☆☆

The price of a coat is reduced by \(20\%\) in a sale, and then the sale price is reduced by a further \(25\%\). The final price is what percentage of the original price?

  1. A\(50\%\)
  2. B\(55\%\)
  3. C\(65\%\)
  4. D\(60\%\)
  5. E\(40\%\)
  6. F\(45\%\)
Worked solution

\(0.8 \times 0.75 = 0.6\), so \(60\%\).

Full method — how and why
Successive percentage changes multiply. Convert each to a multiplier and combine.
  1. A \(20\%\) reduction multiplies by \(0.8\).
    You keep \(80\%\) of the price: \(1 - 0.20 = 0.8\).
  2. A further \(25\%\) reduction multiplies by \(0.75\) - applied to the sale price, not the original.
    The second discount acts on whatever the price already is; that is why the changes multiply.
  3. Combine: \(0.8 \times 0.75 = 0.6\).
    Multiplying the multipliers gives the overall multiplier.
  4. So the final price is \(60\%\) of the original.
    Equivalently a \(40\%\) reduction overall - not \(45\%\).
Where it goes wrong
  • Adding the percentages to get \(45\%\) off, hence \(55\%\) - an offered option.
  • Applying the second reduction to the original price.
Takeaway Percentage changes compose by multiplying multipliers, never by adding percentages.
Answer D · Ratio, proportion and percentages (M3.8) · question RAT-01
2Ratio, proportion and percentages · spec M3.8 · Paper 1 style · ★☆☆

After a \(15\%\) increase, a salary is \(\pounds 27\,600\). What was the original salary?

  1. A\(\pounds 23\,000\)
  2. B\(\pounds 31\,740\)
  3. C\(\pounds 23\,460\)
  4. D\(\pounds 24\,500\)
  5. E\(\pounds 24\,000\)
  6. F\(\pounds 25\,000\)
Worked solution

\(1.15 \times S = 27600 \Rightarrow S = 24000\).

Full method — how and why
A reverse-percentage problem: the given figure is the result, so divide by the multiplier.
  1. A \(15\%\) increase means multiplying by \(1.15\).
    \(100\% + 15\% = 115\% = 1.15\).
  2. Write the relation: \(1.15 \times (\text{original}) = 27\,600\).
    The original is the unknown; the multiplier acts on it.
  3. Divide: \(\text{original} = \dfrac{27\,600}{1.15} = 24\,000\).
    \(27600 \div 115 = 240\), so dividing by \(1.15\) gives \(24\,000\).
  4. Check: \(24\,000 \times 1.15 = 27\,600\) ✓.
    Always multiply back - it catches the commonest error instantly.
Where it goes wrong
  • Taking \(15\%\) off the new salary: \(27\,600 \times 0.85 = 23\,460\) - an offered option.
  • Dividing by \(0.15\) or by \(1.5\).
Takeaway Reverse percentage: divide by the multiplier, then check by multiplying back.
Answer E · Ratio, proportion and percentages (M3.8) · question RAT-02
3Ratio, proportion and percentages · spec M3.4 · Paper 1 style · ★☆☆

\(\pounds 600\) is divided between \(A\), \(B\) and \(C\) in the ratio \(2 : 3 : 7\). How much more does \(C\) receive than \(A\)?

  1. A\(\pounds 300\)
  2. B\(\pounds 250\)
  3. C\(\pounds 150\)
  4. D\(\pounds 200\)
  5. E\(\pounds 350\)
  6. F\(\pounds 100\)
Worked solution

Each part is \(\pounds 50\); \(C - A = 5\) parts \(= \pounds 250\).

Full method — how and why
Sharing in a ratio: find the value of one part, then answer the question actually asked.
  1. Add the ratio parts: \(2 + 3 + 7 = 12\).
    The total is divided into this many equal parts.
  2. One part: \(\pounds 600 \div 12 = \pounds 50\).
    Divide the total by the number of parts.
  3. The difference between \(C\) and \(A\) is \(7 - 2 = 5\) parts.
    Work with the difference in parts rather than computing both shares.
  4. So the difference is \(5 \times \pounds 50 = \pounds 250\).
    Check: \(A\) gets \(\pounds100\), \(C\) gets \(\pounds350\) ✓.
Where it goes wrong
  • Giving \(C\)'s share (\(\pounds350\)) instead of the difference - an offered option.
  • Dividing by \(3\) (the number of people) instead of \(12\) (the number of parts).
Takeaway Find the value of one part first; then read carefully whether a share, a difference or a total is wanted.
Answer B · Ratio, proportion and percentages (M3.4) · question RAT-03
4Ratio, proportion and percentages · spec M3.9 · Paper 1 style · ★☆☆

\(y\) is inversely proportional to the square of \(x\). When \(x = 2\), \(y = 9\). Find \(y\) when \(x = 6\).

  1. A\(3\)
  2. B\(4\)
  3. C\(1\)
  4. D\(81\)
  5. E\(27\)
  6. F\(\dfrac{1}{3}\)
Worked solution

\(y = \frac{k}{x^2}\), \(k = 36\). \(y = \frac{36}{36} = 1\).

Full method — how and why
Inverse proportion to a square. Find the constant, then substitute.
  1. Write the relationship: \(y = \dfrac{k}{x^2}\).
    'Inversely proportional to the square of \(x\)' puts \(x^2\) in the denominator.
  2. Find \(k\) from the given pair: \(9 = \dfrac{k}{4}\), so \(k = 36\).
    Substitute \(x = 2\), \(y = 9\); note \(x^2 = 4\), not \(2\).
  3. Substitute \(x = 6\): \(y = \dfrac{36}{36} = 1\).
    \(6^2 = 36\).
Where it goes wrong
  • Using \(y = \frac kx\) and getting \(3\) - an offered option.
  • Reasoning '\(x\) triples so \(y\) thirds', which ignores the squaring (it should be a ninth).
Takeaway Inverse square: tripling \(x\) divides \(y\) by \(9\). Always find \(k\) explicitly.
Answer C · Ratio, proportion and percentages (M3.9) · question RAT-04
5Ratio, proportion and percentages · spec M3.11 · Paper 1 style · ★☆☆

A bacteria population doubles every \(3\) hours. By what factor does it grow in \(24\) hours?

  1. A\(16\)
  2. B\(48\)
  3. C\(64\)
  4. D\(128\)
  5. E\(8\)
  6. F\(256\)
Worked solution

\(24/3 = 8\) doublings: \(2^8 = 256\).

Full method — how and why
Repeated doubling. Count the doublings, then raise \(2\) to that power.
  1. Count the doubling periods: \(24 \div 3 = 8\).
    Each \(3\)-hour block doubles the population.
  2. Each doubling multiplies by \(2\), so eight of them multiply by \(2^8\).
    Repeated multiplication is exponentiation - not multiplication by \(8\).
  3. \(2^8 = 256\).
    Worth knowing the powers of \(2\) up to \(2^{10} = 1024\).
Where it goes wrong
  • Multiplying by \(8\) or computing \(2 \times 8 = 16\) - both offered.
  • Using \(2^6 = 64\) by dividing \(24\) by \(4\).
Takeaway Growth factor \(= (\text{factor per period})^{\text{number of periods}}\).
Answer F · Ratio, proportion and percentages (M3.11) · question RAT-05
6Ratio, proportion and percentages · spec M3.10 · Paper 1 style · ★☆☆

Two similar solids have surface areas in the ratio \(4 : 9\). Find the ratio of their volumes.

  1. A\(12 : 27\)
  2. B\(16 : 81\)
  3. C\(2 : 3\)
  4. D\(8 : 27\)
  5. E\(64 : 729\)
  6. F\(4 : 9\)
Worked solution

Length ratio \(2 : 3\), so volume ratio \(2^3 : 3^3 = 8 : 27\).

Full method — how and why
Similar solids: areas scale as the square of lengths, volumes as the cube. Work back to the length ratio first.
  1. Areas are in ratio \(4 : 9\), so the length ratio is \(\sqrt4 : \sqrt9 = 2 : 3\).
    Area scales with the square of length, so undo it with a square root.
  2. Volumes scale with the cube of length: \(2^3 : 3^3\).
    Every linear dimension is scaled, so three dimensions give the cube.
  3. So the volume ratio is \(8 : 27\).
    Check: the ratios \(2:3\), \(4:9\), \(8:27\) are the classic length/area/volume triple.
Where it goes wrong
  • Cubing the area ratio to get \(64 : 729\) - an offered option.
  • Leaving the answer as the length ratio \(2:3\).
Takeaway Length \(k\), area \(k^2\), volume \(k^3\): always return to the length ratio before converting.
Answer D · Ratio, proportion and percentages (M3.10) · question RAT-06
7Ratio, proportion and percentages · spec M3.5 · Paper 1 style · ★☆☆

A mixture contains sand and cement in the ratio \(5 : 2\). How much cement must be added to \(35\) kg of the mixture to make the ratio \(5 : 3\)?

  1. A\(2.5\) kg
  2. B\(15\) kg
  3. C\(3\) kg
  4. D\(7\) kg
  5. E\(5\) kg
  6. F\(10\) kg
Worked solution

\(35\) kg contains \(25\) kg sand and \(10\) kg cement. For \(5 : 3\) with \(25\) kg sand, cement must be \(15\) kg: add \(5\) kg.

Full method — how and why
A changing ratio where one quantity stays fixed. Anchor on the unchanged amount.
  1. Split the \(35\) kg in the ratio \(5:2\): seven parts of \(5\) kg, so \(25\) kg sand and \(10\) kg cement.
    \(35 \div 7 = 5\) kg per part.
  2. Only cement is added, so the sand stays at \(25\) kg.
    Identify what does not change - that is the anchor for the new ratio.
  3. In the new ratio \(5 : 3\), the \(5\) parts correspond to \(25\) kg, so one part is \(5\) kg and the cement must be \(3 \times 5 = 15\) kg.
    Scale the new ratio to the fixed quantity.
  4. Cement to add: \(15 - 10 = 5\) kg.
    The question asks for the amount added, not the final amount.
Where it goes wrong
  • Answering \(15\) kg (the new total cement) - an offered option.
  • Rescaling the sand as well, as though both quantities changed.
Takeaway In 'add to change the ratio' problems, anchor on the quantity that stays the same.
Answer E · Ratio, proportion and percentages (M3.5) · question RAT-07
8Ratio, proportion and percentages · spec M3.11 · Paper 1 style · ★☆☆

\(\pounds 2000\) is invested at \(5\%\) compound interest per year. Which expression gives the amount after \(n\) years?

  1. A\(2000 \times 5^{n}\)
  2. B\(2000 + 100n\)
  3. C\(2000 \times 1.05^{n}\)
  4. D\(2000 \times 1.5^{n}\)
  5. E\(2000(1 + 0.05n)\)
  6. F\(2000 \times 0.05^{n}\)
Worked solution

Compound growth multiplies by \(1.05\) each year.

Full method — how and why
Compound interest is repeated multiplication, which gives an exponential, not a linear, expression.
  1. Each year the amount is multiplied by \(1 + 0.05 = 1.05\).
    Interest is added to the running balance, so next year's interest is larger.
  2. After \(n\) years the original has been multiplied \(n\) times: \(2000 \times 1.05^n\).
    Repeated multiplication gives a power.
Where it goes wrong
  • Choosing \(2000 + 100n\) or \(2000(1 + 0.05n)\) - those are simple interest.
  • Choosing \(2000 \times 0.05^n\), which would shrink the investment to almost nothing.
Takeaway Compound: \(P(1+r)^n\). Simple: \(P(1 + rn)\). The word 'compound' means the multiplier repeats.
Answer C · Ratio, proportion and percentages (M3.11) · question RAT-10
9Ratio, proportion and percentages · spec M3.8 · Paper 1 style · ★☆☆

The value of a car falls by \(20\%\) each year. After how many complete years is it first worth less than half its original value?

  1. A\(5\)
  2. B\(6\)
  3. C\(4\)
  4. D\(10\)
  5. E\(2\)
  6. F\(3\)
Worked solution

\(0.8^3 = 0.512 > 0.5\), \(0.8^4 = 0.4096 < 0.5\). So \(4\) years.

Full method — how and why
Repeated decay compared with a threshold. Test successive powers - no logarithms needed.
  1. Each year multiplies the value by \(0.8\).
    A \(20\%\) fall leaves \(80\%\).
  2. After \(n\) years the multiplier is \(0.8^n\); we need the first \(n\) with \(0.8^n < 0.5\).
    'Less than half' is a condition on the multiplier.
  3. Compute: \(0.8^2 = 0.64\), \(0.8^3 = 0.512\), \(0.8^4 = 0.4096\).
    Multiply successively rather than using logs - three multiplications is quick.
  4. \(0.8^3 = 0.512 > 0.5\) but \(0.8^4 < 0.5\), so the answer is \(4\) complete years.
    Note how close year \(3\) is - checking it is essential.
Where it goes wrong
  • Stopping at \(3\) because \(0.512\) 'is about a half'.
  • Reasoning '\(20\%\) per year, so half gone after \(2.5\) years' - decay is multiplicative.
Takeaway Test successive powers against the threshold, and always check the year before your answer.
Answer C · Ratio, proportion and percentages (M3.8) · question RAT-13
10Ratio, proportion and percentages · spec M3.5 · Paper 1 style · ★☆☆

A recipe for \(12\) biscuits uses \(150\) g of flour and \(100\) g of butter. Ann has \(1\) kg of flour and \(500\) g of butter. What is the greatest number of biscuits she can make?

  1. A\(80\)
  2. B\(60\)
  3. C\(50\)
  4. D\(48\)
  5. E\(66\)
  6. F\(72\)
Worked solution

Flour allows \(\frac{1000}{150} \times 12 = 80\); butter allows \(\frac{500}{100} \times 12 = 60\). Butter is the limit: \(60\).

Full method — how and why
A limiting-factor problem. Work out how many each ingredient allows, then take the smaller.
  1. Flour: \(1\) kg \(= 1000\) g, and \(150\) g makes \(12\) biscuits, so flour allows \(\dfrac{1000}{150} \times 12 = 80\) biscuits.
    Convert to the same units first.
  2. Butter: \(500\) g at \(100\) g per \(12\) biscuits allows \(\dfrac{500}{100} \times 12 = 60\) biscuits.
    Same method for the second ingredient.
  3. Take the smaller: butter runs out first, so \(60\) biscuits.
    You can only make as many as the scarcest ingredient permits.
Where it goes wrong
  • Taking the larger number (\(80\)) - an offered option.
  • Averaging the two, or adding them.
Takeaway With several ingredients, the answer is the minimum over all of them - the limiting factor.
Answer B · Ratio, proportion and percentages (M3.5) · question RAT-14
11Ratio, proportion and percentages · spec M3.8 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IIncreasing a quantity by \(50\%\) and then decreasing the result by \(50\%\) returns it to its original value.
  2. IIA \(10\%\) increase followed by a \(10\%\) increase is a \(21\%\) increase overall.
  3. IIIIf \(A\) is \(25\%\) more than \(B\), then \(B\) is \(20\%\) less than \(A\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is false: \(1.5 \times 0.5 = 0.75\). II: \(1.1^2 = 1.21\), true. III: \(A = 1.25B \Rightarrow B = 0.8A\), true.

Full method — how and why
Three percentage claims. Each is settled by converting to multipliers.
  1. I: \(+50\%\) then \(-50\%\) gives \(1.5 \times 0.5 = 0.75\), i.e. \(75\%\) of the original. False.
    The decrease acts on the larger amount, so it removes more than the increase added.
  2. II: \(+10\%\) twice gives \(1.1^2 = 1.21\), a \(21\%\) increase. True.
    The extra \(1\%\) is the interest on the first year's interest.
  3. III: \(A = 1.25B\) means \(B = \dfrac{A}{1.25} = 0.8A\), which is \(20\%\) less than \(A\). True.
    Percentage changes are not symmetric because the base changes.
  4. II and III only.
    Match to the options.
Where it goes wrong
  • Accepting I because the percentages 'cancel'.
  • Rejecting III on the grounds that \(25\%\) up should mean \(25\%\) down.
Takeaway Always convert to multipliers: percentage changes are asymmetric because the base changes.
Answer G · Ratio, proportion and percentages (M3.8) · question RAT-15
12Ratio, proportion and percentages · spec M3.8 · Paper 1 style · ★★☆

A quantity increases by \(x\%\) and then decreases by \(x\%\). The overall change is a decrease of \(4\%\). Find \(x\).

  1. A\(2\)
  2. B\(20\)
  3. C\(10\)
  4. D\(40\)
  5. E\(4\)
  6. F\(25\)
Worked solution

\((1 + \frac{x}{100})(1 - \frac{x}{100}) = 1 - \frac{x^2}{10000} = 0.96 \Rightarrow x^2 = 400 \Rightarrow x = 20\).

Full method — how and why
An increase then an equal-percentage decrease. The multipliers form a difference of two squares.
  1. Write the multipliers: \(\left(1 + \frac{x}{100}\right)\) then \(\left(1 - \frac{x}{100}\right)\).
    Both act on the running value, so they multiply.
  2. Multiply: \(\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = 1 - \dfrac{x^2}{10\,000}\).
    Difference of two squares - which shows the result is always less than \(1\).
  3. A \(4\%\) overall decrease means the multiplier is \(0.96\): \(1 - \dfrac{x^2}{10\,000} = 0.96\).
    Convert the stated overall change into a multiplier.
  4. Solve: \(\dfrac{x^2}{10\,000} = 0.04 \Rightarrow x^2 = 400 \Rightarrow x = 20\).
    Take the positive root, since \(x\) is a percentage size.
Where it goes wrong
  • Assuming \(x = 4\) because the net change is \(4\%\) - an offered option.
  • Arithmetic: \(0.04 \times 10\,000 = 400\), not \(40\).
Takeaway Up then down by \(x\%\) always loses: the net multiplier is \(1 - \frac{x^2}{10\,000}\).
Answer B · Ratio, proportion and percentages (M3.8) · question RAT-08
13Ratio, proportion and percentages · spec M3.9 · Paper 1 style · ★★☆

\(p\) is directly proportional to \(\sqrt{q}\), and \(q\) is inversely proportional to \(r\). Which of the following is true?

  1. A\(p\) is directly proportional to \(\sqrt{r}\)
  2. B\(p\) is inversely proportional to \(r\)
  3. C\(p\) is inversely proportional to \(r^2\)
  4. D\(p\) is inversely proportional to \(\sqrt{r}\)
  5. E\(p\) is directly proportional to \(r\)
Worked solution

\(p = a\sqrt q\) and \(q = \frac{b}{r}\), so \(p = a\sqrt{\frac br} = \frac{a\sqrt b}{\sqrt r}\).

Full method — how and why
Chaining two proportionality statements. Write each with a constant and substitute.
  1. Write them algebraically: \(p = a\sqrt q\) and \(q = \dfrac{b}{r}\) for constants \(a, b > 0\).
    'Directly proportional' means a constant multiple; 'inversely' means a constant over.
  2. Substitute: \(p = a\sqrt{\dfrac br} = \dfrac{a\sqrt b}{\sqrt r}\).
    The square root distributes over the quotient.
  3. Since \(a\sqrt b\) is a constant, \(p = \dfrac{\text{constant}}{\sqrt r}\): \(p\) is inversely proportional to \(\sqrt r\).
    Collect all the constants into one; only the \(r\)-dependence matters.
Where it goes wrong
  • Forgetting the square root and concluding \(p \propto \frac1r\) - an offered option.
  • Losing the 'inversely' and choosing \(p \propto \sqrt r\).
Takeaway Chain proportionalities by substituting; the accumulated constants never affect the final relationship.
Answer D · Ratio, proportion and percentages (M3.9) · question RAT-09
14Ratio, proportion and percentages · spec M3.1 · Paper 1 style · ★★☆

On a map with scale \(1 : 25\,000\), a lake has area \(8\) cm\(^2\). What is the actual area of the lake in km\(^2\)?

  1. A\(0.05\) km\(^2\)
  2. B\(5\) km\(^2\)
  3. C\(2\) km\(^2\)
  4. D\(50\) km\(^2\)
  5. E\(0.2\) km\(^2\)
  6. F\(0.5\) km\(^2\)
Worked solution

\(1\) cm represents \(25\,000\) cm \(= 0.25\) km, so \(1\) cm\(^2\) represents \(0.0625\) km\(^2\). \(8 \times 0.0625 = 0.5\) km\(^2\).

Full method — how and why
A map scale applies to lengths. For areas the scale factor is squared - and the unit change must be handled too.
  1. Convert the length scale: \(1\) cm on the map is \(25\,000\) cm in reality \(= 250\) m \(= 0.25\) km.
    \(25\,000\) cm \(= 250\) m (divide by \(100\)), then \(= 0.25\) km (divide by \(1000\)).
  2. Square it for areas: \(1\) cm\(^2\) represents \((0.25)^2 = 0.0625\) km\(^2\).
    Area scales with the square of the length factor.
  3. Multiply: \(8 \times 0.0625 = 0.5\) km\(^2\).
    \(8 \times \frac{1}{16} = \frac12\).
Where it goes wrong
  • Using the length factor directly: \(8 \times 0.25 = 2\) km\(^2\) - an offered option.
  • Mishandling the cm-to-km conversion by a factor of \(10\), giving \(0.05\) or \(5\).
Takeaway Map areas use the square of the linear scale; convert units before squaring.
Answer F · Ratio, proportion and percentages (M3.1) · question RAT-11
15Ratio, proportion and percentages · spec M3.6 · Paper 1 style · ★★☆

In a year group the ratio of boys to girls is \(3 : 4\). Two boys leave and two girls join, and the ratio becomes \(2 : 3\). How many students are in the year group?

  1. A\(42\)
  2. B\(56\)
  3. C\(63\)
  4. D\(35\)
  5. E\(70\)
  6. F\(84\)
Worked solution

With \(3k\) boys and \(4k\) girls: \(\dfrac{3k - 2}{4k + 2} = \dfrac23 \Rightarrow 9k - 6 = 8k + 4 \Rightarrow k = 10\). Total \(7k = 70\).

Full method — how and why
A ratio before and after a change. Introduce a multiplier \(k\) so both quantities are expressed with one unknown.
  1. Let the numbers be \(3k\) boys and \(4k\) girls.
    The ratio \(3:4\) fixes only the proportion, so a scale factor \(k\) is needed.
  2. After the change: \(3k - 2\) boys and \(4k + 2\) girls, with ratio \(2 : 3\).
    Two boys leave, two girls join.
  3. Write the equation: \(\dfrac{3k-2}{4k+2} = \dfrac23\), so \(3(3k - 2) = 2(4k + 2)\).
    Cross-multiply.
  4. Solve: \(9k - 6 = 8k + 4 \Rightarrow k = 10\), so the total is \(7k = 70\).
    Check: \(30\) boys and \(40\) girls to start; after the change, \(28 : 42 = 2 : 3\) ✓.
Where it goes wrong
  • Forgetting to answer the total and giving \(k\) or one of the groups.
  • Writing the new ratio upside down.
Takeaway Introduce \(k\) for a ratio, apply the change, cross-multiply, then convert back to the quantity asked for.
Answer E · Ratio, proportion and percentages (M3.6) · question RAT-12

How this topic has been examined

Real TMUA questions on Ratio, proportion and percentages, newest first. Links open the official paper at the question and the online marking page.

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