Compound interest, growth & depreciation – GCSE Maths exam questions with mark scheme

1Grade 4–5Calculator3 marks

Leah invests £4000 in a savings account for 3 years. The account pays 2.5% compound interest per year.

Work out the value of her investment at the end of 3 years. [3]

for $1.025$ used as a multiplier, or $4000 \times 0.025 = 100$ then $4100$M1
for $4000 \times 1.025^3$ or a complete year-by-year methodM1
£4307.56 (accept £4307.5625 or £4307.562…)A1
Answer: £4307.56
Examiner tip: Compound interest is not 3 × £100; each year's interest is worked out on the new total.

2Grade 4–5Calculator4 marks

Ben has £2500 to invest for 4 years. He considers two accounts.

Account A
3% simple interest per year

Account B
2.8% compound interest per year

Which account will give Ben more interest after 4 years? You must show your working. [4]

for $2500 \times 0.03 \times 4$ or $75 \times 4$ or $300$ seenM1
for $2500 \times 1.028^4$ (or $2791.98…$ seen)M1
for £300 and £291.98 (interest), or £2800 and £2791.98 (totals)A1
Account A, supported by correct figuresC1
Answer: Account A (£300 vs £291.98 interest)
Examiner tip: Compare like with like – either both interests or both final totals – and state the account clearly.

3Grade 4–5Calculator3 marks

A new car costs £18 500. The value of the car depreciates by 12% each year.

Work out the value of the car after 4 years. Give your answer to the nearest pound. [3]

for $0.88$ used as a multiplier, or $18\,500 \times 0.12 = 2220$ then $16\,280$M1
for $18\,500 \times 0.88^4$ or a complete year-by-year methodM1
£11 094A1
Answer: £11 094
Examiner tip: Depreciation multiplier is 1 − 0.12 = 0.88; taking 4 × 12% = 48% off gives £9620, which is wrong.

4Grade 6–7HigherCalculator3 marks

At the start of an experiment there are 500 bacteria. The number of bacteria increases by 8% every hour.

After how many whole hours will the number of bacteria first be more than 1000? [3]

for $500 \times 1.08^n$ evaluated for at least one value of $n \geq 2$M1
for $500 \times 1.08^9 = 999.5…$ and $500 \times 1.08^{10} = 1079.4…$ (or $1.08^9 = 1.999$ and $1.08^{10} = 2.159$)M1
$10$ (hours)A1
Answer: 10 hours
Examiner tip: After 9 hours there are 999.5 bacteria – just under 1000 – so check both sides of the boundary before deciding.

5Grade 6–7HigherCalculator3 marks

Mo invests £6000 in an account paying $r\%$ compound interest per year. After 3 years the investment is worth £6749.18

Work out the value of $r$. [3]

for $\frac{6749.18}{6000}$ $(= 1.1248…)$ or $6000 \times m^3 = 6749.18$M1
for $\sqrt[3]{\frac{6749.18}{6000}}$ $(= 1.04)$M1
$r = 4$A1
Answer: r = 4
Examiner tip: The cube root gives the multiplier 1.04, not the rate – subtract 1 and convert to a percentage.

6Grade 6–7HigherCalculator4 marks

At the start of 2023 a house was worth £240 000.

During 2023 its value increased by 5%.
During 2024 its value decreased by 3%.
During 2025 its value increased by $x\%$.

At the end of 2025 the house was worth £254 217.60

Work out the value of $x$. [4]

for $240\,000 \times 1.05 \times 0.97$ or $252\,000$ then $\times 0.97$M1
for £244 440A1
for $254\,217.60 \div 244\,440$ $(= 1.04)$ or $(254\,217.60 - 244\,440) \div 244\,440 \times 100$M1
$x = 4$A1
Answer: x = 4
Examiner tip: Each percentage change applies to the value at the end of the previous year, not to £240 000.

7Grade 6–7HigherCalculator3 marks

The value, £$V$, of a machine $t$ years after it is bought is given by

$$V = 25\,000 \times 0.85^t$$

(a) Write down the percentage by which the value decreases each year. [1]

(b) Work out the number of complete years after which the value of the machine is first less than half of its value when new. [2]

(a) $15\%$B1
(b) for $25\,000 \times 0.85^n$ (or $0.85^n$) evaluated for $n = 4$ and $n = 5$, giving $13\,050.15…$ and $11\,092.63…$ (or $0.522$ and $0.4437$), compared with $12\,500$ (or $0.5$)M1
(b) $5$ (years)A1
Answer: (a) 15% (b) 5 years
Examiner tip: Half of the original value is £12 500 – compare with that, not with £25 000.

8Grade 1–3FoundationNon-calculator4 marks

(a) Write down the decimal multiplier that increases an amount by 6%. [1]

(b) Write down the decimal multiplier that decreases an amount by 4%. [1]

Work out the value of the bike after 2 years. [2]

(a) $1.06$ (cao – the question asks for a decimal, so do not accept $106\%$)B1
(b) $0.96$ (cao – do not accept $96\%$)B1
(c) for a correct first year: $400 \times 0.9$ or $400 - 40$ or $360$ seenM1
(c) £324 (do not accept £320, which comes from taking 20% of 400)A1
Answer: (a) 1.06 (b) 0.96 (c) £324
Examiner tip: The second year's 10% is 10% of £360, not of £400 – taking 20% off in one go gives the common wrong answer £320.

9Grade 4–5Non-calculator2 marks

Priya says,

“If a price increases by 10% and then the new price increases by another 10%, that is the same as a single increase of 20%.”

Is Priya correct? You must show how you get your answer. [2]

for $1.1 \times 1.1$ or $1.1^2$ or $1.21$ seen, or a correct worked example, e.g. $100 \to 110 \to 121$ (compared with $120$), or $200 \to 220 \to 242$ (compared with $240$)M1
No (Priya is not correct), with a correct comparison, e.g. the overall increase is 21% not 20%, or 121 ≠ 120 (accept a correct explanation that the second 10% is taken of a larger amount, if supported by working)C1
Answer: No – the overall multiplier is 1.21, a 21% increase
Examiner tip: A bare “No” scores nothing – you must show a calculation (a multiplier or a worked example) that proves the total is 21%.

10Grade 4–5Calculator5 marks

The graph shows the value of a phone over the first 4 years after it was bought.

The value of the phone decreases by the same percentage each year.

(a) Use the graph to work out the percentage by which the value decreases each year. [2]

(b) Work out the value of the phone 7 years after it was bought. Give your answer to the nearest penny. [2]

(a) for reading £800 (at 0 years) and £640 (at 1 year) and calculating $\frac{800-640}{800}$ or $\frac{640}{800}$ $(=0.8)$ (allow use of any pair of consecutive whole-year readings, e.g. 640 and 512)M1
(a) $20\%$A1
(b) for $800 \times 0.8^7$ ft their percentage from (a), or $640 \times 0.8^6$, or year-by-year method with at least 5 correct values after year 2 (allow a fully correct repeated method)M1
(b) £167.77 (accept 167.77216… rounded or truncated to 167.77; ft $800 \times (1 - p/100)^7$ for their $p$)A1
(c) e.g. each year the value is multiplied by 0.8 (it keeps 80% of its value), so it always stays above 0 / it only ever loses a fraction of what is left (do not accept just “the curve doesn’t touch the axis” without reference to the percentage/multiplier)C1
Answer: (a) 20% (b) £167.77 (c) the value is multiplied by 0.8 each year, so it is always a positive amount
Examiner tip: Read the graph at whole years (0 and 1) where the curve crosses gridlines – then use the multiplier 0.8, not 800 − 160 × 7.

11Grade 6–7HigherCalculator3 marks

A farmer bought a tractor. The value of the tractor depreciated by 7% each year.

After 3 years the tractor was worth £38 609.14

Work out the value of the tractor when the farmer bought it. Give your answer to the nearest pound. [3]

for using the multiplier $0.93$ (e.g. $0.93^3$ or $0.804357$ seen)M1
for $38\,609.14 \div 0.93^3$ oe, or $38\,609.14 \div 0.93 \div 0.93 \div 0.93$M1
£48 000 (accept 48 000.00…)A1
Answer: £48 000
Examiner tip: Undo a decrease by dividing by 0.93 – multiplying by 1.07 three times gives the wrong answer £47 298.

12Grade 8–9HigherCalculator5 marks

A lake is stocked with fish. A model for the number of fish is

$$P_{n+1} = 1.2\,P_n - 300$$

where $P_n$ is the number of fish $n$ years after the lake is stocked.

The lake is stocked with 2000 fish, so $P_0 = 2000$.

(a) Work out the number of fish 3 years after the lake is stocked. [2]

(b) Work out the number of years after which the number of fish is first more than 3000. [2]

Describe what the model predicts for the number of fish in this lake. Give a reason for your answer. [1]

(a) for at least one correct iteration, e.g. $P_1 = 1.2 \times 2000 - 300 = 2100$, or $P_2 = 2220$M1
(a) $2364$A1
(b) for continuing the iteration to show values either side of 3000, e.g. $P_6 = 2992.99…$ (or $2993$) and $P_7 = 3291.59…$ (or $3292$) (allow values rounded to whole fish at each stage)M1
(b) $7$ (years)A1
(c) the number of fish stays at 1500 (does not change), because $1.2 \times 1500 - 300 = 1500$C1
Answer: (a) 2364 (b) 7 years (c) stays at 1500, since 1.2 × 1500 − 300 = 1500
Examiner tip: Use each new value as the next input – and in (b) check the year either side, because year 6 (≈ 2993) is only just below 3000.

13Grade 6–7HigherCalculator4 marks

Sam invested some money in an account paying compound interest at $r\%$ per year.

At the end of 2 years the investment was worth £5408
At the end of 3 years the investment was worth £5624.32

(a) Work out the value of $r$. [2]

(b) Work out how much money Sam invested. [2]

(a) for $5624.32 \div 5408$ $(=1.04)$, or $5624.32 - 5408 = 216.32$ and $216.32 \div 5408$M1
(a) $r = 4$A1
(b) for $5408 \div 1.04^2$ or $5624.32 \div 1.04^3$ (ft their multiplier from (a))M1
(b) £5000A1
Answer: (a) r = 4 (b) £5000
Examiner tip: The interest rate comes from one year’s growth (year 2 to year 3) – then divide by the multiplier twice, not by 1.08.

14Grade 6–7HigherCalculator3 marks

Two savings accounts both pay compound interest.

Account A
0.4% per month, added at the end of each month

Account B
4.9% per year, added at the end of each year

Which account pays more interest over one year? You must show your working. [3]

for $1.004^{12}$ $(=1.04907…)$, or for any amount $x$: $x \times 1.004^{12}$ (e.g. $1000 \times 1.004^{12} = 1049.07…$)M1
for $4.907…\%$ (accept 4.91%) compared with $4.9\%$, or correct comparable values for the same amount (e.g. £1049.07 and £1049, or £49.07 and £49 interest on £1000)A1
Account A, supported by correct figures (do not accept a comparison of $0.4 \times 12 = 4.8\%$ with 4.9%, which leads to the wrong account)C1
Answer: Account A (equivalent to 4.907…% per year, more than 4.9%)
Examiner tip: Monthly interest compounds – use $1.004^{12}$, not $0.4 \times 12 = 4.8\%$, which would wrongly choose Account B.

15Grade 4–5Calculator4 marks

Jess invests £1500 in a savings account. The account pays 3.5% compound interest per year.

(a) Which expression gives the value of Jess’s investment after 5 years? Circle your answer. [1]

$1500 \times 0.035^5$ $1500 \times 1.35^5$ $1500 \times 1.035^5$ $1500 \times 1.035 \times 5$

(b) Work out the total amount of interest Jess earns in the 5 years. Give your answer to the nearest penny. [3]

(a) $1500 \times 1.035^5$B1
(b) for $1500 \times 1.035^5$ oe, or a complete year-by-year methodM1
(dep) for subtracting 1500 from their total, e.g. $1781.53 - 1500$ (or for $1500 \times (1.035^5 - 1)$)M1
(b) £281.53A1
Answer: (a) $1500 \times 1.035^5$ (b) £281.53
Examiner tip: The question asks for the interest, not the final amount – subtract the £1500 Jess put in.

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