Elite Mathematics / Linear / Topics / Compound Interest and Depreciation IGCSE · 4MA1 · Topic Compound Interest and Depreciation: past paper questions & worked solutions
Every Compound Interest and Depreciation question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
35 questions
128 marks in total
61% of papers (33 of 54) include it
0/28/7 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
1. Jun 2026 P1HR · Q10 Medium 3 marksShow worked solution Solution | Step 1 \[\text{annual multiplier}=1+\frac{1.4}{100}=1.014.\]
An increase of 1.4% retains the original amount and adds 1.4%.
Solution | Step 2 \[A=200000(1.014)^3=208518.1488\text{ yen}.\]
Apply the multiplier once for each of the 3 years.
Solution | Step 3 \[A\approx208518\text{ yen}.\]
Round to the nearest whole yen.
Final answer \(208518\text{ yen}\)
2. Jun 2026 P1H · Q14 Medium 3 marksShow worked solution Solution | Step 1 Let the original sales be 100 units. After September they are \(100\times1.20=120\); after both months they are \(100\times0.90=90\).
Both overall percentages refer to the original sales.
Solution | Step 2 \[x=\frac{120-90}{120}\times100=25.\]
The October decrease is measured relative to the September sales of 120.
3. Nov 2025 4WM2H · Q12 Medium 3 marksShow worked solution Calculate percentage Decrease by 5% ⟹ Multiplier = 1 − 0.05 = 0.95
Increase by 20% ⟹ Multiplier = 1 + 0.20 = 1.20
Increase by 6.5% ⟹ Multiplier = 1 + 0.065 = 1.065
Calculate value Overall Multiplier = 0.95 × 1.20 × 1.065 = 1.2141
Calculate percentage 1.2141 − 1 = 0.2141 = 21.41%
4. Jun 2025 P2H · Q7 Medium 3 marksShow worked solution The multiplier for a increase The multiplier for a \(7\%\) increase is
\[
1.07
\]
Use compound interest After 3 years,
\[
4000(1.07)^3=4900.172
\]
Correct to the nearest Swiss Correct to the nearest Swiss franc,
\[
4900.172\approx 4900
\]
Final answer \(4900\) Swiss francs
5. Jun 2025 P2HR · Q12 Medium 3 marksShow worked solution Use compound interest After the first 2 years,
\[
16000(0.88)^2=12390.40
\]
Use compound interest In the third year, the value becomes \(11461.12\), so
\[
12390.40\left(1-\frac{x}{100}\right)=11461.12
\]
\[
1-\frac{x}{100}=\frac{11461.12}{12390.40}=0.925
\]
\[
x=7.5
\]
6. Jun 2025 P1H · Q13 Medium 3 marksShow worked solution Use compound interest After the first two years, the value is
\[
750000(0.96)(0.935)=673200
\]
Use compound interest Then the value increases by \(x\%\) to \(698445\), so
\[
673200\left(1+\frac{x}{100}\right)=698445
\]
\[
1+\frac{x}{100}=\frac{698445}{673200}=1.0375
\]
\[
x=3.75
\]
7. May 2025 4WM2H · Q9 Medium 3 marksShow worked solution The multiplier for a increase The multiplier for a \(7\%\) increase is
\[
1.07
\]
Use compound interest After 3 years,
\[
4000(1.07)^3=4900.172
\]
Correct to the nearest Swiss Correct to the nearest Swiss franc,
\[
4900.172\approx 4900
\]
Final answer \(4900\) Swiss francs
8. May 2025 4WM2HR · Q12 Medium 3 marksShow worked solution Required Car bought for $16 000; depreciates 12%/yr for 2 yrs, then *x*%/yr in year 3. Value after 3 years = $11 461.12. Find *x*.
Value after 2 years at Value after 2 years at 12% depreciation: *V*<sub>2</sub> = 16000 × (0.88)<sup>2</sup> = 16000 × 0.7744 = $12 390.40
Use compound interest Apply year-3 depreciation and set equal to final value: $$12390.40\\times\\!\\left(1 - \\frac{x}{100}\\right) = 11461.12$$
Solve Solve: $$1 - \\frac{x}{100} = \\frac{11461.12}{12390.40} = 0.925 \\implies \\frac{x}{100} = 0.075 \\implies x = 7.5$$
9. May 2025 4WM2H · Q15 Medium 3 marksShow worked solution Use multipliers Use multipliers:
\[
1.12\times 0.85=0.952
\]
Calculate percentage So the final value is \(95.2\%\) of the original.
\[
100-95.2=4.8
\]
Final answer \(4.8\%\) decrease.
10. Nov 2024 P2H · Q9 Medium 3 marksShow worked solution The multiplier for compound interest The multiplier for \(3.5\%\) compound interest is
\[
1.035
\]
Use compound interest After 3 years,
\[
20000(1.035)^3=22174.3575
\]
Use compound interest Correct to the nearest euro,
\[
22174.3575\approx 22174
\]
Final answer \(22174\) euros
11. Jun 2024 P2H · Q9 Medium 3 marksShow worked solution The multiplier for compound interest The multiplier for \(2.4\%\) compound interest is
\[
1.024
\]
Use compound interest After 4 years,
\[
4500(1.024)^4=4947.8023\ldots
\]
Use compound interest Correct to the nearest koruna,
\[
4947.8023\ldots\approx 4948
\]
Final answer \(4948\) koruna
12. May 2024 P1HR · Q5 Medium 3 marksShow worked solution The multiplier for compound interest The multiplier for \(2.5\%\) compound interest is
\[
1.025
\]
Use compound interest After 4 years,
\[
5200(1.025)^4=5739.827\ldots
\]
Use compound interest Correct to the nearest euro,
\[
5739.827\ldots\approx 5740
\]
Final answer \(5740\) euros
13. Nov 2023 P2H · Q7 Medium 3 marksShow worked solution A depreciation of gives multiplier A depreciation of \(8\%\) gives multiplier
\[
0.92
\]
Use compound interest After 3 years,
\[
26800(0.92)^3=20868.8384
\]
Use compound interest Correct to the nearest dollar,
\[
20868.8384\approx 20869
\]
14. Jun 2023 P2H · Q8 Medium 3 marksShow worked solution A increase each year gives A \(4\%\) increase each year gives a multiplier of
\[
1.04
\]
Use compound interest After 3 years, the value is
\[
680(1.04)^3
\]
\[
=680(1.124864)
\]
\[
=764.90752
\]
Correct to the nearest dollar Correct to the nearest dollar:
\[
765
\]
15. Jun 2023 P2HR · Q13 Medium 3 marksShow worked solution The final value is \[
80000+6151.25=86151.25
\]
Let the annual compound interest Let the annual compound interest rate be \(x\%\).
\[
80000\left(1+\frac{x}{100}\right)^3=86151.25
\]
\[
1+\frac{x}{100}=\sqrt[3]{\frac{86151.25}{80000}}=1.025
\]
\[
x=2.5
\]
16. May 2023 P1HR · Q9 Medium 3 marksShow worked solution A depreciation of gives multiplier A depreciation of \(12\%\) gives multiplier
\[
0.88
\]
Use compound interest After 3 years,
\[
700000(0.88)^3=477030.4
\]
Use compound interest Correct to the nearest taka,
\[
477030.4\approx 477030
\]
Final answer \(477030\) taka
17. Jun 2022 P1HR · Q8 Medium 3 marksShow worked solution A depreciation of gives multiplier A depreciation of \(15\%\) gives multiplier
\[
0.85
\]
Use compound interest After 4 years,
\[
18000(0.85)^4=9396.1125
\]
Use compound interest Correct to the nearest dollar,
\[
9396.1125\approx 9396
\]
18. Jun 2022 P2H · Q9 Medium 3 marksShow worked solution The multiplier for compound interest The multiplier for \(1.3\%\) compound interest is
\[
1.013
\]
Use compound interest After 4 years,
\[
50000(1.013)^4=52651.1408\ldots
\]
Use compound interest Correct to the nearest dollar,
\[
52651.1408\ldots\approx 52651
\]
Final answer \(52651\) dollars
19. Jan 2022 P2HR · Q5 Medium 3 marksShow worked solution The multiplier for compound interest The multiplier for \(2.5\%\) compound interest is
\[
1.025
\]
Use compound interest After 3 years,
\[
7200(1.025)^3=7753.6125
\]
Use compound interest Correct to the nearest dollar,
\[
7753.6125\approx 7754
\]
Final answer \(7754\) dollars
20. Jan 2022 P1H · Q11 Medium 3 marksShow worked solution The first year multiplier is The first-year multiplier is
\[
1.018
\]
The second and third years The second and third years both have multiplier \(1+\frac{x}{100}\), so
\[
200000(1.018)\left(1+\frac{x}{100}\right)^2=209754
\]
\[
\left(1+\frac{x}{100}\right)^2=\frac{209754}{200000(1.018)}
\]
\[
1+\frac{x}{100}=1.015000\ldots
\]
\[
x=1.5000\ldots
\]
21. Jan 2021 P1HR · Q7 Medium 3 marksShow worked solution The multiplier for a increase The multiplier for a \(2.4\%\) increase is
\[
1.024
\]
Use compound interest After 3 years,
\[
50000(1.024)^3=53687.0912
\]
Use compound interest Correct to the nearest yen,
\[
53687.0912\approx 53687
\]
Final answer \(53687\) yen
22. Jan 2020 P2HR · Q8 Medium 3 marksShow worked solution A depreciation of gives a A depreciation of \(19\%\) gives a multiplier of
\[
1-0.19=0.81
\]
Use compound interest After 3 years,
\[
20000(0.81)^3=10628.82
\]
Use compound interest Correct to the nearest dollar,
\[
10628.82\approx 10629
\]
23. Jan 2020 P1H · Q11 Medium 3 marksShow worked solution Use compound interest After the first 2 years,
\[
6000(1.015)^2=6181.35
\]
Let the third year interest Let the third-year interest rate be \(r\%\).
\[
6181.35\left(1+\frac{r}{100}\right)=6311.16
\]
\[
1+\frac{r}{100}=\frac{6311.16}{6181.35}=1.021000\ldots
\]
\[
r=2.1000\ldots
\]
24. May-Nov 2020 P1HR · Q8 Medium 3 marksShow worked solution The multiplier for an increase The multiplier for an \(8\%\) increase is
\[
1.08
\]
Value after 3 years Value after 3 years:
\[
1200(1.08)^3=1511.6544
\]
Use compound interest Correct to the nearest pound,
\[
1511.6544\approx 1512
\]
Final answer \(1512\) pounds
25. May-Nov 2020 P1H · Q13 Medium 3 marksShow worked solution Let the compound interest rate Let the compound interest rate be \(x\%\).
\[
8000\left(1+\frac{x}{100}\right)^6=8877.62
\]
\[
1+\frac{x}{100}=\sqrt[6]{\frac{8877.62}{8000}}=1.017500\ldots
\]
\[
x=1.7500\ldots
\]
Use compound interest Correct to 2 decimal places,
\[
x=1.75
\]
26. Jun 2026 P2H · Q10 Medium 4 marksShow worked solution Solution | Step 1 The annual multiplier is \(1-0.15=0.85\). After 3 years the value is \[32000(0.85)^3=19652\text{ euros}.\]
Apply the decrease to each successive year, giving a power of 3.
Solution | Step 2 Lena still needs \(19652-2516=17136\) euros. There are \(3\times12=36\) months.
Subtract her existing savings and convert years to months.
Solution | Step 3 \[\text{Monthly saving}=\frac{17136}{36}=476\text{ euros}.\]
Divide the remaining target equally over the months.
Final answer \(476\text{ euros per month}\)
27. Nov 2021 P2H · Q7 Medium 4 marksShow worked solution Ali s interest from Cyclone Ali's interest from Cyclone Bank is
\[
25000(1.045)^3-25000=3529.153\ldots
\]
Badia s interest from Tornado Badia's interest from Tornado Bank is
\[
1150\times 3=3450
\]
Difference Difference:
\[
3529.153\ldots-3450=79.153\ldots
\]
Use compound interest Correct to the nearest dollar,
\[
79.153\ldots\approx 79
\]
28. Jan 2021 P2HR · Q14 Medium 4 marksShow worked solution Let the original house value Let the original house value be \(V\).
Use compound interest After the \(15\%\) decrease in 2018:
\[
0.85V
\]
Use compound interest Let the percentage increase in each of 2019 and 2020 be \(r\).
The final value is greater The final value is \(2.85\%\) greater than the original value:
\[
1.0285V
\]
Use compound interest So
\[
0.85V(1+r)^2=1.0285V
\]
\[
(1+r)^2=\frac{1.0285}{0.85}=1.21
\]
\[
1+r=\sqrt{1.21}=1.1
\]
\[
r=0.1=10\%
\]
29. Jun 2025 P1HR · Q11 Hard 5 marksShow worked solution Use compound interest For Bank A,
\[
\text{amount invested}:\text{interest}=20:3
\]
Use compound interest So the interest from Bank A is
\[
2500\times \frac{3}{20}=375
\]
Use compound interest For Bank B, the compound interest after 2 years is
\[
3000(1.04)^2-3000
\]
\[
=3000(1.0816)-3000=244.80
\]
Difference Difference:
\[
375-244.80=130.20
\]
30. Nov 2025 P2H · Q5 Hard 6 marksShow worked solution Use compound interest For part (a), the increase is
\[
87-75=12
\]
Percentage increase Percentage increase:
\[
\frac{12}{75}\times 100=16
\]
Use compound interest For part (b), a depreciation of \(20\%\) gives multiplier
\[
0.80
\]
Use compound interest After 3 years,
\[
960(0.80)^3=491.52
\]
Final answer (a) \(16\%\), (b) \(491.52\) Swiss francs
31. Nov 2025 4WM2H · Q6 Hard 6 marksShow worked solution Calculate value Increase = 87 − 75 = 12
Convert standard form $$\\text{\\% Increase} = \\frac{\\text{Increase}}{\\text{Original}} \\times 100 = \\frac{12}{75} \\times 100 = 16\\%$$
Answer 16%
**Step 1: Determine the depreciation multiplier.** A 20% decrease means the multiplier is 1 − 0.20 = 0.80.
Solve inequality Final Value = 960 × (0.80)<sup>3</sup> Final Value = 960 × 0.512 = 491.52
Final answer 491.52 Swiss francs
32. Jun 2022 P1H · Q6 Hard 6 marksShow worked solution Use compound interest For part (a), the value in 2021 was \(4\%\) more than in 2020.
\[
\text{2020 value}=\frac{634400}{1.04}=610000
\]
Use compound interest In 2020, \(600000<x<700000\), so the service charge band was B.
Use compound interest In 2021, \(634400\) is also between \(600000\) and \(700000\), so the band was still B.
Use compound interest For part (b), a depreciation of \(15\%\) has multiplier \(0.85\).
Use compound interest After 2 years, the multiplier is
\[
0.85^2=0.7225
\]
Use compound interest So the value left is \(72.25\%\), meaning the total depreciation is
\[
100\%-72.25\%=27.75\%
\]
Final answer (a) No, it stayed in band B. (b) \(27.75\%\)
33. May 2021 P2H · Q11 Hard 6 marksShow worked solution Use compound interest For Chengbo's house, \(180000\) yuan is \(124\%\) of the price paid.
\[
\text{price paid}=180000\div 1.24=145161.29\ldots
\]
To 3 significant figures To 3 significant figures:
\[
145000
\]
Use compound interest For Zhi's house, the value increases by \(1.8\%\) each year for 3 years.
\[
120000\times 1.018^3=126598.2\ldots
\]
To 3 significant figures To 3 significant figures:
\[
127000
\]
Final answer Chengbo paid \(145000\) yuan. Zhi's house was worth \(127000\) yuan.
34. Jan 2021 P1H · Q7 Hard 6 marksShow worked solution Use compound interest For part (a),
\[
40000(1.035)^3=44348.715
\]
Use compound interest Correct to the nearest yuan,
\[
44348.715\approx 44349
\]
Use compound interest For part (b), after one year the value is \(93.5\%\) of \(P\).
\[
0.935P=30481
\]
\[
P=\frac{30481}{0.935}=32600
\]
Final answer (a) \(44349\) yuan, (b) \(P=32600\)
35. Specimen 4WM2H · Q8 Hard 6 marksShow worked solution Use compound interest For part (a), the 2021 value is \(104\%\) of the 2020 value.
\[
\text{2020 value}=\frac{634400}{1.04}=610000
\]
Use compound interest For part (b), after one year the multiplier is \(0.85\). After two years:
\[
0.85^2=0.7225
\]
Use compound interest So \(72.25\%\) of the value remains, and the total depreciation is
\[
100\%-72.25\%=27.75\%
\]
Final answer \(610000\) euros; \(27.75\%\).
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