Hard GCSE Simple and compound interest Questions

Challenging, exam-style GCSE Simple and compound interest questions with worked solutions. Stretch yourself on the hardest compound interest, year by year, rounding to the penny, interest earned problems.

compound interestyear by yearrounding to the pennyinterest earneddepreciationcompound decay
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Jonah opens an account with £500\pounds 500. The account pays 6%6\% compound interest per year. At the end of each year, immediately after the interest has been added, Jonah pays in a further £500\pounds 500. Work out how much is in the account at the end of 55 years.
Show worked solution

Worked solution

  1. Plan the year-by-year table

    balance×1.06+500\text{balance} \to \times 1.06 \to +\,500

    Interest first, then the £500 payment, five times over. The exact balance carries forward every year. Exam convention: keep the full unrounded balance in the calculator all the way through and round to the nearest penny only at the very end.

  2. Write the yearly percentage as a multiplier

    100%+6%=106%=1.06100\% + 6\% = 106\% = 1.06

    A 6% increase keeps the whole 100% and adds 6% on top, so multiply by 1.06 each year.

  3. Year 1: add the interest

    £500×1.06=£530\pounds 500 \times 1.06 = \pounds 530

    6% interest on £500 is £30.00, giving £530.

  4. Year 1: add the £500 payment

    £530+500=£1030\pounds 530 + 500 = \pounds 1030

    The payment goes in after the interest, so the balance is £1030.

  5. Year 2: add the interest

    £1030×1.06=£1091.80\pounds 1030 \times 1.06 = \pounds 1091.80

    6% interest on £1030 is £61.80, giving £1091.80.

  6. Year 2: add the £500 payment

    £1091.80+500=£1591.80\pounds 1091.80 + 500 = \pounds 1591.80

    The payment goes in after the interest, so the balance is £1591.80.

  7. Year 3: add the interest

    £1591.80×1.06£1687.31\pounds 1591.80 \times 1.06 \approx \pounds 1687.31

    6% interest on £1591.80 is £95.51, giving £1687.31.

  8. Year 3: add the £500 payment

    £1687.31+500£2187.31\pounds 1687.31 + 500 \approx \pounds 2187.31

    The payment goes in after the interest, so the balance is £2187.31.

  9. Year 4: add the interest

    £2187.31×1.06£2318.55\pounds 2187.31 \times 1.06 \approx \pounds 2318.55

    6% interest on £2187.31 is £131.24, giving £2318.55.

  10. Year 4: add the £500 payment

    £2318.55+500£2818.55\pounds 2318.55 + 500 \approx \pounds 2818.55

    The payment goes in after the interest, so the balance is £2818.55.

  11. Year 5: add the interest

    £2818.55×1.06£2987.66\pounds 2818.55 \times 1.06 \approx \pounds 2987.66

    6% interest on £2818.55 is £169.11, giving £2987.66.

  12. Year 5: add the £500 payment

    £2987.66+500£3487.66\pounds 2987.66 + 500 \approx \pounds 3487.66

    The payment goes in after the interest, so the balance is £3487.66.

  13. Round to the nearest penny

    £3487.66£3487.66\pounds 3487.66 \to \pounds 3487.66

    The full-precision balance is rounded to the penny only at the end.

  14. Work out the total she paid in

    500+5×500=£3000500 + 5 \times 500 = \pounds 3000

    She paid in £3000 of her own money altogether.

  15. State the answer

    £3487.66\pounds 3487.66

    After 5 years the account holds £3487.66, of which £487.66 is interest.

Answer
£3487.66\pounds 3487.66
Question 2
6 markschallenging
Ellie pays £2500\pounds 2500 into an account that gives 3.5%3.5\% compound interest per year. At the end of every year, immediately after the interest has been added, she pays in a further £100\pounds 100. Work out the smallest number of complete years before the account is worth more than £3200\pounds 3200.
Show worked solution

Worked solution

  1. Plan the year-by-year table

    balance×1.035+100\text{balance} \to \times 1.035 \to +\,100

    Each year: interest first, then the £100 payment. Keep going until the balance passes £3200. Exam convention: keep the full unrounded balance in the calculator all the way through and round to the nearest penny only at the very end.

  2. Write the yearly percentage as a multiplier

    100%+3.5%=103.5%=1.035100\% + 3.5\% = 103.5\% = 1.035

    A 3.5% increase keeps the whole 100% and adds 3.5% on top, so multiply by 1.035 each year.

  3. Year 1: interest, then the payment

    £2500×1.035+100=£2587.50+100=£2687.50\pounds 2500 \times 1.035 + 100 = \pounds 2587.50 + 100 = \pounds 2687.50

    Interest of £87.50 is added to £2500, then the £100 payment goes in: £2687.50. The exact balance carries forward.

  4. Year 2: interest, then the payment

    £2687.50×1.035+100£2781.56+100£2881.56\pounds 2687.50 \times 1.035 + 100 \approx \pounds 2781.56 + 100 \approx \pounds 2881.56

    Interest of £94.06 is added to £2687.50, then the £100 payment goes in: £2881.56. The exact balance carries forward.

  5. Year 3: interest, then the payment

    £2881.56×1.035+100£2982.42+100£3082.42\pounds 2881.56 \times 1.035 + 100 \approx \pounds 2982.42 + 100 \approx \pounds 3082.42

    Interest of £100.85 is added to £2881.56, then the £100 payment goes in: £3082.42. The exact balance carries forward.

  6. Year 4: interest, then the payment

    £3082.42×1.035+100£3190.30+100£3290.30\pounds 3082.42 \times 1.035 + 100 \approx \pounds 3190.30 + 100 \approx \pounds 3290.30

    Interest of £107.88 is added to £3082.42, then the £100 payment goes in: £3290.30. The exact balance carries forward.

  7. Year 5: interest, then the payment

    £3290.30×1.035+100£3405.46+100£3505.46\pounds 3290.30 \times 1.035 + 100 \approx \pounds 3405.46 + 100 \approx \pounds 3505.46

    Interest of £115.16 is added to £3290.30, then the £100 payment goes in: £3505.46. The exact balance carries forward.

  8. Check year 3 against the target

    £3082.42<£3200\pounds 3082.42 < \pounds 3200

    After 3 years the balance is £3082.42, which is still £117.58 SHORT of £3200. So 3 years is not enough.

  9. Check year 4 against the target

    £3290.30>£3200\pounds 3290.30 > \pounds 3200

    After 4 years the balance is £3290.30, which is £90.30 OVER £3200. Year 4 is the first year the target is passed.

  10. Check the total she has paid in

    2500+4×100=£29002500 + 4 \times 100 = \pounds 2900

    She has paid in £2900 of her own money, so £390.30 of the final balance is interest.

  11. Say why the payments alone are not enough

    2500+4×100=2900<32002500 + 4 \times 100 = 2900 < 3200

    Without any interest the account would hold only £2900 after 4 years, which is under the target. The interest is doing real work here.

  12. Say why the interest alone is not enough either

    2500×1.0354£2868.81<£32002500 \times 1.035^{4} \approx \pounds 2868.81 < \pounds 3200

    And with no payments at all, £2500 would grow to only £2868.81 in 4 years. It takes BOTH the payments and the interest to pass £3200 in year 4.

  13. Work out the interest earned in year 4 alone

    0.035×£3082.42£107.880.035 \times \pounds 3082.42 \approx \pounds 107.88

    Year 4 earns £107.88 of interest -- more than year 1 earned (£87.50), because the balance it is worked out on is bigger.

  14. Confirm year 5 stays above

    £3505.46>£3200\pounds 3505.46 > \pounds 3200

    The balance keeps growing (£3505.46 after 5 years), so once past £3200 it stays past it.

  15. State the answer

    n=4n = 4

    After 4 complete years the account is worth more than £3200 for the first time.

Answer
n=4n = 4
Question 3
6 markschallenging
Amir can invest £3000\pounds 3000 in Account A, which pays 5%5\% compound interest per year, or £3200\pounds 3200 in Account B, which pays 3%3\% compound interest per year. Which account is worth more after 44 years, and by how much?
Show worked solution

Worked solution

  1. Plan the comparison

    A: 3000×1.054vsB: 3200×1.034\text{A: } 3000 \times 1.05^{4} \quad\text{vs}\quad \text{B: } 3200 \times 1.03^{4}

    Account B starts with more money but grows more slowly. Work out both after 4 years. Exam convention: keep the full unrounded balance in the calculator all the way through and round to the nearest penny only at the very end.

  2. Write the yearly percentage as a multiplier

    100%+5%=105%=1.05100\% + 5\% = 105\% = 1.05

    A 5% increase keeps the whole 100% and adds 5% on top, so multiply by 1.05 each year.

  3. Balance after year 1

    £3000×1.051=£3150\pounds 3000 \times 1.05^{1} = \pounds 3150

    Year 1: £3000×1.05=£3150£3000 \times 1.05 = £3150. This is worked out from the exact starting amount as P×multiplier1P \times \text{multiplier}^{1}, so no rounding error can build up.

  4. Balance after year 2

    £3000×1.052=£3307.50\pounds 3000 \times 1.05^{2} = \pounds 3307.50

    Year 2: £3150×1.05=£3307.50£3150 \times 1.05 = £3307.50. This is worked out from the exact starting amount as P×multiplier2P \times \text{multiplier}^{2}, so no rounding error can build up.

  5. Balance after year 3

    £3000×1.053£3472.88\pounds 3000 \times 1.05^{3} \approx \pounds 3472.88

    Year 3: £3307.50×1.05=£3472.88£3307.50 \times 1.05 = £3472.88. This is worked out from the exact starting amount as P×multiplier3P \times \text{multiplier}^{3}, so no rounding error can build up.

  6. Balance after year 4

    £3000×1.054£3646.52\pounds 3000 \times 1.05^{4} \approx \pounds 3646.52

    Year 4: £3472.88×1.05=£3646.52£3472.88 \times 1.05 = £3646.52. This is worked out from the exact starting amount as P×multiplier4P \times \text{multiplier}^{4}, so no rounding error can build up.

  7. Round Account A to the nearest penny

    £3646.52£3646.52\pounds 3646.52 \to \pounds 3646.52

    Account A finishes on £3646.52.

  8. Write the yearly percentage as a multiplier

    100%+3%=103%=1.03100\% + 3\% = 103\% = 1.03

    A 3% increase keeps the whole 100% and adds 3% on top, so multiply by 1.03 each year.

  9. Balance after year 1

    £3200×1.031=£3296\pounds 3200 \times 1.03^{1} = \pounds 3296

    Year 1: £3200×1.03=£3296£3200 \times 1.03 = £3296. This is worked out from the exact starting amount as P×multiplier1P \times \text{multiplier}^{1}, so no rounding error can build up.

  10. Balance after year 2

    £3200×1.032=£3394.88\pounds 3200 \times 1.03^{2} = \pounds 3394.88

    Year 2: £3296×1.03=£3394.88£3296 \times 1.03 = £3394.88. This is worked out from the exact starting amount as P×multiplier2P \times \text{multiplier}^{2}, so no rounding error can build up.

  11. Balance after year 3

    £3200×1.033£3496.73\pounds 3200 \times 1.03^{3} \approx \pounds 3496.73

    Year 3: £3394.88×1.03=£3496.73£3394.88 \times 1.03 = £3496.73. This is worked out from the exact starting amount as P×multiplier3P \times \text{multiplier}^{3}, so no rounding error can build up.

  12. Balance after year 4

    £3200×1.034£3601.63\pounds 3200 \times 1.03^{4} \approx \pounds 3601.63

    Year 4: £3496.73×1.03=£3601.63£3496.73 \times 1.03 = £3601.63. This is worked out from the exact starting amount as P×multiplier4P \times \text{multiplier}^{4}, so no rounding error can build up.

  13. Round Account B to the nearest penny

    £3601.63£3601.63\pounds 3601.63 \to \pounds 3601.63

    Account B finishes on £3601.63.

  14. Compare the two accounts and find the difference

    £3646.52£3601.63=£44.89\pounds 3646.52 - \pounds 3601.63 = \pounds 44.89

    Account A ends up worth more -- by £44.89 -- even though it started with £200 less.

  15. Explain the result

    1.054=1.21550625vs1.034=1.125508811.05^{4} = 1.21550625 \quad\text{vs}\quad 1.03^{4} = 1.12550881

    Over 4 years Account A multiplies its money by 1.2155 and Account B only by 1.1255. A x 1.2155 on £3000 beats B x 1.1255 on £3200, so the higher rate wins here -- but note it is close, and with fewer years B would have won.

Answer
Account A, by £44.89\text{Account A, by } \pounds 44.89
Question 4
6 markschallenging
A van costs £30000\pounds 30\,000 when new. It loses 22%22\% of its value in the first year and 12%12\% of its value in each year after that. Work out the value of the van after 55 years, to the nearest penny.
Show worked solution

Worked solution

  1. Plan the calculation

    30000×0.78×0.88430000 \times 0.78 \times 0.88^{4}

    The first year uses a different multiplier from the other four, so it must be done separately and the remaining years compounded on top of it. Exam convention: keep the full unrounded balance in the calculator all the way through and round to the nearest penny only at the very end.

  2. Write the yearly percentage loss as a multiplier

    100%22%=78%=0.78100\% - 22\% = 78\% = 0.78

    Losing 22% each year leaves 78% of the value, so multiply by 0.78 each year.

  3. Value after year 1

    £30000×0.78=£23400\pounds 30000 \times 0.78 = \pounds 23400

    A 22% drop in the first year takes the van from £30 000 to £23400 -- a loss of £6600 in one year.

  4. Write the yearly percentage loss as a multiplier

    100%12%=88%=0.88100\% - 12\% = 88\% = 0.88

    Losing 12% each year leaves 88% of the value, so multiply by 0.88 each year.

  5. Value after year 2

    £23400×0.881=£20592\pounds 23400 \times 0.88^{1} = \pounds 20592

    Year 2: £23400 x 0.88 = £20592. Every year is worked out from the exact year-1 value £23400, so no rounding error can creep in.

  6. Value after year 3

    £23400×0.882=£18120.96\pounds 23400 \times 0.88^{2} = \pounds 18120.96

    Year 3: £20592 x 0.88 = £18120.96. Every year is worked out from the exact year-1 value £23400, so no rounding error can creep in.

  7. Value after year 4

    £23400×0.883£15946.44\pounds 23400 \times 0.88^{3} \approx \pounds 15946.44

    Year 4: £18120.96 x 0.88 = £15946.44. Every year is worked out from the exact year-1 value £23400, so no rounding error can creep in.

  8. Value after year 5

    £23400×0.884£14032.87\pounds 23400 \times 0.88^{4} \approx \pounds 14032.87

    Year 5: £15946.44 x 0.88 = £14032.87. Every year is worked out from the exact year-1 value £23400, so no rounding error can creep in.

  9. Round to the nearest penny

    £14032.87£14032.87\pounds 14032.87 \to \pounds 14032.87

    The exact value is rounded to the penny only here, at the end.

  10. Check with a single calculation

    30000×0.78×0.884£14032.8730000 \times 0.78 \times 0.88^{4} \approx \pounds 14032.87

    One calculation gives the same value as the table, which confirms it.

  11. Reject the wrong method

    30000×0.885£15831.9630000 \times 0.88^{5} \approx \pounds 15831.96

    Using 12% for all five years gives £15831.96, which is far too big -- it ignores the extra 10 percentage points lost in the first year.

  12. Compare the yearly losses

    year 1: £6600year 2: £2808\text{year 1: } \pounds 6600 \quad\text{year 2: } \pounds 2808

    The van loses £6600 in year 1 but only £2808 in year 2. The percentage is applied to a smaller value each year, so the cash loss shrinks every year -- which is why the value never reaches zero.

  13. Work out the total value lost

    £30000£14032.87=£15967.13\pounds 30000 - \pounds 14032.87 = \pounds 15967.13

    The van has lost £15967.13 of its value in 5 years.

  14. Work out the loss as a fraction of the price

    £15967.13£30000×10053.22%\frac{\pounds 15967.13}{\pounds 30000} \times 100 \approx 53.22\%

    The van has lost about 53.2% of its value -- just over half -- which is a sensible size for a five-year-old van.

  15. State the answer

    £14032.87\pounds 14032.87

    After 5 years the van is worth £14032.87.

Answer
£14032.87\pounds 14032.87
Question 5
6 markschallenging
£8000\pounds 8000 is invested at 4.5%4.5\% compound interest per year. Work out the interest earned in the fifth year alone. Give your answer to the nearest penny.
Show worked solution

Worked solution

  1. Understand what is being asked

    interest in year 5=(balance after 5 years)(balance after 4 years)\text{interest in year 5} = (\text{balance after 5 years}) - (\text{balance after 4 years})

    The interest earned in year 5 alone is 4.5% of the balance at the START of year 5, which is the balance at the end of year 4. Exam convention: keep the full unrounded balance in the calculator all the way through and round to the nearest penny only at the very end.

  2. Write the yearly percentage as a multiplier

    100%+4.5%=104.5%=1.045100\% + 4.5\% = 104.5\% = 1.045

    A 4.5% increase keeps the whole 100% and adds 4.5% on top, so multiply by 1.045 each year.

  3. Balance after year 1

    £8000×1.0451=£8360\pounds 8000 \times 1.045^{1} = \pounds 8360

    Year 1: £8000×1.045=£8360£8000 \times 1.045 = £8360. This is worked out from the exact starting amount as P×multiplier1P \times \text{multiplier}^{1}, so no rounding error can build up.

  4. Balance after year 2

    £8000×1.0452=£8736.20\pounds 8000 \times 1.045^{2} = \pounds 8736.20

    Year 2: £8360×1.045=£8736.20£8360 \times 1.045 = £8736.20. This is worked out from the exact starting amount as P×multiplier2P \times \text{multiplier}^{2}, so no rounding error can build up.

  5. Balance after year 3

    £8000×1.0453£9129.33\pounds 8000 \times 1.045^{3} \approx \pounds 9129.33

    Year 3: £8736.20×1.045=£9129.33£8736.20 \times 1.045 = £9129.33. This is worked out from the exact starting amount as P×multiplier3P \times \text{multiplier}^{3}, so no rounding error can build up.

  6. Balance after year 4

    £8000×1.0454£9540.15\pounds 8000 \times 1.045^{4} \approx \pounds 9540.15

    Year 4: £9129.33×1.045=£9540.15£9129.33 \times 1.045 = £9540.15. This is worked out from the exact starting amount as P×multiplier4P \times \text{multiplier}^{4}, so no rounding error can build up.

  7. Note the exact balance at the start of year 5

    £9540.15 (exact: 9540.148805000000)\pounds 9540.15 \text{ (exact: } 9540.148805000000 \text{)}

    This exact value is what year 5 interest is worked out on -- rounding it to the penny first would give a slightly different answer.

  8. Balance after year 5

    £8000×1.0455£9969.46\pounds 8000 \times 1.045^{5} \approx \pounds 9969.46

    Year 5: £9540.15×1.045=£9969.46£9540.15 \times 1.045 = £9969.46. This is worked out from the exact starting amount as P×multiplier5P \times \text{multiplier}^{5}, so no rounding error can build up.

  9. Subtract to find the year 5 interest

    £9969.46£9540.15£429.31\pounds 9969.46 - \pounds 9540.15 \approx \pounds 429.31

    The interest earned in year 5 alone is £429.31.

  10. Check it directly as a percentage

    0.045×£9540.15£429.310.045 \times \pounds 9540.15 \approx \pounds 429.31

    4.5% of the start-of-year-5 balance gives the same answer, which confirms it.

  11. Compare with year 1

    0.045×£8000=£3600.045 \times \pounds 8000 = \pounds 360

    In year 1 the account earned only £360. Year 5 earns £429.31 -- more, because the balance it is worked out on is bigger.

  12. Explain the growth in the yearly interest

    £429.31£3601.0454\frac{\pounds 429.31}{\pounds 360} \approx 1.045^{4}

    Each year's interest is 1.045 times the year before, because the balance it is charged on is 1.045 times bigger. After 4 years of that, year 5 interest is 1.04541.045^4 times the year 1 interest.

  13. Reject the simple-interest answer

    4.5100×8000=£360\frac{4.5}{100} \times 8000 = \pounds 360

    A simple-interest account would pay exactly £360 in year 5 as well. The compound answer is bigger, so the two are genuinely different -- do not confuse them.

  14. Round to the nearest penny

    £429.31£429.31\pounds 429.31 \to \pounds 429.31

    The exact difference is rounded to the penny once, at the end.

  15. State the answer

    £429.31\pounds 429.31

    The interest earned in year 5 alone is £429.31.

Answer
£429.31\pounds 429.31

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