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Exam questions · Maths · Ratio and Proportion

Repeated Percentage Change and Compound Interest

  • 6 exam questions
  • 17 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    Increase \(\pounds 60\) by 15%.

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    Model answer

    10% of 60 is 6 and 5% is 3, so 15% is 9. The new amount is \(60 + 9 = \pounds 69\).

    Mark scheme

    • 15% of 60 \(= 9\), or \(60 \times 1.15\) — M1
    • \(\pounds 69\) — A1
  2. 2 Work out [3 marks]

    \(\pounds 4000\) is invested at 3% per year compound interest. Work out the value of the investment after 2 years.

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    Model answer

    After year 1: \(4000 \times 1.03 = 4120\). After year 2: \(4120 \times 1.03 = 4243.60\). The value is \(\pounds 4243.60\).

    Mark scheme

    • \(4000 \times 1.03\) or \(4000 \times 1.03^2\) — M1
    • \(4120 \times 1.03\) — M1
    • \(\pounds 4243.60\) — A1
  3. 3 Work out [3 marks]

    A tractor is worth \(\pounds 30\,000\). Each year its value falls by 20% of its value at the start of that year. Work out the value of the tractor after 2 years.

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    Model answer

    After year 1: \(30\,000 \times 0.8 = 24\,000\). After year 2: \(24\,000 \times 0.8 = 19\,200\). The value is \(\pounds 19\,200\).

    Mark scheme

    • \(30\,000 \times 0.8\) — M1
    • \(24\,000 \times 0.8\) — M1
    • \(\pounds 19\,200\) — A1
  4. 4 Work out [4 marks]

    Zoe invests \(\pounds 500\) for 2 years at 10% per year simple interest. Yan invests \(\pounds 500\) for 2 years at 10% per year compound interest. How much more money does Yan have than Zoe at the end of the 2 years?

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    Model answer

    Zoe: \(10\%\) of 500 is 50, so after 2 years she has \(500 + 2 \times 50 = \pounds 600\). Yan: \(500 \times 1.1 = 550\) and \(550 \times 1.1 = 605\). The difference is \(605 - 600 = \pounds 5\).

    Mark scheme

    • \(\pounds 600\) for simple interest — M1
    • \(500 \times 1.1\) or \(\pounds 550\) — M1
    • \(\pounds 605\) — A1
    • \(\pounds 5\) — A1
  5. 5 Work out [2 marks]

    After a 10% increase, the price of a meal is \(\pounds 44\). Work out the price before the increase.

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    Model answer

    The multiplier is 1.1, so the original price is \(44 \div 1.1 = \pounds 40\).

    Mark scheme

    • \(44 \div 1.1\) — M1
    • \(\pounds 40\) — A1
  6. 6 Work out [3 marks]

    The number of rabbits on an island increases by 20% each year. There are 500 rabbits at the start of the first year. Work out the number of rabbits after 2 years.

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    Model answer

    After year 1: \(500 \times 1.2 = 600\). After year 2: \(600 \times 1.2 = 720\). There are 720 rabbits.

    Mark scheme

    • \(500 \times 1.2\) — M1
    • \(600 \times 1.2\) or \(500 \times 1.44\) — M1
    • 720 — A1

Quick check

  1. 1

    What is the multiplier for an increase of 15%?

    1. A0.15
    2. B1.15
    3. C0.85
    4. D1.015
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    B: 1.15

    An increase of 15% gives \(100\% + 15\% = 115\%\), which is 1.15.

  2. 2

    What is the multiplier for a decrease of 12%?

    1. A0.88
    2. B0.12
    3. C1.12
    4. D0.012
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    A: 0.88

    \(100\% - 12\% = 88\%\), which is 0.88.

  3. 3

    A phone costs \(\pounds 400\). The price is reduced by 15%. What is the new price?

    1. A\(\pounds 385\)
    2. B\(\pounds 60\)
    3. C\(\pounds 460\)
    4. D\(\pounds 340\)
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    D: \(\pounds 340\)

    \(400 \times 0.85 = 340\).

  4. 4

    \(\pounds 1000\) increases by 10% each year for 2 years. What is it worth after 2 years?

    1. A\(\pounds 1200\)
    2. B\(\pounds 1100\)
    3. C\(\pounds 1210\)
    4. D\(\pounds 1110\)
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    C: \(\pounds 1210\)

    \(1000 \times 1.1 = 1100\) and \(1100 \times 1.1 = 1210\). The second increase is 10% of the new amount.

  5. 5

    \(\pounds 2000\) is invested at 5% compound interest per year. What is it worth after 2 years?

    1. A\(\pounds 2200\)
    2. B\(\pounds 2205\)
    3. C\(\pounds 2100\)
    4. D\(\pounds 2210\)
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    B: \(\pounds 2205\)

    \(2000 \times 1.05^2 = 2000 \times 1.1025 = 2205\). The simple interest answer would be \(\pounds 2200\).

  6. 6

    A car worth \(\pounds 12\,000\) loses 20% of its value each year. What is it worth after 2 years?

    1. A\(\pounds 7680\)
    2. B\(\pounds 7200\)
    3. C\(\pounds 9600\)
    4. D\(\pounds 6144\)
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    A: \(\pounds 7680\)

    \(12\,000 \times 0.8 = 9600\) and \(9600 \times 0.8 = 7680\). Taking 40% off in one go would give \(\pounds 7200\).

  7. 7

    Which statement about interest is correct?

    1. ASimple interest grows faster than compound interest
    2. BSimple and compound interest are always equal
    3. CCompound interest adds the same amount every year
    4. DAfter the first year, compound interest is greater than simple interest
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    D: After the first year, compound interest is greater than simple interest

    Compound interest is paid on interest already earned, so it grows faster than simple interest, which is always paid on the original amount.

  8. 8

    After a 25% increase, a house is worth \(\pounds 150\,000\). What was it worth before the increase?

    1. A\(\pounds 112\,500\)
    2. B\(\pounds 187\,500\)
    3. C\(\pounds 120\,000\)
    4. D\(\pounds 125\,000\)
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    C: \(\pounds 120\,000\)

    The multiplier is 1.25, so the original is \(150\,000 \div 1.25 = 120\,000\).

  9. 9

    A population of 8000 falls by 10% each year. What is the population after 2 years?

    1. A6400
    2. B6480
    3. C7200
    4. D6600
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    B: 6480

    \(8000 \times 0.9^2 = 8000 \times 0.81 = 6480\).