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Exam questions · Maths · Ratio and Proportion

Repeated Percentage Change and Compound Interest

  • 6 exam questions
  • 18 marks
  • 9 quick checks
  1. 1 Calculate [2 marks]

    Decrease \(\pounds 250\) by 12%.

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    Model answer

    10% of 250 is 25 and 1% is 2.50, so 12% is \(25 + 5 = 30\). The new amount is \(250 - 30 = \pounds 220\).

    Mark scheme

    • 12% of 250 \(= 30\), or \(250 \times 0.88\) — M1
    • \(\pounds 220\) — A1
  2. 2 Calculate [3 marks]

    Ed invests \(\pounds 6000\) for 2 years at 5% per year compound interest. Calculate the value of the investment after 2 years.

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    Model answer

    After year 1: \(6000 \times 1.05 = 6300\). After year 2: \(6300 \times 1.05 = 6615\). The value is \(\pounds 6615\).

    Mark scheme

    • \(6000 \times 1.05\) or \(6000 \times 1.05^2\) — M1
    • \(6300 \times 1.05\) — M1
    • \(\pounds 6615\) — A1
  3. 3 Calculate [3 marks]

    A laptop is worth \(\pounds 800\). Its value falls by 25% each year. Calculate the value of the laptop after 2 years.

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    Model answer

    After year 1: \(800 \times 0.75 = 600\). After year 2: \(600 \times 0.75 = 450\). The value is \(\pounds 450\).

    Mark scheme

    • \(800 \times 0.75\) — M1
    • \(600 \times 0.75\) — M1
    • \(\pounds 450\) — A1
  4. 4 Calculate [4 marks]

    \(\pounds 2000\) is invested for 2 years at 4% per year. Calculate how much more interest is earned with compound interest than with simple interest.

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    Model answer

    Simple interest: \(4\%\) of 2000 is 80, so 2 years gives \(\pounds 160\). Compound interest: \(2000 \times 1.04 = 2080\) and \(2080 \times 1.04 = 2163.20\), so the interest is \(\pounds 163.20\). The difference is \(163.20 - 160 = \pounds 3.20\).

    Mark scheme

    • Simple interest \(\pounds 160\) — M1
    • \(2000 \times 1.04\) and \(2080 \times 1.04\) — M1
    • \(\pounds 163.20\) as the compound interest or \(\pounds 2163.20\) — A1
    • \(\pounds 3.20\) — A1
  5. 5 Calculate [3 marks]

    In a sale, the price of a coat is reduced by 20%. The sale price is \(\pounds 48\). Calculate the original price of the coat.

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    Model answer

    The sale price is 80% of the original, so the multiplier is 0.8. The original price is \(48 \div 0.8 = \pounds 60\).

    Mark scheme

    • 80% or 0.8 seen — M1
    • \(48 \div 0.8\) — M1
    • \(\pounds 60\) — A1
  6. 6 Calculate [3 marks]

    A colony of bacteria increases in number by 50% every hour. There are 800 bacteria at the start. Calculate the number of bacteria after 3 hours.

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    Model answer

    After 1 hour: \(800 \times 1.5 = 1200\). After 2 hours: \(1200 \times 1.5 = 1800\). After 3 hours: \(1800 \times 1.5 = 2700\).

    Mark scheme

    • \(800 \times 1.5\) or \(800 \times 1.5^3\) — M1
    • \(1200 \times 1.5\) and \(1800 \times 1.5\) — M1
    • 2700 — 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

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