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Exam questions · Maths · Ratio and Proportion

Repeated Percentage Change and Compound Interest

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    A coat costs \(\pounds 80\). In a sale the price is reduced by 15%. Work out the sale price of the coat.

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    Model answer

    10% of 80 is 8 and 5% is 4, so 15% is 12. The sale price is \(80 - 12 = \pounds 68\).

    Mark scheme

    • 10% \(= 8\) or 5% \(= 4\) or 15% \(= 12\) seen — M1
    • \(80 - 12\) or \(80 \times 0.85\) — M1
    • \(\pounds 68\) — A1
  2. 2 Work out [3 marks]

    Sam invests \(\pounds 5000\) for 2 years at 4% per year compound interest. Work out the value of his investment at the end of 2 years.

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    Model answer

    After year 1: \(5000 \times 1.04 = 5200\). After year 2: \(5200 \times 1.04 = 5408\). The value is \(\pounds 5408\).

    Mark scheme

    • \(5000 \times 1.04\) or \(5000 \times 1.04^2\) — M1
    • \(5200 \times 1.04\) or \(5200 + 208\) — M1
    • \(\pounds 5408\) — A1
  3. 3 Work out [3 marks]

    The value of a car is \(\pounds 9000\). Each year the value of the car falls by 10% of its value at the start of that year. Work out the value of the car after 2 years.

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    Model answer

    After year 1: \(9000 \times 0.9 = 8100\). After year 2: \(8100 \times 0.9 = 7290\). The value is \(\pounds 7290\).

    Mark scheme

    • \(9000 \times 0.9\) or \(9000 - 900\) — M1
    • \(8100 \times 0.9\) or \(8100 - 810\) — M1
    • \(\pounds 7290\) — A1
  4. 4 Work out [4 marks]

    Mia invests \(\pounds 1000\) for 3 years at 5% per year simple interest. Ben invests \(\pounds 1000\) for 3 years at 5% per year compound interest. Who has more money at the end of the 3 years? You must show your working.

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    Model answer

    Mia: \(5\%\) of 1000 is 50, so after 3 years she has \(1000 + 3 \times 50 = \pounds 1150\). Ben: \(1000 \times 1.05 = 1050\), then \(1050 \times 1.05 = 1102.50\), then \(1102.50 \times 1.05 = 1157.625\), so about \(\pounds 1157.63\). Ben has more.

    Mark scheme

    • Simple interest \(\pounds 50\) a year or \(\pounds 1150\) — M1
    • \(1000 \times 1.05\) and \(1050 \times 1.05\) or \(1000 \times 1.05^3\) — M1
    • \(\pounds 1157.63\) or \(\pounds 1157.625\) — A1
    • Ben, with a correct comparison — C1
  5. 5 Work out [3 marks]

    In a sale, all prices are reduced by 20%. The sale price of a bag is \(\pounds 68\). Work out the original price of the bag.

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    Model answer

    The sale price is 80% of the original, so the multiplier is 0.8. The original price is \(68 \div 0.8 = \pounds 85\).

    Mark scheme

    • 80% or 0.8 seen — M1
    • \(68 \div 0.8\) or \(68 \div 80 \times 100\) — M1
    • \(\pounds 85\) — A1
  6. 6 Work out [4 marks]

    The population of a town increases by 10% each year. At the start of 2024 the population was 20 000. (a) Work out the population at the start of 2026. (3 marks) (b) Show that the population increased by 21% over the two years. (1 mark)

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    Model answer

    (a) After 1 year: \(20\,000 \times 1.1 = 22\,000\). After 2 years: \(22\,000 \times 1.1 = 24\,200\). (b) \(1.1^2 = 1.21\), which is an increase of 21%. Check: \(24\,200 \div 20\,000 = 1.21\).

    Mark scheme

    • (a) \(20\,000 \times 1.1\) — M1
    • (a) \(22\,000 \times 1.1\) — M1
    • (a) \(24\,200\) — A1
    • (b) \(1.1^2 = 1.21\) or \(24\,200 \div 20\,000 = 1.21\) — C1

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

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