Mastering Compound Interest and Depreciation for your Maths Exam

What the examiner is testing

The examiner is testing your ability to calculate a value after repeated percentage increases or decreases over multiple periods. Marks are awarded for correctly identifying the multiplier, applying it the correct number of times, and presenting the final answer with appropriate units and rounding.

The method

1. Identify the initial amount. This is the starting value before any changes.
2. Determine the percentage change per period (interest rate or depreciation rate).
3. Calculate the multiplier:
   • For an increase (compound interest), the multiplier is \(1 + \frac{\text{percentage increase}}{100}\).
   • For a decrease (depreciation), the multiplier is \(1 - \frac{\text{percentage decrease}}{100}\).
4. Identify the number of periods (e.g., years, months).
5. Raise the multiplier to the power of the number of periods.
6. Multiply the initial amount by the result from step 5.

Worked example

A bank offers an interest rate of 3% per year. If £500 is invested, how much will be in the account after 4 years? Give your answer to the nearest penny.

Initial amount = £500
Percentage increase = 3%
Multiplier = \(1 + \frac{3}{100} = 1 + 0.03 = 1.03\)
Number of periods = 4 years

$$ \text{Final amount} = \text{Initial amount} \times (\text{Multiplier})^{\text{Number of periods}} $$

$$ \text{Final amount} = 500 \times (1.03)^4 $$

$$ \text{Final amount} = 500 \times 1.12550881 $$

$$ \text{Final amount} = 562.754405 $$

To the nearest penny:
£562.75

Sanity check: The amount should be more than £500 because it's an investment earning interest. £562.75 is greater than £500, which makes sense.

Worked example: a harder one

A car was bought for £18,000. It depreciates by 15% in the first year, and then by 10% for each of the next two years. What is the value of the car after 3 years?

Initial amount = £18,000

First year depreciation: 15%
Multiplier for first year = \(1 - \frac{15}{100} = 0.85\)

Value after 1 year = \(18000 \times 0.85 = 15300\)

For the next two years, the depreciation is 10% per year.
Multiplier for subsequent years = \(1 - \frac{10}{100} = 0.90\)
Number of subsequent periods = 2 years

Value after 3 years = \(15300 \times (0.90)^2\)
Value after 3 years = \(15300 \times 0.81\)
Value after 3 years = \(12393\)

The obvious first move might be to calculate an average depreciation or combine the percentages:
Incorrect approach: average depreciation \(\frac{15\% + 10\% + 10\%}{3} = \frac{35\%}{3} \approx 11.67\%\) or simply add them \(15\% + 10\% + 10\% = 35\%\).
If we tried to use \(1 - \frac{35}{100} = 0.65\) as a single multiplier:
\(18000 \times 0.65 = 11700\). This is incorrect because the percentages apply to the current value, not the original value each time, and the rates are different.

The value of the car after 3 years is £12,393.

Practice

1. A painting increases in value by 5% each year. If it was bought for £2,000, what is its value after 3 years? Give your answer to the nearest pound.
2. A company's profits decrease by 8% each year. If the profits were £120,000 in 2020, what will they be in 2023?
3. A population of bacteria grows at a rate of 12% per hour. If there are initially 500 bacteria, how many will there be after 6 hours? Give your answer to the nearest whole number.
4. A house was bought for £250,000. In the first year, its value increased by 4%. In the second year, its value decreased by 2%. In the third year, its value increased by 3%. What is the value of the house after 3 years?

Answers:
1. £2,315
2. £94,005.12
3. 987 bacteria
4. Working for question 4:
Initial value = £250,000
Year 1 increase: 4%. Multiplier = \(1 + \frac{4}{100} = 1.04\)
Value after 1 year = \(250000 \times 1.04 = 260000\)

Year 2 decrease: 2%. Multiplier = \(1 - \frac{2}{100} = 0.98\)
Value after 2 years = \(260000 \times 0.98 = 254800\)

Year 3 increase: 3%. Multiplier = \(1 + \frac{3}{100} = 1.03\)
Value after 3 years = \(254800 \times 1.03 = 262444\)

The value of the house after 3 years is £262,444.

The three mistakes that lose marks

1. Using simple interest instead of compound interest: Calculating the percentage change on the original amount each time, rather than the new amount.
   • Wrong answer produced: For £100 at 10% for 2 years, calculating \(100 + (0.10 \times 100) + (0.10 \times 100) = 120\) instead of \(100 \times (1.10)^2 = 121\).
2. Incorrectly calculating the multiplier: Adding or subtracting the percentage directly, or using the percentage itself as the multiplier.
   • Wrong answer produced: For a 5% increase, using \(0.05\) or \(1.05\%\) instead of \(1.05\). For a 10% decrease, using \(0.10\) or \(0.90\%\) instead of \(0.90\).
3. Incorrect number of periods: Using the wrong power for the multiplier, especially when the question involves dates or multiple stages.
   • Wrong answer produced: For a calculation from 2020 to 2023, using a power of 4 (including 2020) instead of 3 (2021, 2022, 2023).

30-second recap

Compound interest and depreciation involve applying a percentage change repeatedly to the new value. Calculate the multiplier (\(1 \pm \text{percentage}/100\)) and raise it to the power of the number of periods. Multiply this by the initial amount. Pay close attention to whether the value is increasing or decreasing and the number of times the change occurs.