Compound Interest and Depreciation

Compound interest and depreciation questions are a key part of the Edexcel GCSE Mathematics (1MA1) specification. They build on the percentage multiplier skills from the previous lesson. These questions appear on calculator papers (Papers 2 and 3) and, in simpler forms, on Paper 1 (non-calculator).

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Key Vocabulary

Term	Meaning
Simple interest	Interest calculated only on the original amount (principal)
Compound interest	Interest calculated on the principal plus previously earned interest
Depreciation	A decrease in value over time (e.g. a car losing value)
Principal (P)	The initial amount invested or borrowed
Rate (r)	The percentage rate per time period
Per annum	Per year

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Simple Interest

With simple interest, the interest is calculated on the original amount only each time period.

Formula: Interest $= P \times \frac{r}{100} \times n$=P×100r​×n

where P = principal, r = rate (%), n = number of years.

Worked Example 1

Calculate the simple interest on 3,000 pounds at 4% per annum for 5 years.

• Interest per year = 3,000 x 4/100 = 120 pounds
• Total interest over 5 years = 120 x 5 = 600 pounds
• Total amount = 3,000 + 600 = 3,600 pounds

Edexcel Exam Tip: Simple interest questions are relatively rare at GCSE but do appear. The key difference from compound interest is that the amount of interest is the same each year.

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Compound Interest

With compound interest, the interest each year is calculated on the running total, not just the original amount.

Formula: $A = P(1 + \frac{r}{100})^n$A=P(1+100r​)n

where A = final amount, P = principal, r = annual rate (%), n = number of years.

Worked Example 2

5,000 pounds is invested at 3% compound interest per annum. What is the value after 4 years?

• Multiplier = 1.03
• A = 5,000 x 1.03^4
• A = 5,000 x 1.12550881
• A = 5,627.54 pounds (to the nearest penny)

Worked Example 3

Calculate compound interest year by year for 2,000 pounds at 5% for 3 years.

Year	Start of Year	Interest (5%)	End of Year
1	2,000.00	100.00	2,100.00
2	2,100.00	105.00	2,205.00
3	2,205.00	110.25	2,315.25

Using the formula: A = 2,000 x 1.05^3 = 2,000 x 1.157625 = 2,315.25 pounds

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Comparing Simple and Compound Interest

Worked Example 4

Compare the simple and compound interest earned on 4,000 pounds at 6% per annum over 3 years.

Simple interest:

• Interest = 4,000 x 0.06 x 3 = 720 pounds
• Total = 4,720 pounds

Compound interest:

• A = 4,000 x 1.06^3 = 4,000 x 1.191016 = 4,764.06 pounds
• Interest earned = 764.06 pounds

Difference: 764.06 - 720 = 44.06 pounds more with compound interest.

Edexcel Exam Tip: The difference between simple and compound interest increases with the number of years and the interest rate. On Edexcel papers, you may be asked to calculate the difference over a specific number of years.

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Depreciation

Depreciation works like compound interest but with a decrease. The multiplier is (1 - r/100).

Formula: $A = P(1 - \frac{r}{100})^n$A=P(1−100r​)n

Worked Example 5

A car is bought for 18,000 pounds. It depreciates at 20% per year. What is it worth after 3 years?

• Multiplier = 1 - 20/100 = 0.80
• A = 18,000 x 0.80^3
• A = 18,000 x 0.512
• A = 9,216 pounds

Worked Example 6

A computer costs 1,200 pounds and depreciates by 25% each year. After how many complete years will it first be worth less than 400 pounds?

Year	Value (pounds)
0	1,200.00
1	900.00
2	675.00
3	506.25
4	379.69

It first falls below 400 pounds after 4 years.

Using the formula: 1,200 x 0.75^n < 400, so 0.75^n < 1/3. Trial: 0.75^4 = 0.31640625 < 0.3333..., confirming 4 years.

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Repeated Percentage Change

Any situation involving the same percentage change applied repeatedly uses the same compound formula.

Worked Example 7

A town's population is 50,000. It grows by 2% each year. What will the population be in 10 years? (Give to the nearest whole number.)

• A = 50,000 x 1.02^10
• A = 50,000 x 1.21899...
• A = 60,950 (nearest whole number)

Worked Example 8

A radioactive substance has a mass of 800 g. It decays at 12% per hour. What mass remains after 5 hours?

• A = 800 x 0.88^5
• A = 800 x 0.52773...
• A = 422.19 g (to 2 d.p.)

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Finding the Original Amount

If you know the final amount after compound growth or depreciation, you can work backwards.

Worked Example 9

After 2 years of 8% compound interest per annum, an investment is worth 5,832 pounds. What was the original investment?

• 5,832 = P x 1.08^2
• 5,832 = P x 1.1664
• P = 5,832 / 1.1664
• P = 5,000 pounds

Worked Example 10

A motorbike depreciates by 15% per year. After 3 years it is worth 4,913 pounds. What was the original price? (Nearest pound.)

• 4,913 = P x 0.85^3
• 4,913 = P x 0.614125
• P = 4,913 / 0.614125
• P = 8,000 pounds

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Practice Problems