Financial applications of maths use geometric growth to model savings, loans, price rises and falling values. This note covers compound interest with different compounding periods, depreciation, and how to use your GDC finance solver to find unknowns. These skills help you judge real offers from banks and shops.
Key ideas
With compound interest, interest is added to the balance at the end of each period, and the next interest payment is calculated on the new, larger balance. This is why money grows faster than it does with simple interest, where interest is only ever calculated on the original amount.
If you invest a principal (present value) \(PV\) at an annual interest rate of \(r\%\), compounded \(k\) times per year for \(n\) years, then the future value is
\[FV = PV\left(1 + \frac{r}{100k}\right)^{kn}\]
Here \(r\) is the nominal annual rate. In each period the rate applied is \(\frac{r}{k}\%\), and there are \(kn\) periods in total. Common values are \(k = 1\) (annually), \(k = 2\) (half-yearly), \(k = 4\) (quarterly) and \(k = 12\) (monthly).
Depreciation and inflation. The same structure applies to anything that changes by a fixed percentage each period. A car that loses 12% of its value each year is multiplied by \(0.88\) each year. Prices rising at 2.5% inflation are multiplied by \(1.025\) each year. So a value \(V_0\) changing by a rate of \(p\%\) per period for \(n\) periods becomes \(V_0\left(1 \pm \frac{p}{100}\right)^{n}\), using plus for growth and minus for decay.
Unknowns. Any of the quantities \(FV\), \(PV\), \(r\), \(k\) or \(n\) can be the unknown. For a numerical calculation, substitute into the formula. For an unknown rate or time, use the finance (TVM) solver on your GDC, or the equation solver, and enter the known values. For the number of periods, you can also use a table of values or the graph of \(FV\) against \(n\). Read the question carefully for whether it asks for the amount, the interest earned (\(FV - PV\)), or the number of years.
Money answers are normally given to two decimal places (the nearest cent or penny), unless the question specifies something else. Do not round intermediate values.
Compounding more often. For the same nominal rate, more frequent compounding gives a slightly larger final amount, because interest starts earning interest sooner. The difference between annual and monthly compounding is small at low rates but becomes noticeable over many years or at high rates. When comparing two accounts, calculate the final amount for each over the same time and compare directly.
Real value. Inflation reduces what money can buy. If an amount grows at the same rate as prices, its buying power stays the same. If the interest rate is lower than the inflation rate, the real value of the savings falls even though the balance is rising.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(PV\) | Present value, the principal or starting amount. |
| \(FV\) | Future value, the amount after \(n\) years. |
| \(FV = PV\left(1 + \dfrac{r}{100k}\right)^{kn}\) | Compound interest with nominal annual rate \(r\%\) compounded \(k\) times per year for \(n\) years. |
| \(V = V_0\left(1 - \dfrac{p}{100}\right)^{n}\) | Depreciation by \(p\%\) per period. |
| Interest earned | \(FV - PV\). |
| Nominal rate | The quoted annual rate before splitting it across the compounding periods. |
Worked examples
Example 1 A student invests \(£8000\) in an account paying a nominal annual rate of 3.6%, compounded quarterly. Find the value of the account after 5 years, and the interest earned.
\(PV = 8000\), \(r = 3.6\), \(k = 4\), \(n = 5\).
\(FV = 8000\left(1 + \dfrac{3.6}{400}\right)^{20} = 8000 \times 1.009^{20}\).
\(1.009^{20} = 1.196254\ldots\), so \(FV = 9570.03\) to the nearest cent.
Interest \(= 9570.03 - 8000 = 1570.03\).
Answer: \(£9570.03\), with \(£1570.03\) interest.
Example 2 A new delivery van costs \(£24\,000\) and loses 12% of its value each year. Find its value after 4 years. After how many whole years is it first worth less than half its original cost?
Each year the value is multiplied by \(1 - 0.12 = 0.88\).
\(V = 24\,000 \times 0.88^{4} = 24\,000 \times 0.599695\ldots = 14\,392.69\).
For the second part we need \(24\,000 \times 0.88^{n} \lt 12\,000\), that is \(0.88^{n} \lt 0.5\).
Use the table feature on your GDC: \(0.88^{5} = 0.5277\), which is not below 0.5. \(0.88^{6} = 0.4644\), which is below 0.5.
Answer: \(£14\,392.69\) after 4 years, and it falls below half its cost during year 6, so after 6 whole years.
Example 3 A fund grows from \(£15\,000\) to \(£18\,000\) in 3 years with interest compounded annually. Find the annual interest rate, to 3 significant figures. Then find how long, in whole years, \(£5000\) takes to exceed \(£7000\) at 4% compounded annually.
First part: \(15\,000\left(1 + \dfrac{r}{100}\right)^{3} = 18\,000\), so \(\left(1 + \dfrac{r}{100}\right)^{3} = 1.2\).
You could use the solver on your GDC. By hand, take the cube root: \(1 + \dfrac{r}{100} = 1.2^{1/3} = 1.062658\ldots\), so \(r = 6.27\) to 3 s.f.
Second part: we need \(5000 \times 1.04^{n} \gt 7000\), so \(1.04^{n} \gt 1.4\).
Use the table feature: \(1.04^{8} = 1.3686\), which is too small. \(1.04^{9} = 1.4233\), which is greater than 1.4.
Answer: \(r \approx 6.27\%\) per year, and it takes 9 years (the value is then \(£7116.56\)).
Common errors
- Common error: using \(1 + \frac{r}{k}\) instead of \(1 + \frac{r}{100k}\), forgetting that \(r\) is a percentage.
- Common error: using \(n\) years as the power instead of \(kn\) compounding periods.
- Common error: rounding \(1 + \frac{r}{100k}\) before raising it to a power. This can change the third or fourth significant figure.
- Common error: giving the final amount when the question asks for the interest earned, which is \(FV - PV\).
- Common error: using \(1.12\) instead of \(0.88\) for a 12% decrease.
Exam tip: Write down the values of \(PV\), \(r\), \(k\), \(n\) and \(FV\) (labelling the unknown) before you touch the calculator. If you use the finance solver, also write the equivalent equation, so your method is clear.
Check yourself
- Find the value after 6 years of \(£12\,000\) invested at a nominal annual rate of 4.5%, compounded monthly. Give the interest earned.
Answer
\(FV = 12\,000\left(1 + \dfrac{4.5}{1200}\right)^{72} = 12\,000 \times 1.00375^{72} = 15\,711.64\). Interest \(= 15\,711.64 - 12\,000 = £3711.64\). - A phone bought for \(£900\) loses 15% of its value each year. Find its value after 3 years.
Answer
\(900 \times 0.85^{3} = 900 \times 0.614125 = £552.71\). - How many whole years will it take for \(£2000\) at 5% compounded annually to exceed \(£3000\)?
Answer
Need \(1.05^{n} \gt 1.5\). \(1.05^{8} = 1.477\) (too small) and \(1.05^{9} = 1.551\). So 9 years. - A meal costs \(£40\) today. With inflation of 2.5% per year, what is the expected price in 10 years?
Answer
\(40 \times 1.025^{10} = 40 \times 1.28008 = £51.20\).