Compound Interest Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Compound Interest.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Interest calculated on both the initial principal and the accumulated interest from previous periods. The formula A = P\left(1 + \frac{r}{n}\right)^{nt} gives the amount after t years, and A = Pe^{rt} gives the continuously compounded amount.

Simple interest pays you only on your original deposit. Compound interest pays you interest on your interest—your money earns money on the money it already earned. The more frequently you compound, the more you earn, because each tiny interest payment starts earning its own interest sooner. The ultimate limit of compounding more and more frequently is continuous compounding: A = Pe^{rt}.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Compound interest is exponential growth applied to money. The key insight is that as compounding frequency n increases, the amount approaches the continuous limit Pe^{rt}—this is exactly how Euler's number e was discovered.

Common stuck point: The rate r must be a decimal, not a percentage: 6% means r = 0.06, not r = 6. Also, r and n must use the same time unit—if r is annual, n is compoundings per year.

Sense of Study hint: Label every variable in the formula before plugging in: P = initial amount, r = rate as decimal, n = times per year, t = years. Then substitute carefully.

Worked Examples

Example 1

easy
You invest \5{,}000 at 6\%$ annual interest compounded quarterly. Find the amount after 3 years.

Solution

  1. 1
    Use the formula A = P\left(1 + \frac{r}{n}\right)^{nt} with P = 5000, r = 0.06, n = 4, t = 3.
  2. 2
    Substitute: A = 5000\left(1 + \frac{0.06}{4}\right)^{4 \cdot 3} = 5000(1.015)^{12}.
  3. 3
    Compute (1.015)^{12} \approx 1.19562, so A \approx 5000 \times 1.19562 \approx 5978.09.

Answer

A \approx \$5{,}978.09
Compound interest applies the interest rate multiple times per year. The key parameters are the principal P, annual rate r, compounding frequency n, and time in years t.

Example 2

medium
How long does it take for an investment to double at 8\% annual interest compounded monthly?

Example 3

medium
Find the amount when \2{,}000 is invested at 5\%$ compounded continuously for 4 years.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
You deposit \1{,}200 in a savings account at 4\%$ compounded annually. What is the balance after 5 years?

Example 2

hard
What annual interest rate, compounded monthly, is needed to grow \3{,}000 to \4{,}500 in 6 years?
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Background Knowledge

These ideas may be useful before you work through the harder examples.

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