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Level 9Grade 9skill: compound_interest· 9 min read

Compound Interest Explained

In short: Compound interest multiplies your money by the factor (1 + r) once per period, so you earn interest on past interest. The amount is A = P(1 + r)ⁿ and the interest is A − P — don't report one when asked for the other. Over exactly 2 years, compound beats simple interest by P·r².

How to calculate compound interest

Compound interest is what happens when the interest you earn starts earning interest of its own. Instead of adding the same amount every year, you multiply by the same factor every year. That one word — multiply, not add — is the whole idea, and it's exactly where careless answers go wrong.

Write the yearly growth factor as 1 + r, where r is the rate written as a decimal. A rate of 5% gives a factor of 1.05; 10% gives 1.10. To grow money for n years you apply that factor n times, so the final amount is A = P(1 + r)n, where P is the principal you started with.

The final trap is reading the question. A = P(1 + r)n gives you the amount — the whole balance. The interest is only the growth on top: I = A − P. Mixing those two up is the single most common way to lose an easy mark. And one nuance to carry forward: the exponent n really counts compounding periods, not calendar years — when interest is added more than once a year, you split the rate across those extra periods, which is exactly what a 'half-yearly' question is testing.

  1. Turn the rate into a decimal and write the growth factor 1 + r (e.g. 6% becomes 1.06).
  2. Raise the factor to the power of the number of periods: multiply it by itself once for each year.
  3. Multiply the principal by that result to get the amount: A = P(1 + r)n.
  4. If it compounds more than once a year, halve the rate and double the periods for half-yearly (quarterly: rate ÷ 4, periods × 4).
  5. Subtract the principal for the interest: I = A − P. Then reread the question — did it ask for amount or interest?

Why the growth factor beats adding the rate Why it works

Multiply, don't add
Each year multiplies the balance by (1 + r). Adding r every year is simple interest, a different (smaller) thing.

$1000 at 10% for 2 years: 1000 × 1.1 × 1.1 = $1210, not 1000 + 100 + 100 = $1200.

Amount vs interest
A = P(1 + r)n is the total balance; the interest is A − P.

If the balance grows to $1210 from $1000, the interest is $210 — not $1210.

The 2-year shortcut
Over exactly 2 years, compound interest beats simple interest by exactly P·r² — nothing else needed.

$1000 at 10%: P·r² = 1000 × 0.01 = $10, which is exactly 1210 − 1200.

Here's the insight worth keeping: the gap between compound and simple interest is 'interest on the interest,' and for two years it collapses to the tidy formula P·r². If a two-year problem asks for the difference between compound and simple interest, you often don't need to compute either amount — just square the rate and multiply by the principal.

Worked examples

$8,000 is invested at 5% per year compounded annually for 2 years. Find the compound interest. = $820
  1. Growth factor: 1 + 0.05 = 1.05.
  2. Two years, so square it: 1.05 × 1.05 = 1.1025.
  3. Amount: 8000 × 1.1025 = $8,820.
  4. Interest asked, not amount: 8820 − 8000 = $820.
  5. (Check: simple interest would be 8000 × 0.05 × 2 = $800, and the extra P·r² = 8000 × 0.0025 = $20 accounts for the difference.)
$12,000 is invested at 10% per year compounded annually for 3 years. How much interest is earned? = $3,972
  1. Growth factor: 1.10.
  2. Three years: 1.10 × 1.10 × 1.10 = 1.331.
  3. Amount: 12000 × 1.331 = $15,972.
  4. The question asks for interest: 15972 − 12000 = $3,972.
  5. (Simple interest would give only 12000 × 0.10 × 3 = $3,600 — the extra $372 is interest earning interest.)
$15,000 is invested at 8% per year for 2 years. How much MORE does compound interest earn than simple interest? = $96
  1. This is a 2-year difference, so use P·r² directly.
  2. r = 0.08, so r² = 0.0064.
  3. Difference = 15000 × 0.0064 = $96.
  4. (Full check: compound amount 15000 × 1.1664 = $17,496, interest $2,496; simple interest 15000 × 0.08 × 2 = $2,400; difference 2496 − 2400 = $96.)
$10,000 is invested at 8% per year compounded HALF-YEARLY for 1 year. Find the amount. = $10,816
  1. Half-yearly means rate ÷ 2 and periods × 2: rate 4% per half-year, 2 periods.
  2. Factor per half-year: 1.04.
  3. Amount: 10000 × 1.04 × 1.04 = 10000 × 1.0816 = $10,816.
  4. (Compounding twice a year beats once: annual compounding would give 10000 × 1.08 = $10,800, so the extra half-year of compounding adds $16.)

Traps that catch strong students

Treating compound growth as simple interest

✗ $3,600

On $12,000 at 10% for 3 years, this student multiplied 12000 × 0.10 × 3 = $3,600, adding the same $1,200 each year. That ignores the interest earned on previous interest — it's simple interest wearing a compound costume.

Fix: Multiply by the factor once per year: 12000 × 1.1 × 1.1 × 1.1 = $15,972, so the interest is $3,972. The $372 gap is precisely the interest-on-interest that simple interest leaves out.

si_ci_confusion

Answering the amount when asked the interest

✗ $15,972

On the same $12,000-at-10%-for-3-years problem, this student computed the amount A = 12000 × 1.331 = $15,972 correctly, then stopped and reported it. But the question asked for the interest, not the total balance.

Fix: Always finish with I = A − P. Here that's 15972 − 12000 = $3,972. Reread the last line of the question: 'amount' and 'interest' are different targets.

partial_computation

Squaring only the rate, not the whole factor

✗ $10,400

On $10,000 at 20% for 2 years, this student typed 1 + 0.2² = 1 + 0.04 = 1.04 and got $10,400. The exponent landed on the rate alone instead of on the whole growth factor (1 + r).

Fix: Bracket first: the factor is (1 + 0.2) = 1.2, and it's the factor that gets squared: 1.2² = 1.44. Amount 10000 × 1.44 = $14,400, interest $4,400.

order_of_ops_lr

Adding the rate each year instead of multiplying

✗ $8,800

On $8,000 at 5% for 2 years, this student used a factor of 1 + 0.05 + 0.05 = 1.10, getting 8000 × 1.10 = $8,800. Adding the rate across years is the same as simple interest and never gives compound growth.

Fix: Multiply the factor, don't stack the rate: 1.05 × 1.05 = 1.1025, so 8000 × 1.1025 = $8,820. The missing $20 is the interest the first year's interest should have earned.

wrong_operation

Practice compound interest

Work through these mixed problems. For each one, ask yourself two questions before you compute: is the factor being multiplied or added, and does the question want the amount or just the interest? And watch for the 2-year difference shortcut.

Keep building

First, make sure of

Frequently asked questions

What is the difference between compound and simple interest?

Simple interest adds the same fixed amount every year, calculated only on the original principal. Compound interest multiplies the whole balance by the factor (1 + r) each year, so you also earn interest on interest already added. Over time compound interest always pulls ahead — and for exactly 2 years the gap equals P·r².

How do I find just the compound interest instead of the total amount?

First find the amount with A = P(1 + r)n, then subtract the principal: interest I = A − P. For example, $8,000 at 5% for 2 years gives A = 8000 × 1.05² = $8,820, so the interest is 8820 − 8000 = $820. Reporting the $8,820 amount when the question asked for interest is a classic dropped mark.

What does 'compounded half-yearly' actually change?

You halve the rate and double the number of periods. $10,000 at 8% per year compounded half-yearly for 1 year uses 4% over 2 periods: 10000 × 1.04² = $10,816. Because interest is added more often, it starts earning interest sooner, so more frequent compounding always yields slightly more than annual compounding on the same headline rate.

Is there a fast way to compare compound and simple interest over 2 years?

Yes. For exactly 2 years, compound interest beats simple interest by P·r², where r is the decimal rate. So $15,000 at 8% for 2 years differs by 15000 × 0.08² = $96 — no need to compute either full amount if the question only asks for the difference.

Why is A = P(1 + r)n written with a power?

The power counts how many times you apply the growth factor. Each year multiplies the balance by (1 + r), so after n years you've multiplied by (1 + r) a total of n times, which is (1 + r)n. The exponent is the number of compounding periods, not the rate — squaring only the rate is a common slip.