Compound interest - Year 10 Mathematics

What you will learn

distinguish simple interest from compound interest and explain why compound interest grows faster,
apply the compound interest formula $A = P\!\left(1 + \dfrac{r}{n}\right)^{nt}$ ,
calculate depreciation using $A = P(1 - r)^n$ ,
solve financial modelling problems including comparing investment options and loan repayments.

Worked example 0 Real-world example: savings account comparison

Zara invests $5000 for $3$ years. Bank A offers $6\%$ p.a. simple interest. Bank B offers $6\%$ p.a. compound interest (compounded annually). Which gives more?

Bank A (simple): $I = 5000 \times 0.06 \times 3 = 900$ . Total: $A = 5900$ .
Bank B (compound): $A = 5000 \times 1.06^3 = 5000 \times 1.191016 = 5955.08$ .
Compound interest earns $55.08 more because interest in years 2 and 3 is calculated on a growing balance.

Key idea: simple interest is linear (same amount each year); compound interest is exponential (grows faster over time).

1. Simple vs compound interest

With simple interest, the interest each period is calculated on the original principal only:

A_{\text{simple}} = P(1 + rT)

With compound interest, interest each period is calculated on the current balance (principal plus previously earned interest):

A_{\text{compound}} = P(1 + r)^T \quad \text{(compounded annually)}

After one year both give the same result. After two or more years compound interest always yields more (assuming $r > 0$ ).

Simple interest (straight line) vs compound interest (curve) on the same principal over 20 years. The gap widens as time increases.

2. The compound interest formula

Formula reference

A = P\!\left(1 + \frac{r}{n}\right)^{nt}

$A$ = final amount,
$P$ = principal (initial investment or loan),
$r$ = annual interest rate (as a decimal),
$n$ = number of compounding periods per year,
$t$ = time in years.

For annual compounding ( $n = 1$ ): $A = P(1 + r)^t$ .

Worked example 1 Annual compounding

Find the value of $3000 invested at $5\%$ p.a. compounded annually for $4$ years.

$P = 3000$ , $r = 0.05$ , $n = 1$ , $t = 4$ .
$A = 3000 \times 1.05^4 = 3000 \times 1.21551 = 3646.52$ dollars.
Interest earned: $3646.52 - 3000 = 646.52$ dollars.

Worked example 2 Monthly compounding

Find the value of $2000 invested at $6\%$ p.a. compounded monthly for $3$ years.

$P = 2000$ , $r = 0.06$ , $n = 12$ , $t = 3$ .
$A = 2000\!\left(1 + \dfrac{0.06}{12}\right)^{36} = 2000 \times 1.005^{36}$ .
$1.005^{36} \approx 1.19668$ . $A \approx 2393.36$ dollars.

3. Depreciation

Depreciation is the decrease in value of an asset over time. When the rate of depreciation is constant, the model is exponential decay.

Formula reference

A = P(1 - r)^n

$A$ = value after $n$ periods,
$P$ = original value,
$r$ = rate of depreciation per period (as a decimal),
$n$ = number of periods.

Worked example 3 Car depreciation

A car costs $35000 and depreciates at $18\%$ per year. Find its value after $5$ years.

$A = 35000 \times (1 - 0.18)^5 = 35000 \times 0.82^5$ .
$0.82^5 \approx 0.37043$ .
$A \approx 12965$ dollars.

Worked example 4 Finding the depreciation rate

A laptop worth $2400 is valued at $1200 after $3$ years. Find the annual depreciation rate.

$1200 = 2400 \times (1 - r)^3$ .
$(1 - r)^3 = 0.5$ .
$1 - r = 0.5^{1/3} \approx 0.7937$ .
$r \approx 0.2063$ , so the depreciation rate is approximately $20.6\%$ per year.

4. Financial modelling problems

Financial modelling often combines the compound interest formula with additional conditions such as regular deposits, comparison of options, or target amounts.

Worked example 5 Reaching a savings goal

Ethan wants to have $10000 in $5$ years. His account earns $4\%$ p.a. compounded annually. How much must he invest now?

$10000 = P \times 1.04^5$ .
$P = \dfrac{10000}{1.04^5} = \dfrac{10000}{1.21665} \approx 8219.27$ dollars.
He needs to invest approximately $8219.27 today.

Worked example 6 Comparing two investments

Option A: $6000 at $5\%$ p.a. compounded annually for $10$ years. Option B: $6000 at $4.8\%$ p.a. compounded monthly for $10$ years.

Option A: $A = 6000 \times 1.05^{10} = 6000 \times 1.62889 \approx 9773.37$ .
Option B: $A = 6000\!\left(1 + \dfrac{0.048}{12}\right)^{120} = 6000 \times 1.004^{120}$ .
$1.004^{120} \approx 1.61222$ . $A \approx 9673.35$ .
Option A gives about $100 more despite a lower compounding frequency, because the annual rate is higher.

Practice

Fluency

Tier 1: basic skills

Calculate simple interest on $4000 at $5\%$ p.a. for $3$ years.
Find the total amount when $4000 is invested at $5\%$ p.a. compounded annually for $3$ years.
State the difference between your answers to Q1 and Q2.
Find the value of $10000 invested at $3\%$ p.a. compounded annually for $6$ years.
A computer worth $1800 depreciates at $25\%$ per year. Find its value after $2$ years.
Find the value of $5000 at $8\%$ p.a. compounded quarterly for $2$ years. (Hint: $n = 4$ .)
A painting is bought for $3000 and appreciates at $10\%$ per year. What is it worth after $4$ years?
How much interest is earned on $7000 at $6\%$ p.a. compounded annually for $5$ years?

Reasoning

Tier 2: mixed practice

$12000 is invested at $4.5\%$ p.a. compounded monthly. Find the amount after $3$ years.
A car worth $28000 depreciates at $15\%$ per year. After how many whole years is it first worth less than $10000?
Which is better over $5$ years: $8000 at $6\%$ p.a. compounded annually, or $8000 at $5.8\%$ p.a. compounded monthly? Show both calculations.
Ava invests $P at $7\%$ p.a. compounded annually. After $10$ years she has $15000. Find $P$ .
A motorbike depreciates from $9000 to $4500 in $4$ years. Find the annual depreciation rate.
Calculate the total interest earned on $20000 at $3.2\%$ p.a. compounded quarterly over $8$ years.

Reasoning

Tier 3: explain and apply

Explain why the gap between simple interest and compound interest widens as time increases. Use the formulas to support your answer.
A credit card charges $1.5\%$ interest per month on unpaid balances. What is the effective annual interest rate? (Hint: $(1.015)^{12}$ .)
Mia has $20000 to invest for $10$ years. Bank A offers $5\%$ p.a. compounded annually. Bank B offers $4.9\%$ p.a. compounded daily (assume $365$ days). Which should she choose?
A factory machine costs $50000 and depreciates at $20\%$ per year. The company plans to replace it when its value drops below $10\%$ of the original cost. After how many whole years should they replace it?

Challenge

Reasoning

Harder reasoning

Show that for a principal $P$ invested at rate $r$ p.a. compounded annually for $2$ years, the compound interest exceeds the simple interest by exactly $Pr^2$ .
Jake borrows $15000 at $8\%$ p.a. compounded annually. He makes no repayments. After how many whole years does the debt first exceed $25000? How does this compare with the Rule of 70 estimate for doubling?
An investment of $10000 grows to $10000 \times 1.06^t $after$ t $years. A second investment of \$ 15000 grows to $15000 \times 1.03^t$ after $t$ years. After how many whole years does the first investment first exceed the second? Justify your answer.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 10 core - answers

Fluency

Tier 1: basic skills

$I = 4000 \times 0.05 \times 3 = 600$ dollars. Total: $4600$ .
$A = 4000 \times 1.05^3 = 4000 \times 1.157625 = 4630.50$ dollars.
Compound gives $4630.50 - 4600 = 30.50$ dollars more.
$A = 10000 \times 1.03^6 = 10000 \times 1.19405 \approx 11940.52$ dollars.
$A = 1800 \times 0.75^2 = 1800 \times 0.5625 = 1012.50$ dollars.
$A = 5000\!\left(1 + \dfrac{0.08}{4}\right)^{8} = 5000 \times 1.02^8 = 5000 \times 1.17166 \approx 5858.30$ dollars.
$A = 3000 \times 1.1^4 = 3000 \times 1.4641 = 4392.30$ dollars.
$A = 7000 \times 1.06^5 = 7000 \times 1.33823 \approx 9367.58$ . Interest $= 9367.58 - 7000 = 2367.58$ dollars.

Reasoning

Tier 2: mixed practice

$A = 12000\!\left(1 + \dfrac{0.045}{12}\right)^{36} = 12000 \times 1.00375^{36} \approx 12000 \times 1.14396 \approx 13727.55$ dollars.
$28000 \times 0.85^t < 10000$ . $0.85^t < 0.3571$ . By trial: $0.85^6 \approx 0.3771$ , $0.85^7 \approx 0.3206$ . First below $10000 after 7 whole years.
Option 1: $8000 \times 1.06^5 = 8000 \times 1.33823 \approx 10705.80$ . Option 2: $8000 \times 1.004833^{60} \approx 8000 \times 1.33556 \approx 10684.48$ . Option 1 ( $6\%$ annually) gives about $21 more.
$15000 = P \times 1.07^{10}$ . $1.07^{10} \approx 1.96715$ . $P = \dfrac{15000}{1.96715} \approx 7625.85$ dollars.
$4500 = 9000 \times (1-r)^4$ . $(1-r)^4 = 0.5$ . $1 - r = 0.5^{0.25} \approx 0.8409$ . $r \approx 0.159$ , so approximately $15.9\%$ per year.
$A = 20000\!\left(1 + \dfrac{0.032}{4}\right)^{32} = 20000 \times 1.008^{32} \approx 20000 \times 1.29098 \approx 25819.63$ . Interest $= 25819.63 - 20000 = 5819.63$ dollars.

Reasoning

Tier 3: explain and apply

Simple interest after $T$ years: $A_S = P(1 + rT)$ , which is linear in $T$ . Compound interest: $A_C = P(1+r)^T$ , which is exponential. For small $T$ , $(1+r)^T \approx 1 + rT$ (nearly equal). As $T$ increases, the exponential term grows faster than the linear term because each year’s interest is applied to an ever-larger base. The gap is $P[(1+r)^T - (1 + rT)]$ , which increases with $T$ .
Effective annual rate $= (1.015)^{12} - 1 \approx 1.19562 - 1 = 0.19562$ , or about $19.6\%$ per year.
Bank A: $20000 \times 1.05^{10} = 20000 \times 1.62889 \approx 32577.89$ . Bank B: $20000\!\left(1 + \dfrac{0.049}{365}\right)^{3650} \approx 20000 \times e^{0.49} \approx 20000 \times 1.63232 \approx 32646.33$ . Bank B gives slightly more (about $68 extra). Mia should choose Bank B.
Need $50000 \times 0.8^n < 5000$ . $0.8^n < 0.1$ . By trial: $0.8^{10} \approx 0.10737$ , $0.8^{11} \approx 0.08590$ . Replace after 11 whole years.

Reasoning

Challenge

Simple interest for $2$ years: $I_S = P \times r \times 2 = 2Pr$ . Compound interest for $2$ years: $A_C = P(1+r)^2 = P(1 + 2r + r^2)$ . So $I_C = A_C - P = 2Pr + Pr^2$ . Difference: $I_C - I_S = (2Pr + Pr^2) - 2Pr = Pr^2$ .
$15000 \times 1.08^t > 25000$ . $1.08^t > \tfrac{5}{3} \approx 1.6667$ . By trial: $1.08^6 \approx 1.5869$ , $1.08^7 \approx 1.7138$ . Debt exceeds $25000 after 7 whole years. Rule of 70: doubling time $\approx \tfrac{70}{8} = 8.75$ years; the debt reaches $\tfrac{5}{3}$ of the original (not double) so it happens sooner than the doubling time, which is consistent.
Need $10000 \times 1.06^t > 15000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.5$ . $1.02913^t > 1.5$ . By trial: $1.02913^{14} \approx 1.494$ , $1.02913^{15} \approx 1.538$ . After 15 whole years the first investment first exceeds the second.

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