Compound Interest

Work out compound interest and depreciation with a multiplier, compare it with simple interest, and find the rate from the final amount.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 107 real past-paper questions below

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What Compound Interest Is

With compound interest, the interest is added on at the end of each year. Next year you earn interest on the new total, so the interest grows.

Year	Start	Interest 5%	End
1	$2000	$100	$2100
2	$2100	$105	$2205
3	$2205	$110.25	$2315.25

Key idea

Simple interest pays $100 every year, because it only uses the first $2000. Compound interest pays interest on the interest too.

Key idea

Quick trick: how long until money doubles? Divide $72$ by the yearly rate. At 6%, $72 \div 6 = 12$ years, and indeed $1.0 6^{12} \approx 2.01$ . This rule of 72 was already printed by Luca Pacioli in Italy in 1494.

Amount After n Years: The Multiplier

The 3-mark question: "invests ... at r% per year compound interest. How much after n years?" Do not work year by year. Use one multiplier.

Formula

A = P (1 + \frac{r}{100})^{n}

$P$ is the money at the start, $r$ is the rate each year, $n$ is the number of years and $A$ is the final amount.

Worked example

1
$$7500 at 3.2%, 3 years$ — The question. The interest is added on each year, so each year earns 3.2% of a bigger total.
2
$1 + \frac{3.2}{100} = 1.032$ — Find the multiplier for one year: you keep 100% and gain 3.2%, so you have 103.2%, which is 1.032.
3
$7500 \times 1.032^{3}$ — Multiply by 1.032 once for each year, so the power is the number of years, 3.
4
$1.03 2^{3} = 1.099104768$ — Work out the power first on your calculator. Do not round it yet.
5
$7500 \times 1.099104768 = 8243.28576$ — Now multiply by the starting $7500.
6
$\approx $8243.29$ — Round only at the very end. Money is given to the nearest cent, so 2 decimal places.

Common mistake

Using 1.32 or 0.032 as the multiplier. For 3.2% the multiplier is 1.032. Also, the power is the years: do not multiply by 3.

Key idea

Always write the line $7500 \times 1.03 2^{3}$ before the answer. Most of the marks are for that line, so you still score if you slip on the calculator.

Key idea

If the rate is per month, the power counts months. 2 years at 0.5% per month is $P \times 1.00 5^{24}$ , because 2 years is 24 months.

Depreciation: Value Going Down

The 3-mark question: "a car depreciates by 15% each year". Depreciate means it loses value. The multiplier is less than 1.

Worked example

1
$$18 000, loses 15% a year, 3 years$ — The question. Depreciates means it loses value, so the multiplier will be less than 1.
2
$1 - \frac{15}{100} = 0.85$ — Losing 15% leaves 85% of the value each year, and 85% is 0.85.
3
$18000 \times 0.85^{3}$ — Multiply by 0.85 once for each of the 3 years, so the power is 3.
4
$0.8 5^{3} = 0.614125$ — Work out the power first. The car keeps about 61% of its value after 3 years.
5
$18000 \times 0.614125 = 11054.25$ — Multiply by the starting value of $18 000.
6
$$11 054.25$ — Check: it is less than $18 000, as a value going down must be.

Common mistake

Taking off 45% in one go. That gives $9900, which is wrong: each year loses 15% of a smaller value.

Key idea

Asked for a formula? Write the start × the multiplier to the power n. A $25 000 machine losing 20% a year is worth $V = 25000 \times 0. 8^{n}$ after n years. The same works for a population or a mass that decreases by the same percentage each year or each day.

Simple or Compound: How Much More?

The 4 or 5-mark question compares two banks and asks how much more interest one pays. Work out the interest for each bank, then subtract.

Worked example

1
$A: 5000 \times 1.0 3^{2}$ — Bank A pays 3% compound, so use the multiplier 1.03 once for each of the 2 years.
2
$= 5304.50$ — 1.03² = 1.0609, and 5000 × 1.0609 = 5304.50. This is Bank A's total, not its interest.
3
$5304.50 - 5000 = 304.50$ — Bank A's interest: take away the $5000 that was put in.
4
$B: 5000 \times \frac{3.2}{100} = 160$ — Bank B pays 3.2% simple interest: 3.2% of the first $5000, every year.
5
$160 \times 2 = 320$ — Simple interest is the same each year, so Bank B's interest for 2 years is 2 × 160.
6
$320 - 304.50 = $15.50$ — Both numbers are interests now, so subtract. Bank B pays $15.50 more.

Rule

Interest is the final amount take away the starting amount.

Common mistake

Comparing a total with an interest. Get both interests, or both totals, before you subtract.

Finding the Rate (Extended/Higher)

The 3-mark question: you know the start, the end and the years. Find the rate $x$ . Undo the power with a root.

Worked example

1
$4000 \times m^{3} = 4370.91$ — Call the yearly multiplier m. 3 years means m is used 3 times, so it is cubed.
2
$m^{3} = 4370.91 \div 4000$ — Divide both sides by the starting $4000, to get m³ on its own.
3
$m^{3} = 1.0927275$ — This is the growth over all 3 years together.
4
$m = 3 1.0927275 = 1.03$ — A cube is undone by a cube root. Do not divide by 3: the years are a power, not a times.
5
$1.03 = 1 + \frac{3}{100}$ — A multiplier of 1.03 means 103%, so the money grows by 3% a year. x = 3.

If the first year has a different rate, take that year out first.

Worked example

1
$2000 \times 1.05 \times m^{2} = 2271.36$ — Year 1 pays 5%, so it uses 1.05. The next 2 years use the unknown m, so m is squared.
2
$2100 \times m^{2} = 2271.36$ — Do the known part: 2000 × 1.05 = 2100, the value after year 1.
3
$m^{2} = 2271.36 \div 2100 = 1.0816$ — Divide both sides by 2100, to get m² on its own.
4
$m = 1.0816 = 1.04$ — A square is undone by a square root.
5
$x = 4$ — 1.04 means 104%, so the rate for those 2 years is 4%.

Common mistake

Dividing 1.0927 by 3. The years are a power, so undo them with a root.

Key idea

Money doubles in 12 years? Then $m^{12} = 2$ , so $m = 122 = 1.0595$ and the rate is 5.95%. You do not need the amount.

Finding the Start (Extended/Higher)

Sometimes you know the value at the end and must find the start. Multiplying got you there, so divide to go back.

Worked example

1
$P \times 1.0 4^{3} = 5624.32$ — After 3 years at 4% compound, the account holds $5624.32. Call the unknown start P. 4% gives the multiplier 1.04.
2
$1.0 4^{3} = 1.124864$ — Work out the power first.
3
$P = 5624.32 \div 1.124864$ — P was multiplied by 1.124864, so divide by 1.124864 to undo it.
4
$P = $5000$ — Check: 5000 × 1.124864 = 5624.32. It matches.

Value one year ago: a car loses 10% a year and is worth $8100 now.

Worked example

1
$V \times 0.9 = 8100$ — Losing 10% multiplied last year's value V by 0.9.
2
$V = 8100 \div 0.9$ — Undo the × 0.9 by dividing by 0.9.
3
$V = $9000$ — Check: 10% of 9000 is 900, and 9000 − 900 = 8100.

Common mistake

Adding 10% to $8100 to get $8910. The 10% was taken from last year's bigger value, not from $8100.

Different Rates and Overall Change

When the rate changes from year to year, use each year's own multiplier, one after another.

Worked example

1
$$6000 : 2% then 3.5% for 2 years$ — The question. Year 1 pays 2%. Years 2 and 3 pay 3.5% each.
2
$1.02 and 1.035$ — The multipliers: 2% gives 1.02 and 3.5% gives 1.035.
3
$6000 \times 1.02 \times 1.035^{2}$ — One multiplier per year: 1.02 once, then 1.035 twice.
4
$6000 \times 1.02 = 6120$ — The value at the end of year 1.
5
$6120 \times 1.071225 = 6555.897$ — 1.035² = 1.071225 covers years 2 and 3 together.
6
$\approx $6555.90$ — Round at the end, to the nearest cent.

Overall change: a price falls 20% one year, then 10% the next. What is the overall percentage decrease?

Worked example

1
$0.8 \times 0.9$ — Losing 20% keeps 0.8 of the value. Losing 10% keeps 0.9.
2
$= 0.72$ — Multiply the multipliers to get the change over both years.
3
$1 - 0.72 = 0.28$ — 0.72 means 72% is kept, so 28% is lost.
4
$28% decrease$ — Not 30%: the second 10% is taken from a smaller value.

Common mistake

Adding the percentages: 20% + 10% = 30%. Multiply the multipliers instead.

How Many Years? (Extended)

Some questions ask after how many whole years a value will first pass a target. Try values of n until you cross it.

Worked example

1
$40000 \times 1.0 6^{n} > 60000$ — A town of 40 000 grows by 6% a year. Write what you want: the start × 1.06, n times, must be more than 60 000.
2
$1.0 6^{n} > 1.5$ — Divide both sides by 40 000 to make the trials quicker: 60 000 ÷ 40 000 = 1.5.
3
$n = 6 : 1.0 6^{6} = 1.4185$ — Try a value. 1.4185 is less than 1.5, so after 6 years it is not there yet.
4
$n = 7 : 1.0 6^{7} = 1.5036$ — Try the next one. 1.5036 is more than 1.5, so it has passed 60 000.
5
$n = 7 years$ — The first whole number of years that works. 7 years.

Key idea

Write both trials, one each side of the target. They earn the method marks even if the final answer slips.

Common mistake

Giving 6.96 years. The question asks for whole years, and after 6 years it is still short, so the answer is 7.

Key idea

A value going down works the same way: keep trying n until it first drops below the target.

Which Method? Read the Question

Look for the words "each year", "compound" or "depreciates".

You see	You do
compound interest for n years	P × multiplier to the power n
depreciates by r% each year	multiplier 1 − r/100
total interest, how much more interest	final amount − start, for each
find x, the rate	divide, then take the root
final amount given, find the start	divide by the multiplier to the power n
a different rate in some years	one multiplier per year, multiplied together
how many years until it passes a value	try values of n, one each side
simple interest, same amount each year	Simple Interest
one increase or decrease only	Percentage Increases and Decreases
value after one change is given, find the start	Reverse Percentages
graph of y = a × b to the power x	Exponential Functions

Key idea

Check your answer: growth must end above the start, depreciation below it.

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