Daily Maths

Compound interest and repeated percentage change

Compound Interest Core + Extended 8 min read

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In this note
Formula bank
Interest on the interest
Working backwards
Simple or compound?
Ways they test it
Mistakes to avoid
Summary
Practice
Can I do this?
Quiz

In this note

Syllabus 1.13: compound interest and repeated change

1.13Find the value of an investment with compound interest, and the interest earned.
1.13Use a multiplier for repeated increase or decrease (depreciation).
1.13Work backwards: the amount invested, the rate $r$, the number of years.

Before you start: simple interest and multipliers. The formula is NOT given in the exam.

Formula bank

Compound interestLearn it$A=P\left(1+\dfrac r{100}\right)^{n}$$A$ = value after $n$ years, $P$ = amount invested, $r$ = % rate.Use it when "compound interest", "each year".

DepreciationLearn it$A=P\left(1-\dfrac r{100}\right)^{n}$a value falling by r% each year.Use it when cars, machines, population falling.

Interest earnedLearn it$A-P$the formula gives the VALUE.Use it when "how much interest".

Amount investedLearn it$P=A\div(\text{multiplier})^{n}$Use it when you know the final value.

RateLearn it$1+\dfrac r{100}=\sqrt[n]{\dfrac AP}$Use it when "find r".

Interest on the interest

With compound interest each year's interest is added on, and next year's interest is worked out on the NEW, bigger amount. Every year the amount is multiplied by the same multiplier.

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

After 3 years: $2000\times1.05^{3}=2315.25$. Simple interest would give only $2300.

General methodHow to find any compound interest value

Write the multiplier: increase r% → 1 + r/100; decrease r% → 1 − r/100.
Value = P × multiplier$^{n}$ (use the power key).
Interest = value − P (only if they ask for the interest).
Rate per month? Then n must be in months too.

Example $3500 at 3.6% per year compound interest. Value after 6 years?

Multiplier 1.036, to the power 6.

$3500\times1.036^{6}=4327.395\ldots$

Answer $4327.40

Try it yourself

$2000 at 5% compound interest. Value after 4 years?

Show answer

Answer $2431.01

A car worth $18 000 loses 12% each year. Value after 3 years?

Show answer

Answer $12 266.50

$800 at 3% for 5 years. Interest earned?

Hint

Value − 800.

Show answer

Answer $127.42

Working backwards

General methodHow to find the amount invested, the rate or the number of years

Amount invested: divide the value by multiplier$^{n}$.
Rate: $\dfrac AP$ = multiplier$^{n}$; take the n-th root; subtract 1; × 100.
Number of years: try n = 1, 2, 3 … and show the last value BELOW the target and the first ABOVE it.

Example Rosa invests $7500 at r% compound interest; after 4 years it is $9000. Find r to 2 d.p.

Divide, take the 4th root, subtract 1.

$9000\div7500=1.2$
$\sqrt[4]{1.2}=1.04663\ldots$

Answer $r=4.66$

Try it yourself

After 4 years at 3% an investment is worth $6000. Amount invested?

Show answer

Answer $5330.92

$5000 at 4%. After how many complete years is it first worth more than $6000?

Hint

Try n = 4 and n = 5.

Show answer

Answer 5 years

A price rises 5% every year for 3 years. Overall percentage increase (2 d.p.)?

Show answer

Answer 15.76%

Simple or compound?

	Simple	Compound
Interest worked out on	the original amount	the amount at the start of each year
Interest each year	the same	grows
$1000 at 6% for 2 years	$1120	$1123.60

Ways they test it

Type 1 Value after n years

"$3500 at 3.6% compound interest. Value after 6 years." (2 marks)

Spot it "compound interest" and a time.

P × multiplier$^{n}$.

Type 2 Interest earned

"Work out the interest earned at the end of 4 years." (3 marks)

Spot it "interest" (not "value").

Value first.
Subtract P.

For example $5200 at 2.8% for 4 years

$5200\times1.028^{4}=5807.32$

Answer $607.32

Type 3 Depreciation

"A machine loses 15% of its value each year..."

Spot it "decreases by r% each year".

Multiplier 1 − r/100, to the power n.

For example $25 000, −15% for 4 years

$25\,000\times0.85^{4}$

Answer $13 050.16

Type 4 Find r

"$7500 is worth $9000 after 4 years. Find r." (3 marks)

Spot it an unknown rate.

A ÷ P.
n-th root, − 1, × 100.

Type 5 Find n

"After how many complete years is it first worth more than $9000?" (3-4 marks)

Spot it "complete years", "first".

Trial: n = 1, 2, 3 …
Show both sides of the target.

Type 6 Overall % change

"5% a year for 3 years: overall percentage increase?"

Spot it "overall".

multiplier$^{n}$ − 1, as a %.

Mistakes to avoid

✗ $5200\times1.028\times4$✓ $5200\times1.028^{4}$Multiply by the multiplier 4 times.
✗ Gave the value when asked for the interest✓ subtract PRead the question.
✗ Depreciation 12%: × 1.12✓ × 0.88A decrease multiplier is below 1.
✗ 5% for 3 years = 15%✓ 15.76%Compound changes multiply.
✗ Rate: $1.2\div4$✓ $\sqrt[4]{1.2}$Undo a power with a root.

Summary

Value $=P\times(\text{multiplier})^{n}$; interest = value − P.
Increase: 1 + r/100; decrease: 1 − r/100.
Amount invested: ÷ multiplier$^{n}$. Rate: n-th root. Years: trial.
Compound beats simple after year 1.

If you see …

"each year" with a %multiplier to the power n.
"interest earned"value − P.
"complete years"trial, show both sides.

Practice

Practice: easy, medium, hard

Easy

$800 at 3% compound interest for 5 years. Value?

Show working

$800\times1.03^{5}$

Answer $927.42

A $950 phone loses 10% a year. Value after 2 years?

Show working

$950\times0.9^{2}$

Answer $769.50

Rising 10% a year for 2 years. Overall percentage increase?

Show working

$1.1^{2}=1.21$

Answer 21%

Medium

Nadia invests $4000 at 3% compound interest. Interest earned after 2 years?

Guide me

What is it asking? the interest

What do we know? P = 4000, r = 3, n = 2

Plan value, then − 4000

Show working

$4000\times1.03^{2}=4243.60$
$-4000$

Answer $243.60

Interest on $1200 at 2% compound for 10 years?

Guide me

What is it asking? the interest

What do we know? n = 10

Plan $1200\times1.02^{10}-1200$

Show working

$1200\times1.02^{10}=1462.79$

Answer $262.79

A machine costs $25 000 and loses 15% a year. Overall percentage decrease over 4 years (1 d.p.)?

Guide me

What is it asking? overall % decrease

What do we know? × 0.85 four times

Plan $1-0.85^{4}$

Show working

$0.85^{4}=0.522$
$1-0.522=0.478$

Answer 47.8%

Hard

$2400 is worth $2660.92 after 3 years of compound interest. Find r.

Guide me

What is it asking? the rate

What do we know? A/P = multiplier³

Plan cube root, − 1, × 100

Show working

$2660.92\div2400=1.10872$
$\sqrt[3]{1.10872}=1.035$

Answer 3.5

Part	Answer / working	Marks
	$\left(1+\tfrac r{100}\right)^{3}=\tfrac{2660.92}{2400}$	M1
	cube root	M1
	3.5	A1
Total: 3 marks

Kamal invests $7500 at 4.5% compound interest. First complete year it is worth more than $9000?

Guide me

What is it asking? smallest whole n

What do we know? $7500\times1.045^{n}$

Plan try n = 4, 5

Show working

n = 4: 8943.89
n = 5: 9346.36

Answer 5

Part	Answer / working	Marks
	8943.89	M1
	9346.36	M1
	5	A2
Total: 4 marks

A town of 48 000 decreases by 2.5% each year. Population after 6 years, to the nearest hundred?

Guide me

What is it asking? the population

What do we know? × 0.975 six times

Plan $48\,000\times0.975^{6}$

Show working

$48\,000\times0.975^{6}=41\,235.3$

Answer 41 200

Part	Answer / working	Marks
	$48\,000\times0.975^{6}$	M2
	41 200	A1
Total: 3 marks

Can I do this?

I can explain why compound interest grows faster than simple interest.
I can find the value and the interest, including depreciation.
I can find the amount invested and the rate.
I can find the number of complete years by trial.

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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Part	Answer / working	Marks
	\(\left(1+\tfrac r{100}\right)^{3}=\tfrac{2660.92}{2400}\)	M1
	cube root	M1
	3.5	A1
Total: 3 marks