OxfordAQA International GCSE Mathematics: Percentage change, interest and exponential growth and decay (9260) – Revision Notes

These revision notes are aligned to the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. They cover N15 (section 3.1.2) and the repeated proportional change and exponential growth and decay parts of N20 (section 3.1.3). Core content applies at both tiers. Content marked Extension only is for the Extension tier.

For full explanations and longer worked examples, read the study guide. When you are ready to test yourself, use the practice questions. Find your weak spots with the free diagnostics, track coverage on the course checklist and see the course hub for the other units. The Number revision notes cover the rest of the strand.

Formulas and results

Idea	Result	Tier
Multiplier, increase of r%	1 + r/100	Core
Multiplier, decrease of r%	1 − r/100	Core
New amount	original × multiplier	Core
Percentage change	(change ÷ original) × 100	Core
Simple interest	P × r/100 × t	Core
Compound interest, working year by year	multiply by the yearly multiplier once per year	Core
Repeated change	multiply the multipliers together	Core
Original amount (reverse percentage)	new amount ÷ multiplier	Extension only
Compound interest formula	value = P(1 + r/100)ⁿ	Extension only
Exponential growth and decay	amount = start × kᵗ	Extension only

In the formula, P is the amount invested, r is the percentage rate of interest and n is the number of years of compound interest.

Method boxes

Percentage change

1. Find the change: new − old.
2. Divide by the original (the starting value).
3. Multiply by 100 and say whether it is an increase or a decrease.

Simple versus compound

1. Simple: the interest is the same every year, so interest = P × r/100 × t.
2. Compound: the balance is multiplied by 1 + r/100 each year, so the interest grows.
3. After one year the two are the same. They differ from year 2 onwards.

Reverse percentage (Extension only)

1. Write the multiplier for the change that happened.
2. Set up: original × multiplier = new amount.
3. Divide the new amount by the multiplier.
4. Check by multiplying forwards.

Growth or decay (Extension only)

1. Write k = 1 + r/100 for growth, or k = 1 − r/100 for decay.
2. Write the model: amount = start × kᵗ.
3. To find a time, try whole values of t until the amount crosses the target. State the first value that works.

Small worked reminders

Percentage change. A jacket is £380 and is reduced by 15%.

380 × 0.85 = £323

A bike is £70 in a shop and £91 online. The online price is a 30% increase, because (91 − 70)/70 × 100 = 30.

Simple interest. £1,200 at 4% a year for 3 years.

1200 × 0.04 × 3 = £144

Compound interest (Core). £2,000 at 5% a year for 4 years.

2000 × 1.05 × 1.05 × 1.05 × 1.05 = £2,431.01

Compound interest formula (Extension only). The same sum is 2000 × 1.05⁴ = £2,431.01 (nearest penny).

Reverse percentage (Extension only). A shop price after a 16% increase is £1,566.

1566 ÷ 1.16 = £1,350

A price after a 28% decrease is £396.

396 ÷ 0.72 = £550

Changing rate (Core). £800 earns 2% in year 1 and 3% in year 2.

800 × 1.02 × 1.03 = £840.48

Finding the rate (Extension only). £4,000 becomes £4,368.10 after 2 years of compound interest.

(1 + r/100)² = 4368.10 ÷ 4000 = 1.09203 (rounded)
1 + r/100 = √1.09203 = 1.045
r = 4.5

The rate is 4.5% a year.

Two reverse steps (Extension only). A price is cut by 10%, and the reduced price is cut by a further 20%. The final price is £144.

Overall multiplier = 0.9 × 0.8 = 0.72
Original = 144 ÷ 0.72 = £200

Trial for time (Extension only). A 250 g sample loses 20% of its mass each hour. When does it first fall below 100 g?

Mass after t hours = 250 × 0.8ᵗ
t = 4: 102.4 g   (not yet)
t = 5: 81.92 g   (below 100 g)

It first falls below 100 g during the 5th hour, so the answer is 5 hours.

Repeated change (Core). A quantity rises by 10% and then falls by 10%.

1.1 × 0.9 = 0.99

The overall change is a 1% decrease.

Decay (Extension only). A tank of 6,400 litres loses 10% of its contents each day. After 3 days:

6400 × 0.9³ = 4,665.6 litres

Growth (Extension only). A population of 150 grows by 20% a year. After 5 years:

150 × 1.2⁵ = 373 (nearest whole number)

Reading the wording

The question says	What to do
“Increase by 8%” or “decrease by 8%”	Multiply by 1.08 or by 0.92
“Find the percentage change”	(new − old) ÷ old × 100, then state increase or decrease
“Simple interest”	Same interest each year, P × r/100 × t
“Compound interest”	Multiply by 1 + r/100 once for every year
“Before the increase” or “original price”	Divide by the multiplier (Extension only)
“After t years” with a repeated percentage	Start × kᵗ (Extension only)
“How many years until it exceeds…”	Trial of whole values of t (Extension only)
“Overall change” after two or more changes	Multiply all the multipliers, then convert back to a percentage

To convert an overall multiplier back to a percentage, subtract 1 and multiply by 100. A multiplier of 1.07016 is a 7.016% increase. A multiplier of 0.83 is a 17% decrease.

Must-know distinctions

• Simple and compound. Simple interest uses the original sum every year. Compound interest uses the balance at the start of each year.
• Original and new. Percentage change is divided by the original. A reverse percentage divides the final amount by the multiplier.
• Adding and multiplying. A rise of 10% then a fall of 10% is 1.1 × 0.9, not zero change.
• Growth and decay. Growth has k above 1. Decay has k between 0 and 1.
• Interest earned and total value. “Find the interest” means subtract the starting sum. “Find the value” means the whole amount.
• Percentage points and percent. A rate moving from 3% to 4% is a rise of one percentage point, but a rise of 33.3% in the rate itself. Say which one you mean.
• Core method and Extension formula. The formula P(1 + r/100)ⁿ is Extension only. At Core, show each year’s multiplication.

Quick self-test

1. Increase £520 by 18%.
2. Decrease £520 by 18%.
3. A price goes from £40 to £46. Find the percentage increase.
4. Find the simple interest on £900 at 3% a year for 6 years.
5. Find the total multiplier for three years of compound interest at 7% a year. (Extension only)
6. Find the multiplier for two successive 12% decreases. (Core)
7. After a 15% increase a bill is £345. Find the original bill. (Extension only)
8. Find the value of £2,500 after 2 years at 4% compound interest, using the formula. (Extension only)
9. A radioactive sample of 60 g halves every day. Find the mass after 3 days. (Extension only)
10. A 10% rise followed by a 10% fall: state the single overall percentage change. (Core)

Answers

1. 520 × 1.18 = £613.60
2. 520 × 0.82 = £426.40
3. 6/40 × 100 = 15% increase
4. 900 × 0.03 × 6 = £162
5. 1.07³ = 1.225043
6. 0.88 × 0.88 = 0.7744
7. 345 ÷ 1.15 = £300
8. 2500 × 1.04² = £2,704
9. 60 × 0.5³ = 7.5 g
10. 1.1 × 0.9 = 0.99, so a 1% decrease

Where marks are usually lost

• Choosing the finishing value as the divisor in a percentage change. The starting value is the one to use.
• Using 1.5 instead of 1.05 for a 5% increase, or 0.5 instead of 0.95 for a 5% decrease.
• Taking a percentage of the final amount when the question asks for the starting amount. Divide by the multiplier.
• Answering with the total when the interest was asked for, or with the interest when the total was asked for.
• Adding percentages in repeated change problems instead of multiplying the multipliers.
• Rounding intermediate values, which can move the final answer in the third significant figure or the last penny.
• Leaving the answer to a money question as £2431.0125 instead of rounding to the nearest penny.
• Counting the wrong number of years: interest for 3 years means n = 3.
• Stopping a trial-and-improvement search at the first value that looks close, without checking the value either side of the target.
• Writing a decrease as a negative multiplier or forgetting to say “increase” or “decrease”.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Spec references: N15 (3.1.2) and N20 (3.1.3).