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

This study guide is aligned to the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. It covers N15 in section 3.1.2 (percentage change, interest, reverse percentages and the compound interest formula) and the repeated proportional change and exponential growth and decay statements of N20 in section 3.1.3. N15 and the repeated proportional change in N20 are Core content, so they can be assessed at both tiers. Reverse percentages, the compound interest formula and exponential growth and decay are marked Extension only below.

Use it with the revision notes and the practice questions. The wider Number study guide covers the rest of the strand. Check where you stand with the free diagnostics, and tick off progress on the course checklist. The course hub is at /boards/oxfordaqa/igcse/mathematics/.

What this unit covers

Spec	What you must be able to do	Tier
N15	Solve problems involving percentage increase and decrease	Core
N15	Solve problems involving simple interest and compound interest	Core
N15	Solve reverse percentage problems	Extension only
N15	Use the compound interest formula P(1 + r/100)ⁿ	Extension only
N20	Solve problems involving repeated proportional change	Core
N20	Solve problems on exponential growth and decay	Extension only

Direct and inverse proportion, also in N20, belong to a separate unit.

1. The multiplier method

A percentage change is a multiplication. Write the new amount as a percentage of the old amount, then turn it into a decimal.

Change	New amount is	Multiplier
Increase by 12%	112% of the old	1.12
Decrease by 12%	88% of the old	0.88
Increase by 3.5%	103.5% of the old	1.035
Decrease by 0.5%	99.5% of the old	0.995

Increase: multiplier = 1 + r/100. Decrease: multiplier = 1 − r/100.

Worked example 1. A laptop costs £845. A shop takes 12% off. Find the sale price.

Multiplier for a 12% decrease = 1 − 0.12 = 0.88
Sale price = 845 × 0.88 = £743.60

Working out 12% and subtracting also works. The multiplier is quicker, and you will need it for the later sections.

2. Percentage change

Percentage change = (change ÷ original) × 100. The original is always the amount you started with, not the larger or smaller of the two.

Worked example 2. A flat’s rent rises from £540 to £603 a month. Find the percentage increase.

Change = 603 − 540 = 63
Percentage change = 63/540 × 100 = 11.67...

The increase is 11.7% (3 s.f.).

Worked example 3. A tank holds 64 litres. After a leak it holds 52 litres. Find the percentage decrease.

Change = 64 − 52 = 12
12/64 × 100 = 18.75

The decrease is 18.75%, or 18.8% to 3 s.f. Say “decrease” in your answer, because the question asks for the direction as well as the size.

3. Simple interest

With simple interest, only the sum you first invested earns anything, so each year adds an identical amount.

Simple interest = P × r/100 × t, where P is the sum invested, r is the yearly rate (%) and t is the time in years.

Worked example 4. Find the simple interest on £2,400 invested for 4 years at 3.5% a year, and the total at the end.

Interest = 2400 × 0.035 × 4 = £336
Total = 2400 + 336 = £2,736

Worked example 5. £1,800 earns £198 simple interest in 2.5 years. Find the yearly rate.

198 = 1800 × (r/100) × 2.5
198 = 45r
r = 4.4

The rate is 4.4% a year.

4. Compound interest (Core: repeated multipliers)

Compound interest is paid on the original sum plus the interest already earned. Each year the balance is multiplied by the same multiplier, so you multiply once per year.

Worked example 6. £3,200 earns 2.5% compound interest a year for 3 years. Find the final amount and the total interest.

Multiplier = 1.025
After year 1: 3200 × 1.025 = 3280
After year 2: 3280 × 1.025 = 3362
After year 3: 3362 × 1.025 = 3446.05
Interest = 3446.05 − 3200 = £246.05

If the rate changes from year to year, multiply by a different number each time.

Worked example 7. £5,000 earns 3% in year 1, 2.5% in year 2 and 1.8% in year 3.

5000 × 1.03 × 1.025 × 1.018 = £5,373.77 (nearest penny)

For Core questions, show the multiplier for every year. A single number with no working gives nothing to follow if it is wrong.

5. Compound interest formula (Extension only)

For a fixed rate, the repeated multiplication shortens to a formula.

Value of investment = P(1 + r/100)ⁿ

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

Worked example 8. Find the value of £7,500 invested for 6 years at 3.2% compound interest.

Value = 7500 × (1 + 3.2/100)⁶
      = 7500 × 1.032⁶
      = £9,060.23

To find the rate, divide the final amount by P, then take the nth root.

Worked example 9. £7,000 grows to £9,005 in 4 years at a fixed compound rate. Find the rate.

7000 × (1 + r/100)⁴ = 9005
(1 + r/100)⁴ = 9005/7000 = 1.28643
1 + r/100 = 1.28643^(1/4) = 1.06499...
r = 6.5

The rate is 6.5% a year.

6. Reverse percentages (Extension only)

In a reverse percentage problem you are given the amount after the change and must find the original. Divide by the multiplier. Do not take a percentage of the new amount.

Worked example 10. After a 15% increase, a monthly fee is £1,380. Find the fee before the increase.

Multiplier = 1.15
Original × 1.15 = 1380
Original = 1380 ÷ 1.15 = £1,200

Worked example 11. A coat is reduced by 20% in a sale. The sale price is £68. Find the original price.

Sale price is 80% of the original, so multiplier = 0.8
Original = 68 ÷ 0.8 = £85

Check by working forwards: 85 × 0.8 = 68.

7. Repeated proportional change (Core)

When a quantity changes by one percentage and then by another, multiply the multipliers. Do not add the percentages.

Worked example 12. A price rises by 5% and then falls by 4%. Find the overall percentage change.

Overall multiplier = 1.05 × 0.96 = 1.008

The price rises by 0.8% overall, not 1%. The 4% fall is taken from the new, higher price.

8. Exponential growth and decay (Extension only)

Repeated proportional change with the same multiplier k each time gives a model of the form

amount after t steps = starting amount × kᵗ

If k > 1 you have exponential growth. If 0 < k < 1 you have exponential decay. The interest formula in section 5 is an example with k = 1 + r/100.

Worked example 13. A town has 18,000 people. The population falls by 2.4% a year.

k = 1 − 0.024 = 0.976
Population after t years = 18000 × 0.976ᵗ
t = 10: 18000 × 0.976¹⁰ = 14,118 (nearest whole number)

To find when the population first falls below 14,000, try values of t:

t = 10: 14,118   (still above 14,000)
t = 11: 13,779   (below)

It falls below 14,000 during the 11th year.

Worked example 14. A van is worth £21,500 and loses 14% of its value each year. In which year does it first fall below half its value?

Half of 21,500 = 10,750
Value after t years = 21500 × 0.86ᵗ
t = 4: 11,760.68
t = 5: 10,114.18

The value first falls below £10,750 after 5 years.

A quick way to choose a method

Read the wording before you reach for a formula.

• If the question gives the old and new amounts and asks for a percentage, use change ÷ original.
• If it gives the original and a percentage, multiply by the multiplier.
• If it gives the final amount and a percentage, and asks for the start, divide by the multiplier (Extension only).
• If the same percentage is applied over and over, the power form kᵗ is the shortest route (Extension only). At Core, multiply year by year.

Common errors

• Dividing by the new amount when finding a percentage change. Divide by the original.
• Adding percentages in repeated change. A 20% rise followed by a 20% fall is not “no change”; the multiplier is 1.2 × 0.8 = 0.96.
• Using the simple interest method when the question says compound, or the reverse.
• Reverse percentages: finding 15% of £1,380 and subtracting. The 15% was taken from the original, not from £1,380.
• Using the wrong power in the formula. If interest is paid for 3 years, n = 3, not 4.
• Rounding too soon. Keep the full unrounded value and round money to the nearest penny only at the end.
• Writing 1.5 as the multiplier for a 5% rise. A 5% rise is 1.05.

Next steps

Work through the practice questions, then condense your notes using the revision notes. The Number practice set tests the rest of the strand.

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).