IGCSE Maths · Number · Core & Extended

Percentages: Increase, Decrease, Interest and Reverse Problems

You already know percent means “per hundred.” This note is about using that fact fluently — finding a percentage of an amount, describing how much something has grown or shrunk, and eventually working backwards from a final amount to find where you started. Reverse percentages and general repeated-change problems are Extended-only; everything else applies to both tiers.

Why this matters Sales tax, discounts, tips, exam grade boundaries, salary raises, loan interest, population growth, inflation — percentages are how the real world describes change. Getting comfortable here pays off directly outside the exam room too.

1. Finding a Percentage of a Quantity

Convert the percentage to a decimal (divide by 100), then multiply.

Worked example Find 15% of 240.
15% = 0.15
0.15 × 240 = 36
Worked example — percentage over 100% Find 130% of 60.
130% = 1.3
1.3 × 60 = 78 (more than the original 60, which makes sense — 130% is more than the whole amount)
Try it — Percentage of a Quantity
  1. Find 25% of 160.
  2. Find 8% of 450.
  3. Find 62.5% of 96.
  4. Find 150% of 40.
Show answers
  1. 40
  2. 36
  3. 60
  4. 60

2. Expressing One Quantity as a Percentage of Another

(Part ÷ Whole) × 100
Worked example A student scores 18 out of 40 on a test. Express this as a percentage.
(18 ÷ 40) × 100 = 45%
Worked example — different units Express 45 cm as a percentage of 3 m.
Convert to the same unit first: 3 m = 300 cm.
(45 ÷ 300) × 100 = 15%
Try it — Expressing as a Percentage
  1. Express 21 out of 60 as a percentage.
  2. Express 130 out of 200 as a percentage.
  3. Express 250 g as a percentage of 2 kg.
Show answers
  1. 35%
  2. 65%
  3. 250 g out of 2000 g = 12.5%

3. Percentage Increase and Decrease

The fastest method uses a multiplier — a single number that does the whole calculation in one step.

ChangeMultiplier
Increase by 15%× 1.15
Decrease by 20%× 0.80
Increase by 8%× 1.08
Decrease by 35%× 0.65
Worked example — increase A jacket costs $80. It is increased in price by 15%. Find the new price.
80 × 1.15 = $92
Worked example — decrease A laptop costs $150. It is reduced by 20% in a sale. Find the sale price.
150 × 0.80 = $120
Try it — Increase & Decrease
  1. Increase 90 by 10%.
  2. Decrease 250 by 30%.
  3. A salary of $2,400 increases by 6%. Find the new salary.
  4. A population of 84,000 falls by 5%. Find the new population.
Show answers
  1. 90 × 1.1 = 99
  2. 250 × 0.7 = 175
  3. 2400 × 1.06 = $2,544
  4. 84,000 × 0.95 = 79,800

4. Simple Interest

Simple interest is calculated on the original amount only, every time period — it doesn’t grow on top of itself.

I = (P × R × T) ÷ 100

where P = principal (starting amount), R = rate per year (%), T = time in years.

Worked example $2,000 is invested at a simple interest rate of 5% per year for 3 years. Find the total interest earned, and the final amount.
I = (2000 × 5 × 3) ÷ 100 = $300 interest
Final amount = 2000 + 300 = $2,300
Try it — Simple Interest
  1. Find the simple interest on $1,500 at 4% per year for 5 years.
  2. Find the simple interest on $3,000 at 6% per year for 2 years, and the final amount.
  3. $800 is invested at a simple interest rate of 3.5% per year. Find the interest earned after 4 years.
Show answers
  1. I = 1500×4×5/100 = $300
  2. I = 3000×6×2/100 = $360; final amount = $3,360
  3. I = 800×3.5×4/100 = $112

5. Compound Interest

Compound interest is calculated on the growing total each year, not just the original amount — the interest itself earns interest. Use the multiplier method, applied repeatedly (which is exactly the same as raising the multiplier to a power).

A = P × (1 + R/100)ⁿ

where A = final amount, P = principal, R = rate per year (%), n = number of years.

Worked example $1,500 is invested at a compound interest rate of 4% per year for 3 years. Find the final amount, and the total interest earned.
A = 1500 × (1.04)³ = 1500 × 1.124864 = $1,687.30 (to the nearest cent)
Interest earned = 1687.30 − 1500 = $187.30
Try it — Compound Interest
  1. Find the value of $2,000 invested at 5% compound interest for 2 years.
  2. Find the value of $5,000 invested at 3% compound interest for 4 years, to the nearest dollar.
  3. Compare: $1,000 at 6% simple interest for 3 years vs. $1,000 at 6% compound interest for 3 years. Which earns more, and by how much?
Show answers
  1. 2000 × (1.05)² = 2000 × 1.1025 = $2,205
  2. 5000 × (1.03)⁴ = 5000 × 1.12550881 ≈ $5,628
  3. Simple: I = 1000×6×3/100 = $180, total $1,180. Compound: 1000×(1.06)³ = 1000×1.191016 = $1,191.02. Compound earns $11.02 more.

6. Repeated Percentage Change (Beyond Interest) Extended only

The same “multiply repeatedly” idea from compound interest applies to any situation where something changes by the same percentage, over and over — not just money in a bank.

Worked example — depreciation A car is bought for $18,000. It depreciates (loses value) by 10% each year. Find its value after 3 years.
18,000 × (0.9)³ = 18,000 × 0.729 = $13,122
Worked example — population growth A town’s population is 50,000 and grows by 3% per year. Find the population after 5 years, to the nearest whole number.
50,000 × (1.03)⁵ = 50,000 × 1.159274… ≈ 57,964
Try it — Repeated Percentage Change Extended
  1. A machine worth $12,000 depreciates by 15% per year. Find its value after 2 years.
  2. A bacteria culture of 4,000 cells grows by 20% every hour. Find the population after 3 hours.
  3. A phone originally priced at $900 loses 12% of its value every year. After how many whole years will it first be worth less than $500? (Try each year in turn.)
Show answers
  1. 12,000 × (0.85)² = 12,000 × 0.7225 = $8,670
  2. 4,000 × (1.2)³ = 4,000 × 1.728 = 6,912 cells
  3. Year 1: 900×0.88=792; Year 2: ≈697; Year 3: ≈613; Year 4: ≈540; Year 5: ≈475 — so after 5 years.

7. Reverse Percentages Extended only

Sometimes you’re given the amount after a percentage change and asked to find the original amount. The trap: you cannot simply reverse the percentage and apply it to the new amount — you must divide by the multiplier, not subtract the same percentage back off.

Worked example — after an increase After a 20% increase, the price of an item is $96. Find the original price.
The multiplier for a 20% increase is 1.20, so: original × 1.20 = 96
Original = 96 ÷ 1.20 = $80
Worked example — after a decrease After a 15% decrease, the price of an item is $102. Find the original price.
The multiplier for a 15% decrease is 0.85, so: original × 0.85 = 102
Original = 102 ÷ 0.85 = $120
Why you can’t just “reverse” the percentage directly It’s tempting to think: “if $96 is after a 20% increase, then decreasing $96 by 20% should give back the original.” Let’s test that: 96 × 0.80 = $76.80 — not $80. The percentages don’t cancel like that, because 20% of the original price is a different amount than 20% of the new price. This is exactly why reverse percentage questions always divide by the multiplier instead.
Try it — Reverse Percentages Extended
  1. After a 25% increase, an item costs $75. Find the original price.
  2. After a 10% decrease, an item costs $63. Find the original price.
  3. A shop’s revenue this year is $184,000, which is 15% higher than last year. Find last year’s revenue.
  4. A shirt is on sale for $38.25 after a 15% discount. Find the original price.
Show answers
  1. 75 ÷ 1.25 = $60
  2. 63 ÷ 0.90 = $70
  3. 184,000 ÷ 1.15 = $160,000
  4. 38.25 ÷ 0.85 = $45
Common mistakes to watch for
  • For reverse percentages, subtracting/adding the percentage back onto the new amount instead of dividing by the correct multiplier.
  • Using the wrong multiplier — a 15% decrease is × 0.85, not × 0.15 (0.15 would only give you the amount of the decrease, not what’s left).
  • Applying simple interest’s formula to a compound interest question, or vice versa — check the wording carefully for “simple” vs “compound.”
  • Forgetting to round money answers sensibly (to the nearest cent) only at the very end, not partway through a calculation.
  • For repeated percentage change, using n × (percentage) instead of raising the multiplier to the power n — repeated percentage change is multiplicative, not additive.

8. Quick-Fire Challenge Round

True or False?
  1. To increase an amount by 8%, you multiply by 1.08.
  2. To decrease an amount by 8%, you multiply by 0.8.
  3. Compound interest always gives a bigger final amount than simple interest, for the same rate and time (beyond year 1).
  4. For a reverse percentage question, you always divide the new amount by the multiplier.
  5. 150% of a number is always bigger than the number itself.
Show answers
  1. True.
  2. False — decreasing by 8% means multiplying by 0.92, not 0.8 (0.8 would be a 20% decrease).
  3. True — beyond the first year, compound interest earns interest on the growing balance, so it overtakes simple interest.
  4. True (for reverse percentages — this is Extended-only content). Ext.
  5. True — since 150% > 100%.
Spot the Error
  1. “To find the original price before a 20% increase, I took the new price and decreased it by 20%.”
  2. “Simple interest on $500 at 4% for 3 years is 500 × 1.04 × 3 = $1,560.”
  3. “A 25% decrease means multiplying by 0.25.”
Show answers
  1. This doesn’t reverse an increase correctly — you must divide by the multiplier (1.20), not decrease the new price by the same percentage.
  2. Simple interest uses I = PRT/100, not P × (1+R) × T. Correct: I = 500×4×3/100 = $60.
  3. A 25% decrease means multiplying by 0.75 (keeping 75% of the amount), not 0.25.

9. Mixed Practice — Bringing It All Together

Core & Extended
  1. Find 35% of 220.
  2. Express 54 out of 90 as a percentage.
  3. Increase 320 by 12%.
  4. Find the simple interest on $2,500 at 3% per year for 4 years.
  5. Find the value of $4,000 invested at 5% compound interest for 2 years.
Show answers
  1. 77
  2. 60%
  3. 320 × 1.12 = 358.4
  4. 2500×3×4/100 = $300
  5. 4000×(1.05)² = 4000×1.1025 = $4,410
Extended-level mixed set Extended only
  1. After a 30% increase, an item costs $91. Find the original price.
  2. A car worth $20,000 depreciates by 12% per year. Find its value after 3 years.
  3. A shop’s sale price is $68 after a 20% discount. Find the original price.
  4. A savings account starts at $3,000 and grows by 4% each year. Find the balance after 3 years, to the nearest dollar.
Show answers
  1. 91 ÷ 1.30 = $70
  2. 20,000 × (0.88)³ = 20,000 × 0.681472 = $13,629.44
  3. 68 ÷ 0.80 = $85
  4. 3000 × (1.04)³ ≈ $3,374.59 → $3,375

Practice at Home

A large set again — take it in stages.

Part A — Percentage of a quantity, and expressing as a percentage
  1. Find 45% of 320.
  2. Find 12.5% of 88.
  3. Express 42 out of 56 as a percentage.
  4. Express 3.5 kg as a percentage of 10 kg.
Part B — Increase and decrease
  1. Increase 150 by 24%.
  2. Decrease 480 by 35%.
  3. A rent of $850 per month increases by 4%. Find the new rent.
  4. A crowd of 12,000 people falls by 8% the following week. Find the new crowd size.
Part C — Simple and compound interest
  1. Find the simple interest on $1,800 at 5% per year for 3 years.
  2. Find the value of $2,500 at 6% compound interest for 2 years.
  3. Find the value of $6,000 at 2.5% compound interest for 4 years, to the nearest dollar.
Part D — Repeated percentage change Extended
  1. A boat worth $45,000 depreciates by 8% per year. Find its value after 3 years, to the nearest dollar.
  2. A colony of 2,500 insects grows by 25% every week. Find the population after 4 weeks.
Part E — Reverse percentages Extended
  1. After a 40% increase, an item costs $112. Find the original price.
  2. After a 25% decrease, an item costs $84. Find the original price.
  3. A company’s profit this year is $276,000, which is 20% higher than last year. Find last year’s profit.
  4. A jacket is on sale for $59.40 after a 10% discount. Find the original price.
Part F — Applied / challenge
  1. An investment of $5,000 earns 4% compound interest per year. After how many whole years will it first exceed $6,000? (Try each year in turn.)
  2. A phone’s price is reduced by 10%, then reduced by a further 10% in a second sale. Is this the same as a single 20% reduction? Show your working to justify your answer.
  3. A test score of 78 out of 120 is scaled up by 15% for a curve. Find the new score as a percentage of 120, and then as a percentage rounded to 1 decimal place.

Key Vocabulary Recap

Multiplier — a single number that increases or decreases an amount in one step
Principal — the original amount of money invested or borrowed
Simple interest — interest calculated on the original amount only
Compound interest — interest calculated on the growing total each period
Depreciation — a repeated percentage decrease in value over time
Reverse percentage — working backwards from a final amount to find the original Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C1.13 (Core subject content) and section E1.13 (Extended subject content) — Percentages. Reverse percentages, and repeated-percentage-change problems beyond interest (such as depreciation or population growth), are Extended-only, marked above. Simple interest and compound interest are shared by both tiers.

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