Decimals and Percentages

This lesson covers calculating with decimals, finding percentages of amounts, percentage increase and decrease, and expressing one quantity as a percentage of another. These are among the most heavily used skills in OCR GCSE Mathematics (J560) because they connect to money, measures, ratio and proportion, and they appear on every paper. We focus on reliable non-calculator methods as well as efficient multiplier methods for the calculator papers.

This lesson mainly builds AO1 fluency in percentage calculation, with strong AO3 problem-solving in the real-life contexts (pay rises, sale prices, test scores) and AO2 reasoning when you interpret what a percentage change means.

Key Vocabulary

Term	Meaning
Percentage	A number of parts per hundred, written with the % sign
Multiplier	The decimal you multiply by to apply a percentage change, e.g. $1.15$1.15 for $+15\%$+15%
Percentage increase	A rise expressed as a percentage of the original
Percentage decrease	A fall expressed as a percentage of the original
Original (or 100%)	The starting amount a percentage is calculated from
Profit / loss	The increase / decrease compared with the cost price

Calculating with Decimals

Decimal arithmetic on Paper 1 relies on aligning place value. For multiplication, ignore the points, multiply as integers, then put back the total number of decimal places.

Worked Example 1

Work out $3.7 \times 0.6$3.7×0.6.

Ignore the points: $37 \times 6 = 222$37×6=222. The numbers have $1 + 1 = 2$1+1=2 decimal places in total, so the answer has $2$2 decimal places.

$3.7 \times 0.6 = 2.22.$3.7×0.6=2.22.

Answer: $2.22$2.22.

Common error: Forgetting to count all the decimal places. $3.7 \times 0.6$3.7×0.6 is not $22.2$22.2.

Worked Example 2

Work out $4.8 \div 0.4$4.8÷0.4.

Make the divisor a whole number by multiplying both numbers by $10$10: $4.8 \div 0.4 = 48 \div 4 = 12$4.8÷0.4=48÷4=12.

Answer: $12$12.

Percentages of Amounts

A percentage simply means "out of $100$100", so $35\%$35% means $35$35 parts in every $100$100, or the fraction $\dfrac{35}{100}$10035​, or the decimal $0.35$0.35. Holding all three views of a percentage in mind lets you pick the easiest route for a given question. To find a percentage of an amount without a calculator, build it from the easy building blocks: $10\%$10% (divide by $10$10), $1\%$1% (divide by $100$100), $50\%$50% (halve) and $25\%$25% (quarter). Almost any percentage can be assembled from these — for example $35\% = 25\% + 10\%$35%=25%+10%, or $30\% + 5\%$30%+5%.

Worked Example 3

Work out $35\%$35% of £240.

$10\%$10% of £240 is £24, so $30\%$30% is £72. $5\%$5% is half of $10\%$10%, which is £12. Total $= 72 + 12 = 84$=72+12=84, so £84.

Answer: £84.

With a calculator, multiply by the decimal: $0.35 \times 240 = 84$0.35×240=84.

Worked Example 4

Work out $17.5\%$17.5% of £80.

$10\%$10% of £80 is £8, $5\%$5% is £4, and $2.5\%$2.5% is £2. Total $= 8 + 4 + 2 = 14$=8+4+2=14, so £14.

Answer: £14.

Percentage Increase and Decrease

The efficient method uses a multiplier. For an increase of $r\%$r%, multiply by $1 + \dfrac{r}{100}$1+100r​; for a decrease, multiply by $1 - \dfrac{r}{100}$1−100r​.

Worked Example 5

A salary of £28{,}000 is increased by $4\%$4%. Work out the new salary.

Multiplier $= 1.04$=1.04. New salary $= 28{,}000 \times 1.04 = 29{,}120$=28,000×1.04=29,120, so £29,120.

Answer: £29,120.

Worked Example 6

A coat costing £85 is reduced by $20\%$20% in a sale. Work out the sale price.

Multiplier $= 1 - 0.20 = 0.80$=1−0.20=0.80. Sale price $= 85 \times 0.80 = 68$=85×0.80=68, so £68.

Answer: £68.

OCR Exam Tip: The multiplier method is fastest and least error-prone, especially for several steps. For a $20\%$20% decrease, multiply by $0.8$0.8 in one go rather than finding $20\%$20% and subtracting.

Expressing One Quantity as a Percentage of Another

Write the two quantities as a fraction (the part over the whole), then multiply by $100$100. Make sure both are in the same units.

Worked Example 7

A student scores $48$48 out of $60$60 in a test. Work out the percentage score.

$\dfrac{48}{60} \times 100 = 80\%.$6048​×100=80%.

Answer: $80\%$80%.

Worked Example 8

A shopkeeper buys an item for £40 and sells it for £52. Work out the percentage profit.

Profit $= 52 - 40 = 12$=52−40=12, i.e. £12. As a percentage of the cost:

$\dfrac{12}{40} \times 100 = 30\%.$4012​×100=30%.

Answer: $30\%$30% profit.

Worked Example 8a (matching units)

Express $45$45p as a percentage of £2.

Both quantities must be in the same unit. Convert £2 to $200$200p first. Then:

$\dfrac{45}{200} \times 100 = 22.5\%.$20045​×100=22.5%.

Answer: $22.5\%$22.5%.

Common error: Writing $\dfrac{45}{2} \times 100$245​×100 without converting, which gives the nonsensical $2250\%$2250%. Always match the units before forming the fraction.

Common error: Dividing by the selling price. Percentage profit is always taken as a fraction of the original (cost) price.

Reverse Percentages — Finding the Original

A reverse percentage question gives you the amount after a change and asks for the original. The key idea is that the figure you are given is not $100\%$100% — it is the result of multiplying the original by some multiplier — so you must divide by that multiplier to get back to $100\%$100%.

Worked Example 8b

A price includes $20\%$20% VAT and is £54. Work out the price before VAT.

The £54 is $120\%$120% of the original (the original plus $20\%$20%), so the multiplier is $1.2$1.2. Divide:

$\text{original} = \dfrac{54}{1.2} = 45.$original=1.254​=45.

So the price before VAT is £45. Check: $45 \times 1.2 = 54$45×1.2=54 ✓.

Common error: Finding $20\%$20% of £54 (which is £10.80) and subtracting. That is wrong because the $20\%$20% was added to the original (£45), not to £54; the correct reduction is £9, recovered only by dividing by the multiplier.

OCR Exam Tip: The phrases "before the increase", "original price" or "price before VAT/tax" almost always signal a reverse percentage. Set up "amount $=$= multiplier $\times$× original" and divide.

Extended Worked Examples

Worked Example 9: A two-stage percentage change

A laptop costs £600. Its price is increased by $10\%$10%, then the new price is reduced by $10\%$10%. Work out the final price and explain why it is not £600.

Increase: $600 \times 1.1 = 660$600×1.1=660 (£660). Decrease: $660 \times 0.9 = 594$660×0.9=594 (£594).

Answer: £594. It is not £600 because the $10\%$10% decrease is taken from the larger amount (£660), so the reduction (£66) is bigger than the original increase (£60). Successive percentage changes are not additive.

Worked Example 10: Finding the percentage change

A town's population rises from $24{,}000$24,000 to $25{,}800$25,800. Work out the percentage increase.

Increase $= 25{,}800 - 24{,}000 = 1{,}800$=25,800−24,000=1,800.

$\dfrac{1{,}800}{24{,}000} \times 100 = 7.5\%.$24,0001,800​×100=7.5%.

Answer: $7.5\%$7.5% increase.

Worked Example 11: Combined multiplier

A £45 ticket is increased by $15\%$15% for booking fees, then a $\dfrac{1}{3}$31​ student discount is applied to that total. Work out the price the student pays.

After the fee: $45 \times 1.15 = 51.75$45×1.15=51.75 (£51.75). A $\dfrac{1}{3}$31​ discount means paying $\dfrac{2}{3}$32​: $51.75 \times \dfrac{2}{3} = 34.50$51.75×32​=34.50, so £34.50.

Answer: £34.50.

Worked Example 12: Percentage of a percentage

$60\%$60% of the members of a club are adults. Of these adults, $25\%$25% are over $60$60. What percentage of the whole club are adults over $60$60?

$25\%$25% of $60\%$60% is $0.25 \times 60\% = 15\%$0.25×60%=15%.

Answer: $15\%$15% of the whole club.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: A jacket is reduced by $30\%$30% in a sale to £63. Work out the original price.

Grades 3–4 response: £63 is $70\%$70% of the original. $1\%$1% is $63 \div 70 = 0.9$63÷70=0.9. Original $= 0.9 \times 100 = 90$=0.9×100=90, so £90.

Grades 5–6 response: After a $30\%$30% reduction, the sale price is $70\%$70% of the original, so original $\times 0.7 = 63$×0.7=63. Original $= 63 \div 0.7 = 90$=63÷0.7=90, so £90.