Fractions, Decimals and Percentages · Study Notes · Edexcel IGCSE 4MA1 Higher

How to convert between forms, do arithmetic with fractions, work with percentage multipliers, and solve reverse-percentage problems. The biggest source of marks on Paper 1H Number questions.

At a glance

Convert: fraction $\to$ decimal (divide top by bottom); decimal $\to$ percentage (multiply by $100$ ).
Mixed $\to$ improper FIRST before any arithmetic.
Divide fractions: keep, change, flip.
Multiplier for percentage change: increase $\to \times (1 + p/100)$ ; decrease $\to \times (1 - p/100)$ .
Compound changes — multiply multipliers; never add percentages.
Reverse: original $=$ final $\div$ multiplier.
Recurring decimal $\to$ fraction: let $x =$ decimal, multiply by $10^n$ , subtract.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.4 — Convert between fractions, decimals and percentages.
1.4 — Order fractions, decimals and percentages.
1.4 — Identify recurring and terminating decimals.
1.4 — Convert recurring decimals to fractions.
1.5 — Use percentages, including percentage change, increase and decrease.
1.5 — Solve reverse percentage problems.
1.5 — Apply compound interest and growth/decay.

Converting between fractions, decimals and percentages

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Three forms of the same number. Always be ready to convert.

Fraction → Decimal. Divide numerator by denominator (long division or calculator).

$\dfrac{3}{8} = 3 \div 8 = 0.375$ .
$\dfrac{5}{6} = 5 \div 6 = 0.8\overline{3}$ (recurring).

Decimal → Percentage. Multiply by $100$ .

$0.375 \to 37.5\%$ .
$0.8\overline{3} \to 83.\overline{3}\%$ .

Percentage → Decimal. Divide by $100$ .

$25\% = 0.25$ .
$7.5\% = 0.075$ .

Percentage → Fraction. Write as $\dfrac{p}{100}$ and simplify.

$25\% = \dfrac{25}{100} = \dfrac{1}{4}$ .
$12.5\% = \dfrac{12.5}{100} = \dfrac{1}{8}$ .

Decimal → Fraction. Write digits over an appropriate power of $10$ , then simplify.

$0.625 = \dfrac{625}{1000} = \dfrac{5}{8}$ .

A fraction, its decimal and its percentage are the same shaded amount written three ways.

Common conversions to memorise:

Fraction	Decimal	Percentage
$\dfrac{1}{2}$	$0.5$	$50\%$
$\dfrac{1}{3}$	$0.\overline{3}$	$33.\overline{3}\%$
$\dfrac{1}{4}$	$0.25$	$25\%$
$\dfrac{1}{5}$	$0.2$	$20\%$
$\dfrac{1}{8}$	$0.125$	$12.5\%$
$\dfrac{1}{10}$	$0.1$	$10\%$
$\dfrac{2}{3}$	$0.\overline{6}$	$66.\overline{6}\%$
$\dfrac{3}{4}$	$0.75$	$75\%$

Worked qualitative. Order $\dfrac{3}{8}$ , $0.4$ , $35\%$ , $\dfrac{2}{5}$ from smallest to largest.

All as decimals: $\dfrac{3}{8} = 0.375$ , $0.4 = 0.4$ , $35\% = 0.35$ , $\dfrac{2}{5} = 0.4$ .
$0.35 < 0.375 < 0.4 = 0.4$ .
Order: $35\%, \dfrac{3}{8}, 0.4 = \dfrac{2}{5}$ .

Edexcel tip. Always convert to ONE common form (decimals are easiest) before comparing or ordering.

Fraction $\to$ decimal: divide top by bottom.
Decimal $\to$ percentage: $\times 100$ .
Common ones: $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{3}$ .
To compare: convert all to decimals.

Fraction arithmetic — adding, subtracting, multiplying, dividing

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Convert mixed to improper FIRST. Common denominator for + and -. Keep-change-flip for division.

Rule 0. Convert mixed numbers to IMPROPER FRACTIONS before any operation.

$2\frac{1}{4} = \frac{9}{4}$ , $1\frac{1}{2} = \frac{3}{2}$ .

Addition and subtraction. Need a common denominator. Find LCM of denominators.

$\dfrac{1}{4} + \dfrac{1}{6}$ :

LCM of $4$ and $6$ is $12$ .
$\dfrac{1}{4} = \dfrac{3}{12}$ and $\dfrac{1}{6} = \dfrac{2}{12}$ .
$\dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}$ .

Multiplication. Multiply numerators; multiply denominators. Cancel common factors first if possible.

$\dfrac{2}{3} \times \dfrac{9}{8} = \dfrac{2 \times 9}{3 \times 8} = \dfrac{18}{24} = \dfrac{3}{4}$ .

(Or cancel: $\dfrac{2}{3} \times \dfrac{9}{8} = \dfrac{1}{1} \times \dfrac{3}{4} = \dfrac{3}{4}$ .)

Division — keep, change, flip.

$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ .

$\dfrac{9}{4} \div \dfrac{3}{2} = \dfrac{9}{4} \times \dfrac{2}{3} = \dfrac{18}{12} = \dfrac{3}{2} = 1\dfrac{1}{2}$ .

Final answer.

If question asks for a mixed number, convert improper $\to$ mixed at the end.
Always simplify to lowest terms.

Worked qualitative. Calculate $3\dfrac{1}{2} - 1\dfrac{2}{3}$ .

Convert: $\dfrac{7}{2} - \dfrac{5}{3}$ .
LCM of $2, 3$ is $6$ : $\dfrac{7}{2} = \dfrac{21}{6}$ , $\dfrac{5}{3} = \dfrac{10}{6}$ .
Subtract: $\dfrac{21 - 10}{6} = \dfrac{11}{6} = 1\dfrac{5}{6}$ .

Edexcel tip. Mark schemes have separate marks for: (1) converting to improper, (2) common denominator (or keep-change-flip for division), (3) final answer. Show every step.

Mixed $\to$ improper FIRST.
- and -: common denominator (LCM).
× : multiply tops and bottoms; cancel.
÷ : keep, change, flip.
Simplify final answer.

Percentages — multipliers and compound changes

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Multipliers are the cleanest way. Compound = multiply multipliers, never add percentages.

The multiplier method is the most reliable approach for ALL percentage problems.

Increase by $p\%$ : multiplier $= 1 + \dfrac{p}{100}$ .

Increase by $8\%$ : $\times 1.08$ .
Increase by $25\%$ : $\times 1.25$ .

Decrease by $p\%$ : multiplier $= 1 - \dfrac{p}{100}$ .

Decrease by $15\%$ : $\times 0.85$ .
Decrease by $4\%$ : $\times 0.96$ .

Single percentage change.

New value = Original × multiplier

Example: £ $650$ increased by $8\%$ : $650 \times 1.08 = £702$ .

Compound percentage changes.

Apply multipliers in sequence — DON'T add percentages.

Example: £ $240{,}000$ rises $5\%$ then falls $3\%$ : $240{,}000 \times 1.05 \times 0.97 = £244{,}440$ .

Each multiplier acts on the running total — the second percentage is taken from a new base.

That's NOT a $2\%$ increase ( $240{,}000 \times 1.02 = £244{,}800$ ). The actual change is $1.05 \times 0.97 - 1 = 0.0185 = 1.85\%$ .

Compound interest. Same idea applied repeatedly.

Value after $n$ years $= P \times \left(1 + \dfrac{r}{100}\right)^{n}$

where $P$ is the principal and $r\%$ is the annual rate.

Example: £ $5{,}000$ at $4\%$ compound interest for $3$ years: $5{,}000 \times 1.04^{3} = 5{,}000 \times 1.124864 = £5{,}624.32$ .

Percentage change formula.

When given before/after and need the change:

$\%\,\text{change} = \dfrac{\text{New} - \text{Original}}{\text{Original}} \times 100$

Example: from $40$ to $50$ : $\dfrac{50 - 40}{40} \times 100 = 25\%$ .

Worked qualitative. Why is $20\%$ off followed by $20\%$ on NOT the same as the original?

Original: £ $100$ .
After $20\%$ off: $100 \times 0.8 = £80$ .
After $20\%$ on: $80 \times 1.2 = £96$ .
$\ne £100$ — it's £ $4$ less.
Reason: the second $20\%$ is calculated from a smaller base.

Edexcel tip. When questions show successive percentages, always use multipliers. Adding/subtracting raw percentages is the most common error in compound problems.

Multiplier $= 1 \pm p/100$ .
Compound: multiply multipliers in sequence.
Compound interest: $P(1 + r/100)^n$ .
Adding percentages is wrong for compound.

Reverse percentage problems

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Given the FINAL value after a change. Find the ORIGINAL. Always: final = multiplier × original.

The trap: when given a price AFTER a discount, students compute the discount on the FINAL price. Wrong.

The setup:

Final value = multiplier × Original

So:

Original = $\dfrac{\text{Final}}{\text{multiplier}}$

To find the original, divide the final value by the multiplier — never take the percentage off the final price.

Example 1. After $15\%$ off, price is £ $170$ . Find the original.

$15\%$ off means $85\%$ remains. Multiplier $= 0.85$ .
Original $= 170 / 0.85 = £200$ .

Example 2. After $25\%$ VAT added, price is £ $300$ . Find the original (pre-VAT).

$25\%$ added means $\times 1.25$ .
Original $= 300 / 1.25 = £240$ .

Example 3. A house's value INCREASED by $20\%$ over 2 years. It's now £ $360{,}000$ . What was it 2 years ago?

Multiplier $= 1.2$ .
Original $= 360{,}000 / 1.2 = £300{,}000$ .

Why it's hard: the natural instinct is to apply the percentage to the value you're given. Always read carefully — is the percentage of the ORIGINAL or the FINAL?

Worked qualitative. A coat is on sale at £ $180$ after a $10\%$ discount. The shopkeeper claims this is a 'great saving'. What was the saving in £?

Original: $180 / 0.9 = £200$ .
Saving: $200 - 180 = £20$ .
That's $10\%$ of £ $200$ , NOT $10\%$ of £ $180$ (which would be £ $18$ ).

Edexcel tip. Always set up the equation: final = multiplier × original. Then divide. Showing this earns the M1 mark even if your arithmetic slips.

Final = multiplier × Original.
Original = Final / multiplier.
$15\%$ off means $\times 0.85$ .
$25\%$ on means $\times 1.25$ .
Always set up the equation explicitly.

Recurring decimals and converting to fractions

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Use the algebraic technique: $x$ , $10^n x$ , subtract.

Notation:

$0.\overline{3} = 0.333\ldots$ (the $3$ recurs).
$0.\overline{27} = 0.272727\ldots$ (the block " $27$ " recurs).
$0.1\overline{6} = 0.16666\ldots$ (only $6$ recurs after the $1$ ).

Single-digit recurring.

Let $x = 0.\overline{3}$ . Then $10x = 3.\overline{3}$ .

Subtract: $10x - x = 3.\overline{3} - 0.\overline{3} = 3$ .

So $9x = 3 \implies x = \dfrac{3}{9} = \dfrac{1}{3}$ .

Two-digit recurring.

Let $x = 0.\overline{27}$ . Then $100x = 27.\overline{27}$ .

Subtract: $100x - x = 27 \implies 99x = 27 \implies x = \dfrac{27}{99} = \dfrac{3}{11}$ .

Mixed (some non-recurring, then recurring).

$0.16\overline{6}$ :

Let $x = 0.16\overline{6}$ .
$10x = 1.6\overline{6}$ (shifts past the non-recurring part).
$100x = 16.\overline{6}$ (shifts past one recurring digit).
$100x - 10x = 16.\overline{6} - 1.6\overline{6} = 15$ .
$90x = 15 \implies x = \dfrac{15}{90} = \dfrac{1}{6}$ .

Pattern. Number of $9$ s in denominator (before simplifying) = length of recurring block. Number of $0$ s = length of non-recurring part after the decimal point.

Worked qualitative. Convert $0.\overline{142857}$ to a fraction.

6 recurring digits → multiply by $10^{6} = 1{,}000{,}000$ .
$1{,}000{,}000x - x = 142857$ .
$999{,}999 x = 142{,}857$ .
$x = \dfrac{142{,}857}{999{,}999} = \dfrac{1}{7}$ (after simplifying).

Edexcel tip. Mark scheme awards M1 for setting up $x$ and $10^n x$ , M1 for subtracting, A1 for the simplified fraction. Always show the algebra.

Let $x =$ decimal.
Multiply by $10^n$ ( $n$ = recurring block length).
Subtract; solve for $x$ .
Simplify to lowest terms.
$n$ recurring digits $\to n$ nines in denominator (pre-simplification).

Quick recap

Convert via decimal as common ground.
Mixed $\to$ improper FIRST for arithmetic.
Divide fractions: keep, change, flip.
Multipliers for percentages: $1 + p/100$ or $1 - p/100$ .
Compound: multiply multipliers; never add.
Reverse: original $=$ final $\div$ multiplier.
Recurring $\to$ fraction: $x$ , $10^n x$ , subtract.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Multiplier: increase $\to 1 + p/100$ ; decrease $\to 1 - p/100$ .
Reverse percentage: original $=$ final $\div$ multiplier.
Compound interest: $P(1 + r/100)^n$ .
Recurring: $x$ , $10^n x$ , subtract.
Common conversions: $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{8}, \frac{2}{3}$ .

How it’s examined

FDP appears every Paper 1H AND Paper 2H. Typical: percentage change/multiplier (3-5 marks), reverse percentage (3-4 marks), recurring decimal proof (3 marks Paper 1H), fraction arithmetic (3-4 marks). Examiner reports flag (1) using addition for compound percentages, (2) applying the percentage to the wrong base in reverse problems, and (3) not converting mixed to improper first.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H May/Jun 2023, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on fractions, decimals and percentages , written exactly the way a tutor would explain them at the board.

1Convert a recurring decimal to a fraction
Higher• Adapted from 4MA1/1H May/Jun 2024 Q14• recurring decimal, fraction
▼
Question
Show that $0.\overline{27} = \dfrac{3}{11}$ .
Step-by-step solution
1. Step 1
  Let $x = 0.\overline{27} = 0.272727\ldots$
2. Step 2
  Two recurring digits, so multiply by $100$ :
  $100x = 27.272727\ldots$
3. Step 3
  Subtract the first equation from the second:
  $100x - x = 27.272727\ldots - 0.272727\ldots = 27$
4. Step 4
  Simplify:
  $99x = 27 \implies x = \frac{27}{99} = \frac{3}{11}$
Answer
$0.\overline{27} = \dfrac{3}{11}$
Examiner tip
Edexcel awards M1 for setting up $x$ and $100x$ correctly, M1 for the subtraction, A1 for the answer. Always show the algebra — bare answers score 0 or 1.
2Percentage increase and decrease
Foundation• percentage change
▼
Question
A laptop costs £ $650$ . Its price is INCREASED by $8\%$ . What is the new price?
Step-by-step solution
1. Step 1
  Method 1 (multiplier). $8\%$ increase means multiplying by $1.08$ :
  $650 \times 1.08 = 702$
2. Step 2
  Method 2 (find then add). $8\%$ of $650 = 0.08 \times 650 = 52$ . Then $650 + 52 = 702$ .
Answer
£ $702$
Examiner tip
The multiplier method is the most marker-friendly approach. Increase $\to$ multiply by $1 + \frac{p}{100}$ . Decrease $\to$ multiply by $1 - \frac{p}{100}$ .
3Reverse percentage problem
Higher• Adapted from 4MA1/2H May/Jun 2023 Q9• reverse percentage
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Question
After a $15\%$ discount, a coat costs £ $170$ . Find the ORIGINAL price.
Step-by-step solution
1. Step 1
  $£170$ represents $100\% - 15\% = 85\%$ of the original.
2. Step 2
  $85\% = 0.85$ . Set up:
  $0.85 \times \text{original} = 170$
3. Step 3
  Divide:
  $\text{original} = \frac{170}{0.85} = 200$
Answer
Original price = £ $200$
Examiner tip
The classic mistake: students compute $170 + 15\% = 195.50$ . But $15\%$ of WHAT? It must be $15\%$ of the ORIGINAL, not the discounted. Always solve via 'final = multiplier × original'.
4Mixed-fraction arithmetic with division
Foundation• fraction arithmetic
▼
Question
Calculate $2\dfrac{1}{4} \div 1\dfrac{1}{2}$ . Give your answer as a mixed number in its simplest form.
Step-by-step solution
1. Step 1
  Convert to improper fractions: $2\frac{1}{4} = \frac{9}{4}$ and $1\frac{1}{2} = \frac{3}{2}$ .
2. Step 2
  Division: keep, change, flip:
  $\frac{9}{4} \div \frac{3}{2} = \frac{9}{4} \times \frac{2}{3}$
3. Step 3
  Cancel common factors: $\dfrac{9}{4} \times \dfrac{2}{3} = \dfrac{9 \times 2}{4 \times 3} = \dfrac{18}{12} = \dfrac{3}{2}$ .
4. Step 4
  Convert back to mixed: $\dfrac{3}{2} = 1\dfrac{1}{2}$ .
Answer
$1\dfrac{1}{2}$
Examiner tip
Edexcel mark schemes credit conversion to improper fractions (M1) and the keep-change-flip step (M1). Working with mixed numbers throughout almost always loses marks.
5Compound percentage problem
Higher• Adapted from 4MA1/2H Jan 2024 Q12• compound interest, percentage
▼
Question
A house is worth £ $240{,}000$ . Its value rises by $5\%$ in year 1, then falls by $3\%$ in year 2. What is its value at the end of year 2? Give your answer to the nearest pound.
Step-by-step solution
1. Step 1
  Apply both multipliers in sequence:
  $\text{Value} = 240{,}000 \times 1.05 \times 0.97$
2. Step 2
  Compute: $1.05 \times 0.97 = 1.0185$ .
3. Step 3
  Final: $240{,}000 \times 1.0185 = 244{,}440$ .
Answer
£ $244{,}440$
Examiner tip
Always multiply, never add the percentages. $+5\%$ then $-3\%$ is NOT the same as $+2\%$ — it's $\times 1.0185$ . Edexcel reports note 30%+ of candidates use the wrong logic here.
6Compare fractions, decimals and percentages
Foundation• conversion, ordering
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Question
Order these from smallest to largest: $\dfrac{3}{8},\ 0.4,\ 35\%,\ \dfrac{2}{5}$ .
Step-by-step solution
1. Step 1
  Convert all to decimals (the easiest common form):
2. Step 2
  $\dfrac{3}{8} = 0.375$ , $0.4 = 0.4$ , $35\% = 0.35$ , $\dfrac{2}{5} = 0.4$ .
3. Step 3
  Order: $0.35 < 0.375 < 0.4 = 0.4$ . So smallest to largest: $35\%, \dfrac{3}{8}, 0.4 = \dfrac{2}{5}$ .
Answer
$35\%,\ \dfrac{3}{8},\ 0.4 = \dfrac{2}{5}$
Examiner tip
Use decimals as the common ground for comparison. Don't try to use a common denominator with mixed types — it's slower and more error-prone.
7Converting a fraction to a terminating decimal [2 marks]
Getting started• Paper 1H/2H style — 2 marks• terminating decimals, conversion
▼
Question
Write $\dfrac{13}{16}$ as a decimal.

Show your working clearly. [2]
Step-by-step solution
1. Step 1
  M1 $16 = 2^{4}$ , and $16 \times 625 = 10\,000$ , so the fraction can be written over a power of ten.
  $\frac{13}{16} = \frac{13 \times 625}{16 \times 625} = \frac{8125}{10\,000}$
2. Step 2
  A1 Four decimal places for ten-thousandths.
  $\frac{8125}{10\,000} = 0.8125$
Answer
$0.8125$
Examiner tip
Examiner tip: A written short division, $13.0000 \div 16$ , is an equally acceptable method. What matters is that the working is on the page — with "Show your working clearly", the answer $0.8125$ alone scores 0.

Common mistake: Dividing the wrong way, $16 \div 13 = 1.2307\ldots$ — a fraction less than $1$ must give a decimal less than $1$ . Writing only the calculator value with no working also scores 0 here.
8Fraction and decimal parts of a quantity [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• fraction of a quantity, decimals
▼
Question
A bag contains $360$ marbles. The marbles are red, blue or green.

$\dfrac{5}{12}$ of the marbles are red. $0.35$ of the marbles are blue.

Work out the number of green marbles. [3]
Step-by-step solution
1. Step 1
  M1 Red marbles: divide by the denominator, multiply by the numerator.
  $\frac{5}{12} \times 360 = 5 \times 30 = 150$
2. Step 2
  M1 Blue marbles.
  $0.35 \times 360 = 126$
3. Step 3
  A1 Green marbles are the rest.
  $360 - 150 - 126 = 84$
Answer
$84$
Examiner tip
Examiner tip: An alternative is to find the green fraction first: $1 - \frac{5}{12} - \frac{7}{20} = \frac{60}{60} - \frac{25}{60} - \frac{21}{60} = \frac{14}{60} = \frac{7}{30}$ , then $\frac{7}{30} \times 360 = 84$ .

Common mistake: Stopping at $150$ and $126$ , or giving the answer as the fraction $\frac{7}{30}$ . The question asks for the NUMBER of green marbles.
9Adding, then dividing, mixed numbers [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• mixed numbers, fraction arithmetic
▼
Question
Work out $\left(1\dfrac{3}{4} + 2\dfrac{2}{3}\right) \div 2\dfrac{1}{2}$

Give your answer as a mixed number in its simplest form. Show your working clearly. [4]
Step-by-step solution
1. Step 1
  M1 Write every mixed number as an improper fraction.
  $1\tfrac{3}{4} = \frac{7}{4}, \quad 2\tfrac{2}{3} = \frac{8}{3}, \quad 2\tfrac{1}{2} = \frac{5}{2}$
2. Step 2
  M1 Brackets first: add using the common denominator $12$ .
  $\frac{7}{4} + \frac{8}{3} = \frac{21}{12} + \frac{32}{12} = \frac{53}{12}$
3. Step 3
  M1 Divide by $\frac{5}{2}$ by multiplying by its reciprocal.
  $\frac{53}{12} \times \frac{2}{5} = \frac{106}{60} = \frac{53}{30}$
4. Step 4
  A1 Convert to a mixed number.
  $\frac{53}{30} = 1\frac{23}{30}$
Answer
$1\dfrac{23}{30}$
Examiner tip
Examiner tip: Cancel before multiplying where you can — the $2$ on top and the $12$ below share a factor $2$ , giving $\frac{53}{6} \times \frac{1}{5} = \frac{53}{30}$ directly.

Common mistake: Dividing the whole-number parts and the fraction parts separately, e.g. $4 \div 2 = 2$ and then working with $\frac{5}{12} \div \frac{1}{2}$ . Mixed numbers can only be multiplied or divided once they are improper fractions.
10Reverse percentage from the increase, then percentage change [5 marks]
Stretch• Paper 1H/2H style — 5 marks• reverse percentage, percentage change
▼
Question
The number of visitors to a museum in 2025 was $18\%$ more than the number in 2024.

There were $1062$ more visitors in 2025 than in 2024.

(a) Work out the number of visitors in 2024. [2]

In 2026 the number of visitors was $6136$ .

(b) Work out the percentage decrease in the number of visitors from 2025 to 2026. Give your answer correct to $1$ decimal place. [3]
Step-by-step solution
1. Step 1
  M1 (a) The increase of $1062$ is $18\%$ of the 2024 number, so divide by $0.18$ .
  $1062 \div 0.18$
2. Step 2
  A1 Evaluate.
  $1062 \div 0.18 = 5900$
3. Step 3
  M1 (b) Find the 2025 number.
  $5900 + 1062 = 6962 \quad (\text{or } 5900 \times 1.18)$
4. Step 4
  M1 Percentage decrease, measured against 2025 — the year the change starts from.
  $\frac{6962 - 6136}{6962} \times 100 = \frac{826}{6962} \times 100$
5. Step 5
  A1 Evaluate and round only at the end.
  $11.864\ldots \approx 11.9\%$
Answer
(a) $5900$ (b) $11.9\%$
Examiner tip
Examiner tip: In (a) the $1062$ is the INCREASE, not the new total, so it is $18\%$ of the original, not $118\%$ . Read which quantity each number represents before choosing a multiplier.

Common mistake: In (a), $1062 \div 1.18 = 900$ — an answer smaller than the increase itself, which should prompt a check. In (b), dividing by the 2026 value, $\frac{826}{6136} \times 100 = 13.5\%$ , measures the change from the wrong year.

Question 1
Paper 1H/2H style — 1 mark1 mark
Which is larger, $\dfrac{2}{3}$ or $0.67$ ? Give a reason for your answer. [1]
Model answer
$\frac{2}{3} = 0.666\ldots = 0.\dot{6}$ , and $0.666\ldots < 0.670$

Answer: $0.67$ is larger.
Why this scores
Mark scheme:
- B1 for $0.67$ with a correct reason, e.g. $\frac{2}{3} = 0.666\ldots$ which is less than $0.67$
Why this scores full marks: The reason is a comparison of like with like — both numbers as decimals to three places. Writing $\frac{2}{3} \approx 0.67$ and calling them equal is the common error; $0.67$ is only a rounded value of $\frac{2}{3}$ , and a bare " $0.67$ " with no reason scores 0.
Question 2
Paper 1H/2H style — 2 marks2 marks
Write $45$ minutes as a fraction of $2$ hours. Give your fraction in its simplest form. [2]
Model answer
$2$ hours $= 120$ minutes

$\frac{45}{120} = \frac{3}{8}$

Answer: $\dfrac{3}{8}$
Why this scores
Mark scheme:
- M1 for $\frac{45}{120}$ oe, with both quantities in the same units
- A1 for $\frac{3}{8}$ cao
Why this scores full marks: The script converts to one unit before forming the fraction. Scripts that write $\frac{45}{2}$ or $\frac{45}{200}$ compare minutes with hours, or treat an hour as $100$ minutes, and score 0. The fraction is then cancelled fully by the HCF, $15$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
(a) Explain why $12 \div \dfrac{1}{4}$ gives the same answer as $12 \times 4$ [1]

(b) Work out $2.4 \div \dfrac{3}{8}$ [1]
Model answer
(a) $12 \div \frac{1}{4}$ asks how many quarters there are in $12$ . There are $4$ quarters in each whole, so there are $12 \times 4 = 48$ . Dividing by $\frac{1}{4}$ is the same as multiplying by its reciprocal, $4$ .

(b) $2.4 \div \frac{3}{8} = 2.4 \times \frac{8}{3} = \frac{19.2}{3} = 6.4$

Answer: $6.4$
Why this scores
Mark scheme:
- B1 for a correct explanation, e.g. there are $4$ quarters in every whole one, or $\frac{1}{4}$ and $4$ are reciprocals (multiplicative inverses) so dividing by one is multiplying by the other
- B1 for $6.4$ oe
Why this scores full marks: Part (a) gives a reason, not just a restatement — "because it is the rule" scores 0. Part (b) applies the same idea to a non-unit fraction: the reciprocal of $\frac{3}{8}$ is $\frac{8}{3}$ . Weaker scripts drop the division sign and multiply, $2.4 \times \frac{3}{8} = 0.9$ , an answer smaller than $2.4$ , which cannot be right: dividing by a number less than $1$ makes the answer larger.
Question 4
Paper 1H/2H style — 3 marks3 marks
Show that $1\dfrac{5}{6} \times 2\dfrac{2}{11} = 4$

Show your working clearly. [3]
Model answer
$1\frac{5}{6} \times 2\frac{2}{11} = \frac{11}{6} \times \frac{24}{11}$

$= \frac{11 \times 24}{6 \times 11} = \frac{264}{66}$

$= 4$
Why this scores
Mark scheme:
- M1 for both mixed numbers written as improper fractions, $\frac{11}{6}$ and $\frac{24}{11}$
- M1 for $\frac{11 \times 24}{6 \times 11}$ oe, or correct cancelling, e.g. $\frac{1}{6} \times \frac{24}{1}$
- A1 for completing to $4$ with correct working throughout
Why this scores full marks: Every stage from the mixed numbers to $4$ is written down, which is all a "show that" rewards — the $4$ is printed, so it earns nothing alone. Multiplying the whole numbers and the fractions separately, $1 \times 2 + \frac{5}{6} \times \frac{2}{11}$ , is the error weaker scripts make.
Question 5
Paper 1H/2H style — 3 marks3 marks
A gym membership costs £ $42$ per month. The monthly cost increases by $7.5\%$ .

Work out the total cost of $12$ months of membership at the new monthly price. [3]
Model answer
$42 \times 1.075 = 45.15$

$45.15 \times 12 = 541.80$

Answer: £ $541.80$
Why this scores
Mark scheme:
- M1 for $42 \times 1.075$ oe, or $42 + 0.075 \times 42$
- M1 for their $45.15 \times 12$
- A1 for $541.80$
Why this scores full marks: The multiplier $1.075$ is written for a $7.5\%$ increase — not $1.75$ or $1.0075$ , the two slips seen most often. The new monthly price is found first and only then multiplied by $12$ , and the money answer is written to two decimal places, £ $541.80$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
The price of a bicycle is £ $378$ . This price includes VAT at $20\%$ .

Work out the price of the bicycle before VAT was added. [3]
Model answer
Price with VAT $= 120\%$ of the price before VAT.

$378 \div 1.2 = 315$

Check: $315 \times 1.2 = 378$

Answer: £ $315$
Why this scores
Mark scheme:
- M1 for recognising £ $378$ as $120\%$ , e.g. $1.2$ or $\frac{120}{100}$ used
- M1 for $378 \div 1.2$ oe
- A1 for $315$
Why this scores full marks: The script names the percentage the given price represents, then divides by the multiplier and checks forwards. Taking $20\%$ off the price with VAT, $378 \times 0.8 = 302.40$ , is the reverse-percentage error the examiner report highlights, and scores 0.
Question 7
Paper 1H/2H style — 4 marks4 marks
Owen buys $40$ T-shirts for a total of £ $240$ .

He sells $28$ of the T-shirts for £ $9.50$ each and the rest for £ $5$ each.

Calculate his percentage profit. Give your answer correct to $1$ decimal place. [4]
Model answer
$28 \times 9.50 = 266$

$40 - 28 = 12$ , and $12 \times 5 = 60$

Total takings $= 266 + 60 = 326$

Profit $= 326 - 240 = 86$

$\frac{86}{240} \times 100 = 35.833\ldots$

Answer: $35.8\%$
Why this scores
Mark scheme:
- M1 for $28 \times 9.50$ or $12 \times 5$
- M1 for $326 - 240$ oe (profit $86$ )
- M1 for $\frac{86}{240} \times 100$ oe, or $\frac{326}{240} \times 100 - 100$
- A1 for awrt $35.8$
Why this scores full marks: The profit is divided by the amount Owen PAID, $240$ — the value the change starts from. Dividing by the takings, $\frac{86}{326} \times 100 = 26.4\%$ , loses the last two marks. The full display $35.833\ldots$ is written before rounding, so the rounding can be seen to be correct.
Question 8
Paper 1H/2H style — 4 marks4 marks
The price of a television is reduced from £ $480$ to £ $408$

(a) Work out the percentage decrease. [2]

Later, the price goes back up from £ $408$ to £ $480$ . Sam says, "The price has gone up by $15\%$ ."

(b) Is Sam correct? Show working to support your answer. [2]
Model answer
(a) Decrease $= 480 - 408 = 72$

$\frac{72}{480} \times 100 = 15$

Answer: $15\%$

(b) The increase is still £ $72$ , but it is now a percentage of £ $408$ .

$\frac{72}{408} \times 100 = 17.64\ldots$

Sam is not correct. The increase is $17.6\%$ (to 1 d.p.), because the $15\%$ was of the larger amount, £ $480$ .
Why this scores
Mark scheme:
- M1 for $\frac{72}{480} \times 100$ oe, or $\frac{408}{480} = 0.85$
- A1 for $15$
- M1 for $\frac{72}{408} \times 100$ oe, or $\frac{480}{408}$
- A1 for "No" with awrt $17.6$
Why this scores full marks: The script states that a percentage change is measured from the value the change starts at, so the same £ $72$ is a larger percentage of the smaller price. "No" on its own, or a decision based on $15\%$ of $408 = 61.20$ without comparing it to $72$ , is not a supported conclusion.
Question 9
Paper 1H/2H style — 5 marks5 marks
(a) Prove algebraically that the recurring decimal $0.1\dot{2}\dot{3}$ can be written as $\dfrac{61}{495}$ [3]

(b) Using algebra, write $0.8\dot{7}\dot{6}$ as a fraction and hence show that $0.1\dot{2}\dot{3} + 0.8\dot{7}\dot{6} = 1$ [2]
Model answer
(a) Let $x = 0.12323\ldots$

$10x = 1.2323\ldots$

$1000x = 123.2323\ldots$

$990x = 122$

$x = \frac{122}{990} = \frac{61}{495}$

(b) Let $y = 0.87676\ldots$

$10y = 8.7676\ldots, \quad 1000y = 876.7676\ldots$

$990y = 868 \quad\Rightarrow\quad y = \frac{868}{990} = \frac{434}{495}$

$\frac{61}{495} + \frac{434}{495} = \frac{495}{495} = 1$
Why this scores
Mark scheme:
- M1 for two recurring decimals whose difference is a terminating decimal, e.g. $10x = 1.2323\ldots$ and $1000x = 123.2323\ldots$
- M1 for $990x = 122$ oe, or $\frac{122}{990}$
- A1 for completing the proof to $\frac{61}{495}$ with no errors
- M1 for $990y = 868$ oe, or $\frac{868}{990}$
- A1 for $\frac{434}{495}$ oe and a correct sum equal to $1$
Why this scores full marks: Only the $23$ recurs, so the script uses $10x$ and $1000x$ , whose decimal parts are both $.2323\ldots$ and the subtraction gives the whole number $122$ . ( $100x - x = 12.2$ also terminates and is credited, but leaves a decimal to clear.) Part (b) uses algebra as instructed; adding the decimals column by column to get $0.999\ldots$ is not the method asked for.
Question 10
Paper 1H/2H style — 5 marks5 marks
At a concert, $\dfrac{5}{8}$ of the audience are adults and the rest are children.

$36\%$ of the adults are men. The rest of the adults are women.

There are $135$ men in the audience.

(a) Work out the number of children in the audience. [3]

(b) Work out the percentage of the whole audience who are women. [2]
Model answer
(a) Adults: $135 \div 0.36 = 375$

Whole audience: $375 \div \frac{5}{8} = 375 \times \frac{8}{5} = 600$

Children: $600 - 375 = 225$

Answer: $225$

(b) Women $= 375 - 135 = 240$

$\frac{240}{600} \times 100 = 40$

Answer: $40\%$
Why this scores
Mark scheme:
- M1 for $135 \div 0.36$ oe (adults $375$ )
- M1 for their $375 \div \frac{5}{8}$ oe, or their $375 \times \frac{3}{5}$
- A1 for $225$
- M1 for $\frac{375 - 135}{600} \times 100$ oe (ft their $375$ and $600$ )
- A1 for $40$
Why this scores full marks: Each division is backed by a statement of what it finds. The $36\%$ is of the ADULTS, not of the whole audience, so $135 \div 0.36$ gives adults; weaker scripts divide by $0.36$ and treat $375$ as the whole audience. In (b) the women are divided by the whole audience, $600$ , as the question asks — not by the $375$ adults, which gives $64\%$ .

4MA1 Fractions, Decimals and Percentages

At a glance

What you’ll learn

Converting between fractions, decimals and percentages

Fraction arithmetic — adding, subtracting, multiplying, dividing

Percentages — multipliers and compound changes

Reverse percentage problems

Recurring decimals and converting to fractions

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Convert a recurring decimal to a fraction

2Percentage increase and decrease

3Reverse percentage problem

4Mixed-fraction arithmetic with division

5Compound percentage problem

6Compare fractions, decimals and percentages

7Converting a fraction to a terminating decimal [2 marks]

8Fraction and decimal parts of a quantity [3 marks]

9Adding, then dividing, mixed numbers [4 marks]

10Reverse percentage from the increase, then percentage change [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10