OCR GCSE Maths: Number, Ratio & Proportion Revision Guide (J560)

Number is the bedrock of GCSE Mathematics. Whether you are working towards a grade 4 pass or pushing for a grade 9, your number skills decide how reliably you can earn marks everywhere else on the paper. A shaky grasp of fractions, percentages or rounding does not just cost you marks on number questions -- it leaks into algebra, geometry, statistics and probability too, because every one of those topics asks you to calculate with confidence.

This guide covers the Number strand and the closely related Ratio, Proportion and Rates of Change strand of the OCR GCSE Mathematics (J560) specification. Between them, these two areas account for a very large share of the available marks across the three papers, and -- crucially -- much of the content is accessible, practical and highly learnable. The methods are consistent, the question types are predictable once you have practised enough of them, and the marks are genuinely there for the taking.

We will work through the key sub-topics in turn, flag the Higher-tier-only material with a [H] marker, walk through worked examples in full prose, point out the errors examiners see year after year, and explain how OCR likes to phrase its questions. If you would like the bigger picture across the whole subject first, start with our OCR GCSE Maths revision guide hub, then come back here to drill into Number and Ratio.

How the OCR J560 Papers Work

Before we dive into content, it helps to understand the exam structure, because it shapes how you should revise -- and in one important respect, OCR differs from some other boards.

OCR GCSE Mathematics is assessed by three papers per tier. Each paper is worth 100 marks and lasts 1 hour 30 minutes, giving a total of 300 marks across the qualification. There are two tiers of entry:

Tier	Grades available	Papers
Foundation	1 to 5	Paper 1, Paper 2, Paper 3
Higher	4 to 9 (with an allowed 3)	Paper 4, Paper 5, Paper 6

The detail that catches students out is which paper bans the calculator. On OCR, the non-calculator paper is the middle one of the three: that is Paper 2 at Foundation and Paper 5 at Higher. The two outer papers in each tier (Papers 1 and 3 at Foundation; Papers 4 and 6 at Higher) both allow a calculator. So you cannot simply assume "the first paper is the hard arithmetic one" -- on OCR it sits in the middle.

The assessment objectives are weighted differently by tier as well:

Assessment objective	What it tests	Foundation	Higher
AO1	Use and apply standard techniques	50%	40%
AO2	Reason, interpret and communicate mathematically	25%	30%
AO3	Solve problems within mathematics and in context	25%	30%

The practical message is that Higher tier asks you to do proportionally more reasoning and problem-solving -- the AO2 and AO3 strands -- and proportionally less plug-and-chug. Number and ratio questions are a favourite vehicle for AO3 problem-solving, because they map so naturally onto real contexts: money, recipes, maps, speeds and growth.

OCR uses a consistent set of command words, and learning to read them precisely is worth easy marks. "Work out" and "Calculate" both ask for a numerical answer with working shown. "Show that" gives you the target answer and asks you to demonstrate the route to it. "Estimate" expects rounded values and an approximate result. "Describe" and "Compare" want written interpretation. "Give a reason for your answer" tells you that a bare number will not score full marks -- you must justify it.

For paper-by-paper tactics, see our OCR GCSE Maths exam technique guide. For now, keep one rule in mind: because the non-calculator paper is the middle paper, your mental and written arithmetic has to be rock solid.

The Number Strand

Number covers everything from fundamental arithmetic to Higher-tier work with surds and bounds. Let us take the key areas one at a time.

Place Value, Ordering and Rounding

You must be able to order integers, decimals and negative numbers, and to read and write the value of any digit according to its position. Rounding then appears constantly, in two forms: rounding to a given number of decimal places and rounding to a given number of significant figures.

The rule is the same for both. Look at the digit immediately after the place you are rounding to. If it is 5 or more, round up; if it is 4 or less, round down. The most common error is miscounting significant figures when a number starts with zeros. In $0.00374$0.00374, the first significant figure is the 3, not a leading zero -- so rounded to 2 significant figures this is $0.0037$0.0037. Leading zeros are placeholders; trailing zeros after a decimal point can be significant.

OCR also expects you to estimate by rounding each value to 1 significant figure and then calculating. A question might say: "Estimate the value of $\frac{49.2 \times 6.1}{0.48}$0.4849.2×6.1​." Round to $\frac{50 \times 6}{0.5} = \frac{300}{0.5} = 600$0.550×6​=0.5300​=600. Always show the rounded values before you combine them -- that is where the method marks live, and the word "Estimate" is a clear signal that an exact answer is neither needed nor rewarded.

The Four Operations

You need complete fluency with addition, subtraction, multiplication and division of integers, decimals and negative numbers. On the non-calculator paper this means reliable written methods -- long multiplication, long division (the "bus stop" method), and careful column addition and subtraction.

For negative numbers, hold the sign rules firmly. Two numbers with the same sign multiply or divide to give a positive result; two different signs give a negative result. Subtracting a negative is the same as adding: $7 - (-3) = 7 + 3 = 10$7−(−3)=7+3=10. These rules resurface throughout algebra, so getting them automatic now pays off repeatedly.

Factors, Multiples, Primes, HCF and LCM

You should be able to list factors and multiples, recognise prime numbers, and write any number as a product of its prime factors using a factor tree or repeated division.

From the prime factorisations you can build the highest common factor (HCF) and lowest common multiple (LCM). The HCF multiplies together the prime factors common to both numbers, taking the lowest power of each. The LCM multiplies together every prime factor that appears in either number, taking the highest power of each.

For example, $60 = 2^2 \times 3 \times 5$60=22×3×5 and $90 = 2 \times 3^2 \times 5$90=2×32×5. The shared factors at their lowest powers give $\text{HCF} = 2 \times 3 \times 5 = 30$HCF=2×3×5=30. Taking the highest power of each factor gives $\text{LCM} = 2^2 \times 3^2 \times 5 = 180$LCM=22×32×5=180.

A reliable sense-check stops most mistakes here: the HCF is never larger than either number, and the LCM is never smaller than either number. If your "HCF" is bigger than both, or your "LCM" is smaller than both, you have swapped them.

Powers, Roots and Indices

Memorise the square numbers up to $15^2 = 225$152=225 and the cube numbers up to $5^3 = 125$53=125 -- you cannot reach for a calculator on the middle paper.

The laws of indices apply at both tiers:

• $a^m \times a^n = a^{m+n}$am×an=am+n
• $a^m \div a^n = a^{m-n}$am÷an=am−n
• $(a^m)^n = a^{mn}$(am)n=amn
• $a^0 = 1$a0=1
• $a^{-n} = \frac{1}{a^n}$a−n=an1​

At Higher tier [H] you also need fractional indices: $a^{1/n}$a1/n means the $n$n-th root of $a$a, and $a^{m/n}$am/n means take the $n$n-th root and raise it to the power $m$m. For example, $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$82/3=(38​)2=22=4. A neat exam trick is to do the root first and the power second -- the numbers stay small and manageable.

Standard Form

Standard form writes very large or very small numbers as a value between 1 and 10 multiplied by a power of 10. So $4{,}500{,}000 = 4.5 \times 10^6$4,500,000=4.5×106 and $0.00032 = 3.2 \times 10^{-4}$0.00032=3.2×10−4.

You must convert in both directions and calculate with numbers in standard form. On a calculator paper, enter the power of ten using your calculator's standard-form key (often labelled $\times 10^x$×10x or EXP) -- never type the literal digits "$\times 10$×10", as that is a frequent source of slips. On the non-calculator paper you manipulate the powers by hand: to work out $(3 \times 10^4) \times (2 \times 10^5)$(3×104)×(2×105), multiply the numbers ($3 \times 2 = 6$3×2=6) and add the powers ($4 + 5 = 9$4+5=9) to get $6 \times 10^9$6×109. If the front number comes out as 10 or more -- say $6 \times 8 = 48$6×8=48 giving $48 \times 10^9$48×109 -- rewrite it as $4.8 \times 10^{10}$4.8×1010 so the answer is in proper standard form.

Fractions

Fraction arithmetic is one of the most heavily tested areas on the non-calculator paper. You need all four operations cold:

• Adding and subtracting: find a common denominator, rewrite both fractions, then add or subtract the numerators and simplify.
• Multiplying: multiply numerators together and denominators together, then simplify.
• Dividing: flip the second fraction (its reciprocal) and multiply.

A worked example: to add $\frac{2}{3} + \frac{3}{5}$32​+53​, the common denominator is 15, so $\frac{10}{15} + \frac{9}{15} = \frac{19}{15} = 1\frac{4}{15}$1510​+159​=1519​=1154​. Mixed numbers must become improper fractions before you multiply or divide: $2\frac{1}{3} = \frac{7}{3}$231​=37​.

You should also convert fluently between fractions, decimals and percentages. Knowing the staples by heart -- $\frac{1}{4} = 0.25 = 25\%$41​=0.25=25%, $\frac{1}{3} = 0.\dot{3} = 33.\dot{3}\%$31​=0.3˙=33.3˙%, $\frac{3}{8} = 0.375 = 37.5\%$83​=0.375=37.5% -- saves time right across the paper.

Percentages

Percentages appear on almost every paper and in several guises:

• Percentage of an amount. To find $15\%$15% of 240 without a calculator, build it from tens: $10\%$10% is 24, $5\%$5% is 12, so $15\%$15% is 36. On a calculator, multiply by the decimal: $240 \times 0.15 = 36$240×0.15=36.
• Percentage increase and decrease. To increase 80 by $12\%$12%, multiply by the multiplier $1.12$1.12 to get $89.6$89.6. To decrease by $12\%$12%, multiply by $0.88$0.88. The multiplier method is faster and far less error-prone than finding the part and adding or subtracting it separately.
• Reverse percentages. Suppose a coat costs £69 after a $15\%$15% reduction, and you must find the original price. The sale price represents $85\%$85% of the original, so the original is $69 \div 0.85 = 81.18$69÷0.85=81.18, that is £81.18 to the nearest penny. The classic mistake is to find $15\%$15% of £69 and add it on -- but the $15\%$15% was taken from the original price, not the reduced one, so that gives the wrong answer.

Compound Interest, Growth and Decay

Compound interest, depreciation and population change all use the same engine: repeated percentage change.

$\text{final value} = \text{initial value} \times (\text{multiplier})^n$final value=initial value×(multiplier)n

where $n$n is the number of periods. For $3\%$3% annual growth the multiplier is $1.03$1.03; for $5\%$5% annual depreciation it is $0.95$0.95.

For example, £2000 invested at $4\%$4% compound interest for 3 years grows to $2000 \times 1.04^3 = 2000 \times 1.124864 = 2249.73$2000×1.043=2000×1.124864=2249.73, that is £2249.73. Note that this is more than the £2240 you would get from simple interest ($2000 \times 1.12 = 2240$2000×1.12=2240) -- the difference is the "interest on the interest", which is exactly what makes it compound.

At Higher tier [H] you may meet iterative growth and decay dressed up in less familiar contexts -- the activity of a radioactive sample, the cooling of a drink, the spread of a rumour -- but the underlying calculation is identical: identify the multiplier and raise it to the appropriate power. On the non-calculator paper you might be asked for two or three iterations by hand, so practise multiplying decimals carefully.

Surds [H]

A surd is an irrational root that does not simplify to a whole number, such as $\sqrt{3}$3​ or $\sqrt{50}$50​. You simplify a surd by factoring out the largest perfect square: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$50​=25×2​=52​.

Rationalising the denominator means clearing a surd from the bottom of a fraction. To rationalise $\frac{1}{\sqrt{3}}$3​1​, multiply top and bottom by $\sqrt{3}$3​ to get $\frac{\sqrt{3}}{3}$33​​. For a denominator of the form $a + \sqrt{b}$a+b​, multiply by its conjugate $a - \sqrt{b}$a−b​, which uses the difference of two squares to remove the root.

Surd questions sit in the latter part of Higher papers and reward students who keep answers exact rather than converting to rounded decimals. If a Higher question says "give your answer in the form $a\sqrt{b}$ab​" or "give an exact answer", that is your cue not to touch the calculator's decimal button.

Bounds and Error Intervals [H]

When a measurement is rounded, the true value lies within an interval. If a length is given as 12 cm to the nearest centimetre, the lower bound is 11.5 cm and the upper bound is 12.5 cm. We write the error interval as $11.5 \leq \text{length} < 12.5$11.5≤length<12.5. Notice the strict inequality on the upper bound: 12.5 itself would round up to 13, so it is the limit of the interval rather than a possible value.

When you combine bounds in a calculation, think about which combination produces the extreme you want. For a sum, the upper bound comes from adding the two upper bounds. For a division, the maximum result comes from dividing the largest numerator by the smallest denominator -- so $\frac{\text{upper}}{\text{lower}}$lowerupper​ -- and the minimum from $\frac{\text{lower}}{\text{upper}}$upperlower​. A frequent error is mixing these up and dividing two upper bounds together.

If you want structured practice on every one of these number topics, our OCR GCSE Mathematics: Number course works through each one with examples and questions pitched at both tiers.

Ratio, Proportion and Rates of Change

This strand connects pure number skills to real-world problem-solving. It is heavily examined precisely because it models everyday situations -- and it is a rich source of the AO3 problem-solving marks that weigh more heavily at Higher tier.

Simplifying Ratios and Sharing in a Ratio

Simplify a ratio exactly as you would a fraction: divide every part by the highest common factor. So $12:18$12:18 simplifies to $2:3$2:3. Before you simplify, make sure both quantities are in the same units -- if one is in metres and the other in centimetres, convert first or your ratio will be nonsense.

To share an amount in a given ratio, add the parts to find the total number of shares, divide the amount by that total to find the value of one share, then multiply out. To share £200 in the ratio $3:5$3:5: there are $3 + 5 = 8$3+5=8 shares, one share is $200 \div 8 = 25$200÷8=25 (that is £25), so the split is £75 and £125. A quick check: the two parts should add back to the original total.

More demanding "ratio problem" questions give you a difference or a single part rather than the total. For example: "The ratio of red to blue counters is $5:3$5:3. There are 12 more red counters than blue. How many counters are there altogether?" The difference is $5 - 3 = 2$5−3=2 parts. If 2 parts represent 12 counters, then 1 part is 6, and the total is $8 \times 6 = 48$8×6=48 counters. These are textbook AO3 problems -- the maths is simple, but you have to decode which quantity the numbers attach to.

Ratio Change Problems [H]

Higher tier likes ratios that change. A typical question: "The ratio of sweets Amir has to sweets Beth has is $4:3$4:3. Amir gives Beth 6 sweets, and now the ratio is $1:1$1:1. How many did Amir start with?" Set up the parts algebraically -- start with $4x$4x and $3x$3x -- then form an equation from the new situation: $4x - 6 = 3x + 6$4x−6=3x+6, which solves to $x = 12$x=12, so Amir began with $4 \times 12 = 48$4×12=48 sweets. The skill being tested is translating a changing ratio into an equation, which is why these appear at Higher tier.

Direct and Inverse Proportion

Two quantities are in direct proportion if doubling one doubles the other. The relationship has the form $y = kx$y=kx, where $k$k is the constant of proportionality. Find $k$k by substituting a known pair of values, then use the equation for any other value.

Two quantities are in inverse proportion if doubling one halves the other -- think of the time taken to dig a hole against the number of workers. The relationship is $y = \frac{k}{x}$y=xk​. Again, substitute a known pair to find $k$k, then apply the equation.

At Higher tier [H] the relationship may involve powers or roots: $y$y proportional to $x^2$x2, or $y$y proportional to $\sqrt{x}$x​. The method does not change. For instance, if $y$y is proportional to $x^2$x2 and $y = 18$y=18 when $x = 3$x=3, then $18 = k \times 3^2 = 9k$18=k×32=9k, so $k = 2$k=2 and $y = 2x^2$y=2x2. To find $y$y when $x = 5$x=5, substitute: $y = 2 \times 25 = 50$y=2×25=50.

Compound Measures and Rates of Change

Speed, density and pressure are the three compound measures you must know:

Measure	Formula	Typical units
Speed	$\text{speed} = \dfrac{\text{distance}}{\text{time}}$speed=timedistance​	m/s, km/h
Density	$\text{density} = \dfrac{\text{mass}}{\text{volume}}$density=volumemass​	g/cm³, kg/m³
Pressure	$\text{pressure} = \dfrac{\text{force}}{\text{area}}$pressure=areaforce​	N/m² (pascals)

You need to rearrange each to find any of the three quantities. A "formula triangle" works as a memory aid, but understanding the rearrangement algebraically is far more reliable when a problem has several steps.

A worked example: a car travels 150 km in 2 hours 30 minutes. Convert the time to hours first -- 2 hours 30 minutes is 2.5 hours -- then $\text{speed} = \frac{150}{2.5} = 60$speed=2.5150​=60 km/h. The single most common error here is mismatched units: if you leave the time as "2.30" or work in minutes, your answer collapses. Always make units consistent before you calculate, and sanity-check that the units of your answer make sense.

Gradient and Area as Rates of Change

On a distance-time graph the gradient represents speed. On a velocity-time graph the gradient represents acceleration, and the area under the graph represents distance travelled. More generally, the gradient of a real-life graph tells you the rate at which one quantity changes with respect to another -- the rate water flows into a tank, or temperature rises over time.

At Higher tier [H] you may be asked to estimate the gradient of a curve at a point by drawing a tangent and finding its gradient, or to estimate the area under a curve by counting squares or splitting it into trapezia. These skills point forward to the calculus of A-Level Maths, and OCR uses them to check that you can interpret graphs, not merely plot them.

Unit Conversions

You must convert fluently between metric units of length, mass and capacity (millimetres, centimetres, metres, kilometres; grams and kilograms; millilitres and litres), and handle compound unit conversions such as m/s to km/h. Converting a speed of 20 m/s to km/h is a two-stage job: there are 3600 seconds in an hour and 1000 metres in a kilometre, so $20 \text{ m/s} = 20 \times 3600 \div 1000 = 72$20 m/s=20×3600÷1000=72 km/h. You should also work with conversions of area and volume, where the scale factor is squared or cubed respectively: 1 m² is $100^2 = 10{,}000$1002=10,000 cm², and 1 m³ is $100^3 = 1{,}000{,}000$1003=1,000,000 cm³. Forgetting to square or cube the conversion factor is one of the most common slips on these questions.

Everything in this strand -- from simple sharing to Higher-tier proportion and rates of change -- is covered step by step in our OCR GCSE Mathematics: Ratio & Proportion course.

Non-Calculator Techniques for the Middle Paper

Because Foundation Paper 2 and Higher Paper 5 ban the calculator, strong arithmetic is not optional. The following techniques deserve dedicated practice.

Long multiplication. Use a column or grid method, whichever you find more secure. To multiply $347 \times 26$347×26, split it into $347 \times 20$347×20 and $347 \times 6$347×6, work out each part, and add. Practise until the method is automatic.

Long division. This is the method students avoid most -- and avoidance costs marks, because division appears on the non-calculator paper. Use the bus-stop method: to divide $952 \div 14$952÷14, ask how many 14s fit into 95 (six, with 11 left over), bring down the 2 to make 112, then ask how many 14s fit into 112 (eight). The answer is 68.

Fraction arithmetic by hand. Adding fractions with different denominators is a core skill. Build the habit of finding the common denominator first, every time, rather than guessing.

Percentages from 10%. Build awkward percentages from a $10\%$10% block. To find $37.5\%$37.5% of 160: $10\%$10% is 16, so $30\%$30% is 48; $5\%$5% is 8; $2.5\%$2.5% is 4; total $48 + 8 + 4 = 60$48+8+4=60. This is easy to show as working and banks method marks even if a small slip creeps in.

Estimate before you calculate. On any multi-step calculation, estimate first. If you are evaluating $489 \times 31$489×31, a quick $500 \times 30 = 15{,}000$500×30=15,000 tells you the answer should be around fifteen thousand. If your worked answer comes out as 1,515 or 151,590, a place-value error has crept in and you can catch it before it costs you.

Common Mistakes Examiners See

The same errors recur on number and ratio questions year after year. Knowing them is half the cure.

• Swapping HCF and LCM. Remember: HCF is never bigger than either number; LCM is never smaller. Check your answer against that rule.
• Reverse-percentage confusion. Finding the percentage of the new value instead of the original in a reverse-percentage question. Set up the multiplier and divide.
• Wrong multiplier. Multiplying by $0.2$0.2 instead of $1.2$1.2 for a $20\%$20% increase, or by $1.15$1.15 instead of $0.85$0.85 for a $15\%$15% decrease.
• Sign errors with negatives. Forgetting that subtracting a negative adds, or muddling the sign when multiplying. These cascade through multi-step problems.
• Not simplifying a ratio fully. Writing $6:10$6:10 when the question wants simplest form loses the final mark, even with perfect working before it.
• Mismatched units in compound measures. Calculating a speed in metres per minute when km/h was asked, or omitting the units altogether.
• Forgetting to square or cube an area/volume conversion. Treating 1 m² as 100 cm² rather than 10,000 cm².
• Rounding too early. On a calculator paper, carry the full value through with the answer memory and round only at the end. On the non-calculator paper, keep an extra digit in intermediate steps. Premature rounding is a leading cause of inaccurate final answers.

How These Topics Are Examined

A few habits convert knowledge into marks on OCR papers specifically.

Match your method to the command word. "Work out" and "Calculate" want a number with working. "Estimate" wants rounded values -- do not waste time on an exact answer. "Show that" gives you the destination, so every step must be visible and you may not use the given answer as part of your reasoning. "Give a reason for your answer" means a bare number scores at most partial credit; add the justifying sentence.

Show all working on multi-mark questions. OCR mark schemes split credit into method marks and accuracy marks. A correct, visible method with a slip at the end still earns most of the marks; an unsupported wrong answer earns nothing. This is the single most valuable exam habit in GCSE Maths.

Respect the accuracy instruction. If a question says "to 3 significant figures", give exactly that. If it says "exact" -- common on Higher surd and proportion questions -- leave surds and fractions unconverted. If nothing is specified, 3 significant figures is a safe default.

Sanity-check every answer. A "number of people" that comes out as 3.7, a "percentage increase" that lands below the original, or a speed of 4,000 km/h for a cyclist all signal an error. A two-second reasonableness check rescues marks.

Prepare with LearningBro

LearningBro's OCR GCSE Mathematics courses are built around the J560 specification and designed to grow both your mathematical fluency and your exam technique. The Number course covers every number topic from place value and fractions through to Higher-tier surds and bounds, while the Ratio & Proportion course takes you through sharing, direct and inverse proportion, compound measures, growth and decay, and unit conversions -- with practice questions that mirror the format and demand of the real papers.

Once individual topics feel secure, the OCR GCSE Maths exam preparation course shifts you onto mixed-topic practice under timed conditions, with particular attention to the non-calculator middle paper. Combine topic-by-topic study, regular timed practice and the strategies in this guide, and the Number and Ratio strands will become two of your most reliable sources of marks.

Related Reading

• OCR GCSE Maths revision guide -- the hub overview of the whole J560 course.
• OCR GCSE Maths: Statistics & Probability revision guide -- the companion topic guide to this one.
• OCR GCSE Maths: Algebra revision guide -- where your number skills get put to work.
• OCR GCSE Maths exam technique guide -- paper-by-paper tactics, including the non-calculator middle paper.

Good luck with your revision.