1. Number
- Syllabus
- 0580–2028–2029
- Section
- 1
- Level
- Core
Number types describe different properties. First test the definition, then use factor pairs or prime factors when the task asks for common factors, common multiples, HCF or LCM.
| Type | Defining test | Examples and boundary |
|---|---|---|
| natural number | counting number | 1,2,3,… |
| integer | whole number, including zero and negatives | …,−2,−1,0,1,2,… |
| prime number | integer greater than 1 with exactly two positive factors | 2,3,5,7; 2 is the only even prime and 1 is not prime |
| square number | n2 for an integer n | 0,1,4,9,16,… |
| cube number | n3 for an integer n | …,−8,−1,0,1,8,27,… |
| rational number | can be written a/b for integers a,b with b=0 | integers, fractions, terminating and recurring decimals |
| irrational number | cannot be written as such a fraction | 2,7,π; its decimal is non-terminating and non-recurring |
| reciprocal of x | number that multiplies by x to give 1 | 1/x for x=0; zero has no reciprocal |
For figures and words, group digits in threes from the right and preserve zero place holders: six billion is 6000000000, while 10007 is ten thousand and seven.
| Idea | Meaning | Method |
|---|---|---|
| factor of n | divides n with no remainder | list factor pairs or use prime factors |
| multiple of n | n× an integer | generate n,2n,3n,… |
| common factor / HCF | factor shared by all numbers / greatest such factor | multiply shared prime factors using the lowest powers |
| common multiple / LCM | multiple shared by all numbers / least positive such multiple | multiply every required prime factor using the highest powers |
24=23×3,36=22×32⇒HCF=22×3=12,LCM=23×32=72
A prime factorisation contains only prime factors. For example, 72=23×32; stopping at 8×9 is not complete because both factors are composite.
Do not confuse factor and multiple: factors divide a fixed number, whereas its multiples continue without end. A square root such as 9=3 is rational; a root symbol does not automatically make a number irrational.
A set is a collection whose members are fixed by a rule or a list. A two-set Venn diagram sorts every member of the universal set into four regions: in both sets, in only one of them, or in neither.
| Notation | Meaning | Region in a two-set diagram |
|---|---|---|
| n(A) | number of elements in A | count all of circle A |
| A′ | complement of A | everything in the universal set but outside A |
| A∪B | union: in A or B or both | both circles, including the overlap |
| A∩B | intersection: in both A and B | the overlap only |
The rectangle represents the universal set, written E. A complement is meaningful only after this universe is fixed: changing E can change A′ even when A stays the same.
| Form | What it says | Example |
|---|---|---|
| list or roster | write the elements between braces | B={a,b,c,…} |
| set-builder | name a variable and give its rule | A={x:x is a natural number} |
| bounded set-builder | give the allowed range | C={x:a≤x≤b}; both endpoints are included |
Place the intersection first. Then place members of A that are not in B in the A-only region, members of B that are not in A in the B-only region, and all remaining members of E outside both circles.
n(A∪B)=n(A)+n(B)−n(A∩B)
The intersection is subtracted once because it was counted once in n(A) and again in n(B). Also, n(A′)=n(E)−n(A) when every element is counted within the same universal set.
In set language, ‘or’ is inclusive: an element in both sets belongs to A∪B. Do not omit the overlap, and do not treat A′ as everything imaginable outside A—it means outside A but still inside the stated universal set. This Core objective uses no more than two sets.
A power repeats multiplication, while a matching root undoes that power. For a positive number a, a2 and a3 are its square and cube; a and 3a ask which numbers produce a when squared or cubed.
(a2)1/2=a(a≥0),(a3)1/3=a
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| n2 | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 | 225 |
Read the square table in either direction: 132=169 and 169=13. The radical 169 means the principal, non-negative square root.
| n | 1 | 2 | 3 | 4 | 5 | 10 |
|---|---|---|---|---|---|---|
| n3 | 1 | 8 | 27 | 64 | 125 | 1000 |
Read the cube table in either direction: 43=64 and 364=4. Cubes and cube roots also preserve sign, so (−3)3=−27 and 3−27=−3.
For another power or root, identify the index before calculating. For example, 54=5×5×5×5=625, and 4625=5 because 54=625.
| Expression | Safe entry and check | Result |
|---|---|---|
| 53.29 | square-root key; check 7.32 | 7.3 |
| 30.729 | cube-root template; check 0.93 | 0.9 |
| 45−54 | enter each complete power before subtracting | 1024−625=399 |
Do not halve a number to find its square root, and do not multiply the base by the exponent: 63 is 6×6×6=216, not 18. Keep this calculation objective separate from the index laws used to simplify algebraic powers in C1.4.
Fractions, decimals and percentages can all describe the same part of one whole. The useful form depends on the context: fractions show equal parts exactly, decimals use place value, and percentages compare with 100.
| Form | Meaning | Example |
|---|---|---|
| proper fraction | numerator is smaller than denominator, so its value is less than 1 | 53 |
| improper fraction | numerator is at least the denominator, so its value is at least 1 | 47 |
| mixed number | whole-number part plus a proper fraction | 143=1+43 |
| decimal | digits after the point represent tenths, hundredths, and so on | 0.25 is 25 hundredths |
| percentage | a number of parts per 100 | 25%=25 out of 100 |
In ba, the denominator b states how many equal parts make one whole and the numerator a states how many of those parts are taken. The denominator cannot be zero.
A fraction of a quantity must compare like units and use the whole as the denominator. For example, 32 cm as a fraction of 2 m is 20032=254 after converting metres to centimetres.
A percentage may exceed 100% and a decimal may exceed 1; neither is restricted to a proper fraction. A mixed number is a sum, so 231 means 2+31, not 2×31.
Equivalent forms have the same value even though their notation differs. Convert by using division or the meaning ‘per 100’, then simplify any fraction by dividing numerator and denominator by their highest common factor.
| Conversion | Method | Example |
|---|---|---|
| fraction → decimal | numerator ÷ denominator | 83=3÷8=0.375 |
| decimal → fraction | write over its place-value denominator, then simplify | 0.724=1000724=250181 |
| percentage → decimal | divide by 100 | 15%=0.15 |
| decimal → percentage | multiply by 100 and attach % | 0.25=25% |
| percentage ↔ fraction | use p%=100p, then simplify if needed | 72%=10072=2518 |
rac{a}{b}=\left(rac{a}{b} imes100 ight)\%
To convert a mixed number to an improper fraction, multiply the whole number by the denominator and add the numerator: 253=52×5+3=513. Reverse this by division: 13÷5=2 remainder 3, so 513=253.
Multiplying or dividing both parts of a fraction by the same non-zero number preserves its value. Thus 127=8449 because both numerator and denominator were multiplied by 7.
A simplest-form fraction has no common factor greater than 1. Do not move the decimal point by guesswork: multiplying by 100 moves from a decimal to a percentage, while dividing by 100 moves from a percentage to a decimal. Recurring-decimal notation and conversions are outside this Core objective.
To compare quantities, express them on one common scale and then read their positions from smallest to largest. On a number line, values increase to the right; this remains true for negatives, fractions, decimals, percentages, roots and standard form.
| Symbol | Meaning | Example |
|---|---|---|
| = | equal to | 0.5=50% |
| = | not equal to | 0.5=5% |
| > | greater than | 0.5>5% |
| < | less than | −7<−5 |
| ≥ | greater than or equal to | x≥3 includes 3 |
| ≤ | less than or equal to | x≤3 includes 3 |
Choose a common form that preserves enough accuracy: convert percentages to decimals, divide numerator by denominator for fractions, and evaluate roots or standard-form values. Keep extra decimal places until the order is secure; do not round two close values to the same comparison value.
| Original value | Comparable decimal |
|---|---|
| 58% | 0.58 |
| 127 | 0.5833… |
| 0.6 | 0.6 |
| 138 | 0.6153… |
| 32 | 0.6666… |
Therefore 58%<127<0.6<138<32. A chain uses a comparison between every neighbouring pair; each inequality must point consistently from smaller to larger.
To find a fraction between two fractions, a common denominator can expose a gap. Since 253=506 and 254=508, the value 507 lies between them.
The symbols ≥ and ≤ include equality, while > and < do not. For negative values, the number with the greater absolute size can be smaller: −8<−3 because −8 lies further left on the number line.
The four operations combine quantities: addition joins, subtraction finds a change or difference, multiplication scales, and division shares or asks how many groups fit. Choose the operation from the relationship, then apply the agreed calculation order.
| Priority | What to calculate | Boundary |
|---|---|---|
| 1 | brackets | work from the innermost brackets outward |
| 2 | powers and roots | evaluate the complete powered or rooted value |
| 3 | multiplication and division | work left to right when both occur |
| 4 | addition and subtraction | work left to right when both occur |
For 9+5×7−4÷2, calculate multiplication and division first: 9+35−2=42. Brackets change the structure: (9+5)×7−4÷2=96.
| Operation with signed numbers | Reliable rule | Example |
|---|---|---|
| add or subtract | treat subtraction as adding the opposite; track movement on the number line | −7−5+8=−12+8=−4 |
| multiply or divide | same signs give positive; different signs give negative | −18÷(−4)=4.5 |
| difference | subtract one value from the other and use the positive distance when the context asks how far apart | 5−(−7)=12 |
| Fraction operation | Method |
|---|---|
| add or subtract | use a common denominator, then combine numerators |
| multiply | convert mixed numbers to improper fractions, multiply, then simplify |
| divide | multiply by the reciprocal of the divisor, then simplify |
\left(2rac13-rac78 ight) imesrac6{25}=\left(rac{56}{24}-rac{21}{24} ight) imesrac6{25}=rac{35}{24} imesrac6{25}=rac7{20}
For decimal addition and subtraction, align place values. For multiplication or division, calculate with the correct place-value scale and estimate first: 1.6×0.02 is close to 2×0.02=0.04, so the exact result 0.032 has a sensible size.
Translate practical wording before calculating. ‘5∘C lower than −7∘C’ means −7−5=−12, while ‘8∘C higher’ then means −12+8=−4. When a whole item is required, such as coaches or bags, interpret the remainder and round up rather than reporting a partial item.
Do not automatically work from left to right across unlike operations, and do not add denominators when adding fractions. A negative sign belongs to its number; subtracting a negative changes the operation to addition.
In an, a is the base and n is the index. A positive integer index counts repeated factors; zero and negative indices extend the same pattern so that moving one index step down always divides by the base.
| Integer index | Meaning | Example |
|---|---|---|
| n>0 | multiply n copies of the base | 24=2×2×2×2=16 |
| 0 | value is 1 when the base is non-zero | 190=1 |
| −n | reciprocal of the matching positive power | 2−4=241=161=0.0625 |
a^0=1,\qquad a^{-n}=rac{1}{a^n}\quad(a e0,\ n>0)
The sequence 23=8, 22=4, 21=2, 20=1, 2−1=21 divides by 2 at every step. This explains both the zero-index and negative-index definitions rather than treating them as disconnected rules.
Brackets decide whether a negative sign belongs to the base: (−3)2=9, but −32=−(32)=−9. A negative index does not make the value negative; it creates a reciprocal.
The expressions 00 and 0−n are not defined here because the negative-index rule would require division by zero. This objective uses integer indices only; fractional indices are outside C1.7 Core scope.
Index laws compress repeated multiplication. They apply when the bases match, or when one complete product, quotient or power is raised to an index; the operation tells you what to do with the indices.
| Structure | Index law | Why |
|---|---|---|
| multiply same base | am×an=am+n | join the two groups of factors |
| divide same base | am÷an=am−n | cancel matching factors; a=0 |
| power of a power | (am)n=amn | repeat a group of m factors, n times |
| power of a product | (ab)n=anbn | every repeated factor contains both a and b |
| power of a quotient | (a/b)n=an/bn | apply the power to numerator and denominator; b=0 |
| Expression | Apply the law | Value |
|---|---|---|
| 2−3×24 | 2−3+4=21 | 2 |
| (23)2 | 23×2=26 | 64 |
| 23÷24 | 23−4=2−1 | 21 |
27imes812=33imes(34)2=33imes38=311
When the bases already match, equate indices after simplifying. For 5n÷54=56, the quotient law gives 5n−4=56, so n−4=6 and n=10.
Do not use the addition law when bases differ, and do not multiply indices when multiplying powers: am×an adds indices, whereas (am)n multiplies them. Also, (a+b)n does not become an+bn.
Standard form writes a non-zero number as one coefficient multiplied by an integer power of 10. The coefficient shows the significant digits, while the power of 10 records the number's scale.
Aimes10n,1≤A<10,n∈Z
| Part | Requirement | Meaning |
|---|---|---|
| A | at least 1 but less than 10 | contains exactly one non-zero digit before the decimal point |
| 10n | n is an integer | positive n scales to a large value; negative n scales to a value between 0 and 1 |
85.1×104 has the correct value but is not in standard form because 85.1 is not less than 10. Renormalising gives 8.51×105.
For positive standard-form numbers, compare powers of 10 first. Thus 2.04×109>9.78×108 because 109 is ten times the scale of 108. If powers match, compare the coefficients.
The power of 10 alone does not make a representation standard form: 0.3×10−2 is not valid because its coefficient is below 1. Standard form preserves the exact value; it is not automatically a rounded approximation.
To convert into standard form, reposition the decimal point so the coefficient is at least 1 and less than 10, then use the power of 10 that restores the original place value.
| Ordinary number | Valid coefficient | Decimal-point movement | Standard form |
|---|---|---|---|
| 153000000 | 1.53 | 8 places left | 1.53×108 |
| 0.0605 | 6.05 | 2 places right | 6.05×10−2 |
| 0.0000000347 | 3.47 | 8 places right | 3.47×10−8 |
A positive exponent restores a large number by moving the decimal point right. A negative exponent restores a small number by moving it left. The exponent records the reverse of the movement used to create the coefficient.
| Standard form | Apply the scale | Ordinary number |
|---|---|---|
| 4.73×106 | move 6 places right | 4730000 |
| 2.06×10−2 | move 2 places left | 0.0206 |
| 3.47×10−8 | move 8 places left | 0.0000000347 |
Check both value and format: the coefficient must satisfy 1≤A<10, and converting back must reproduce every zero and significant digit of the ordinary number.
Do not choose the exponent from the number of visible zeros alone; count place-value moves from the original decimal point. Leading zeros in a small decimal are place holders, not significant digits.
In standard-form calculations, operate on the coefficients and powers separately, then renormalise the result so its coefficient returns to the interval 1≤A<10.
| Operation | Method | Example before normalising |
|---|---|---|
| multiply | multiply coefficients; add exponents | (4.1×10−3)(8.9×107)=36.49×104 |
| divide | divide coefficients; subtract exponents | (6.4×105)÷(2.5×10−7)=2.56×1012 |
| add or subtract | first rewrite both terms with the same power of 10 | 3×10199+2×10201=0.03×10201+2×10201 |
36.49imes104=3.649imes105
0.03imes10201+2imes10201=2.03imes10201
For a power, apply it to both parts: (3×10−3)3=33×10−9=27×10−9=2.7×10−8.
Keep full calculator precision during the calculation, normalise first, and round only the final coefficient when a degree of accuracy is requested. For example, 4.6×102×6.7×105=3.082×108, which is 3.1×108 to 2 significant figures.
Do not add coefficients until the powers match, and do not add exponents when adding numbers. For Core candidates, calculation with standard form is expected only on Paper 3; conversion and recognition remain part of the Topic generally.
Rounding replaces a value with the nearest value at a stated accuracy. Identify the final digit to keep, inspect the next digit, and increase the kept digit by 1 only when the next digit is 5 or more.
| Accuracy instruction | Where counting starts | Example |
|---|---|---|
| nearest 10, 100, 1000, … | named place in the whole-number part | 11678→11700 to nearest 100 |
| decimal places (dp) | first digit after the decimal point | 3.72194→3.722 to 3 dp |
| significant figures (sf) | first non-zero digit | 0.03682→0.037 to 2 sf |
Keep all digits before the rounding position unchanged. If rounding up creates a 10, carry left through place values: 9876 to the nearest thousand is 10000.
Zeros can communicate accuracy. The value 57.3997 to 4 significant figures is 57.40: the final zero must remain because it is the fourth significant figure. Similarly, 0.0050 has 2 significant figures.
The rounded result alone may not reveal the instruction: 4896→4900 could be rounding to the nearest hundred or nearest ten. State the requested accuracy with the result when context does not already specify it.
An estimate replaces input values with nearby values that make the calculation quick while preserving its original structure. When instructed, round every input to 1 significant figure before calculating.
Use this order: round each input separately; rewrite the complete expression with the rounded values and the same brackets, powers and operations; calculate that simpler expression; then compare its scale with the original values.
rac{41.3}{9.79 imes0.765}pproxrac{40}{10 imes0.8}=5
For 23.5423.8−78.4, rounding to 1 significant figure gives 20400−80=16. The subtraction remains in the numerator; removing its grouping would estimate a different calculation.
An estimate is not always above or always below the exact answer. Its direction depends on how each rounded input affects the operation: rounding both positive factors down makes their estimated product smaller, but rounding a denominator down can make a quotient larger.
Do not round an intermediate result again unless instructed. Estimation simplifies inputs before calculation; it is different from calculating accurately and rounding only the final answer.
A reasonable final accuracy communicates what the context and input data can support. Keep guard digits during working, then round once at the end using any explicit instruction or the practical meaning of the result.
| Context | Sensible final form | Reason |
|---|---|---|
| counted objects or whole items | whole number, with direction chosen by context | partial people, buses or packs may be impossible |
| money in ordinary currency units | usually 2 decimal places | records the smallest common currency unit |
| measured quantity | usually no more significant figures than the least precise input | avoids claiming unsupported measurement precision |
| specified dp or sf | follow the stated instruction exactly | the requested accuracy controls the final digit |
Context can control rounding direction rather than nearest rounding. If 5.25 bags are needed and only whole bags can be bought, the answer is 6 bags; rounding to 5 would leave too little material.
For a calculator value 4.285714… requested to 4 significant figures, retain the full value until the end and report 4.286. Early rounding of intermediate values can change the final digit.
Do not add decimal places merely because a calculator displays them. A result such as 12 people, 8.40,or3.7$ cm can each be appropriately precise in its own context; one universal number of decimal places is not sensible.
A rounded value represents an interval of possible original values. Find the rounding step, take half of it, then subtract and add that half-step to locate the lower and upper boundaries.
x-rac{r}{2}\le v<x+rac{r}{2}
Here x is the stated rounded value, r is one unit at the stated accuracy, and v is the original value. The lower bound is included; the upper bound is excluded because an exact upper-bound value rounds to the next stated value.
| Stated accuracy | Rounding step r | Half-step | Example interval |
|---|---|---|---|
| nearest kilogram | 1 kg | 0.5 kg | 428.5≤m<429.5 for 429 kg |
| nearest 5 g | 5 g | 2.5 g | 112.5≤m<117.5 for 115 g |
| 1 decimal place | 0.1 unit | 0.05 unit | 76.25≤h<76.35 for 76.3 m |
| 2 decimal places | 0.01 unit | 0.005 unit | 37.835≤h<37.845 for 37.84 m |
Convert the stated accuracy into the variable's unit before halving. If p is in kilograms and 12.4 kg is correct to the nearest 100 g, then r=0.1 kg, so 12.35≤p<12.45.
For significant figures, use the place value of the final significant digit as the rounding step. For example, 350 correct to 2 significant figures has step 10, giving 345≤v<355.
Do not use the full rounding step on each side, and do not write ≤ at the upper bound. This Core objective asks for bounds of rounded data only; it does not require bounds for results calculated from rounded inputs.
A ratio compares quantities as multiplicative parts. In a:b:c, each quantity is the same scale factor times a, b or c. Keep the quantities in the stated order and convert them to the same units before comparing.
| Learning job | Ratio-parts method | Example |
|---|---|---|
| Simplify a ratio | Divide every term by the same highest common factor. | 20:30:40=2:3:4 |
| Share a total T in a:b:c | Find p=a+b+c, then one part is T÷p. Multiply by a, b and c. | Share 190 in 12:5:2: one part =190÷19=10; shares are 120, 50 and 20. |
| Use a known difference | Subtract the corresponding ratio parts, then divide the quantity difference by that part difference. | Red:green:blue =12:5:2 and red exceeds green by 112. Seven parts =112, so one part =16 and blue =2×16=32. |
one part=sum of ratio partsknown totalorone part=difference of ratio partsknown difference
For a direct proportion, every quantity changes by the same scale factor. A recipe using 550 g for 8 cakes needs 550×(360÷8)=24750 g, or 24.75 kg, for 360 cakes. Scale from a known pair; do not add the same amount.
For map scales, first use matching units. A scale of 1 cm to 8 km is 1:800000 because 8 km is 800 000 cm. At a scale of 1:250000, 3.5 cm represents 3.5×250000=875000 cm, or 8.75 km.
To determine best value, compare like with like: calculate the cost for one common unit, or scale every option to the same quantity. The lowest cost per common unit is the best value, provided the units and quantities are equivalent.
A ratio of 2:3 does not mean 2/3 of the whole. There are 2+3=5 parts, so the shares are 2/5 and 3/5. Ratio order matters, and a context involving whole items may require a final whole-number decision only after the proportional calculation.
A rate compares quantities with different units. The unit tells you both the division and the calculation direction: 15.20perhourmeans$15.20$ for each hour, while 20 litres per minute means 20 litres for each minute.
rate=reference quantityquantityquantity=rate×reference quantity
| Context | Read the units | Typical calculation |
|---|---|---|
| Hourly pay | dollars/hour | pay = hours × hourly rate |
| Exchange rate | new currency/old currency | old amount × rate = new amount |
| Flow rate | litres/minute | volume = flow rate × time |
| Fuel use | litres/100 km or km/litre | follow the stated unit; the two forms are not interchangeable |
At \15.20perhour,40hoursgives40\times15.20=$608.If1dollarexchangesfor1.25euros,80\times1.25=€100$; converting back divides by 1.25. Write the units beside the numbers so the unwanted unit cancels.
Convert the reference unit before applying the rate. For example, 84 km/h is 84000÷60=1400 m/min. Multiplying by 1000 changes kilometres to metres; dividing by 60 changes 'per hour' to 'per minute'.
Do not multiply automatically. Use the rate's written units to decide whether to multiply or divide, and do not assume that every fuel-consumption figure is expressed in the same direction.
Pressure, material density and population density are rates: one quantity is measured per unit of another. In this objective, the required formula is supplied in the question; your job is to match each value and unit to that formula.
| Measure | Common given formula | Meaning of the unit |
|---|---|---|
| Pressure | P=F/A | force per unit area, such as N/m² |
| Density | ρ=m/V | mass per unit volume, such as g/cm³ |
| Population density | d=N/A | people per unit area, such as people/km² |
A metal cuboid has mass 4 kg and volume 600 cm³. Convert 4 kg to 4000 g, then ρ=4000÷600=6.67 g/cm³ (3 s.f.). If density and volume are known, rearrange ρ=m/V to m=ρV.
For a population of 735 000 across 1 477 300 km², population density is 735000÷1477300≈0.498 people/km². A value below 1 is possible: it means fewer than one person per square kilometre on average, not a fraction of a person at each location.
Do not memorise extra physics formulas as syllabus requirements here. Use the formula supplied, keep numerator and denominator in the correct order, and convert mass, area or volume units before dividing.
Average speed describes a whole journey. It is the total distance travelled divided by the total elapsed time, including any stops when the question treats them as part of the journey.
average speed=total timetotal distanced=stt=sd
A cyclist travels 45 km in 3 hours 45 minutes. Since 45 minutes is 45/60=0.75 hours, the total time is 3.75 hours. The average speed is 45÷3.75=12 km/h.
For a two-stage run, first find missing totals. Running for 45 minutes at 9.5 km/h covers 9.5×0.75=7.125 km. Running 8.1 km at 7.5 km/h takes 8.1÷7.5=1.08 h. So the whole-run average is (7.125+8.1)÷(0.75+1.08)≈8.32 km/h.
To convert m/s to km/h, multiply by 60×60÷1000=3.6. To convert km/h to m/s, divide by 3.6. A decimal hour is not decimal minutes: 2.4 hours is 2 hours 24 minutes.
Do not average two speeds unless their travel times are equal. Average speed always uses total distance divided by total time; it is not generally the arithmetic mean of the stage speeds.
A percentage is a number of hundredths. To find p percent of a quantity Q, turn p percent into the multiplier p/100 and multiply.
percentage amount=100p×Q
To find 57% of 45, calculate 0.57×45=25.65. To find a 16% discount amount on \12,400,calculate0.16\times12,400=$1984$.
| Percentage | Useful multiplier or split | Example of the quantity |
|---|---|---|
| 10% | divide by 10 | 10% of 168 is 16.8 |
| 5% | half of 10% | 5% of 168 is 8.4 |
| 25% | divide by 4 | 25% of 240 is 60 |
| 125% | multiply by 1.25 | 125% of 80 is 100 |
The percentage amount is not always the final quantity. A discount amount must be subtracted from the original; a tax or increase amount must be added. Percentages greater than 100% are valid and produce more than the original quantity.
To express one quantity as a percentage of another, compare the part with the whole. The word “of” identifies the reference whole, which belongs in the denominator.
percentage=wholepart×100
A student scores 58 out of 80. The whole is 80, so 58÷80×100=72.5, giving a score of 72.5%. If 10 650 of 15 000 seats are occupied, 10650÷15000×100=71, so occupancy is 71%.
The part can exceed the reference whole. For example, 150÷120×100=125, so 150 is 125% of 120.
Do not reverse the fraction. Ask “what is the reference whole?” before calculating. A percentage has no physical unit, but the two quantities must use matching units before division.
Percentage change compares the change with the original value. The original—not the new value—is the reference whole.
percentage change=original∣new−original∣×100
| Learning job | Calculation |
|---|---|
| Find an increase percentage | (new−original)÷original×100, then attach % |
| Find a decrease percentage | (original−new)÷original×100, then attach % |
| Increase by p% | multiply by 1+p/100 |
| Decrease by p% | multiply by 1−p/100 |
Cyclist numbers rise from 3546 to 4067. The increase is 521, and 521÷3546×100≈14.7, so the increase is 14.7%. Decreasing \3450by 18% uses multiplier 0.82, giving3450\times0.82=$2829$.
For profit or loss, the original cost price is the denominator. Buying for \2.50andsellingfor$4.20givesprofit$1.70.Since1.70\div2.50\times100=68$, the percentage profit is 68%.
A 20% increase followed by a 20% decrease does not return to the start because the second percentage uses a different base. Keep increase/decrease amount, percentage change and final value as distinct quantities.
Simple interest adds the same amount each year because it is always calculated from the original principal. Compound interest applies each year's percentage to the current balance, so the interest itself earns interest.
| Interest type | Formula to know | What changes each year? |
|---|---|---|
| Simple | I=P×(r/100)×n; A=P+I | yearly interest stays P×r/100 |
| Compound | A=P(1+r/100)n | balance is multiplied by 1+r/100 each year |
Here P is the principal, r is the annual percentage rate, n is the number of years, I is total interest and A is the final amount. These formulas are not supplied in the Core examination.
At 1.7% simple interest for 4 years on \8500,totalinterestis8500\times0.017\times4=$578.Thefinalamountwouldbe8500+578=$9078$.
At 2.5% compound interest for 7 years on \30,000,A=30,000(1.025)^7=35,660.57\ldots,whichis$35,661tothenearestdollar.TotalinterestwouldbeA-P$.
Do not multiply a compound rate by the number of years; that is the simple-interest pattern. Keep full calculator precision until the final answer, and distinguish interest earned from the final account value.
Efficient calculator use preserves the structure and precision of a calculation. Plan the expression, enter it in one controlled sequence, and check that the display is reasonable.
For 345.96, enter the whole value under the square-root operation. The display gives 18.6, and the check 18.62=345.96 confirms it.
If an intermediate result is 7.428571…, do not replace it with 7.43 before the next operation. Reuse the full value; early rounding can change the final digit.
A calculator evaluates what was entered, not what was intended. An unexpected sign or order of magnitude is a reason to inspect brackets and keys, not to force the display toward an expected answer.
A calculator entry must represent the same mathematical object as the written expression. Group complete numerators, denominators, powers and time values before evaluating.
| Written value | Safe entry idea | Why |
|---|---|---|
| 9.79×0.76541.3 | 41.3 ÷ (9.79 × 0.765) or a fraction template |
keeps the full product in the denominator |
| (2.4−0.7)3 | (2.4 − 0.7) then power 3 |
applies the power to the whole difference |
| 5.2×10−4 | use the standard-form exponent entry with exponent −4 | avoids treating 10−4 as subtraction |
Enter 2 hours 30 minutes as 2.5 hours when the calculation uses decimal hours, because 30/60=0.5. On a calculator with a degrees–minutes–seconds key, the equivalent entry is 2∘30′0′′.
Read the calculator's expression line before pressing equals. Check that every opening bracket has a closing bracket and that a negative sign belongs to the intended number or exponent.
Do not type 2.30 to mean 2 hours 30 minutes in an ordinary decimal calculation: 2.30 hours is 2 hours 18 minutes. Decimal notation and hours–minutes notation use different place-value systems.
The display is a numerical value; the context decides how it must be written. Convert or format the value without changing what it means.
| Display | Context | Appropriate interpretation |
|---|---|---|
| 4.8 | money | $4.80, because currency uses two decimal places |
| 3.25 | hours | 3 hours 15 minutes, because 0.25×60=15 |
| 2.4 | hours | 2 hours 24 minutes, because 0.4×60=24 |
| 0.333333… | answer requested to 3 s.f. | 0.333 |
To convert a decimal-hour display, keep the whole number as hours and multiply only the decimal part by 60. If seconds are needed, multiply the remaining fraction of a minute by 60 again.
For money, write the correct currency symbol and two decimal places: a display of 12 means 12.00,while12.5means12.50. Apply the rounding rule before adding the trailing zero.
A decimal point does not automatically separate hours from minutes, and trailing zeros can be necessary communication even when they do not change the numerical value. Interpret first, then format for the context.
Time units do not use one place-value system. Convert through the exact relationship between neighbouring units, multiplying toward smaller units and dividing toward larger units.
| Relationship | Toward the smaller unit | Toward the larger unit |
|---|---|---|
| 1 minute = 60 seconds | minutes × 60 | seconds ÷ 60 |
| 1 hour = 60 minutes | hours × 60 | minutes ÷ 60 |
| 1 day = 24 hours | days × 24 | hours ÷ 24 |
| 1 week = 7 days | weeks × 7 | days ÷ 7 |
| 1 year = 12 months | years × 12 | months ÷ 12 |
For day calculations in this syllabus, use 1 year = 365 days unless the question supplies other information.
July has 31 days, so it contains 31×24×60×60=2678400 seconds. For 2.15 hours, multiply the whole decimal by 60: 2.15×60=129 minutes.
For a mixed-unit answer, convert to the smallest useful unit first. Eighteen trips of 23 minutes total 414 minutes; 414÷60=6 remainder 54, so the time is 6 hours 54 minutes.
A decimal hour is not hours and minutes: 2.15 hours is 2 hours 9 minutes, not 2 hours 15 minutes. Months have different numbers of days, so do not invent a fixed month-to-day conversion unless the question provides one.
Clock times name positions in a day; durations measure the interval between them. Convert both clock formats consistently, then add, subtract or bridge midnight in hours and minutes.
| 12-hour time | 24-hour time | Rule |
|---|---|---|
| 3.25 a.m. | 03 25 | keep the hour and add a leading zero |
| 3.25 p.m. | 15 25 | add 12 to the hour |
| 12.00 noon | 12 00 | noon starts the p.m. half of the day |
| 12.00 midnight | 00 00 | midnight starts a new day |
To add a duration, add minutes first and exchange every 60 minutes for 1 hour. Starting at 19 50, adding 2 hours 42 minutes gives 21 92, which normalises to 22 32.
For an interval across midnight, split at 24 00. From 21 15 to 24 00 is 2 hours 45 minutes; from 00 00 to 04 33 is 4 hours 33 minutes. The total is 7 hours 18 minutes.
To find a start time, reverse the process and borrow 1 hour as 60 minutes when necessary. A film ending at 23 05 after 2 hours 50 minutes started at 20 15.
Do not subtract clock digits as ordinary base-10 numbers. There are 60 minutes in an hour, and 24 00 is the same boundary instant as 00 00 on the next day.
A timetable links places or events to clock times. Read the correct row and column first; then distinguish travel time, waiting time and local-time differences.
For each stage, pair its departure with its arrival. A boat departing Millwater at 11 45 and arriving Westbridge at 13 07 travels for 1 hour 22 minutes. A wait between an arrival and the next departure is not travel time, but it is included if the question asks for the whole journey.
| Statement | Conversion at the same instant |
|---|---|
| destination is k hours ahead | destination time = source time + k hours |
| destination is k hours behind | destination time = source time − k hours |
For a flight: 1. Start with the departure day and local time. 2. Add the flight duration in the departure time zone. 3. Apply the destination's ahead/behind offset. 4. Move the day or date whenever the running time crosses 24 00 or 00 00.
A plane leaves Seattle at 07 30 on Tuesday, flies for 10 hours 55 minutes, and Seoul is 16 hours ahead. In Seattle time it lands at 18 25 Tuesday; adding 16 hours gives 10 25 Wednesday in Seoul.
A time-zone offset changes the local clock label, not the flight duration. Do not add the offset to the elapsed flying time, and always state the new day or date after a midnight crossing.
Money calculations combine quantities, unit prices and payment rules. Keep every amount in one currency and one unit, model the whole bill, then format the final amount appropriately.
| Money job | Calculation structure |
|---|---|
| Cost of several items | quantity × unit price |
| Total bill | add every item or charge once |
| Change | amount paid − total bill |
| Fixed fee plus usage | fixed fee + number of additional units × extra-unit price |
| Maximum whole items | divide budget by unit price, then take the whole-number part |
With 20 dollars and pineapples costing 1.45 dollars each, 20÷1.45=13.79…, so at most 13 can be bought. Their cost is 13×1.45=18.85 dollars, leaving 20−18.85=1.15 dollars change.
Convert cents and dollars before combining them: 47 cents is 0.47 dollars, while 5 dollars is 500 cents. For mass or volume prices, also match the quantity unit; 125 ml is 0.125 litres before multiplying by a price per litre.
Read special pricing literally. “Buy 3 for the price of 2” means every complete group of 3 costs 2 unit prices. “First hour 15.50 dollars, each additional hour 7.25 dollars” means the first hour is not charged again at 7.25 dollars.
Round a final money answer to the smallest stated currency unit, normally two decimal places for dollars. Do not round a unit price or intermediate total early, and never round a maximum item count upward beyond the available budget.
An exchange rate is a unit rate. Read its direction before calculating: if 1 unit of currency A equals r units of currency B, then r converts one A into B.
1 A=r B
| Conversion | Operation | Unit check |
|---|---|---|
| A to B | multiply by r | A × B/A = B |
| B to A | divide by r | B ÷ B/A = A |
If 1 rupee = 0.016 dollars, a 20-dollar ticket costs 20÷0.016=1250 rupees. Division is required because the given rate tells how many dollars one rupee is worth, not how many rupees one dollar buys.
For two currencies with a common bridge, convert in two labelled stages. If 1 dollar = 0.615 euros and 1 krona = 0.087 euros, then 2000 dollars becomes 2000×0.615=1230 euros, then 1230÷0.087=14137.93… krona, or 14 138 krona to the nearest krona.
To compare prices in different currencies, convert both prices into the same currency first, subtract to find the difference, and round only the final requested amount.
Do not choose multiply or divide from whether the number should become larger. Currency values vary; let the written rate and units determine the operation. Keep full precision until the final money rounding.