1 Types of numberNatural numbers : 1, 2, 3, … Integers : whole numbers including 0 and negatives (…, −2, −1, 0, 1, 2, …).Prime : exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, … 1 is not prime; 2 is the only even prime.Factor of 12: divides into it (1, 2, 3, 4, 6, 12). Multiple of 12: in its times table (12, 24, 36, …).Rational : can be written as a fraction a/b (all integers, terminating and recurring decimals). Irrational : cannot — π, √2, √3, √5 and the square root of any non-square number.Reciprocal of x is 1/x: reciprocal of 4 is 1/4, of 2/5 is 5/2. A number times its reciprocal is 1.n 1 2 3 4 5 6 7 8 9 10 n2 1 4 9 16 25 36 49 64 81 100 n3 1 8 27 64 125 216 343 512 729 1000
Worked example From −3, 0, 1/2, √9, √5, 7 and 0.333…, list (a) the irrational numbers, (b) the prime numbers.
√9 = 3 and 0.333… = 1/3, so both are rational. √5 is not a whole number and does not recur: irrational. Primes are positive whole numbers with exactly two factors: √9 = 3 and 7. (a) √5 (b) √9 and 7
Exam tip: Also know 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225 for the non-calculator paper.
2 Order of operations and negative numbersBIDMAS : Brackets → Indices → Division and Multiplication (left to right) → Addition and Subtraction (left to right).Two signs next to each other: same signs give +, different signs give −. So 5 − (−3) = 8 and 4 × (−2) = −8. (−3)2 = 9 but −32 = −9: without brackets only the 3 is squared. A fraction line acts like brackets: (8 + 4) ÷ (5 − 2) = 4. On a calculator, put brackets round the top and the bottom. Order to work in Work out anything in brackets. Then powers and roots. Then × and ÷ from left to right. Finally + and − from left to right. Worked example Work out (−3)2 − 4 × (5 − 7).
Brackets: 5 − 7 = −2. Power: (−3)2 = 9. Multiply: 4 × (−2) = −8. 9 − (−8) = 9 + 8. 17
3 FractionsTo add or subtract, use a common denominator (the LCM of the bottoms), then add the tops. To multiply, multiply tops and bottoms (cancel first to keep numbers small). To divide, flip the second fraction and multiply : a/b ÷ c/d = a/b × d/c. Change mixed numbers to improper fractions before multiplying or dividing: 2¾ = 11/4. Fraction of an amount: divide by the bottom, multiply by the top. 3/8 of 240 = 240 ÷ 8 × 3 = 90. Always give the answer in its simplest form. Formulas a/b + c/d = (ad + bc)/bd then simplify
a/b × c/d = ac/bd multiply across
a/b ÷ c/d = a/b × d/c flip the second, then multiply
Worked example Work out 2¾ − 1⅚.
Improper fractions: 11/4 − 11/6. Common denominator 12: 33/12 − 22/12. = 11/12. 11/12
Exam tip: On the non-calculator paper write every step: the method marks need the common denominator to be seen.
4 Decimals and orderingMultiplying decimals: ignore the points, multiply, then give the answer as many decimal places as the question had in total. 0.3 × 0.04 = 0.012. Dividing by a decimal: multiply both numbers by 10, 100, … until you divide by a whole number. 4.8 ÷ 0.06 = 480 ÷ 6 = 80. Fraction → decimal: divide the top by the bottom. 3/8 = 0.375. To order a mixture of fractions, decimals and percentages, change them all to decimals (to 3 or 4 places). Symbols: = equal, ≠ not equal, > greater than, < less than, ⩾ greater than or equal to, ⩽ less than or equal to. Worked example Write in order, smallest first: 0.7, 2/3, 68%, 0.67.
2/3 = 0.6667, 68% = 0.68. Compare: 0.6667, 0.67, 0.68, 0.7. 2/3, 0.67, 68%, 0.7
5 Halfway values and differencesThe number halfway between a and b is (a + b) ÷ 2. The difference between two numbers is the larger minus the smaller; across zero, add the distances: from −12 to 9 is 12 + 9 = 21. Temperature questions: a rise from −7 °C to 5 °C is 12 degrees; a fall of 9 degrees from 4 °C gives −5 °C. Formulas halfway = (a + b) ÷ 2 works with negatives and decimals
Worked example Find the number halfway between −2.6 and 4.2.
Add: −2.6 + 4.2 = 1.6. Halve: 1.6 ÷ 2. 0.8
6 Recurring decimals to fractionsA dot over a digit means it repeats: 0.4̇ = 0.444…, 0.2̇7̇ = 0.272727… Every recurring decimal is a fraction. One repeating digit → over 9; two repeating digits → over 99. If some digits do not repeat (0.3666…), multiply twice so the repeating parts line up, then subtract. How to do it Let x = the decimal. Multiply by 10 (one repeating digit) or 100 (two) so the repeating part lines up. Subtract to remove the repeating part. Divide and simplify. Fraction 1/3 2/3 1/9 1/11 1/6 Decimal 0.3̇ 0.6̇ 0.1̇ 0.0̇9̇ 0.16̇
Worked example Write 0.3666… as a fraction in its simplest form.
x = 0.3666…, 10x = 3.666…, 100x = 36.666… 100x − 10x: 90x = 33. x = 33/90. 11/30
Exam tip: 'Show that 0.2̇7̇ = 3/11' needs the algebra (x = …, 100x = …, subtract). Just dividing 3 by 11 gets no marks.
✗ Common mistakesDon’t write Write instead ✗ Calling 1 a prime number✓ 1 has only one factor, so it is not prime✗ 3 + 4 × 2 = 14✓ Multiply first: 3 + 8 = 11✗ −32 = 9✓ Only the 3 is squared: −9. (−3)2 = 9✗ 1/2 + 1/3 = 2/5✓ Common denominator: 3/6 + 2/6 = 5/6✗ 2¾ × 2 = 4¾✓ Improper fraction first: 11/4 × 2 = 11/2 = 5½✗ Ordering 0.7 before 0.67 because 7 < 67✓ Compare place by place: 0.70 > 0.67
★ In the examNon-calculator questions want every step shown — a correct answer with no working may lose marks. Give fractions in their simplest form unless the question asks otherwise. Check an answer is sensible: dividing by a number less than 1 makes it bigger.
1 The laws of indicesThe index (power) says how many times a number is multiplied by itself: 25 = 2 × 2 × 2 × 2 × 2 = 32. The laws only work when the base is the same : 23 × 24 = 27 , but 23 × 34 cannot be combined. Numbers in front are dealt with normally: 3x2 × 4x5 = 12x7 . Formulas am × an = am+n multiply → add the powers
am ÷ an = am−n divide → subtract the powers
(am )n = amn power of a power → multiply
(ab)n = an bn everything inside is raised
Worked example Simplify 25 × 23 ÷ 24 and find its value.
Add: 5 + 3 = 8. Subtract: 8 − 4 = 4. 24 = 16. 24 = 16
2 Zero, negative and fractional powersa0 = 1 for any a except 0: 70 = 1, (−5)0 = 1. A negative power means 'one over': a−n = 1/an . It never makes the answer negative: 2−3 = 1/8. A fraction raised to a negative power flips: (2/3)−2 = (3/2)2 = 9/4. A fractional power is a root: a1/2 = √a, a1/3 = ∛a, and am/n = (ⁿ√a)m — do the root first. Negative fractional: do all three — flip, root, power. 8−2/3 = 1/(∛8)2 = 1/4. For a−m/n Negative sign? Flip the number (write 1 over it). Bottom of the fraction? Take that root. Top of the fraction? Raise to that power. Formulas a0 = 1
a−n = 1/an one over
a1/n = ⁿ√a root
am/n = (ⁿ√a)m root first, then power
Worked example Find the value of (27/8)−2/3 .
Flip: (8/27)2/3 . Cube root: ∛8 = 2 and ∛27 = 3, giving 2/3. Square: (2/3)2 . 4/9
Exam tip: Know the powers of 2 (2, 4, 8, 16, 32, 64, 128), 3 (3, 9, 27, 81, 243) and 5 (5, 25, 125, 625) by heart.
3 Indices in algebraDeal with the numbers and each letter separately. (3x2 )3 = 33 × x6 = 27x6 — the number in front is cubed too. 12a5 ÷ 3a−2 = 4a5−(−2) = 4a7 . (2x3/5 )2 = 4x6/25 : square the top and the bottom. Worked example Simplify (2x3 y)4 ÷ 8x5 y2 .
Expand the bracket: 16x12 y4 . Divide the numbers: 16 ÷ 8 = 2. Subtract powers: x12−5 = x7 , y4−2 = y2 . 2x7 y2
4 Equations with the unknown in the powerWrite both sides as powers of the same base , then set the powers equal. Useful bases: 4 = 22 , 8 = 23 , 16 = 24 , 32 = 25 ; 9 = 32 , 27 = 33 , 81 = 34 ; 25 = 52 , 125 = 53 . 1/8 = 2−3 and √2 = 21/2 — negative and fractional powers often appear here. No logarithms are needed (or expected) at this level. How to do it Choose a common base (usually 2, 3 or 5). Rewrite each side as basepower , using (am )n = amn . Set the powers equal and solve the linear equation. Worked example Solve 4x+1 = 8x .
Base 2: (22 )x+1 = (23 )x . 22x+2 = 23x . 2x + 2 = 3x. x = 2
Exam tip: If one side is a fraction like 1/27, write it as a negative power first: 3−3 .
✗ Common mistakesDon’t write Write instead ✗ 23 × 24 = 212 ✓ Add the powers: 27 ✗ 23 × 24 = 47 ✓ The base stays 2: 27 ✗ 4−2 = −16✓ 4−2 = 1/16✗ (3x2 )3 = 3x6 ✓ Cube the 3 too: 27x6 ✗ 50 = 0✓ 50 = 1✗ 82/3 = (82 )1/3 worked out the long way✓ Root first is easier: (∛8)2 = 4
★ In the examNon-calculator questions on indices are common — practise without a calculator. For 'find n' questions, write every number as a power of the same base before comparing. Check by substituting your answer back in.
1 Function notationf(x) = 3x − 4 is a rule: f(2) means put 2 in place of x: f(2) = 2. Solve f(x) = 11 by writing 3x − 4 = 11. f(x + 1) means replace x by (x + 1): 3(x + 1) − 4 = 3x − 1. 2f(x) = 2(3x − 4) = 6x − 8 but f(2x) = 6x − 4. Worked example g(x) = x2 − 2x. Find g(−3) and solve g(x) = 8.
g(−3) = 9 + 6 = 15. x2 − 2x − 8 = 0 → (x − 4)(x + 2) = 0. g(−3) = 15; x = 4 or x = −2
2 Inverse functionsf⁻¹ undoes f: if f(3) = 11 then f⁻¹(11) = 3. f⁻¹(x) is not 1/f(x). How to do it Write y = f(x). Swap x and y. Make y the subject. Write f⁻¹(x) = … Worked example f(x) = (2x − 5)/3. Find f⁻¹(x).
y = (2x − 5)/3; swap: x = (2y − 5)/3. 3x = 2y − 5. 2y = 3x + 5. f⁻¹(x) = (3x + 5)/2
Exam tip: Check: f⁻¹(f(x)) must give x back. Try a number.
3 Composite functionsfg(x) means f(g(x)): do g first , then f. Replace the x in f with the whole expression g(x), in brackets. ff(x) means apply f twice. fg and gf are usually different. Worked example f(x) = 2x + 1 and g(x) = x2 . Find fg(x), gf(x) and fg(3).
fg(x) = f(x2 ) = 2x2 + 1. gf(x) = g(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1. fg(3) = 2 × 9 + 1 = 19. fg(x) = 2x2 + 1, gf(x) = 4x2 + 4x + 1, fg(3) = 19
4 Domain, range and asymptotesDomain : the x values allowed. Leave out any value that makes a denominator zero.Range : the y values you get out. x2 is never negative, so the range of f(x) = x2 is f(x) ⩾ 0.y = k/(x − a) + b has asymptotes x = a (vertical) and y = b (horizontal). y = ax + b has the horizontal asymptote y = b. Worked example f(x) = 3/(x − 2) + 1. State the value excluded from the domain and the asymptotes.
x − 2 = 0 when x = 2. As x gets large, 3/(x − 2) → 0, so y → 1. x = 2 excluded; asymptotes x = 2 and y = 1
✗ Common mistakesDon’t write Write instead ✗ fg(2): working out f first✓ fg means f(g(2)): do g first✗ f⁻¹(x) = 1/f(x)✓ The inverse undoes f; it is not the reciprocal✗ gf(x) = 2x + 12 ✓ Use brackets: (2x + 1)2 ✗ 2f(x) and f(2x) treated as the same✓ 2f(x) doubles the output; f(2x) doubles the input
★ In the examUse brackets every time you substitute an expression. Check an inverse with a number: f(2) then f⁻¹ of the answer should give 2. For 'solve ff(x) = …' find ff(x) as an expression first.
1 Common factorsTake out the highest common factor of every term — numbers and letters. 'Factorise fully' means nothing more can be taken out. Check by expanding back. Worked example Factorise fully 12x2 y − 18xy2 .
HCF of 12 and 18 is 6; of x2 y and xy2 is xy. 12x2 y ÷ 6xy = 2x and 18xy2 ÷ 6xy = 3y. 6xy(2x − 3y)
2 Difference of two squaresa2 − b2 = (a − b)(a + b). Both terms must be squares and there must be a minus between them. 16x2 − 25y2 = (4x − 5y)(4x + 5y). Take out a common factor first: 2x2 − 50 = 2(x2 − 25) = 2(x − 5)(x + 5). a2 + b2 does not factorise. Formulas a2 − b2 = (a − b)(a + b)
Worked example Factorise fully 3x2 − 75.
Common factor 3: 3(x2 − 25). x2 − 25 = (x − 5)(x + 5). 3(x − 5)(x + 5)
3 Quadratics x2 + bx + cFind two numbers that multiply to c and add to b . If c is positive, both numbers have the sign of b. If c is negative, the signs are different. Expression Numbers Factorised x2 + 7x + 12 3, 4 (x + 3)(x + 4) x2 − 7x + 12 −3, −4 (x − 3)(x − 4) x2 + x − 12 4, −3 (x + 4)(x − 3) x2 − x − 12 −4, 3 (x − 4)(x + 3)
Worked example Factorise x2 + 2x − 15.
Multiply to −15, add to 2: 5 and −3. (x + 5)(x − 3)
4 Quadratics ax2 + bx + cUse the ac method : find two numbers that multiply to a × c and add to b. Split the middle term with them, then factorise in pairs. ac method Multiply a × c. Find two numbers with that product and sum b. Rewrite bx as the two terms. Factorise the first pair and the second pair. Take out the common bracket. Worked example Factorise 6x2 + x − 2.
ac = −12; numbers 4 and −3. 6x2 + 4x − 3x − 2. 2x(3x + 2) − 1(3x + 2). (2x − 1)(3x + 2)
5 Four terms: groupingGroup into two pairs, factorise each pair, then take out the common bracket. If the brackets don't match, try grouping the terms differently. Worked example Factorise xy − 3x + 2y − 6.
x(y − 3) + 2(y − 3). (x + 2)(y − 3)
6 Solving by factorisingRearrange to = 0, factorise, then set each factor equal to 0. x(x − 5) = 0 gives x = 0 or x = 5 — never divide both sides by x. (2x + 3)(x − 3) = 0 gives x = −3/2 or x = 3. Worked example Solve 2x2 − 3x − 9 = 0.
ac = −18: −6 and 3. 2x2 − 6x + 3x − 9 = 2x(x − 3) + 3(x − 3). (2x + 3)(x − 3) = 0. x = 3 or x = −1.5
✗ Common mistakesDon’t write Write instead ✗ 16x2 − 25 = (4x − 5)2 ✓ That expands to 16x2 − 40x + 25. Use (4x − 5)(4x + 5)✗ 6x2 + 9x = 3(2x2 + 3x)✓ Take out 3x as well: 3x(2x + 3)✗ (x + 5)(x − 3) = 0 → x = 5 or −3✓ Change the signs: x = −5 or 3✗ Dividing x2 = 5x by x✓ Rearrange: x(x − 5) = 0, so x = 0 or 5
★ In the examAlways check a factorisation by expanding it in your head. If a quadratic will not factorise, the question probably wants the formula. 'Hence solve' means use the factorisation you have just found.
1 Averages and rangeMean = total ÷ number of values. Median = middle value in order (average the middle two if there is an even count).Mode = most common value. Range = largest − smallest (a measure of spread).From a frequency table: mean = Σfx ÷ Σf; median position = (n + 1) ÷ 2. Missing value from a mean: total = mean × number of values. Formulas mean = Σfx ÷ Σf frequency table
median position = (n + 1) ÷ 2
Worked example Scores 1, 2 and 3 have frequencies 4, 5 and 1. Find the mean and the median.
Σfx = 4 + 10 + 3 = 17; Σf = 10. Mean = 1.7. Median: position 5.5 — the 5th and 6th values are both 2. Mean 1.7, median 2
2 Grouped dataYou don't know the exact values, so use each class's midpoint : the mean is an estimate . The modal class is the class with the highest frequency (for unequal widths, the highest frequency density). The class containing the median: find the (n/2)th value by adding up frequencies. Worked example Heights 150–160 cm (f = 4), 160–170 (f = 10), 170–180 (f = 6). Estimate the mean.
Midpoints 155, 165, 175. Σfx = 620 + 1650 + 1050 = 3320; Σf = 20. Estimated mean 166 cm
3 Charts and diagramsBar charts (gaps, equal widths), pictograms (with a key), pie charts (angle = frequency ÷ total × 360°). Stem-and-leaf: always give a key (3 | 4 = 34); leaves in order; read the median by counting. Compare two distributions with an average AND a measure of spread, in context. Worked example A stem-and-leaf diagram has key 3 | 4 = 34. What does 5 | 1 mean?
Stem = tens, leaf = units. 51
4 HistogramsWith unequal class widths the vertical axis is frequency density = frequency ÷ class width. The area of each bar is the frequency. To read a frequency: frequency density × class width. For part of a class, take that fraction of the bar. Formulas frequency density = frequency ÷ class width
frequency = density × width
Worked example A bar for 10 < x ⩽ 25 has frequency density 2. A bar for 25 < x ⩽ 30 has frequency 15. Find the first frequency and the second density.
Width 15: 2 × 15 = 30. Width 5: 15 ÷ 5 = 3. Frequency 30; density 3
5 Cumulative frequency, quartiles and IQRCumulative frequency is a running total. Plot each total at the upper end of its class and join with a smooth curve. Read the median at n/2, the lower quartile at n/4, the upper quartile at 3n/4 (on the cumulative frequency axis). Interquartile range = UQ − LQ: the spread of the middle half, not affected by extremes. 'How many more than x?' — read the cumulative frequency at x and subtract from the total. How to do it Make a running total column. Plot (upper class boundary, running total). Join with a smooth curve. Read across from n/4, n/2, 3n/4 and down to the x-axis. Worked example 80 students: the graph gives LQ = 31 marks and UQ = 46 marks. Find the IQR. 52 students scored 40 or less: how many scored more than 40?
IQR = 46 − 31 = 15. 80 − 52 = 28. IQR 15 marks; 28 students
6 Scatter diagramsCorrelation: positive (both increase), negative (one up, one down), or none. Line of best fit: one ruled line following the trend, with points roughly balanced either side; it need not pass through (0, 0). Use it to estimate within the data (interpolation). Estimates outside the data (extrapolation) are unreliable. An outlier is a point that doesn't fit the pattern. Worked example As temperature rises, sales of hot drinks fall. Describe the correlation.
One goes up while the other goes down. Negative correlation
✗ Common mistakesDon’t write Write instead ✗ Median of 3, 4, 5, 7, 9, 12 is 5✓ Even count: average the middle two, 6✗ Giving the frequency as the modal class✓ Give the class, e.g. 10–20✗ Plotting cumulative frequency at midpoints✓ Plot at the upper end of each class✗ Histogram bar height = frequency with unequal widths✓ Height = frequency density✗ Dividing Σfx by the number of classes✓ Divide by Σf, the total frequency
★ In the examShow Σfx and Σf in a table — they earn method marks. When comparing data, compare an average AND a spread, and say what it means in context. Read graphs carefully — check the scale on each axis first.
1 Pythagoras' theoremIn a right-angled triangle: a2 + b2 = c2 , where c is the hypotenuse (opposite the right angle, the longest side). Finding the hypotenuse: square, add , square root. A shorter side: square, subtract , square root. Pythagorean triples to recognise: 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and multiples like 6-8-10). Formulas a2 + b2 = c2 c = hypotenuse
Worked example A ladder 6.5 m long leans against a wall with its foot 2.5 m from the wall. How high up the wall does it reach?
Hypotenuse 6.5: h2 = 6.52 − 2.52 . = 42.25 − 6.25 = 36. 6 m
2 Sin, cos and tan (SOH CAH TOA)Label the sides from the angle: O pposite, A djacent (next to the angle), H ypotenuse. Choose the ratio that uses the two sides involved (one known, one wanted). Finding an angle: use the inverse (sin⁻¹, cos⁻¹, tan⁻¹). Calculator in degrees. 0° 30° 45° 60° 90° sin 0 ½ √2/2 √3/2 1 cos 1 √3/2 √2/2 ½ 0 tan 0 √3/3 1 √3 —
Formulas sin θ = O/H
cos θ = A/H
tan θ = O/A
Worked example In a right-angled triangle the angle is 52° and the adjacent side is 8 cm. Find the opposite side.
O and A → tan. O = 8 × tan 52° = 10.239… 10.2 cm (3 s.f.)
3 Elevation, depression and bearingsAngle of elevation : measured up from the horizontal. Angle of depression : down from the horizontal.The angle of depression from A to B equals the angle of elevation from B to A (alternate angles). Bearings questions: draw north lines, mark the bearings, find a right-angled (or other) triangle. The shortest distance from a point to a line is the perpendicular distance. Worked example From a point 20 m from the foot of a tower, the angle of elevation of the top is 35°. Find the height.
tan 35° = h ÷ 20. h = 20 × tan 35° = 14.0… 14.0 m (3 s.f.)
4 Sine rule and cosine ruleFor triangles without a right angle. Label sides a, b, c opposite angles A, B, C. Sine rule : use when you know a side and its opposite angle (plus one more).Cosine rule : use with two sides and the angle between them (to find the third side) or all three sides (to find an angle).A negative cosine means an obtuse angle. sin x = sin(180° − x): the sine rule may give an obtuse answer (the ambiguous case). Formulas a/sin A = b/sin B = c/sin C sine rule
a2 = b2 + c2 − 2bc cos A cosine rule (side)
cos A = (b2 + c2 − a2 ) / 2bc cosine rule (angle)
Worked example Triangle PQR: PQ = 6 cm, PR = 9 cm, angle P = 50°. Find QR.
QR2 = 62 + 92 − 2 × 6 × 9 × cos 50°. = 117 − 69.42 = 47.58. QR = 6.90 cm (3 s.f.)
5 Area of any triangleArea = ½ab sin C, where C is the angle between sides a and b. Use it backwards to find an angle or a side when the area is given. Worked example A triangle has sides 8 cm and 10 cm with 30° between them. Find its area.
½ × 8 × 10 × sin 30°. = 40 × 0.5. 20 cm²
✗ Common mistakesDon’t write Write instead ✗ Using SOH CAH TOA without a right angle✓ Use the sine or cosine rule✗ Forgetting to square root at the end of Pythagoras or the cosine rule✓ c2 = 49 → c = 7✗ Adding in Pythagoras when finding a shorter side✓ Shorter side: subtract✗ Calculator in radians✓ Set it to degrees✗ Giving only the acute angle when the triangle is obtuse✓ The obtuse angle is 180° − acute
★ In the examDraw and label the triangle, marking O, A, H or a, b, c, A, B, C. Keep full calculator values between steps; round only at the end (angles to 1 d.p.). Non-calculator paper: use the exact-value table.
At a glance Units, areas and perimeters, circles and sectors, surface area and volume of every solid, and working backwards.
Volumes of a cone, pyramid and sphere, and the curved surface areas of a cone and sphere, are on the formula list; the rest are not.
Units Area and perimeter of 2D shapes Circles, arcs, sectors and segments Prisms, cuboids and cylinders Pyramids, cones and spheres Working backwards 1 UnitsLength: 1 km = 1000 m; 1 m = 100 cm; 1 cm = 10 mm. Area: square the conversion — 1 m² = 1002 = 10 000 cm²; 1 cm² = 100 mm². Volume: cube it — 1 m³ = 1003 = 1 000 000 cm³. Capacity: 1 litre = 1000 cm³ = 1000 ml; 1 cm³ = 1 ml. Convert Length Area Volume m → cm × 100 × 10 000 × 1 000 000 cm → mm × 10 × 100 × 1000 km → m × 1000 × 1 000 000 —
Worked example Convert 3.5 m² to cm².
1 m² = 10 000 cm². 3.5 × 10 000. 35 000 cm²
2 Area and perimeter of 2D shapesPerimeter = distance all the way round. Area = space inside. Use the perpendicular height for triangles, parallelograms and trapezia. Compound shapes: split into rectangles, triangles and semicircles, then add (or subtract a hole). Formulas rectangle A = lw
triangle A = ½bh
parallelogram A = bh
trapezium A = ½(a + b)h
Worked example A semicircle of diameter 8 cm sits on top of an 8 cm × 5 cm rectangle. Find the total area.
Rectangle: 8 × 5 = 40 cm². Semicircle: ½ × π × 42 = 8π = 25.13 cm². 65.1 cm² (3 s.f.)
3 Circles, arcs, sectors and segmentsCircumference = πd = 2πr. Area = πr2 . A sector is θ/360 of the circle: arc length = θ/360 × 2πr; sector area = θ/360 × πr2 . Perimeter of a sector = arc + 2 radii. Segment area = sector area − triangle area (triangle = ½r2 sin θ). Formulas C = 2πr
A = πr2
arc = (θ/360) × 2πr
sector = (θ/360) × πr2
Worked example A sector has radius 10 cm and angle 90°. Find the area of the minor segment.
Sector: ¼ × π × 100 = 78.54 cm². Triangle: ½ × 10 × 10 = 50 cm². 78.54 − 50. 28.5 cm² (3 s.f.)
Exam tip: 'In terms of π' means leave π in: the sector area 12π, not 37.7.
4 Prisms, cuboids and cylindersVolume of a prism = area of cross-section × length. A cylinder is a prism with a circular cross-section. Surface area = total area of all faces (draw the net if unsure). Cylinder curved surface = 2πrh (the rectangle that wraps round); total = 2πrh + 2πr2 . Formulas prism V = cross-section area × length
cylinder V = πr2 h
cylinder curved SA = 2πrh
Worked example A cylinder has radius 3 cm and height 10 cm. Find its volume and total surface area in terms of π.
V = π × 9 × 10 = 90π cm³. Curved: 2π × 3 × 10 = 60π; two ends: 2 × 9π = 18π. V = 90π cm³; SA = 78π cm²
5 Pyramids, cones and spheresPyramid and cone volume = ⅓ × base area × height. Cone curved surface = πrl (l = slant height; l2 = r2 + h2 ). Sphere: V = ⁴⁄₃πr3 , SA = 4πr2 . Hemisphere: half of each, plus the flat circle πr2 for total SA. Frustum = big cone − small cone. Formulas pyramid/cone V = ⅓ × base × h
cone curved SA = πrl
sphere V = ⁴⁄₃πr3
sphere SA = 4πr2
Worked example A cylinder (radius 2 cm, height 5 cm) has a hemisphere of radius 2 cm on top. Find the volume.
Cylinder: π × 4 × 5 = 20π. Hemisphere: ½ × ⁴⁄₃ × π × 8 = 16π/3. Total = 76π/3 = 79.6. 79.6 cm³ (3 s.f.)
Exam tip: The formulas for cone, pyramid and sphere are given on the formula list — but you must know which to use.
6 Working backwardsGiven the volume or area, substitute and solve for the missing length. Rearrange before using the calculator: r2 = V ÷ (πh), then square root. Equal volumes: liquid poured from one container into another keeps its volume. Worked example A cylinder has volume 500 cm³ and height 8 cm. Find its radius.
500 = πr2 × 8. r2 = 500 ÷ (8π) = 19.89… r = √19.89… 4.46 cm (3 s.f.)
✗ Common mistakesDon’t write Write instead ✗ Using the diameter as the radius✓ Halve the diameter first✗ Cone volume πr2 h✓ It's ⅓πr2 h✗ 1 m² = 100 cm²✓ 1 m² = 10 000 cm²✗ Sector perimeter = arc length only✓ Add the two radii✗ Using the slant height as h in a cone's volume✓ Volume uses the vertical height
★ In the examWrite the formula, substitute, then calculate — each step can earn a mark. Keep π in until the end (or use the π key), then round. Check units: area in cm², volume in cm³.
1 12- and 24-hour clocks24-hour times use four digits: 07 45, 19 45. Midnight is 00 00; noon is 12 00. pm → 24-hour: add 12 to the hours (except 12 pm). am times stay the same (12 am = 00 00). 24-hour → 12-hour: subtract 12 from hours above 12 and write pm. 12-hour 24-hour 12:30 am 00 30 7:45 am 07 45 12:15 pm 12 15 7:45 pm 19 45 11:59 pm 23 59
Worked example Write 4:05 pm in the 24-hour clock.
pm and after 1 pm → add 12. 4 + 12 = 16. 16 05
2 Time intervals and decimal hoursCount up on a time line: to the next whole hour, then whole hours, then the minutes left. Never subtract times like decimals: 21 15 − 18 50 is not 2.65. Decimal hours → minutes: multiply the decimal part by 60. 2.35 h = 2 h 21 min. Minutes → decimal hours: divide by 60. 45 min = 0.75 h; 48 min = 0.8 h. Time line method From the start time, count to the next o'clock. Count the whole hours. Add the minutes at the end. Add the pieces together. Worked example A bus leaves at 22 40 and arrives at 03 15 the next day. How long is the journey?
22 40 → 23 00: 20 min. 23 00 → 03 00: 4 h. 03 00 → 03 15: 15 min. 4 h 35 min
3 Timetables and time zonesRead a timetable along the row for the stop and down the column for one bus or train. Arrival = departure + journey time. A place that is ahead is later there: add the difference. Behind: subtract. Flight problems: work in one time zone first, then convert at the end. Worked example Karachi is 5 h ahead of London. A flight leaves London at 22 00 (London time) and takes 8 h. What is the local time in Karachi on landing?
Landing in London time: 22 00 + 8 h = 06 00. Karachi is 5 h ahead: 06 00 + 5 h. 11 00 (next day)
Exam tip: Watch for journeys that cross midnight — the date changes too.
✗ Common mistakesDon’t write Write instead ✗ 2.35 h = 2 h 35 min✓ 0.35 × 60 = 21 min → 2 h 21 min✗ 21 15 − 18 50 = 2 h 65 min✓ Count up: 2 h 25 min✗ 12:30 am = 12 30✓ 12:30 am is just after midnight: 00 30✗ Using 100 minutes in an hour✓ There are 60
★ In the examShow your time line — it earns marks even if you slip in adding. Change minutes to hours (÷ 60) before using them in a speed formula. Give answers in the form asked: 24-hour clock, hours and minutes, or just minutes.
1 Speed, distance and timespeed = distance ÷ time; distance = speed × time; time = distance ÷ speed. Average speed = total distance ÷ total time, including stops. It is not the average of the speeds.Convert units first: 45 minutes = 0.75 h; km/h ÷ 3.6 = m/s. Formulas speed = distance ÷ time
km/h → m/s ÷ 3.6
m/s → km/h × 3.6
Worked example A cyclist rides 30 km in 1 h, rests for 30 min, then rides 20 km in 1 h. Find the average speed.
Total distance 50 km. Total time 2.5 h (the rest counts). 50 ÷ 2.5. 20 km/h
2 Distance–time graphsGradient = speed. Steeper = faster. Horizontal line = stationary. Line sloping down = returning towards the start. A curve means the speed is changing; use a tangent for the speed at an instant. Worked example A straight line on a distance–time graph goes from (0 s, 0 m) to (40 s, 200 m). Find the speed.
Gradient = 200 ÷ 40. 5 m/s
3 Speed–time graphsGradient = acceleration (m/s²). Negative gradient = deceleration. Horizontal line = constant speed (not stopped!). Area under the graph = distance travelled. Split into triangles, rectangles and trapezia.Formulas acceleration = change in speed ÷ time gradient
distance = area under the graph
trapezium area = ½(a + b)h
Worked example A bus accelerates from 0 to 12 m/s in 8 s, keeps 12 m/s for 20 s, then stops in 6 s. How far does it go?
The shape is a trapezium with parallel sides 34 s and 20 s, height 12. Area = ½ × (34 + 20) × 12. 324 m
4 Conversion graphs and ratesA straight-line conversion graph through the origin means direct proportion: find the rate from one point. Use it to convert in either direction; for values off the graph, scale up. The gradient of any real-life graph is a rate (e.g. litres per minute, cost per kg). Worked example 10 miles = 16 km. Convert 25 miles to km.
1 mile = 1.6 km. 25 × 1.6. 40 km
✗ Common mistakesDon’t write Write instead ✗ 2 h 30 min = 2.3 h✓ 30 min = 0.5 h, so 2.5 h✗ Averaging the speeds of two parts✓ Average speed = total distance ÷ total time✗ Reading distance from a speed–time graph's axis✓ Distance is the AREA under it✗ A flat line on a speed–time graph means stopped✓ It means constant speed
★ In the examWrite the units: m/s, m/s², km/h. Split areas carefully and show each part. If the graph has different time units (minutes and hours), convert before calculating.
1 Circle wordsRadius : centre to edge. Diameter : across through the centre (2 × radius).Chord : a line joining two points on the circle. Tangent : a line touching the circle at one point.Arc : part of the circumference. Sector : a 'pizza slice'. Segment : area between a chord and the arc.Worked example What is the line from the centre to the point where a tangent touches?
It is a radius. A radius (it meets the tangent at 90°)
2 Right angles in circlesThe angle in a semicircle is 90° (angle at the circumference standing on a diameter). A tangent meets the radius at 90°. Tangents from an external point are equal in length (look for an isosceles triangle or a kite).The perpendicular from the centre to a chord bisects the chord. Worked example AB is a diameter and C is on the circle. Angle CAB = 35°. Find angle CBA.
Angle ACB = 90° (angle in a semicircle). 180 − 90 − 35. 55°
3 Angles at the centre and circumferenceThe angle at the centre is twice the angle at the circumference standing on the same arc. Angles in the same segment (standing on the same chord, same side) are equal.Opposite angles of a cyclic quadrilateral add to 180°.Worked example O is the centre. Angle AOC = 140°. B is on the major arc and D on the minor arc. Find angle ADC.
Angle ABC = 70° (angle at the centre is twice the angle at the circumference). ABCD is cyclic, so angle ADC = 180 − 70. 110°
4 Alternate segment theoremThe angle between a tangent and a chord equals the angle in the alternate segment (on the other side of the chord). Spot it: a tangent, a chord from the point of contact, and a triangle drawn in the circle. Worked example A tangent at T makes 58° with chord TA. B is on the circle in the alternate segment. Find angle TBA.
Alternate segment theorem. 58°
5 Writing reasonsName the theorem every time: 'angle in a semicircle', 'angle at the centre is twice the angle at the circumference', 'opposite angles of a cyclic quadrilateral add to 180°', 'tangent perpendicular to radius', 'alternate segment theorem'. Radii are equal → isosceles triangles: use them to find base angles. Where to start Mark radii and tangents on the diagram. Look for right angles first. Look for isosceles triangles. Then centre/circumference, same segment, cyclic quadrilateral, alternate segment. Worked example Tangents from P touch the circle at A and B. Angle APB = 50°. Find angle PAB.
PA = PB (tangents from a point). Triangle PAB is isosceles: (180 − 50) ÷ 2. 65°
✗ Common mistakesDon’t write Write instead ✗ Angle at circumference = 2 × angle at centre✓ It's the other way: centre = 2 × circumference✗ Cyclic quadrilateral rules with the centre as a corner✓ All four corners must be on the circle✗ Giving the angle without a reason✓ Each step needs the theorem named✗ Assuming a line is a diameter✓ Only if it goes through the centre
★ In the examDiagrams are not to scale — use theorems, never measure. Find the easy angles first (right angles, isosceles), then the theorems. Learn the exact wording of each theorem for the 'give a reason' marks.
1 EliminationMake the coefficient of one letter the same in both equations (multiply one or both equations). Same signs → subtract ; different signs → add .Solve for the remaining letter, then substitute back to find the other. Check both values in the equation you didn't substitute into. How to do it Label the equations (1) and (2). Multiply to match one coefficient. Add or subtract to eliminate it. Solve; substitute back; check. Worked example Solve 3x + 2y = 19 and 2x − 5y = −19.
(1) × 5: 15x + 10y = 95. (2) × 2: 4x − 10y = −38. Add: 19x = 57, so x = 3. 3(3) + 2y = 19 → y = 5. Check (2): 6 − 25 = −19 ✓ x = 3, y = 5
2 SubstitutionBest when one equation already says y = … or x = … Replace that letter in the other equation, solve, then substitute back. Worked example Solve y = 3x − 4 and 2x + y = 11.
2x + (3x − 4) = 11. 5x = 15, x = 3. y = 3(3) − 4 = 5. x = 3, y = 5
3 Forming simultaneous equationsChoose a letter for each unknown and say what it means. Write one equation for each fact in the question. Answer in words with units at the end. Worked example 2 adult tickets and 3 child tickets cost $31. 1 adult and 1 child ticket cost $12. Find the price of each.
2a + 3c = 31 and a + c = 12. From the second, a = 12 − c. 2(12 − c) + 3c = 31 → c = 7, a = 5. Adult $5, child $7
4 One linear and one non-linear equationRearrange the linear equation to y = … (or x = …) and substitute into the non-linear one. This gives a quadratic: solve it to get two x values. Find the matching y for each x using the linear equation, and pair them. Graphically, the solutions are where the line meets the curve (0, 1 or 2 points). Worked example Solve y = x + 1 and x2 + y2 = 25.
x2 + (x + 1)2 = 25 → 2x2 + 2x − 24 = 0. x2 + x − 12 = 0 → (x + 4)(x − 3) = 0. x = 3 → y = 4; x = −4 → y = −3. (3, 4) and (−4, −3)
✗ Common mistakesDon’t write Write instead ✗ Subtracting when the signs are different✓ Different signs: add the equations✗ −2y − (−2y) = −4y✓ −2y − (−2y) = 0✗ Giving only the x values✓ Find each y and pair it with its x✗ Squaring (x + 1)2 as x2 + 1✓ (x + 1)2 = x2 + 2x + 1
★ In the examWrite what you do to each equation: (1) × 2, (2) × 3. Check your pair in the original equations — a quick way to catch sign errors. In word problems, state the answer in words with units.
1 Coordinates and straight-line graphs(x, y): across first, then up. Points can be in any of the four quadrants. Draw a line from its equation with a table of three x values (three points catch mistakes). x = a is a vertical line; y = b is a horizontal line. A line crosses the y-axis where x = 0 and the x-axis where y = 0. Worked example Find where 3x + 2y = 12 crosses the axes.
x = 0: 2y = 12 → y = 6. y = 0: 3x = 12 → x = 4. (0, 6) and (4, 0)
2 GradientGradient = change in y ÷ change in x (rise over run). Uphill left to right → positive; downhill → negative; horizontal → 0; vertical → undefined. From a graph: pick two points exactly on grid corners, far apart. Formulas m = (y₂ − y₁) / (x₂ − x₁) gradient between two points
x y (1, 2) (4, 8) run 3 rise 6 Rise 6, run 3: gradient 6 ÷ 3 = 2.
Worked example Find the gradient of the line through (−1, 7) and (3, −5).
Change in y: −5 − 7 = −12. Change in x: 3 − (−1) = 4. −12 ÷ 4. −3
3 Length and midpointMidpoint: average the x's and average the y's. Length: draw the right-angled triangle and use Pythagoras on the run and the rise. Formulas M = ((x₁ + x₂)/2, (y₁ + y₂)/2) midpoint
d = √((x₂ − x₁)2 + (y₂ − y₁)2 ) length
Worked example A is (−2, 1) and B is (4, 9). Find the midpoint and the length of AB.
Midpoint: ((−2 + 4)/2, (1 + 9)/2) = (1, 5). Run 6, rise 8: √(36 + 64) = √100. Midpoint (1, 5); length 10
4 The equation y = mx + cm is the gradient; c is where the line cuts the y-axis. Rearrange to y = … before reading m: 2y = 6x + 4 → y = 3x + 2, so m = 3. Through a point with gradient m: substitute the point into y = mx + c to find c (or use y − y₁ = m(x − x₁)). Through two points: find m first, then c. How to do it Find m (from two points if needed). Substitute one point into y = mx + c. Solve for c. Write y = mx + c. Worked example Find the equation of the line through (2, 7) and (5, 13).
m = (13 − 7) ÷ (5 − 2) = 2. 7 = 2(2) + c → c = 3. y = 2x + 3
5 Parallel and perpendicular linesParallel lines have the same gradient . Perpendicular gradients multiply to −1: flip the fraction and change the sign (2/3 → −3/2). Perpendicular bisector of AB: passes through the midpoint of AB, with gradient perpendicular to AB. Formulas m₁ = m₂ parallel
m₁ × m₂ = −1 perpendicular
Worked example A(1, 2) and B(5, 4). Find the equation of the perpendicular bisector of AB.
Midpoint (3, 3). Gradient of AB = 2/4 = 1/2, so perpendicular gradient −2. 3 = −2(3) + c → c = 9. y = −2x + 9
✗ Common mistakesDon’t write Write instead ✗ Run over rise: (x₂ − x₁)/(y₂ − y₁)✓ Always y on top: change in y ÷ change in x✗ Reading m from 2y = 6x + 4 as 6✓ Make y the subject first: y = 3x + 2, so m = 3✗ Perpendicular to m = 2 written as −2✓ Flip it too: −1/2✗ 3 − (−2) = 1✓ 3 − (−2) = 5. Put negatives in brackets
★ In the examNone of the gradient, midpoint or length formulas are given — learn them. Sketch the points roughly first: it shows whether the gradient should be positive or negative. Leave a length as a surd if the question asks for an exact answer.
1 Term-to-term rulesThe term-to-term rule says how to get from one term to the next: 3, 7, 11, 15 → add 4. Arithmetic (linear) sequences add the same each time; geometric ones multiply by the same each time. Look at the differences: constant first difference → linear; constant second difference → quadratic. Worked example Find the next two terms of 2, 6, 18, 54, …
Each term × 3. 162, 486
2 Linear sequences: the nth termnth term = dn + (first term − d) , where d is the common difference. To test whether a number is a term, set the nth term equal to it: a whole-number n means yes. To find which term a value is, solve nth term = value. Formulas nth term = dn + (a − d) a = first term, d = difference
Worked example Find the nth term of 7, 11, 15, 19, … Is 101 a term?
d = 4: 4n. 4n gives 4, 8, 12…; add 3. nth term 4n + 3. 4n + 3 = 101 → n = 24.5, not a whole number. 4n + 3; 101 is not a term
3 Special sequencesRecognise these and their shifts (+ or −) and multiples: Sequence Terms nth term Square 1, 4, 9, 16, 25 n2 Cube 1, 8, 27, 64, 125 n3 Triangle 1, 3, 6, 10, 15 n(n + 1)/2 Powers of 2 2, 4, 8, 16, 32 2n Square + 1 2, 5, 10, 17, 26 n2 + 1 Twice square 2, 8, 18, 32, 50 2n2
Worked example Find the nth term of 0, 3, 8, 15, 24, …
Compare with squares 1, 4, 9, 16, 25: each is 1 less. n2 − 1
4 Quadratic sequencesIf the second difference is constant, the nth term is an2 + bn + c. a = second difference ÷ 2. How to do it Find the first and second differences. a = second difference ÷ 2. Subtract an2 from each term. Find the nth term of what is left (linear) and add it on. Worked example Find the nth term of 4, 10, 18, 28, 40, …
Differences 6, 8, 10, 12; second difference 2, so a = 1. Subtract n2 (1, 4, 9, 16, 25): 3, 6, 9, 12, 15 = 3n. n2 + 3n
5 Cubic and exponential sequencesConstant third difference → cubic; often a simple form like n3 + 1. A constant ratio → exponential: nth term = a × rn−1 (a = first term, r = ratio). Combinations: compare with the basic sequence and spot the change. Formulas nth term = a × rn−1 geometric (exponential)
Worked example Find the nth term of 5, 10, 20, 40, …
Ratio r = 2, first term 5. 5 × 2n−1
6 Patterns of shapesCount the items in the first few patterns and write them as a sequence. Find the nth term, then use it to predict any pattern or to find which pattern uses a given number. Worked example Pattern n uses 3n + 1 matchsticks. Which pattern uses 100 matchsticks?
3n + 1 = 100. 3n = 99. Pattern 33
✗ Common mistakesDon’t write Write instead ✗ nth term of 5, 8, 11 is n + 3✓ The difference is 3, so 3n + 2✗ Using the second difference as a✓ a is HALF the second difference✗ Saying 101 is the 24.5th term✓ n must be a whole number, so 101 is not a term✗ 5 × 2n for 5, 10, 20, 40✓ Check n = 1: 5 × 2n−1 gives 5
★ In the examAlways check your nth term with n = 1 and n = 2. Show the differences — they earn method marks. Give the nth term in simplest form (3n + 2, not 3n + 5 − 3).
1 Solving equations from a graphSolutions of f(x) = 0 are the x values where the curve crosses the x-axis . Solutions of f(x) = k: draw the horizontal line y = k and read the x values where it meets the curve. The number of crossing points is the number of solutions. Worked example The graph of y = x2 − x − 6 is drawn. Solve x2 − x − 6 = 4.
Draw y = 4. Read the x values where the line meets the curve. about x = −2.7 and x = 3.7
2 Finding the line to drawYou have the graph of one curve and must solve a different equation. Rearrange the new equation so one side is exactly the curve you have; the other side is the line to draw. How to do it Write the curve's expression (e.g. x2 − 3x). Rearrange the new equation to: that expression = something. Draw y = something. Read the x values of the crossing points. Worked example The graph of y = x2 − 3x is drawn. What line should you draw to solve x2 − 5x + 2 = 0?
Add 2x − 2 to both sides: x2 − 3x = 2x − 2. Draw y = 2x − 2
Exam tip: Check by subtracting: x2 − 3x − (2x − 2) should give back x2 − 5x + 2.
3 Simultaneous equations on a graphThe solution is the point where the two graphs cross: read both x and y. A line and a curve can cross 0, 1 or 2 times (a cubic up to 3). Worked example Where do y = 2x + 1 and y = 7 − x meet?
2x + 1 = 7 − x → x = 2. y = 2(2) + 1 = 5. (2, 5)
4 Gradient of a curveDraw a tangent at the point: a straight line that touches the curve there without cutting it. Pick two points far apart on the tangent; gradient = rise ÷ run. On a distance–time graph the gradient is speed; on a speed–time graph it is acceleration. How to do it Draw the tangent with a ruler. Choose two points on it, far apart. Gradient = change in y ÷ change in x. Worked example A tangent passes through (1, 3) and (4, 12). Find its gradient.
Rise = 12 − 3 = 9. Run = 4 − 1 = 3. Gradient 3
✗ Common mistakesDon’t write Write instead ✗ Reading y values when x values are asked✓ Solutions of an equation in x are x values✗ Drawing y = x − 3 for x2 + x − 3 = 0 with y = x2 + 2x✓ Add the same to both sides: y = x + 3✗ Using two points on the curve for a gradient✓ Use two points on the tangent✗ A tangent that cuts through the curve✓ It must touch at one point only
★ In the examDraw lines long enough to meet the curve everywhere it can. State the equation of any line you draw. Gradient estimates are given a range of accepted answers — draw carefully and read accurately.
1 Symbols and number lines< less than, > greater than, ⩽ less than or equal to, ⩾ greater than or equal to. On a number line: open circle for < or >; filled circle for ⩽ or ⩾. Double inequality −2 < x ⩽ 3 is drawn between the two values: open at −2, filled at 3. Inequality Circle Arrow x > 2 open at 2 right x ⩾ 2 filled at 2 right x < −1 open at −1 left −2 < x ⩽ 3 open at −2, filled at 3 between
Worked example List the integers n with −2 < n ⩽ 3.
−2 is not included; 3 is. −1, 0, 1, 2, 3
2 Solving inequalitiesSolve like an equation: add, subtract, multiply or divide both sides. Multiplying or dividing by a negative number reverses the sign. Double inequalities: do the same to all three parts. Worked example Solve −3 ⩽ 2x + 1 < 7 and list the integer solutions.
Subtract 1: −4 ⩽ 2x < 6. Divide by 2: −2 ⩽ x < 3. x = −2, −1, 0, 1, 2
Exam tip: To avoid flipping, move the x term to the side where it is positive: 4 − 2x ⩽ 10 → −6 ⩽ 2x → x ⩾ −3.
3 Regions on a graphDraw each boundary line: solid for ⩽ or ⩾, broken for < or >. Pick a test point not on the line (usually (0, 0)) to decide which side you want. Read the question: shade the region you want, or the region you don't want (unwanted region). y ⩾ 1 is above the line y = 1; x ⩽ 3 is to the left of x = 3. How to do it Draw the line (make a table or use y = mx + c). Decide solid or broken. Test a point: does it make the inequality true? Shade the correct side. Worked example The region R satisfies x ⩾ 1, y ⩾ 2 and x + y ⩽ 8. Find the greatest value of 3x + 2y for integer points in R.
Corners of R: (1, 2), (6, 2), (1, 7). 3x + 2y: 7, 22 and 17. 22, at (6, 2)
Exam tip: Maximum and minimum values of an expression in a region are always at or near a corner.
✗ Common mistakesDon’t write Write instead ✗ −2x ⩽ 6 → x ⩽ −3✓ Dividing by −2 flips the sign: x ⩾ −3✗ Including −2 in −2 < n✓ < means −2 is not included✗ Drawing y < x + 3 with a solid line✓ Strict inequality → broken line✗ Shading the wanted region when the question says shade the unwanted one✓ Read which region to shade
★ In the examLabel every boundary line with its equation. Use (0, 0) as a test point when it is not on the line — the arithmetic is easy. When listing integers, check the end values against the symbols.
1 Line symmetryA line of symmetry splits a shape into two mirror images. Test by folding (or with a mirror or tracing paper). A regular polygon with n sides has n lines of symmetry. Worked example How many lines of symmetry does a regular hexagon have?
Regular n-gon: n lines. 6
2 Rotational symmetryThe order is how many times a shape looks the same in one full turn. Order 1 means no rotational symmetry (it only matches itself after 360°). A regular n-gon has rotational symmetry of order n. Worked example What is the order of rotational symmetry of a parallelogram?
It looks the same after a half turn (180°) and a full turn. Order 2
3 Symmetry of triangles and quadrilateralsKnow these well — they are used to find missing angles and lengths. Shape Lines Order Equilateral triangle 3 3 Isosceles triangle 1 1 Square 4 4 Rectangle 2 2 Rhombus 2 (the diagonals) 2 Parallelogram 0 2 Kite 1 1 Isosceles trapezium 1 1
Worked example Which quadrilateral has rotational symmetry of order 2 but no lines of symmetry?
Rectangle and rhombus have lines; parallelogram has none. Parallelogram
4 Symmetry in solidsA plane of symmetry cuts a solid into two mirror halves. Cuboid (all edges different): 3 planes. Cube: 9 planes. Prisms, cylinders, pyramids and cones: picture slicing through the middle. Worked example How many planes of symmetry does a cuboid with three different edge lengths have?
One plane through the middle of each pair of opposite faces. 3
✗ Common mistakesDon’t write Write instead ✗ A parallelogram has 2 lines of symmetry✓ It has none; its order of rotation is 2✗ Order 0 rotational symmetry✓ The smallest order is 1✗ A rhombus's lines go through the midpoints of its sides✓ Its lines are the diagonals
★ In the examUse tracing paper in the exam to check lines and order. Complete-the-pattern questions: reflect each square across the given line(s). Answer with a number for 'order' — 'order 2', not 'twice'.
1 Basic angle factsAngles on a straight line add to 180°. Angles at a point add to 360°. Vertically opposite angles are equal. Angles in a triangle add to 180°; in a quadrilateral, 360°. Isosceles triangle: two equal sides and the two base angles equal. Equilateral: all angles 60°. The exterior angle of a triangle equals the sum of the two opposite interior angles. Worked example Three angles on a straight line are 2x, 3x and 4x. Find x.
2x + 3x + 4x = 180. 9x = 180. x = 20
2 Angles and parallel linesAlternate angles (Z-shape) are equal.Corresponding angles (F-shape) are equal.Co-interior angles (C-shape) add to 180°.Write the reason in words: 'alternate angles are equal' — not 'Z angles'. Pair Shape Rule Alternate Z equal Corresponding F equal Co-interior C / U add to 180° Vertically opposite X equal
Worked example Two parallel lines are crossed by a transversal. One co-interior angle is 115°. Find the other.
Co-interior angles add to 180°. 180 − 115. 65°
3 Angles in polygonsSum of interior angles of an n-sided polygon = (n − 2) × 180°. Exterior angles of any polygon add to 360°. Regular polygon: exterior angle = 360° ÷ n; interior = 180° − exterior. Number of sides of a regular polygon = 360° ÷ exterior angle. Polygon Sides Angle sum Regular interior Triangle 3 180° 60° Quadrilateral 4 360° 90° Pentagon 5 540° 108° Hexagon 6 720° 120° Octagon 8 1080° 135° Decagon 10 1440° 144°
Formulas (n − 2) × 180° interior angle sum
360° ÷ n exterior angle of a regular polygon
interior + exterior = 180°
Worked example The interior angle of a regular polygon is 5 times its exterior angle. How many sides does it have?
Interior + exterior = 180, so 6 × exterior = 180. Exterior = 30°. n = 360 ÷ 30. 12 sides
4 Giving reasonsUse three-letter notation: angle ABC is the angle at B. Each step: the angle, then its reason. For example: 'angle ABD = 40° (alternate angles are equal)'. Common reasons: angles on a straight line, angles in a triangle, isosceles triangle, alternate / corresponding / co-interior angles, angles in a polygon. Worked example In triangle ABC, AB = AC and angle BAC = 40°. Find angle ABC, giving a reason.
Base angles of an isosceles triangle are equal. (180 − 40) ÷ 2 = 70. 70° (base angles of an isosceles triangle are equal; angles in a triangle add to 180°)
✗ Common mistakesDon’t write Write instead ✗ Regular hexagon interior angle = 360 ÷ 6 = 60°✓ That's the exterior angle; interior = 120°✗ Using n × 180 for the angle sum✓ (n − 2) × 180✗ Writing 'Z angles' as a reason✓ Write 'alternate angles are equal'✗ Saying co-interior angles are equal✓ They add to 180°
★ In the examMark angles on the diagram as you find them. 'Give reasons' questions lose marks without the words — write every reason. Diagrams are not drawn to scale: never measure; calculate.
1 PercentagesPercentage of an amount: change the % to a decimal and multiply. 35% of 240 = 0.35 × 240 = 84. One quantity as a percentage of another: (part ÷ whole) × 100. Use a multiplier for increases and decreases: +15% → × 1.15; −15% → × 0.85. Percentage change = change ÷ original × 100 (profit and loss % work the same way). Formulas percentage change = (change ÷ original) × 100
new = original × multiplier
Worked example A price rises from $40 to $46. Find the percentage increase.
Change = 6. 6 ÷ 40 = 0.15. 15%
2 Reverse percentagesYou know the amount after a percentage change and want the original. Original = new amount ÷ multiplier. After a 20% discount the multiplier is 0.8. Never take the percentage of the new amount. How to do it Write the multiplier for the change (e.g. +15% → 1.15). Divide the new amount by it. Worked example A phone costs $345 after a 15% price rise. What did it cost before?
Multiplier = 1.15. 345 ÷ 1.15. $300
3 Simple and compound interestSimple interest is the same every year: I = PRT ÷ 100 (P = principal, R = rate %, T = years).Compound interest is added to the amount each year, so it grows: A = P(1 + r/100)n .Total amount with simple interest = P + I. Compound interest earned = A − P. Formulas I = PRT/100 simple interest
A = P(1 + r/100)n compound: amount after n years
Worked example $2000 is invested at 5% per year compound interest for 3 years. Find the interest earned.
A = 2000 × 1.053 = 2315.25. Interest = 2315.25 − 2000. $315.25
4 Ratio and proportionSimplify a ratio by dividing all parts by the HCF; use the same units first. Sharing in a ratio: add the parts, find one part, multiply. Map scale 1 : 50 000 means 1 cm on the map = 50 000 cm = 500 m in real life. Best value: compare the price of the same amount (per 100 g, per litre) or the amount per $1. Recipes: find the amount for one person (or one unit), then scale up. Worked example Share $360 in the ratio 4 : 5.
4 + 5 = 9 parts. One part = 360 ÷ 9 = 40. 4 × 40 = 160 and 5 × 40 = 200. $160 and $200
5 Rates, currency and unitsExchange rates: multiply to change into the currency the rate is quoted in, divide to change back. Rates: speed (km/h), hourly pay ($/h), flow (litres/min), fuel use (km/litre), density (g/cm³), population density (people/km²). Length: 1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm. Mass: 1 kg = 1000 g; 1 tonne = 1000 kg. Area and volume scale with the square and cube: 1 m² = 10 000 cm²; 1 m³ = 1 000 000 cm³; 1 litre = 1000 cm³. Formulas density = mass ÷ volume
population density = population ÷ area
Worked example $1 = 280 rupees. Convert 7000 rupees to dollars.
Rupees → dollars: divide by 280. 7000 ÷ 280. $25
6 Pie chartsThe whole pie is 360°. Angle for a group = frequency ÷ total × 360°. From a pie chart: frequency = angle ÷ 360 × total. If you know one sector's frequency and angle, find 'degrees per item' and scale. Worked example In a survey of 72 students, 18 walk to school. What angle represents walking?
18 ÷ 72 = 1/4. 1/4 × 360°. 90°
✗ Common mistakesDon’t write Write instead ✗ Reverse %: 64 + 20% of 64✓ Divide by the multiplier: 64 ÷ 0.8 = 80✗ Percentage change ÷ the new value✓ Always divide by the original value✗ Compound interest worked as simple interest✓ Multiply by 1.05 each year: P × 1.05n ✗ 1 m² = 100 cm²✓ 1 m² = 100 × 100 = 10 000 cm²✗ Sharing in 4 : 5 by dividing by 4 and 5✓ Divide by the total number of parts, 9
★ In the examMoney answers: give 2 decimal places ($315.25, not $315.3). Say clearly what you are comparing in best-value questions (price per 100 g…). Read whether the question wants the interest or the total amount.
1 Direct proportiony ∝ x means y = kx: when x doubles, y doubles. The graph is a straight line through the origin. Other direct forms: y ∝ x2 (y = kx2 ), y ∝ x3 , y ∝ √x, y ∝ ∛x. Find k from a given pair of values, then use the formula. Method for every proportion question Write the formula with k. Substitute the pair you are given to find k. Write the formula again with the number for k. Substitute the new value. Formulas y ∝ x ⇒ y = kx
y ∝ x2 ⇒ y = kx2
y ∝ √x ⇒ y = k√x
Worked example y is directly proportional to the square of x. y = 18 when x = 3. Find y when x = 5.
y = kx2 . 18 = k × 9, so k = 2. y = 2x2 . y = 2 × 25. y = 50
2 Inverse proportiony ∝ 1/x means y = k/x: when x doubles, y halves. xy is always the same. Other inverse forms: y ∝ 1/x2 , y ∝ 1/√x. The graph of y = k/x is a curve that gets close to both axes. Formulas y ∝ 1/x ⇒ y = k/x
y ∝ 1/x2 ⇒ y = k/x2
Worked example The force F is inversely proportional to the square of the distance d. F = 20 when d = 3. Find F when d = 6.
F = k/d2 . 20 = k/9, so k = 180. F = 180/36. F = 5
3 The effect of a changeIf y ∝ xn and x is multiplied by m, then y is multiplied by mn . Inverse: if y ∝ 1/xn and x is multiplied by m, y is divided by mn . Percentage changes work the same way: x up 10% means x × 1.1. Relationship x doubled → y x tripled → y y ∝ x × 2 × 3 y ∝ x2 × 4 × 9 y ∝ x3 × 8 × 27 y ∝ 1/x ÷ 2 ÷ 3 y ∝ 1/x2 ÷ 4 ÷ 9
Worked example y ∝ x3 . x is increased by 10%. By what percentage does y increase?
x is multiplied by 1.1. y is multiplied by 1.13 = 1.331. 33.1%
✗ Common mistakesDon’t write Write instead ✗ Using y = kx for 'proportional to the square'✓ Square it: y = kx2 ✗ Inverse proportion written y = x/k✓ It is y = k/x✗ Not squaring when substituting into y = kx2 ✓ 18 = k × 32 , not 18 = k × 3✗ Forgetting to write the formula with k✓ Write y = 2x2 — it is often worth a mark
★ In the examRead the wording carefully: 'square', 'square root', 'cube', 'inversely'. Always write the formula with your value of k in it before the final step. Check: a larger x should make y bigger (direct) or smaller (inverse).
1 Bounds of a rounded valueA measurement rounded to some accuracy could have been anything within half a unit either side. Lower bound = value − half the unit; upper bound = value + half the unit. The upper bound itself would round up, so the error interval is lower ⩽ x < upper . Given as Unit Lower bound Upper bound 7.4 cm (1 d.p.) 0.1 7.35 7.45 3400 (nearest 100) 100 3350 3450 12 kg (nearest kg) 1 11.5 12.5 0.36 (2 s.f.) 0.01 0.355 0.365
Worked example A length is 7.4 cm to the nearest mm. Write its error interval.
Nearest mm = nearest 0.1 cm. Half of 0.1 = 0.05. 7.35 ⩽ l < 7.45
2 Bounds of calculationsTo make an answer as large as possible, choose the bounds that push it up; for as small as possible, the opposite. Adding and multiplying: upper with upper gives the upper bound. Subtracting and dividing: one of each. The biggest difference is upper − lower; the biggest quotient is upper ÷ lower. Calculation Upper bound Lower bound a + b Ua + Ub La + Lb a × b Ua × Ub La × Lb a − b Ua − Lb La − Ub a ÷ b Ua ÷ Lb La ÷ Ub
Worked example A car travels 120 m (nearest 10 m) in 15.2 s (1 d.p.). Find the lower bound of its speed.
Lower distance = 115 m; upper time = 15.25 s. Lower speed = 115 ÷ 15.25 = 7.540… 7.54 m/s (3 s.f.)
Exam tip: Write the bounds you are using before calculating — they earn method marks.
3 Giving an answer to a suitable accuracyWork out both the upper and lower bound of the result. Round both to fewer and fewer figures until they agree: that is the accuracy you can trust. Example: bounds 7.54 and 7.81 agree only to 1 s.f. → answer 8. Worked example The bounds of a length are 4.4637 m and 4.4762 m. Give the length to a suitable accuracy.
To 3 s.f.: 4.46 and 4.48 — different. To 2 s.f.: 4.5 and 4.5 — the same. 4.5 m
✗ Common mistakesDon’t write Write instead ✗ Upper bound of 7.4 (1 d.p.) is 7.49✓ It is 7.45 — the boundary✗ Error interval 11.5 ⩽ x ⩽ 12.5✓ Upper end is strict: 11.5 ⩽ x < 12.5✗ Upper bound of a ÷ b = Ua ÷ Ub✓ Divide by the smallest: Ua ÷ Lb✗ Upper bound of a − b = Ua − Ub✓ Subtract the smallest: Ua − Lb
★ In the examFind the unit first ('nearest 5' → half unit 2.5; '2 s.f.' on 360 → nearest 10). Keep full calculator values for bounds; round only the final answer. For a mixed calculation (like speed = d/t or density = m/V) decide each bound separately.
1 Prime factorisationEvery whole number above 1 can be written as a product of primes in exactly one way. Use a factor tree or keep dividing by the smallest prime (2, 3, 5, 7, 11, …). Write the answer in index form: 360 = 23 × 32 × 5. Repeated division Divide by 2 until you can't. Then by 3, then 5, then 7… Stop at 1; collect equal primes as powers. Worked example Write 360 as a product of its prime factors.
360 ÷ 2 = 180, ÷ 2 = 90, ÷ 2 = 45. 45 ÷ 3 = 15, ÷ 3 = 5, ÷ 5 = 1. Primes used: 2, 2, 2, 3, 3, 5. 23 × 32 × 5
2 HCF and LCMHCF (highest common factor): the biggest number that divides into both.LCM (lowest common multiple): the smallest number both divide into.From prime factors: HCF = primes they share , lowest power of each. LCM = every prime, highest power of each. A Venn diagram of the prime factors works too: HCF = the overlap; LCM = everything in the diagram. Formulas HCF shared primes, lowest powers
LCM all primes, highest powers
HCF × LCM = a × b for two numbers a and b
Worked example Find the HCF and LCM of 84 and 120.
84 = 22 × 3 × 7; 120 = 23 × 3 × 5. HCF: 22 × 3 = 12. LCM: 23 × 3 × 5 × 7 = 840. HCF 12, LCM 840
3 Word problems'Together again', 'at the same time', 'smallest number of each' → LCM . 'Largest possible', 'cut into equal pieces', 'share into identical groups' → HCF . Algebraic HCF: HCF of the numbers × the lowest power of each shared letter. HCF of 12x2y and 18xy3 is 6xy. Worked example Two lighthouses flash every 12 s and every 20 s. They flash together at 9:00. When do they next flash together?
Together again → LCM. 12 = 22 × 3, 20 = 22 × 5, LCM = 22 × 3 × 5 = 60. 60 s later, at 9:01
4 Square roots, cube roots and the smallest nA square number has every prime to an even power; a cube has every power a multiple of 3. √: halve every power. ∛: divide every power by 3. √(22 × 32 × 72 ) = 2 × 3 × 7 = 42. Smallest number to multiply by to make a square: supply whatever makes every power even. Worked example Find the smallest whole number n such that 120n is a cube number.
120 = 23 × 3 × 5. For a cube every power must be a multiple of 3. Need 32 × 52 more. n = 225
✗ Common mistakesDon’t write Write instead ✗ LCM of 18 and 24 = 18 × 24 = 432✓ That is a common multiple, but the lowest is 72✗ Writing 2 × 2 × 2 × 3 × 3 × 5 when 'index form' is asked✓ Write 23 × 32 × 5✗ Including 1 in a prime factorisation✓ 1 is not prime — leave it out✗ Taking the highest powers for the HCF✓ HCF uses the LOWEST powers of shared primes
★ In the examAlways check the HCF divides both numbers and both numbers divide the LCM. In word problems, say which one you are finding (HCF or LCM) before calculating. Show the prime factorisations — they earn method marks even if a multiplication slips.
1 Shapes of graphsRecognise each family from its equation and sketch its shape. Type Example Shape and features Linear y = 2x + 1 straight line Quadratic y = x2 − 4 ∪ if x2 term positive, ∩ if negative; symmetrical Cubic y = x3 − 3x S-shaped; up to two turning points Reciprocal y = 2/x two branches; asymptotes x = 0 and y = 0 Exponential y = 2x through (0, 1); asymptote y = 0; grows fast
Worked example Which type is y = 5 − x2 , and where is its turning point?
Quadratic with a negative x2 term: ∩-shaped. At x = 0, y = 5. ∩-shaped parabola, maximum at (0, 5)
2 Tables of values and plottingUse brackets for negatives: x = −2 in x2 − 3x + 1 gives 4 + 6 + 1 = 11. Use the calculator's TABLE mode to check every value. Plot with small crosses, then draw one smooth curve through all points — no straight segments, no joining across an asymptote. How to do it Complete the table (check with TABLE mode). Plot each point accurately. Draw a smooth curve; extend only as far as the table. Worked example For y = 12/x + x, find y when x = 4 and x = −2.
x = 4: 3 + 4 = 7. x = −2: −6 + (−2) = −8. 7 and −8
3 Key featuresy-intercept : put x = 0. Roots (x-intercepts): solve y = 0.Turning point of a quadratic: halfway between the roots, or from completed-square form (x − p)2 + q → (p, q).Cubic in factorised form y = (x − a)(x − b)(x − c) crosses the x-axis at a, b and c. Worked example For y = x2 − 4x − 5, find the intercepts and the turning point.
x = 0: y = −5. x2 − 4x − 5 = (x − 5)(x + 1) = 0 → x = 5 or −1. Halfway: x = 2, y = 4 − 8 − 5 = −9. (0, −5); (5, 0) and (−1, 0); minimum (2, −9)
4 Reading from a graphRead solutions of 'curve = 0' where the curve crosses the x-axis. Read a y value: go up from x to the curve, then across. Give values to the accuracy of the grid (often to the nearest half small square). Worked example The graph of y = x3 − 3x crosses the x-axis three times. Where?
x(x2 − 3) = 0. x = 0 or x = ±√3 ≈ ±1.73. x = −1.7, 0 and 1.7
✗ Common mistakesDon’t write Write instead ✗ (−2)2 entered as −22 = −4✓ Use brackets: (−2)2 = 4✗ Joining points with straight lines✓ Draw one smooth curve✗ Joining the two branches of y = k/x✓ They never meet — break the curve at x = 0✗ Putting the turning point of (x − 3)2 + 4 at (−3, 4)✓ It is at (3, 4)
★ In the examDraw curves in pencil with one confident stroke. Use the symmetry of a quadratic to check your table. Label the curve with its equation when there is more than one on the grid.
1 Expressions and substitutionCollect like terms only: 3x + 2y − x = 2x + 2y; x2 and x are not like terms. Substitute with brackets round negative numbers: if a = −2, then a2 = (−2)2 = 4 and 3a = −6. Write terms in a standard way: 3 × a × a × b = 3a2 b. Worked example Find the value of 2p2 − 5q when p = 4 and q = −3.
2 × 42 = 2 × 16 = 32. 5 × (−3) = −15. 32 − (−15) = 32 + 15. 47
2 Expanding bracketsSingle bracket: multiply every term inside. −3(2x − 5) = −6x + 15 (watch the signs). Two brackets: every term in the first × every term in the second (FOIL or a grid), then collect like terms. Three brackets: expand two of them first, then multiply the result by the third. Perfect squares: (a + b)2 = a2 + 2ab + b2 — never just a2 + b2 . Formulas (a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b)(a − b) = a2 − b2
Worked example Expand and simplify (x + 2)(x − 3)(2x + 1).
(x + 2)(x − 3) = x2 − x − 6. (x2 − x − 6)(2x + 1) = 2x3 + x2 − 2x2 − x − 12x − 6. Collect like terms. 2x3 − x2 − 13x − 6
3 Solving linear equationsDo the same to both sides until the letter is on its own. Brackets: expand first. Letters on both sides: collect them on the side with more. Fractions: multiply every term by the lowest common denominator first. Algebraic denominators: 3/(x − 1) = 2/(x + 2) → cross-multiply: 3(x + 2) = 2(x − 1) → x = −8. How to do it Clear fractions (× LCD). Expand brackets. Collect x terms on one side, numbers on the other. Divide by the number in front of x. Check by substituting. Worked example Solve (x + 1)/3 − (x − 2)/4 = 2.
Multiply every term by 12: 4(x + 1) − 3(x − 2) = 24. 4x + 4 − 3x + 6 = 24. x + 10 = 24. x = 14
Exam tip: Put the second numerator in brackets: −3(x − 2) = −3x + 6, not −3x − 6.
4 Forming expressions and equationsTurn words into algebra: '3 more than x' = x + 3, 'twice x' = 2x, 'x squared' = x2 . Perimeter adds all the sides; area multiplies; angles in a triangle add to 180°. Define your letter clearly, form the equation, solve it, then answer the question asked. Worked example A rectangle's length is 3 cm more than its width. Its perimeter is 34 cm. Find its area.
Width w, length w + 3. 2(w + w + 3) = 34 → 4w + 6 = 34. w = 7, length 10. 70 cm²
5 Changing the subjectUndo the operations on the new subject in reverse order, doing the same to both sides. Square ↔ square root; × ↔ ÷; + ↔ −. If the subject appears twice: collect those terms on one side, factorise it out, then divide. When the subject appears twice Clear fractions and roots. Expand any brackets. Get every term with the subject on one side. Factorise the subject out. Divide. Worked example Make t the subject of 3t + a = bt − 5.
Collect t terms: 3t − bt = −5 − a. Factorise: t(3 − b) = −5 − a. Divide by (3 − b). t = (a + 5)/(b − 3)
Exam tip: Example with a root: V = ⅓πr2 h gives r = √(3V/(πh)).
6 Algebraic fractionsSimplify : factorise top and bottom completely, then cancel whole brackets (never single terms).Add/subtract : common denominator (often the product of the bottoms), then simplify the top.Multiply/divide : like number fractions — flip the second when dividing; factorise and cancel first.Worked example Write 2/(x + 1) + 3/(x − 2) as a single fraction.
Common denominator (x + 1)(x − 2). Top: 2(x − 2) + 3(x + 1) = 2x − 4 + 3x + 3. = 5x − 1. (5x − 1)/((x + 1)(x − 2))
Exam tip: You can cancel (x + 3) from top and bottom, but not the x in (x + 3)/x.
7 Quadratic equationsRearrange to ax2 + bx + c = 0 first. Factorise when it works; each bracket = 0 gives a solution.Use the formula when the question asks for decimals ('2 d.p.', '3 s.f.'). Completing the square : x2 + bx + c = (x + b/2)2 − (b/2)2 + c. It gives the turning point and exact solutions.If b2 − 4ac is negative there are no real solutions. Formulas x = (−b ± √(b2 − 4ac)) / 2a the quadratic formula
x2 + bx + c = (x + b/2)2 − (b/2)2 + c completing the square
Worked example Solve 2x2 − 5x − 1 = 0, giving your answers to 2 decimal places.
a = 2, b = −5, c = −1. x = (5 ± √(25 + 8)) ÷ 4 = (5 ± √33) ÷ 4. √33 = 5.7446… x = 2.69 or x = −0.19
Exam tip: Completing the square on x2 − 8x + 11 gives (x − 4)2 − 5: minimum point (4, −5), and solutions x = 4 ± √5.
✗ Common mistakesDon’t write Write instead ✗ (x + 3)2 = x2 + 9✓ There is a middle term: x2 + 6x + 9✗ −3(x − 2) = −3x − 6✓ Negative times negative: −3x + 6✗ x(x + 3) = 40, so x = 40 or x + 3 = 40✓ Only '= 0' lets you split: x2 + 3x − 40 = 0✗ Cancelling x in (x + 3)/x to get 3✓ Only whole factors cancel✗ Dividing only part of the top by 2a in the formula✓ Divide the whole of −b ± √(b2 − 4ac) by 2a
★ In the examShow the substitution into the quadratic formula — it is worth a mark on its own. Give both solutions of a quadratic unless the context rules one out (a negative length). Check a linear equation's answer by substituting it back in.
1 Set notationLearn the symbols — questions use them without explanation. Symbol Meaning ξ the universal set (everything) x ∈ A x is in A x ∉ A x is not in A A′ everything not in A A ∪ B in A or B or both (union) A ∩ B in both A and B (intersection) ∅ the empty set A ⊆ B A is a subset of B n(A) the number of elements in A {x : x > 3} the set of x such that x > 3
Worked example A = {2, 4, 6, 8, 10}, B = {3, 6, 9}. Find A ∩ B and n(A ∪ B).
Both: 6. A ∪ B = {2, 3, 4, 6, 8, 9, 10}. A ∩ B = {6}; n(A ∪ B) = 7
2 Venn diagrams and shadingTwo sets make four regions: A only, both, B only, neither. A ∩ B′ = in A but not B. (A ∪ B)′ = outside both. A′ ∪ B = everything except 'A only'. For three sets, there are eight regions; fill from the centre outwards. Region Notation A only A ∩ B′ B only A′ ∩ B Both A ∩ B Neither (A ∪ B)′ or A′ ∩ B′ Exactly one (A ∩ B′) ∪ (A′ ∩ B)
Worked example Describe the region 'in B but not in A or C' using set notation.
In B, not in A, not in C. B ∩ A′ ∩ C′
3 Venn diagram problemsStart with the overlap, then subtract it from each set's total. Remember the region outside the circles. With an unknown, write each region in terms of x and set the total equal to n(ξ). Probability from a Venn diagram = number in the region ÷ the total (or ÷ the given set for 'given that'). Formulas n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Worked example In a group of 50, 32 like tea, 25 like coffee and 6 like neither. How many like only tea?
Let x like both: (32 − x) + x + (25 − x) + 6 = 50. 63 − x = 50, x = 13. Only tea = 32 − 13. 19
✗ Common mistakesDon’t write Write instead ✗ n(A ∪ B) = n(A) + n(B)✓ Subtract the overlap✗ Putting 32 in the 'tea only' region✓ 32 is all of tea; tea only = 32 − both✗ Forgetting the region outside the circles✓ It is part of ξ — include it in the total✗ Listing 1 as a prime in a set✓ 1 is not prime
★ In the examDraw the Venn diagram even when not asked — it organises the information. Check that all the regions add up to n(ξ). Shading questions: shade carefully and clearly; mark the region boundaries.
1 Single eventsProbability is from 0 (impossible) to 1 (certain). Write it as a fraction, decimal or percentage. P(event) = number of ways it happens ÷ total number of equally likely outcomes. P(not A) = 1 − P(A). The probabilities of all possible outcomes add to 1. Formulas P(A) = favourable ÷ total
P(not A) = 1 − P(A)
Worked example The only outcomes A, B and C have probabilities 0.2, 0.45 and x. Find x.
0.2 + 0.45 + x = 1. x = 0.35
2 Relative and expected frequencyRelative frequency = number of times it happened ÷ number of trials. It estimates the probability; more trials = better estimate. Expected frequency = probability × number of trials. Worked example A spinner landed on 6 in 45 of 300 spins. How many 6s would you expect in 1000 spins?
Relative frequency = 45/300 = 0.15. 0.15 × 1000. 150
3 Sample spaces and combined eventsA sample space diagram (grid) lists every outcome of two events — useful for two dice or two spinners. AND (independent events): multiply. OR (events that can't both happen): add.'At least one' = 1 − P(none). Formulas P(A and B) = P(A) × P(B) independent
P(A or B) = P(A) + P(B) mutually exclusive
P(at least one) = 1 − P(none)
Worked example Two fair dice are thrown. Find the probability that the total is 7.
36 outcomes in the grid. Totals of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes. 6/36 = 1/6
4 Tree diagramsEach set of branches adds to 1. Multiply along a path; add the paths you want. With replacement : the second-stage probabilities are the same as the first.Without replacement : one fewer item in total, and one fewer of the colour already taken.How to do it Draw the branches with probabilities. Multiply along each wanted path. Add the results of all wanted paths. Worked example A bag has 4 red and 6 blue sweets. Two are eaten without replacement. Find P(one of each colour).
P(R then B) = 4/10 × 6/9 = 24/90. P(B then R) = 6/10 × 4/9 = 24/90. Add: 48/90. 8/15
5 Conditional probability'Given that' narrows the total to the group you are told about. From a table or Venn diagram: count only within the given group. From a tree: the second branches are already conditional on the first. Formulas P(A given B) = P(A and B) ÷ P(B)
Worked example 12 of 20 girls and 6 of 10 boys wear glasses. A student who wears glasses is chosen. Find the probability it is a boy.
18 students wear glasses. 6 of them are boys. 6/18 = 1/3
✗ Common mistakesDon’t write Write instead ✗ Adding along the branches✓ Multiply along a path; add the results✗ Same denominator on both stages without replacement✓ 4/10 then 3/9 (or 6/9)✗ A probability bigger than 1✓ Probabilities are between 0 and 1✗ Writing a probability as '3 out of 5' or 3 : 5✓ Write 3/5, 0.6 or 60%✗ Forgetting one of the two orders for 'one of each'✓ Red-then-blue AND blue-then-red
★ In the examDraw the tree or grid even when not asked. Fractions don't need simplifying unless asked — but keep them exact. Check: all end probabilities of a tree add to 1.
1 Column vectorsA column vector has a top number (x, across) and a bottom number (y, up). Add or subtract top with top and bottom with bottom. Multiply by a number: multiply both parts. 3 × (−2, 4) = (−6, 12). Written a, a (by hand) or AB with an arrow. Worked example a = (3, −2) and b = (−1, 5). Find 2a − b.
2a = (6, −4). (6 − (−1), −4 − 5). (7, −9)
2 MagnitudeThe magnitude |a| is the length of the vector: Pythagoras on its two parts. Formulas |(x, y)| = √(x2 + y2 )
Worked example Find the magnitude of (5, −12).
√(25 + 144) = √169. 13
3 Position vectorsThe position vector of A is OA (from the origin). AB = OB − OA = b − a. Midpoint M of AB has position vector ½(a + b). Formulas AB = b − a end minus start
OM = ½(a + b) midpoint
Worked example A is (2, 3) and B is (7, −1). Find the vector AB.
B − A = (7 − 2, −1 − 3). (5, −4)
4 Vectors in diagramsFind a route along known vectors; going against an arrow changes the sign. A point dividing AB in the ratio m : n: AP = m/(m + n) × AB. Parallel : one vector is a multiple of the other.Collinear A, B, C: AB is a multiple of BC and they share the point B.How to do it Write the route (e.g. O → A → P). Replace each step by a vector, sign included. Simplify by collecting a's and b's. Worked example OA = a, OB = b. P is on AB with AP : PB = 1 : 3. Find OP.
AB = b − a. AP = ¼(b − a). OP = a + ¼(b − a). OP = ¾a + ¼b
✗ Common mistakesDon’t write Write instead ✗ AB = a − b✓ Start at A, go back to O, then to B: b − a✗ |(5, −12)| = 5 − 12✓ Use Pythagoras: √(25 + 144) = 13✗ Using ½ for 'ratio 1 : 2'✓ AP is 1/3 of AB✗ Proving collinear with the multiple only✓ Also say the vectors share a point
★ In the examUnderline vectors in handwriting: a . Simplify answers fully: collect the a terms and the b terms. For 'show that' questions, write each route step by step.
1 Similar shapes and lengthsSimilar shapes have equal angles and sides in the same ratio. Length scale factor k = a length on the new shape ÷ the matching length on the old one. Two triangles are similar if two pairs of angles are equal (the third then matches too). How to do it Identify matching sides (opposite equal angles). Find k from a pair of known matching sides. Multiply (or divide) to find the missing length. Worked example Two similar triangles have matching sides 4 cm and 10 cm. A side of 6 cm on the small one matches which length?
k = 10 ÷ 4 = 2.5. 6 × 2.5. 15 cm
2 Nested trianglesA line parallel to one side of a triangle cuts off a smaller similar triangle. Use the whole sides of the big triangle, not just the extra piece. Redraw the two triangles separately the same way round. Worked example In triangle ABC, DE ∥ BC with D on AB, E on AC. AD = 3, DB = 6, DE = 4. Find BC.
AB = 3 + 6 = 9. k = 9 ÷ 3 = 3. BC = 4 × 3. 12
3 Areas and volumes of similar shapesLength factor k → area factor k2 → volume factor k3 . Surface area scales like area (k2 ); mass and capacity scale like volume (k3 ). Work backwards with roots: area ratio → k = √(area ratio); volume ratio → k = ∛(volume ratio). Formulas area factor = k2
volume factor = k3
k = √(area ratio) = ∛(volume ratio)
Worked example Two similar jugs have surface areas 180 cm² and 405 cm². The small one holds 0.8 litres. How much does the large one hold?
Area factor = 405 ÷ 180 = 2.25, so k = 1.5. Volume factor = 1.53 = 3.375. 0.8 × 3.375. 2.7 litres
✗ Common mistakesDon’t write Write instead ✗ Area doubles when the lengths double✓ Area × 4 (k2 = 22 )✗ Using DB instead of AB for the big triangle✓ Use the whole side AB✗ Cubing the volume ratio to get a length✓ Take the cube root✗ Showing similarity with one pair of equal angles✓ You need two pairs of equal angles
★ In the examWrite k clearly as a fraction or decimal before using it. For area or volume questions, find the length factor first. To prove similarity, state the equal angles with reasons.
1 What congruent meansCongruent shapes are identical in shape and size . They may be rotated or reflected. Corresponding sides are equal and corresponding angles are equal. Similar shapes have equal angles but sides in proportion — different sizes are allowed. Worked example Two triangles have the same three angles but different sides. Are they congruent?
Same angles only shows they are similar. No — they are similar, not necessarily congruent
2 The four conditionsTwo triangles are congruent if they match on one of these: Condition What must be equal SSS all three sides SAS two sides and the angle BETWEEN them ASA (or AAS) two angles and a corresponding side RHS right angle, hypotenuse and one other side
Worked example Triangles have AB = DE, angle B = angle E and BC = EF. Are they congruent? Which condition?
Two sides equal; the equal angle is between them. Yes, SAS
3 Using congruenceMatch the vertices in order: ABC ≅ PQR means A ↔ P, B ↔ Q, C ↔ R. Then the side AB = PQ, BC = QR, angle B = angle Q, and so on. AAA and SSA do not prove congruence. Worked example ABC ≅ PQR. AB = 7 cm, BC = 9 cm, angle B = 48°. Find QR and angle Q.
BC ↔ QR. Angle B ↔ angle Q. QR = 9 cm, angle Q = 48°
✗ Common mistakesDon’t write Write instead ✗ Using AAA to show congruence✓ AAA shows similarity only✗ Matching sides by their position on the page✓ Match by letter order or equal angles✗ Calling any two sides and an angle SAS✓ The angle must be between the two sides
★ In the examName the condition (SSS, SAS, ASA, RHS) in your answer. Redraw the triangles the same way round if they are turned or flipped. Mark equal sides and angles on the diagram first.
1 Simplifying surdsA surd is a root that is irrational, like √2 or √12. It is exact — keep it as a surd unless told to round. Find the largest square factor : √72 = √(36 × 2) = 6√2. Square factors to look for: 4, 9, 16, 25, 36, 49, 64, 81, 100. Formulas √(ab) = √a × √b
√(a/b) = √a / √b
√a × √a = a
Worked example Simplify √75 + √12.
√75 = √25 × √3 = 5√3. √12 = √4 × √3 = 2√3. 5√3 + 2√3. 7√3
2 Adding, multiplying and expandingOnly like surds add or subtract: 6√2 − 4√2 = 2√2, but √2 + √3 cannot be combined. Multiply: √3 × √12 = √36 = 6; 2√5 × 3√2 = 6√10. Expand brackets just as in algebra, then simplify each term. (a + √b)(a − √b) = a2 − b: the surds cancel. This is used to rationalise. Worked example Expand and simplify (3 + √2)2 .
(3 + √2)(3 + √2) = 9 + 3√2 + 3√2 + 2. Collect: 11 + 6√2. 11 + 6√2
Exam tip: (a + √b)2 is NOT a2 + b — there is a middle term 2a√b.
3 Rationalising the denominator'Rationalise' means: no surd on the bottom of the fraction. a/√b: multiply top and bottom by √b. 10/√5 = 10√5/5 = 2√5. a/(b + √c): multiply top and bottom by (b − √c). a/(b − √c): multiply by (b + √c). How to do it Look at the bottom. Single surd √b → multiply top and bottom by √b. b ± √c → multiply top and bottom by the same with the opposite sign. Expand and simplify; cancel any common factor. Worked example Rationalise 6/(3 − √3).
Multiply by (3 + √3)/(3 + √3). Bottom: 9 − 3 = 6. Top: 6(3 + √3). 6(3 + √3)/6. 3 + √3
✗ Common mistakesDon’t write Write instead ✗ √9 + √16 = √25✓ √9 + √16 = 3 + 4 = 7✗ Leaving √72 as 2√18✓ Use the largest square factor: 6√2✗ (3 + √2)2 = 9 + 2✓ Middle term: 9 + 6√2 + 2 = 11 + 6√2✗ Multiplying only the top when rationalising✓ Multiply top and bottom by the same surd
★ In the exam'Give your answer in the form a + b√c' — make sure a, b, c are integers and the surd is simplified. Surd questions are on the non-calculator paper; check by squaring your answer mentally. Exact answers (like in Pythagoras) are often wanted as simplified surds.
1 Lengths in 3DFind a right-angled triangle that contains the length you want, then use Pythagoras. Usually two steps: the diagonal of the base first, then the vertical triangle. Space diagonal of a cuboid a × b × c: √(a2 + b2 + c2 ). Formulas d = √(a2 + b2 + c2 ) space diagonal of a cuboid
Worked example A box is 12 cm long, 4 cm wide and 3 cm high. Find the length of the longest rod that fits inside.
Base diagonal: √(144 + 16) = √160. Space diagonal: √(160 + 9) = √169. 13 cm
2 The angle between a line and a planeDrop a perpendicular from the top of the line to the plane; the foot gives the line's projection . The angle is between the line and its projection — not the vertical. The right-angled triangle is: line (hypotenuse), projection (adjacent), vertical height (opposite). How to do it Identify the line and the plane. Find the projection of the line on the plane. Draw the right-angled triangle flat on its own. Use SOH CAH TOA. Worked example A cuboid is 6 cm by 8 cm by 5 cm high. Find the angle between the space diagonal and the base.
Base diagonal (projection) = 10 cm. tan θ = 5 ÷ 10. 26.6°
3 PyramidsThe vertical height goes from the apex to the centre of the base. For a square or rectangular base, the centre is the midpoint of the base diagonal. The angle between a sloping edge and the base uses half the base diagonal; the angle between a sloping face and the base uses half the base width. Worked example A square-based pyramid has base 8 cm × 8 cm and vertical height 6 cm. Find the angle between a sloping edge and the base.
Half the base diagonal = 4√2 = 5.657 cm. tan θ = 6 ÷ 5.657. 46.7°
4 Prisms and wedgesLook for the right angles between the faces. Redraw each triangle you need as a flat 2D triangle with its lengths. You may need the sine or cosine rule in a non-right-angled face. Worked example A wedge has a horizontal base 12 cm by 5 cm and a vertical back edge 9 cm high above one 5 cm side. Find the length from a front bottom corner to the opposite top corner.
Base diagonal: √(144 + 25) = 13. Then with the 9 cm height: √(169 + 81) = √250. 15.8 cm (3 s.f.)
✗ Common mistakesDon’t write Write instead ✗ Using the base side instead of the base diagonal✓ The projection of a space diagonal is the base diagonal✗ Measuring the angle from the vertical✓ The angle with a plane is measured from the plane✗ Rounding √160 to 12.6 early✓ Keep it exact or keep all digits
★ In the examDraw every triangle you use separately and label it. Name the angle with letters (e.g. angle AGE) so the examiner can follow. Give angles to 1 decimal place.
1 Growth and decay multipliersRepeated percentage change multiplies by the same number each period. Growth of r% → multiplier (1 + r/100). Decay of r% → multiplier (1 − r/100). After n periods: amount = starting amount × multipliern . Typical contexts: population, bacteria, compound interest (growth); depreciation, radioactive decay, cooling (decay). Change Multiplier +4% 1.04 +12.5% 1.125 −15% 0.85 −2.5% 0.975 doubles 2 halves 0.5
Formulas A = P(1 + r/100)n growth
A = P(1 − r/100)n decay
Worked example A car worth $12 000 loses 15% of its value each year. Find its value after 3 years.
Multiplier 0.85. 12 000 × 0.853 = 7369.5. $7369.50
2 Working backwards and finding the rateStarting amount = final amount ÷ multipliern . To find the rate: (final ÷ start) = multipliern , so take the n-th root to get the multiplier. Multiplier 1.1 means 10% growth; 0.92 means 8% decay. Worked example A value grows from 400 to 484 in 2 years at a constant percentage rate. Find the rate.
484 ÷ 400 = 1.21. Multiplier = √1.21 = 1.1. 10% per year
3 Finding the number of periodsNo logarithms needed: try n = 1, 2, 3, … on the calculator until the amount passes the target. Give the first whole period that passes it, and show the values either side. Half-life questions: each half-life multiplies the amount by 0.5. How to do it Write amount = P × multipliern . Try values of n. Write the value just below and just above the target. Answer with the first n that passes it. Worked example $800 is invested at 3% compound interest. After how many whole years is it first worth more than $1000?
800 × 1.037 = 983.9… 800 × 1.038 = 1013.4… 8 years
✗ Common mistakesDon’t write Write instead ✗ 15% depreciation for 2 years = 30% off✓ Multiply by 0.85 twice: 0.7225 (27.75% off)✗ Using 1.15 for a 15% decrease✓ Decrease: 0.85✗ Giving the year just before the target is reached✓ Give the first whole period that passes it✗ Dividing by 2 for the rate from a 2-year change✓ Take the square root of the overall multiplier
★ In the examWrite the multiplier first; most method marks come from it. Show the trial values when finding the number of years. Read whether the question wants the amount, the increase/decrease, or the percentage.
1 Measuring and drawingMeasure lengths to the nearest millimetre and angles to the nearest degree. Use the correct scale on the protractor (inner or outer) — start from the 0 on the arm you measure from. Leave all construction lines and arcs showing. Worked example A line measures 7.4 cm on a drawing with scale 1 cm to 5 m. How long is it in real life?
7.4 × 5. 37 m
2 Triangles from three sidesOnly a ruler and compasses are allowed. How to do it Draw the longest side as the base. Set the compasses to the second length; draw an arc from one end. Set them to the third length; draw an arc from the other end. Join the crossing point to both ends. Worked example Construct a triangle with sides 7 cm, 5 cm and 6 cm. What do you draw after the 7 cm base?
Arc of radius 5 cm from one end. Arc of radius 6 cm from the other end. Join the point where the arcs cross to both ends
3 BisectorsPerpendicular bisector of AB: arcs of the same radius (more than half AB) from A and from B, above and below; join the crossings. Every point on it is equally far from A and B.Angle bisector : arc from the vertex cutting both arms; equal arcs from those two points; join the vertex to the crossing. Every point on it is equally far from both arms.Worked example Which construction finds the points equally distant from two towns A and B?
Equal distance from two points. The perpendicular bisector of AB
4 NetsA net is the flat shape that folds into a solid. Cube: 6 squares. Cuboid: 3 pairs of rectangles. Triangular prism: 2 triangles + 3 rectangles. Square-based pyramid: 1 square + 4 triangles. Cylinder: 1 rectangle + 2 circles. The surface area is the area of the net. Worked example How many faces does the net of a triangular prism have?
2 triangles and 3 rectangles. 5
5 Scale drawings and bearingsScale 1 : n: real length = drawing length × n (watch the units). Bearings: measured from north, clockwise , written with three figures (045°, 120°, 300°). 'Bearing of B from A': stand at A, face north, turn clockwise to B. Back bearing (A from B) = bearing ± 180°. Formulas back bearing = bearing ± 180° add if less than 180°, subtract if more
Worked example The bearing of B from A is 070°. Find the bearing of A from B.
070 + 180. 250°
Exam tip: Always draw a north line at the point you measure FROM.
✗ Common mistakesDon’t write Write instead ✗ Writing a bearing as 45°✓ Three figures: 045°✗ Bearing of A from B = bearing of B from A✓ Add or subtract 180°✗ Rubbing out construction arcs✓ Leave them — they earn the marks✗ Using a protractor in a 'compasses only' construction✓ Only arcs and a straight edge
★ In the examUse a sharp pencil; accuracy marks allow about ±2 mm and ±2°. Check the scale units (cm to km?) before converting. For bearings, sketch the diagram with north lines before calculating.