Paper 22, worked through. Cambridge IGCSE Mathematics, May/June 2026: worked solutions

A hundred marks in two hours, so a little over a minute a mark. There is no calculator, which means the numbers have been chosen to come out neatly: if you meet an ugly one, look back for a slip before pressing on. The formula list is on page 2 of the paper, so nothing on it needs to be learned by heart.

1. 1(a)Adding one more car to a scatter diagram of CO2 emissions against fuel consumption.[1]
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   1. Work out what one small square is worth on each axis before plotting anything. Across, ten small squares cover 0.5, so each is 0.05. Up, ten small squares cover 20, so each is 2.
   2. Across: 6.2 is 0.2 past 6.0, which is four small squares.
   3. Up: 162 is 2 above 160, which is one small square.
   4. Mark a small, neat cross where the two meet.

   AnswerA cross at (6.2, 162): four small squares to the right of 6.0 and one small square above 160.

   The two axes have different scales. One small square is worth 0.05 across but 2 up. Most of the marks lost here go to people who assume they match.

   Revise this: Charts and scatter diagrams
2. 1(b)Naming the kind of correlation a scatter diagram shows.[1]
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   1. Look at the overall drift of the crosses from left to right.
   2. Here they climb: cars that use more fuel also give out more CO2. When both go up together, the correlation is positive.

   AnswerPositive

   One word does it. "Strong positive" is fine too. A plus sign with no word, or "it goes up", is not.

   Revise this: Charts and scatter diagrams
3. 1(c)Drawing a line of best fit through the crosses.[1]
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   1. Use a ruler. A line of best fit is one straight line: not a curve, and not dot to dot.
   2. Lay the ruler along the trend and shift it until there are about as many crosses above the edge as below it, all the way along.
   3. Draw it across the whole spread of the crosses. A good line here passes close to 125 above 4.35 on the fuel axis and close to 167 above 6.5.

   AnswerA single ruled straight line sloping upwards through the middle of the crosses, with roughly equal numbers of crosses on each side.

   It does not have to pass through any particular cross, and it does not have to start at the corner of the grid. Forcing it through the first and last crosses is the usual mistake.

   Revise this: Charts and scatter diagrams
4. 1(d)Using the line of best fit to estimate one car's fuel consumption from its emissions.[1]
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   1. Find 140 on the emissions axis and go straight across to your line.
   2. From that point on the line, go straight down to the fuel axis and read the scale. Each small square across is 0.05.
   3. With a line through the middle of the crosses, the reading comes out close to 5.1.

   AnswerAbout 5.1 litres/100 km. The mark is for reading your own line correctly, to within one small square.

   Read from the line, not from the nearest cross. And draw the two guide lines on the diagram, so the marker can see where your number came from.

   Revise this: Charts and scatter diagrams
5. 2(a)Measuring an accurately drawn net of an open box and finding the area of the whole net.[3]
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   1. The word "accurate" is the instruction to pick up your ruler. On the printed paper, measure each edge of the net. Every length comes out as 2 cm, 3 cm or 5 cm.
   2. The net is five rectangles, because the box has no lid: a base and four walls.
   3. Two of them are 2 by 5, two are 2 by 3 and one is 3 by 5.
   4. Areas: 2 × (2 × 5) + 2 × (2 × 3) + 3 × 5 = 20 + 12 + 15.

   Answer47 cm²

   62 is the trap: that is the surface area of a closed box, with six faces. This one is open, so count the rectangles in the net. There are five.

   On a screen the net is not life size. Measure it on the printed paper.

   Revise this: Area and perimeter
6. 2(b)Finding the volume of the box the net folds into.[2]
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   1. Picture it folded. The largest rectangle, 3 by 5, is the base. The four rectangles around it fold up into walls 2 cm high.
   2. Volume of a cuboid = length × width × height = 5 × 3 × 2.

   Answer30 cm³

   Having no lid changes the area of the net, not the space inside. The volume is the same as a closed box with these three edges.

   Revise this: Volume and surface area
7. 3Finding the reciprocal of a mixed number.[2]
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   1. Turn the mixed number into a single fraction first. Three and a third is (3 × 3 + 1) thirds, which is 10/3.
   2. The reciprocal is that fraction turned upside down.

   Answer3/10

   Flip it only once it is a single fraction. Flipping just the fraction part of the mixed number is the common mistake.

   The question asks for a fraction. 0.3 is the same number, but it loses a mark.

   Check: a number times its reciprocal is always 1, and 10/3 × 3/10 = 1.

   Revise this: Fractions, decimals and the order of operations
8. 4Dividing one fraction by another.[2]
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   1. To divide by a fraction, keep the first fraction, change ÷ to ×, and turn the second fraction upside down.
   2. 1/6 ÷ 2/7 becomes 1/6 × 7/2.
   3. Multiply the tops and multiply the bottoms: 7/12. Nothing cancels, so that is the answer.

   Answer7/12

   Only the second fraction is turned over. Turn the first one over by mistake and you get 12/7.

   Revise this: Fractions, decimals and the order of operations
9. 5(a)(i)Sliding shape A by a column vector.[2]
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   1. Shape A has corners at (−4, 2), (−2, 2), (−2, 0) and (−3, 0).
   2. In a column vector the top number moves the shape across and the bottom number moves it up. Top number −2: two squares left. Bottom number 5: five squares up.
   3. Move every corner the same way: take 2 from each x and add 5 to each y.
   4. The corners land on (−6, 7), (−4, 7), (−4, 5) and (−5, 5). Join them with a ruler.

   AnswerThe same shape, the same way up, with corners at (−6, 7), (−4, 7), (−4, 5) and (−5, 5).

   A translation never turns or flips a shape. If yours looks different from A, a corner has gone astray.

   A negative number on top means left, not down.

   Revise this: Transformations
10. 5(a)(ii)Reflecting shape A in the line y = x.[2]
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    1. Draw the mirror line first. y = x is the diagonal through (0, 0), (1, 1), (2, 2) and so on.
    2. Reflecting in y = x swaps the two coordinates of every point: (x, y) becomes (y, x).
    3. So (−4, 2) goes to (2, −4), (−2, 2) goes to (2, −2), (−2, 0) goes to (0, −2) and (−3, 0) goes to (0, −3).
    4. Plot the four new corners and join them.

    AnswerA shape with corners at (2, −4), (2, −2), (0, −2) and (0, −3).

    y = x is not the x-axis and not the y-axis. Drawing the line on the grid before you start stops you reflecting in the wrong one.

    Check a corner by counting squares diagonally: it should be as far from the mirror line as the corner it came from, on the other side.

    Revise this: Transformations
11. 5(b)Describing the one transformation that takes shape A to shape B.[3]
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    1. B is the same size as A but turned round, so it is a rotation. A rotation needs three things said: the word rotation, the angle with its direction, and the centre.
    2. Angle: the top edge of A is horizontal. In B that edge has become vertical, so it is a quarter turn, 90°.
    3. Direction and centre: use tracing paper. Trace A, hold the pencil point on a likely centre and turn the paper. Only a quarter turn anticlockwise about (3, 3) puts the tracing exactly on B.
    4. Check with a corner. (−4, 2) is 7 left and 1 down from (3, 3). A quarter turn anticlockwise turns that into 1 right and 7 down, which is (4, −4): a corner of B.

    AnswerRotation, 90° anticlockwise, centre (3, 3).

    One mark for each of the three parts. "Rotation 90°" with no direction and no centre scores one out of three.

    Give one transformation only. A rotation followed by a translation scores nothing, even if it does get A to B.

    Write "rotation", not "turn".

    Revise this: Transformations
12. 6(a)Putting a square root and a whole number into a formula that squares them both.[2]
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    1. Square each value before doing anything else. Squaring a square root undoes it: (√18)² = 18. And 2² = 4.
    2. E = 5 × 18 − 4 = 90 − 4.

    AnswerE = 86

    5c² means 5 × (c²): square first, then multiply by 5. It is not (5c)².

    There is no need to simplify √18 first. It disappears the moment it is squared.

    Revise this: Squares, cubes and roots
13. 6(b)Rearranging a formula to make a squared letter the subject.[3]
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    1. The aim is c alone on one side. Undo what has been done to it, one step at a time.
    2. Add d² to both sides: E + d² = 5c².
    3. Divide both sides by 5: (E + d²) ÷ 5 = c².
    4. Square root both sides: c = √((E + d²) ÷ 5).

    Answerc = √((E + d²)/5), with the root sign covering the whole fraction.

    Each of the three moves is a mark: adding d², dividing by 5, taking the square root.

    The root must cover everything, the top and the bottom of the fraction. A root over the top alone is a different formula.

    Do not "simplify" √(E + d²) to √E + d. A root cannot be split across a plus sign.

    Revise this: Changing the subject of a formula
14. 7Taking a common factor out of two terms.[2]
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    1. Look for what both terms share. Numbers: 6 and 4 share 2. Letters: both have an x.
    2. So 2x comes out. Divide each term by 2x to see what is left: 6wx ÷ 2x = 3w, and 4xy ÷ 2x = 2y.

    Answer2x(3w − 2y)

    Taking out only the 2, or only the x, earns one mark of the two. Look inside the bracket: if what is there still shares a factor, you have not finished.

    Multiply it back out to check: 2x × 3w = 6wx and 2x × 2y = 4xy.

    Revise this: Expanding and factorising
15. 8Recalling two exact trigonometric values.[2]
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    1. These are learned, not worked out on the day. But two triangles give you all of them.
    2. For 45°: a right-angled triangle with two short sides of 1 has a long side of √2. Cosine is adjacent over hypotenuse, so cos 45° = 1/√2.
    3. For 60°: cut an equilateral triangle of side 2 in half. The half has sides 1, √3 and 2, with the 60° angle next to the side of 1. Tangent is opposite over adjacent, so tan 60° = √3 ÷ 1 = √3.

    Answercos 45° = 1/√2 and tan 60° = √3

    The list has near misses in it on purpose. 1/√3 is tan 30°, 1/2 is cos 60° and sin 30°, and √3/2 is sin 60° and cos 30°.

    Sketch those two triangles in the margin at the start of every non-calculator paper.

    Revise this: Exact values and trigonometric graphs
16. 9(a)Writing a number as a product of primes.[2]
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    1. Divide by the smallest prime that goes in, and keep going. 63 is odd, so 2 is out. 63 ÷ 3 = 21.
    2. 21 ÷ 3 = 7, and 7 is prime, so stop.

    Answer3 × 3 × 7, or 3² × 7

    A product means the primes written with multiplication signs between them. The list "3, 3, 7" earns only one of the two marks.

    Revise this: Types of number
17. 9(b)Finding the lowest common multiple of two numbers from their prime factors.[2]
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    1. Write both as products of primes: 63 = 3 × 3 × 7 and 42 = 2 × 3 × 7.
    2. The LCM needs every prime that appears, as many times as it appears in whichever number has more of it: one 2, two 3s and one 7.
    3. 2 × 3 × 3 × 7 = 2 × 63.

    Answer126

    Multiplying 63 by 42 gives a common multiple, but nowhere near the lowest one.

    Quick check: 126 = 63 × 2 and 126 = 42 × 3, so both go into it.

    Revise this: Types of number
18. 9(c)Finding the highest common factor of two algebraic terms.[2]
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    1. Deal with the numbers and each letter separately.
    2. Numbers: 63 = 3 × 3 × 7 and 42 = 2 × 3 × 7. They share one 3 and one 7, so the HCF of the numbers is 21.
    3. Letters: take the lower power of each. a⁵ and a² share a². c² and c⁵ share c².

    Answer21a²c²

    For an HCF take the smaller power of each letter. The larger powers, a⁵ and c⁵, belong to the LCM.

    Revise this: Types of number
19. 10Turning a recurring decimal into a fraction, when only the last digit repeats.[3]
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    1. The dot is over the 3 only, so the number is 0.7333… Call it x.
    2. Multiply by 10 and by 100, so that the repeating 3s line up after the decimal point: 10x = 7.333… and 100x = 73.333…
    3. Subtract one from the other and the recurring part vanishes: 100x − 10x = 73.333… − 7.333…, so 90x = 66.
    4. x = 66/90. Divide top and bottom by 6.

    Answer11/15

    Check where the dot is before you start. With dots over both digits it would be 0.737373…, which is 73/99: a different number.

    The fraction must be cancelled right down. 66/90 and 22/30 earn two of the three marks.

    Revise this: Recurring decimals
20. 11Finding how many sides a regular polygon has from its interior angle.[2]
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    1. Go through the exterior angle. An interior angle and its exterior angle sit on a straight line, so the exterior angle is 180 − 140 = 40°.
    2. The exterior angles of any polygon add up to 360°. In a regular polygon they are all equal, so the number of sides is 360 ÷ 40.

    Answer9

    Dividing 360 by 140 is the trap. 360 goes with exterior angles, never interior ones.

    The other route: (n − 2) × 180 = 140n gives 40n = 360, so n = 9. Same answer, more algebra.

    Revise this: Angles in polygons
21. 12(a)Drawing four straight lines on a grid from their equations.[5]
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    1. x = 8 is every point whose x-coordinate is 8: a vertical line through 8 on the x-axis.
    2. y = 2 is a horizontal line through 2 on the y-axis.
    3. x + y = 8: find where it crosses each axis. When x = 0, y = 8. When y = 0, x = 8. Rule a line through (0, 8) and (8, 0).
    4. 2y − x = 4: when x = 0, 2y = 4, so y = 2. When x = 8, 2y = 12, so y = 6. Rule a line through (0, 2) and (8, 6). A third point checks it: x = 4 gives y = 4.

    AnswerFour ruled, solid lines: a vertical line at x = 8, a horizontal line at y = 2, a line through (0, 8) and (8, 0), and a line through (0, 2), (4, 4) and (8, 6).

    x = 8 is vertical and y = 2 is horizontal. It feels the wrong way round, and swapping them is the commonest slip.

    Draw the lines solid, not dashed. Every inequality in the next part includes "or equal to", and a solid line is how that is shown.

    The last line carries two of the five marks. A line with the right slope in the wrong place, or one that only passes through (0, 2), keeps one of them.

    Revise this: Equations of straight lines
22. 12(b)Shading out the unwanted side of each line to leave the region that satisfies four inequalities.[2]
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    1. Take one inequality at a time and shade the side you do not want.
    2. x ≤ 8: the wanted points are on or to the left of the vertical line, so shade to the right of it.
    3. y ≥ 2: the wanted points are on or above the horizontal line, so shade below it.
    4. x + y ≥ 8: test the point (0, 0). 0 + 0 is not 8 or more, so the origin's side is unwanted. Shade below and to the left of that line.
    5. 2y − x ≤ 4: test (0, 0) again. 0 − 0 = 0, which is no more than 4, so the origin's side is wanted. Shade above that line.
    6. The patch left clean is R. It is a four-sided shape with corners at (4, 4), (6, 2), (8, 2) and (8, 6).

    AnswerR is the unshaded four-sided region with corners at (4, 4), (6, 2), (8, 2) and (8, 6), with the letter R written inside it.

    Read the instruction: here you shade what you do not want. Shading R itself is the classic slip.

    Testing a point settles every doubt. Put (0, 0) into the inequality, unless the line passes through it, and see whether it works.

    Write the letter R inside the region. The question asks for it.

    Revise this: Inequalities
23. 13Finding the length of the arc of a quarter-circle sector, leaving π in the answer.[2]
    Show how to do itHide the working

    1. An arc is a fraction of the whole circumference. The fraction is the sector angle over 360: 90/360 = 1/4.
    2. Whole circumference = 2 × π × r = 2 × π × 12 = 24π.
    3. A quarter of that: 24π ÷ 4.

    Answer6π cm

    "In terms of π" means leave π as a symbol. Multiplying it out to 18.8 throws away the last mark.

    Arc length uses the circumference, 2πr. Sector area uses πr². Mixing up the two is the usual error.

    Revise this: Circles, arcs and sectors
24. 14(a)Using an nth-term rule to write out the start of a sequence.[2]
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    1. n is the position of the term. Put in n = 1, then 2, then 3.
    2. n = 1: 1² − 3 = −2. n = 2: 2² − 3 = 4 − 3 = 1. n = 3: 3² − 3 = 9 − 3 = 6.

    Answer−2, 1, 6

    Start at n = 1, not n = 0. Starting at zero gives −3, −2, 1, which is one place out all the way along.

    Square first, then take away 3.

    Revise this: Sequences
25. 14(b)Finding the nth term of a sequence that is multiplied by the same number each time.[2]
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    1. Check the differences first: 8, 24, 72. They are not equal, so this is not a sequence that goes up in equal steps.
    2. Check the ratios instead: 12 ÷ 4 = 3, 36 ÷ 12 = 3, 108 ÷ 36 = 3. Each term is 3 times the one before. That makes it a geometric sequence.
    3. The nth term of a geometric sequence is the first term × the ratio to the power (n − 1). The power is n − 1 because the first term has not been multiplied by 3 at all yet.
    4. Here the first term is 4 and the ratio is 3.

    Answer4 × 3ⁿ⁻¹, which is 4 times 3 to the power (n − 1)

    Test it: n = 1 gives 4 × 3⁰ = 4, and n = 3 gives 4 × 3² = 36. Both match.

    4 × 3ⁿ is one term ahead, because it starts at 12. And the power belongs to the 3 only: 12 to the power (n − 1) is something else.

    Revise this: Sequences
26. 15Finding the angle between a tangent and a chord, using two circle theorems, with a reason for every step.[5]
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    1. ABCD is a cyclic quadrilateral: all four of its corners are on the circle. Opposite angles of a cyclic quadrilateral add up to 180°. Angle BAD is opposite angle BCD, so angle BAD = 180 − 75 = 105°.
    2. Now look at triangle ABD. Angles in a triangle add up to 180°, so angle ABD = 180 − 105 − 20 = 55°.
    3. Last, the tangent. The angle between a tangent and a chord equals the angle in the alternate segment. The tangent is EA and the chord is AD, and the angle that chord AD makes at the far side of the circle is angle ABD.
    4. So angle EAD = angle ABD = 55°.

    AnswerAngle EAD = 55°. Reasons: opposite angles of a cyclic quadrilateral add up to 180°; angles in a triangle add up to 180°; alternate segment theorem.

    Three of the five marks are for the angles and two are for the reasons. The two reasons that have to be there are the cyclic quadrilateral one and the alternate segment theorem, each in words.

    Write "alternate segment theorem" in full. Abbreviations, and near misses such as "alternate angles", do not get the mark.

    Another way round: the alternate segment theorem also gives angle BAF = angle ADB = 20°. Then angles on a straight line give 180 − 105 − 20 = 55°.

    Revise this: Circle theorems
27. 16Estimating the radius of a sphere from its volume, with π taken as 3.[2]
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    1. The volume of a sphere is 4/3 × π × r³. It is in the formula list on page 2.
    2. With π as 3, the 4/3 × 3 becomes 4, so the volume is simply 4 × r³.
    3. 4 × r³ = 500, so r³ = 125.
    4. The cube root of 125 is 5, because 5 × 5 × 5 = 125.

    Answer5 cm

    You are told to use 3 for π so that the numbers fall out neatly without a calculator. Using 3.14 only makes the arithmetic harder.

    r³ = 125 is not the answer: take the cube root. And a length cannot be negative, so ±5 is wrong.

    Revise this: Volume and surface area
28. 17(a)Multiplying out two brackets that contain surds.[2]
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    1. Multiply each term in the first bracket by each term in the second: four products in all.
    2. 4 × 3 = 12. 4 × 2√7 = 8√7. −√7 × 3 = −3√7. −√7 × 2√7 = −2 × 7 = −14, because √7 × √7 = 7.
    3. Collect the whole numbers and the surds separately: 12 − 14 = −2, and 8√7 − 3√7 = 5√7.

    Answer−2 + 5√7, which can also be written 5√7 − 2

    √7 × √7 is exactly 7. Writing 49, or leaving it as √49, is where this goes wrong.

    Revise this: Surds
29. 17(b)Clearing a surd from the bottom of a fraction when the bottom has two terms.[3]
    Show how to do itHide the working

    1. Multiply the top and the bottom by the bottom bracket with its middle sign changed: 3 + √5. Top and bottom are multiplied by the same thing, so the value of the fraction does not change.
    2. Bottom: (3 − √5)(3 + √5) = 9 + 3√5 − 3√5 − 5 = 4. The two surd terms cancel each other, which is the whole point.
    3. Top: 2 × (3 + √5) = 6 + 2√5.
    4. So the fraction is (6 + 2√5)/4. Every term divides by 2.

    Answer(3 + √5)/2

    Stopping at (6 + 2√5)/4 costs the last mark. "Simplify" means cancel the 2.

    When you cancel, divide every term by 2: the 6, the 2√5 and the 4. Cancelling only one of the terms on top is a common slip.

    Multiplying top and bottom by 3 − √5, with the same sign, does not get rid of the surd.

    Revise this: Surds
30. 18(a)The probability that two tiles, picked one after the other with the first put back, show the same even number.[3]
    Show how to do itHide the working

    1. The even numbers on the tiles are 2 and 4. So there are two ways to succeed: a 2 then a 2, or a 4 then a 4.
    2. Count the tiles. Two of the ten are 2s and five of the ten are 4s. The first tile goes back, so the second pick has the same chances as the first.
    3. 2 then 2: 2/10 × 2/10 = 4/100. 4 then 4: 5/10 × 5/10 = 25/100.
    4. Either one will do, so add them: 4/100 + 25/100.

    Answer29/100

    "And" means multiply, "or" means add. Here it is (2 and 2) or (4 and 4).

    "The same even number" rules out a 2 with a 4. The chance of any two evens, 7/10 × 7/10, answers a different question.

    Replaced means the bottom of the fraction stays 10 both times.

    Revise this: Probability of combined events
31. 18(b)The probability that three tiles, picked without putting any back, add up to 11.[3]
    Show how to do itHide the working

    1. First find which tiles can make 11. The biggest tiles are 4s. 4 + 4 + 4 = 12 is too many, and 4 + 4 + 3 = 11 works. Anything smaller falls short, so the only way is two 4s and a 3.
    2. Take one order: 4, then 4, then 3. Nothing goes back, so the numbers shrink each time: 5/10 × 4/9 × 2/8 = 40/720.
    3. The 3 could come first, second or third. That is three orders, and each has the same probability.
    4. 3 × 40/720 = 120/720.

    Answer1/6

    Forgetting the three orders gives 40/720 = 1/18, the most common wrong answer.

    Without replacement, both numbers in each fraction change: one fewer 4 on top, one fewer tile underneath.

    Revise this: Probability of combined events
32. 19Multiplying out three brackets.[3]
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    1. Do two of the brackets first, then multiply the result by the third.
    2. (x + 4)(x − 2) = x² − 2x + 4x − 8 = x² + 2x − 8.
    3. Now multiply every term of that by 3x, and then by 1. 3x × (x² + 2x − 8) = 3x³ + 6x² − 24x. And 1 × (x² + 2x − 8) = x² + 2x − 8.
    4. Collect like terms: 3x³, then 6x² + x² = 7x², then −24x + 2x = −22x, then −8.

    Answer3x³ + 7x² − 22x − 8

    Tidy up the first pair before bringing in the third bracket. Three terms times two is six products; skip the tidying and it is eight.

    Check with a number. Put x = 1 into the question: 4 × 5 × (−1) = −20. Into the answer: 3 + 7 − 22 − 8 = −20. They agree.

    Revise this: Expanding and factorising
33. 20Finding a height in one of two similar bottles from their volumes.[3]
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    1. For similar shapes, volumes go with the cube of the lengths. So work backwards: the ratio of the lengths is the cube root of the ratio of the volumes.
    2. The volumes are 216 and 27. Their cube roots are 6 and 3, because 6 × 6 × 6 = 216 and 3 × 3 × 3 = 27.
    3. So lengths are in the ratio 6 : 3, which is 2 : 1. The larger bottle is twice as tall as the smaller one.
    4. h = 10 ÷ 2.

    Answerh = 5

    Dividing 10 by 8 (because 216 ÷ 27 = 8) is the trap. 8 is how many times bigger the volume is, not the height.

    Lengths use the scale factor, areas use its square, volumes use its cube.

    Revise this: Similar and congruent shapes
34. 21(a)Completing the square.[2]
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    1. Halve the number in front of x: half of 8 is 4. That goes in the bracket: (x + 4)².
    2. Multiplied out, (x + 4)² is x² + 8x + 16. That is 16 more than wanted, so take the 16 away again.
    3. x² + 8x − 7 = (x + 4)² − 16 − 7.

    Answer(x + 4)² − 23

    The number to take away is the square of what is in the bracket, 4² = 16. Not 4, and not 8.

    Check by expanding: x² + 8x + 16 − 23 = x² + 8x − 7.

    Revise this: Quadratic equations
35. 21(b)Reading the turning point of a quadratic curve from its completed-square form.[1]
    Show how to do itHide the working

    1. Use the answer to part (a): y = (x + 4)² − 23.
    2. A square is never negative, so the smallest that (x + 4)² can be is 0. That happens when x = −4.
    3. At that point y = 0 − 23 = −23.

    Answer(−4, −23)

    The x-coordinate has the opposite sign to the number in the bracket: + 4 in the bracket means x = −4. The y-coordinate keeps its sign.

    One mark, and the words "write down": no new working is needed if part (a) is right.

    Revise this: Graphs of curves
36. 22(a)(i)Finding the vector along part of a parallelogram's diagonal, given how a point divides it.[2]
    Show how to do itHide the working

    1. Find the whole diagonal from A to C first, by a route you know: A to O, then O to C. Going from A to O is against the arrow of a, so it is −a. Then comes + c. So the vector AC is c − a.
    2. P divides AC in the ratio 1 : 4. That is 5 equal parts with AP as 1 of them, so AP is one fifth of AC.

    AnswerAP = 1/5 (c − a), which is the same as 1/5 c − 1/5 a

    A ratio of 1 : 4 means fifths, not quarters. Add the two parts of the ratio to get the bottom of the fraction.

    Going against an arrow changes the sign of the vector.

    Revise this: Vectors
37. 22(a)(ii)Finding the vector from O to the point on the diagonal.[1]
    Show how to do itHide the working

    1. Get to P by a route you already know: O to A, then A to P.
    2. OP = a + 1/5 (c − a) = a − 1/5 a + 1/5 c.
    3. a − 1/5 a is 4/5 a.

    AnswerOP = 4/5 a + 1/5 c, which can also be written 1/5 (4a + c)

    Simplest form means collecting the a terms together. Leaving a + 1/5 (c − a) does not get the mark.

    Revise this: Vectors
38. 22(b)Proving that three points are in a straight line, using vectors.[2]
    Show how to do itHide the working

    1. Find the vector from O to Q by going O to A, then A to Q: OQ = a + 1/4 c.
    2. Write both vectors so that the same bracket shows. OP = 1/5 (4a + c) and OQ = 1/4 (4a + c).
    3. So OQ is a multiple of OP: OQ = 5/4 OP. Vectors that are multiples of each other are parallel.
    4. Both start at O. Two parallel lines through the same point are one and the same line, so O, P and Q all lie on it.

    AnswerOP = 1/5 (4a + c) and OQ = 1/4 (4a + c), so OQ = 5/4 OP. The two vectors are parallel and share the point O, so O, P and Q lie in a straight line.

    Two marks, two things: the vectors shown to be multiples of each other, and a sentence saying they are parallel with a point in common. "They are parallel" alone leaves the proof unfinished, because parallel lines need not be the same line.

    Taking a factor out of each vector, so the same bracket appears in both, makes the multiple obvious.

    Revise this: Vectors
39. 23(a)Finding the gradient of a curve at a point by differentiating.[3]
    Show how to do itHide the working

    1. The gradient of the tangent is the value of dy/dx at that point.
    2. Differentiate one term at a time: multiply by the power, then take one off the power. 2x² becomes 4x. −11x becomes −11. The 12 standing alone becomes 0.
    3. dy/dx = 4x − 11.
    4. At x = 2: 4 × 2 − 11 = 8 − 11.

    Answer−3

    Put x = 2 into dy/dx, not into the original equation. The original gives the height of the curve there, not its slope.

    A number standing alone differentiates to nothing. Leaving the 12 in is a common slip.

    Revise this: Differentiation
40. 23(b)Finding the equation of the line through a point on the curve, at right angles to the tangent there.[4]
    Show how to do itHide the working

    1. A straight line needs a point and a gradient.
    2. Point: put x = 2 into the curve. y = 2 × 2² − 11 × 2 + 12 = 8 − 22 + 12 = −2. So the line passes through (2, −2).
    3. Gradient: the tangent's gradient is −3, from part (a). Perpendicular gradients multiply to −1, so turn it upside down and change the sign: 1/3.
    4. Use y = mx + c with m = 1/3 and the point (2, −2): −2 = 1/3 × 2 + c, so c = −2 − 2/3 = −8/3.

    Answery = 1/3 x − 8/3, which can also be written 3y = x − 8

    Turn it upside down and change the sign: −3 becomes + 1/3. Using −3 again gives the tangent itself, and −1/3 or 3 gives a line that is not perpendicular.

    The point comes from the curve's equation and the gradient comes from dy/dx. Keep the two jobs apart.

    Leave the fractions as fractions. 0.33 and 2.67 are rounded, and they lose the last mark.

    Revise this: Parallel and perpendicular lines
41. 24Solving an equation with the unknown in the powers, by writing both sides as powers of the same number.[3]
    Show how to do itHide the working

    1. 49 is 7², so the right-hand side can be rewritten with 7 as its base: 49 to the power (2 − x) is 7² to the power (2 − x).
    2. A power of a power multiplies, so that is 7 to the power 2(2 − x), which is 7 to the power (4 − 2x).
    3. Now both sides are powers of 7, so the powers themselves must be equal: 2x + 1 = 4 − 2x.
    4. Add 2x to both sides: 4x + 1 = 4. Take away 1: 4x = 3.

    Answerx = 3/4

    Multiply the whole of (2 − x) by 2. Writing 4 − x in place of 4 − 2x is the usual slip.

    Check: the power on the left is 2 × 3/4 + 1 = 2.5. On the right, 49 to the power 1.25 is 7 to the power 2.5. They match.

    Revise this: Indices
42. 25Finding where a straight line crosses a curve: a pair of simultaneous equations, one of them a quadratic.[5]
    Show how to do itHide the working

    1. At a crossing point both equations give the same y, so set them equal: 7 − 4x = 4 + 3x − 2x².
    2. Move everything to one side, so that the x² term is positive. Add 2x², take away 3x, take away 4: 2x² − 7x + 3 = 0.
    3. Factorise. Two numbers that multiply to 2 × 3 = 6 and add to −7 are −6 and −1. So 2x² − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3).
    4. Either bracket can be zero. 2x − 1 = 0 gives x = 1/2, and x − 3 = 0 gives x = 3.
    5. Find each y from the line, the easier of the two equations. x = 1/2: y = 7 − 2 = 5. x = 3: y = 7 − 12 = −5.

    Answer(1/2, 5) and (3, −5)

    Finding the two x values is not the end. The question asks for coordinates, so each x needs its y.

    Pair them up correctly: 1/2 goes with 5, and 3 goes with −5.

    Check one point in the curve's equation: x = 3 gives 4 + 9 − 18 = −5. It agrees.

    Revise this: Simultaneous equations

These explanations are Papermunch’s own, written to teach the method. The answers have been checked against the exam board’s mark scheme, which has the final say: open it above. The question paper and mark scheme belong toCambridge University Press & Assessment: each question shown here is drawn from the paper itself as you read, and is not kept on this site.