INT Math — Worked Solutions

Cambridge IGCSE Core · 0580/12 & 0580/33

Paper 1 (0580/12) — Question 1

Question 1

Write the number thirty thousand and fifty in figures.

Thirty thousand = 30 000, and fifty = 50.

Combine: \(30\,000 + 50 = 30\,050\).

Answer: 30 050
Paper 1 (0580/12) — Question 2

Question 2

Write 5926 correct to the nearest 10.

The units digit is 6 (\(\ge 5\)), so round the tens up: the 2 tens become 3 tens and the units become 0.

Answer: 5930
Paper 1 (0580/12) — Question 3

Question 3

Mark the midpoint of the line ST.
S T midpoint
Line ST — the midpoint (centre of the segment) is shown in colour.

Measure the full length of ST, halve it, and mark the point at that distance from S (equivalently from T).

Answer: midpoint of ST marked (the centre point of the segment)
Paper 1 (0580/12) — Question 4

Question 4

(a) Shade \(\tfrac{2}{9}\) of the shape.   (b) Write \(\tfrac{2}{9}\) as a percentage.
6 × 6 grid (36 squares); \(\tfrac{2}{9}\times 36 = 8\) squares shaded.

(a) Shading

The grid is a \(6\times 6 = 36\)-square array.

\[\frac{2}{9}\times 36 = \frac{72}{9} = 8 \text{ squares}\]

(a) Shade 8 squares

(b) Percentage

\[\frac{2}{9} = 2 \div 9 = 0.2222\ldots = 22.22\ldots\%\]

(b) 22.2% (22.22…%)
Paper 1 (0580/12) — Question 5

Question 5

A night bus runs from 21:50 to 05:18 the next day. Work out the running time.

Split the interval at midnight:

  • 21:50 → 24:00 = 2 h 10 min
  • 00:00 → 05:18 = 5 h 18 min

\[2\text{ h }10\text{ min} + 5\text{ h }18\text{ min} = 7\text{ h }28\text{ min}\]

Answer: 7 h 28 min
Paper 1 (0580/12) — Question 6

Question 6

(a) From 34, 55, 76, 83, 111, 121 write all multiples of 11.   (b) Show 918 271 937 is a multiple of 11 using the alternating digit method.

(a)

\(55 = 5\times 11\) and \(121 = 11\times 11\); the others are not multiples of 11.

(a) 55 and 121

(b)

Alternately subtract and add the digits 9, 1, 8, 2, 7, 1, 9, 3, 7:

\[9 - 1 + 8 - 2 + 7 - 1 + 9 - 3 + 7 = 33\]

Since \(33 = 3\times 11\) is a multiple of 11, the number is a multiple of 11.

(b) 33 = 3 × 11 ⇒ multiple of 11
Paper 1 (0580/12) — Question 7

Question 7

The range of eight numbers is 31. Seven of them are 28, 36, 42, 24, 38, 16, 21. Find the two possible values of the eighth number.

For the seven known numbers: maximum = 42, minimum = 16. Let the eighth number be \(x\).

Case 1 — \(x\) is the new maximum: \(\;x - 16 = 31 \Rightarrow x = 47\)  (47 > 42 ✓)

Case 2 — \(x\) is the new minimum: \(\;42 - x = 31 \Rightarrow x = 11\)  (11 < 16 ✓)

Answer: 11 or 47
Paper 1 (0580/12) — Question 8

Question 8

Calculate \(\sqrt{5.76} + 2.8^{3}\).

\[\sqrt{5.76} = 2.4\]

\[2.8^{3} = 2.8\times 2.8\times 2.8 = 21.952\]

\[2.4 + 21.952 = 24.352\]

Answer: 24.352
Paper 1 (0580/12) — Question 9

Question 9

Simplify \(4m + 7k - m + 3k\).

Collect like terms: \(4m - m = 3m\) and \(7k + 3k = 10k\).

Answer: 3m + 10k
Paper 1 (0580/12) — Question 10

Question 10

List: −9, −7, −3, −1, 0, 2, 5, 6, 8. (a) Highest product of two numbers. (b) Lowest product of three numbers.

(a)

Two large negatives give a large positive: \((-9)\times(-7) = 63\). Compare \(6\times 8 = 48\).

(a) 63

(b)

Most negative result: largest negative × two largest positives: \((-9)\times 8\times 6 = -432\). (e.g. \((-9)(-7)(-3) = -189\) is larger.)

(b) −432
Paper 1 (0580/12) — Question 11

Question 11

Data (14 values): 28, 46, 54, 71, 70, 65, 49, 50, 64, 77, 68, 72, 45, 58. (a) Complete the stem-and-leaf diagram. (b) Find the median.

(a) Stem-and-leaf (Key 2 | 8 = 28)

2 | 8
3 |
4 | 5 6 9
5 | 0 4 8
6 | 4 5 8
7 | 0 1 2 7

(b) Median

Ordered: 28, 45, 46, 49, 50, 54, 58, 64, 65, 68, 70, 71, 72, 77.

With 14 values the median is the mean of the 7th and 8th: \[\frac{58 + 64}{2} = \frac{122}{2} = 61\]

(b) 61
Paper 1 (0580/12) — Question 12

Question 12

Net of a cuboid with dimensions 10 cm × 4 cm × 5 cm. (a) Surface area. (b) Volume.
10 cm 4 cm 5 cm
Net of the cuboid (10 cm × 4 cm × 5 cm).

(a) Surface area

\[2(lw + lh + wh) = 2(10\times4 + 10\times5 + 4\times5) = 2(40+50+20) = 220\ \text{cm}^2\]

(a) 220 cm²

(b) Volume

\[10\times 4\times 5 = 200\ \text{cm}^3\]

(b) 200 cm³
Paper 1 (0580/12) — Question 13

Question 13

20 cars, 3 are blue. (a) Pie-chart angle for blue. (b) P(not blue).

(a)

\[\frac{3}{20}\times 360^\circ = \frac{1080}{20} = 54^\circ\]

(a) 54°

(b)

\[P(\text{not blue}) = \frac{20-3}{20} = \frac{17}{20}\]

(b) 17/20
Paper 1 (0580/12) — Question 14

Question 14

Factorise \(3x^{3} - 7xy\).

Common factor \(x\): \(\;3x^{3} - 7xy = x(3x^{2} - 7y)\).

Answer: x(3x² − 7y)
Paper 1 (0580/12) — Question 15

Question 15

A = (7, 1), B = (−3, 4). Write \(\vec{AB}\) as a column vector.
x y −4−3−2−1 1234 5678 123 45 × A × B
A = (7, 1) and B = (−3, 4) plotted on the grid.

\[\vec{AB} = \binom{x_B - x_A}{y_B - y_A} = \binom{-3-7}{4-1} = \binom{-10}{3}\]

Answer: (−10, 3)ᵀ
Paper 1 (0580/12) — Question 16

Question 16

1 euro = 1.05 dollars, 1 rupee = 0.013 dollars. Vani changes \(x\) euros to dollars, then to 17 850 rupees. Find \(x\).

Rupees → dollars: \[17850 \times 0.013 = 232.05 \text{ dollars}\]

Dollars → euros: \[x = \frac{232.05}{1.05} = 221\]

Answer: 221
Paper 1 (0580/12) — Question 17

Question 17

The line \(y = 2x - 5\) meets \(y = 3\) at P. Find P.

Set \(y = 3\): \(\;3 = 2x - 5 \Rightarrow 2x = 8 \Rightarrow x = 4\).

Answer: P = (4, 3)
Paper 1 (0580/12) — Question 18

Question 18

(a) Describe the single transformation mapping A onto B. (b) Reflect A in the line \(x = -1\).
x y A B −8−7−6 −5−4−3 −2−1 123 456 4321 −1−2−3−4 −5−6−7−8
Shape A (shaded) and shape B, exactly as positioned on the paper. A → B is a 90° rotation about the origin.

(a)

A (upper-left) maps onto B (lower-middle) by a quarter turn about the origin.

(a) Rotation, 90° anticlockwise, centre (0, 0)

(b)

Under reflection in \(x = -1\), each point \((x, y)\mapsto(-2 - x,\ y)\). Apply to every vertex of A and draw the mirror image on the right of the line.

(b) Image of A reflected in x = −1 drawn correctly
Paper 1 (0580/12) — Question 19

Question 19

Small circle radius 7 cm, large radius \(R\) cm. Area of 16 small circles = area of 1 large circle. Find \(R\).
7 cm R cm
Small circle (radius 7 cm) and large circle (radius \(R\) cm). Not to scale.

\[16 \times \pi \times 7^{2} = \pi R^{2}\]

\[R^{2} = 16\times 49 = 784 \Rightarrow R = \sqrt{784} = 28\]

Answer: R = 28
Paper 1 (0580/12) — Question 20

Question 20

(a) \(n\)th term = \(n^{2} - 3\); find first three terms. (b) Find the \(n\)th term of 2, 9, 16, 23, 30.

(a)

\(n=1: 1-3=-2;\quad n=2: 4-3=1;\quad n=3: 9-3=6\).

(a) −2, 1, 6

(b)

Common difference 7, so the term contains \(7n\): values \(7,14,21,28,35\), each 5 more than the sequence.

(b) 7n − 5
Paper 1 (0580/12) — Question 21

Question 21

\(l = 18.7\) m correct to the nearest 10 cm (= 0.1 m). Complete \(\ldots \le l < \ldots\)

Half of 0.1 m is 0.05 m: \(\;18.7 - 0.05 = 18.65\) and \(18.7 + 0.05 = 18.75\).

Answer: 18.65 ≤ l < 18.75
Paper 1 (0580/12) — Question 22

Question 22

\(6.5\times 10^{19}\times n = 5.46\times 10^{23}\). Find \(n\) in standard form.

\[n = \frac{5.46\times 10^{23}}{6.5\times 10^{19}} = \frac{5.46}{6.5}\times 10^{4} = 0.84\times 10^{4} = 8.4\times 10^{3}\]

Answer: 8.4 × 10³
Paper 1 (0580/12) — Question 23

Question 23

Triangle ABC similar to DEF. AB = \(h\), AC = 118.9, DE = 97.5, DF = 159.9. Find \(h\).
B A C h cm 118.9 cm E D F 97.5 cm 159.9 cm
Triangle ABC is mathematically similar to triangle DEF. Not to scale.

\[\frac{h}{97.5} = \frac{118.9}{159.9}\]

\[h = 97.5\times\frac{118.9}{159.9} = 97.5\times 0.74359\ldots = 72.5\]

Answer: h = 72.5
Paper 1 (0580/12) — Question 24

Question 24

Without a calculator, work out \(1\tfrac{1}{4} - \tfrac{5}{6}\), as a fraction in simplest form.

\[1\tfrac{1}{4} = \frac{5}{4}\]

Common denominator 12: \(\;\dfrac{5}{4} = \dfrac{15}{12},\quad \dfrac{5}{6} = \dfrac{10}{12}\).

\[\frac{15}{12} - \frac{10}{12} = \frac{5}{12}\]

Answer: 5/12
Paper 1 (0580/12) — Question 25

Question 25

HCF of two numbers is 6, LCM is 90, both numbers > 6. Find them.

Write the numbers as \(6a\) and \(6b\) with \(a,b\) coprime. Then \(\text{LCM} = 6ab = 90 \Rightarrow ab = 15\).

Coprime pairs of 15: \((1,15)\) → 6 and 90 (rejected, 6 is not > 6); \((3,5)\) → 18 and 30.

Check: HCF(18, 30) = 6 ✓, LCM(18, 30) = 90 ✓.

Answer: 18 and 30
Paper 3 (0580/33) — Question 1

Question 1

Write down all the factors of 26.

\(26 = 1\times 26 = 2\times 13\).

Answer: 1, 2, 13, 26
Paper 3 (0580/33) — Question 2

Question 2

A and B lie on a circle, centre O. (a) Name the line AB. (b) Draw a radius.
A B O
A and B on a circle, centre O; AB is a chord (it does not pass through O).

(a) AB joins two points on the circle without passing through the centre → it is a chord.

(b) A radius is a straight line from centre O to any point on the circle.

(a) Chord  ·  (b) Radius drawn
Paper 3 (0580/33) — Question 3

Question 3

(a) Write 0.25 as a fraction. (b) Write \(\tfrac{1}{2}\) as a percentage. (c) Write 7% as a decimal.

(a) \(0.25 = \dfrac{25}{100} = \dfrac{1}{4}\)

(b) \(\dfrac{1}{2} = 50\%\)

(c) \(7\% = \dfrac{7}{100} = 0.07\)

(a) 1/4 · (b) 50% · (c) 0.07
Paper 3 (0580/33) — Question 4

Question 4

(a) Name the solid (circular ends, uniform tube). (b) Name the triangle with all angles 60°.
60° 60° 60°
(a) The solid is a cylinder.   (b) A triangle with three 60° angles is equilateral.

(a) Cylinder.   (b) Equilateral triangle.

(a) Cylinder · (b) Equilateral
Paper 3 (0580/33) — Question 5

Question 5

Fair 5-sided spinner: red, white, white, blue, red (red ×2, white ×2, blue ×1).
red white white blue red
The fair 5-sided spinner: red ×2, white ×2, blue ×1.

(a) Least likely = smallest count = blue.

(b) \(P(\text{white}) = \dfrac{2}{5}\).

(c) \(P(\text{not red}) = \dfrac{5-2}{5} = \dfrac{3}{5}\).

(a) Blue · (b) 2/5 · (c) 3/5
Paper 3 (0580/33) — Question 6

Question 6

(a) Reciprocal of 8. (b) Work out \(19^{3}\).

(a) Reciprocal of 8 = \(\dfrac{1}{8} = 0.125\).

(b) \(19^{3} = 361\times 19 = 6859\).

(a) 1/8 · (b) 6859
Paper 3 (0580/33) — Question 7

Question 7

Give two reasons why the bar chart is incorrect.
0102030 40506070 8090100125 150175200 Number of cars sold MarAprMay Months
The newspaper's bar chart, reproduced from the paper — bar widths are unequal and the vertical scale is not linear (steps of 10 up to 100, then steps of 25).

1. The bars are not of equal width.

2. The vertical scale is not linear (uneven spacing of values).

Bar widths unequal; scale not linear
Paper 3 (0580/33) — Question 8

Question 8

Work out the number of seconds in six and a half hours.

\[6.5\times 60\times 60 = 6.5\times 3600 = 23\,400 \text{ seconds}\]

Answer: 23 400 seconds
Paper 3 (0580/33) — Question 9

Question 9

Find the area of the compound (step) shape.
23 cm 5 cm 15 cm 7 cm 8 cm 12 cm
Compound shape (right angles at each corner). Not to scale.

Split into two rectangles (bottom 23 × 5, upper 8 × 7):

\[23\times 5 + 8\times 7 = 115 + 56 = 171\ \text{cm}^2\]

(Equivalently \(12\times 8 + 5\times 15 = 96 + 75 = 171\).)

Answer: 171 cm²
Paper 3 (0580/33) — Question 10

Question 10

(a) Scores 16, 7, 21, 8, 20, 9, 11, 7, 14, 3 — find mode and range. (b) Ed: mean of 5 games = 28; after a 6th game mean = 26. Find the 6th score.

(a)

Mode = value occurring most = 7 (twice). Range = \(21 - 3 = 18\).

(a) Mode 7 · Range 18

(b)

Total after 5 = \(5\times 28 = 140\); total after 6 = \(6\times 26 = 156\). 6th score = \(156 - 140 = 16\).

(b) 16
Paper 3 (0580/33) — Question 11

Question 11

Calculate 19% of $46.25 (to the nearest cent).

\[\frac{19}{100}\times 46.25 = 0.19\times 46.25 = 8.7875 \approx 8.79\]

Answer: $8.79
Paper 3 (0580/33) — Question 12

Question 12

(a) Simplify \(5b - 8c + 2b - 3c\). (b) \(q = 3r + 5t\); find \(t\) when \(q = 37, r = 4\).

(a)

\((5b + 2b) + (-8c - 3c) = 7b - 11c\).

(a) 7b − 11c

(b)

\(37 = 3(4) + 5t = 12 + 5t \Rightarrow 5t = 25 \Rightarrow t = 5\).

(b) t = 5
Paper 3 (0580/33) — Question 13

Question 13

Divide $136 in the ratio 3 : 5.

Total parts \(= 8\); one part \(= 136 \div 8 = 17\). So \(3\times 17 = 51\) and \(5\times 17 = 85\).

Answer: $51 and $85
Paper 3 (0580/33) — Question 14

Question 14

Each box holds 68 pens; Carl has 980 pens and 14 boxes. Does he have enough?

Capacity of 14 boxes: \(14\times 68 = 952\) pens. Since \(952 < 980\), not enough. (\(980\div 68 = 14.4\ldots\), so 15 boxes needed; 28 pens remain.)

Answer: No — 14 boxes hold only 952 pens (needs 15 boxes)
Paper 3 (0580/33) — Question 15

Question 15

Angles round a point: 68°, \(x°\), 43°, 135°. Find \(x\).
68° x° 43° 135°
Four rays meeting at a point (not to scale); the angles sum to 360°.

\[68 + x + 43 + 135 = 360 \Rightarrow x = 360 - 246 = 114\]

Answer: x = 114
Paper 3 (0580/33) — Question 16

Question 16

Isosceles triangle: height 5 cm, base 4 cm, slant side \(x\) cm. (a) Find \(x\). (b) Draw the net of the square-based pyramid.
5 cm x cm 4 cm
Isosceles triangle: height 5 cm, base 4 cm, slant side \(x\) cm. Not to scale.

(a)

The height meets the base at its midpoint → right triangle with legs 5 and \(\tfrac{4}{2}=2\):

\[x = \sqrt{5^{2} + 2^{2}} = \sqrt{29} = 5.385\ldots \approx 5.39\]

(a) 5.39 (√29)

(b)

Draw a 4 × 4 square (base) with one identical isosceles triangle (height 5, base 4) on each side.

(b) 4×4 square + 4 identical isosceles triangles
Paper 3 (0580/33) — Question 17

Question 17

Write 46.179 correct to 4 significant figures.

First four s.f. are 4, 6, 1, 7; next digit 9 (\(\ge 5\)) rounds the 7 up to 8.

Answer: 46.18
Paper 3 (0580/33) — Question 18

Question 18

(a) \(\sqrt{\dfrac{8.4^{2} + 9.3}{26.5}}\). (b) \(4^{-3}\).

(a)

\(8.4^{2} = 70.56;\; 70.56 + 9.3 = 79.86;\; \dfrac{79.86}{26.5} = 3.0135\ldots\)

\[\sqrt{3.0135\ldots} = 1.7359\ldots \approx 1.74\]

(a) 1.74

(b)

\[4^{-3} = \frac{1}{4^{3}} = \frac{1}{64} = 0.015625\]

(b) 1/64 (0.015625)
Paper 3 (0580/33) — Question 19

Question 19

Expand and simplify \(3(5x + 2) - 4(x + 1)\).

\(3(5x+2) = 15x + 6;\quad -4(x+1) = -4x - 4\).

\[15x + 6 - 4x - 4 = 11x + 2\]

Answer: 11x + 2
Paper 3 (0580/33) — Question 20

Question 20

Due to an issue with question 20, it was removed from the paper.

No solution required — full marks (2) awarded automatically.

Removed — 2 marks awarded
Paper 3 (0580/33) — Question 21

Question 21

Transformation question. Diagram not reproduced in the provided pages; answer from the mark scheme.

(a) Rotation, 90° clockwise, centre (0, 0).

(b) Reflection mapping points to (−4, 2), (−2, −2), (−3, −3), (−1, −3).

(a) Rotation 90° cw, centre (0,0) · (b) points as above
Paper 3 (0580/33) — Question 22

Question 22

(a) Complete the table for \(y = x^{2} - 4\). (b) Draw the graph.

(a) Table

x−4−3−2−101234
y1250−3−4−30512

Calculations: \((-3)^2-4=5,\; 0^2-4=-4,\; 3^2-4=5\).

(b) Graph

x y −4−3−2−1 1234 1211109 8765 4321 −1−2−3−4 y = x² − 4
Graph of \(y = x^{2} - 4\) for \(-4 \le x \le 4\) (vertex at \((0,-4)\)).
(a) 5, −4, 5 · (b) Correct parabola
Paper 3 (0580/33) — Question 23

Question 23

(a) Candle cost $3.20, sold $4.64 — percentage profit. (b) Cone of volume 900 cm³, height 8 cm — find radius.

(a)

Profit = \(4.64 - 3.20 = 1.44\). \[\frac{1.44}{3.20}\times 100 = 45\%\]

(a) 45%

(b)

\[900 = \frac{1}{3}\pi r^{2}(8) \Rightarrow r^{2} = \frac{900\times 3}{8\pi} = 107.43\ldots\]

\[r = \sqrt{107.43\ldots} = 10.36\ldots \approx 10.4\ \text{cm}\]

(b) 10.4 cm
Paper 3 (0580/33) — Question 24

Question 24

Make \(p\) the subject of \(m = 9p + 32\).

\(m - 32 = 9p \Rightarrow p = \dfrac{m - 32}{9}\).

Answer: p = (m − 32)/9
Paper 3 (0580/33) — Question 25

Question 25

Find the LCM of 20 and 36.

\(20 = 2^{2}\times 5,\quad 36 = 2^{2}\times 3^{2}\). Take the highest power of each prime:

\[\text{LCM} = 2^{2}\times 3^{2}\times 5 = 4\times 9\times 5 = 180\]

Answer: 180
Paper 3 (0580/33) — Question 26

Question 26

Find the equation of line L in the form \(y = mx + c\).
x y L 22201816 1412108 642 −2−4−6 −5−4−3−2−1 1234
Line L crosses the \(y\)-axis at 9 and the \(x\)-axis at \(-3\).

From the graph the \(y\)-intercept is 9, so \(c = 9\). The line also passes through \((-3, 0)\):

\[m = \frac{9 - 0}{0 - (-3)} = \frac{9}{3} = 3\]

Answer: y = 3x + 9
Paper 3 (0580/33) — Question 27

Question 27

Due to an issue with question 27, it was removed from the paper.

No solution required — full marks (4) awarded automatically.

Removed — 4 marks awarded
Paper 3 (0580/33) — Question 28

Question 28

Speed question. Diagram/data not reproduced in the provided pages; answer from the mark scheme.

Using \(\text{speed} = \dfrac{\text{distance}}{\text{time}}\) with distance 5.4 and time 36 min:

\[\frac{5.4}{36}\times 60 = 9\]

Answer: 9 (km/h)
Paper 3 (0580/33) — Question 29

Question 29

Similar-triangle / trigonometry question. Diagram not reproduced in the provided pages; answers from the mark scheme.

(a) Similar triangles: \(\dfrac{20.8}{x} = \dfrac{9.1}{1.4} \Rightarrow x = 3.2\).

(b) \(\cos 38° = \dfrac{y}{5.4} \Rightarrow y = 5.4\cos 38° = 4.255\ldots \approx 4.26\).

(a) 3.2 · (b) 4.26
Verification against the official mark scheme

Paper 1 (0580/12) — Answer Check

QMy answerMark schemeMatch
130 05030 050✓
259305930✓
3Midpoint of ST markedMidpoint of ST marked✓
4(a)8 squares shaded8 squares shaded✓
4(b)22.2%22.2 / 22.22…✓
57 h 28 min7h 28min✓
6(a)55 and 12155 121✓
6(b)33 = 3×1133 = 3×11✓
711 or 4711, 47✓
824.35224.352✓
93m + 10k3m + 10k✓
10(a)6363✓
10(b)−432−432✓
11(a)Stem-leaf as shownsame✓
11(b)6161✓
12(a)220 cm²220✓
12(b)200 cm³200✓
13(a)54°54✓
13(b)17/2017/20✓
14x(3x² − 7y)x(3x² − 7y)✓
15(−10, 3)ᵀcolumn (−10, 3)✓
16221221✓
17(4, 3)(4, 3)✓
18(a)Rotation 90° acw, (0,0)same✓
18(b)Reflection in x = −1Shape drawn correctly✓
192828✓
20(a)−2, 1, 6−2, 1, 6✓
20(b)7n − 57n − 5✓
2118.65 ≤ l < 18.7518.65 … 18.75✓
228.4 × 10³8.4 × 10³✓
2372.572.5✓
245/125/12✓
2518 and 3018 30✓

Paper 1 result: 25/25 questions match.

Verification against the official mark scheme

Paper 3 (0580/33) — Answer Check

QMy answerMark schemeMatch
11, 2, 13, 261 2 13 26✓
2(a)ChordChord✓
2(b)Radius drawnRadius drawn✓
3(a)1/41/4✓
3(b)50%50✓
3(c)0.070.07✓
4(a)CylinderCylinder✓
4(b)EquilateralEquilateral✓
5(a)BlueBlue✓
5(b)2/52/5✓
5(c)3/53/5✓
6(a)1/81/8 / 0.125✓
6(b)68596859✓
7Unequal widths; non-linear scalesame✓
823 40023 400✓
9171 cm²171✓
10(a)(i)77✓
10(a)(ii)1818✓
10(b)1616✓
11$8.798.79✓
12(a)7b − 11c7b − 11c✓
12(b)55✓
13$51 and $8551 85✓
14No — 952 < 980No, correct statement✓
15114114✓
16(a)5.395.39 / 5.385…✓
16(b)4×4 square + 4 trianglesCorrect net✓
1746.1846.18✓
18(a)1.741.74✓
18(b)1/640.015625 / 1/64✓
1911x + 211x + 2✓
20Removed (2 marks)Removed — award 2✓
21(a)Rotation 90° cw, (0,0)same✓
21(b)(−4,2)(−2,−2)(−3,−3)(−1,−3)same✓
22(a)5, −4, 55 −4 5✓
22(b)Parabola vertex (0,−4)Correct curve✓
23(a)45%45✓
23(b)10.4 cm10.4 / 10.36…✓
24(m − 32)/9(m − 32)/9✓
25180180✓
26y = 3x + 9y = 3x + 9✓
27Removed (4 marks)Removed — award 4✓
2899✓
29(a)3.23.2✓
29(b)4.264.26 / 4.255…✓

Paper 3 result: all reproduced questions match.

* The question text/diagrams for Q21, Q28 and Q29 were not included in the question-paper pages of the PDF (only their mark-scheme entries were available). Their solutions are reconstructed from the mark-scheme method and final answers. Q20 and Q27 were officially removed with marks awarded automatically.