IGCSE Maths Past Paper Solutions: 0580/13 May/June 2024 (Paper 1 Core)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 1 Core exam paper from May/June 2024 (0580/13/M/J/24) — 23 questions, 56 marks, 1 hour. Attempt each question yourself first, then click “Show solution” to check your working step by step.
Write the number two million two thousand and two in figures.
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2 002 002
Put one pair of brackets into this calculation to make it correct: 5 + 4 × 3 + 9 = 53
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Testing 5 + 4 × (3+9): this gives 5 + 4×12 = 5+48 = 53. Correct!
5 + 4 × (3 + 9) = 53
Simplify. 7x − 8y − x − y
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Collecting x-terms: 7x−x=6x. Collecting y-terms: −8y−y=−9y.
6x − 9y
- (a) Write 164 703 correct to the nearest thousand. [1]
- (b) Write 16.983 correct to 1 decimal place. [1]
- (c) Write 0.037665 correct to 2 significant figures. [1]
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- (a) 164703 lies between 164000 and 165000; the hundreds digit (7) rounds up: 165 000
- (b) The second decimal digit (8) rounds the first decimal place up: 17.0
- (c) The first two significant figures are 3 and 7; the next digit (6) rounds up: 0.038
- (a) The diagram shows a kite. On the diagram, draw any lines of symmetry. [1]
- (b) The diagram shows an equilateral triangle. Write down the order of rotational symmetry of this shape. [1]
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- (a) A kite has 1 line of symmetry: the vertical line running from the top vertex straight down through the bottom vertex.
- (b) An equilateral triangle maps onto itself after rotating by 120°, 240°, or 360° about its centre: order 3
Write these numbers in order, starting with the smallest: 0.45, 42%, &frac{4}{11}, ⅖
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Converting each to a decimal: 0.45 stays 0.45. 42% = 0.42. &frac{4}{11} = 0.364. ⅖ = 0.4
4/11 < 2/5 < 42% < 0.45
The base of a cuboid measures 10 cm by 7 cm. The volume of the cuboid is 280 cm³. Calculate the height of the cuboid.
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Base area = 10 × 7 = 70 cm². Height = volume ÷ base area = 280÷70 = 4 cm
In a city, the probability that it will rain today is 0.15. Find the probability that it will not rain today in this city.
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1 − 0.15 = 0.85
One day the temperature in Tokyo is −5°C and the temperature in Manila is 18°C.
- (a) Work out the difference between these two temperatures. [1]
- (b) The temperature in Tokyo rises by 4°C. Find the new temperature in Tokyo. [1]
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- (a) 18 − (−5) = 18+5 = 23°C
- (b) −5 + 4 = −1°C
- (a) These are the first four terms of a sequence: 3, 10, 17, 24.
- Write down the next term. [1]
- Write down the term to term rule for continuing the sequence. [1]
- (b) These are the first four terms of another sequence: 16, 14, 11, 7. Write down the next two terms of this sequence. [2]
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- (a)(i) The sequence increases by 7 each time (10−3=7, etc.): 24+7 = 31
- (a)(ii) Add 7 to the previous term
- (b) The differences between terms are −2, −3, −4 (14−16=−2, 11−14=−3, 7−11=−4) — each difference is one more negative than the last. Continuing this pattern, the next differences are −5 and −6: 7−5=2, then 2−6=−4.
2, −4
The diagram shows an isosceles triangle with one base angle 36° and the two marked (tick) sides equal, with the angle at the top vertex labelled x°.
Find the value of x.
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Since the two marked sides are equal, the base angles are equal, so the other base angle is also 36°. The angles of a triangle sum to 180°: x = 180 − 36 − 36 = 108
The diagram shows a cuboid with dimensions 6 cm × 2 cm × 3 cm. On the 1 cm² grid, complete a net of this cuboid. One face (6 cm × 2 cm) has been drawn for you.
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A cuboid net needs six rectangular faces: two of 6×2, two of 6×3, and two of 2×3, arranged so that they fold up correctly with matching edges. Starting from the given 6×2 face, attach a 2×3 face to each of the two short (2 cm) ends, and a 6×3 face along one of the long (6 cm) edges above or below, with the final 6×2 face attached beyond that 6×3 face (or similarly arranged so every edge that must join in the folded cuboid lines up in the net).
Factorise completely. 4x²y − 5xy²
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The highest common factor of 4x²y and 5xy² is xy: xy(4x − 5y)
The scale of a map is 1:40 000. On the map the distance between two villages is 37 cm. Calculate the actual distance between the two villages. Give your answer in kilometres.
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Actual distance = 37 × 40000 = 1 480 000 cm.
Converting to kilometres (100 000 cm = 1 km): 1480000÷100000 = 14.8 km
Without using a calculator, work out &frac37; − &frac{1}{14}. You must show all your working and give your answer as a fraction in its simplest form.
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Common denominator of 7 and 14 is 14: &frac37; = &frac{6}{14}
&frac{6}{14} − &frac{1}{14} = 5/14
The price of a game increases from $48 to $56.40. Calculate the percentage increase in the price.
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Increase = 56.40 − 48 = $8.40. Percentage increase = (8.40÷48) × 100 = 17.5%
The diagram shows a right-angled triangle ABC, with the right angle at A, AC=8 cm, and angle ABC=37°.
Calculate AB.
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AC is opposite angle B, and AB is adjacent to angle B, so: tan(37°) = AC/AB → AB = AC÷tan(37°) = 8÷0.7536
AB = 10.6 cm (3 s.f.)
The length, s metres, of a ship is 83 m, correct to the nearest metre. Complete this statement about the value of s.
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Rounding to the nearest metre means s could be up to half a metre above or below 83:
82.5 ≤ s < 83.5
Solve the simultaneous equations.
5t − 2w = 19
3t + 2w = 5
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Adding the two equations eliminates w: (5t−2w)+(3t+2w) = 19+5 → 8t = 24 → t=3
Substituting into the second equation: 3(3)+2w=5 → 9+2w=5 → 2w=−4 → w=−2
t = 3 w = −2
The diagram shows the positions of three towns A, B and C. Angle ABC=103°. The bearing of town B from town A is 048°. Town C is due east of town A.
Find the bearing of town C from town B.
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Since C is due east of A, the bearing of C from A is 090°. So angle BAC (between AB and AC) = 90−48 = 42°.
In triangle ABC: angle BAC=42°, angle ABC=103°, so angle BCA = 180−42−103 = 35°.
The bearing of A from B is the back bearing of 048°, i.e. 048+180=228°. Since C lies to the other side of B from this direction by the interior angle at B (103°, rotating towards the east where C lies): bearing of C from B = 228−103 = 125°
- (a) 𝒰 = {1, 4, 5, 8, 9, 12, 16, 64}. C = {cube numbers}. S = {square numbers}.
- Complete the Venn diagram. [2]
- Find n(C∪S). [1]
- (b) On a separate Venn diagram with sets A and B, shade the region A∩B. [1]
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- (a)(i) From 𝒰={1,4,5,8,9,12,16,64}: cube numbers C={1,8,64} (1=1³, 8=2³, 64=4³). Square numbers S={1,4,9,16,64} (1=1², 4=2², 9=3², 16=4², 64=8²).
C∩S (in both) = {1, 64}. C only = {8}. S only = {4, 9, 16}. Outside both = {5, 12} (the only elements of 𝒰 left over).
C only: 8 C∩S: 1, 64 S only: 4, 9, 16 Outside: 5, 12 - (a)(ii) n(C∪S) counts every element in C or S (or both): {1, 8, 64, 4, 9, 16} = 6
- (b) A∩B means “in both A and B” — shade only the lens-shaped overlap where the two circles cross.
- (a) Write these numbers in standard form.
- 0.007 [1]
- 700 000 000 [1]
- (b) Calculate (3200 × 5.4 × 10⁻³) ÷ (4.8 × 10⁻⁴). Give your answer in standard form. [2]
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- (a)(i) 7 × 10⁻³
- (a)(ii) 7 × 10⁸
- (b) 3200 × 5.4 ÷ 4.8 = 17280÷4.8 = 3600. Powers of 10: 10⁻³÷10⁻⁴ = 10¹. So the result is 3600 × 10 = 36000. In standard form: 3.6 × 10⁴
The diagram shows a spherical tank with radius 0.5 m and a cylindrical jug with diameter 24 cm and height 32 cm. The tank is full of water.
Calculate how many times the jug can be completely filled with water from the tank. [The volume, V, of a sphere with radius r is V=&frac43;πr³.]
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Working entirely in centimetres: radius of tank = 0.5 m = 50 cm.
Volume of tank = &frac43;π(50)³ = &frac43;π(125000) = 166666.7π = 523598.8 cm³
Volume of jug (radius 12 cm, height 32 cm) = πr²h = π(12)²(32) = 4608π = 14476.5 cm³
Number of complete fills = 523598.8 ÷ 14476.5 = 36.17…
Since only a whole number of complete fills counts (the leftover water isn’t enough for a full 37th jug): 36 times