IGCSE Maths Past Paper Solutions: 0580/11 May/June 2024 (Paper 1 Core)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 1 Core exam paper from May/June 2024 (0580/11/M/J/24) — 25 questions, 56 marks, 1 hour. Attempt each question yourself first, then click “Show solution” to check your working step by step.
- (a) Write 0.8 as a fraction. [1]
- (b) Write 28% as a decimal. [1]
- (c) Write 4876 correct to the nearest hundred. [1]
Show solution
- (a) 0.8 = 8/10, which simplifies by dividing top and bottom by 2: 4/5
- (b) 28% means 28 per 100: 28÷100 = 0.28
- (c) 4876 lies between 4800 and 4900; since the tens digit (7) is 5 or more, it rounds up: 4900
The diagram shows a straight line from point A to point B.
Measure the length of line AB in millimetres. [1]
Show solution
This is a direct ruler-measurement question. Placing a ruler along the printed line from A to B gives a length of approximately 107 mm (accept 105–109 mm to allow for small differences in printing/scaling).
The travel graph shows the journey of a bus. The bus travels from 1400, rising steadily to cover 12 km by 1500, remains stationary for a period, then returns to a distance of 0 km by 1600.
- (a) Find the distance the bus travels in the first 50 minutes. [1]
- (b) Find how long, in minutes, the bus is stationary. [1]
Show solution
- (a) The graph rises in a straight line from (1400, 0) to (1500, 12), a rate of 12 km per 60 minutes, i.e. 0.2 km per minute. In the first 50 minutes: 0.2 × 50 = 10 km
- (b) The flat (stationary) section of the graph runs from 1500 to 1540, a duration of 40 minutes
Write down the order of rotational symmetry of a rhombus.
Show solution
A rhombus maps onto itself after a 180° rotation about its centre, but not after 90° (unless it’s a square). So it has rotational symmetry of order 2
The diagram shows a shape on a 1 cm² grid: a rectangle-like top section with a triangular notch pointing downward from the middle of its base.
Find the area of this shape. [1]
Show solution
The shape can be read off the grid as a hexagon with vertices (using the top-left corner of the shape as (0,0), across then down): a 5 cm × 2 cm rectangle, with a triangular flap hanging 1 cm further down from the middle of its base.
Rectangle area = 5 × 2 = 10 cm². The triangular flap has a 5 cm base and 1 cm height: area = ½ × 5 × 1 = 2.5 cm².
Total area = 10 + 2.5 = 12.5 cm²
- (a) Work out. 28 − 16÷2 [1]
- (b) Find the reciprocal of ⅘. [1]
Show solution
- (a) Following order of operations, division first: 16÷2 = 8. Then 28−8 = 20
- (b) The reciprocal of a fraction swaps numerator and denominator: 5/4 (1.25)
The temperature on Monday is −27°C. The temperature on Tuesday is 15°C higher than on Monday. Work out the temperature on Tuesday.
Show solution
−27 + 15 = −12°C
The diagram shows a cross-shaped figure, made of a central square with a rectangular arm attached on each of its four sides. The horizontal (left and right) arms are noticeably longer than the vertical (top and bottom) arms, though each pair of opposite arms is equal in length.
Draw all the lines of symmetry on this shape. [2]
Show solution
Because the horizontal arms are longer than the vertical arms, this cross is not a perfectly square (Greek) cross, so it does not have diagonal symmetry or 4-fold rotational symmetry. It does, however, still look the same when reflected left-right or top-to-bottom. So there are exactly 2 lines of symmetry: one vertical line straight down the centre, and one horizontal line straight across the centre.
The diagram shows two sides of a parallelogram ABCD, plotted on a coordinate grid: A = (−7, 5), B = (−1, 1), and C = (3, 3).
Find the coordinates of point D. [2]
Show solution
In parallelogram ABCD, the diagonals AC and BD bisect each other, so A + C = B + D, giving D = A + C − B.
D = (−7+3−(−1), 5+3−1) = (−7+3+1, 7) = (−3, 7)
Check: vector AB = (6,−4). Vector DC = C−D = (3−(−3), 3−7) = (6,−4) — matches AB, confirming ABCD is indeed a parallelogram.
A cuboid has length 15 cm, width 6 cm and height h cm. The total surface area of this cuboid is 369 cm².
Work out the value of h. [4]
Show solution
Surface area of a cuboid = 2(lw + lh + wh) = 2(15×6 + 15h + 6h) = 2(90 + 21h) = 180 + 42h
Setting this equal to 369: 180 + 42h = 369 → 42h = 189 → h = 4.5 cm
Geetha has a box of toys. She picks a toy at random from the box. The probability that she picks a wooden toy is 0.6.
- (a) Work out the probability that she does not pick a wooden toy. [1]
- (b) The box contains three types of toys, wooden, plastic or metal.
| Type of toy | Wooden | Plastic | Metal |
|---|---|---|---|
| Number of toys | ? | 14 | 14 |
| Probability | 0.6 | ? | ? |
Complete the table. [2]
Show solution
- (a) Since picking a wooden toy and not picking one are complementary events: 1 − 0.6 = 0.4
- (b) Plastic and metal toys together = 14+14 = 28, and together they make up 1−0.6 = 0.4 of the total. So total toys = 28÷0.4 = 70. Number of wooden toys = 0.6 × 70 = 42. Probability of plastic = 14÷70 = 0.2. Probability of metal = 14÷70 = 0.2.
Wooden: 42 toys Plastic: probability 0.2 Metal: probability 0.2
The table shows some information about two sequences.
| nth term | 5th term | |
|---|---|---|
| Sequence A | 60 − 4n | ? |
| Sequence B | n² − 300 | ? |
Complete the table. [2]
Show solution
Sequence A: substitute n=5 into 60−4n: 60−20 = 40
Sequence B: substitute n=5 into n²−300: 25−300 = −275
Find the coordinates of the point where the line y = 3x − 5 crosses the y-axis.
Show solution
A line crosses the y-axis where x=0: y = 3(0)−5 = −5. (0, −5)
By writing each number in the calculation correct to 1 significant figure, find an estimate for the value of (28.2−5.6) / (4.2×1.68). You must show all your working.
Show solution
Rounding each number to 1 significant figure: 28.2→30, 5.6→6, 4.2→4, 1.68→2.
(30−6) ÷ (4×2) = 24÷8 = 3
Factorise completely. 36x² + 40x
Show solution
The highest common factor of 36x² and 40x is 4x: 4x(9x + 10)
The diagram shows a rectangle with length 3x−12 and width x+7.
Find an expression for the perimeter of the rectangle. Give your answer in its simplest form.
Show solution
Perimeter = 2(length + width) = 2[(3x−12)+(x+7)] = 2(4x−5) = 8x − 10
The diagram shows a circle, centre O. P lies on the circle.
- (a) Write down the mathematical name of the line OP. [1]
- (b) Draw a tangent to the circle at P. [1]
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- (a) A line from the centre of a circle to a point on the circle is a radius
- (b) A tangent at P is a straight line touching the circle at that single point, drawn perpendicular to the radius OP at P.
Find the greatest odd number that is a factor of 140 and a factor of 210.
Show solution
140 = 2² × 5 × 7, so its odd part (after removing all factors of 2) is 5×7 = 35.
210 = 2 × 3 × 5 × 7, so its odd part is 3×5×7 = 105.
The greatest odd common factor of 140 and 210 is the highest common factor of these odd parts, HCF(35, 105) = 35
Check: 140÷35 = 4 and 210÷35 = 6, both whole numbers, confirming 35 is a common factor.
Calculate.
- (a) ∛343 − √40.96 [1]
- (b) (192 + 4×16)1.25 [1]
Show solution
- (a) Cube root of 343 = 7 (since 7³=343). Square root of 40.96 = 6.4. 7−6.4 = 0.6
- (b) Inside the brackets: 192+4×16 = 192+64 = 256. Then 2561.25 = 2565/4 = (2561/4)&sup5; = 4⁵ = 1024
- (a) Find the value of 137⁰. [1]
- (b) 7¹² ÷ 7p = 7¹&sup7;. Find the value of p. [1]
Show solution
- (a) Any non-zero number raised to the power 0 equals 1: 1
- (b) Using the law 712−p = 7¹&sup7;, so 12−p = 17 → p = −5
Calculate 1.827 × 10⁶ ÷ 9000. Give your answer in standard form.
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1.827×10⁶ ÷ 9000 = 1.827×10⁶ ÷ (9×10³) = (1.827÷9) × 10³ = 0.203 × 10³ = 203
In standard form: 2.03 × 10²
Solve the simultaneous equations. You must show all your working.
6x + 2y = 29
3x − 4y = 17
Show solution
Multiply the first equation by 2: 12x + 4y = 58. Add this to the second equation to eliminate y: 12x+4y+3x−4y = 58+17 → 15x = 75 → x = 5
Substitute x=5 into the first equation: 6(5)+2y = 29 → 30+2y = 29 → 2y = −1 → y = −0.5
x = 5 y = −0.5
Change 9.6 km/h into m/s.
Show solution
9.6 km/h = 9.6×1000 m per 3600 s = 9600÷3600 m/s = 2.67 m/s (8/3, 3 s.f.)
These are the first five terms of a sequence: 11, 18, 25, 32, 39. Find an expression for the nth term of the sequence.
Show solution
The common difference between consecutive terms is 7 (18−11=7, 25−18=7, etc.), so the nth term has the form 7n+c. Using the first term (n=1, value 11): 7(1)+c=11 → c=4.
nth term = 7n + 4
The diagram shows a shape made from a triangle JKL and a semicircle with diameter JL. JKL is an isosceles right-angled triangle with JK = JL = 12.8 cm (the right angle is at J).
- (a) Calculate the area of this shape. [3]
- (b) Calculate the perimeter of this shape. [4]
Show solution
- (a) Area of triangle JKL = ½ × JK × JL = ½ × 12.8 × 12.8 = 81.92 cm². The semicircle has diameter JL=12.8 cm, so radius = 6.4 cm: area = ½πr² = ½ × π × 6.4² = ½ × π × 40.96 = 64.34 cm². Total area = 81.92+64.34 = 146 cm² (3 s.f.)
- (b) The perimeter is made up of side JK, side KL, and the curved arc of the semicircle (the diameter JL is internal, where the triangle and semicircle join, so it is not part of the outer perimeter). Since the triangle is right-angled at J with JK=JL=12.8: KL = √(12.8²+12.8²) = 12.8√2 = 18.10 cm. Arc length = πr = π × 6.4 = 20.11 cm. Perimeter = JK + KL + arc = 12.8 + 18.10 + 20.11 = 51.0 cm (3 s.f.)