IGCSE Maths (0580) · Core · Paper 1 Non-calculator

IGCSE Maths Past Paper Solutions: 0580/01 Specimen Paper 2025 (Paper 1 Core)

Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 1 Core Specimen Paper for examination from 2025 (0580/01/SP/25) — 23 questions, 80 marks, non-calculator. Attempt each question yourself first, then click “Show solution” to check your working step by step.

13 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Kim takes part in a race that covers a total distance of 20 000 m. She cycles 17 875 m and runs the remaining distance.

  1. (a) Work out the distance Kim runs.
  2. (b) Write the number 17 875 in words.
  3. (c) Write the number 17 875 correct to the nearest hundred.
Show solution
  1. (a) Distance run = 20 000 − 17 875 = 2125 m
  2. (b) 17 875 in words: seventeen thousand, eight hundred and seventy-five
  3. (c) The tens digit (7) rounds the hundreds digit up: 17 875 → 17 900
22 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The diagram shows the net of a solid: a central square with a triangle attached to each of its four sides.

Question 2 — net of the solid diagram
  1. (a) What is the mathematical name of the solid?
  2. (b) For this solid, write down the number of vertices.
Show solution
  1. (a) A square base with four triangular faces that fold up to meet at a single point folds into a square-based pyramid
  2. (b) The four base corners plus the apex where the triangles meet → 5 vertices
32 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The number N is both a multiple of 12 and a square number. Find the smallest possible value of N.

Show solution

List square numbers and check which is also a multiple of 12: 1, 4, 9, 16, 25, 36, … — 36 ÷ 12 = 3 exactly, and 36 = 6². No smaller square number (1, 4, 9, 16, 25) divides exactly by 12.

N = 36

42 marksCoreSpecimen 2025 · Paper 1 (0580/01)

A coin is made from a mixture of tin, copper and zinc. Tin = 0.4%, Copper = 96.5%, Zinc = k%. Work out the value of k.

Show solution

All percentages must add to 100%: 0.4 + 96.5 = 96.9, so k = 100 − 96.9 = 3.1

51 markCoreSpecimen 2025 · Paper 1 (0580/01)

Here are four number cards: 0, 1, 3, 5. Using each card once, write down one number between 3020 and 3200.

Show solution

The number must start with 3 (to stay near 3000–3200). Testing arrangements of the remaining digits 0, 1, 5:

  • 3051 = 3051 — too low (below 3020)… wait: 3051 is between 3020 and 3200 ✓
  • 3105 = 3105 ✓ and 3150 = 3150 ✓ (both start 31–, automatically in range)

Any of these works. Answer: 3105 (3150 or 3051 also accepted)

61 markCoreSpecimen 2025 · Paper 1 (0580/01)

Write the ratio 90 : 120 in its simplest form.

Show solution

The highest common factor of 90 and 120 is 30. Dividing both parts by 30: 90÷30=3, 120÷30=4.

Simplest form: 3 : 4

71 markCoreSpecimen 2025 · Paper 1 (0580/01)

The diagram shows a hexagon divided into 6 triangular sections, 5 of which are shaded. Shade one more section so the diagram has rotational symmetry of order 3.

Question 7 — hexagon rotational symmetry diagram
Show solution
Question 7 solution — hexagon rotational symmetry diagram (with the correct triangle shaded)

Method: with 6 equal triangular sections around a centre point, rotational symmetry of order 3 means the pattern must look identical every 120° — that is, every 2nd section around the hexagon must match. Since 5 of the 6 sections are already shaded, shading the one remaining unshaded section makes every section shaded, which is symmetric under a 120° turn (and indeed any turn).

Answer: Shade the one remaining unshaded triangle

Please double-check against the original diagram: this question depends on exactly which section is currently unshaded in the printed figure. Confirm the position against the PDF before sharing this with students.
82 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The scale drawing shows the position of a rock, R. The scale is 1 centimetre represents 30 metres. A lighthouse, L, is 210 m from R, on a bearing of 125°. On the scale drawing, mark the position of L.

Question 8 — scale drawing (rock R and North line)
Show solution
Question 8 solution — scale drawing with lighthouse L marked

Convert the real distance to a drawing length: 210 ÷ 30 = 7 cm

Method: at R, use a protractor to measure 125° clockwise from the North line. Draw a straight line 7 cm long in that direction using a ruler, and label the end point L.

92 marksCoreSpecimen 2025 · Paper 1 (0580/01)

A cake has a mass of 600 g. Joe eats ⅕ of the cake. Find the mass of the cake that is left.

Show solution

Joe eats ⅕ of 600 = 120 g, so the fraction left is ⅘ of the cake.

⅘ × 600 = 480 g

106 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Lines AB and CD are parallel. EF and EG are straight lines. A 60° angle is marked, together with a°, b° and c° as shown in the diagram.

Question 10 — parallel lines and angles diagram
  1. (a) Find the value of a, with a geometrical reason.
  2. (b) Find the value of b, with a geometrical reason.
  3. (c) Find the value of c, with a geometrical reason.
Show solution
  1. (a) a = 60 — alternate angles (AB is parallel to CD)
  2. (b) a and b lie on the straight line EG, so they’re supplementary: b = 180 − 60 = 120 — angles on a straight line
  3. (c) Using the angle sum of the triangle (or angles on the straight line through E): c = 180 − 60 − 46 = 74 — angle sum of a triangle is 180°
Please double-check against the original diagram: this is a multi-line parallel-angle figure where the exact position of each labelled angle affects which specific angle rule (alternate/corresponding/co-interior) applies. The values above (a=60, b=120, c=74) are internally consistent, but please verify the reasons written match the exact diagram before use.
114 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Work out:

  1. (a) 7 + 9 × 3
  2. (b) −6 − (−12)
  3. (c) 10−2
Show solution
  1. (a) Multiplication first: 9×3=27, then 7+27 = 34
  2. (b) Subtracting a negative is the same as adding: −6+12 = 6
  3. (c) 10−2 = 1/10² = 1/100 = 0.01
123 marksCoreSpecimen 2025 · Paper 1 (0580/01)
  1. (a) Factorise: 9x + 12
  2. (b) Solve: 6x − 5 = 2x + 13
Show solution
  1. (a) HCF of 9 and 12 is 3: 9x+12 = 3(3x + 4)
  2. (b) 6x−2x = 13+5 → 4x = 18 → x = 4.5
133 marksCoreSpecimen 2025 · Paper 1 (0580/01)

A plane flies from London to Colombo. The time in London when the plane leaves is 08:20 on Saturday. The time in Colombo when the plane arrives is 02:15 on Sunday. The flight time is 13 hours 25 minutes. Find the time difference between London and Colombo, and state whether Colombo is ahead of or behind London.

Show solution

Find what the London time would be at the moment of arrival: 08:20 + 13h 25m = 21:45 (Saturday, London time)

The plane actually lands at 02:15 on Sunday, Colombo time. Comparing 21:45 Saturday to 02:15 Sunday: that’s 4 hours 30 minutes later.

Time difference = 4 hours 30 minutes, and Colombo is ahead of London.

144 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The diagram shows a shape made from two parallelograms, sharing a base length of 15 cm, with heights x cm and 4 cm. The shape has a total area of 210 cm². Find the value of x.

Question 14 — two parallelograms diagram
Show solution

Area of a parallelogram = base × height. Total area = (15 × x) + (15 × 4) = 210

15x + 60 = 210
15x = 150
x = 10

159 marksCoreSpecimen 2025 · Paper 1 (0580/01)
  1. (a) A scatter diagram shows 100m sprint time vs swimming race time for 11 athletes. Three more athletes’ times are given in a table: (10.20, 23.5), (10.86, 25.4), (11.04, 24.9).
    Question 15 — scatter diagram (blank/given points)
    1. Plot these three points on the scatter diagram.
    2. State the type of correlation shown.
    3. Draw a line of best fit.
    4. Another athlete completes the swim in 23.8 seconds. Use the line of best fit to estimate their 100m time.
  2. (b) Nine medals have diameters (cm): 85, 85, 70, 60, 68, 70, 70, 60, 66, and masses (g): 500, 412, 200, 135, 180, 181, 231, 152, 102.
    1. Write down the mode of the diameters.
    2. Find the median of the masses.
Show solution
Question 15 solution — scatter diagram with new points and line of best fit
  1. (a)
    1. Plot (10.20, 23.5), (10.86, 25.4) and (11.04, 24.9) on the grid.
    2. As 100m time increases, swimming time generally increases too → positive correlation
    3. Draw a straight line through the middle of the data, with roughly equal points on each side.
    4. Reading from a line of best fit through this data, a swim time of 23.8s corresponds to a 100m time of approximately 10.3 seconds (accept any value your line of best fit reasonably gives, typically 10.2–10.4)
  2. (b)
    1. Diameters: 85, 85, 70, 60, 68, 70, 70, 60, 66 — 70 appears three times, more than any other value → mode = 70 cm
    2. Masses sorted: 102, 135, 152, 180, 181, 200, 231, 412, 500 (9 values). The median is the 5th value → 181 g
169 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The line L is shown on a grid, passing through approximately (−2, −2), (0, −1), (2, 0) and (4, 1).

Question 16 — grid showing line L
  1. (a) Find the equation of line L in the form y = mx + c.
  2. (b) The table shows values for y = x² − 2x − 3, with some values missing.
    1. Complete the table.
    2. Draw the graph of y = x² − 2x − 3 for −2 ≤ x ≤ 4.
  3. (c) Write down the equation of the line of symmetry of the graph of y = x² − 2x − 3.
  4. (d) Write down the negative value of x where line L and the graph of y = x² − 2x − 3 intersect.
Show solution
Question 16 solution — grid showing line L and the parabola
  1. (a) Using two points on the line, e.g. (−2,−2) and (4,1): gradient m = (1−(−2))/(4−(−2)) = 3/6 = 0.5. Substituting (4,1): 1 = 0.5(4)+c → c = −1.
    Equation: y = 0.5x − 1
  2. (b) (i) Substituting each x-value into y=x²−2x−3:
    x−2−101234
    y50−3−4−305
    (ii) Plot these seven points and join with a smooth curve (a parabola/U-shape).
  3. (c) The curve is symmetric about its turning point. Since y(0)=y(2)=−3 and y(−1)=y(3)=0, the middle is at x=1 → x = 1
  4. (d) Solving 0.5x−1 = x²−2x−3 gives x²−2.5x−2=0, i.e. 2x²−5x−4=0. Using the quadratic formula, x = (5±√57)/4, giving x≈3.14 or x≈−0.64. The negative solution: x ≈ −0.6 (read from the graph; small variation from the graphical reading is expected)
174 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Two bags, A and B, each contain blue and white beads only. P(blue from A) = 0.8. P(blue from B) = 0.3.

  1. (a) Complete the tree diagram.
  2. (b) A student takes one bead at random from bag A and one from bag B. Find the probability that both beads are white.
Show solution
  1. (a) Since probabilities on each pair of branches must add to 1:
    • P(White from A) = 1 − 0.8 = 0.2
    • P(White from B), on the “Blue from A” branch = 1 − 0.3 = 0.7
    • P(Blue from B), on the “White from A” branch = 0.3
    • P(White from B), on the “White from A” branch = 1 − 0.3 = 0.7
  2. (b) P(both white) = P(White from A) × P(White from B) = 0.2 × 0.7 = 0.14
187 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Shape A is shown on a grid (a quadrilateral around x=2–3, y=1–4), and shape B is a congruent shape rotated into the region x=−4 to −1, y=2–3.

Question 18 — coordinate grid with shapes A and B
  1. (a) Describe fully the single transformation that maps shape A onto shape B.
  2. (b) On the grid, draw the image of:
    1. shape A after a translation by the vector (−5, −6).
    2. shape A after an enlargement by scale factor 3, centre (1, 4).
Show solution
Question 18 solution — coordinate grid with translated and enlarged shapes
  1. (a) Shape A is “tall” (roughly 1 wide by 3 high) while shape B is “wide” (roughly 3 wide by 1 high) — swapping width and height like this is the signature of a 90° rotation.
    Answer: Rotation, 90°, about a centre found by construction
  2. (b)(i) Translating each vertex of A by (−5,−6): a vertex at (2,1) →(−3,−5); (3,1)→(−2,−5); (3,4)→(−2,−2); (2,3)→(−3,−3). Plot these and join in order to draw the translated shape.
  3. (b)(ii) Enlarging about centre (1,4) with scale factor 3: for each vertex, new point = centre + 3×(vertex − centre). E.g. (2,1)→(4,−5); (3,1)→(7,−5); (3,4)→(7,4); (2,3)→(4,1). Plot these and join in order.
Please double-check against the original diagram: the exact centre and direction (clockwise/anticlockwise) of the rotation in part (a) depends on the precise grid coordinates of shapes A and B. To find the centre exactly: join two pairs of corresponding vertices (A to its image in B), construct the perpendicular bisector of each, and their intersection point is the centre of rotation. Please verify this against the printed grid before sharing with students.
192 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Rearrange the formula to make t the subject: w = 7t − 5

Show solution

Add 5 to both sides: w+5 = 7t. Divide both sides by 7:

t = (w + 5) / 7

203 marksCoreSpecimen 2025 · Paper 1 (0580/01)
  1. (a) Write down the smallest even integer that satisfies y > 2.5.
  2. (b) Write an inequality, in terms of x, for the interval shown on a number line: an open circle at −2 and a filled/closed circle at 4, with the segment between them shaded.
Show solution
  1. (a) Even integers greater than 2.5: 4, 6, 8, … The smallest is y = 4
  2. (b) An open circle means that value is NOT included; a closed/filled circle means it IS included. Open at −2, closed at 4: −2 < x ≤ 4
214 marksCoreSpecimen 2025 · Paper 1 (0580/01)

𝒰 = {a, b, d, e, f, h, i, m, p, t, u}, X = {a, e, i, u}, Y = {d, e, m, p, t, u}

  1. (a) Use this information to complete the Venn diagram.
  2. (b) List the elements of X ∩ Y.
  3. (c) Find n(X′).
Show solution
  1. (a) X only (in X, not Y): a, i. Overlap (in both): e, u. Y only (in Y, not X): d, m, p, t. Outside both circles (in 𝒰 but neither set): b, f, h.
  2. (b) Elements in both X and Y: e, u
  3. (c) X′ means everything in 𝒰 that is NOT in X. 𝒰 has 11 elements, X has 4, so X′ has 11−4 = 7
222 marksCoreSpecimen 2025 · Paper 1 (0580/01)

The length, L, of a road is 39 700 m, correct to the nearest 50 m. Complete this statement about the value of L: ………… ≤ L < …………

Show solution

“Correct to the nearest 50” means the true value is within half of 50 (i.e. 25) of the rounded value: 39 700 − 25 = 39 675, and 39 700 + 25 = 39 725.

39 675 ≤ L < 39 725

233 marksCoreSpecimen 2025 · Paper 1 (0580/01)

Solve the simultaneous equations: 3x − 5y = 22,   7x + 10y = 8

Show solution

Multiply the first equation by 2 so the y-terms cancel when added: 6x − 10y = 44

Add to the second equation: (6x−10y) + (7x+10y) = 44+8 → 13x = 52 → x = 4

Substitute x=4 into 3x−5y=22: 12−5y=22 → −5y=10 → y=−2

x = 4   y = −2

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