October/November 2025 Paper 31 Worked Answers (IGCSE Maths 0580 Core)

32 questions · 32 marks · 90 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1a

   1 marksPrime Factorisation
   126 = 2 × 63 = 2 × 9 × 7 = 2 × 3² × 7. 2 × 63 and 6 × 21 use non-primes. 2² × 3 × 7 has wrong power.
   Examiner tips

   • Use a factor tree or systematic division to find all prime factors.
   • Always express your answer using index notation for repeated primes.
   • Check your answer by multiplying the factors back together.

2. Question 1b

   1 marksHCF
   56 = 2³ × 7, 84 = 2² × 3 × 7. HCF = 2² × 7 = 28. 14 is half. 168 is LCM. 7 is too small.
   Examiner tips

   • Write out the prime factorisation of both numbers clearly.
   • Circle or highlight the common prime factors.
   • Take the lowest power of each common prime factor.

3. Question 1c

   1 marksLCM
   15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 60. 5 is HCF. 300 is product. 35 adds.
   Examiner tips

   • List all prime factors from both numbers.
   • Take the highest power of each prime that appears in either factorisation.
   • Multiply these together to get the LCM.

4. Question 2a

   1 marksMultiplying Mixed Numbers
   $1\frac{2}{3} = \frac{5}{3}$, $2\frac{2}{5} = \frac{12}{5}$. $\left(\frac{5}{3}\right) \times \left(\frac{12}{5}\right) = \frac{60}{15} = 4$. $3\frac{2}{3}$ adds. $2 \frac{1}{15}$ and $4 \frac{1}{15}$ miscalculate.
   Method:

   1⅔ = 5/3, 2⅖ = 12/5. (5/3) × (12/5) = 60/15 = 4. 3⅔ adds. 2 1/15 and 4 1/15 miscalculate.
   Examiner tips

   • Always convert mixed numbers to improper fractions before multiplying.
   • Cancel common factors before multiplying to simplify calculation.
   • Convert back to a mixed number if required.

5. Question 2b

   1 marksDividing Mixed Numbers
   $4\frac{1}{2} = \frac{9}{2}$, $1\frac{2}{3} = \frac{5}{3}$. $\left(\frac{9}{2}\right) \div \left(\frac{5}{3}\right) = \left(\frac{9}{2}\right) \times \left(\frac{3}{5}\right) = \frac{27}{10} = 2\frac{7}{10}$. $7\frac{1}{2}$ multiplies. $2\frac{2}{3}$ and $3$ miscalculate.
   Method:

   4½ = 9/2, 1⅔ = 5/3. (9/2) ÷ (5/3) = (9/2) × (3/5) = 27/10 = 2 7/10. 7½ multiplies. 2⅔ and 3 miscalculate.
   Examiner tips

   • Remember: dividing by a fraction means multiplying by its reciprocal.
   • Convert mixed numbers to improper fractions first.
   • Always simplify your final answer.

6. Question 3a

   1 marksSimplifying Expressions
   $5a - 3a = 2a$, $2b + 4b = 6b$. Answer: $2a + 6b$. $8a + 6b$ adds a terms. $2a - 2b$ subtracts b terms. $8ab$ combines illegally.
   Method:

   5a − 3a = 2a, 2b + 4b = 6b. Answer: 2a + 6b. 8a + 6b adds a terms. 2a − 2b subtracts b terms. 8ab combines illegally.
   Examiner tips

   • Underline or highlight like terms in different colours.
   • Be very careful with positive and negative signs.
   • Only combine terms with exactly the same letter(s).

7. Question 3b

   1 marksExpanding and Simplifying
   $12x - 8 + 3x + 15 = 15x + 7$. $15x - 7$ has wrong sign. $9x + 7$ misses $12x$. $15x + 23$ adds wrong.
   Method:

   12x − 8 + 3x + 15 = 15x + 7. 15x − 7 has wrong sign. 9x + 7 misses 12x. 15x + 23 adds wrong.
   Examiner tips

   • Expand each bracket separately first.
   • Write out all terms before collecting like terms.
   • Double-check signs when multiplying negatives.

8. Question 3c

   1 marksFactorisation
   HCF = $4a$. So $4a(2a - 3b)$. $4(2a^2 - 3ab)$, $2a(4a - 6b)$, and $a(8a - 12b)$ don't fully factorise.
   Method:

   HCF = 4a. So 4a(2a − 3b). 4(2a² − 3ab), 2a(4a − 6b), and a(8a − 12b) don't fully factorise.
   Examiner tips

   • Find the HCF of the numbers (8 and 12).
   • Find the common variables (both terms have 'a').
   • Check by expanding - you should get the original expression.

9. Question 3d

   1 marksSolving Linear Equations
   $4x - 8 = 3x + 15$, $x = 23$. $x = 7$ doesn't expand right. $x = 3$ uses wrong numbers. $x = -23$ has wrong sign.
   Method:

   4x − 8 = 3x + 15, x = 23. x = 7 doesn't expand right. x = 3 uses wrong numbers. x = −23 has wrong sign.
   Examiner tips

   • Expand brackets on both sides first.
   • Collect all x terms on one side, all numbers on the other.
   • Remember: change the sign when moving terms across the equals sign.

10. Question 4a

    1 marksCompound Interest
    4000 × 1.03² = 4000 × 1.0609 = $\$4243.60$. $\$4240.00$ uses simple interest. $\$4120.00$ uses 1 year. $\$4360.00$ uses wrong rate.
    Examiner tips

    • Convert the percentage to a multiplier (3% = 1.03).
    • Raise the multiplier to the power of the number of years.
    • Compound interest = Principal × (1 + r)^n.

11. Question 4b

    1 marksDepreciation
    20000 × 0.9³ = 20000 × 0.729 = $\$14580$. $\$14000$ uses simple depreciation. $\$18000$ is 1 year. $\$16200$ is 2 years.
    Examiner tips

    • Depreciation uses a multiplier less than 1 (100% - 10% = 90% = 0.9).
    • Apply the multiplier once for each year.
    • Value after n years = Initial × (multiplier)^n.

12. Question 5a

    1 marksStandard Form
    $5.08 \times 10^4 = 50800$ (4 places). $508000$ uses $10^5$. $5080$ uses $10^3$. $0.000508$ uses negative power.
    Method:

    5.08 × 10⁴ = 50800 (4 places). 508000 uses 10⁵. 5080 uses 10³. 0.000508 uses negative power.
    Examiner tips

    • Positive power of 10 = move decimal right = larger number.
    • Count the power to know how many places to move.
    • Fill in zeros as needed after moving the decimal.

13. Question 5b

    1 marksStandard Form
    $0.00062 = 6.2 \times 10^{-4}$ (4 places). $6.2 \times 10^{-3}$ counts 3. $62 \times 10^{-5}$ not standard form. $6.2 \times 10^{4}$ positive power.
    Method:

    0.00062 = 6.2 × 10⁻⁴ (4 places). 6.2 × 10⁻³ counts 3. 62 × 10⁻⁵ not standard form. 6.2 × 10⁴ positive power.
    Examiner tips

    • For small numbers (less than 1), the power will be negative.
    • Count how many places you move the decimal to get a number between 1 and 10.
    • The first number must be between 1 and 10 (e.g., 6.2, not 62 or 0.62).

14. Question 5c

    1 marksStandard Form Calculations
    $4.5 \times 2 = 9$. $10^{-2} \times 10^{5} = 10^{3}$. Answer: $9 \times 10^{3}$. $9 \times 10^{7}$ adds powers. $9 \times 10^{-7}$ makes negative. $6.5 \times 10^{3}$ adds numbers.
    Method:

    4.5 × 2 = 9. 10⁻² × 10⁵ = 10³. Answer: 9 × 10³. 9 × 10⁷ adds powers. 9 × 10⁻⁷ makes negative. 6.5 × 10³ adds numbers.
    Examiner tips

    • Multiply the decimal parts separately: 4.5 × 2 = 9.
    • Add the powers: -2 + 5 = 3.
    • Combine to get 9 × 10³ (already in standard form as 9 is between 1 and 10).

15. Question 6a

    1 marksCumulative Frequency
    Cumulative at 3rd = 8 + 12 + 15 = 35. 15 is just 3rd frequency. 20 is first two. 40 is total.
    Examiner tips

    • Cumulative means 'running total' - add each frequency to the previous cumulative total.
    • The last cumulative frequency should equal the total frequency.
    • Write out the running totals step by step.

16. Question 6b

    1 marksCumulative Frequency Diagrams
    Cumulative frequency is plotted at upper class boundaries. At class midpoints is for frequency polygons. At lower class boundaries and At any point in the class are incorrect.
    Examiner tips

    • Cumulative frequency represents 'less than or equal to' the upper boundary.
    • Always plot at the end (upper boundary) of each class interval.
    • Join points with a smooth curve, not straight lines.

17. Question 6c

    1 marksMedian from Cumulative Frequency
    Median is at n/2 = 60/2 = 30. 60 is total. 15 is Q1 position. 45 is Q3 position.
    Examiner tips

    • Median position = n/2, where n is the total frequency.
    • Draw a horizontal line from the median position to the curve.
    • Then draw a vertical line down to read the value on the x-axis.

18. Question 6d

    1 marksInterquartile Range
    IQR = Q3 − Q1 = 28 − 12 = 16. 40 adds. 20 is average. 8 is Q1 − 4.
    Examiner tips

    • IQR = Q3 - Q1 (upper quartile minus lower quartile).
    • The IQR measures the spread of the middle 50% of data.
    • A larger IQR means more spread in the data.

19. Question 7a

    1 marksAngles in Triangles
    Angles in triangle sum to $180°$. $ACB = 180° - 48° - 67° = 65°$. $115°$ is $180 - 65$. $19°$ subtracts angles. $180°$ is sum of all angles in a triangle.
    Method:

    Angles in triangle sum to 180°. ACB = 180° − 48° − 67° = 65°. 115° is 180 − 65. 19° subtracts angles. 180° is sum.
    Examiner tips

    • Always remember: angles in a triangle sum to 180°.
    • Subtract both known angles from 180° to find the third.
    • Show your working: 180° - 48° - 67° = 65°.

20. Question 7b

    1 marksSine Rule
    Angle $R = 80°$. $\frac{QR}{\sin(40°)} = \frac{10}{\sin(80°)}$. $QR = 10 \times \frac{\sin(40°)}{\sin(80°)} \approx 6.5$ cm.
    Method:

    Angle R = 80°. QR/sin(40°) = 10/sin(80°). QR = 10 × sin(40°)/sin(80°) ≈ 6.5 cm. Closest answer depends on calculation.
    Examiner tips

    • Find the third angle first: 180° - 40° - 60° = 80°.
    • Set up sine rule with the side opposite to its angle.
    • Make sure your calculator is in degree mode.

21. Question 8a

    1 marksSubstitution
    $y = 9 - 15 + 4 = -2$. $4$ just uses constant. $2$ subtracts wrong. $-8$ miscalculates.
    Method:

    y = 9 − 15 + 4 = −2. 4 just uses constant. 2 subtracts wrong. −8 miscalculates.
    Examiner tips

    • Work out each term separately: x² = 9, then 5x = 15.
    • Combine carefully with the correct signs: 9 - 15 + 4.
    • Check your answer makes sense on the graph.

22. Question 8b

    1 marksQuadratic Graphs
    $x^2 - 4x + 3 = 0$, $(x-1)(x-3) = 0$, $x = 1$ or $x = 3$. $x = -1$ and $x = -3$ has wrong signs. $x = 0$ and $x = 3$ includes $0$. $x = 4$ and $x = -3$ has wrong values.
    Method:

    x² − 4x + 3 = 0, (x−1)(x−3) = 0, x = 1 or x = 3. x = −1 and x = −3 has wrong signs. x = 0 and x = 3 includes 0. x = 4 and x = −3 has wrong values.
    Examiner tips

    • X-intercepts are where the graph crosses the x-axis (where y = 0).
    • Factorise the quadratic: find two numbers that multiply to +3 and add to -4.
    • Those numbers are -1 and -3, so factors are (x-1)(x-3).

23. Question 8c

    1 marksSolving Equations Graphically
    $x^2 = 4$, $x = \pm 2$. $x = 4$ is the $y$-value. $x = 2$ only forgets negative. $x = 16$ squares $4$.
    Method:

    x² = 4, x = ±2. x = 4 is the y-value. x = 2 only forgets negative. x = 16 squares 4.
    Examiner tips

    • To find where curves meet, set the equations equal to each other.
    • Remember: √4 = ±2 (both positive and negative).
    • A parabola y = x² is symmetric, so it crosses y = 4 at two points.

24. Question 9a

    1 marksArea of Triangle
    Area = $\frac{1}{2} \times 5 \times 12 = 30$ cm$^2$. $60$ cm$^2$ forgets $\frac{1}{2}$. $17$ cm$^2$ adds sides. $65$ cm$^2$ uses Pythagoras.
    Method:

    Area = ½ × 5 × 12 = 30 cm². 60 cm² forgets ½. 17 cm² adds sides. 65 cm² uses Pythagoras.
    Examiner tips

    • Triangle area = ½ × base × height.
    • For right-angled triangles, the two shorter sides are base and height.
    • Don't forget the ½ in the formula!

25. Question 9b

    1 marksVolume of Prism
    $\text{Volume} = \text{area} \times \text{length} = 15 \times 8 = 120 \text{ cm}^3$. $23 \text{ cm}^3$ adds. $1.875 \text{ cm}^3$ divides. $120 \text{ cm}^2$ has wrong units.
    Method:

    Volume = area × length = 15 × 8 = 120 cm³. 23 cm³ adds. 1.875 cm³ divides. 120 cm² has wrong units.
    Examiner tips

    • Volume of prism = cross-sectional area × length.
    • This formula works for ALL prisms (triangular, rectangular, etc.).
    • Volume is measured in cubic units (cm³).

26. Question 9c

    1 marksSurface Area
    TSA = $6 \times 5^2 = 6 \times 25 = 150$ cm$^2$. $125$ cm$^3$ is volume. $25$ cm$^2$ is one face. $30$ cm$^2$ is perimeter of face.
    Method:

    TSA = 6 × 5² = 6 × 25 = 150 cm². 125 cm³ is volume. 25 cm² is one face. 30 cm² is perimeter of face.
    Examiner tips

    • Surface area of a cube = 6 × (side)².
    • Count all the faces and add their areas.
    • Surface area uses square units (cm²), volume uses cubic units (cm³).

27. Question 10a

    1 marksProbability
    P(green) = 7/11. 4/11 is P(yellow). 7/4 is ratio not probability. 4/7 inverts.
    Examiner tips

    • Probability = favourable outcomes / total outcomes.
    • Total = 7 + 4 = 11 balls.
    • Probability must be between 0 and 1.

28. Question 10b

    1 marksProbability Without Replacement
    P(RR) = (4/6) × (3/5) = 12/30 = 2/5. 4/9 uses replacement. 16/36 uses wrong denominator. 6/15 not simplified.
    Examiner tips

    • Without replacement: reduce both numerator AND denominator for the second pick.
    • First pick: P(R) = 4/6. Second pick (given first was red): P(R) = 3/5.
    • Multiply along the branches: (4/6) × (3/5) = 12/30 = 2/5.

29. Question 10c

    1 marksCombined Probability
    P(same) = P(RR) + P(BB) = 2/5 + 1/15 = 6/15 + 1/15 = 7/15. 2/75 multiplies. 3/20 wrong calculation. 8/15 is P(different).
    Examiner tips

    • For 'OR' (either event), ADD the probabilities.
    • For 'AND' (both events), MULTIPLY the probabilities.
    • P(same colour) = P(RR) + P(BB).

30. Question 11a

    1 marksEnlargements
    SF 3 from origin: multiply both coordinates by 3. (2,1) → (6,3). (5, 4) adds 3. (6, 1) only multiplies x. (2, 3) only multiplies y.
    Examiner tips

    • For enlargement from origin: new coordinates = scale factor × old coordinates.
    • Both x and y are multiplied by the same scale factor.
    • If centre is not the origin, the method is more complex.

31. Question 11b

    1 marksReflections
    Reflection in y = x swaps coordinates: (3, 7) → (7, 3). (−3, −7) reflects in origin. (−7, −3) combines. (3, −7) reflects in x-axis.
    Examiner tips

    • Reflection in y = x: swap x and y coordinates.
    • Reflection in x-axis: change sign of y.
    • Reflection in y-axis: change sign of x.

32. Question 11c

    1 marksRotations
    90° clockwise about origin: (x, y) → (y, −x). (4, 2) → (2, −4). (−2, 4) is anticlockwise. (−4, −2) is 180°. (2, 4) just swaps.
    Examiner tips

    • 90° clockwise about origin: (x, y) → (y, −x).
    • 90° anticlockwise about origin: (x, y) → (−y, x).
    • 180° about origin: (x, y) → (−x, −y).