October/November 2025 Paper 12 Worked Answers (IGCSE Maths 0580 Core)

32 questions · 32 marks · 90 minutes

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1. Question 1

   1 marksPlace Value
   Three hundred and fifteen thousand = 315000, eight = 8. Combined: 315008. 31508 drops a zero. 315080 misplaces 8. 3150008 adds extra zeros.
   Examiner tips

   • Write large numbers in stages - thousands first, then the remaining digits.
   • Count your digits to verify you have the correct number of places.

2. Question 2a

   1 marksPercentages and Fractions
   35% = 35/100 = 7/20 (dividing by 5). 35/100 not simplified. 7/10 wrong simplification. 1/35 inverts.
   Examiner tips

   • Always check if your fraction can be simplified further.
   • Remember: percentage means 'out of 100', so start with the number over 100.

3. Question 2b

   1 marksFractions and Decimals
   5/8 = 5 ÷ 8 = 0.625. 0.58 concatenates. 1.6 divides wrong way. 0.0625 decimal in wrong place.
   Examiner tips

   • Know common fraction-decimal equivalents: 1/8 = 0.125, so 5/8 = 5 × 0.125.
   • Check your answer by converting back: 0.625 × 8 = 5.

4. Question 3

   1 marksOrdering Decimals
   0.307 < 0.37 < 3.07 < 3.7 < 37. 0.37, 0.307, 3.07, 3.7, 37 swaps first two. 3.07, 3.7, 0.307, 0.37, 37 mixes order. 37, 3.7, 3.07, 0.37, 0.307 is reversed.
   Examiner tips

   • Write all decimals with the same number of decimal places (add trailing zeros) to compare easily.
   • Separate numbers into groups: less than 1, between 1 and 10, greater than 10.

5. Question 4a

   1 marksSquare Roots
   $\sqrt{144} = 12$ because $12 \times 12 = 144$. $72$ is $\frac{144}{2}$. $14$ is close but $14^2 = 196$. $11$ is $\sqrt{121}$.
   Method:

   √144 = 12 because 12 × 12 = 144. 72 is 144/2. 14 is close but 14² = 196. 11 is √121.
   Examiner tips

   • Memorize perfect squares up to 15² = 225.
   • Check your answer by squaring it: 12 × 12 = 144.

6. Question 4b

   1 marksPowers
   $3^4 = 3 \times 3 \times 3 \times 3 = 81$. $12$ is $3 \times 4$. $64$ is $4^4 \div 4$. $27$ is $3^3$.
   Method:

   3⁴ = 3 × 3 × 3 × 3 = 81. 12 is 3 × 4. 64 is 4⁴ ÷ 4. 27 is 3³.
   Examiner tips

   • Work in stages: 3² = 9, then 3⁴ = 9 × 9 = 81.
   • The exponent tells you HOW MANY times to multiply, not what to multiply by.

7. Question 5

   1 marksNegative Numbers
   −12 − (−7) = −12 + 7 = −5. −19 subtracts. 5 ignores first negative. 19 adds positives.
   Examiner tips

   • Remember: minus a minus equals plus. Two negatives make a positive.
   • Use a number line to visualize: start at −12, move 7 to the right.

8. Question 6

   1 marksPrime Factorisation
   60 = 4 × 15 = 2² × 3 × 5. 4 × 15 uses non-primes. 2 × 3 × 10 has 10. 2³ × 5 misses 3.
   Examiner tips

   • Check your answer by multiplying the prime factors: 4 × 3 × 5 = 60.
   • Use a factor tree systematically, always dividing by the smallest prime first.

9. Question 7

   1 marksProbability
   P(H) = 1/2. P(H and H) = 1/2 × 1/2 = 1/4. 1/2 is one flip. 1 is certainty. 1/8 is three flips.
   Examiner tips

   • For 'AND' probabilities (both events happening), multiply the individual probabilities.
   • Draw a tree diagram or sample space if unsure: HH, HT, TH, TT - one out of four is HH.

10. Question 8

    1 marksSimplifying Expressions
    6x − 3x = 3x, 4y + 2y = 6y. Answer: 3x + 6y. 9x + 6y adds x terms. 3x + 2y subtracts y terms. 9xy combines illegally.
    Examiner tips

    • Only combine like terms - terms with the same variable(s).
    • Be careful with signs: −3x means subtracting 3x from 6x.

11. Question 9

    1 marksSubtracting Mixed Numbers
    $3\frac{1}{4} = \frac{13}{4}$, $1\frac{2}{3} = \frac{5}{3}$. $\frac{13}{4} - \frac{5}{3} = \frac{39}{12} - \frac{20}{12} = \frac{19}{12} = 1\frac{7}{12}$. $2\frac{7}{12}$ adds. $1\frac{5}{12}$ uses wrong denominator. $4\frac{11}{12}$ adds.
    Method:

    3¼ = 13/4, 1⅔ = 5/3. 13/4 − 5/3 = 39/12 − 20/12 = 19/12 = 1 7/12. 2 7/12 adds. 1 5/12 uses wrong denominator. 4 11/12 adds.
    Examiner tips

    • Always find the LCM of denominators (4 and 3 → 12) for a common denominator.
    • Check: 1 7/12 + 1 2/3 should equal 3 1/4.

12. Question 10

    1 marksRatio
    Total parts = 5. Each part = 12. Larger share = 3 × 12 = 36. 24 is smaller share. 40 uses 2:1. 30 halves.
    Examiner tips

    • The larger share corresponds to the larger number in the ratio.
    • Always check that your two shares add up to the total (36 + 24 = 60).

13. Question 11a

    1 marksNegative Numbers - Comparing
    −8 < −2 < 0 < 3. Most negative = coldest. 0°C is freezing point. −2°C is less cold. 3°C is warmest.
    Examiner tips

    • Think of a thermometer: temperatures below zero are colder, and the more negative, the colder.
    • Visualize a number line with 0 in the middle.

14. Question 11b

    1 marksNegative Numbers - Difference
    Change = 12 − (−5) = 12 + 5 = 17°C. 7°C subtracts. −17°C has wrong sign. −7°C subtracts with wrong sign.
    Examiner tips

    • A temperature rise should give a positive answer.
    • Count the steps on a number line from −5 to 12: 5 steps to 0, then 12 more to 12 = 17 steps.

15. Question 12

    1 marksStandard Form
    0.00042 = 4.2 × 10⁻⁴ (4 places). 4.2 × 10⁻³ counts 3. 42 × 10⁻⁵ not standard form. 4.2 × 10⁴ has positive power.
    Examiner tips

    • For numbers less than 1, the power of 10 is negative.
    • Standard form: A × 10ⁿ where 1 ≤ A < 10.

16. Question 13a

    1 marksProbability
    Total = 15. P(blue) = 6/15 = 2/5. 6/15 not simplified. 6/10 uses wrong total. 1/3 is wrong.
    Examiner tips

    • Always find the total number of outcomes first.
    • Simplify fractions to their lowest terms.

17. Question 13b

    1 marksProbability Without Replacement
    P(RB) + P(BR) = (2/5 × 3/4) + (3/5 × 2/4) = 6/20 + 6/20 = 12/20 = 3/5. 6/25 uses replacement. 12/20 not simplified. 1/5 is wrong.
    Examiner tips

    • Without replacement means the denominator decreases by 1 for the second pick.
    • 'One of each' means consider both possible orders.

18. Question 14

    1 marksSolving Linear Equations
    7x − 4x = 9 + 3, 3x = 12, x = 4. x = 2 uses wrong arithmetic. x = 6 uses 18÷3. x = −4 has sign error.
    Examiner tips

    • Check your answer by substituting back: 7(4) − 3 = 25, 4(4) + 9 = 25.
    • When moving terms, remember to change the sign.

19. Question 15

    1 marksExpanding Double Brackets
    (x + 5)(x − 2) = x² − 2x + 5x − 10 = x² + 3x − 10. x² − 3x − 10 has wrong middle sign. x² + 7x − 10 adds coefficients. x² + 3x + 10 has wrong constant sign.
    Examiner tips

    • Write out all four terms before simplifying.
    • Check signs carefully: positive × negative = negative.

20. Question 16

    1 marksFactorisation
    HCF = 5a. So 5a(2a − 3b). 5(2a² − 3ab), a(10a − 15b), and 5a(2a − 3) don't fully factorise.
    Examiner tips

    • Find HCF of coefficients (10 and 15 → 5) and HCF of variables (a² and ab → a).
    • Check by expanding: 5a(2a − 3b) = 10a² − 15ab.

21. Question 17

    1 marksSequences
    4n + 5 = 37, 4n = 32, n = 8. So 8th term. 9th term is 4n+5=41. 7th term is 4n+5=33. 32nd term uses 37−5.
    Examiner tips

    • Verify: 4(8) + 5 = 32 + 5 = 37.
    • This is essentially solving a linear equation.

22. Question 18a

    1 marksRotational Symmetry
    Regular hexagon maps onto itself 6 times in 360°. 3 is half. 2 is too few. 12 is twice.
    Examiner tips

    • For regular polygons, order of rotational symmetry = number of sides.
    • A shape has rotational symmetry of order n if it maps onto itself n times in one full turn.

23. Question 18b

    1 marksInterior Angles of Polygons
    Sum = (9−2) × 180° = 1260°. Each = 1260° ÷ 9 = 140°. 120° is hexagon. 160° is 18-gon. 135° is octagon.
    Examiner tips

    • Memorize the formula: interior angle sum = (n−2) × 180°.
    • For ONE angle of a regular polygon, divide the sum by n.

24. Question 19

    1 marksSine Rule
    Angle $PRQ = 70°$. $\frac{QR}{\sin(50°)} = \frac{10}{\sin(70°)}$. $QR = 10 \times \frac{\sin(50°)}{\sin(70°)} \approx 8.2$ cm, closest to A. $11.5$ cm uses wrong angle. $7.7$ cm and $13.1$ cm miscalculate.
    Method:

    Angle PRQ = 70°. QR/sin(50°) = 10/sin(70°). QR = 10 × sin(50°)/sin(70°) ≈ 8.2 cm, closest to A. 11.5 cm uses wrong angle. 7.7 cm and 13.1 cm miscalculate.
    Examiner tips

    • Always find the third angle first: 180° − 50° − 60° = 70°.
    • Match sides with opposite angles in the sine rule.

25. Question 20a

    1 marksReflections
    Point is 3 above y=2. Image is 3 below, at y = −1. x stays same: (4, −1). (0, 5) reflects in x=2. (4, 1) uses wrong distance. (−4, 5) reflects in y-axis.
    Examiner tips

    • Distance from point to mirror line = distance from mirror line to image.
    • For y = k reflections, x doesn't change; for x = k reflections, y doesn't change.

26. Question 20b

    1 marksRotations
    90° clockwise about origin: (x, y) → (y, −x). (2, 3) → (3, −2). 90° anticlockwise gives (−3, 2). 180° gives (−2, −3). 270° clockwise same as B.
    Examiner tips

    • Know the rotation rules: 90° CW: (x,y)→(y,−x), 90° ACW: (x,y)→(−y,x), 180°: (x,y)→(−x,−y).
    • A full description needs: angle, direction, centre.

27. Question 21

    1 marksSurface Area of Cylinder
    $TSA = 2\pi r^2 + 2\pi rh = 2\pi(9) + 2\pi(3)(8) = 18\pi + 48\pi = 66\pi$ cm$^2$. $48\pi$ cm$^2$ is curved surface only. $72\pi$ cm$^2$ uses wrong formula. $57\pi$ cm$^2$ miscalculates.
    Method:

    TSA = 2πr² + 2πrh = 2π(9) + 2π(3)(8) = 18π + 48π = 66π cm². 48π cm² is curved surface only. 72π cm² uses wrong formula. 57π cm² miscalculates.
    Examiner tips

    • TSA of cylinder = 2πr² (two circles) + 2πrh (rectangle when unrolled).
    • Leave answer in terms of π unless asked otherwise.

28. Question 22a

    1 marksMap Scales
    8 × 25000 = 200000 cm = 2000 m = 2 km. 0.32 km divides wrong. 20 km uses wrong conversion. 200 km uses 25 km per cm.
    Examiner tips

    • 1 km = 1000 m = 100,000 cm. Set up your conversion carefully.
    • Scale 1:25000 means 1 cm on map = 25000 cm actual = 250 m = 0.25 km.

29. Question 22b

    1 marksMap Scales and Area
    2 km² = 2 × 10¹⁰ cm². Area scale = 50000² = 2.5 × 10⁹. Map area = 2 × 10¹⁰ ÷ 2.5 × 10⁹ = 80 cm². 8 cm², 800 cm², and 4 cm² use wrong scale.
    Examiner tips

    • For area calculations, SQUARE the linear scale factor.
    • 1 km² = (100,000 cm)² = 10¹⁰ cm².

30. Question 23

    1 marksSimultaneous Equations
    From eq2: $y = 4x - 3$. Substituting into eq1: $2x + 3(4x - 3) = 11$, which gives $14x - 9 = 11$, so $14x = 20$ and $x = \frac{10}{7}$. Then $y = 4\left(\frac{10}{7}\right) - 3 = \frac{40}{7} - \frac{21}{7} = \frac{19}{7}$. Verification: $2\left(\frac{10}{7}\right) + 3\left(\frac{19}{7}\right) = \frac{20}{7} + \frac{57}{7} = \frac{77}{7} = 11$ ✓ and $4\left(\frac{10}{7}\right) - \frac{19}{7} = \frac{40}{7} - \frac{19}{7} = \frac{21}{7} = 3$ ✓. Therefore $x = \frac{10}{7}$ and $y = \frac{19}{7}$.
    Examiner tips

    • Always verify your solution by substituting back into BOTH original equations.
    • Choose the method (substitution or elimination) based on which is easier for the given equations.

31. Question 24

    1 marksQuadratic Graphs
    $x = -\frac{b}{2a} = \frac{4}{2} = 2$. Or: $y = (x-1)(x-3)$, vertex midway at $x = 2$. $x = 4$ is $-\frac{b}{a}$. $x = 1$ and $x = 3$ are roots not the minimum point.
    Method:

    x = −b/(2a) = 4/2 = 2. Or: y = (x−1)(x−3), vertex midway at x = 2. x = 4 is −b/a. x = 1 and x = 3 are roots.
    Examiner tips

    • Formula for vertex: x = −b/(2a).
    • Alternatively, the vertex is midway between the roots.

32. Question 25

    1 marksQuadratic Applications
    Max at $t = \frac{24}{2 \times 4} = 3$. $h = 24(3) - 4(9) = 72 - 36 = 36$ m. 24 m is coefficient. 48 m is $2 \times 24$. 6 m is $t$ value.
    Method:

    Max at t = 24/(2×4) = 3. h = 24(3) − 4(9) = 72 − 36 = 36 m. 24 m is coefficient. 48 m is 2×24. 6 m is t value.
    Examiner tips

    • For h = at² + bt + c, maximum is at t = −b/(2a).
    • Substitute this t value back into h to find the maximum height.