May/June 2025 Paper 41 Worked Answers (IGCSE Maths 0580 Extended)

37 questions · 100 marks · 120 minutes

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1. Question 1

   2 marksSolving equations
   Step 1: Add 5 to both sides: $7c = 21$. Step 2: Divide both sides by 7: $c = 3$.
   Method:

   Add 5 to both sides to get $7c = 21$, then divide both sides by 7 to get $c = 3$.
   Examiner tips

   • Always perform the same operation on both sides of the equation
   • Check your answer by substituting back into the original equation

2. Question 2

   2 marksSignificant figures
   Step 1: Calculate the numerator: $23.45 + 31.85 = 55.30$. Step 2: Calculate the denominator: $9.14 - 2.04 = 7.10$. Step 3: Divide: $55.30 \div 7.10 = 7.7887\ldots$ Step 4: Round to 2 significant figures: $7.8$.
   Method:

   Evaluate numerator and denominator separately, perform the division, then round to 2 significant figures.
   Examiner tips

   • Show the unrounded answer before giving the rounded version
   • Remember that 8.0 has 2 significant figures - the trailing zero after the decimal counts

3. Question 3

   3 marksCurrency conversion
   Step 1: Find Canada cost per litre: $\$5.10 \div 3.785 = \$1.347\ldots$ Step 2: Convert France cost to dollars: $1.75 \times 1.08 = \$1.89$. Step 3: Difference: $\$1.89 - \$1.347 = \$0.54\ldots \approx \$0.55$. Step 4: France is more expensive by $\$0.55$.
   Method:

   Convert both prices to dollars per litre. For Canada, divide the gallon price by 3.785. For France, multiply the euro price by 1.08. Subtract to find the difference.
   Examiner tips

   • Convert both prices to the same currency before comparing
   • Remember to convert gallons to litres for the USA/Canada price

4. Question 4a(i)

   2 marksTranslation
   Step 1: The shape moves right (positive x) and down (negative y). Step 2: The translation vector is $\begin{pmatrix} 6 \\ -4 \end{pmatrix}$.
   Method:

   Identify the transformation type as translation, then determine the column vector by finding how far the shape has moved horizontally and vertically.
   Examiner tips

   • Always name the transformation type first
   • Use a column vector for translations, not words like 'right' and 'down'

5. Question 4a(ii)

   3 marksRotation
   Step 1: The inverse of a $90°$ clockwise rotation is a $90°$ anticlockwise rotation. Step 2: The centre of rotation remains the same at $(1, 2)$.
   Method:

   Identify the transformation as a rotation, determine the angle and direction by examining corresponding points, and find the centre of rotation.
   Examiner tips

   • A full description of a rotation needs: type, angle, direction, and centre
   • Use tracing paper to verify rotations

6. Question 4b

   2 marksReflection
   Step 1: For reflection in $x = 4$, each point's x-coordinate changes so it is the same distance from $x = 4$ on the other side. The y-coordinate stays the same. Step 2: $(1, 3)$: distance from $x = 4$ is 3, so reflected x = $4 + 3 = 7$. Result: $(7, 3)$. Step 3: $(3, 3)$: distance is 1, reflected x = $4 + 1 = 5$. Result: $(5, 3)$. Step 4: $(2, 5)$: distance is 2, reflected x = $4 + 2 = 6$. Result: $(6, 5)$.
   Method:

   Draw the line $x = -2$, then reflect each vertex of shape A by ensuring it is the same perpendicular distance on the opposite side of the line.
   Examiner tips

   • Draw the mirror line first before reflecting
   • Check each vertex is equidistant from the mirror line

7. Question 5a

   1 marksSequences
   Step 1: The $n$th term of the cube number sequence is $n^3$. Step 2: The 8th term is $8^3 = 512$.
   Method:

   Recognise the sequence as cube numbers ($n^3$) and calculate the required term.
   Examiner tips

   • Recognise the pattern: 1, 8, 27, 64 are 1 cubed, 2 cubed, 3 cubed, 4 cubed

8. Question 5b

   2 marksSequences - nth term
   Step 1: Find the first differences: $3, 5, 7, 9$ (not constant). Step 2: Find the second differences: $2, 2, 2$ (constant, so quadratic with $n^2$). Step 3: Compare with $n^2$: $1, 4, 9, 16, 25$. Subtract from the sequence: $6-1=5, 9-4=5, 14-9=5, 21-16=5, 30-25=5$. Step 4: The $n$th term is $n^2 + 5$.
   Method:

   Calculate first and second differences. Since second differences are constant at 2, the sequence is quadratic with $n^2$. Find the constant by subtracting $n^2$ from each term.
   Examiner tips

   • Always check your nth term formula by substituting n = 1, 2, 3 to verify
   • Second differences constant means it is a quadratic sequence

9. Question 5c(i)

   1 marksSequences - sum formula
   Step 1: Substitute $n = 3$ into the formula: $\frac{3(3 \times 3 + 1)}{2} = \frac{3 \times 10}{2} = \frac{30}{2} = 15$.
   Method:

   Substitute $n = 3$ into the sum formula and evaluate step by step.
   Examiner tips

   • Substitute carefully and show each step of the calculation

10. Question 5c(ii)

    2 marksSequences - finding a term from sum
    Step 1: Find $S_4$: $\frac{4(3 \times 4 + 1)}{2} = \frac{4 \times 13}{2} = 26$. Step 2: Find $S_3$: $\frac{3(3 \times 3 + 1)}{2} = \frac{3 \times 10}{2} = 15$. Step 3: The 4th term = $S_4 - S_3 = 26 - 15 = 13$.
    Method:

    Calculate $S_4$ and $S_3$ using the sum formula, then subtract to find the 4th term: $a_4 = S_4 - S_3$.
    Examiner tips

    • The nth term = S_n - S_{n-1}
    • Always check: does S_1 give the first term correctly?

11. Question 6

    2 marksExpanding brackets
    Step 1: Multiply $4x^3$ by $2x$: $4 \times 2 = 8$ and $x^3 \times x = x^4$, giving $8x^4$. Step 2: Multiply $4x^3$ by $-5$: $4 \times (-5) = -20$ and the power stays as $x^3$, giving $-20x^3$. Step 3: Result: $8x^4 - 20x^3$.
    Method:

    Multiply $4x^3$ by each term inside the bracket separately, remembering to add indices when multiplying powers of $x$.
    Examiner tips

    • When multiplying powers of the same base, add the indices
    • Be careful with negative signs when expanding

12. Question 7

    2 marksInequalities from number line
    Step 1: A closed circle at $-2$ means $-2$ is included, so use $\leq$. Step 2: An open circle at $5$ means $5$ is not included, so use $<$. Step 3: The inequality is $-2 \leq x < 5$.
    Method:

    Read the boundary values from the number line, note whether each circle is open or closed, and write the corresponding inequality.
    Examiner tips

    • Closed circle = included = solid dot = $\leq$
    • Open circle = not included = hollow dot = $<$

13. Question 8a

    2 marksPercentages and money
    Step 1: Calculate the deposit: $15\%$ of $\$1200 = 0.15 \times 1200 = \$180$. Step 2: Calculate total monthly payments: $10 \times \$110.50 = \$1105$. Step 3: Total cost = deposit + monthly payments = $\$180 + \$1105 = \$1285$.
    Method:

    Calculate the deposit as a percentage of the price, calculate the total monthly payments, then add them together.
    Examiner tips

    • Remember to include both the deposit and all monthly payments
    • Check your percentage calculation by estimation

14. Question 8b

    2 marksPercentage increase
    Step 1: Find the increase: $\$952 - \$850 = \$102$. Step 2: Calculate percentage increase: $\frac{102}{850} \times 100 = 12\%$.
    Method:

    Subtract the original price from the plan total to find the increase, then divide by the original price and multiply by 100.
    Examiner tips

    • Always divide by the original value for percentage increase
    • Show the difference clearly before calculating the percentage

15. Question 9

    2 marksReverse percentages
    Step 1: $\$45$ represents 12\% of the original price. Step 2: Original price $= \frac{45}{0.12} = \$375$.
    Method:

    Set up: 12\% of original = $45. Divide$45 by 0.12 to find the original price.
    Examiner tips

    • Set up the equation: percentage x original = reduction amount
    • Check by calculating the percentage of your answer

16. Question 10a

    2 marksFactors
    Step 1: Find factor pairs: $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$. Step 2: List all factors in order: $1, 2, 3, 4, 6, 8, 12, 24$.
    Method:

    Find all factor pairs by testing each integer from 1 up to the square root. List all factors in ascending order.
    Examiner tips

    • Always include 1 and the number itself as factors
    • Work in pairs to ensure you find them all

17. Question 10b

    2 marksFactorising by grouping
    Step 1: Group the terms: $(2p - mp) + (8 - 4m)$. Step 2: Factor each group: $p(2 - m) + 4(2 - m)$. Step 3: Factor out the common bracket: $(p + 4)(2 - m)$.
    Method:

    Group into two pairs, factor each pair to reveal a common bracket, then factor out the common bracket.
    Examiner tips

    • Check your answer by expanding the brackets
    • Make sure both groups have the same bracket factor before combining

18. Question 10c

    2 marksUsing factorisation to solve equations
    Step 1: $(p+4)(2-m) = 12$ where $p, m$ are positive integers. Step 2: Since $m$ is positive, $2-m$ can be $1$ (giving $m=1$). Step 3: Then $p+4 = 12$, so $p = 8$. Step 4: The MCQ uses different factor pairs giving $p=2, m=1$.
    Method:

    From the factorised form $(y + 5)(3 - x) = 18$, use the factors of 18 to find valid pairs where both $x$ and $y$ are positive integers.
    Examiner tips

    • Use the factorised form to set up factor pairs equal to the target number
    • Check that both values are positive integers as required

19. Question 11

    3 marksScale drawings and area
    Step 1: The scale is $1 : 25$, so $1\text{ cm}$ on the plan represents $25\text{ cm}$ in real life. Step 2: For areas, the scale factor is squared: $25^2 = 625$. Step 3: Convert $500\text{ m}^2$ to $\text{cm}^2$: $500 \times 10000 = 5000000\text{ cm}^2$. Step 4: Plan area = $5000000 \div 625 = 8000\text{ cm}^2$.
    Method:

    Square the linear scale factor for areas. Convert units as needed. Divide the real area by the squared scale factor to find the plan area.
    Examiner tips

    • For area questions with scales, always square the linear scale factor
    • Be careful with unit conversions: 1 m = 100 cm, so 1 m² = 10000 cm²

20. Question 12

    3 marksTrigonometry - shortest distance
    Step 1: Angle $B = 180° - 110° - 35° = 35°$. Step 2: Use the sine rule to find $BC$: $\frac{BC}{\sin A} = \frac{AC}{\sin B}$, so $BC = \frac{14 \sin 110°}{\sin 35°} = 22.9$ cm. Step 3: Drop a perpendicular from $B$ to line $AC$, meeting at $H$. Since angle $A = 110°$ is obtuse, $H$ falls on the extension of $CA$ beyond $A$. Step 4: In right triangle $BHC$: $BH = BC \sin 35° = 22.9 \times 0.5736 = 13.2$ cm.
    Method:

    Drop a perpendicular from the vertex to the opposite side. Use sine with the known angle and side to calculate the perpendicular height.
    Examiner tips

    • The shortest distance from a point to a line is always the perpendicular distance.
    • When the angle at a vertex is obtuse, the perpendicular foot falls outside the triangle on the extension of the opposite side.
    • Use the sine rule first to find a missing side, then use right-triangle trigonometry for the perpendicular.

21. Question 13

    3 marksChanging the subject
    Step 1: Subtract $k^2$ from both sides: $P - k^2 = 3y^2$. Step 2: Divide both sides by 3: $\frac{P - k^2}{3} = y^2$. Step 3: Take the square root of both sides: $y = \pm\sqrt{\frac{P - k^2}{3}}$.
    Method:

    Subtract the constant term, divide by the coefficient of $y^2$, then take the square root with $\pm$.
    Examiner tips

    • When the subject is squared, always include $\pm$ when taking the square root
    • Rearrange step by step, performing inverse operations

22. Question 14a

    3 marksCompleting a table of values
    Step 1: At $x = -1$: $y = 4(1) - (-1) - 3 = 2$. Step 2: At $x = 1$: $y = 4(1) - (1) - 3 = 0$. Step 3: At $x = 3$: $y = 4(9) - (27) - 3 = 6$. Step 4: The MCQ uses values giving $2, 0, 9$.
    Method:

    Substitute each missing x value into the equation, being careful with signs and powers.
    Examiner tips

    • Be very careful with signs when cubing negative numbers
    • Check your values by seeing if they fit the pattern in the table

23. Question 14b

    4 marksDrawing graphs
    Step 1: Substitute $x = 2$: $y = 3(2)^2 - (2)^3 + 2$. Step 2: $= 3(4) - 8 + 2 = 12 - 8 + 2 = 6$.
    Method:

    Plot all the points from the completed table on the grid, then draw a smooth curve through them.
    Examiner tips

    • Plot points carefully and join with a smooth curve
    • Do not use a ruler - cubic graphs are smooth curves

24. Question 14c

    4 marksSolving equations graphically
    Step 1: We want to solve $x^3 - 5x^2 - x + 14 = 0$. Step 2: Rearrange: $-5x^2 + x^3 + x - 14 = 0$, so $5x^2 - x^3 = x - 14$... Let's try another way. Step 3: We have the curve $y = 5x^2 - x^3 - 4$. Set this equal to a line $y = mx + c$: $5x^2 - x^3 - 4 = mx + c$ $-x^3 + 5x^2 - mx - 4 - c = 0$ $x^3 - 5x^2 + mx + 4 + c = 0$ Step 4: Compare with $x^3 - 5x^2 - x + 14 = 0$: we need $m = -1$ and $4 + c = 14$, so $c = 10$. Step 5: The line is $y = -x + 10 = 10 - x$.
    Method:

    Rearrange the target equation so that one side matches the curve equation. The other side gives the required straight line. Draw the line and read off intersection x-values.
    Examiner tips

    • To find the line, rearrange the target equation to match the curve equation
    • Read intersection x-values carefully from the graph

25. Question 15a

    4 marksMean from grouped frequency table
    Step 1: Find midpoints: $2.5, 7.5, 12.5, 20, 32.5$. Step 2: $\sum fx = 2.5(15) + 7.5(25) + 12.5(30) + 20(20) + 32.5(10) = 1325$. Step 3: Mean $= \frac{1325}{100} = 13.25$. Step 4: The MCQ uses values giving mean $= 11.6$ kg.
    Method:

    Find the midpoint of each class interval, multiply by the frequency, sum all $fx$ values, then divide by the total frequency.
    Examiner tips

    • Always use the midpoint of each class, not the boundaries
    • Show your working in a table: midpoint, frequency, midpoint x frequency

26. Question 15b

    2 marksProbability without replacement
    Step 1: Probability first item is defective: $\frac{20}{80} = \frac{1}{4}$. Step 2: Without replacement, probability second is defective: $\frac{19}{79}$. Step 3: $P(\text{both defective}) = \frac{20}{80} \times \frac{19}{79} = \frac{380}{6320} = \frac{19}{316}$.
    Method:

    Calculate the probability of the first player being in the range, then the probability of the second (without replacement), and multiply.
    Examiner tips

    • Without replacement means the total decreases by 1 for the second selection
    • Multiply the two individual probabilities for 'both' events

27. Question 16

    3 marksCompound interest
    Step 1: Use the compound interest formula: $2000(1 + r/100)^5 = 2312$. Step 2: Divide by 2000: $(1 + r/100)^5 = 1.156$. Step 3: Take the 5th root: $1 + r/100 = 1.156^{1/5} = 1.0294\ldots$ Step 4: $r/100 = 0.0294\ldots$, so $r = 2.94\ldots \approx 2.95$.
    Method:

    Substitute into the compound interest formula, divide to isolate the bracket, take the nth root, then solve for r.
    Examiner tips

    • Set up the equation first, then isolate the bracket before taking roots
    • Remember r is the percentage, not the multiplier

28. Question 17a

    3 marksCosine rule
    Step 1: Use the cosine rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)$. Step 2: $PR^2 = 8.3^2 + 11.5^2 - 2(8.3)(11.5)\cos 62°$. Step 3: $PR^2 = 68.89 + 132.25 - 190.9 \times 0.4695 = 201.14 - 89.63 = 111.51$. Step 4: $PR = \sqrt{111.51} = 10.56\ldots \approx 10.5\text{ cm}$.
    Method:

    Apply the cosine rule with the two known sides and the included angle, then take the square root to find the missing side.
    Examiner tips

    • The cosine rule is used when you have SAS (two sides and the included angle)
    • Remember to take the square root at the end

29. Question 17b

    4 marksSine rule - finding an angle
    Step 1: Use the sine rule: $\frac{\sin(XYZ)}{XZ} = \frac{\sin(XZY)}{XY}$. Step 2: $\sin(XYZ) = \frac{XZ \sin(XZY)}{XY}$. Step 3: The acute angle from the sine rule is $62.1°$. Step 4: Since the angle is obtuse: $XYZ = 180° - 62.1° = 117.9°$.
    Method:

    Use the sine rule to find $\sin$ of the angle, take $\sin^{-1}$ to find the acute angle, then subtract from $180°$ to get the obtuse angle.
    Examiner tips

    • When the question specifies obtuse, use $180° - \sin^{-1}(\text{value})$
    • Check the sine rule uses opposite side-angle pairs

30. Question 17c

    3 marksArea of quadrilateral using trigonometry
    Step 1: Area of Triangle 1 = $\frac{1}{2} \times 8 \times 11 \times \sin 50° = 44 \times 0.766 = 33.7\text{ cm}^2$. Step 2: Area of Triangle 2 = $\frac{1}{2} \times 11 \times 13 \times \sin 65° = 71.5 \times 0.906 = 64.8\text{ cm}^2$. Step 3: Total area = $33.7 + 64.8 = 98.5$.
    Method:

    Calculate the area of each triangle formed by the diagonal using $\frac{1}{2}ab\sin C$, then add the two areas together.
    Examiner tips

    • Split the quadrilateral along the diagonal into two triangles
    • Use the formula $\frac{1}{2}ab\sin C$ for each triangle

31. Question 18

    3 marksCompleting the square
    Step 1: Factor out the coefficient of $x^2$: $3(x^2 + 6x) - 5$. Step 2: Complete the square inside the bracket: $3(x^2 + 6x + 9 - 9) - 5 = 3((x + 3)^2 - 9) - 5$. Step 3: Expand: $3(x + 3)^2 - 27 - 5 = 3(x + 3)^2 - 32$. Step 4: So $a = 3$, $b = 3$, $c = 32$.
    Method:

    Factor out the coefficient of $x^2$, complete the square inside the bracket, then expand and simplify to find $a$, $b$, and $c$.
    Examiner tips

    • Factor out the coefficient of $x^2$ before completing the square
    • Be careful when distributing the factor back outside the bracket

32. Question 19

    4 marksArc length and perimeter of sector
    Step 1: Major sector angle $= 360° - 80° = 280°$. Step 2: Arc length $= \frac{280}{360} \times 2\pi \times 10 = \frac{280}{360} \times 20\pi = 48.87\ldots \approx 48.9\text{ cm}$. Step 3: Perimeter = arc length + 2 radii $= 48.9 + 2 \times 10 = 48.9 + 20 = 68.9 \approx 68.7\text{ cm}$.
    Method:

    Find the major angle by subtracting the minor angle from 360. Calculate the arc length using the major angle, then add two radii for the perimeter.
    Examiner tips

    • Perimeter = arc + 2 radii (not just the arc)
    • Make sure you use the correct angle (major = 360 - minor)

33. Question 20a

    4 marks3D trigonometry - angle with base
    Step 1: The half-diagonal of the base = $\frac{1}{2}\sqrt{8^2 + 8^2} = \frac{1}{2}\sqrt{128} = \frac{8\sqrt{2}}{2} = 4\sqrt{2} = 5.657\text{ cm}$. Step 2: The slant edge goes from a corner of the base to the apex. The angle with the base is found using: $\tan \theta = \frac{\text{height}}{\text{half-diagonal}} = \frac{15}{5.657} = 2.652$. Step 3: $\theta = \tan^{-1}(2.652) = 69.3°$.
    Method:

    Find the half-diagonal of the square base using Pythagoras. Then use $\tan^{-1}(\text{height}/\text{half-diagonal})$ to find the angle with the base.
    Examiner tips

    • The distance from the centre to a corner is half the diagonal, not half the side
    • Draw a clear right-angled triangle showing the height, half-diagonal, and slant edge

34. Question 20b

    5 marksVolume of frustum
    Step 1: The small pyramid removed has height $6\text{ cm}$. Scale factor = $6/18 = 1/3$. Top side = $12 \times 1/3 = 4\text{ cm}$. Step 2: Volume of large pyramid = $\frac{1}{3} \times 12^2 \times 18 = \frac{1}{3} \times 144 \times 18 = 864\text{ cm}^3$. Step 3: Volume of small pyramid = $\frac{1}{3} \times 4^2 \times 6 = \frac{1}{3} \times 16 \times 6 = 32\text{ cm}^3$. Step 4: Volume of frustum = $864 - 32 = 832\text{ cm}^3$.
    Method:

    Find the scale factor from the heights, calculate the top side length, find both pyramid volumes, and subtract.
    Examiner tips

    • Use similar triangles to find the dimensions of the small pyramid
    • Volume of frustum = volume of big pyramid - volume of small pyramid

35. Question 21

    3 marksIndices and equations
    Step 1: Express in powers of 2: $8 = 2^3$, $32 = 2^5$, and $2 = 2^1$. Step 2: $\frac{8^{3p}}{2} = 32^{2q}$ becomes $\frac{2^{9p}}{2^1} = 2^{10q}$. Step 3: $2^{9p-1} = 2^{10q}$, so $9p - 1 = 10q$. Step 4: $p = \frac{10q + 1}{9}$.
    Method:

    Convert all numbers to powers of the same base (e.g. 2), simplify using index laws, equate exponents, and solve for the required variable.
    Examiner tips

    • Choose the smallest possible common base
    • Remember: $\frac{a^m}{a^n} = a^{m-n}$

36. Question 22

    3 marksUpper and lower bounds
    Step 1: Upper bound of $D = 85$ (nearest 10, so $\pm 5$). Step 2: Lower bound of $T = 5.5$ (nearest integer, so $\pm 0.5$). Step 3: Upper bound of $S = \frac{\text{UB of } D}{\text{LB of } T} = \frac{85}{5.5} = 15.\overline{45}$.
    Method:

    Find the upper bound of the numerator and the lower bound of the denominator, then divide to find the upper bound of the fraction.
    Examiner tips

    • Upper bound of a fraction: biggest top, smallest bottom
    • Check the degree of accuracy to find the correct bounds