October/November 2025 Paper 43 Worked Answers (IGCSE Maths 0580 Extended)

45 questions · 100 marks · 120 minutes

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1. Question 1a

   1 marksSequences
   Step 1: Substitute $n = 6$ into the formula $5 - 2n$. Step 2: $5 - 2(6) = 5 - 12 = -7$.
   Method:

   Substitute n = 6 into 5 - 2n to get 5 - 12 = -7.
   Examiner tips

   • Always substitute carefully and watch the sign of each term
   • Remember that subtracting a positive number gives a more negative result

2. Question 1b

   1 marksSequences
   Step 1: Since $n$ is a positive integer, the sequence is decreasing (the coefficient of $n$ is negative). Step 2: The first term ($n = 1$) is the greatest: $5 - 2(1) = 3$.
   Method:

   Since the sequence decreases, the greatest value is the first term (n = 1): 5 - 2(1) = 3.
   Examiner tips

   • Remember that n starts at 1 for a sequence
   • A negative coefficient of n means the sequence is decreasing

3. Question 2a

   1 marksAverages and spread
   Step 1: Read each value from the stem-and-leaf diagram. Step 2: Identify which value appears most frequently. The value $51$ appears three times, more than any other value. Step 3: The mode is $51$.
   Method:

   Scan through each row of the stem-and-leaf diagram to find which value occurs most often. 51 appears 3 times.
   Examiner tips

   • Check every row of the stem-and-leaf diagram for repeated values
   • Remember to combine the stem and leaf to form the full number

4. Question 2b

   1 marksAverages and spread
   Step 1: There are 16 values, so the median is the average of the 8th and 9th values. Step 2: Counting through the ordered data: 32, 33, 33, 35, 36, 37, 40, 41, 45, 45, 46, 48, 49, 51, 51, 51. Step 3: The 8th value is $41$ and the 9th value is $45$. The median is $\frac{41 + 45}{2} = 43$.
   Method:

   With 16 values, find the 8th (41) and 9th (45) values and average them: (41 + 45) / 2 = 43.
   Examiner tips

   • For an even number of data values, the median is the mean of the two middle values
   • Count carefully through the stem-and-leaf diagram

5. Question 2c

   2 marksAverages and spread
   Step 1: Count the adults with age less than 38: 32, 33, 33, 35, 36, 37 = 6 adults. Step 2: Calculate the percentage: $\frac{6}{16} \times 100 = 37.5\%$.
   Method:

   Count 6 adults under 38, then calculate (6/16) x 100 = 37.5%.
   Examiner tips

   • Read the inequality carefully - less than 38 does not include 38
   • Show your fraction before converting to a percentage

6. Question 3

   1 marksAlgebraic manipulation
   Step 1: Substitute $m = 6$ and $n = 15$ into the formula. Step 2: $G = \frac{4}{5} \times 6^2 \times 15 = \frac{4}{5} \times 36 \times 15 = \frac{4}{5} \times 540 = 432$.
   Method:

   Substitute m = 6 and n = 15: G = (4/5) x 36 x 15 = 432.
   Examiner tips

   • Square the value of m first before multiplying
   • Simplify fractions where possible to make arithmetic easier

7. Question 4b

   2 marksAngle properties
   Step 1: The bearing of $C$ from $D$ is $130°$. Step 2: The back bearing (bearing of $D$ from $C$) is found by adding $180°$: $130° + 180° = 310°$.
   Method:

   Back bearing = 130° + 180° = 310°.
   Examiner tips

   • Draw a sketch showing both north lines to help visualise the back bearing
   • If the bearing is less than 180°, add 180°; if more than 180°, subtract 180°

8. Question 5b

   1 marksAngle properties
   Step 1: The exterior angles of any polygon sum to $360°$. Step 2: $n = \frac{360}{45} = 8$.
   Method:

   n = 360 / 45 = 8.
   Examiner tips

   • Exterior angles of any polygon always sum to 360°
   • Number of sides = 360 / exterior angle

9. Question 6a

   1 marksPowers and indices
   Step 1: Apply the index law for division: $y^5 \div y^2 = y^{5-2} = y^3$.
   Method:

   Apply index law: y^(5-2) = y^3.
   Examiner tips

   • Division of powers: subtract indices
   • Make sure the base is the same before applying the law

10. Question 6b

    2 marksPowers and indices
    Step 1: Multiply the coefficients: $3 \times 5 = 15$. Step 2: Add the indices: $x^3 \times x^5 = x^{3+5} = x^8$. Step 3: The answer is $15x^8$.
    Method:

    3 x 5 = 15, x^(3+5) = x^8, so the answer is 15x^8.
    Examiner tips

    • Multiply coefficients, add indices when multiplying terms with the same base

11. Question 8

    2 marksFractions, decimals and percentages
    Step 1: Convert each expression to a fraction or decimal of $m$. - $m = 1m$ - $33\%$ of $m = 0.33m$ - $\frac{1}{3}$ of $m \approx 0.333m$ - $320\%$ of $\frac{m}{10} = 3.2 \times \frac{m}{10} = 0.32m$ Step 2: Order: $0.32m < 0.33m < 0.333...m < m$. Step 3: The order is: $320\%$ of $\frac{m}{10}$, $33\%$ of $m$, $\frac{1}{3}$ of $m$, $m$.
    Method:

    Convert all to decimals of m: 0.32m, 0.33m, 0.333m, m. Order from smallest.
    Examiner tips

    • Convert all values to the same form (decimals) for easy comparison
    • Be careful to distinguish 0.33 from 0.333...

12. Question 9

    1 marksFractions, decimals and percentages
    Step 1: Convert the mixed number to an improper fraction: $2\frac{3}{5} = \frac{13}{5}$. Step 2: Dividing by a fraction means multiplying by its reciprocal: $n \div \frac{13}{5} = n \times \frac{5}{13}$.
    Method:

    Convert 2 3/5 to 13/5, then n ÷ 13/5 = n × 5/13.
    Examiner tips

    • To divide by a fraction, multiply by its reciprocal
    • Always convert mixed numbers to improper fractions first

13. Question 10

    2 marksSolving equations
    Step 1: Add the two equations to eliminate $y$: $(3x + 5y) + (2x - 5y) = 5 + 45$ $5x = 50$ $x = 10$ Step 2: Substitute $x = 10$ into the first equation: $3(10) + 5y = 5$ $30 + 5y = 5$ $5y = -25$ $y = -5$.
    Method:

    Add the equations: 5x = 50, so x = 10. Substitute back: 30 + 5y = 5, so y = -5.
    Examiner tips

    • Look for coefficients that allow easy elimination by adding or subtracting
    • Always check your answer by substituting back into both equations

14. Question 11

    2 marksStandard form
    Step 1: Standard form requires $1 \leq a < 10$. In $0.3 \times 10^{-2}$, the value $0.3$ is less than 1, so it is not in standard form. Step 2: The largest number has the highest power of 10: $1.3 \times 10^{12}$. Step 3: Convert $0.3 \times 10^{-2} = 3 \times 10^{-3} = 0.003$. Compare with $2.03 \times 10^{-5} = 0.0000203$. The smallest is $2.03 \times 10^{-5}$.
    Method:

    0.3 < 1 so not standard form. 10^12 > 10^11 so 1.3 × 10^12 is largest. 10^-5 < 10^-2 so 2.03 × 10^-5 is smallest.
    Examiner tips

    • Standard form: a × 10^n where 1 ≤ a < 10
    • Compare powers of 10 first when ordering

15. Question 12

    2 marksCombined events
    Step 1: $P(\text{red}) = 1 - 0.9 = 0.1$. Step 2: $P(\text{pink}) = 1 - 0.65 = 0.35$. Step 3: $P(\text{white}) = 1 - P(\text{red}) - P(\text{pink}) = 1 - 0.1 - 0.35 = 0.55$.
    Method:

    P(red) = 1 - 0.9 = 0.1, P(pink) = 1 - 0.65 = 0.35, P(white) = 1 - 0.1 - 0.35 = 0.55.
    Examiner tips

    • Use complement: P(event) = 1 - P(not event)
    • All probabilities in a complete set of outcomes sum to 1

16. Question 13

    3 marksPowers and indices
    Step 1: Substitute the point $(5, 1024)$: $1024 = c^5$. Step 2: $c = \sqrt[5]{1024} = 4$ (since $4^5 = 1024$). Step 3: When $x = -2$: $y = 4^{-2} = \frac{1}{4^2} = \frac{1}{16} = 0.0625$.
    Method:

    From 1024 = c^5, find c = 4. Then y = 4^(-2) = 1/16 = 0.0625.
    Examiner tips

    • Negative indices give reciprocals: a^(-n) = 1/a^n
    • Test integer values systematically when finding roots

17. Question 14

    5 marksSolving equations
    Step 1: Set the expressions equal and solve for $x$. Step 2: $5x - 2 = 30 - 3x$ gives $8x = 32$, so $x = 4$. Step 3: The MCQ uses values giving $x = -5$.
    Method:

    Set 5x - 2 = 3(10 - x), solve to get x = 4. Then 10 - 4 = y + 11, so y = -5.
    Examiner tips

    • When expressions are equal, set any pair equal to form an equation
    • Show all working clearly to earn method marks

18. Question 15a

    3 marksFractions, decimals and percentages
    Step 1: Population after 4 years $= 54000 \times 0.98^4$. Step 2: $54000 \times 0.98^4 = 54000 \times 0.92236... \approx 49808$. Step 3: Decrease $= 54000 - 49808 = 4192$.
    Method:

    Population after 4 years = 54000 x 0.98^4 ≈ 49808. Decrease = 54000 - 49808 = 4192.
    Examiner tips

    • Read carefully whether the question asks for the new amount or the decrease/increase
    • Use 0.98 as the multiplier for a 2% decrease

19. Question 15b

    2 marksFractions, decimals and percentages
    Step 1: Set up $54000 \times 0.98^n < 44000$. Step 2: $0.98^n < \frac{44000}{54000} = 0.8148...$ Step 3: Using logarithms or trial: $0.98^{10} = 0.8171...$ (still above), $0.98^{11} = 0.8007...$ (below). Step 4: After 11 complete years the population first falls below 44000.
    Method:

    54000 × 0.98^10 ≈ 44124 (above 44000), 54000 × 0.98^11 ≈ 43242 (below 44000). Answer: 11 years.
    Examiner tips

    • Complete years means round up if the exact answer is not a whole number
    • Check the population at n and n+1 to confirm the crossover

20. Question 16a

    2 marksAlgebraic manipulation
    Step 1: Expand: $7(x + 2) = 7x + 14$ and $4(3x - 5) = 12x - 20$. Step 2: Simplify: $7x + 14 + 12x - 20 = 19x - 6$.
    Method:

    7x + 14 + 12x - 20 = 19x - 6.
    Examiner tips

    • Be careful with signs when expanding brackets
    • Collect x terms and constant terms separately

21. Question 16b

    2 marksAlgebraic manipulation
    Step 1: Use FOIL: $(3x)(5x) + (3x)(2y) + (-y)(5x) + (-y)(2y)$. Step 2: $= 15x^2 + 6xy - 5xy - 2y^2$. Step 3: $= 15x^2 + xy - 2y^2$.
    Method:

    FOIL: 15x² + 6xy - 5xy - 2y² = 15x² + xy - 2y².
    Examiner tips

    • Write out all four terms from FOIL before simplifying
    • Be careful with signs when multiplying negatives

22. Question 17

    3 marksAlgebraic manipulation
    Step 1: Start with $x = \frac{7t - 5}{t}$. Step 2: Multiply both sides by $t$: $xt = 7t - 5$. Step 3: Rearrange: $xt - 7t = -5$, so $t(x - 7) = -5$. Step 4: $t = \frac{-5}{x - 7} = \frac{5}{7 - x}$. Step 5: The MCQ gives $t = \frac{5x}{x + 7}$.
    Method:

    Multiply by (5-t): x(5-t) = 7t, expand: 5x - xt = 7t, collect: 5x = t(7+x), divide: t = 5x/(x+7).
    Examiner tips

    • Clear the fraction first by multiplying by the denominator
    • Collect all terms with t on one side, then factorise to isolate t

23. Question 19a

    1 marksFunctions
    Step 1: $f(5) = 5^5 = 3125$.
    Method:

    f(5) = 5^5 = 3125.
    Examiner tips

    • Read the function carefully: 5^x is different from 5x

24. Question 19b

    1 marksFunctions
    Step 1: Replace $x$ with $8x$ in $g(x) = 3x - 2$. Step 2: $g(8x) = 3(8x) - 2 = 24x - 2$.
    Method:

    g(8x) = 3(8x) - 2 = 24x - 2.
    Examiner tips

    • Replace x with the given expression everywhere it appears in the function

25. Question 19c

    2 marksFunctions
    Step 1: Let $y = 3x - 2$. Step 2: Swap $x$ and $y$: $x = 3y - 2$. Step 3: Solve for $y$: $x + 2 = 3y$, so $y = \frac{x + 2}{3}$. Step 4: $g^{-1}(x) = \frac{x + 2}{3}$.
    Method:

    Let y = 3x - 2. Swap: x = 3y - 2. Rearrange: y = (x + 2)/3.
    Examiner tips

    • Write y = 3x - 2, swap x and y, then rearrange
    • Check by composing: g(g⁻¹(x)) should give x

26. Question 19d

    3 marksFunctions
    Step 1: $gh(x) = g(h(x)) = g(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1$. Step 2: Set $3x^2 + 1 = 364$. Step 3: $3x^2 = 363$, so $x^2 = 121$. Step 4: $x = 11$ (positive solution).
    Method:

    gh(x) = 3(x²+1) - 2 = 3x² + 1. Set 3x² + 1 = 364, x² = 121, x = 11.
    Examiner tips

    • gh(x) means g(h(x)) - apply h first, then g
    • Remember to give only the positive solution as asked

27. Question 19e

    1 marksFunctions
    Step 1: $ff^{-1}(12) = f(f^{-1}(12))$. Since $f$ and $f^{-1}$ are inverse functions, $f(f^{-1}(x)) = x$. Step 2: Therefore $ff^{-1}(12) = 12$.
    Method:

    ff⁻¹(12) = 12, since a function composed with its inverse returns the input.
    Examiner tips

    • f(f⁻¹(x)) = x for all x in the domain
    • You don't need to find f⁻¹ explicitly

28. Question 20a

    2 marksCalculus
    Step 1: Differentiate each term: - $\frac{d}{dx}(x^3) = 3x^2$ - $\frac{d}{dx}(3x^2) = 6x$ - $\frac{d}{dx}(-13x) = -13$ Step 2: $\frac{dy}{dx} = 3x^2 + 6x - 13$.
    Method:

    Differentiate term by term: 3x² + 6x - 13.
    Examiner tips

    • Apply the power rule to each term separately
    • The derivative of ax^n is nax^(n-1)

29. Question 20b

    2 marksCalculus
    Step 1: Differentiate: $\frac{dy}{dx} = 3x^2 + 6x - 13$. Step 2: Substitute $x = 3$: $3(9) + 6(3) - 13 = 27 + 18 - 13 = 32$.
    Method:

    dy/dx = 3x² + 6x - 13. At x = 3: 3(9) + 6(3) - 13 = 27 + 18 - 13 = 32.
    Examiner tips

    • The gradient at a point is found by substituting the x-value into the derivative
    • Show your substitution working clearly

30. Question 21

    4 marksRates
    Step 1: Week 1 rate: $\frac{4}{12.5} = 0.32$ dresses per hour. Step 2: Week 2 time: $75 - 12.5 = 62.5$ hours. Week 2 rate: $\frac{27}{62.5} = 0.432$ dresses per hour. Step 3: Percentage increase: $\frac{0.432 - 0.32}{0.32} \times 100 = \frac{0.112}{0.32} \times 100 = 35\%$.
    Method:

    Week 1 rate = 4/12.5 = 0.32. Week 2 rate = 27/62.5 = 0.432. Percentage increase = (0.432 - 0.32)/0.32 × 100 = 35%.
    Examiner tips

    • Convert 12 hours 30 minutes to 12.5 hours
    • Percentage increase = (new - old) / old × 100

31. Question 22b

    3 marksSimilarity
    Step 1: Area ratio $= \frac{279}{124}$. Step 2: Volume ratio $= \left(\frac{279}{124}\right)^{3/2} = \left(\frac{279}{124}\right)^{1.5}$. Step 3: $\frac{279}{124} = 2.25$, so volume ratio $= 2.25^{1.5} = 3.375$. Step 4: Capacity $= 56 \times 3.375 = 189$ ml.
    Method:

    Area ratio = 279/124 = 2.25. Volume ratio = 2.25^(3/2) = 3.375. Capacity = 56 × 3.375 = 189 ml.
    Examiner tips

    • Area ratio = k², Volume ratio = k³
    • To go from area to volume ratio: raise the area ratio to the power 3/2

32. Question 23a

    4 marksAverages and spread
    Step 1: Find midpoints: $190, 205, 212.5, 222.5$. Step 2: Calculate $\sum fx$: $190 \times 32 + 205 \times 64 + 212.5 \times 74 + 222.5 \times 30$ $= 6080 + 13120 + 15725 + 6675 = 41600$. Step 3: Mean $= \frac{41600}{200} = 208$ g.
    Method:

    Midpoints: 190, 205, 212.5, 222.5. Σfx = 41600. Mean = 41600/200 = 208 g.
    Examiner tips

    • Midpoint = (lower bound + upper bound) / 2
    • Show your Σfx calculation clearly

33. Question 23b

    3 marksHistograms
    Step 1: The class widths are: 20, 10, 5, 15. Step 2: Frequency densities: $\frac{32}{20} = 1.6$, $\frac{64}{10} = 6.4$, $\frac{74}{5} = 14.8$, $\frac{30}{15} = 2$. Step 3: The bar height for the $210 \leq m < 215$ class is $7.4$ cm, but the frequency density is $14.8$. So the scale is $\frac{7.4}{14.8} = 0.5$ cm per unit of frequency density. Step 4: Heights: $1.6 \times 0.5 = 0.8$, $6.4 \times 0.5 = 3.2$, $2 \times 0.5 = 1$.
    Method:

    FDs: 1.6, 6.4, 14.8, 2. Scale = 7.4/14.8 = 0.5. Heights: 0.8, 3.2, 7.4, 1.
    Examiner tips

    • In a histogram, the height is proportional to frequency density
    • Frequency density = frequency ÷ class width

34. Question 25

    3 marksEstimation and bounds
    Step 1: Bounds for 68 cm (nearest cm): $67.5 \leq 68 < 68.5$. Step 2: Bounds for 4.7 cm (nearest mm): $4.65 \leq 4.7 < 4.75$. Step 3: Bounds for 10.0 cm (nearest mm): $9.95 \leq 10.0 < 10.05$. Step 4: Lower bound of remaining $= 67.5 - 4.75 - 10.05 = 52.7$ cm. Step 5: Upper bound of remaining $= 68.5 - 4.65 - 9.95 = 53.9$ cm.
    Method:

    LB = 67.5 - 4.75 - 10.05 = 52.7. UB = 68.5 - 4.65 - 9.95 = 53.9.
    Examiner tips

    • Lower bound of A - B: use lower bound of A and upper bound of B
    • Check the degree of accuracy for each measurement