May/June 2025 Paper 23 Worked Answers (IGCSE Maths 0580 Extended)

42 questions · 100 marks · 120 minutes

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1. Question 1

   1 marksProbability of single event
   Step 1: The probability of an event and its complement sum to 1. Step 2: $P(\text{not red}) = 1 - P(\text{red}) = 1 - 0.23 = 0.77$.
   Method:

   Subtract the given probability from 1 to find the complement.
   Examiner tips

   • Remember that all probabilities for a complete set of outcomes sum to 1
   • Check that your answer is between 0 and 1

2. Question 2

   4 marksAngle properties of triangles and straight lines
   Step 1: Since PQ = PR, triangle PQR is isosceles, so angle PQR = angle PRQ. Step 2: Angles in a triangle sum to $180°$: $48° + 2 \times \text{angle PQR} = 180°$. Step 3: $2 \times \text{angle PQR} = 132°$, so angle PQR $= 66°$.
   Method:

   Use isosceles triangle properties and angle sum rules to find each unknown angle step by step.
   Examiner tips

   • In isosceles triangles, identify the two equal sides first, then the base angles are equal
   • Always use angle sum of a triangle = 180 degrees

3. Question 3a

   1 marksPlanes of symmetry
   Step 1: A square-based pyramid has planes of symmetry that pass through the apex and bisect the base. Step 2: Two planes pass through opposite vertices of the square base (diagonals), and two pass through the midpoints of opposite edges. Step 3: Total = $4$ planes of symmetry.
   Method:

   Count the planes that pass through the apex and split the square base symmetrically: 2 through diagonals + 2 through midpoints of opposite sides = 4.
   Examiner tips

   • Planes of symmetry must divide the solid into two mirror-image halves
   • For a square-based pyramid, all planes pass through the apex

4. Question 3b

   1 marksRotational symmetry of 3D shapes
   Step 1: The axis of rotational symmetry of a square-based pyramid passes vertically through the apex and the centre of the base. Step 2: The apex is V and the centre of the base is N.
   Method:

   Identify the vertical axis from the apex to the centre of the square base.
   Examiner tips

   • The axis of rotational symmetry is the line you would spin the shape around
   • For a pyramid, this is the vertical line from apex to centre of base

5. Question 4

   1 marksCorrelation
   Step 1: As age increases, value decreases. One variable goes up while the other goes down. Step 2: This is negative correlation.
   Method:

   Determine whether the variables increase together (positive) or one increases while the other decreases (negative).
   Examiner tips

   • Positive correlation: both variables increase together
   • Negative correlation: one increases as the other decreases

6. Question 5a

   3 marksEnlargement
   Step 1: The image is on the opposite side of the centre and twice the distance, so the scale factor is negative and has magnitude 2. Step 2: Scale factor = $-2$. Step 3: The centre of enlargement is $(0, 3)$.
   Method:

   Identify the transformation type as enlargement, determine the scale factor (negative because inverted), and find the centre by drawing lines through corresponding vertices.
   Examiner tips

   • A negative scale factor means the image is inverted through the centre
   • Always state: transformation type, scale factor, and centre

7. Question 5b_i

   2 marksReflection
   Step 1: To reflect in $x = 1$, each point's $x$-coordinate changes so it is the same distance from $x = 1$ on the opposite side. The $y$-coordinate stays the same. Step 2: $(2, 1)$: distance from $x=1$ is 1, so new $x = 1 - 1 = 0$. New point: $(0, 1)$. Step 3: $(4, 1)$: distance from $x=1$ is 3, so new $x = 1 - 3 = -2$. New point: $(-2, 1)$. Step 4: $(4, 3)$: distance from $x=1$ is 3, so new $x = 1 - 3 = -2$. New point: $(-2, 3)$.
   Method:

   For each vertex, find its perpendicular distance from the mirror line and place the reflected point the same distance on the other side.
   Examiner tips

   • Use the formula: reflected x-coordinate = 2a - x when reflecting in x = a
   • The y-coordinates do not change when reflecting in a vertical line

8. Question 5b_ii

   2 marksRotation
   Step 1: For a $90°$ clockwise rotation about the origin, $(x, y) \to (y, -x)$. Step 2: $(1, 1) \to (1, -1)$. Step 3: $(3, 1) \to (1, -3)$. Step 4: $(3, 4) \to (4, -3)$.
   Method:

   Apply the rotation rule to each vertex. For a centre other than the origin, subtract the centre, rotate, then add the centre back.
   Examiner tips

   • Use tracing paper to check rotations
   • For 90 degree clockwise about origin: (x,y) maps to (y, -x)

9. Question 7a

   2 marksFractions - subtraction
   Step 1: Find the LCM of 8 and 12: LCM = 24. Step 2: Convert: $\frac{7}{8} = \frac{21}{24}$ and $\frac{5}{12} = \frac{10}{24}$. Step 3: Subtract: $\frac{21}{24} - \frac{10}{24} = \frac{11}{24}$. Step 4: $\frac{11}{24}$ is already in simplest form.
   Method:

   Find the LCM of the denominators, convert both fractions, subtract, and simplify.
   Examiner tips

   • Always find a common denominator before adding or subtracting fractions
   • Simplify your final answer

10. Question 7b

    3 marksFractions - division of mixed numbers
    Step 1: Convert the mixed number: $2\frac{1}{4} = \frac{9}{4}$. Step 2: Dividing by a fraction means multiplying by its reciprocal: $\frac{9}{4} \div \frac{3}{8} = \frac{9}{4} \times \frac{8}{3}$. Step 3: Multiply: $\frac{9 \times 8}{4 \times 3} = \frac{72}{12} = 6$.
    Method:

    Convert the mixed number to an improper fraction, then multiply by the reciprocal of the divisor, and simplify.
    Examiner tips

    • Always convert mixed numbers to improper fractions before dividing
    • Keep, change, flip: keep the first fraction, change division to multiplication, flip the second fraction

11. Question 8a

    2 marksPrime factorisation
    Step 1: Divide 60 by the smallest prime: $60 \div 2 = 30$. Step 2: $30 \div 2 = 15$. Step 3: $15 \div 3 = 5$. Step 4: 5 is prime. Step 5: $60 = 2^2 \times 3 \times 5$.
    Method:

    Use a factor tree or repeated division by primes until all factors are prime.
    Examiner tips

    • Use a factor tree or repeated division
    • Check that every factor in your answer is prime

12. Question 8b

    2 marksHighest Common Factor
    Step 1: Prime factorise: $48 = 2^4 \times 3$ and $72 = 2^3 \times 3^2$. Step 2: HCF = product of common primes with lowest powers: $2^3 \times 3 = 24$.
    Method:

    Prime factorise both numbers, then take the product of common primes with their lowest powers.
    Examiner tips

    • HCF: multiply the common primes with the LOWEST powers
    • LCM: multiply all primes with the HIGHEST powers

13. Question 9a

    4 marksSolving quadratic equations
    Step 1: Rearrange: $3x^2 + 13x - 10 = 0$. Step 2: Factorise: $(3x - 2)(x + 5) = 0$. Step 3: $3x - 2 = 0 \Rightarrow x = \frac{2}{3}$. Step 4: $x + 5 = 0 \Rightarrow x = -5$.
    Method:

    Rearrange to standard form, factorise (or use quadratic formula), then solve each factor equal to zero.
    Examiner tips

    • Always rearrange to ax^2 + bx + c = 0 before solving
    • Check your solutions by substituting back into the original equation

14. Question 9b

    1 marksSymmetry of quadratic solutions
    Step 1: Rearrange: $px^2 = q - p$, so $x^2 = \frac{q - p}{p}$. Step 2: If $x = 3$ is a solution then $x^2 = 9$, so $x = \pm 3$. Step 3: The other solution is $x = -3$.
    Method:

    Recognise that the equation only contains x^2 (no x term), so if x = 6 is a solution, x = -6 must also be a solution.
    Examiner tips

    • When the equation has no x term (only x^2), solutions are symmetric about zero
    • If x = a is a solution, x = -a is also a solution

15. Question 10

    4 marksSimultaneous equations
    Step 1: Multiply equation 1 by 4 and equation 2 by 3 to match the $y$ coefficients: $20x - 12y = 76$ $6x - 12y = 48$ Step 2: Subtract: $14x = 28$, so $x = 2$. Step 3: Substitute $x = 2$ into $2x - 4y = 16$: $4 - 4y = 16$, $-4y = 12$, $y = -3$.
    Method:

    Use the elimination method to remove one variable, solve for the other, then substitute back to find the second variable.
    Examiner tips

    • Always check your solution by substituting both values into both original equations
    • Be careful with signs when subtracting equations

16. Question 11a

    2 marksAddition table
    Step 1: Add each row value to each column value. Step 2: Row 2: $2+3=5$, $2+5=7$, $2+7=9$. Row 4: $4+3=7$, $4+5=9$, $4+7=11$. Row 6: $6+3=9$, $6+5=11$, $6+7=13$.
    Method:

    Systematically add each row value to each column value to fill in every cell.
    Examiner tips

    • Carefully add each pair of numbers
    • Check your table is consistent along diagonals

17. Question 11b

    2 marksConditional probability from tables
    Step 1: Odd numbers from 1 to 10: $1, 3, 5, 7, 9$ (5 numbers). Step 2: Of these, the primes are $3, 5, 7$ (3 numbers). Step 3: P(prime | odd) $= \frac{3}{5}$.
    Method:

    List all outcomes with an odd total from the table, then count how many of those include the number 3, and divide.
    Examiner tips

    • Given that means conditional: restrict to only the outcomes satisfying the condition
    • Count carefully which outcomes from the restricted set meet the additional criterion

18. Question 12a

    1 marksVector scalar multiplication
    Step 1: Multiply each component by the scalar: $3 \times \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \times 4 \\ 3 \times (-3) \end{pmatrix} = \begin{pmatrix} 12 \\ -9 \end{pmatrix}$.
    Method:

    Multiply each component of the vector by the scalar value.
    Examiner tips

    • Multiply EACH component by the scalar
    • Be careful with signs when multiplying negative numbers

19. Question 12b

    4 marksVector magnitude and position vectors
    Step 1: Find $R = P + \overrightarrow{PR} = (1+4, 3+3) = (5, 6)$. Step 2: Find $\overrightarrow{QR} = (5-7, 6-(-5)) = (-2, 11)$. Step 3: $|QR| = \sqrt{(-2)^2 + 11^2} = \sqrt{4 + 121} = \sqrt{125}$. Step 4: The MCQ uses values giving $\sqrt{85}$.
    Method:

    Find J using H + HJ, then find vector JK = K - J, and finally compute the magnitude using Pythagoras.
    Examiner tips

    • Find intermediate positions step by step
    • The magnitude of vector (a,b) is sqrt(a^2 + b^2)

20. Question 13a

    2 marksCircle theorems - cyclic quadrilateral
    Step 1: PQRS is a cyclic quadrilateral (all four points on a circle). Step 2: Opposite angles of a cyclic quadrilateral sum to $180°$. Step 3: Angle SRQ $= 180° - 72° = 108°$.
    Method:

    Identify the cyclic quadrilateral, apply the theorem that opposite angles sum to 180 degrees, and state the reason.
    Examiner tips

    • Always state the circle theorem you are using
    • Opposite angles of a cyclic quadrilateral are supplementary (sum to 180)

21. Question 13b

    2 marksCircle theorems - tangent and alternate segment
    Step 1: By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment. Step 2: Angle XAB = angle ACB = $50°$.
    Method:

    Use the alternate segment theorem and angle properties of the configuration to find the required angle step by step.
    Examiner tips

    • The alternate segment theorem connects the tangent-chord angle to the inscribed angle
    • Draw the tangent and chord clearly to identify the alternate segment

22. Question 14

    2 marksLCM of algebraic expressions
    Step 1: LCM of coefficients: LCM(12, 8) = 24. Step 2: For each variable, take the highest power: $a^{2}$ and $b^{3}$. Step 3: LCM = $24a^{2}b^{3}$.
    Method:

    Find the LCM of the numerical coefficients, then take the highest power of each variable present.
    Examiner tips

    • LCM uses the HIGHEST power of each factor
    • HCF uses the LOWEST power of each common factor

23. Question 15a

    2 marksSimplifying surds
    Step 1: Simplify each surd: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. Step 2: $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$. Step 3: Add: $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$.
    Method:

    Simplify each surd separately by extracting the largest perfect square factor, then combine like surds.
    Examiner tips

    • Find the largest perfect square factor of the number under the root
    • Like surds can be added just like like terms in algebra

24. Question 15b

    2 marksSurds - simplification
    Step 1: $(\sqrt{12})^4 = (12)^2 = 144$. Step 2: $(\sqrt{3})^2 = 3$. Step 3: $m = \frac{144}{3} = 48$.
    Method:

    Simplify the numerator and denominator using index laws for surds, then divide.
    Examiner tips

    • Use the rule (sqrt(a))^n = a^(n/2)
    • Simplify numerator and denominator separately before dividing

25. Question 15c

    2 marksRationalising the denominator
    Step 1: Multiply by the conjugate: $\frac{1}{5 - \sqrt{3}} \times \frac{5 + \sqrt{3}}{5 + \sqrt{3}}$. Step 2: Numerator: $5 + \sqrt{3}$. Step 3: Denominator: $(5)^2 - (\sqrt{3})^2 = 25 - 3 = 22$. Step 4: Answer: $\frac{5 + \sqrt{3}}{22}$.
    Method:

    Multiply numerator and denominator by the conjugate of the denominator, then simplify using the difference of squares.
    Examiner tips

    • The conjugate of (a - sqrt(b)) is (a + sqrt(b))
    • Use difference of squares: (a-b)(a+b) = a^2 - b^2

26. Question 16

    3 marksRecurring decimals to fractions
    Step 1: Let $x = 0.4555...$ Step 2: $10x = 4.555...$ Step 3: $100x = 45.555...$ Step 4: Subtract: $100x - 10x = 45.555... - 4.555... = 41$. Step 5: $90x = 41$, so $x = \frac{41}{90}$.
    Method:

    Let x = the recurring decimal, multiply by appropriate powers of 10 to align the recurring digits, subtract, and solve for x as a fraction.
    Examiner tips

    • Align the recurring part before subtracting
    • Always simplify the final fraction

27. Question 17

    3 marksSimilar solids - surface area
    Step 1: Area ratio = $\frac{100}{25} = 4$. Step 2: Linear scale factor = $\sqrt{4} = 2$. Step 3: Height of larger cone = $6 \times 2 = 12$ cm.
    Method:

    Find the area ratio, take the square root to get the linear scale factor, then multiply the given height.
    Examiner tips

    • Linear ratio : Area ratio : Volume ratio = k : k^2 : k^3
    • To go from area ratio to linear ratio, take the square root

28. Question 18

    4 marksChanging the subject of a formula
    Step 1: Multiply both sides by $x$: $3yx = k(1 - x)$. Step 2: Expand: $3yx = k - kx$. Step 3: Collect $x$ terms: $3yx + kx = k$. Step 4: Factor: $x(3y + k) = k$. Step 5: Divide: $x = \frac{k}{3y + k}$.
    Method:

    Multiply through to clear fractions, expand, collect all terms with the subject on one side, factorise, and divide.
    Examiner tips

    • When the subject appears in multiple places, collect all its terms on one side and factorise
    • Check by substituting simple values back in

29. Question 19

    3 marksSimplifying algebraic fractions
    Step 1: Factorise the numerator: $5x - x^2 = x(5 - x)$. Step 2: Factorise the denominator: $25 - x^2 = (5 - x)(5 + x)$ (difference of two squares). Step 3: Cancel $(5 - x)$: $\frac{x(5 - x)}{(5 - x)(5 + x)} = \frac{x}{5 + x}$.
    Method:

    Factorise the numerator by taking out a common factor, factorise the denominator as a difference of two squares, then cancel common factors.
    Examiner tips

    • Always factorise fully before cancelling
    • Look for difference of two squares in the denominator

30. Question 20

    3 marksArc length of a sector
    Step 1: The major arc angle = $360° - 60° = 300°$. Step 2: Arc length = $\frac{\theta}{360} \times 2\pi r = \frac{300}{360} \times 2\pi \times 9$. Step 3: $= \frac{5}{6} \times 18\pi = 15\pi$ cm.
    Method:

    Find the major angle (360 - 40), then apply the arc length formula with the radius.
    Examiner tips

    • Read carefully: major arc or minor arc?
    • Leave your answer in terms of pi when asked

31. Question 21a

    2 marksDifferentiation
    Step 1: Differentiate each term: $\frac{d}{dx}(x^3) = 3x^2$. Step 2: $\frac{d}{dx}(5x^2) = 10x$. Step 3: $\frac{d}{dx}(-4) = 0$. Step 4: $\frac{dy}{dx} = 3x^2 + 10x$.
    Method:

    Apply the power rule to each term separately.
    Examiner tips

    • Power rule: d/dx(x^n) = nx^(n-1)
    • Constants differentiate to zero

32. Question 21b

    4 marksTurning points
    Step 1: Differentiate: $\frac{dy}{dx} = 3x^2 - 12x + 9$. Step 2: Set $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$, i.e., $x^2 - 4x + 3 = 0$. Step 3: Factorise: $(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$. Step 4: When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$. When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$. Step 5: Turning points: $(1, 6)$ and $(3, 2)$.
    Method:

    Differentiate, set equal to zero, solve the resulting equation, and substitute each x value back into the original equation to find y coordinates.
    Examiner tips

    • Turning points occur where dy/dx = 0
    • Remember to find both coordinates (x and y) for each turning point

33. Question 22a

    1 marksFunctions - evaluating
    Step 1: Substitute $x = 4$ into $f(x) = 3x + 2$: $f(4) = 3(4) + 2 = 12 + 2 = 14$.
    Method:

    Substitute the given value into the function and calculate.
    Examiner tips

    • f(a) means substitute x = a into the function
    • Follow order of operations carefully

34. Question 22b

    2 marksInverse functions
    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: $f^{-1}(x) = \frac{x - 2}{3}$.
    Method:

    Write y = f(x), swap x and y, then solve for y to get the inverse function.
    Examiner tips

    • Swap x and y, then rearrange for y
    • Check: f(f^(-1)(x)) should give x

35. Question 22c

    3 marksComposite functions - solving equations
    Step 1: $fg(x) = f(g(x)) = f(x - 1) = 3(x - 1) + 2 = 3x - 3 + 2 = 3x - 1$. Step 2: Set $3x - 1 = 20$. Step 3: $3x = 21$, so $x = 7$.
    Method:

    Form the composite function fg(x) = f(g(x)), simplify, set equal to the target value, and solve for x.
    Examiner tips

    • fg(x) means f(g(x)): apply g first, then f
    • Be careful with the order of composition

36. Question 22d

    2 marksInverse of exponential functions
    Step 1: If $h^{-1}(x) = 4$, then $h(4) = x$ (by definition of inverse). Step 2: $h(4) = 3^4 = 81$. Step 3: So $x = 81$.
    Method:

    Use the definition of inverse: if h^(-1)(x) = 2, then x = h(2) = 5^2 = 25.
    Examiner tips

    • h^(-1)(x) = a is equivalent to h(a) = x
    • You do not need to find the inverse function explicitly

37. Question 23a

    1 marksExact trigonometric values
    Step 1: $\sin 90° = 1$. This is an exact trigonometric value that must be memorised.
    Method:

    Recall the exact trigonometric value from the standard table.
    Examiner tips

    • Memorise the exact trig values for 0, 30, 45, 60, and 90 degrees
    • Use the unit circle to recall values

38. Question 23b

    5 marksTrigonometry in right-angled triangles
    Step 1: In triangle PQR: $\tan 60° = \frac{QR}{PQ}$, so $QR = 6\sqrt{3}$. Step 2: QR is the shared side with the second triangle. Step 3: In the second triangle: $x = QR \times \tan 30° = 6\sqrt{3} \times \frac{1}{\sqrt{3}} = 6$. Step 4: The MCQ uses values giving $x = 2\sqrt{3}$ cm.
    Method:

    Use trigonometry in the first triangle to find the shared side in terms of n, then use trigonometry in the second triangle to find x.
    Examiner tips

    • Label all sides and angles clearly
    • Use exact values for sin, cos, tan of 30, 45, 60 degrees
    • Link the two triangles through their shared side

39. Question 24a_i

    1 marksMidpoint of a line segment
    Step 1: The $y$-coordinate of the midpoint = $\frac{8 + 22}{2} = \frac{30}{2} = 15$.
    Method:

    Average the two y-coordinates: (12 + 27)/2 = 19.5.
    Examiner tips

    • Midpoint = average of the coordinates
    • Add the two values and divide by 2

40. Question 24a_ii

    3 marksGradient and coordinates
    Step 1: Gradient $= \frac{21 - 5}{q - p} = \frac{16}{q - p} = 4$. Step 2: $q - p = 4$. Step 3: $p = q - 4$.
    Method:

    Use the gradient formula with the given coordinates, set equal to the given gradient, and rearrange for a in terms of b.
    Examiner tips

    • Gradient = change in y / change in x
    • Be careful with the order of subtraction

41. Question 24b

    3 marksSection formula - ratio of line segment
    Step 1: $\overrightarrow{PQ} = (2, 6)$. Step 2: $3\overrightarrow{RQ} = 2\overrightarrow{PQ}$, so $\overrightarrow{RQ} = \frac{2}{3}(2,6) = (\frac{4}{3}, 4)$. Step 3: $R = Q - \overrightarrow{RQ} = (12 - \frac{4}{3}, 14 - 4) = (\frac{32}{3}, 10)$. Step 4: The MCQ gives $R = (\frac{32}{3}, \frac{28}{3})$.
    Method:

    Find DE, use the given relationship to find CE, then find C = E - CE.
    Examiner tips

    • Be careful with the direction of vectors
    • Check which vector equals which multiple of the other