IGCSE Mathematics 0580/23 May/June 2025 Past Paper with Answers

Question 1

A bag contains coloured counters. A counter is picked at random. The probability that it is red is $0.23$. Find the probability that it is not red.

1. $0.77$
2. $0.23$
3. $0.33$
4. $1.23$

Question 2

In triangle PQR, angle QPR = $48°$ and PQ = PR. Find angle PQR.

1. $66°$
2. $48°$
3. $132°$
4. $84°$

Question 3

A square-based pyramid has apex V and base PQRS. How many planes of symmetry does it have?

1. $4$
2. $2$
3. $5$
4. $8$

Question 4 · Answer A

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.

Question 5 · Answer A

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

Question 6 · Answer A

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 $(0, 3)$.

Question 7 · Answer A

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: $(-2, 3)$.

Question 8 · Answer A

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: $(3, 4) \to (4, -3)$.

Question 9 · Answer A

Step 1: There are 12 values (even number), so the median is the average of the 6th and 7th values. Step 2: The 6th value is $29$ and the 7th value is $31$. Step 3: Median $= \frac{29 + 31}{2} = \frac{60}{2} = 30$ kg.

Question 10 · Answer A

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.

Question 11 · Answer A

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: $\frac{9 \times 8}{4 \times 3} = \frac{72}{12} = 6$.

Question 12 · Answer A

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: $60 = 2^2 \times 3 \times 5$.

Question 13 · Answer A

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: $2^3 \times 3 = 24$.

Question 14 · Answer A

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: $x + 5 = 0 \Rightarrow x = -5$.

Question 15 · Answer A

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 $x = -3$.

Question 16 · Answer A

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 $2x - 4y = 16$: $4 - 4y = 16$, $-4y = 12$, $y = -3$.

Question 17 · Answer A

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: $6+3=9$, $6+5=11$, $6+7=13$.

Question 18 · Answer A

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) $= \frac{3}{5}$.

Question 19 · Answer A

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

Question 20 · Answer A

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 $\sqrt{85}$.

Question 21 · Answer A

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 $= 180° - 72° = 108°$.

Question 22 · Answer A

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 = $50°$.

Question 23 · Answer A

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 = $24a^{2}b^{3}$.

Question 24 · Answer A

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: $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$.

Question 25 · Answer A

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

Question 26 · Answer A

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: $\frac{5 + \sqrt{3}}{22}$.

Question 27 · Answer A

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 $x = \frac{41}{90}$.

Question 28 · Answer A

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.

Question 29 · Answer A

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: $x = \frac{k}{3y + k}$.

Question 30 · Answer A

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 $(5 - x)$: $\frac{x(5 - x)}{(5 - x)(5 + x)} = \frac{x}{5 + x}$.

Question 31 · Answer A

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.

Question 32 · Answer A

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: $\frac{dy}{dx} = 3x^2 + 10x$.

Question 33 · Answer A

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 $(3, 2)$.

Question 34 · Answer A

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

Question 35 · Answer A

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

Question 36 · Answer A

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 $x = 7$.

Question 37 · Answer A

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 $x = 81$.

Question 38 · Answer A

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

Question 39 · Answer A

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.

Question 40 · Answer A

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

Question 41 · Answer A

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

Question 42 · Answer A

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 $R = (\frac{32}{3}, \frac{28}{3})$.