IGCSE Mathematics 0580/22 May/June 2025 Past Paper with Answers

Question 1

A $4 \times 4$ grid has some squares shaded. Shade one more square so that the dotted line is a line of symmetry.

1. Row 2, Column 3
2. Row 1, Column 4
3. Row 3, Column 1
4. Row 4, Column 2

Question 2

A $4 \times 4$ grid has some squares shaded. Shade one more square so that the pattern has rotational symmetry of order 2.

1. Row 3, Column 4
2. Row 1, Column 1
3. Row 2, Column 3
4. Row 4, Column 1

Question 3

On a map, the scale is 1 cm represents 2 km. The distance between two villages on the map is 6.5 cm. What is the actual distance between the two villages?

1. 13 km
2. 3.25 km
3. 8.5 km
4. 6.5 km

Question 4 · Answer A

Step 1: The bearing of B from A is $135°$. Step 2: To find the back bearing (A from B), add $180°$: $135° + 180° = 315°$. Step 3: Since $315° < 360°$, the answer is $315°$.

Question 5 · Answer A

Step 1: Using angles in a triangle formed by the two transversals and the parallel line: $z = 180° - 50° - 70° = 60°$. Step 2: Alternatively, the exterior angle at the lower parallel line between the two transversals is $50° + 70° = 120°$. The angle at the top is $180° - 120° = 60°$.

Question 6 · Answer A

Step 1: The even numbers from 1 to 9 are: 2, 4, 6, 8. That is 4 even numbers. Step 2: Total number of cards = 9. Step 3: Probability = $\frac{4}{9}$.

Question 7 · Answer A

Step 1: Numbers greater than 4 on a die are 5 and 6, so $P(>4) = \frac{2}{6} = \frac{1}{3}$. Step 2: Expected number = probability $\times$ number of trials = $\frac{1}{3} \times 60 = 20$.

Question 8 · Answer A

Step 1: Add the translation vector to the point: $(2 + 3, 5 + (-4)) = (5, 1)$.

Question 9 · Answer A

Step 1: Check the mapping of vertices. $(1,2) \to (2,-1)$. Under $90°$ clockwise rotation about the origin: $(x,y) \to (y,-x)$. So $(1,2) \to (2,-1)$. ✓ Step 2: Check another vertex: $(3,2) \to (2,-3)$. ✓ Step 3: The transformation is a rotation of $90°$ clockwise about the origin.

Question 10 · Answer A

Step 1: Subtract 9 from both sides: $5x = 34 - 9 = 25$. Step 2: Divide both sides by 5: $x = \frac{25}{5} = 5$.

Question 11 · Answer A

Step 1: Expand the brackets: $12y - 15 = 33$. Step 2: Add 15 to both sides: $12y = 48$. Step 3: Divide by 12: $y = 4$.

Question 12 · Answer A

Step 1: Find the common difference: $11 - 15 = -4$. Step 2: The next term is $3 + (-4) = -1$.

Question 13 · Answer A

Step 1: The common difference is $13 - 17 = -4$. Step 2: The $n$th term of an arithmetic sequence is $a + (n-1)d = 17 + (n-1)(-4) = 17 - 4n + 4 = 21 - 4n$.

Question 14 · 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 15 · Answer A

Step 1: $\overrightarrow{PR} = 3\overrightarrow{PQ} = 3\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}$. Step 2: $R = P + \overrightarrow{PR} = (2 + 9, 5 + (-6)) = (11, -1)$.

Question 16 · Answer A

Step 1: Length $= \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45}$. Step 2: Simplify: $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$. Step 3: So $k = 3$.

Question 17 · Answer A

Step 1: $MP:PN = 1:3$ means $P$ is $\frac{1}{4}$ of the way from $M$ to $N$. Step 2: $\overrightarrow{MN} = (9-1, 6-2) = (8, 4)$. Step 3: $\overrightarrow{MP} = \frac{1}{4}(8, 4) = (2, 1)$. Step 4: $P = M + \overrightarrow{MP} = (1+2, 2+1) = (3, 3)$.

Question 18 · Answer A

Step 1: Arc length $= \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2\pi \times 12$. Step 2: $= \frac{1}{6} \times 24\pi = 4\pi$. Step 3: So $n = 4$.

Question 19 · Answer A

Step 1: Move the decimal point so the number is between 1 and 10: $3.56$. Step 2: Count the places moved: 4 places to the right, so the power is $-4$. Step 3: $0.000356 = 3.56 \times 10^{-4}$.

Question 20 · Answer A

Step 1: Rewrite with the same power of 10: $0.25 \times 10^{16} + 2.5 \times 10^{16}$. Step 2: Add: $(0.25 + 2.5) \times 10^{16} = 2.75 \times 10^{16}$. Step 3: $2.75$ is between 1 and 10, so this is already in standard form.

Question 21 · Answer A

Step 1: AC is a diameter, so angle $ABC$ is an angle in a semicircle. Step 2: The angle in a semicircle is always $90°$. Step 3: Therefore $ABC = 90°$.

Question 22 · Answer A

Step 1: Find midpoints: $5, 20, 40$. Step 2: Calculate $\Sigma fx = 5 \times 20 + 20 \times 35 + 40 \times 25 = 100 + 700 + 1000 = 1800$. Step 3: Mean $= \frac{1800}{80} = 22.5$ km. Step 4: The MCQ uses values giving mean $= 20$ km.

Question 23 · Answer A

Step 1: Class width for $10 < x \leq 25$ is $25 - 10 = 15$. Step 2: Frequency density $= \frac{\text{frequency}}{\text{class width}} = \frac{30}{15} = 2$.

Question 24 · Answer A

Step 1: Read the cumulative frequencies: at 10 it is 8, at 20 it is 25, at 30 it is 50. Step 2: The cumulative frequency first reaches (or exceeds) 50 at the value 30.

Question 25 · Answer A

Step 1: The median is the middle value. Step 2: For 120 values, the median is at the $\frac{120}{2} = 60$th value. Step 3: Read across at cumulative frequency 60.

Question 26 · Answer A

Step 1: The lower quartile is at $\frac{n}{4}$ on the cumulative frequency axis. Step 2: $\frac{80}{4} = 20$. Step 3: Read across at cumulative frequency 20.

Question 27 · Answer A

Step 1: Let $x = 0.1\overline{6} = 0.16666...$ Step 2: $10x = 1.6666...$ Step 3: $100x = 16.6666...$ Step 4: $100x - 10x = 16.6666... - 1.6666... = 15$. Step 5: $90x = 15$, so $x = \frac{15}{90} = \frac{1}{6}$.

Question 28 · Answer A

Step 1: At the $x$-axis, $y = 0$. Step 2: Set $\frac{4}{x} - 2 = 0$, so $\frac{4}{x} = 2$, giving $x = 2$. Step 3: The point is $(2, 0)$.

Question 29 · Answer A

Step 1: The function $y = \frac{5}{x} + 3$ is undefined when $x = 0$, so the vertical asymptote is $x = 0$. Step 2: As $x \to \pm\infty$, $\frac{5}{x} \to 0$, so $y \to 3$. The horizontal asymptote is $y = 3$.

Question 30 · Answer A

Step 1: We want to solve $\frac{3}{x} - x - 1 = 0$, i.e. $\frac{3}{x} - 1 = x$. Step 2: The left side is the graph already drawn: $y = \frac{3}{x} - 1$. Step 3: Setting this equal to $x$ means we need the line $y = x$.

Question 31 · Answer A

Step 1: Curved surface of hemisphere $= 2\pi r^2 = 2\pi(16) = 32\pi$. Step 2: Curved surface of cylinder $= 2\pi rh = 2\pi(4)(7) = 56\pi$. Step 3: Flat base of cylinder $= \pi r^2 = 16\pi$. Step 4: Total $= 32\pi + 56\pi + 16\pi = 104\pi$ cm$^2$. Note: The flat circle where hemisphere meets cylinder is NOT included (it is internal).

Question 32 · Answer A

Step 1: $64^{\frac{2}{3}} = (\sqrt[3]{64})^2$. Step 2: $\sqrt[3]{64} = 4$. Step 3: $4^2 = 16$.

Question 33 · Answer A

Step 1: $9^{-\frac{3}{2}} = \frac{1}{9^{\frac{3}{2}}}$. Step 2: $9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$. Step 3: So $9^{-\frac{3}{2}} = \frac{1}{27}$.

Question 34 · Answer A

Step 1: Multiply numerator and denominator by $\sqrt{2}$: $\frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2}$. Step 2: Simplify: $\frac{8\sqrt{2}}{2} = 4\sqrt{2}$.

Question 35 · Answer A

Step 1: Expand: $(3 + \sqrt{5})(2 - \sqrt{5}) = 6 - 3\sqrt{5} + 2\sqrt{5} - (\sqrt{5})^2$. Step 2: Simplify: $= 6 - 3\sqrt{5} + 2\sqrt{5} - 5 = 1 - \sqrt{5}$. Step 3: So $a = 1$ and $b = -1$.

Question 36 · Answer A

Step 1: Multiply numerators and denominators: $\frac{4x \times 3y}{9 \times 2x} = \frac{12xy}{18x}$. Step 2: Cancel $x$: $\frac{12y}{18}$. Step 3: Simplify: $\frac{12y}{18} = \frac{2y}{3}$.

Question 37 · Answer A

Step 1: Find the LCD of 3 and 5: LCD = 15. Step 2: $\frac{m}{3} = \frac{5m}{15}$ and $\frac{2n}{5} = \frac{6n}{15}$. Step 3: Add: $\frac{5m + 6n}{15}$.

Question 38 · Answer A

Step 1: Common denominator is $(x+3)(x-1)$. Step 2: Numerator: $5(x-1) - 2(x+3) = 5x - 5 - 2x - 6 = 3x - 11$. Step 3: Result: $\frac{3x - 11}{(x+3)(x-1)}$.

Question 39 · Answer A

Step 1: $y \propto \frac{1}{\sqrt{x}}$, so $y = \frac{k}{\sqrt{x}}$. Step 2: When $x = 4$, $y = 6$: $6 = \frac{k}{\sqrt{4}} = \frac{k}{2}$, so $k = 12$. Step 3: When $x = 16$: $y = \frac{12}{\sqrt{16}} = \frac{12}{4} = 3$.

Question 40 · Answer A

Step 1: $y = \frac{k}{\sqrt{x}}$. If $x$ is multiplied by 9, new $y = \frac{k}{\sqrt{9x}} = \frac{k}{3\sqrt{x}} = \frac{1}{3} \cdot \frac{k}{\sqrt{x}}$. Step 2: So $y$ is multiplied by $\frac{1}{3}$.

Question 41 · Answer A

Step 1: Set $y = 0$: $5x - x^3 = 0$. Step 2: Factorise: $x(5 - x^2) = 0$. Step 3: So $x = 0$ or $5 - x^2 = 0$, giving $x^2 = 5$, so $x = \pm\sqrt{5}$.

Question 42 · Answer A

Step 1: Differentiate $5x$: $\frac{d}{dx}(5x) = 5$. Step 2: Differentiate $-x^3$: $\frac{d}{dx}(-x^3) = -3x^2$. Step 3: $\frac{dy}{dx} = 5 - 3x^2$.

Question 43 · Answer A

Step 1: $\frac{dy}{dx} = 5 - 3x^2 = 0$. Step 2: $3x^2 = 5$, so $x^2 = \frac{5}{3}$, giving $x = \pm\sqrt{\frac{5}{3}}$. Step 3: Substitute back into $y = 5x - x^3$ to find the $y$-coordinates. Step 4: For $x = \sqrt{\frac{5}{3}}$: $y = 5\sqrt{\frac{5}{3}} - \left(\sqrt{\frac{5}{3}}\right)^3 = \frac{10\sqrt{\frac{5}{3}}}{3}$.

Question 44 · Answer A

Step 1: From the exact values table or the equilateral triangle: $\cos 60° = \frac{1}{2}$.

Question 45 · Answer A

Step 1: $2\cos x = \sqrt{3}$, so $\cos x = \frac{\sqrt{3}}{2}$. Step 2: The principal value is $x = 30°$. Step 3: Cosine is positive in the 1st and 4th quadrants, so $x = 30°$ and $x = 360° - 30° = 330°$.

Question 46 · Answer A

Step 1: $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$. Step 2: $P$ is the midpoint of $AB$, so $\overrightarrow{AP} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$. Step 3: $\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$.

Question 47 · Answer A

Step 1: Set equal: $3x + 1 = x^2 + x - 5$. Step 2: Rearrange: $x^2 - 2x - 6 = 0$. Step 3: Factorise: $(x-3)(x+2) = 0$. Step 4: $x = 3$ or $x = -2$.