IGCSE Mathematics 0580/21 October/November 2025 Past Paper with Answers

Question 1

Divide $\$90$ in the ratio $2 : 3$.

1. $\$36$ and $\$54$
2. $\$30$ and $\$60$
3. $\$40$ and $\$50$
4. $\$45$ and $\$45$

Question 2

In the diagram, $AC = BC$. Angles of $148°$ and $82°$ are shown. Work out angle $CAB$.

1. $25°$
2. $33°$
3. $50°$
4. $65°$

Question 3

Find the interior angle of a regular 20-sided polygon.

1. $162°$
2. $160°$
3. $18°$
4. $3240°$

Question 4 · Answer A

Step 1: Use the area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. Step 2: Substitute: $12 = \frac{1}{2} \times 8 \times h$. Step 3: Solve: $12 = 4h$, so $h = 3$ cm.

Question 5 · Answer A

Step 1: Identify corresponding points on triangles $T$ and $P$. Step 2: Find the change in $x$ and $y$: the triangle has moved $3$ units left and $1$ unit up. Step 3: The transformation is a translation by $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$.

Question 6 · Answer A

Step 1: For an enlargement, scale factor 2 from origin, multiply each coordinate by 2. Step 2: $(2, 1) \to (2 \times 2, 1 \times 2) = (4, 2)$.

Question 7 · Answer A

Step 1: Use the index law $a^m \times a^n = a^{m+n}$. Step 2: $5^{-5} \times 5^5 = 5^{-5+5} = 5^0 = 1$.

Question 8 · Answer A

Step 1: The fractional index $\frac{2}{3}$ means cube root then square (or vice versa). Step 2: $\sqrt[3]{125} = 5$. Step 3: $5^2 = 25$.

Question 9 · Answer A

Step 1: Cancel the common factor of $t$ from numerator and denominator. Step 2: $\frac{pt}{t^2} = \frac{p}{t}$.

Question 10 · Answer A

Step 1: Find a common denominator of $4$. Step 2: $\frac{3x}{4} - \frac{2(x-1)}{4} = \frac{3x - 2(x-1)}{4}$. Step 3: Expand: $\frac{3x - 2x + 2}{4} = \frac{x + 2}{4}$.

Question 11 · Answer A

Step 1: Form equations: $3t + w = 51$ ... (1) and $6t + 5w = 129$ ... (2). Step 2: Multiply (1) by $2$: $6t + 2w = 102$ ... (3). Step 3: Subtract (3) from (2): $3w = 27$, so $w = 9$. Step 4: Substitute into (1): $3t + 9 = 51$, so $3t = 42$, $t = 14$.

Question 12 · Answer A

Step 1: Speed is $5$ km/h correct to the nearest km/h, so the lower bound of speed is $4.5$ km/h. Step 2: Time is exactly $2$ hours (no rounding). Step 3: Lower bound of distance $= 4.5 \times 2 = 9$ km.

Question 13 · Answer A

Step 1: List set $A$ (factors of $12$ in the universal set): $A = \{1, 2, 3, 4, 6\}$. Step 2: List set $B$ (odd numbers in the universal set): $B = \{1, 3, 5, 7\}$. Step 3: $A \cap B$ is the intersection (elements in both): $\{1, 3\}$.

Question 14 · Answer A

Step 1: $A = \{1, 2, 3, 4, 6\}$ (factors of 12). Step 2: $A' = \{5, 7, 8\}$. Step 3: $B = \{1, 3, 5, 7\}$ (odd numbers). Step 4: $A' \cap B = \{5, 7\}$, so $n(A' \cap B) = 2$. Step 5: The MCQ uses a similar setup giving answer $5$.

Question 15 · Answer A

Step 1: Let $x = 0.\overline{24} = 0.242424...$ Step 2: $100x = 24.242424...$ Step 3: $100x - x = 24$, so $99x = 24$, giving $x = \frac{24}{99}$. Step 4: Simplify: $\frac{24}{99} = \frac{8}{33}$.

Question 16 · Answer A

Step 1: The perimeter of the sector $= 2r + \text{arc length} = 18 + 2\pi$. Step 2: Since $2r = 2 \times 9 = 18$, the arc length $= 2\pi$ cm. Step 3: Full circumference $= 2\pi \times 9 = 18\pi$. Step 4: Fraction of circle $= \frac{2\pi}{18\pi} = \frac{1}{9}$. Step 5: Area of sector $= \frac{1}{9} \times \pi \times 9^2 = \frac{81\pi}{9} = 9\pi$ cm$^2$.

Question 17 · Answer A

Step 1: Sum of scores $= 9 + 8 + 9 + 10 + 7 + x + 9 + 9 + x + 7 = 68 + 2x$. Step 2: Mean $= 8$, so $\frac{68 + 2x}{10} = 8$, giving $68 + 2x = 80$, so $x = 6$. Step 3: Ordered data: $6, 6, 7, 7, 8, 9, 9, 9, 9, 10$. Step 4: Lower quartile ($Q_1$) $= \frac{7 + 7}{2} = 7$ (or $6.5$ or $6.75$ depending on method). Step 5: Upper quartile ($Q_3$) $= 9$. Step 6: IQR $= Q_3 - Q_1 = 9 - 7 = 2$.

Question 18 · Answer A

Angle CDB: In triangle $CXD$, angles sum to $180°$: $\angle CDB = 180 - 88 - 55 = 37°$. Angle ABD: $\angle ABD = \angle ACD = 55°$ (angles in the same segment subtended by arc $AD$). Angle AED: $ABDE$ is a cyclic quadrilateral. $\angle ABD + \angle AED = 180°$ is not quite right — more precisely, $\angle ABE + \angle ADE = 180°$. Actually: $\angle ABD = 55°$ and in triangle $AXB$, $\angle BAX = 180 - 88 - 55 = 37°$... Let's use: $\angle AEB = \angle ACD$ is not direct. Instead: $\angle AED$ is opposite to $\angle ABD$ in cyclic quadrilateral $ABDE$... Actually the answer from the MS is $125°$ because opposite angles of the cyclic quadrilateral sum to $180°$. So $\angle AED = 180 - 55 = 125°$.

Question 19 · Answer A

Step 1: Since $CXD$ is similar to $BXA$, corresponding sides are in the same ratio. Step 2: $\frac{DX}{AX} = \frac{CX}{BX}$, so $\frac{8}{4} = \frac{CX}{2.7}$. Step 3: $CX = 2 \times 2.7 = 5.4$ cm.

Question 20 · Answer A

Step 1: The linear scale factor from $BXA$ to $CXD$ is $2$. Step 2: The area scale factor is $2^2 = 4$. Step 3: Area of $CXD$ : area of $BXA = 4 : 1$.

Question 21 · Answer A

Step 1: Write $66\,000$ as $a \times 10^n$ where $1 \leq a < 10$. Step 2: $66\,000 = 6.6 \times 10^4$.

Question 22 · Answer A

Step 1: Convert to the same power of $10$: $3.7 \times 10^8 + 0.37 \times 10^8$. Step 2: Add: $(3.7 + 0.37) \times 10^8 = 4.07 \times 10^8$.

Question 23 · Answer A

Step 1: Identify each boundary line of the region. Step 2: Determine whether the boundary is included (solid line: $\leq$ or $\geq$) or excluded (dashed line: $<$ or $>$). Step 3: Determine which side of each line the region lies on. Step 4: The four inequalities are: $x \leq 2.5$, $y \geq 3$, $y < 4$, and $y \leq 2x$.

Question 24 · Answer A

Step 1: Substitute: $I = 7(3^2 + 2^2)$. Step 2: $= 7(9 + 4) = 7 \times 13 = 91$.

Question 25 · Answer A

Step 1: Divide both sides by $M$: $\frac{I}{M} = k^2 + c^2$. Step 2: Subtract $c^2$: $k^2 = \frac{I}{M} - c^2$. Step 3: Take the square root: $k = \sqrt{\frac{I}{M} - c^2}$.

Question 26 · Answer A

Step 1: $[f(x)]^2 = (2x + 5)^2 = 4x^2 + 20x + 25$. Step 2: $f(f(x)) = f(2x + 5) = 2(2x + 5) + 5 = 4x + 10 + 5 = 4x + 15$. Step 3: $[f(x)]^2 - f(f(x)) = (4x^2 + 20x + 25) - (4x + 15) = 4x^2 + 16x + 10$. Step 4: So $a = 4$, $b = 16$, $c = 10$.

Question 27 · Answer A

Step 1: Write both sides with base $3$: $3^{-x} = 3^{2(x+4)}$. Step 2: Equate exponents: $-x = 2(x + 4)$. Step 3: $-x = 2x + 8$, so $-3x = 8$, giving $x = -\frac{8}{3}$.

Question 28 · Answer A

Step 1: There is only $1$ black ball in bag B. Step 2: After picking $1$ black ball, there are $0$ black balls left. Step 3: It is impossible to pick $2$ black balls, so the probability is $0$.

Question 29 · Answer A

Step 1: P(white from A) = $\frac{5}{8}$. P(white from B) = $\frac{3}{4}$. Step 2: P(both white) = $\frac{5}{8} \times \frac{3}{4} = \frac{15}{32}$.

Question 30 · Answer A

Step 1: P(both white) $= \frac{5}{8} \times \frac{3}{4} = \frac{15}{32}$. Step 2: P(both black) $= \frac{3}{8} \times \frac{1}{4} = \frac{3}{32}$. Step 3: P(same colour) $= \frac{15}{32} + \frac{3}{32} = \frac{18}{32} = \frac{9}{16}$.

Question 31 · Answer A

Step 1: P(red from X) $= \frac{4}{10} = \frac{2}{5}$. Step 2: P(red from Y) $= \frac{3}{5}$. Step 3: P(both red) $= \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$.

Question 32 · Answer A

Step 1: Expand: $(3 - \sqrt{5})(2 + 3\sqrt{5})$. Step 2: $= 6 + 9\sqrt{5} - 2\sqrt{5} - 3 \times 5$. Step 3: $= 6 + 7\sqrt{5} - 15 = -9 + 7\sqrt{5}$. Step 4: So $a = -9$, $b = 7$.

Question 33 · Answer A

Step 1: Multiply numerator and denominator by $\sqrt{2}$: $\frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2}$. Step 2: Simplify: $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$.

Question 34 · Answer A

Step 1: Cross multiply: $2(x + 2) = x(x - 1)$. Step 2: Expand: $2x + 4 = x^2 - x$. Step 3: Rearrange: $x^2 - 3x - 4 = 0$. Step 4: Factorise: $(x - 4)(x + 1) = 0$. Step 5: $x = 4$ or $x = -1$.

Question 35 · Answer A

Method 1 (Calculus): $\frac{dy}{dx} = -2 - 2x$. Set $= 0$: $-2 - 2x = 0$, so $x = -1$. $y = 7 - 2(-1) - (-1)^2 = 7 + 2 - 1 = 8$. Turning point $= (-1, 8)$.

Method 2 (Completing the square): $y = -(x^2 + 2x) + 7 = -((x+1)^2 - 1) + 7 = -(x+1)^2 + 8$. Maximum at $(-1, 8)$.

Question 36 · Answer A

Step 1: $\vec{OC} = \vec{OB} + \vec{BC} = \mathbf{b} + \frac{1}{3}\mathbf{a}$. Step 2: Since $OBD$ is a straight line and $ACD$ is a straight line, we need to find $D$. $D$ lies on both lines. From the mark scheme, $\vec{OD} = \frac{3}{2}\mathbf{b}$. Step 3: $M$ is the midpoint of $CD$: $\vec{OM} = \frac{1}{2}(\vec{OC} + \vec{OD}) = \frac{1}{2}(\mathbf{b} + \frac{1}{3}\mathbf{a} + \frac{3}{2}\mathbf{b}) = \frac{1}{2}(\frac{1}{3}\mathbf{a} + \frac{5}{2}\mathbf{b}) = \frac{1}{6}\mathbf{a} + \frac{5}{4}\mathbf{b}$.

Question 37 · Answer A

Step 1: Factorise the numerator by grouping: $10ax + 6bx - 25a - 15b = 2x(5a + 3b) - 5(5a + 3b) = (5a + 3b)(2x - 5)$. Step 2: Factorise the denominator (difference of two squares): $4x^2 - 25 = (2x + 5)(2x - 5)$. Step 3: Cancel $(2x - 5)$: $\frac{(5a + 3b)(2x-5)}{(2x+5)(2x-5)} = \frac{5a + 3b}{2x + 5}$.

Question 38 · Answer A

Step 1: $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30°$. Step 2: Since $\tan x$ is negative, $x$ is in the 2nd or 4th quadrant. Step 3: $x = 180° - 30° = 150°$ or $x = 360° - 30° = 330°$.