IGCSE Mathematics 0580/41 October/November 2025 Past Paper with Answers

Question 1

A quadrilateral has rotational symmetry of order 2 and two diagonals that are the only lines of symmetry. What is the geometrical name of this quadrilateral?

1. Rhombus
2. Rectangle
3. Parallelogram
4. Square

Question 2

Solve $11 - 2x = 4$.

1. $x = 3.5$
2. $x = -3.5$
3. $x = 7.5$
4. $x = 7$

Question 3

The diagram shows two parallel lines cut by a transversal. One angle is $55°$ and another is $x°$. Find the value of $x$.

1. $125$
2. $55$
3. $135$
4. $35$

Question 4 · Answer A

Step 1: Add the hours: $22\text{:}16 + 5\text{ hours} = 03\text{:}16$ (next day, since $22 + 5 = 27$, and $27 - 24 = 3$). Step 2: Add the minutes: $03\text{:}16 + 52\text{ minutes} = 04\text{:}08$ (since $16 + 52 = 68$ minutes $= 1$ hour $8$ minutes).

Question 5 · Answer A

Step 1: Check the triangle inequality: the sum of any two sides must be greater than the third. Step 2: $8 + 6 = 14 > 11$ ✓, $11 + 6 = 17 > 8$ ✓, $11 + 8 = 19 > 6$ ✓. Step 3: The triangle can be constructed.

Question 6 · Answer A

Step 1: Angles in a triangle sum to $180°$. Step 2: Angle ACB = $180° - 52° - 45° = 83°$.

Question 7 · Answer A

Step 1: $AB = 8$ cm on the drawing. Step 2: Actual distance $= 8 \times 10\,000 = 80\,000$ cm $= 800$ m $= 0.8$ km.

Question 8 · Answer A

Step 1: If $B$ is due east of $A$, then $A$ is due west of $B$. Step 2: The bearing of due west is $270°$.

Question 9 · Answer A

Step 1: Multiply $g$ by each term inside the bracket: $g \times 3 = 3g$ $g \times (-2g) = -2g^{2}$ Step 2: Result: $3g - 2g^{2}$.

Question 10 · Answer A

Step 1: An open circle at $-3$ means $x > -3$ (not equal to $-3$). Step 2: A closed circle at $2$ means $x \leq 2$ (can equal $2$). Step 3: Combined: $-3 < x \leq 2$.

Question 11 · Answer A

Step 1: Divide all parts by 2: $-2 < x \leq 4$. Step 2: List integers satisfying this: $x = -1, 0, 1, 2, 3, 4$. Note: $x = -2$ is not included because the inequality is strict ($<$, not $\leq$). $x = 4$ is included because the inequality is inclusive ($\leq$).

Question 12 · Answer A

Step 1: $3.5^{2} = 12.25$. Step 2: $2.2^{3} = 10.648$. Step 3: $12.25 - 10.648 = 1.602$. Step 4: $1.602^{0.25} = 1.125\ldots \approx 1.13$.

Question 13 · Answer A

Step 1: Convert m/s to km/h by multiplying by $\frac{3600}{1000} = 3.6$. Step 2: $9.5 \times 3.6 = 34.2$ km/h.

Question 14 · Answer A

Step 1: $B = A + \vec{AB}$. Step 2: $B = (2 + 4,\; 1 + 2) = (6, 3)$.

Question 15 · Answer A

Step 1: Calculate the discount: $15\%$ of $\$60 = 0.15 \times 60 = \$9$. Step 2: Sale price $= \$60 - \$9 = \$51$. Alternatively: $60 \times 0.85 = \$51$.

Question 16 · Answer A

Step 1: The sale price is 85% of the original price. Step 2: Original price $= \frac{58.14}{0.85} = \$68.40$.

Question 17 · Answer A

Step 1: Students who do not like rugby are those in $F$ only and those outside both sets. Step 2: $33 + 4 = 37$.

Question 18 · Answer A

Step 1: 'Like rugby' means in set $R$. Step 2: 'But not football' means not in set $F$, i.e., in $F'$. Step 3: The intersection gives $F' \cap R$.

Question 19 · Answer A

Step 1: The vertical difference between $T$ and $E$ is $12.6 - 1.5 = 11.1$ m. Step 2: The horizontal distance is $33$ m. Step 3: $\tan(\theta) = \frac{11.1}{33}$. Step 4: $\theta = \tan^{-1}\left(\frac{11.1}{33}\right) = 18.6°$.

Question 20 · Answer A

Step 1: The sphere touches all faces of the cube, so the diameter equals the side length: $d = 8$ cm, thus $r = 4$ cm. Step 2: $V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi(4)^{3} = \frac{4}{3}\pi \times 64 = \frac{256}{3}\pi$ cm$^{3}$.

Question 21 · Answer A

Step 1: Volume of cube $= 8^{3} = 512$ cm$^{3}$. Step 2: Volume of sphere $= \frac{256}{3}\pi \approx 268.08$ cm$^{3}$. Step 3: Volume not occupied $= 512 - 268.08 = 243.92$ cm$^{3}$. Step 4: Percentage $= \frac{243.92}{512} \times 100 \approx 47.6\%$.

Question 22 · Answer A

Step 1: Mass $=$ density $\times$ volume $= 7.86 \times \frac{256}{3}\pi \approx 7.86 \times 268.08 \approx 2107$ g. Step 2: Convert to kg: $2107 \div 1000 = 2.11$ kg.

Question 23 · Answer A

Step 1: Volume of cylinder $=$ volume of sphere: $\pi(3.1)^{2}h = \frac{256}{3}\pi$. Step 2: $h = \frac{256}{3 \times 3.1^{2}} = \frac{256}{3 \times 9.61} = \frac{256}{28.83} = 8.880\ldots$ cm. Step 3: Total surface area $= 2\pi r^{2} + 2\pi r h = 2\pi(3.1)^{2} + 2\pi(3.1)(8.880)$. Step 4: $= 2\pi(9.61) + 2\pi(27.528) = 60.38 + 172.97 \approx 233$ cm$^{2}$.

Question 24 · Answer A

Step 1: Angle at centre $= 2 \times$ angle at circumference: $\angle AOB = 2 \times 67 = 134°$. Step 2: Tangent meets radius at $90°$: $\angle OAT = \angle OBT = 90°$. Step 3: In quadrilateral $OATB$: $\angle ATB = 360 - 90 - 90 - 134 = 46°$. Alternatively: $\angle ATB = 180 - 2 \times 67 = 46°$.

Question 25 · Answer A

Step 1: Mid-values: $7.5$, $11$, $16$. Step 2: $\Sigma fx = 3 \times 7.5 + 24 \times 11 + 23 \times 16 = 22.5 + 264 + 368 = 654.5$. Step 3: Mean $= \frac{654.5}{50} = 13.09$ cm.

Question 26 · Answer A

Step 1: Gradient $m = \frac{4 - 0}{0 - 2} = \frac{4}{-2} = -2$. Step 2: The $y$-intercept is $c = 4$ (since the line passes through $(0, 4)$). Step 3: $y = -2x + 4$.

Question 27 · Answer A

Step 1: Distance = area under the speed-time graph. Step 2: Triangle (0 to 2 s): $\frac{1}{2} \times 2 \times 8 = 8$ m. Step 3: Rectangle (2 to 10 s): $8 \times 8 = 64$ m. Step 4: Total $= 8 + 64 = 72$ m.

Question 28 · Answer A

Step 1: Distance remaining $= 100 - 72 = 28$ m. Step 2: Time for remaining distance $= \frac{28}{8} = 3.5$ s. Step 3: Total time $= 10 + 3.5 = 13.5$ s.

Question 29 · Answer A

Step 1: After $n$ days: mass $= 20 \times 0.9^{n}$. Step 2: Solve $20 \times 0.9^{n} < 1$, so $0.9^{n} < 0.05$. Step 3: $n > \frac{\ln(0.05)}{\ln(0.9)} = \frac{-2.996}{-0.1054} = 28.43\ldots$ Step 4: So $n = 29$ whole days.

Question 30 · Answer A

Step 1: Each term is half the previous term (common ratio $= \frac{1}{2}$). Step 2: Next term $= 3 \times \frac{1}{2} = 1.5$.

Question 31 · Answer A

Step 1: The $n$th term of a geometric sequence is $ar^{n-1}$ where $a = 48$ and $r = \frac{1}{2}$. Step 2: $T_{n} = 48 \times \left(\frac{1}{2}\right)^{n-1} = 48 \times \frac{1}{2^{n-1}} = \frac{48}{2^{n-1}} = \frac{96}{2^{n}} = 96 \times \left(\frac{1}{2}\right)^{n}$.

Question 32 · Answer A

Step 1: Area $= \frac{1}{2}ab\sin C = \frac{1}{2} \times 8 \times 9 \times \sin 50°$. Step 2: $= \frac{1}{2} \times 72 \times 0.766 = 27.6$ cm$^{2}$.

Question 33 · Answer A

Step 1: $200\left(1 + \frac{r}{100}\right)^{25} = 301.10$. Step 2: $\left(1 + \frac{r}{100}\right)^{25} = \frac{301.10}{200} = 1.5055$. Step 3: $1 + \frac{r}{100} = 1.5055^{1/25} = 1.01649\ldots$ Step 4: $\frac{r}{100} = 0.01649\ldots$, so $r = 1.65$.

Question 34 · Answer A

Step 1: $y \propto \frac{1}{\sqrt{x+1}}$, so $y = \frac{k}{\sqrt{x+1}}$. Step 2: When $x = 8$: $3 = \frac{k}{\sqrt{9}} = \frac{k}{3}$, so $k = 9$. Step 3: $y = \frac{9}{\sqrt{x+1}}$.

Question 35 · Answer A

Step 1: Time $=$ distance $\div$ speed $= \frac{10}{x}$ hours.

Question 36 · Answer A

Step 1: $\frac{10}{x} + \frac{5}{x+4} = \frac{7}{2}$. Step 2: Multiply through by $2x(x+4)$: $20(x+4) + 10x = 7x(x+4)$. Step 3: $20x + 80 + 10x = 7x^{2} + 28x$. Step 4: $30x + 80 = 7x^{2} + 28x$. Step 5: $0 = 7x^{2} - 2x - 80$.

Question 37 · Answer A

Step 1: Using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(7)(-80)}}{2(7)}$. Step 2: $x = \frac{2 \pm \sqrt{4 + 2240}}{14} = \frac{2 \pm \sqrt{2244}}{14}$. Step 3: $\sqrt{2244} = 47.37\ldots$ Step 4: $x = \frac{2 + 47.37}{14} = 3.53$ or $x = \frac{2 - 47.37}{14} = -3.24$.

Question 38 · Answer A

Step 1: Time at 80 km/h $= \frac{120}{80} = 1.5$ hours. Step 2: Time at 60 km/h $= \frac{120}{60} = 2$ hours. Step 3: Difference $= 2 - 1.5 = 0.5$ hours $= 30$ minutes.

Question 39 · Answer A

Step 1: Frequency $=$ frequency density $\times$ class width $= 20 \times 2 = 40$.

Question 40 · Answer A

Step 1: Frequency density $= \frac{\text{frequency}}{\text{class width}} = \frac{102}{5 - 2} = \frac{102}{3} = 34$. Step 2: Draw a bar from $m = 2$ to $m = 5$ with height $34$.

Question 41 · Answer A

Step 1: Set equal: $2x^{2} - 3x - 7 = 2x - 7$. Step 2: Simplify: $2x^{2} - 5x = 0$. Step 3: Factorise: $x(2x - 5) = 0$. Step 4: $x = 0$ or $x = 2.5$. Step 5: When $x = 0$: $y = 2(0) - 7 = -7$. When $x = 2.5$: $y = 2(2.5) - 7 = -2$.

Question 42 · Answer A

Step 1: $G$ is directly above a point on the base. Since $ABFE$ is the base, $G$ is directly above a point that is $7$ cm from $F$ along $FE$ (since $DC = 7$) and $5$ cm above the base (since $AD = 5$). Step 2: The projection of $G$ onto the base is the point $G'$ where $G'$ is at position $(7, 0, 15)$ relative to $A$ at origin. So $AG' = \sqrt{7^{2} + 15^{2}} = \sqrt{49 + 225} = \sqrt{274}$. Step 3: The vertical height from $G'$ to $G$ is $5$ cm. Step 4: $\tan(\theta) = \frac{5}{\sqrt{274}}$. Step 5: $\theta = \tan^{-1}\left(\frac{5}{\sqrt{274}}\right) = \tan^{-1}(0.3021) = 16.8°$.

Question 43 · Answer A

Step 1: $7^{x-4} = 1$. Step 2: Since $7^{0} = 1$, we need $x - 4 = 0$. Step 3: $x = 4$.

Question 44 · Answer A

Step 1: If $f^{-1}(x) = 1$, then $x = f(1)$. Step 2: $f(1) = 7^{1-4} = 7^{-3} = \frac{1}{343}$.

Question 45 · Answer A

Step 1: Use the cosine rule: $\cos y = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$. Step 2: $\cos y = \frac{10^2 + 14^2 - 13^2}{2 \times 10 \times 14} = \frac{127}{280}$. Step 3: $y = \cos^{-1}(\frac{127}{280}) = 63.0°$.

Question 46 · Answer A

Step 1: Use the cosine rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)$. Step 2: $PR^2 = 11^2 + 8^2 - 2(11)(8)\cos 65° = 121 + 64 - 176 \times 0.4226 = 185 - 74.4 = 110.6$. Step 3: $PR = \sqrt{110.6} = 10.5$ cm.