IGCSE Mathematics 0580/42 October/November 2025 Past Paper with Answers

Question 1

Write the ratio 60 grams : 3 kilograms in the form $1 : n$.

1. $1 : 50$
2. $1 : 5$
3. $1 : 500$
4. $1 : 20$

Question 2

Solve the equation $8x - 17 = 27$.

1. $x = 5.5$
2. $x = 5$
3. $x = 1.25$
4. $x = 6$

Question 3

Write down the order of rotational symmetry of a regular decagon.

1. $10$
2. $5$
3. $8$
4. $12$

Question 4 · Answer A

Step 1: Convert $8$ hours to minutes: $8 \times 60 = 480$ minutes. Step 2: Find the number of $20$-minute intervals: $480 \div 20 = 24$. Step 3: Multiply by the rate: $24 \times 9 = 216$ cards.

Question 5 · Answer A

Step 1: Calculate the profit: $50 - 12 = 38$ cents. Step 2: Percentage profit $= \frac{\text{profit}}{\text{cost}} \times 100 = \frac{38}{12} \times 100 = 316\frac{2}{3}\%$.

Question 6 · Answer A

Step 1: Convert 953 Yuan to dollars: $953 \times 0.152 = 144.856$ dollars. Step 2: Find the difference: $144.856 - 141 = 3.856$ dollars. Step 3: Round to the nearest cent: $\$3.86$.

Question 7 · Answer A

Step 1: Sum of the 8 known values: $1 + 4 + 10 + 12 + 19 + 21 + 25 + 25 = 117$. Step 2: Mean of $10$ patients is $16$, so total $= 16 \times 10 = 160$. Therefore $P + Q = 160 - 117 = 43$. Step 3: The range is $26$. The current minimum is $1$ and maximum is $25$. For the range to be $26$, the maximum must be $27$ (since $27 - 1 = 26$). So $P = 27$. Step 4: $Q = 43 - 27 = 16$. Step 5: Check: $P > Q$ ($27 > 16$) and the range $= 27 - 1 = 26$. Confirmed.

Question 8 · Answer A

Step 1: Circumference of circle $= 2\pi r = 2\pi(4) = 8\pi$. Step 2: Perimeter of parallelogram $= 2(x + 6.51)$. Step 3: Set equal: $2(x + 6.51) = 8\pi$. Step 4: $x + 6.51 = 4\pi = 12.566...$ Step 5: $x = 12.566... - 6.51 = 6.056...$ Step 6: $x = 6.06$ (correct to 2 d.p.).

Question 9 · Answer A

Step 1: Area of parallelogram $= ab\sin C = 6.06 \times 6.51 \times \sin 36°$. Step 2: $6.06 \times 6.51 = 39.4506$. Step 3: $39.4506 \times \sin 36° = 39.4506 \times 0.5878 = 23.19$ m$^2$ (approx). Step 4: Total cost $= 23.19 \times 18 = \$417.42$.

Question 10 · Answer A

Step 1: A translation by $\begin{pmatrix} -1 \\ -7 \end{pmatrix}$ moves every point $1$ unit to the left and $7$ units down. Step 2: Subtract $1$ from each $x$-coordinate and $7$ from each $y$-coordinate of shape A's vertices. Step 3: Plot and connect the translated vertices.

Question 11 · Answer A

Step 1: When reflecting in the line $y = x$, the $x$ and $y$ coordinates are swapped. Step 2: $(3, 2) \to (2, 3)$.

Question 12 · Answer A

Step 1: Compare the orientations of shapes A and B. The shape is rotated (not just translated or reflected). Step 2: Determine the angle of rotation by comparing corresponding sides. The shape is rotated $180°$. Step 3: Find the centre of rotation: the midpoint between corresponding vertices of A and B gives the centre $(0.5, 3.5)$.

Question 13 · Answer A

Step 1: Distance = area under the speed-time graph. Step 2: Trapezium (0 to 15 min): $\frac{1}{2}(80 + 60) \times 15 = 1050$. Step 3: Triangle (15 to 25 min): $\frac{1}{2} \times 60 \times 10 = 300$. Step 4: Total area $= 1050 + 300 = 1350$. But time is in minutes and speed in km/h. Step 5: Convert: $1350 \div 60 = 22.5$ km.

Question 14 · Answer A

Step 1: Common difference $d = 9 - 17 = -8$. Step 2: $n$th term $= a + (n-1)d = 17 + (n-1)(-8) = 17 - 8n + 8 = 25 - 8n$. Step 3: Check: $n = 1$: $25 - 8 = 17$ \checkmark. $n = 2$: $25 - 16 = 9$ \checkmark.

Question 15 · Answer A

Step 1: First differences: $12-3=9$, $27-12=15$, $48-27=21$, $75-48=27$. Step 2: Second differences: $15-9=6$, $21-15=6$, $27-21=6$. Constant, so quadratic. Step 3: Coefficient of $n^2 = \frac{6}{2} = 3$. Try $3n^2$: $3(1)^2=3$, $3(2)^2=12$, $3(3)^2=27$. All match. Step 4: $n$th term $= 3n^2$.

Question 16 · Answer A

Step 1: For $x = -1.5$: $y = (-1.5)^3 - 2(-1.5) + 3 = -3.375 + 3 + 3 = 2.625 \approx 2.63$. Step 2: For $x = 1.5$: $y = (1.5)^3 - 2(1.5) + 3 = 3.375 - 3 + 3 = 3.375 \approx 3.38$.

Question 17 · Answer A

Step 1: Substitute $x = 0.5$: $y = (0.5)^3 - 2(0.5) + 3$. Step 2: $= 0.125 - 1 + 3 = 2.125$.

Question 18 · Answer A

Step 1: Rearrange: $x^3 - 2.5x + 1 = 0$ becomes $x^3 - 2x + 3 = 0.5x + 2$. Step 2: The graph already shows $y = x^3 - 2x + 3$. Draw the line $y = 0.5x + 2$. Step 3: The line $y = 0.5x + 2$ passes through $(0, 2)$ with gradient $0.5$. Step 4: Read off the $x$-coordinates of the intersection points: $x \approx -1.8$, $x \approx 0.4$, $x \approx 1.4$.

Question 19 · Answer A

Step 1: The radius OC meets the tangent DE at C, so angle $OCE = 90°$ (tangent is perpendicular to radius). Step 2: $x = \text{angle } OCB = \text{angle } OCE - \text{angle } BCE = 90° - 65° = 25°$. Reason: The angle between a tangent and a radius at the point of contact is $90°$.

Question 20 · Answer A

Step 1: By the alternate segment theorem, the angle between the tangent CE and chord CB equals the angle in the alternate segment. Step 2: Angle $BCE = 65°$ equals angle $BAC$ (alternate segment theorem). Step 3: Therefore $y = 65°$.

Question 21 · Answer A

Step 1: Find a common denominator. LCM of $2m$ and $8m$ is $8m$. Step 2: $\dfrac{7}{2m} = \dfrac{28}{8m}$. Step 3: $\dfrac{28}{8m} + \dfrac{3}{8m} = \dfrac{31}{8m}$.

Question 22 · Answer A

Step 1: Use the compound interest formula: $A = P(1 + r)^n$. Step 2: $A = 24000 \times 1.032^4$. Step 3: $1.032^4 = 1.13427...$ Step 4: $A = 24000 \times 1.13427... = 27222.62$.

Question 23 · Answer A

Step 1: Selling price $= 1.34 \times$ cost price. So $1.34x = 40870$. Step 2: $x = 40870 \div 1.34 = 30500$. Step 3: Profit $= 40870 - 30500 = \$10370$.

Question 24 · Answer A

Step 1: A $23\%$ decrease means the car retains $100\% - 23\% = 77\% = 0.77$ of its value each year. Step 2: After $n$ years: $V = 32500 \times 0.77^n$.

Question 25 · Answer A

Step 1: In triangle $BDC$, use the cosine rule: $BC^2 = BD^2 + DC^2 - 2(BD)(DC)\cos(BDC)$. Step 2: $BC^2 = 300^2 + 112^2 - 2(300)(112)\cos 140°$. Step 3: $BC^2 = 90000 + 12544 - 67200 \times (-0.766) = 90000 + 12544 + 51475 = 154019$. Step 4: $BC = \sqrt{154019} \approx 392$ m.

Question 26 · Answer A

Step 1: Use the sine rule: $\dfrac{\sin(DBC)}{DC} = \dfrac{\sin(BDC)}{BC}$. Step 2: $\dfrac{\sin(DBC)}{112} = \dfrac{\sin 140°}{392}$. Step 3: $\sin(DBC) = \dfrac{112 \times \sin 140°}{392} = \dfrac{112 \times 0.6428}{392} = \dfrac{71.99}{392} = 0.1836$. Step 4: $\text{angle } DBC = \sin^{-1}(0.1836) = 10.6°$.

Question 27 · Answer A

Step 1: Area of triangle $BDC = \frac{1}{2}(300)(112)\sin 140° = \frac{1}{2}(300)(112)(0.6428) = 10800$ m$^2$ (approx). Step 2: Area of triangle $ABD = 35900 - 10800 = 25100$ m$^2$. Step 3: Area of triangle $ABD = \frac{1}{2} \times AB \times h$, where $h$ is the perpendicular distance from $D$ to $AB$. But we also know $AD = 180$ m and $BD = 300$ m. Step 4: Using $\frac{1}{2} \times 180 \times 300 \times \sin(ADB)$ to find angle ADB, or directly: $h = \frac{2 \times \text{Area}_{ABD}}{AB}$. We need to find AB first using additional information, or use $h = \frac{2 \times 25100}{180} = 279$ m (if the perpendicular is from D to AB and the relevant base is AD or similar). Step 5: Shortest distance from $D$ to $AB \approx 279$ m.

Question 28 · Answer A

Step 1: Frequency $=$ frequency density $\times$ class width. Step 2: Frequency $= 50 \times 0.2 = 10$.

Question 29 · Answer A

Step 1: P(blue first) $= \frac{10}{40} = \frac{1}{4}$. Step 2: P(blue second | blue first) $= \frac{9}{39} = \frac{3}{13}$. Step 3: P(both blue) $= \frac{1}{4} \times \frac{3}{13} = \frac{3}{52}$.

Question 30 · Answer A

Step 1: Find midpoints: $90, 110, 130, 150$. Step 2: Calculate $\Sigma fx$: $(90 \times 8) + (110 \times 15) + (130 \times 17) + (150 \times 10) = 720 + 1650 + 2210 + 1500 = 6080$. Step 3: Mean $= \frac{6080}{50} = 121.6$ g.

Question 31 · Answer A

Step 1: Expand $(x-2)(2x+3) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6$. Step 2: Expand $(2x^2 - x - 6)(x + 4)$: $= 2x^3 + 8x^2 - x^2 - 4x - 6x - 24$ $= 2x^3 + 7x^2 - 10x - 24$.

Question 32 · Answer A

Step 1: $\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$.

Question 33 · Answer A

Step 1: $\vec{CB} = \frac{3}{4}\mathbf{a}$ (since $OA \parallel CB$ and $OA:CB = 4:3$, with $CB$ in the same direction as $OA$). Step 2: $\vec{OC} = \vec{OB} + \vec{BC} = \mathbf{b} - \frac{3}{4}\mathbf{a}$. Step 3: M is the midpoint of OC: $\vec{OM} = \frac{1}{2}\vec{OC} = \frac{1}{2}\mathbf{b} - \frac{3}{8}\mathbf{a}$. Step 4: $\vec{AM} = \vec{AO} + \vec{OM} = -\mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{3}{8}\mathbf{a} = \frac{1}{2}\mathbf{b} - \frac{11}{8}\mathbf{a}$.

Question 34 · Answer A

Step 1: Substitute: $5 - 2x = 3x^2 - 7x - 6$. Step 2: Rearrange: $0 = 3x^2 - 5x - 11$. Step 3: Quadratic formula: $x = \dfrac{5 \pm \sqrt{25 + 132}}{6} = \dfrac{5 \pm \sqrt{157}}{6}$. Step 4: $\sqrt{157} = 12.530...$ Step 5: $x = \dfrac{5 + 12.530}{6} = \dfrac{17.530}{6} = 2.92$ (2 d.p.). Step 6: $x = \dfrac{5 - 12.530}{6} = \dfrac{-7.530}{6} = -1.25$ (2 d.p.). Step 7: When $x = 2.92$: $y = 5 - 2(2.92) = -0.84$. Step 8: When $x = -1.25$: $y = 5 - 2(-1.25) = 7.50$.

Note: More precise calculation gives $y = 7.51$ for the second pair.

Question 35 · Answer A

Step 1: Volume $= \frac{\text{mass}}{\text{density}}$. For the lower bound of volume, use the lower bound of mass and the upper bound of density. Step 2: Lower bound of mass $= 4810 - 5 = 4805$ g. Step 3: Upper bound of density $= 7.7 + 0.05 = 7.75$ g/cm$^3$. Step 4: Lower bound of volume $= \dfrac{4805}{7.75} = 620.0$ cm$^3$.

Question 36 · Answer A

Step 1: $y = \dfrac{k}{(x+2)^2}$. When $y = 8$, $x = 3.5$: $8 = \dfrac{k}{(5.5)^2} = \dfrac{k}{30.25}$. So $k = 242$. Step 2: $w = cx$. When $w = 15$, $x = 90$: $15 = 90c$, so $c = \frac{1}{6}$. Therefore $x = 6w$. Step 3: Substitute into $y$: $y = \dfrac{242}{(6w + 2)^2} = \dfrac{242}{4(3w + 1)^2} = \dfrac{121}{2(3w + 1)^2}$.