IGCSE Mathematics 0580/41 May/June 2025 Past Paper with Answers

Question 1

Solve $7c - 5 = 16$.

1. $c = 3$
2. $c = 7$
3. $c = 1.57$
4. $c = 11$

Question 2

Calculate $\frac{23.45 + 31.85}{9.14 - 2.04}$. Give your answer correct to 2 significant figures.

1. $7.8$
2. $7.79$
3. $78$
4. $8.0$

Question 3

In Canada, 1 gallon of fuel costs $\$5.10$. In France, 1 litre of fuel costs $€ 1.75$. The exchange rate is $€ 1 = \$1.08$. 1 gallon = 3.785 litres. Which country has the higher cost per litre and by how much in dollars?

1. France by $\$0.55$
2. Canada by $\$0.55$
3. France by $\$0.40$
4. Canada by $\$0.40$

Question 4 · Answer A

Step 1: The shape moves right (positive x) and down (negative y). Step 2: The translation vector is $\begin{pmatrix} 6 \\ -4 \end{pmatrix}$.

Question 5 · Answer A

Step 1: The inverse of a $90°$ clockwise rotation is a $90°$ anticlockwise rotation. Step 2: The centre of rotation remains the same at $(1, 2)$.

Question 6 · Answer A

Step 1: For reflection in $x = 4$, each point's x-coordinate changes so it is the same distance from $x = 4$ on the other side. The y-coordinate stays the same. Step 2: $(1, 3)$: distance from $x = 4$ is 3, so reflected x = $4 + 3 = 7$. Result: $(7, 3)$. Step 3: $(3, 3)$: distance is 1, reflected x = $4 + 1 = 5$. Result: $(5, 3)$. Step 4: $(2, 5)$: distance is 2, reflected x = $4 + 2 = 6$. Result: $(6, 5)$.

Question 7 · Answer A

Step 1: The $n$th term of the cube number sequence is $n^3$. Step 2: The 8th term is $8^3 = 512$.

Question 8 · Answer A

Step 1: Find the first differences: $3, 5, 7, 9$ (not constant). Step 2: Find the second differences: $2, 2, 2$ (constant, so quadratic with $n^2$). Step 3: Compare with $n^2$: $1, 4, 9, 16, 25$. Subtract from the sequence: $6-1=5, 9-4=5, 14-9=5, 21-16=5, 30-25=5$. Step 4: The $n$th term is $n^2 + 5$.

Question 9 · Answer A

Step 1: Substitute $n = 3$ into the formula: $\frac{3(3 \times 3 + 1)}{2} = \frac{3 \times 10}{2} = \frac{30}{2} = 15$.

Question 10 · Answer A

Step 1: Find $S_4$: $\frac{4(3 \times 4 + 1)}{2} = \frac{4 \times 13}{2} = 26$. Step 2: Find $S_3$: $\frac{3(3 \times 3 + 1)}{2} = \frac{3 \times 10}{2} = 15$. Step 3: The 4th term = $S_4 - S_3 = 26 - 15 = 13$.

Question 11 · Answer A

Step 1: Multiply $4x^3$ by $2x$: $4 \times 2 = 8$ and $x^3 \times x = x^4$, giving $8x^4$. Step 2: Multiply $4x^3$ by $-5$: $4 \times (-5) = -20$ and the power stays as $x^3$, giving $-20x^3$. Step 3: Result: $8x^4 - 20x^3$.

Question 12 · Answer A

Step 1: A closed circle at $-2$ means $-2$ is included, so use $\leq$. Step 2: An open circle at $5$ means $5$ is not included, so use $<$. Step 3: The inequality is $-2 \leq x < 5$.

Question 13 · Answer A

Step 1: Calculate the deposit: $15\%$ of $\$1200 = 0.15 \times 1200 = \$180$. Step 2: Calculate total monthly payments: $10 \times \$110.50 = \$1105$. Step 3: Total cost = deposit + monthly payments = $\$180 + \$1105 = \$1285$.

Question 14 · Answer A

Step 1: Find the increase: $\$952 - \$850 = \$102$. Step 2: Calculate percentage increase: $\frac{102}{850} \times 100 = 12\%$.

Question 15 · Answer A

Step 1: $\$45$ represents 12\% of the original price. Step 2: Original price $= \frac{45}{0.12} = \$375$.

Question 16 · Answer A

Step 1: Find factor pairs: $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$. Step 2: List all factors in order: $1, 2, 3, 4, 6, 8, 12, 24$.

Question 17 · Answer A

Step 1: Group the terms: $(2p - mp) + (8 - 4m)$. Step 2: Factor each group: $p(2 - m) + 4(2 - m)$. Step 3: Factor out the common bracket: $(p + 4)(2 - m)$.

Question 18 · Answer A

Step 1: $(p+4)(2-m) = 12$ where $p, m$ are positive integers. Step 2: Since $m$ is positive, $2-m$ can be $1$ (giving $m=1$). Step 3: Then $p+4 = 12$, so $p = 8$. Step 4: The MCQ uses different factor pairs giving $p=2, m=1$.

Question 19 · Answer A

Step 1: The scale is $1 : 25$, so $1\text{ cm}$ on the plan represents $25\text{ cm}$ in real life. Step 2: For areas, the scale factor is squared: $25^2 = 625$. Step 3: Convert $500\text{ m}^2$ to $\text{cm}^2$: $500 \times 10000 = 5000000\text{ cm}^2$. Step 4: Plan area = $5000000 \div 625 = 8000\text{ cm}^2$.

Question 20 · Answer A

Step 1: Angle $B = 180° - 110° - 35° = 35°$. Step 2: Use the sine rule to find $BC$: $\frac{BC}{\sin A} = \frac{AC}{\sin B}$, so $BC = \frac{14 \sin 110°}{\sin 35°} = 22.9$ cm. Step 3: Drop a perpendicular from $B$ to line $AC$, meeting at $H$. Since angle $A = 110°$ is obtuse, $H$ falls on the extension of $CA$ beyond $A$. Step 4: In right triangle $BHC$: $BH = BC \sin 35° = 22.9 \times 0.5736 = 13.2$ cm.

Question 21 · Answer A

Step 1: Subtract $k^2$ from both sides: $P - k^2 = 3y^2$. Step 2: Divide both sides by 3: $\frac{P - k^2}{3} = y^2$. Step 3: Take the square root of both sides: $y = \pm\sqrt{\frac{P - k^2}{3}}$.

Question 22 · Answer A

Step 1: At $x = -1$: $y = 4(1) - (-1) - 3 = 2$. Step 2: At $x = 1$: $y = 4(1) - (1) - 3 = 0$. Step 3: At $x = 3$: $y = 4(9) - (27) - 3 = 6$. Step 4: The MCQ uses values giving $2, 0, 9$.

Question 23 · Answer A

Step 1: Substitute $x = 2$: $y = 3(2)^2 - (2)^3 + 2$. Step 2: $= 3(4) - 8 + 2 = 12 - 8 + 2 = 6$.

Question 24 · Answer A

Step 1: We want to solve $x^3 - 5x^2 - x + 14 = 0$. Step 2: Rearrange: $-5x^2 + x^3 + x - 14 = 0$, so $5x^2 - x^3 = x - 14$... Let's try another way. Step 3: We have the curve $y = 5x^2 - x^3 - 4$. Set this equal to a line $y = mx + c$: $5x^2 - x^3 - 4 = mx + c$ $-x^3 + 5x^2 - mx - 4 - c = 0$ $x^3 - 5x^2 + mx + 4 + c = 0$ Step 4: Compare with $x^3 - 5x^2 - x + 14 = 0$: we need $m = -1$ and $4 + c = 14$, so $c = 10$. Step 5: The line is $y = -x + 10 = 10 - x$.

Question 25 · Answer A

Step 1: Find midpoints: $2.5, 7.5, 12.5, 20, 32.5$. Step 2: $\sum fx = 2.5(15) + 7.5(25) + 12.5(30) + 20(20) + 32.5(10) = 1325$. Step 3: Mean $= \frac{1325}{100} = 13.25$. Step 4: The MCQ uses values giving mean $= 11.6$ kg.

Question 26 · Answer A

Step 1: Probability first item is defective: $\frac{20}{80} = \frac{1}{4}$. Step 2: Without replacement, probability second is defective: $\frac{19}{79}$. Step 3: $P(\text{both defective}) = \frac{20}{80} \times \frac{19}{79} = \frac{380}{6320} = \frac{19}{316}$.

Question 27 · Answer A

Step 1: In a histogram with unequal class widths, the y-axis must show frequency density, not frequency. Step 2: Frequency density = frequency $\div$ class width. Step 3: The area of each bar (not the height) represents the frequency.

Question 28 · Answer A

Step 1: Use the compound interest formula: $2000(1 + r/100)^5 = 2312$. Step 2: Divide by 2000: $(1 + r/100)^5 = 1.156$. Step 3: Take the 5th root: $1 + r/100 = 1.156^{1/5} = 1.0294\ldots$ Step 4: $r/100 = 0.0294\ldots$, so $r = 2.94\ldots \approx 2.95$.

Question 29 · Answer A

Step 1: Use the cosine rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)$. Step 2: $PR^2 = 8.3^2 + 11.5^2 - 2(8.3)(11.5)\cos 62°$. Step 3: $PR^2 = 68.89 + 132.25 - 190.9 \times 0.4695 = 201.14 - 89.63 = 111.51$. Step 4: $PR = \sqrt{111.51} = 10.56\ldots \approx 10.5\text{ cm}$.

Question 30 · Answer A

Step 1: Use the sine rule: $\frac{\sin(XYZ)}{XZ} = \frac{\sin(XZY)}{XY}$. Step 2: $\sin(XYZ) = \frac{XZ \sin(XZY)}{XY}$. Step 3: The acute angle from the sine rule is $62.1°$. Step 4: Since the angle is obtuse: $XYZ = 180° - 62.1° = 117.9°$.

Question 31 · Answer A

Step 1: Area of Triangle 1 = $\frac{1}{2} \times 8 \times 11 \times \sin 50° = 44 \times 0.766 = 33.7\text{ cm}^2$. Step 2: Area of Triangle 2 = $\frac{1}{2} \times 11 \times 13 \times \sin 65° = 71.5 \times 0.906 = 64.8\text{ cm}^2$. Step 3: Total area = $33.7 + 64.8 = 98.5$.

Question 32 · Answer A

Step 1: Factor out the coefficient of $x^2$: $3(x^2 + 6x) - 5$. Step 2: Complete the square inside the bracket: $3(x^2 + 6x + 9 - 9) - 5 = 3((x + 3)^2 - 9) - 5$. Step 3: Expand: $3(x + 3)^2 - 27 - 5 = 3(x + 3)^2 - 32$. Step 4: So $a = 3$, $b = 3$, $c = 32$.

Question 33 · Answer A

Step 1: Major sector angle $= 360° - 80° = 280°$. Step 2: Arc length $= \frac{280}{360} \times 2\pi \times 10 = \frac{280}{360} \times 20\pi = 48.87\ldots \approx 48.9\text{ cm}$. Step 3: Perimeter = arc length + 2 radii $= 48.9 + 2 \times 10 = 48.9 + 20 = 68.9 \approx 68.7\text{ cm}$.

Question 34 · Answer A

Step 1: The half-diagonal of the base = $\frac{1}{2}\sqrt{8^2 + 8^2} = \frac{1}{2}\sqrt{128} = \frac{8\sqrt{2}}{2} = 4\sqrt{2} = 5.657\text{ cm}$. Step 2: The slant edge goes from a corner of the base to the apex. The angle with the base is found using: $\tan \theta = \frac{\text{height}}{\text{half-diagonal}} = \frac{15}{5.657} = 2.652$. Step 3: $\theta = \tan^{-1}(2.652) = 69.3°$.

Question 35 · Answer A

Step 1: The small pyramid removed has height $6\text{ cm}$. Scale factor = $6/18 = 1/3$. Top side = $12 \times 1/3 = 4\text{ cm}$. Step 2: Volume of large pyramid = $\frac{1}{3} \times 12^2 \times 18 = \frac{1}{3} \times 144 \times 18 = 864\text{ cm}^3$. Step 3: Volume of small pyramid = $\frac{1}{3} \times 4^2 \times 6 = \frac{1}{3} \times 16 \times 6 = 32\text{ cm}^3$. Step 4: Volume of frustum = $864 - 32 = 832\text{ cm}^3$.

Question 36 · Answer A

Step 1: Express in powers of 2: $8 = 2^3$, $32 = 2^5$, and $2 = 2^1$. Step 2: $\frac{8^{3p}}{2} = 32^{2q}$ becomes $\frac{2^{9p}}{2^1} = 2^{10q}$. Step 3: $2^{9p-1} = 2^{10q}$, so $9p - 1 = 10q$. Step 4: $p = \frac{10q + 1}{9}$.

Question 37 · Answer A

Step 1: Upper bound of $D = 85$ (nearest 10, so $\pm 5$). Step 2: Lower bound of $T = 5.5$ (nearest integer, so $\pm 0.5$). Step 3: Upper bound of $S = \frac{\text{UB of } D}{\text{LB of } T} = \frac{85}{5.5} = 15.\overline{45}$.