IGCSE Mathematics 0580/21 May/June 2025 Past Paper with Answers

Question 1

Simplify $9a - 7b + 3a + 2b$.

1. $12a - 5b$
2. $12a - 9b$
3. $6a - 5b$
4. $12a + 5b$

Question 2

Two parallel lines are cut by two transversals. One transversal makes an angle of $65°$ with the lower parallel line, and the other makes an angle of $140°$ at the same intersection point on the lower parallel line. What is the angle between the two transversals at the point where they meet above the upper parallel line?

1. $75°$
2. $65°$
3. $105°$
4. $40°$

Question 3

Tom invests $\$2000$ at a rate of $4\%$ per year simple interest. What is the total value of his investment at the end of 5 years?

1. $\$2400$
2. $\$2433.31$
3. $\$400$
4. $\$2040$

Question 4 · Answer A

Step 1: Perform the multiplication first (BIDMAS): $\frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10}$. Step 2: Now subtract: $\frac{7}{8} - \frac{3}{10}$. Find a common denominator: LCD = 40. Step 3: $\frac{7}{8} = \frac{35}{40}$ and $\frac{3}{10} = \frac{12}{40}$. Step 4: $\frac{35}{40} - \frac{12}{40} = \frac{23}{40}$.

Question 5 · Answer A

Step 1: The exterior angle is $180° - 160° = 20°$. Step 2: The number of sides is $\frac{360°}{20°} = 18$.

Question 6 · Answer A

Step 1: Substitute $x = 2$: $y = 3(2) - 2 = 6 - 2 = 4$. So $(2, 4)$ is on the line. Step 2: Check $(1, 3)$: $y = 3(1) - 2 = 1 \neq 3$. Not on the line. Step 3: Check $(0, 2)$: $y = 3(0) - 2 = -2 \neq 2$. Not on the line. Step 4: Check $(3, 6)$: $y = 3(3) - 2 = 7 \neq 6$. Not on the line.

Question 7 · Answer A

Step 1: Set the equations equal: $x + 3 = -2x + 9$. Step 2: Solve: $3x = 6$, so $x = 2$. Step 3: Substitute: $y = 2 + 3 = 5$. Step 4: Therefore $a + b = 2 + 5 = 7$.

Question 8 · Answer A

Step 1: Let $x = 0.18888...$ Step 2: $10x = 1.8888...$ Step 3: $100x = 18.8888...$ Step 4: Subtract: $100x - 10x = 18.888... - 1.888... = 17$ Step 5: $90x = 17$, so $x = \frac{17}{90}$.

Question 9 · Answer A

Step 1: Calculate $2\mathbf{a} = 2\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 10 \\ 6 \end{pmatrix}$. Step 2: Subtract $\mathbf{b}$: $\begin{pmatrix} 10 \\ 6 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 10 - (-2) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}$.

Question 10 · Answer A

Step 1: Use the magnitude formula: $\sqrt{3^2 + p^2} = 5$. Step 2: Square both sides: $9 + p^2 = 25$. Step 3: Solve: $p^2 = 16$, so $p = 4$ (since $p > 0$).

Question 11 · Answer A

Step 1: Total frequency: $4 + k + 6 + 3 = 13 + k = 20$, so $k = 7$. Step 2: Verify with mean: $\frac{0(4) + 1(7) + 2(6) + 3(3)}{20} = \frac{0 + 7 + 12 + 9}{20} = \frac{28}{20} = 1.4$. Note: Using the total frequency condition gives $k = 7$ directly. The mean can be used as a check.

Question 12 · Answer A

Step 1: The angle at the centre is twice the angle at the circumference subtended by the same arc. Step 2: Angle at centre $= 2 \times 35° = 70°$.

Question 13 · Answer A

Step 1: The angle at the circumference is half the angle at the centre. Step 2: Angle at circumference $= \frac{110°}{2} = 55°$.

Question 14 · Answer A

Step 1: By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment. Step 2: Therefore the angle between the tangent and chord is $55°$.

Question 15 · Answer A

Step 1: The difference between the two angles is $55° - 40° = 15°$.

Question 16 · Answer A

Step 1: Substitute $y = 5$: $5 = a^2 + 1$. Step 2: Solve: $a^2 = 4$, so $a = 2$ (since $a > 0$).

Question 17 · Answer A

Step 1: Use $y - y_1 = m(x - x_1)$ with $m = -4$ and $(x_1, y_1) = (2, 3)$. Step 2: $y - 3 = -4(x - 2)$. Step 3: $y - 3 = -4x + 8$. Step 4: $y = -4x + 11$.

Question 18 · Answer A

Step 1: Interquartile range = Upper quartile $-$ Lower quartile. Step 2: IQR $= 28 - 12 = 16$ minutes.

Question 19 · Answer A

Step 1: Parcels with mass $\geq m$: $60 - 45 = 15$. Step 2: Fraction $= \frac{15}{60} = \frac{1}{4}$.

Question 20 · Answer A

Step 1: The cones are similar, so the ratio of corresponding lengths is constant. Step 2: Scale factor $= \frac{10}{4} = \frac{5}{2}$. Step 3: Height of larger cone $= 6 \times \frac{5}{2} = 15$ cm.

Question 21 · Answer A

Step 1: The radius, height, and slant height form a right triangle. Step 2: $l^2 = r^2 + h^2 = 6^2 + 8^2 = 36 + 64 = 100$. Step 3: $l = \sqrt{100} = 10$ cm.

Question 22 · Answer A

Step 1: Substitute $x = -1$: $f(-1) = 3(-1)^2 - 7$. Step 2: $(-1)^2 = 1$, so $f(-1) = 3(1) - 7 = 3 - 7 = -4$.

Question 23 · Answer A

Step 1: Let $y = 5x - 2$. Step 2: Swap $x$ and $y$: $x = 5y - 2$. Step 3: Solve for $y$: $x + 2 = 5y$, so $y = \frac{x + 2}{5}$. Step 4: Therefore $g^{-1}(x) = \frac{x + 2}{5}$.

Question 24 · Answer A

Step 1: $fg(x) = f(g(x)) = f(3x - 2)$. Step 2: Substitute into $f$: $2(3x - 2) + 1 = 6x - 4 + 1 = 6x - 3$. Step 3: So $a = 6$ and $b = -3$.

Question 25 · Answer A

Step 1: Common denominator is $(2x + 1)(x - 3)$. Step 2: $\frac{3(x - 3) - 2(2x + 1)}{(2x + 1)(x - 3)}$. Step 3: Expand numerator: $3x - 9 - 4x - 2 = -x - 11$. Step 4: Result: $\frac{-x - 11}{(2x + 1)(x - 3)}$.

Question 26 · Answer A

Step 1: Expand using FOIL: $(4)(3) + (4)(\sqrt{3}) + (-\sqrt{3})(3) + (-\sqrt{3})(\sqrt{3})$. Step 2: $= 12 + 4\sqrt{3} - 3\sqrt{3} - 3$. Step 3: $= (12 - 3) + (4\sqrt{3} - 3\sqrt{3}) = 9 + \sqrt{3}$.

Question 27 · Answer A

Step 1: Multiply by $\frac{\sqrt{6}}{\sqrt{6}}$: $\frac{4}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt{6}}{6}$. Step 2: Simplify: $\frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$.

Question 28 · Answer A

Step 1: First expand $(x + 2)(x - 1) = x^2 + x - 2$. Step 2: Now multiply by $(2x + 3)$: $(x^2 + x - 2)(2x + 3)$. Step 3: $= 2x^3 + 3x^2 + 2x^2 + 3x - 4x - 6$. Step 4: $= 2x^3 + 5x^2 - x - 6$.

Question 29 · Answer A

Step 1: $P(\text{red first}) = \frac{3}{10}$. Step 2: With replacement, $P(\text{red second}) = \frac{3}{10}$. Step 3: $P(\text{both red}) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}$.

Question 30 · Answer A

Step 1: $P(\text{no rain Monday}) = 1 - P(\text{rain Monday}) = 1 - \frac{3}{5} = \frac{2}{5}$.

Question 31 · Answer A

Step 1: $P(\text{pass 1st}) = \frac{2}{3}$. Step 2: $P(\text{fail 1st then pass 2nd}) = \frac{1}{3} \times \frac{3}{4} = \frac{3}{12} = \frac{1}{4}$. Step 3: $P(\text{pass within 2}) = \frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$.

Question 32 · Answer A

Step 1: Time $= \frac{\text{Distance}}{\text{Speed}} = \frac{80}{v}$.

Question 33 · Answer A

Step 1: Time $= \frac{\text{Distance}}{\text{Speed}} = \frac{60}{s + 5}$.

Question 34 · Answer A

Step 1: Set up the equation: $\frac{10}{x-4} - \frac{12}{x} = 1$. Step 2: Multiply through by $x(x-4)$: $10x - 12(x-4) = x(x-4)$. Step 3: Expand: $10x - 12x + 48 = x^2 - 4x$. Step 4: Simplify: $-2x + 48 = x^2 - 4x$, giving $x^2 - 2x - 48 = 0$.

Question 35 · Answer A

Step 1: Find two numbers that multiply to $-28$ and add to $+3$: $7$ and $-4$. Step 2: Factorise: $(x + 7)(x - 4) = 0$. Step 3: $x + 7 = 0$ gives $x = -7$. $x - 4 = 0$ gives $x = 4$.

Question 36 · Answer A

Step 1: Since speed must be positive, $x = 8$ km/h. Step 2: Time $= \frac{12}{8} = 1.5$ hours.

Question 37 · Answer A

Step 1: $27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}}$. Step 2: $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$. Step 3: $27^{\frac{2}{3}} = 3^2 = 9$. Step 4: $27^{-\frac{2}{3}} = \frac{1}{9}$.

Question 38 · Answer A

Step 1: Using $\cos 30° = \frac{\text{adjacent}}{\text{hypotenuse}}$. Step 2: $\cos 30° = \frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2} = \frac{9}{h}$. Step 3: $h = \frac{9 \times 2}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18\sqrt{3}}{3} = 6\sqrt{3}$ cm.

Question 39 · Answer A

Step 1: $\overrightarrow{BA} = \overrightarrow{BO} + \overrightarrow{OA}$. Step 2: $\overrightarrow{BO} = -\overrightarrow{OB} = -\mathbf{b}$. Step 3: $\overrightarrow{BA} = -\mathbf{b} + \mathbf{a} = \mathbf{a} - \mathbf{b}$.

Question 40 · Answer A

Step 1: $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}$. Step 2: $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$. Step 3: $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.

Question 41 · Answer A

Step 1: $\overrightarrow{SR} = 4\mathbf{a} = 2 \times 2\mathbf{a} = 2\overrightarrow{PQ}$. Step 2: Since $\overrightarrow{SR}$ is a scalar multiple of $\overrightarrow{PQ}$, they are parallel. Step 3: $PQ \neq SR$ (different lengths), so exactly one pair of sides is parallel, making $PQRS$ a trapezium.

Question 42 · Answer A

Step 1: The completed square form $y = (x - p)^2 + q$ has its turning point at $(p, q)$. Step 2: Since the turning point is $(4, -5)$, we have $p = 4$ and $q = -5$.

Question 43 · Answer A

Step 1: Set the curve equal to the line: $(x - 3)^2 = -2x + 6$. Step 2: Expand: $x^2 - 6x + 9 = -2x + 6$. Step 3: Rearrange: $x^2 - 4x + 3 = 0$. Step 4: Factorise: $(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$. Step 5: When $x = 1$: $y = -2(1) + 6 = 4$. When $x = 3$: $y = -2(3) + 6 = 0$. Step 6: Intersection points: $(1, 4)$ and $(3, 0)$.

Question 44 · Answer A

Step 1: Factorise the numerator: $3x^2 + 6x = 3x(x + 2)$. Step 2: Factorise the denominator: $3x^2 - 12 = 3(x^2 - 4) = 3(x + 2)(x - 2)$. Step 3: Cancel common factors: $\frac{3x(x + 2)}{3(x + 2)(x - 2)} = \frac{x}{x - 2}$.