2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

First, find the highest common factor of \( 12a^2b \) and \( 8ab^2 \), which is \( 4ab \). Then divide each term by the common factor: \( 12a^2b \div 4ab = 3a \) and \( 8ab^2 \div 4ab = 2b \). Thus, the completely factorised expression is \( 4ab(3a - 2b) \).

M1 for correct partial factorisation, e.g., \( 2ab(6a - 4b) \) or \( 4a(3ab - 2b^2) \). A1 for correct final answer \( 4ab(3a - 2b) \).

Subtract 7 from both sides: \( -3x \le 12 \). Divide both sides by \( -3 \) and reverse the inequality sign: \( x \ge -4 \).

M1 for isolating the \( x \) term, e.g. \( -3x \le 12 \). A1 for \( x \ge -4 \).

Simplify each surd: \( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \) and \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \). Then subtract them: \( 5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3} \).

M1 for simplifying at least one surd correctly, e.g., \( 5\sqrt{3} \) or \( 2\sqrt{3} \). A1 for \( 3\sqrt{3} \).

The terms increase by 4 each time, so the \( n \)th term is in the form \( 4n + c \). Using \( n = 1 \): \( 4(1) + c = 3 \implies c = -1 \). Therefore, the \( n \)th term is \( 4n - 1 \).

M1 for \( 4n + c \) or finding the common difference is 4. A1 for \( 4n - 1 \).

A regular hexagon has 6 lines of symmetry (3 passing through opposite vertices, and 3 passing through the midpoints of opposite sides). It also has rotational symmetry of order 6 as it maps onto itself 6 times during a full \( 360^\circ \) rotation.

B1 for 6 lines of symmetry. B1 for order of rotational symmetry 6.

Write the decimal as a fraction: \( 0.085 = \frac{85}{1000} \). Simplify by dividing the numerator and the denominator by their highest common factor, which is 5: \( \frac{85 \div 5}{1000 \div 5} = \frac{17}{200} \).

M1 for \( \frac{85}{1000} \) or equivalent unsimplified fraction. A1 for \( \frac{17}{200} \).

The cosine of an angle in a right-angled triangle is the ratio of the adjacent side to the hypotenuse: \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{10} = 0.6 \).

M1 for \( \frac{6}{10} \) or \( \cos \theta = \frac{6}{10} \). A1 for \( 0.6 \).

Multiply both sides of the equation by 4: \( 2x - 3 = 20 \). Add 3 to both sides: \( 2x = 23 \). Divide by 2: \( x = 11.5 \) (or \( \frac{23}{2} \)).

M1 for \( 2x - 3 = 20 \) or equivalent. A1 for \( 11.5 \) or \( \frac{23}{2} \) or \( 11\frac{1}{2} \).

First, simplify each surd by finding the largest square factor: \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\) and \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). Subtracting the simplified surds: \(5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3}\).

M1 for \(5\sqrt{3}\) or \(2\sqrt{3}\) seen
A1 for \(3\sqrt{3}\)

Find the highest common factor of both terms, which is \(3ab\). Divide each term by \(3ab\) to find the terms inside the brackets: \(\frac{6a^2b}{3ab} = 2a\) and \(\frac{9ab^2}{3ab} = 3b\). Thus, the factorised expression is \(3ab(2a - 3b)\).

M1 for a correct partial factorisation, e.g. \(3(2a^2b - 3ab^2)\) or \(ab(6a - 9b)\)
A1 for \(3ab(2a - 3b)\)

(a) \(\sqrt{48} = 4\sqrt{3}\), \(2\sqrt{27} = 2(3\sqrt{3}) = 6\sqrt{3}\), and \(\sqrt{75} = 5\sqrt{3}\). Thus, \(4\sqrt{3} - 6\sqrt{3} + 5\sqrt{3} = 3\sqrt{3}\).
(b) \(\frac{6}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{6(3 + \sqrt{3})}{9 - 3} = \frac{6(3 + \sqrt{3})}{6} = 3 + \sqrt{3}\).

(a) M1 for simplifying at least one surd (e.g., \(4\sqrt{3}\) or \(5\sqrt{3}\)), M1 for expressing all terms with \(\sqrt{3}\), A1 for final answer \(3\sqrt{3}\).
(b) M1 for multiplying numerator and denominator by the conjugate \(3 + \sqrt{3}\), A1 for final answer \(3 + \sqrt{3}\).

(a) Gradient of \(PQ = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\).
(b) Midpoint of \(PQ = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1)\). The perpendicular gradient \(m_{\perp} = -\frac{1}{-4/3} = \frac{3}{4}\). Using the point-slope form: \(y - 1 = \frac{3}{4}(x - 1) \implies y = \frac{3}{4}x + \frac{1}{4}\).

(a) M1 for substituting coordinates into the gradient formula, A1 for \(-\frac{4}{3}\) (or equivalent).
(b) B1 for midpoint \((1, 1)\), M1 for using the perpendicular gradient \(\frac{3}{4}\), A1 for the final equation \(y = \frac{3}{4}x + \frac{1}{4}\) (or equivalent).

(a) Angle \(ADC = 90^{\circ}\) because the angle subtended by a diameter at the circumference (angle in a semicircle) is always a right angle.
(b) (i) In triangle \(ACD\), the angles sum to \(180^{\circ}\). Thus, angle \(ACD = 180^{\circ} - 90^{\circ} - 25^{\circ} = 65^{\circ}\).
(ii) Since \(ABCD\) is a cyclic quadrilateral, opposite angles sum to \(180^{\circ}\). Angle \(BAD = angle BAC + angle CAD = 35^{\circ} + 25^{\circ} = 60^{\circ}\). Thus, angle \(BCD = 180^{\circ} - 60^{\circ} = 120^{\circ}\).

(a) B1 for \(90^{\circ}\), B1 for 'angle in a semicircle' (or equivalent).
(b) (i) B1 for \(65^{\circ}\).
(ii) M1 for finding angle \(BAD = 60^{\circ}\) or calculating angle \(BCA = 55^{\circ}\), A1 for \(120^{\circ}\).

(a) \((2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15\). Also, \(-x(x - 2) = -x^2 + 2x\). Adding these together: \(2x^2 + 7x - 15 - x^2 + 2x = x^2 + 9x - 15\).
(b) First factor out the common factor of 3: \(12y^2 - 75 = 3(4y^2 - 25)\). Then apply the difference of two squares: \(3(2y - 5)(2y + 5)\).

(a) M1 for expanding \((2x-3)(x+5)\) with at least 3 correct terms, M1 for \(-x^2 + 2x\), A1 for \(x^2 + 9x - 15\).
(b) M1 for \(3(4y^2 - 25)\), A1 for \(3(2y - 5)(2y + 5)\).

(a) \(y = \frac{k}{\sqrt{x}}\). Substitute \(x = 16\) and \(y = 3\): \(3 = \frac{k}{\sqrt{16}} \implies 3 = \frac{k}{4} \implies k = 12\). So, \(y = \frac{12}{\sqrt{x}}\).
(b) Substitute \(x = 36\) into the equation: \(y = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\).

(a) M1 for writing \(y = \frac{k}{\sqrt{x}}\), M1 for substituting \(x = 16\) and \(y = 3\) to find \(k = 12\), A1 for \(y = \frac{12}{\sqrt{x}}\).
(b) M1 for substituting \(x = 36\) into their equation, A1 for \(2\).

(a) \(\left(\frac{8}{27}\right)^{-\frac{2}{3}} = \left(\frac{27}{8}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\frac{27}{8}}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}\).
(b) \((4 \times 10^5) \times (8 \times 10^{-2}) = 32 \times 10^3 = 3.2 \times 10^4\).

(a) M1 for reciprocation to remove the negative power: \(\left(\frac{27}{8}\right)^{\frac{2}{3}}\), M1 for taking the cube root: \(\left(\frac{3}{2}\right)^2\), A1 for \(\frac{9}{4}\) (or equivalent decimal \(2.25\)).
(b) M1 for multiplying coefficients and adding indices: \(32 \times 10^3\), A1 for \(3.2 \times 10^4\).

(a) \(g(3) = 3^2 + 2 = 11\). Then \(f(11) = 3(11) - 1 = 32\).
(b) Let \(y = 3x - 1 \implies y + 1 = 3x \implies x = \frac{y + 1}{3}\). Thus, \(f^{-1}(x) = \frac{x + 1}{3}\).
(c) \(g(2) = 2^2 + 2 = 6\). Thus, \(3x - 1 = 6 \implies 3x = 7 \implies x = \frac{7}{3}\).

(a) M1 for evaluating \(g(3) = 11\), A1 for \(32\).
(b) M1 for attempting to swap variables or rearrange the equation, A1 for \(\frac{x + 1}{3}\).
(c) B1 for \(\frac{7}{3}\) (or equivalent).

Let the five integers be \(a, b, c, d, e\) in ascending order.
Since the median is 5, \(c = 5\).
Since the mode is 7 and must occur more than once, and \(d, e \ge 5\), we must have \(d = 7\) and \(e = 7\).
The range is 6, so \(e - a = 6 \implies 7 - a = 6 \implies a = 1\).
Since the mean is 4.6, we have \(\frac{1 + b + 5 + 7 + 7}{5} = 4.6 \implies 20 + b = 23 \implies b = 3\).
The five integers in ascending order are 1, 3, 5, 7, 7.

B1 for identifying the largest two numbers are 7, B1 for identifying the smallest number is 1, B1 for identifying the middle number is 5, M1 for setting up the mean equation: \(\frac{1 + b + 5 + 7 + 7}{5} = 4.6\), A1 for the correct list: 1, 3, 5, 7, 7.

(a) Midpoint \( M = \left( \frac{-1+3}{2}, \frac{5+(-3)}{2} \right) = (1, 1) \).

(b) Gradient of \( CD = \frac{-3 - 5}{3 - (-1)} = \frac{-8}{4} = -2 \).

Gradient of the perpendicular bisector \( m = -\frac{1}{-2} = \frac{1}{2} \).

Using point-slope form with midpoint \( M(1, 1) \):
\( y - 1 = \frac{1}{2}(x - 1) \)
\( y = \frac{1}{2}x + \frac{1}{2} \).

(a) [2 marks]
M1 for a correct midpoint formula with substitution.
A1 for (1, 1).

(b) [3 marks]
M1 for finding the gradient of CD (-2).
M1 for finding the negative reciprocal gradient (1/2).
A1 for the correct final equation \( y = 0.5x + 0.5 \) or equivalent.

(a) Since \( AC \) is a diameter, angle \( ABC = 90^\circ \) (angle in a semicircle).
In triangle \( ABC \), angle \( ACB = 180^\circ - 90^\circ - 28^\circ = 62^\circ \).

(b) In triangle \( OBC \), \( OB = OC \) because both are radii of the circle.
Therefore, triangle \( OBC \) is an isosceles triangle, which means angle \( OBC = \text{angle } OCB = 62^\circ \).
Angle \( BOC = 180^\circ - 62^\circ - 62^\circ = 56^\circ \).

(a) [2 marks]
M1 for stating angle ABC = 90 degrees or showing \( 90 - 28 \).
A1 for 62.

(b) [3 marks]
M1 for recognizing OB = OC (isosceles triangle).
M1 for \( 180 - 2 \times \text{their } 62 \).
A1 for 56.

(a) Multiply the entire equation by the common denominator \( 3x \):
\( 3x \left( \frac{x}{3} \right) + 3x \left( \frac{12}{x} \right) = 3x (5) \)
\( x^2 + 36 = 15x \)

Rearranging terms to put all terms on one side gives:
\( x^2 - 15x + 36 = 0 \).

(b) Factorise the quadratic equation:
Find two numbers that multiply to 36 and add to -15. These are -3 and -12.
\( (x - 3)(x - 12) = 0 \)

Therefore, \( x = 3 \) or \( x = 12 \).

(a) [2 marks]
M1 for multiplying by 3x to clear fractions.
A1 for fully correct rearrangement to the given form.

(b) [3 marks]
M2 for correct factorisation to (x - 3)(x - 12) = 0 (M1 for attempting to factorise with correct signs but incorrect numbers).
A1 for x = 3 or x = 12.