2023 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

To factorise completely, find the highest common factor of the coefficients and the variables.
The highest common factor of \(12\) and \(18\) is \(6\).
The highest common factor of \(x^2y\) and \(xy^2\) is \(xy\).
Taking out the common factor \(6xy\):
\(12x^2y - 18xy^2 = 6xy(2x - 3y)\).

M1 for a correct partial factorisation, e.g. \(3xy(4x - 6y)\) or \(6x(2xy - 3y^2)\).
A1 for \(6xy(2x - 3y)\).

Substitute \(n = 8\) into the given expression for the \(n\)-th term:
\(3(8)^2 - 5(8) = 3(64) - 40\)
\(= 192 - 40 = 152\).

M1 for substituting \(n = 8\) into the formula, e.g. \(3(8)^2 - 5(8)\).
A1 for 152.

First, find the principal angle using the inverse cosine function:
\(\cos^{-1}(-0.4) \approx 113.58^\circ\) (which lies in the second quadrant).

Since we require \(\theta\) to be in the range \(180^\circ \le \theta \le 360^\circ\), the cosine function is negative in the third quadrant.
The corresponding value in this quadrant is:
\(360^\circ - 113.58^\circ = 246.42^\circ\).

Rounding to 1 decimal place gives \(246.4^\circ\).

M1 for finding the basic angle or principal angle, e.g. \(113.6^\circ\) or \(66.4^\circ\).
A1 for 246.4 (accept 246.4\(^\circ\)).

The total number of pens is \(5 + 3 = 8\).

The probability of choosing two red pens is:
\(\text{P(Red, Red)} = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\).

The probability of choosing two blue pens is:
\(\text{P(Blue, Blue)} = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\).

The probability that both pens are the same colour is:
\(\frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}\).

M1 for \(\frac{5}{8} \times \frac{4}{7}\) or \(\frac{3}{8} \times \frac{2}{7}\).
A1 for \(\frac{13}{28}\) (or equivalent fraction, or 0.464 or better).

Let \(C\) be the set of students studying Chemistry and \(P\) be the set of students studying Physics.
The total number of students studying at least one of these subjects is:
\(n(C \cup P) = 30 - 5 = 25\).

Using the set union formula:
\(n(C \cup P) = n(C) + n(P) - n(C \cap P)\)
\(25 = 18 + 15 - n(C \cap P)\)
\(25 = 33 - n(C \cap P)\)
\(n(C \cap P) = 33 - 25 = 8\).

M1 for any correct equation involving the intersection, e.g., \(18 + 15 - x = 25\) or a correctly labelled Venn diagram showing the same relationship.
A1 for 8.

The area of a sector of a circle with radius \(r\) and arc length \(s\) is given by:
\(\text{Area} = \frac{1}{2} r s\).

Substituting the given values:
\(\text{Area} = \frac{1}{2} \times 6 \times 4 = 12\text{ cm}^2\).

M1 for a correct formula or method used to find the area (e.g. finding the angle \(\theta = \frac{4}{6}\text{ radians}\) or \(\theta \approx 38.2^\circ\) first, and using it in \(\text{Area} = \frac{\theta}{360} \pi r^2\)).
A1 for 12.

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Rewrite the fraction with the factorised forms: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancel the common factor of \((2x + 1)\) from the numerator and the denominator to obtain the final simplified fraction: \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\). M1 for factorising the denominator to \((2x - 1)(2x + 1)\). M1 for identifying and cancelling the common factor of \((2x + 1)\). A1 for the correct final simplified fraction \(\frac{x - 3}{2x - 1}\).

Multiply every term in the equation by the common denominator \(x(x - 1)\) to eliminate the fractions: \(6(x - 1) - 2x = x(x - 1)\). Expand both sides of the equation: \(6x - 6 - 2x = x^2 - x\). Combine like terms on the left: \(4x - 6 = x^2 - x\). Rearrange the equation into the standard quadratic form \(ax^2 + bx + c = 0\): \(x^2 - 5x + 6 = 0\). Factorise the quadratic expression: \((x - 2)(x - 3) = 0\). Solve for \(x\) to obtain the solutions: \(x = 2\) or \(x = 3\).

M1 for clearing the fractions correctly to get \(6(x - 1) - 2x = x(x - 1)\). M1 for expanding and simplifying to a three-term quadratic equation \(x^2 - 5x + 6 = 0\). M1 for factorising into \((x - 2)(x - 3) = 0\) (or using the quadratic formula). A1 for both correct solutions: \(x = 2\) and \(x = 3\).

Multiply both sides of the formula by \((5 - p)\) to clear the fraction: \(q(5 - p) = 3p + 2\). Expand the left side: \(5q - pq = 3p + 2\). Rearrange the terms so that all terms containing \(p\) are on one side, and all other terms are on the other side: \(5q - 2 = 3p + pq\). Factorise out \(p\) from the right side: \(5q - 2 = p(3 + q)\). Divide both sides by \((3 + q)\) to isolate \(p\): \(p = \frac{5q - 2}{q + 3}\).

M1 for clearing the fraction to obtain \(q(5 - p) = 3p + 2\). M1 for expanding and isolating all terms with \(p\) on one side: \(5q - 2 = 3p + pq\) (allow sign errors). M1 for factorising the side containing \(p\): \(p(3 + q) = 5q - 2\). A1 for the correct final expression \(p = \frac{5q - 2}{q + 3}\) or equivalent (e.g. \(p = \frac{2 - 5q}{-q - 3}\)).

First, solve the corresponding quadratic equation \(3x^2 - 10x - 8 = 0\) to find the critical values. Factorising the quadratic gives \((3x + 2)(x - 4) = 0\). Therefore, the critical values are \(x = -\frac{2}{3}\) and \(x = 4\). Because the inequality is \(< 0\), we are looking for the region where the quadratic graph lies below the x-axis, which is between these two critical values. This gives the solution: \(-\frac{2}{3} < x < 4\).

M1 for factorising the quadratic expression to \((3x + 2)(x - 4)\) (or equivalent method to find roots). A1 for finding the critical values \(x = -\frac{2}{3}\) (or approximately \(-0.67\)) and \(x = 4\). M1 for identifying that the inequality holds between the critical values. A1 for the correct final inequality: \(-\frac{2}{3} < x < 4\).

Use the given terms to set up a system of linear equations in \(a\) and \(b\). For the third term (\(n = 3\)): \(T_3 = a(3)^2 + b(3) - 3 = 18 \implies 9a + 3b - 3 = 18 \implies 9a + 3b = 21\), which simplifies to \(3a + b = 7\) (Equation 1). For the fifth term (\(n = 5\)): \(T_5 = a(5)^2 + b(5) - 3 = 42 \implies 25a + 5b - 3 = 42 \implies 25a + 5b = 45\), which simplifies to \(5a + b = 9\) (Equation 2). Subtract Equation 1 from Equation 2: \((5a + b) - (3a + b) = 9 - 7 \implies 2a = 2 \implies a = 1\). Substitute \(a = 1\) back into Equation 1: \(3(1) + b = 7 \implies 3 + b = 7 \implies b = 4\).

M1 for setting up one correct equation using the term formula: e.g. \(9a + 3b = 21\) or \(25a + 5b = 45\). M1 for setting up a second correct equation and initiating a valid method to solve them simultaneously. A1 for finding \(a = 1\). A1 for finding \(b = 4\).

(a) Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substituting the given values: \(AC^2 = 120^2 + 150^2 - 2 \cdot 120 \cdot 150 \cdot \cos(72^\circ)\) which gives \(AC^2 = 14400 + 22500 - 36000 \cdot 0.309017 = 36900 - 11124.61 = 25775.39\). Thus, \(AC = \sqrt{25775.39} \approx 160.55\text{ m}\). Rounding to 3 significant figures gives \(161\text{ m}\) (or \(160.5\text{ m}\)).

(b) The area of triangle \(ABC\) is given by: \(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC) = \frac{1}{2} \cdot 120 \cdot 150 \cdot \sin(72^\circ) \approx 8559.51\text{ m}^2\). Alternatively, the area is \(\text{Area} = \frac{1}{2} \cdot AC \cdot h\), where \(h\) is the shortest distance from \(B\) to \(AC\). Equating the two expressions: \(8559.51 = \frac{1}{2} \cdot 160.55 \cdot h\), which yields \(h = \frac{2 \cdot 8559.51}{160.55} \approx 106.6\text{ m}\). Rounding to 3 significant figures gives \(107\text{ m}\) (or \(106.6\text{ m}\)).

M1 for correct substitution into the Cosine Rule: \(120^2 + 150^2 - 2(120)(150)\cos(72^\circ)\)
A1 for \(AC^2 \approx 25775\)
A1 for \(AC = 161\) or \(160.5\) to \(160.6\)
M1 for finding the area of the triangle: \(\frac{1}{2}(120)(150)\sin(72^\circ)\)
M1 for equating their area to \(\frac{1}{2} \times \text{their } AC \times h\)
A1 for \(107\) or \(106.6\) to \(106.7\)

(a) The arc length \(AB\) is: \(\text{Arc } AB = \frac{110}{360} \cdot 2\pi \cdot 12 = \frac{22}{3}\pi \approx 23.04\text{ cm}\). The perimeter of the sector is: \(\text{Perimeter} = \text{Arc } AB + 2r = 23.04 + 2(12) = 47.04\text{ cm}\). Rounding to 3 significant figures gives \(47.0\text{ cm}\).

(b) The area of the sector is: \(\text{Area of sector} = \frac{110}{360} \cdot \pi \cdot 12^2 = 44\pi \approx 138.23\text{ cm}^2\). The area of triangle \(OAB\) is: \(\text{Area of triangle} = \frac{1}{2} \cdot 12^2 \cdot \sin(110^\circ) = 72 \cdot \sin(110^\circ) \approx 67.66\text{ cm}^2\). The area of the segment is: \(\text{Area of segment} = \text{Area of sector} - \text{Area of triangle} = 138.23 - 67.66 = 70.57\text{ cm}^2\). Rounding to 3 significant figures gives \(70.6\text{ cm}^2\).

M1 for correct formula/method for arc length: \(\frac{110}{360} \times 2\pi \times 12\) (or \(23.0\))
M1 for adding \(2 \times 12\) to their arc length
A1 for \(47.0\) or \(47.03\) to \(47.04\)
M1 for area of sector \(= \frac{110}{360} \times \pi \times 12^2\) (or \(138\))
M1 for area of triangle \(= \frac{1}{2} \times 12^2 \times \sin(110^\circ)\) (or \(67.7\))
A1 for \(70.6\) or \(70.57\) to \(70.60\)

(a) First, find the length of the base diagonal \(AC\) using Pythagoras' theorem on triangle \(ABC\): \(AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100 \implies AC = 10\text{ cm}\). Next, find the length of the space diagonal \(EC\) using Pythagoras' theorem on the right-angled triangle \(EAC\): \(EC^2 = AC^2 + AE^2 = 10^2 + 15^2 = 100 + 225 = 325\). Thus, \(EC = \sqrt{325} \approx 18.028\text{ cm}\). Rounding to 3 significant figures gives \(18.0\text{ cm}\).

(b) The angle \(\theta\) that \(EC\) makes with the base \(ABCD\) is the angle \(\angle ECA\) in the right-angled triangle \(EAC\). \(\tan(\theta) = \frac{AE}{AC} = \frac{15}{10} = 1.5\). Thus, \(\theta = \tan^{-1}(1.5) \approx 56.31^\circ\). Rounding to 3 significant figures gives \(56.3^\circ\).

M1 for finding the base diagonal: \(AC = \sqrt{8^2 + 6^2}\) (or showing \(10\))
M1 for \(EC = \sqrt{\text{their } AC^2 + 15^2}\)
A1 for \(18.0\) or \(\sqrt{325}\) or \(18.02\) to \(18.03\)
M1 for recognizing the correct triangle \(EAC\) and trig ratio: e.g. \(\tan(\theta) = \frac{15}{\text{their } AC}\)
M1 for \(\theta = \tan^{-1}(1.5)\)
A1 for \(56.3^\circ\) or \(56.30^\circ\) to \(56.31^\circ\)

(a) Let us determine the angle \(\angle PQR\) between the two paths. The bearing back to \(P\) from \(Q\) is \(040^\circ + 180^\circ = 220^\circ\). The bearing from \(Q\) to \(R\) is \(130^\circ\). The angle between the line \(QP\) and \(QR\) is \(220^\circ - 130^\circ = 90^\circ\). Since \(\angle PQR = 90^\circ\), triangle \(PQR\) is a right-angled triangle. Using Pythagoras' theorem: \(PR^2 = PQ^2 + QR^2 = 12^2 + 16^2 = 144 + 256 = 400\), which gives \(PR = \sqrt{400} = 20\text{ km}\).

(b) To find the bearing of \(R\) from \(P\), we first find the angle \(\angle QPR\) inside the right-angled triangle \(PQR\): \(\tan(\angle QPR) = \frac{QR}{PQ} = \frac{16}{12} = \frac{4}{3}\), giving \(\angle QPR = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ\). Since \(R\) lies clockwise from \(Q\) as seen from \(P\), we add this angle to the initial bearing of \(040^\circ\): \(\text{Bearing} = 40^\circ + 53.13^\circ = 93.13^\circ\). Rounding to 1 decimal place (standard for bearings) gives \(093.1^\circ\) (or \(93.1^\circ\)).

M1 for calculating \(\angle PQR = 90^\circ\) (using bearings at \(Q\))
M1 for \(PR = \sqrt{12^2 + 16^2}\)
A1 for \(20\)
M1 for \(\tan(\angle QPR) = \frac{16}{12}\) (or alternative correct trig ratio)
M1 for bearing \(= 40 + \text{their } \angle QPR\)
A1 for \(093.1^\circ\) or \(93.1^\circ\) or \(93.13^\circ\)

(a) To find the local minimum, we find the first derivative of \( f(x) \):
\( f(x) = x^2 + 2x^{-1} \)
\( f'(x) = 2x - 2x^{-2} = 2x - \frac{2}{x^2} \)
Set \( f'(x) = 0 \):
\( 2x - \frac{2}{x^2} = 0 \implies 2x^3 = 2 \implies x^3 = 1 \implies x = 1 \)
Substitute \( x = 1 \) back into \( f(x) \):
\( f(1) = 1^2 + \frac{2}{1} = 3 \)
So, the coordinates of the local minimum point are \( (1, 3) \).

(b) The function \( f(x) = x^2 + \frac{2}{x} \) is undefined at \( x = 0 \). As \( x \to 0^+ \), \( f(x) \to \infty \).
Thus, the equation of the vertical asymptote is \( x = 0 \).

(c)(i) Set \( f(x) = 4 \):
\( x^2 + \frac{2}{x} = 4 \implies x^3 + 2 = 4x \implies x^3 - 4x + 2 = 0 \)
Using a graphic display calculator or numerical solver for \( x > 0 \):
One root is \( x \approx 0.53919 \), which rounds to \( 0.539 \).
The other root is \( x \approx 1.67513 \), which rounds to \( 1.68 \).
Thus, the x-coordinates of \( A \) and \( B \) are \( 0.539 \) and \( 1.68 \).

(c)(ii) The length of the line segment \( AB \) is the difference between the two x-coordinates:
\( AB = 1.67513 - 0.53919 = 1.13594 \approx 1.14 \) (to 3 s.f.).

(a) [3 Marks]
M1 for differentiating to get \( 2x - \frac{2}{x^2} \)
M1 for setting derivative to 0 and solving to get \( x = 1 \)
A1 for finding \( y = 3 \) and presenting the coordinates as \( (1, 3) \)

(b) [1 Mark]
B1 for \( x = 0 \)

(c)(i) [4 Marks]
M1 for setting up the equation \( x^2 + \frac{2}{x} = 4 \) (or equivalent cubic equation)
M1 for equating to zero and using a calculator to locate roots
A1 for \( x = 0.539 \)
A1 for \( x = 1.68 \)

(c)(ii) [2 Marks]
M1 for subtracting the smaller x-coordinate from the larger x-coordinate
A1 for \( 1.14 \) (allow \( 1.13 \) to \( 1.14 \) from correct rounded values)

(a) To find the y-intercept, substitute \( x = 0 \) into \( g(x) \):
\( g(0) = 3 + 2^{0-1} = 3 + 2^{-1} = 3 + 0.5 = 3.5 \).
So, the coordinates are \( (0, 3.5) \).

(b) As \( x \to -\infty \), the exponential term \( 2^{x-1} \to 0 \).
Thus, the function value \( g(x) \to 3 \).
So, the horizontal asymptote is \( y = 3 \).

(c)(i) Step 1: Reflecting \( y = g(x) \) in the y-axis replaces \( x \) with \( -x \):
\( y_1 = 3 + 2^{-x-1} \)
Step 2: Translating by the vector \( \begin{pmatrix} 0 \\ -3 \end{pmatrix} \) shifts the graph down by 3 units:
\( h(x) = y_1 - 3 = (3 + 2^{-x-1}) - 3 = 2^{-x-1} \).

(c)(ii) To find the intersection of the two graphs, we set \( g(x) = h(x) \):
\( 3 + 2^{x-1} = 2^{-x-1} \)
Rewrite using \( 2^x \):
\( 3 + \frac{1}{2}(2^x) = \frac{1}{2}(2^{-x}) \)
Let \( u = 2^x \):
\( 3 + \frac{u}{2} = \frac{1}{2u} \)
Multiply the entire equation by \( 2u \):
\( 6u + u^2 = 1 \implies u^2 + 6u - 1 = 0 \)
Using the quadratic formula to solve for \( u \) (noting that \( u = 2^x > 0 \)):
\( u = \frac{-6 + \sqrt{36 - 4(1)(-1)}}{2} = \frac{-6 + \sqrt{40}}{2} = -3 + \sqrt{10} \approx 0.162278 \)
Now, solve for \( x \):
\( 2^x = -3 + \sqrt{10} \implies x = \log_2(-3 + \sqrt{10}) = \frac{\ln(-3 + \sqrt{10})}{\ln 2} \approx -2.6235 \)
So, \( x \approx -2.62 \) (to 3 s.f.).
Now, find the y-coordinate using \( h(x) \):
\( y = h(-2.6235) = 2^{-(-2.6235)-1} = 2^{1.6235} \approx 3.0811 \)
So, \( y \approx 3.08 \) (to 3 s.f.).
The coordinates of the intersection point are \( (-2.62, 3.08) \).

(a) [2 Marks]
M1 for setting \( x = 0 \) and attempting to evaluate
A1 for \( (0, 3.5) \) or equivalent coordinate format

(b) [1 Mark]
B1 for \( y = 3 \)

(c)(i) [3 Marks]
M1 for substituting \( -x \) into the original equation: \( 3 + 2^{-x-1} \)
M1 for subtracting 3 from their reflected function
A1 for the fully simplified expression \( h(x) = 2^{-x-1} \) (or equivalent, such as \( 0.5 \cdot 2^{-x} \))

(c)(ii) [4 Marks]
M1 for equating functions: \( 3 + 2^{x-1} = 2^{-x-1} \)
M1 for transforming this into a quadratic equation in terms of \( 2^x \) (or directly using GDC to find intersections)
A1 for finding the x-coordinate: \( x \approx -2.62 \)
A1 for finding the y-coordinate: \( y \approx 3.08 \) (accept coordinates written together as \( (-2.62, 3.08) \))

(a) Let \( y = \frac{3x - 1}{x + 2} \).
Multiply both sides by \( x + 2 \):
\( y(x + 2) = 3x - 1 \implies yx + 2y = 3x - 1 \)
Rearrange to isolate the terms with \( x \) on one side:
\( 2y + 1 = 3x - yx \implies x(3 - y) = 2y + 1 \)
Divide by \( 3 - y \):
\( x = \frac{2y + 1}{3 - y} \)
Replacing \( y \) with \( x \) gives:
\( f^{-1}(x) = \frac{2x + 1}{3 - x} \).

(b) To find \( f(g(x)) \), substitute \( g(x) = 2x - 3 \) into \( f(x) \):
\( f(g(x)) = \frac{3(2x - 3) - 1}{(2x - 3) + 2} \)
Simplify the numerator:
\( 3(2x - 3) - 1 = 6x - 9 - 1 = 6x - 10 \)
Simplify the denominator:
\( (2x - 3) + 2 = 2x - 1 \)
Thus:
\( f(g(x)) = \frac{6x - 10}{2x - 1} \).

(c) Set up the inequality:
\( \frac{3x - 1}{x + 2} \le 2x - 3 \)
To find the critical values, we solve the equation:
\( \frac{3x - 1}{x + 2} = 2x - 3 \)
For \( x \neq -2 \), multiply both sides by \( x + 2 \):
\( 3x - 1 = (2x - 3)(x + 2) \implies 3x - 1 = 2x^2 + x - 6 \)
Rearrange into a quadratic equation:
\( 2x^2 - 2x - 5 = 0 \)
Using the quadratic formula:
\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-5)}}{2(2)} = \frac{2 \pm \sqrt{4 + 40}}{4} = \frac{2 \pm \sqrt{44}}{4} = \frac{1 \pm \sqrt{11}}{2} \)
Evaluating the critical boundaries in decimals:
\( x_1 = \frac{1 - \sqrt{11}}{2} \approx -1.158 \approx -1.16 \) (to 3 s.f.)
\( x_2 = \frac{1 + \sqrt{11}}{2} \approx 2.158 \approx 2.16 \) (to 3 s.f.)
We also have a vertical asymptote (discontinuity) at \( x = -2 \).
Testing points in the intervals:
- For \( x < -2 \) (e.g., \( x = -3 \)): \( f(-3) = 10 \) and \( g(-3) = -9 \). Since \( 10 \le -9 \) is false, this interval is not part of the solution.
- For \( -2 < x \le -1.16 \) (e.g., \( x = -1.5 \)): \( f(-1.5) = -11 \) and \( g(-1.5) = -6 \). Since \( -11 \le -6 \) is true, this is a solution interval.
- For \( -1.16 < x < 2.16 \) (e.g., \( x = 0 \)): \( f(0) = -0.5 \) and \( g(0) = -3 \). Since \( -0.5 \le -3 \) is false, this is not a solution interval.
- For \( x \ge 2.16 \) (e.g., \( x = 3 \)): \( f(3) = 1.6 \) and \( g(3) = 3 \). Since \( 1.6 \le 3 \) is true, this is a solution interval.

So the solution set is:
\( -2 < x \le -1.16 \) or \( x \ge 2.16 \).

(a) [3 Marks]
M1 for attempting to swap variables and clear denominator: \( y(x+2) = 3x-1 \)
M1 for isolating the variable \( x \)
A1 for the correct inverse formula: \( f^{-1}(x) = \frac{2x + 1}{3 - x} \)

(b) [3 Marks]
M1 for substituting \( 2x-3 \) into \( f(x) \)
M1 for expanding and simplifying numerator or denominator correctly
A1 for the final simplified fraction: \( \frac{6x - 10}{2x - 1} \)

(c) [4 Marks]
M1 for setting up the equation and obtaining the quadratic equation \( 2x^2 - 2x - 5 = 0 \)
A1 for finding the critical boundary roots: \( x \approx -1.16 \) and \( x \approx 2.16 \)
B1 for identifying the asymptote boundary at \( x = -2 \)
A1 for the correct final solution intervals: \( -2 < x \le -1.16 \) or \( x \ge 2.16 \) (allow equivalent exact expressions using \( \frac{1 \pm \sqrt{11}}{2} \))

(a) (i) The total number of balls in the bag is \(x + 4\).
Probability of drawing a blue ball first is \(\frac{x}{x+4}\).
Probability of drawing a blue ball second is \(\frac{x-1}{x+3}\).
Multiplying these gives:
\text{P(both blue)} = \frac{x(x-1)}{(x+4)(x+3)}

(ii) One of each colour can happen in two ways: Blue then Red (BR) or Red then Blue (RB).
\text{P(BR)} = \frac{x}{x+4} \times \frac{4}{x+3} = \frac{4x}{(x+4)(x+3)}
\text{P(RB)} = \frac{4}{x+4} \times \frac{x}{x+3} = \frac{4x}{(x+4)(x+3)}
Adding these together:
\text{P(one of each)} = \frac{4x}{(x+4)(x+3)} + \frac{4x}{(x+4)(x+3)} = \frac{8x}{(x+4)(x+3)}

(b) (i) We are given:
\frac{x(x-1)}{(x+4)(x+3)} = \frac{1}{3}
Multiplying both sides by \(3(x+4)(x+3)\):
3x(x-1) = (x+4)(x+3)
3x^2 - 3x = x^2 + 7x + 12
Subtracting \(x^2 + 7x + 12\) from both sides:
2x^2 - 10x - 12 = 0
Dividing the entire equation by 2:
x^2 - 5x - 6 = 0 (as required)

(ii) Factorising the quadratic equation:
(x - 6)(x + 1) = 0
This gives \(x = 6\) or \(x = -1\).
Since the number of balls \(x\) must be positive, we have \(x = 6\).

(c) If \(x = 6\), there are 6 blue balls and 4 red balls, giving a total of 10 balls.
\text{P(both same colour)} = \text{P(both blue)} + \text{P(both red)}
We know \text{P(both blue)} = \frac{1}{3} = \frac{30}{90}.
\text{P(both red)} = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}.
\text{P(both same colour)} = \frac{1}{3} + \frac{2}{15} = \frac{5}{15} + \frac{2}{15} = \frac{7}{15} \approx 0.467.

(a) (i) M1 for numerator \(x(x-1)\) or denominator \((x+4)(x+3)\) seen in a fraction. A1 for correct final expression: \(\frac{x(x-1)}{(x+4)(x+3)}\) or \(\frac{x^2-x}{x^2+7x+12}\).
(ii) M1 for identifying BR and RB both needed. M1 for adding \(\frac{4x}{(x+4)(x+3)}\) and \(\frac{4x}{(x+4)(x+3)}\). A1 for correct final expression: \(\frac{8x}{(x+4)(x+3)}\) or \(\frac{8x}{x^2+7x+12}\).

(b) (i) M1 for setting their expression in (a)(i) equal to \(\frac{1}{3}\). M1 for expanding correctly to \(3x^2 - 3x = x^2 + 7x + 12\). A1 for completing show-that correctly with no errors to obtain \(x^2 - 5x - 6 = 0\).
(ii) M1 for factorisation \((x-6)(x+1)\) or using formula. A1 for \(x = 6\) (must reject or not include \(x = -1\) in final answer).

(c) M1 for finding \(\text{P(both red)} = \frac{4}{10} \times \frac{3}{9}\) or using \(1 - \text{P(one of each)}\) with \(x = 6\). A1 for \(\frac{7}{15}\) (or 0.467 or equivalent fraction/decimal).

(a) To estimate the mean, we find the midpoints of each class interval:
- For \(0 < t \le 10\): Midpoint = 5. \(f \times x = 15 \times 5 = 75\)
- For \(10 < t \le 25\): Midpoint = 17.5. \(f \times x = 45 \times 17.5 = 787.5\)
- For \(25 < t \le 30\): Midpoint = 27.5. \(f \times x = 24 \times 27.5 = 660\)
- For \(30 < t \le 50\): Midpoint = 40. \(f \times x = 36 \times 40 = 1440\)

Sum of \(f \times x = 75 + 787.5 + 660 + 1440 = 2962.5\)
Total frequency = 120
\text{Estimated Mean} = \frac{2962.5}{120} = 24.6875 \text{ minutes} (or 24.7 to 3 s.f.)

(b) Frequency density is calculated as:
\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}

(i) For \(10 < t \le 25\): Class width = 15, Frequency = 45.
\text{Frequency Density} = \frac{45}{15} = 3

(ii) For \(25 < t \le 30\): Class width = 5, Frequency = 24.
\text{Frequency Density} = \frac{24}{5} = 4.8

(iii) For \(30 < t \le 50\): Class width = 20, Frequency = 36.
\text{Frequency Density} = \frac{36}{20} = 1.8

(c) The total number of students who took more than 10 minutes is:
45 (from \(10 < t \le 25\)) + 24 (from \(25 < t \le 30\)) + 36 (from \(30 < t \le 50\)) = 105 students.

Within this group of 105 students:
- Those who took more than 30 minutes = 36 students.
- Those who took 30 minutes or less = 45 + 24 = 69 students.

We select two students without replacement.
The probability that one took > 30 mins and the other took \(\le\) 30 mins is:
\text{P}( (>30, \le30) \text{ or } (\le30, >30) )
= \left(\frac{36}{105} \times \frac{69}{104}\right) + \left(\frac{69}{105} \times \frac{36}{104}\right)
= 2 \times \frac{2484}{10920} = \frac{4968}{10920} = \frac{207}{455} \approx 0.455.

(a) M1 for finding at least 3 correct midpoints (5, 17.5, 27.5, 40). M1 for correct calculation of sum of products \(f \times x\) (2962.5). M1 for dividing their sum by 120. A1 for 24.6875 or 24.7.

(b) (i) M1 for dividing 45 by 15. A1 for 3.
(ii) M1 for dividing 24 by 5. A1 for 4.8.
(iii) M1 for dividing 36 by 20. A1 for 1.8.

(c) M1 for identifying denominator 105 and setting up sum of two products: \(\frac{36}{105} \times \frac{69}{104} + \frac{69}{105} \times \frac{36}{104}\) (or equivalent). A1 for \(\frac{207}{455}\) or 0.455 (or 45.5%).