2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Rearranging \(3y - 2x = 9\) into slope-intercept form gives \(3y = 2x + 9\), which simplifies to \(y = \frac{2}{3}x + 3\). The gradient of this line is \(m_1 = \frac{2}{3}\). The gradient of the perpendicular line, \(m_2\), satisfies \(m_1 \times m_2 = -1\), so \(m_2 = -\frac{3}{2} = -1.5\). Using the point-gradient formula with the point \((4, -1)\): \(y - (-1) = -1.5(x - 4)\), which simplifies to \(y + 1 = -1.5x + 6\), and thus \(y = -1.5x + 5\).

M1 for finding the gradient of the perpendicular line is \(-1.5\) or \(-\frac{3}{2}\) (or for using the relation \(m_1 \times m_2 = -1\)). A1 for the correct equation \(y = -1.5x + 5\) or equivalent.

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). Substituting the given values: \(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(43^\circ)\). This gives \(AC^2 = 49 + 81 - 126\cos(43^\circ) \approx 130 - 92.1506 = 37.8494\). Taking the square root: \(AC \approx \sqrt{37.8494} \approx 6.152\text{ cm}\). Correct to 3 significant figures, the length of \(AC\) is \(6.15\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7^2 + 9^2 - 2(7)(9)\cos(43^\circ)\). A1 for \(6.15\) (accept answers in the range \(6.15\) to \(6.152\)).

Multiply the first equation by 2 to align the \(y\)-coefficients: \(10x - 4y = 38\). Now, add this equation to the second equation: \((10x - 4y) + (3x + 4y) = 38 + 1 \implies 13x = 39 \implies x = 3\). Substitute \(x = 3\) back into the first equation to find \(y\): \(5(3) - 2y = 19 \implies 15 - 2y = 19 \implies -2y = 4 \implies y = -2\). The solution is therefore \(x = 3, y = -2\).

M1 for a correct method to eliminate one variable (e.g., multiplying the first equation by 2 to get \(10x - 4y = 38\) and adding). A1 for both \(x = 3\) and \(y = -2\) correct.

Multiply the entire equation by the common denominator \(x(x-1)\) to clear the fractions:
\(6x - 4(x-1) = x(x-1)\)

Expand the terms:
\(6x - 4x + 4 = x^2 - x\)

Rearrange into a standard quadratic equation:
\(2x + 4 = x^2 - x\)
\(x^2 - 3x - 4 = 0\)

Factorise the quadratic expression:
\((x-4)(x+1) = 0\)

This gives \(x = 4\) or \(x = -1\).

Since we are looking for the positive value of \(x\), the solution is \(x = 4\).

M1 for multiplying by \(x(x-1)\) and expanding correctly to form a quadratic equation of the form \(x^2 - 3x - 4 = 0\) or equivalent.
A1 for 4 (with any negative solution discarded or identified as extra).

The area of a non-right-angled triangle is given by:
\(\text{Area} = \frac{1}{2} a c \sin B\)

Substitute the known values into the formula:
\(30 = \frac{1}{2} \times 11 \times 8 \times \sin B\)
\(30 = 44 \sin B\)
\(\sin B = \frac{30}{44} = \frac{15}{22}\)

Find the principal (acute) angle:
\(B \approx \sin^{-1}\left(\frac{15}{22}\right) \approx 43.00^{\circ}\)

Since angle \(ABC\) is given as obtuse (between \(90^{\circ}\) and \(180^{\circ}\)):
\(B = 180^{\circ} - 42.9967^{\circ} = 137.003^{\circ}\)

To 1 decimal place, the angle is \(137.0^{\circ}\).

M1 for setting up the area equation: \(\frac{1}{2} \times 8 \times 11 \times \sin B = 30\) or obtaining \(\sin B = \frac{15}{22}\) (or approximately \(0.682\)) or the acute angle \(43.0^{\circ}\).
A1 for \(137.0\) or \(137\).

The gradient of the given line \(y = 4x - 7\) is \(m_1 = 4\).

The gradient of a line perpendicular to it, \(m_2\), satisfies \(m_1 \times m_2 = -1\):
\(m_2 = -\frac{1}{4} = -0.25\)

Substitute the gradient \(m = -0.25\) and the point \((8, 5)\) into the equation of a straight line, \(y = mx + c\):
\(5 = -0.25(8) + c\)
\(5 = -2 + c\)
\(c = 7\)

Therefore, the equation of the perpendicular line is \(y = -0.25x + 7\) or \(y = -\frac{1}{4}x + 7\).

M1 for finding the perpendicular gradient of \(-\frac{1}{4}\) or \(-0.25\).
A1 for \(y = -0.25x + 7\) or \(y = -\frac{1}{4}x + 7\) (or any equivalent simplified form).

The gradient of the given line is \(-\frac{1}{3}\). Since the perpendicular line has a gradient \(m\) such that \(m \times \left(-\frac{1}{3}\right) = -1\), we find \(m = 3\). Using the equation of a straight line passing through \((4, -1)\) with gradient \(3\): \(y - (-1) = 3(x - 4) \implies y + 1 = 3x - 12 \implies y = 3x - 13\).

M1 for identifying the perpendicular gradient is 3 (or showing \(m \times -1/3 = -1\)). A1 for the correct equation \(y = 3x - 13\).

Using the cosine rule: \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\). Substituting the values: \(BC^2 = 8.4^2 + 6.5^2 - 2(8.4)(6.5)\cos(52^\circ) = 70.56 + 42.25 - 109.2\cos(52^\circ) \approx 112.81 - 67.2302 = 45.5798\). Taking the square root: \(BC = \sqrt{45.5798} \approx 6.7513\text{ cm}\). To 3 significant figures, \(BC = 6.75\text{ cm}\).

M1 for a correct substitution into the cosine rule: \(8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(52^\circ)\). A1 for \(6.75\) or \(6.751...\)

Let \(W_1\) be the event of winning the first stage, and \(W_2\) be the event of winning the second stage. We have \(P(W_1) = 0.4\) and \(P(\text{not } W_1) = 0.6\). The probability of winning both stages is \(P(W_1 \cap W_2) = 0.4 \times 0.7 = 0.28\). The probability of losing the first stage and winning the second stage is \(P(\text{not } W_1 \cap W_2) = 0.6 \times 0.3 = 0.18\). The total probability of winning the second stage is \(P(W_2) = 0.28 + 0.18 = 0.46\). The required conditional probability is \(P(W_1 | W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)} = \frac{0.28}{0.46} = \frac{14}{23}\).

M1 for finding the total probability of winning the second stage: \(0.4 \times 0.7 + 0.6 \times 0.3\) (or \(0.46\)), or for the correct conditional probability expression. A1 for \(14/23\) (or equivalent fraction).

First, rearrange the equation of the given line into gradient-intercept form, \(y = mx + c\): \(3x - 2y = 8\) becomes \(2y = 3x - 8\), which simplifies to \(y = \frac{3}{2}x - 4\). The gradient of this line is \(m_1 = \frac{3}{2}\). The gradient of a perpendicular line, \(m_2\), satisfies the condition \(m_1 \times m_2 = -1\). Therefore, \(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\).

M1 for rearranging the equation to find the gradient of the given line, e.g. \(y = \frac{3}{2}x - 4\) or gradient \(= 1.5\). A1 for \(-\frac{2}{3}\) or \(-0.667\) or \(-0.67\).

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\). Substitute the given values: \(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\). Calculate the terms: \(AC^2 = 49 + 81 - 126 \times 0.5 = 130 - 63 = 67\). Thus, \(AC = \sqrt{67} \approx 8.19\text{ cm}\) (to 3 significant figures).

M1 for correct substitution into the Cosine Rule: \(7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\). A1 for \(8.19\) or \(8.185\dots\) or \(\sqrt{67}\).

Cross-multiply to remove the denominators: \(5(x+4) = 3(x-2)\). Expand both sides: \(5x + 20 = 3x - 6\). Rearrange the equation to solve for \(x\): \(5x - 3x = -6 - 20\), which simplifies to \(2x = -26\). Dividing both sides by 2 gives \(x = -13\).

M1 for \(5(x+4) = 3(x-2)\) or \(5x + 20 = 3x - 6\) (or equivalent algebraic step). A1 for \(-13\).

We start with the given equation:

\(\frac{x + 2}{3} = \frac{5}{x}\)

Multiply both sides by \(3x\) to clear the fractions:

\(x(x + 2) = 3 \times 5\)

\(x^2 + 2x = 15\)

Rearrange the terms into a quadratic equation form:

\(x^2 + 2x - 15 = 0\)

Factorise the quadratic expression:

\((x + 5)(x - 3) = 0\)

This gives the two potential solutions:

\(x = -5\) or \(x = 3\)

Since the question specifies that \(x > 0\), we reject \(x = -5\).

Therefore, the only valid solution is \(x = 3\).

M1 for obtaining a correct quadratic equation, e.g. \(x^2 + 2x - 15 = 0\) (or equivalent)
A1 for \(x = 3\) (the negative root must be excluded)

The gradient of the given line \(y = 3x - 5\) is \(m_1 = 3\).

The gradient \(m_2\) of the perpendicular line satisfies the relation:

\(m_1 \times m_2 = -1\)

\(3 \times m_2 = -1 \implies m_2 = -\frac{1}{3}\)

Now, we find the equation of the line with gradient \(-\frac{1}{3}\) passing through the point \((6, 2)\):

\(y - y_1 = m(x - x_1)\)

\(y - 2 = -\frac{1}{3}(x - 6)\)

\(y - 2 = -\frac{1}{3}x + 2\)

\(y = -\frac{1}{3}x + 4\)

M1 for finding the gradient of the perpendicular line to be \(-\frac{1}{3}\) (or equivalent)
A1 for \(y = -\frac{1}{3}x + 4\) (or any equivalent form with \(y\) as the subject)

To find the length of the side opposite the given angle, we use the Cosine Rule:

\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)\)

Substitute the given values into the formula:

\(AC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(60^\circ)\)

Recall that \( \cos(60^\circ) = 0.5 \):

\(AC^2 = 49 + 81 - 126 \times 0.5\)

\(AC^2 = 130 - 63 = 67\)

Now, calculate the square root of 67:

\(AC = \sqrt{67} \approx 8.18535\text{ cm}\)

Rounding to 3 significant figures, we get \(AC = 8.19\text{ cm}\).

M1 for a correct substitution into the Cosine Rule formula, e.g. \(7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(60^\circ)\)
A1 for \(8.19\) (accept answers in the range \(8.185\) to \(8.19\))

To find the distance \(AC\), we first need to determine the interior angle \(\angle ABC\) of triangle \(ABC\).

1. Find the bearing of \(A\) from \(B\):
\(\text{Bearing of } A \text{ from } B = 040^\circ + 180^\circ = 220^\circ\).

2. Find the interior angle \(\angle ABC\):
Since the bearing of \(C\) from \(B\) is \(115^\circ\), the angle between the line \(BA\) (at bearing \(220^\circ\)) and the line \(BC\) (at bearing \(115^\circ\)) is:
\(\angle ABC = 220^\circ - 115^\circ = 105^\circ\).

3. Apply the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(AC^2 = 7^2 + 9^2 - 2 \cdot 7 \cdot 9 \cdot \cos(105^\circ)\)
\(AC^2 = 49 + 81 - 126 \cdot (-0.258819)\)
\(AC^2 = 130 + 32.611 = 162.611\)
\(AC = \sqrt{162.611} \approx 12.75\text{ km}\).

Rounded to 3 significant figures, the distance is \(12.8\text{ km}\).

M1: For correctly identifying the interior angle \(\angle ABC = 105^\circ\) (or showing a complete valid method to find it).
M1: For substitution of their angle and given lengths into the Cosine Rule: \(7^2 + 9^2 - 2(7)(9)\cos(\theta)\).
A1: For evaluating \(AC^2 \approx 162.6\) or showing \(AC = \sqrt{162.6}\).
A1: For the final answer \(12.8\) (accept answers in the range \(12.75\) to \(12.8\)).

Let \(R_1\) and \(R_2\) represent selecting a red pen first and second, respectively. Let \(B_1\) and \(B_2\) represent selecting a blue pen first and second, respectively.

We need to find the conditional probability \(P(\text{at least one red} \mid B_2)\).

1. Find the probability of outcomes where the second pen is blue:
- First red, second blue:
\(P(R_1 \cap B_2) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)
- First blue, second blue:
\(P(B_1 \cap B_2) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\)

2. Find the total probability that the second pen is blue:
\(P(B_2) = P(R_1 \cap B_2) + P(B_1 \cap B_2) = \frac{35}{132} + \frac{42}{132} = \frac{77}{132}\)

3. Find the probability that at least one pen is red AND the second pen is blue:
This only occurs in the scenario \(R_1 \cap B_2\), which has a probability of \(\frac{35}{132}\).

4. Calculate the conditional probability:
\(P(\text{at least one red} \mid B_2) = \frac{P(R_1 \cap B_2)}{P(B_2)} = \frac{35/132}{77/132} = \frac{35}{77} = \frac{5}{11}\).

This can also be written as a decimal, \(0.455\) (to 3 s.f.).

M1: For calculating \(P(R_1 \cap B_2) = \frac{35}{132}\) (or equivalent).
M1: For calculating \(P(B_1 \cap B_2) = \frac{42}{132}\) (or equivalent).
M1: For dividing their \(P(R_1 \cap B_2)\) by their total \(P(B_2)\) where \(P(B_2) = P(R_1 \cap B_2) + P(B_1 \cap B_2)\).
A1: For the final answer \(5/11\) (or exact decimal equivalent \(0.\dot{4}\dot{5}\), or \(0.455\) correct to 3 s.f.).

To find the coordinates of point \(C\), we first need to determine the equation of the perpendicular bisector of \(AB\).

1. Find the midpoint, \(M\), of line segment \(AB\):
\(M = \left(\frac{2 + 6}{2}, \frac{5 + (-3)}{2}\right) = (4, 1)\).

2. Find the gradient of line segment \(AB\):
\(m_{AB} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2\).

3. Find the gradient, \(m\), of the perpendicular bisector:
Since the lines are perpendicular, \(m = -\frac{1}{m_{AB}} = -\frac{1}{-2} = \frac{1}{2}\).

4. Determine the equation of the perpendicular bisector:
Using the point-slope form with \(M(4, 1)\):
\(y - 1 = \frac{1}{2}(x - 4)\)
\(y - 1 = \frac{1}{2}x - 2\)
\(y = \frac{1}{2}x - 1\).

5. Find the coordinates of the \(y\)-intercept, point \(C\):
Since \(C\) lies on the \(y\)-axis, its \(x\)-coordinate is 0.
\(y = \frac{1}{2}(0) - 1 = -1\).

Thus, the coordinates of point \(C\) are \((0, -1)\).

M1: For finding the correct midpoint of \(AB\) as \((4, 1)\).
M1: For finding the gradient of \(AB\) as \(-2\) and stating that the perpendicular gradient is \(1/2\).
M1: For substituting their midpoint and perpendicular gradient into a linear equation to find the equation of the bisector (e.g. \(y - 1 = 0.5(x - 4)\)).
A1: For the final coordinates \((0, -1)\) (accept \(x = 0, y = -1\)).

1. **Determine the interior angle \(\angle ABC\):**
The bearing from \(A\) to \(B\) is \(054^\circ\), so the back-bearing from \(B\) to \(A\) is \(054^\circ + 180^\circ = 234^\circ\).
The bearing from \(B\) to \(C\) is \(142^\circ\).
Therefore, the interior angle \(\angle ABC = 234^\circ - 142^\circ = 92^\circ\).

2. **Apply the Cosine Rule to find \(AC\):**
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 7.2^2 + 5.5^2 - 2(7.2)(5.5)\cos(92^\circ)\)
\(AC^2 = 51.84 + 30.25 - 79.2 \cos(92^\circ)\)
\(AC^2 \approx 82.09 - 79.2(-0.03490)\)
\(AC^2 \approx 82.09 + 2.764 \approx 84.854\)
\(AC = \sqrt{84.854} \approx 9.2116\text{ km}\).

3. **Rounding:**
Correct to 3 significant figures, the distance is \(9.21\text{ km}\).

M1 for finding the interior angle \(\angle ABC = 92^\circ\) (or equivalent angle calculation).
M1 for correct substitution into the Cosine Rule formula: \(7.2^2 + 5.5^2 - 2(7.2)(5.5)\cos(92^\circ)\).
A1 for \(AC^2 \approx 84.9\) or \(AC \approx 9.211...\)
A1 for final answer \(9.21\) (accept answers in the range \(9.21\) to \(9.22\)).

1. **Find the coordinates of the midpoint, \(M\), of the line segment \(PQ\):**
\(M = \left( \frac{1 + 5}{2}, \frac{2 + 10}{2} \right) = (3, 6)\).

2. **Find the gradient of the line \(PQ\), \(m_{PQ}\):**
\(m_{PQ} = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2\).

3. **Find the gradient of the perpendicular bisector, \(m_{\perp}\):**
\(m_{\perp} = -\frac{1}{m_{PQ}} = -\frac{1}{2} = -0.5\).

4. **Find the equation of the line passing through \(M(3, 6)\) with gradient \(-0.5\):**
\(y - 6 = -0.5(x - 3)\)
\(y - 6 = -0.5x + 1.5\)
\(y = -0.5x + 7.5\).

M1 for finding the midpoint of \(PQ\) as \((3, 6)\).
M1 for finding the gradient of \(PQ\) as \(2\).
M1 for finding the gradient of the perpendicular bisector as \(-0.5\) (or \(-1/2\)) using the product of perpendicular gradients rule.
A1 for the correct equation \(y = -0.5x + 7.5\) (or equivalent fraction form, e.g., \(y = -\frac{1}{2}x + \frac{15}{2}\)).

Let \(R_1\) and \(B_1\) represent selecting a red or blue pen first, respectively.
Let \(R_2\) and \(B_2\) represent selecting a red or blue pen second, respectively.

1. **Calculate the probability of selecting pens of different colours (event \(D\)):**
This can happen in two ways:
- Red first, then Blue:
\(P(R_1 \cap B_2) = \frac{8}{13} \times \frac{5}{12} = \frac{40}{156}\)
- Blue first, then Red:
\(P(B_1 \cap R_2) = \frac{5}{13} \times \frac{8}{12} = \frac{40}{156}\)

Total probability of different colours:
\(P(D) = P(R_1 \cap B_2) + P(B_1 \cap R_2) = \frac{40}{156} + \frac{40}{156} = \frac{80}{156}\).

2. **Find the probability that the second pen is red AND they are of different colours:**
This corresponds only to the sequence Blue then Red:
\(P(R_2 \cap D) = P(B_1 \cap R_2) = \frac{40}{156}\).

3. **Apply the conditional probability formula:**
\(P(R_2 \mid D) = \frac{P(R_2 \cap D)}{P(D)} = \frac{\frac{40}{156}}{\frac{80}{156}} = \frac{40}{80} = 0.5\).

M1 for calculating \(P(\text{Red first, Blue second}) = \frac{8}{13} \times \frac{5}{12} = \frac{40}{156}\) (or equivalent decimal \(\approx 0.256\)).
M1 for calculating \(P(\text{Blue first, Red second}) = \frac{5}{13} \times \frac{8}{12} = \frac{40}{156}\) (or equivalent decimal \(\approx 0.256\)).
M1 for setting up the correct conditional probability fraction: \(\frac{\text{probability of blue-red}}{\text{probability of blue-red + probability of red-blue}}\).
A1 for \(0.5\) (or \(\frac{1}{2}\)).

To find the equation of the perpendicular bisector:

1. **Find the midpoint of \(AB\):**
\[ M = \left(\frac{2 + 8}{2}, \frac{-3 + 5}{2}\right) = (5, 1) \]

2. **Find the gradient of the line segment \(AB\):**
\[ m = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3} \]

3. **Find the gradient of the perpendicular bisector:**
The perpendicular gradient \(m_{\perp}\) is the negative reciprocal of \(m\):
\[ m_{\perp} = -\frac{3}{4} \]

4. **Find the equation of the perpendicular bisector:**
Using the point-gradient formula with the midpoint \((5, 1)\):
\[ y - 1 = -\frac{3}{4}(x - 5) \]
Multiply both sides by 4:
\[ 4(y - 1) = -3(x - 5) \]
\[ 4y - 4 = -3x + 15 \]
Rearrange into the form \(ay + bx = c\):
\[ 4y + 3x = 19 \]

- **M1**: For finding the midpoint of \(AB\) as \((5, 1)\).
- **M1**: For finding the gradient of \(AB\) as \(\frac{4}{3}\).
- **M1**: For using the perpendicular gradient \(-\frac{3}{4}\) in an equation of a line passing through their midpoint.
- **A1**: For the correct final equation \(4y + 3x = 19\) or \(3x + 4y = 19\) (or any integer multiple if specified as equivalent, but must be simplified).

1. **Find the angle \(\angle PQR\) using the area formula:**
\[ \text{Area} = \frac{1}{2} a c \sin B \]
\[ 30 = \frac{1}{2} \times 8 \times 12 \times \sin(\angle PQR) \]
\[ 30 = 48 \sin(\angle PQR) \]
\[ \sin(\angle PQR) = \frac{30}{48} = 0.625 \]

2. **Find the obtuse angle \(\angle PQR\):**
The acute angle is \(\arcsin(0.625) \approx 38.682^\circ\).
Since \(\angle PQR\) is obtuse:
\[ \angle PQR = 180^\circ - 38.682^\circ = 141.318^\circ \]

3. **Use the Cosine Rule to find \(PR\):**
\[ PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR) \]
\[ PR^2 = 8^2 + 12^2 - 2 \times 8 \times 12 \times \cos(141.318^\circ) \]
\[ PR^2 = 64 + 144 - 192 \times (-0.780625) \]
\[ PR^2 = 208 + 149.88 = 357.88 \]
\[ PR = \sqrt{357.88} \approx 18.918\text{ cm} \]

To 3 significant figures, the length is \(18.9\text{ cm}\).

- **M1**: For setting up the area equation \(\frac{1}{2} \times 8 \times 12 \times \sin(\theta) = 30\) to find \(\sin(\theta)\).
- **A1**: For finding the correct obtuse angle \(\theta \approx 141.3^\circ\) (or \(141.318^\circ\)), or for finding the correct exact value of \(\cos(\theta) = -\frac{\sqrt{39}}{8}\).
- **M1**: For applying the Cosine Rule correctly using their angle: \(PR^2 = 8^2 + 12^2 - 2(8)(12)\cos(\theta)\).
- **A1**: For the final answer of \(18.9\) (accept \(18.92\) or \(18.9\) from correct working).

Let \(R_1\) and \(R_2\) be the events of selecting a red pen on the first and second draw, and \(B_1\) and \(B_2\) be the events of selecting a blue pen on the first and second draw.

1. **Calculate the probability of drawing two red pens:**
\[ P(\text{both red}) = P(R_1 R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \]

2. **Calculate the probability of drawing two blue pens:**
\[ P(\text{both blue}) = P(B_1 B_2) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28} \]

3. **Calculate the probability that at least one pen is red:**
This is the complement of drawing two blue pens:
\[ P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - \frac{6}{56} = \frac{50}{56} = \frac{25}{28} \]

4. **Apply the conditional probability formula:**
We want \(P(\text{both red} \mid \text{at least one red})\):
\[ P(\text{both red} \mid \text{at least one red}) = \frac{P(\text{both red} \cap \text{at least one red})}{P(\text{at least one red})} \]
Since "both red" is a subset of "at least one red", the numerator is simply \(P(\text{both red})\):
\[ P(\text{both red} \mid \text{at least one red}) = \frac{\frac{20}{56}}{\frac{50}{56}} = \frac{20}{50} = \frac{2}{5} = 0.4 \]

- **M1**: For calculating the probability of choosing two red pens as \(\frac{20}{56}\) (or equivalent).
- **M1**: For finding the probability of at least one red pen as \(1 - P(BB) = \frac{50}{56}\) (or equivalent, e.g., by adding \(P(RR) + P(RB) + P(BR)\)).
- **M1**: For setting up the division of the two relevant probabilities: \(\frac{20/56}{50/56}\) or equivalent.
- **A1**: For the correct final answer of \(0.4\) or \(\frac{2}{5}\).

To find the equation of the perpendicular bisector of \(AB\): 1. Find the midpoint of \(AB\): The midpoint, \(M\), is given by \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 + 5}{2}, \frac{2 + 10}{2}\right) = (3, 6)\). 2. Find the gradient of the line \(AB\): The gradient of \(AB\), \(m_{AB}\), is \(m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2\). 3. Find the gradient of the perpendicular bisector: Since the line is perpendicular to \(AB\), its gradient \(m\) satisfies \(m \times m_{AB} = -1\), which gives \(m = -\frac{1}{2} = -0.5\). 4. Find the equation of the line: Using the point-slope form with midpoint \((3, 6)\) and gradient \(-0.5\): \(y - 6 = -0.5(x - 3)\), which simplifies to \(y - 6 = -0.5x + 1.5\), and finally \(y = -0.5x + 7.5\).

M1 for finding the midpoint of \(AB\) as \((3, 6)\) (or correct working shown). M1 for finding the gradient of \(AB\) as \(2\) (or correct working shown). M1 for finding the perpendicular gradient as \(-0.5\) (or \(-\frac{1}{\text{their } m_{AB}}\)). A1 for the correct equation \(y = -0.5x + 7.5\) (or equivalent, e.g., \(y = -\frac{1}{2}x + \frac{15}{2}\)).

(a) Using the Cosine Rule to find \(AC\):
\[AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\]
\[AC^2 = 120^2 + 85^2 - 2(120)(85) \cdot \cos(112^\circ)\]
\[AC^2 = 14400 + 7225 - 20400 \cdot (-0.3746)\]
\[AC^2 \approx 21625 + 7641.97 = 29266.97\]
\[AC = \sqrt{29266.97} \approx 171.08\text{ m} \approx 171\text{ m} \text{ (to 3 s.f.)}\]

(b) Using the Sine Rule to find angle \(ACB\):
\[\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}\]
\[\frac{\sin(ACB)}{120} = \frac{\sin(112^\circ)}{171.08}\]
\[\sin(ACB) = \frac{120 \cdot \sin(112^\circ)}{171.08} \approx 0.65038\]
\[ACB = \sin^{-1}(0.65038) \approx 40.57^\circ \approx 40.6^\circ \text{ (to 1 d.p.)}\]

(c) Calculating the area of triangle \(ABC\):
\[\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC)\]
\[\text{Area} = \frac{1}{2} \cdot 120 \cdot 85 \cdot \sin(112^\circ) \approx 5100 \cdot 0.92718 \approx 4728.6\text{ m}^2 \approx 4730\text{ m}^2 \text{ (to 3 s.f.)}\]

(d)(i) In the right-angled triangle \(TAB\) (right-angled at \(A\)):
\[\tan(18^\circ) = \frac{TA}{AB}\]
\[TA = 120 \cdot \tan(18^\circ) \approx 120 \cdot 0.32492 \approx 38.99\text{ m} \approx 39.0\text{ m} \text{ (to 3 s.f.)}\]

(d)(ii) In the right-angled triangle \(TAC\) (right-angled at \(A\)):
\[\tan(TCA) = \frac{TA}{AC}\]
\[\tan(TCA) = \frac{38.99}{171.08} \approx 0.22790\]
\[\text{Angle of elevation } TCA = \tan^{-1}(0.22790) \approx 12.839^\circ \approx 12.8^\circ \text{ (to 1 d.p.)}\]

(a) [3 marks]
M1 for correct Cosine Rule substitution: \(120^2 + 85^2 - 2(120)(85)\cos(112^\circ)\)
A1 for \(AC^2 = 29267\) (or \(29300\))
A1 for \(AC = 171\) (accept 171 to 171.1)

(b) [3 marks]
M1 for correct Sine Rule substitution: \(\frac{\sin(ACB)}{120} = \frac{\sin(112^\circ)}{\text{their } AC}\)
M1 for \(\sin(ACB) = \frac{120 \sin(112^\circ)}{\text{their } AC}\) (or alternative correct rearrangement)
A1 for \(40.6^\circ\) (accept 40.5 to 40.7)

(c) [2 marks]
M1 for correct area formula substitution: \(0.5 \cdot 120 \cdot 85 \cdot \sin(112^\circ)\)
A1 for \(4730\) (accept 4728 to 4730)

(d)(i) [3 marks]
M1 for using \(\tan(18^\circ)\) in triangle \(TAB\)
M1 for \(TA = 120 \tan(18^\circ)\)
A1 for \(39.0\) (accept 38.99 to 39.0)

(d)(ii) [4 marks]
M1 for recognizing the right-angled triangle \(TAC\) with angle \(TCA\)
M1 for substituting \(\tan(TCA) = \frac{\text{their } TA}{\text{their } AC}\)
M1 for \(\tan^{-1}\left(\frac{38.99}{171.08}\right)\)
A1 for \(12.8^\circ\) (accept 12.7 to 12.9)

(a)(i) Outward time = \(\frac{36}{x}\) hours.

(a)(ii) Return time = \(\frac{36}{x+3}\) hours.

(b) Since \(40\text{ minutes} = \frac{40}{60} = \frac{2}{3}\text{ hours}\), we set up the difference:
\[\frac{36}{x} - \frac{36}{x+3} = \frac{2}{3}\]
Multiply the entire equation by the common denominator \(3x(x+3)\):
\[3 \cdot 36(x+3) - 3 \cdot 36x = 2x(x+3)\]
\[108(x+3) - 108x = 2(x^2 + 3x)\]
\[108x + 324 - 108x = 2x^2 + 6x\]
\[324 = 2x^2 + 6x\]
Divide everything by 2:
\[162 = x^2 + 3x\]
Rearranging to make it equal to zero:
\[x^2 + 3x - 162 = 0\]

(c) Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-162)}}{2(1)}\]
\[x = \frac{-3 \pm \sqrt{9 + 648}}{2}\]
\[x = \frac{-3 \pm \sqrt{657}}{2}\]
\[x = \frac{-3 \pm 25.632}{2}\]
\[x_1 = \frac{22.632}{2} = 11.316 \approx 11.32\]
\[x_2 = \frac{-28.632}{2} = -14.316 \approx -14.32\]

(d) Since speed must be positive, we choose \(x = 11.316\text{ km/h}\).
Outward time = \(\frac{36}{11.316} \approx 3.1813\text{ hours}\).
Return time = \(\frac{36}{11.316 + 3} = \frac{36}{14.316} \approx 2.5147\text{ hours}\).
Total time = \(3.1813 + 2.5147 = 5.696\text{ hours}\).

Convert \(5.696\text{ hours}\) to hours and minutes:
Hours = \(5\)
Minutes = \(0.696 \times 60 = 41.76\text{ minutes}\), which rounds to \(42\text{ minutes}\).
Total time = 5 hours 42 minutes.

(a)(i) [1 mark]
B1 for \(\frac{36}{x}\)

(a)(ii) [1 mark]
B1 for \(\frac{36}{x+3}\)

(b) [4 marks]
M1 for converting 40 mins to \(\frac{2}{3}\) hours (or setting up equation in minutes with appropriate conversions)
M1 for writing correct equation: \(\frac{36}{x} - \frac{36}{x+3} = \frac{2}{3}\) (or equivalent)
M1 for algebraic step clearing fractions, e.g., \(108(x+3) - 108x = 2x(x+3)\)
A1 for fully correct working showing simplification to \(x^2 + 3x - 162 = 0\)

(c) [4 marks]
M1 for substitution into quadratic formula (allow sign errors in formula if clear)
A1 for \(\sqrt{657}\) or \(25.6\)
A1 for \(11.32\)
A1 for \(-14.32\)

(d) [5 marks]
M1 for choosing positive root \(x \approx 11.32\) (or \(11.3\))
M1 for substituting their positive value of \(x\) into both time expressions to find total time
M1 for obtaining \(5.696\) hours (or equivalent decimal, e.g., \(5.7\) hours)
M1 for multiplying decimal fraction (e.g. \(0.696\)) by 60 to convert to minutes
A1 for 5 hours 42 minutes (accept 5 hours 41 minutes to 5 hours 43 minutes)