2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Let the initial profits be \(P\).

At the end of 2021, the profits are \(1.12P\).
At the end of 2022, the profits are:
\(1.12P \times (1 - 0.08) = 1.12P \times 0.92 = 1.0304P\).

The overall percentage increase is:
\((1.0304 - 1) \times 100 = 3.04\%\).

M1 for \(1.12 \times 0.92\) or \(1.0304\) seen
A1 for 3.04

The volume of a cylinder is given by:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi\text{ cm}^3\).

The volume of a sphere is given by:
\(V_{\text{sphere}} = \frac{4}{3}\pi R^3\).

Equating the two volumes:
\(\frac{4}{3}\pi R^3 = 160\pi\)
\(\frac{4}{3}R^3 = 160\)
\(R^3 = 120\)
\(R = \sqrt[3]{120} \approx 4.93\text{ cm}\) (correct to 3 significant figures).

M1 for volume of cylinder expression \(\pi \times 4^2 \times 10\) or \(160\pi\) (or 502.7...)
M1 for equating to volume of sphere formula \(\frac{4}{3}\pi R^3 = \text{their Volume}\)
A1 for 4.93 (accept 4.932...)

Multiply both sides by \(3x + 4\):
\(y(3x + 4) = 5 - 2x\)
\(3xy + 4y = 5 - 2x\)

Collect the terms in \(x\) on one side of the equation:
\(3xy + 2x = 5 - 4y\)

Factorise \(x\):
\(x(3y + 2) = 5 - 4y\)

Divide by \(3y + 2\):
\(x = \frac{5 - 4y}{3y + 2}\)

M1 for multiplying by denominator to clear fraction: \(y(3x + 4) = 5 - 2x\)
M1 for collecting \(x\) terms on one side and factorising: \(x(3y + 2) = 5 - 4y\)
A1 for \(x = \frac{5 - 4y}{3y + 2}\) (or equivalent)

The total number of balls is \(5 + 3 + 2 = 10\).

Since selection is without replacement:
\(\text{P(both Red)} = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}\)
\(\text{P(both Blue)} = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}\)
\(\text{P(both Green)} = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90}\)

Total probability of the same colour is:
\(\frac{20}{90} + \frac{6}{90} + \frac{2}{90} = \frac{28}{90} = \frac{14}{45}\) (or approximately \(0.311\)).

M1 for at least one correct product of probabilities without replacement (e.g. \(\frac{5}{10} \times \frac{4}{9}\))
M1 for adding three correct products: \(\frac{5}{10} \times \frac{4}{9} + \frac{3}{10} \times \frac{2}{9} + \frac{2}{10} \times \frac{1}{9}\)
A1 for \(\frac{14}{45}\) or 0.311 (accept 0.311 or 0.3111...)

In the right-angled triangle formed, the ladder is the hypotenuse (6.5 m) and the distance along the ground is the adjacent side (2.5 m).

\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5}\)
\(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right) \approx 67.4^\circ\) (correct to 1 decimal place).

M1 for \(\text{cos}(\theta) = \frac{2.5}{6.5}\) or equivalent correct trig statement
A1 for 67.4 (accept 67.38...)

Find the total goals scored:
\(0 \times 8 + 1 \times 15 + 2 \times 11 + 3 \times 4 + 4 \times 2 = 0 + 15 + 22 + 12 + 8 = 57\text{ goals}\).

The total number of matches is 40.

\(\text{Mean} = \frac{57}{40} = 1.425\text{ goals per match}\).

M1 for showing sum of products (at least 3 correct products shown added, e.g. \(15 + 22 + 12 + 8\))
M1 for dividing their sum by 40
A1 for 1.425

Using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(62^\circ)\)
\(AC^2 = 49 + 81 - 126\cos(62^\circ)\)
\(AC^2 = 130 - 126(0.46947) = 130 - 59.153 = 70.847\)

\(AC = \sqrt{70.847} \approx 8.42\text{ cm}\) (correct to 3 significant figures).

M1 for correct substitution into the Cosine Rule: \(7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(62^\circ)\)
M1 for \(AC^2 \approx 70.8\) or better
A1 for 8.42 (accept 8.417...)

Differentiate the function with respect to \(x\):
\(\frac{dy}{dx} = 6x^2 - 18x - 24\)

At a stationary point, the gradient \(\frac{dy}{dx} = 0\):
\(6x^2 - 18x - 24 = 0\)

Divide by 6:
\(x^2 - 3x - 4 = 0\)

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

This gives \(x = 4\) or \(x = -1\).
Since we require the positive \(x\)-coordinate, \(x = 4\).

M1 for correct differentiation of at least two terms of the cubic expression
M1 for setting their derivative to 0 and solving the resulting quadratic equation
A1 for 4

To find the upper bound for the perimeter, we first determine the upper bounds for both the length and the width.

For the length, $L = 80\text{ m}$ to the nearest metre, so its upper bound is:
$$L_{\text{upper}} = 80 + 0.5 = 80.5\text{ m}$$

For the width, $W = 45\text{ m}$ to the nearest $5\text{ m}$, so its upper bound is:
$$W_{\text{upper}} = 45 + 2.5 = 47.5\text{ m}$$

The perimeter $P$ is given by:
$$P = 2(L + W)$$

Therefore, the upper bound of the perimeter is:
$$P_{\text{upper}} = 2(L_{\text{upper}} + W_{\text{upper}}) = 2(80.5 + 47.5) = 2(128) = 256\text{ m}$$

M1 for $L_{\text{upper}} = 80.5$ or $W_{\text{upper}} = 47.5$ seen
M1 for $2(\text{their } L_{\text{upper}} + \text{their } W_{\text{upper}})$
A1 for 256

Multiply both sides by $t + 2$:
$$w(t + 2) = 3t - 5$$

Expand the bracket:
$$wt + 2w = 3t - 5$$

Rearrange the terms to group all terms with $t$ on one side:
$$2w + 5 = 3t - wt$$

Factorise $t$ on the right-hand side:
$$2w + 5 = t(3 - w)$$

Divide both sides by $3 - w$:
$$t = \frac{2w + 5}{3 - w}$$

M1 for $w(t + 2) = 3t - 5$
M1 for grouping $t$ terms on one side, e.g. $2w + 5 = t(3 - w)$
A1 for $t = \frac{2w+5}{3-w}$ or equivalent

Let the scale factor for the volumes be $\frac{V_2}{V_1}$:
$$\frac{V_2}{V_1} = \frac{128\pi}{54\pi} = \frac{64}{27}$$

The linear scale factor $k$ is the cube root of the volume scale factor:
$$k = \sqrt[3]{\frac{64}{27}} = \frac{4}{3}$$

The area scale factor is $k^2$:
$$k^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$

Therefore, the total surface area of the larger cylinder $A_2$ is:
$$A_2 = A_1 \times k^2 = 36\pi \times \frac{16}{9} = 4\pi \times 16 = 64\pi\text{ cm}^2$$

M1 for volume ratio $\frac{128}{54}$ or linear scale factor $\sqrt[3]{\frac{128}{54}}$ or $\frac{4}{3}$ seen
M1 for area ratio $\left(\text{their } \frac{4}{3}\right)^2 = \frac{16}{9}$ seen
A1 for $64\pi$ or $201$ or $201.06...$

Using the Cosine Rule:
$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)$$

Substitute the given values:
$$AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(64^\circ)$$
$$AC^2 = 49 + 81 - 126\cos(64^\circ)$$
$$AC^2 = 130 - 126(0.43837...)$$
$$AC^2 \approx 130 - 55.235$$
$$AC^2 \approx 74.765$$
$$AC = \sqrt{74.765} \approx 8.6467\text{ cm}$$

Rounding to 3 significant figures gives $8.65\text{ cm}$.

M1 for $7^2 + 9^2 - 2(7)(9)\cos(64^\circ)$
A1 for $74.8$ or $74.76...$ seen as $AC^2$
A1 for $8.65$ or $8.646...$

The total number of pens in the box is $5 + 3 = 8$.

Using the complement rule, the probability of choosing at least one blue pen is:
$$\text{P(at least one blue)} = 1 - \text{P(both red)}$$

The probability that both pens are red (without replacement) is:
$$\text{P(both red)} = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

Therefore:
$$\text{P(at least one blue)} = 1 - \frac{5}{14} = \frac{9}{14}$$

M1 for $\frac{5}{8} \times \frac{4}{7}$ oe
M1 for $1 - \text{their } P(\text{both red})$
A1 for $\frac{9}{14}$

First, reflect the vertex $(1, 4)$ in the line $y = -x$.
The general mapping for reflection in the line $y = -x$ is $(x, y) \rightarrow (-y, -x)$.
Therefore, the image of $(1, 4)$ is:
$$(-4, -1)$$

Next, apply the translation by the vector $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ to $(-4, -1)$:
$$\begin{pmatrix} -4 \\ -1 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} -4 + 2 \\ -1 - 3 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}$$

So, the coordinates of the corresponding vertex of Triangle $R$ are $(-2, -4)$.

M1 for finding coordinates of $Q$ as $(-4, -1)$ or for applying $(x, y) \rightarrow (-y, -x)$
M1 for adding $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ to their coordinates of $Q$
A1 for $(-2, -4)$

First, simplify the numerator by writing both terms with the same power of 10:
$$4.2 \times 10^7 = 42 \times 10^6$$
$$42 \times 10^6 + 1.8 \times 10^6 = 43.8 \times 10^6 = 4.38 \times 10^7$$

Next, divide by the denominator:
$$\frac{4.38 \times 10^7}{2.5 \times 10^{-3}} = \frac{4.38}{2.5} \times 10^{7 - (-3)} = 1.752 \times 10^{10}$$

M1 for $4.38 \times 10^7$ seen in the numerator or $43.8 \times 10^6$
M1 for dividing $4.38$ by $2.5$ and applying index subtraction laws
A1 for $1.752 \times 10^{10}$

The vector $\overrightarrow{AB}$ is given by:
$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

Substitute the given column vectors:
$$\overrightarrow{AB} = \begin{pmatrix} -1 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 - 3 \\ 6 - (-2) \end{pmatrix} = \begin{pmatrix} -4 \\ 8 \end{pmatrix}$$

M1 for $\mathbf{b} - \mathbf{a}$ seen as a vector subtraction, e.g. $\begin{pmatrix} -1 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix}$
B1 for top value $-4$ or bottom value $8$
A1 for $\begin{pmatrix} -4 \\ 8 \end{pmatrix}$

Let the original price of the laptop be \(x\).
Since the price is reduced by 15%, the sale price represents 85% of the original price.
\[0.85 \times x = 646\]
\[x = \frac{646}{0.85}\]
\[x = 760\]
Thus, the original price of the laptop is $760.

M1 for \(0.85 \times x = 646\) or equivalent
M1 for \(\frac{646}{0.85}\)
A1 for 760

First, calculate the volume of the sphere:
\[V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = 36\pi \text{ cm}^3\]

Next, express the volume of the cylinder in terms of its radius \(R\):
\[V_{\text{cylinder}} = \pi R^2 h = \pi R^2 (4) = 4\pi R^2\]

Since the volume remains constant during melting and recasting:
\[4\pi R^2 = 36\pi\]
\[R^2 = 9\]
\[R = 3\text{ cm}\]
Thus, the radius of the cylinder is 3 cm.

M1 for volume of sphere = \(\frac{4}{3}\pi (3)^3\) or \(36\pi\)
M1 for equating their sphere volume to \(\pi R^2 (4)\)
A1 for 3

The total number of balls is \(5 + 4 = 9\).
The two ways to pick balls of different colours are (Red then Blue) or (Blue then Red).

\[P(\text{Red then Blue}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\]
\[P(\text{Blue then Red}) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}\]

Adding these two mutually exclusive probabilities:
\[P(\text{Different colours}) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9}\]

M1 for \(\frac{5}{9} \times \frac{4}{8}\) or \(\frac{4}{9} \times \frac{5}{8}\)
M1 for \(\frac{20}{72} + \frac{20}{72}\) or equivalent
A1 for \(\frac{5}{9}\)

Factorise the numerator:
\[2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\]

Factorise the denominator:
\[x^2 + 3x - 10 = (x + 5)(x - 2)\]

Simplify the fraction by cancelling out the common factor \((x - 2)\):
\[\frac{2(x - 2)(x + 2)}{(x + 5)(x - 2)} = \frac{2(x + 2)}{x + 5}\]

B1 for factorising the numerator to \(2(x - 2)(x + 2)\)
B1 for factorising the denominator to \((x + 5)(x - 2)\)
B1 for final answer \(\frac{2(x+2)}{x+5}\) or equivalent

The ladder, wall, and ground form a right-angled triangle where the ladder is the hypotenuse (5.5 m) and the distance from the foot of the ladder to the wall is the adjacent side (1.8 m).
Let \(\theta\) be the angle between the ladder and the horizontal ground.
\[\cos\theta = \frac{1.8}{5.5}\]
\[\theta = \arccos\left(\frac{1.8}{5.5}\right) \approx 70.898^\circ\]
Rounding to 1 decimal place, we get \(70.9^\circ\).

M1 for \(\cos\theta = \frac{1.8}{5.5}\) or equivalent
M1 for \(\theta = \arccos\left(\frac{1.8}{5.5}\right)\)
A1 for 70.9

Using the cosine rule:
\[AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\]
\[AC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(64^\circ)\]
\[AC^2 = 49 + 81 - 126 \times \cos(64^\circ)\]
\[AC^2 \approx 130 - 126 \times 0.43837 = 130 - 55.2348 = 74.7652\]
\[AC = \sqrt{74.7652} \approx 8.6467\text{ cm}\]
To 3 significant figures, \(AC = 8.65\text{ cm}\).

M1 for substituting correctly into the cosine rule: \(7^2 + 9^2 - 2(7)(9)\cos(64^\circ)\)
A1 for \(AC^2 \approx 74.8\) or better
A1 for 8.65

To find the stationary points, find the derivative \(\frac{dy}{dx}\) and set it to 0:
\[\frac{dy}{dx} = 6x^2 - 18x + 12\]
\[6x^2 - 18x + 12 = 0\]
\[6(x^2 - 3x + 2) = 0\]
\[6(x - 1)(x - 2) = 0\]
So the stationary points are at \(x = 1\) and \(x = 2\).

To determine which point is the local minimum, use the second derivative test:
\[\frac{d^2y}{dx^2} = 12x - 18\]
At \(x = 1\): \(\frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0\) (local maximum).
At \(x = 2\): \(\frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0\) (local minimum).

Find the \(y\)-coordinate at \(x = 2\):
\[y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 16 - 36 + 24 - 5 = -1\]
Thus, the coordinates of the local minimum point are \((2, -1)\).

M1 for \(\frac{dy}{dx} = 6x^2 - 18x + 12\)
M1 for setting \(\frac{dy}{dx} = 0\) and solving to find stationary points at \(x = 1, x = 2\)
A1 for \((2, -1)\)

The diagonal of a cuboid with length \(l\), width \(w\), and height \(h\) is calculated using the 3D Pythagoras' theorem:
\[d = \sqrt{l^2 + w^2 + h^2}\]
\[d = \sqrt{5^2 + 6^2 + 8^2}\]
\[d = \sqrt{25 + 36 + 64} = \sqrt{125} \approx 11.18\text{ cm}\]
To 3 significant figures, \(d = 11.2\text{ cm}\).

M1 for \(5^2 + 6^2 + 8^2\)
M1 for \(\sqrt{125}\) or better
A1 for 11.2

The gradient of the line \(y = 3x - 5\) is \(3\). The gradient of the perpendicular line, \(m\), is given by the negative reciprocal: \(m = -\frac{1}{3}\). Using the equation of a straight line, \(y = mx + c\), we substitute the gradient and the point \((6, 2)\): \(2 = -\frac{1}{3}(6) + c\), which simplifies to \(2 = -2 + c\), giving \(c = 4\). Therefore, the equation of the perpendicular line is \(y = -\frac{1}{3}x + 4\).

M1 for gradient of perpendicular line \(= -\frac{1}{3}\)
M1 for substituting \((6, 2)\) into \(y = mx + c\) with their perpendicular gradient
A1 for \(y = -\frac{1}{3}x + 4\) or equivalent

We apply the fractional power to each part of the term: \(\left(64y^{12}\right)^{\frac{2}{3}} = 64^{\frac{2}{3}} \times \left(y^{12}\right)^{\frac{2}{3}}\). First, evaluate the coefficient: \(64^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\). Next, simplify the index of \(y\): \((y^{12})^{\frac{2}{3}} = y^{12 \times \frac{2}{3}} = y^8\). Combining these terms gives \(16y^8\).

B1 for \(16\) or \(y^8\) in the final answer
or M1 for \(64^{\frac{2}{3}}\) evaluated as \(16\) or \(\left(y^{12}\right)^{\frac{2}{3}} = y^8\)

(a) Value last year = 12150 / 0.9 = $13500.

(b) Value after 3 years = 12150 * (0.92)^3 = 12150 * 0.778688 = $9461.06, which rounds to $9460 (to 3 significant figures).

(c) 8000 * (1 + r/100)^6 = 9850
(1 + r/100)^6 = 1.23125
1 + r/100 = 1.23125^(1/6) ≈ 1.035305
r = 3.53% (to 3 significant figures).

(d)(i) Let M_0 be the initial mass. Mass after 12 hours = M_0 * (1 - 0.045)^12 = M_0 * (0.955)^12 ≈ 0.57413 * M_0.
Overall percentage decrease = (1 - 0.57413) * 100% = 42.587% ≈ 42.6%.

(ii) (0.955)^t < 0.40
t * ln(0.955) < ln(0.40)
t * (-0.046044) < -0.91629
t > 19.9 hours.
Therefore, the number of whole hours is 20.

(a) M1 for 12150 / 0.90, A1 for 13500.
(b) M1 for 12150 * (0.92)^3, A1 for 9460.
(c) M1 for (1 + r/100)^6 = 9850 / 8000, M1 for 1.23125^(1/6), A1 for 3.53.
(d)(i) M1 for (0.955)^12, A1 for 0.574 or 57.4%, A1 for 42.6%.
(d)(ii) M1 for (0.955)^t = 0.40 or trial and error with at least two values, A1 for 19.9, A1 for 20.

(a)(i) Arc length of sector = 2 * pi * 12 * (y / 360).
Circumference of cone base = 2 * pi * 3.2.
2 * pi * 12 * (y / 360) = 2 * pi * 3.2
12 * y / 360 = 3.2
y = (3.2 * 360) / 12 = 96.

(a)(ii) Slant height of cone, l = 12 cm. Base radius, r = 3.2 cm.
Height, h = √(12^2 - 3.2^2) = √(144 - 10.24) = √133.76 ≈ 11.565 cm, which rounds to 11.6 cm.

(a)(iii) Volume = (1/3) * pi * r^2 * h = (1/3) * pi * 3.2^2 * 11.565 ≈ 124.28 cm^3, which rounds to 124 cm^3.

(b) Volume of cylinder = pi * r^2 * h = pi * 6^2 * 15 = 540 * pi ≈ 1696.46 cm^3.
Volume of one sphere = (4/3) * pi * R^3 = (4/3) * pi * 2.5^3 = (50/3) * pi ≈ 65.45 cm^3.
Number of spheres = 540 * pi / ((50/3) * pi) = 1620 / 50 = 32.4.
Maximum number of complete spheres is 32.

(c) Surface area = 4 * pi * R^2 = 4 * pi * 2.5^2 = 25 * pi ≈ 78.5398 cm^2, which rounds to 78.5 cm^2.

(a)(i) M1 for 2 * pi * 12 * (y / 360) = 2 * pi * 3.2, M1 for y = 3.2 * 30, A1 for 96.
(a)(ii) M1 for 12^2 - 3.2^2, A1 for 11.6.
(a)(iii) M1 for (1/3) * pi * 3.2^2 * (their h), A1 for 124.
(b) M1 for volume of cylinder = 540 * pi, M1 for volume of sphere = (50/3) * pi, A1 for 32.
(c) M1 for 4 * pi * 2.5^2, A1 for 78.5.

(a)(i) Number of basic models = 450 / x
Number of premium models = 500 / (x + 10)
Total earbuds = 450 / x + 500 / (x + 10) = 50.
Divide the equation by 50:
9 / x + 10 / (x + 10) = 1
Multiply by x(x + 10):
9(x + 10) + 10x = x(x + 10)
9x + 90 + 10x = x^2 + 10x
19x + 90 = x^2 + 10x
x^2 - 9x - 90 = 0.

(a)(ii) We look for two numbers that multiply to -90 and add to -9. These are -15 and 6.
Thus, x^2 - 9x - 90 = (x - 15)(x + 6).

(a)(iii) x = 15 or x = -6. Since cost must be positive, the cost of a basic model is $15.

(b) 4(y + 2) - 3(y - 3) = (y - 3)(y + 2)
4y + 8 - 3y + 9 = y^2 - y - 6
y + 17 = y^2 - y - 6
y^2 - 2y - 23 = 0.
Using the quadratic formula:
y = (2 ± √((-2)^2 - 4 * 1 * (-23))) / 2
y = (2 ± √(4 + 92)) / 2
y = (2 ± √96) / 2
y = 1 ± √24
y ≈ 5.90 or y ≈ -3.90.

(a)(i) M1 for 450/x + 500/(x+10) = 50, M1 for 9(x+10) + 10x, M1 for clearing fractions to get 19x + 90 = x^2 + 10x, A1 for completing the show that.
(a)(ii) M1 for (x + a)(x + b) where ab = -90 or a + b = -9, A1 for (x - 15)(x + 6).
(a)(iii) B1 for 15.
(b) M1 for 4(y+2) - 3(y-3), M1 for (y-3)(y+2) = y^2 - y - 6, M1 for simplifying to y^2 - 2y - 23 = 0, B1 for √96 seen, A1 for 5.90 and -3.90.

(a) Using the cosine rule on triangle CAD:
CD^2 = 12^2 + 15^2 - 2 * 12 * 15 * cos(42°)
CD^2 = 144 + 225 - 360 * 0.74314
CD^2 = 369 - 267.53 = 101.47
CD = √101.47 ≈ 10.073 m, which rounds to 10.1 m.

(b)(i) AB is vertical, so triangle BAC is right-angled at A.
tan(angle BCA) = AB / AC = 8.4 / 12 = 0.7
angle BCA = tan^-1(0.7) ≈ 34.99°, which rounds to 35.0°.

(b)(ii) AB is vertical, so triangle BAD is right-angled at A.
BD^2 = AB^2 + AD^2 = 8.4^2 + 15^2 = 70.56 + 225 = 295.56
BD = √295.56 ≈ 17.19 m, which rounds to 17.2 m.

(b)(iii) For the wire BP to be as short as possible, BP must be perpendicular to CD. Since AB is vertical, BP^2 = AB^2 + AP^2.
BP is minimised when AP is minimised, which means AP is perpendicular to CD on the ground.
Area of triangle CAD = 0.5 * AC * AD * sin(42°) = 0.5 * 12 * 15 * sin(42°) = 90 * 0.66913 = 60.222 m^2.
Also, Area of triangle CAD = 0.5 * CD * AP = 0.5 * 10.073 * AP.
60.222 = 5.0365 * AP
AP = 11.957 m.
Now, BP^2 = AB^2 + AP^2 = 8.4^2 + 11.957^2 = 70.56 + 142.97 = 213.53
BP = √213.53 ≈ 14.61 m, which rounds to 14.6 m.

(a) M1 for 12^2 + 15^2 - 2 * 12 * 15 * cos(42°), A1 for 101.5, A1 for 10.1.
(b)(i) M1 for tan(theta) = 8.4 / 12, A1 for 35.0.
(b)(ii) M1 for 8.4^2 + 15^2, A1 for 295.56, A1 for 17.2.
(b)(iii) M1 for Area of CAD = 0.5 * 12 * 15 * sin(42°), M1 for AP = Area / (0.5 * 10.073) ≈ 11.96, M1 for BP^2 = 8.4^2 + AP^2, A1 for 14.6.

(a) Using the cosine rule on triangle PQR:
PR^2 = PQ^2 + QR^2 - 2 * PQ * QR * cos(108°)
PR^2 = 45^2 + 62^2 - 2 * 45 * 62 * cos(108°)
PR^2 = 2025 + 3844 - 5580 * (-0.30902)
PR^2 = 5869 + 1724.33 = 7593.33
PR = √7593.33 ≈ 87.14 m, which rounds to 87.1 m.

(b) Area of PQR = 0.5 * PQ * QR * sin(108°)
Area = 0.5 * 45 * 62 * sin(108°) = 1395 * 0.95106 ≈ 1326.7 m^2, which rounds to 1330 m^2.

(c) In triangle PSR, angle PSR = 180° - 38° - 47° = 95°.
Using the sine rule:
SR / sin(38°) = PR / sin(95°)
SR = 87.14 * sin(38°) / sin(95°)
SR = 87.14 * 0.61566 / 0.99619 ≈ 53.85 m, which rounds to 53.9 m.

(d) Using the sine rule on triangle PSR:
PS / sin(47°) = PR / sin(95°)
PS = 87.14 * sin(47°) / sin(95°)
PS = 87.14 * 0.73135 / 0.99619 ≈ 63.97 m, which rounds to 64.0 m.

(e) Perimeter = PQ + QR + SR + PS = 45 + 62 + 53.85 + 63.97 = 224.82 m, which rounds to 225 m.

(a) M1 for 45^2 + 62^2 - 2 * 45 * 62 * cos(108°), A1 for 7593, A1 for 87.1.
(b) M1 for 0.5 * 45 * 62 * sin(108°), A1 for 1330.
(c) M1 for angle PSR = 95°, M1 for SR / sin(38°) = PR / sin(95°), A1 for 53.9.
(d) M1 for PS / sin(47°) = PR / sin(95°), A1 for 64.0.
(e) B1 for 225.

(a)(i) P(both blue) = P(Blue first) * P(Blue second | Blue first) = (8/12) * (7/11) = 56 / 132 = 14/33 ≈ 0.424.

(a)(ii) P(different colours) = P(Blue then Red) + P(Red then Blue)
= (8/12) * (4/11) + (4/12) * (8/11) = 32/132 + 32/132 = 64/132 = 16/33 ≈ 0.485.

(a)(iii) P(at least one red) = 1 - P(no red) = 1 - P(both blue) = 1 - 14/33 = 19/33 ≈ 0.576.

(b) P(three same) = P(B, B, B) + P(R, R, R)
P(B, B, B) = (8/12) * (7/11) * (6/10) = 336 / 1320
P(R, R, R) = (4/12) * (3/11) * (2/10) = 24 / 1320
P(three same) = (336 + 24) / 1320 = 360 / 1320 = 3/11 ≈ 0.273.

(c) We are given the first two balls have different colours: {BR, RB}.
P(First was Blue | Different) = P(BR) / (P(BR) + P(RB)) = (32/132) / (64/132) = 32 / 64 = 0.5.

(a)(i) M1 for (8/12) * (7/11), A1 for 14/33 (or 0.424).
(a)(ii) M1 for (8/12) * (4/11), M1 for (4/12) * (8/11) + (8/12) * (4/11), A1 for 16/33 (or 0.485).
(a)(iii) M1 for 1 - (their (a)(i)), A1 for 19/33 (or 0.576).
(b) M1 for (8/12) * (7/11) * (6/10), M1 for (4/12) * (3/11) * (2/10), A1 for 3/11 (or 0.273).
(c) M1 for (32/132) / (64/132) or equivalent logic, A1 for 0.5 (or 1/2).

(a) dy/dx = d/dx (2x^3 - 9x^2 - 24x + 10) = 6x^2 - 18x - 24.

(b) Stationary points occur when dy/dx = 0:
6x^2 - 18x - 24 = 0
x^2 - 3x - 4 = 0
(x - 4)(x + 1) = 0
So x = 4 or x = -1.

When x = 4:
y = 2(4)^3 - 9(4)^2 - 24(4) + 10 = 2(64) - 9(16) - 96 + 10 = 128 - 144 - 96 + 10 = -102.
The coordinate is (4, -102).

When x = -1:
y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 10 = -2 - 9 + 24 + 10 = 23.
The coordinate is (-1, 23).

(c) Find the second derivative:
d^2y/dx^2 = 12x - 18.
At x = 4:
d^2y/dx^2 = 12(4) - 18 = 30 > 0. Since the second derivative is positive, (4, -102) is a local minimum.
At x = -1:
d^2y/dx^2 = 12(-1) - 18 = -30 < 0. Since the second derivative is negative, (-1, 23) is a local maximum.

(d) At x = 2:
y = 2(2)^3 - 9(2)^2 - 24(2) + 10 = 16 - 36 - 48 + 10 = -58.
The gradient of the tangent, m, is the value of dy/dx at x = 2:
m = 6(2)^2 - 18(2) - 24 = 24 - 36 - 24 = -36.
The equation of the tangent is:
y - y1 = m(x - x1)
y - (-58) = -36(x - 2)
y + 58 = -36x + 72
y = -36x + 14.

(a) B2 for 6x^2 - 18x - 24 (B1 for 6x^2 or -18x or -24).
(b) M1 for setting their dy/dx = 0, A1 for x = 4 and x = -1, A1 for y = -102, A1 for y = 23.
(c) M1 for finding d^2y/dx^2 = 12x - 18, A1 for evaluating d^2y/dx^2 at x = 4 and x = -1, A1 for correctly stating minimum and maximum.
(d) M1 for evaluating y = -58 at x = 2, M1 for evaluating gradient m = -36, A1 for y = -36x + 14.

(a)(i) Since the x-coordinates are negated and y-coordinates remain unchanged, this is a reflection in the y-axis (or the line x = 0).

(a)(ii) Since the y-coordinates are negated and x-coordinates remain unchanged, this is a reflection in the x-axis (or the line y = 0).

(a)(iii) Both x and y coordinates are multiplied by 2. This is an enlargement with scale factor 2 and the centre of enlargement is the origin (0, 0).

(b) A rotation of 90° clockwise about the origin transforms (x, y) to (y, -x).
- (1, 2) becomes (2, -1)
- (4, 2) becomes (2, -4)
- (4, 4) becomes (4, -4)
The vertices of Triangle E are (2, -1), (2, -4), and (4, -4).

(c) A shear with the x-axis invariant and shear factor 2 transforms (x, y) to (x + 2y, y).
- (1, 2) becomes (1 + 2*2, 2) = (5, 2)
- (4, 2) becomes (4 + 2*2, 2) = (8, 2)
- (4, 4) becomes (4 + 2*4, 4) = (12, 4)
The vertices of Triangle F are (5, 2), (8, 2), and (12, 4).

(a)(i) B1 for Reflection, B1 for the y-axis (or line x = 0).
(a)(ii) B1 for Reflection, B1 for the x-axis (or line y = 0).
(a)(iii) B1 for Enlargement, B1 for scale factor 2, B1 for centre (0, 0).
(b) M1 for any two coordinates correct, A1 for all three correct: (2, -1), (2, -4), (4, -4).
(c) M1 for use of (x + 2y, y) or finding two vertices correctly, A1 for all three correct: (5, 2), (8, 2), (12, 4).

(a)(i) Using the cosine rule in triangle $ABC$:
$$\cos \angle ACB = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC}$$
$$\cos \angle ACB = \frac{120^2 + 180^2 - 150^2}{2 \cdot 120 \cdot 180}$$
$$\cos \angle ACB = \frac{14400 + 32400 - 22500}{43200} = \frac{24300}{43200} = 0.5625$$
$$\angle ACB = \arccos(0.5625) \approx 55.771^\circ$$
Rounding to 1 decimal place gives $55.8^\circ$.

(a)(ii) First, find the angle $CAB$ using the cosine rule:
$$\cos \angle CAB = \frac{AC^2 + AB^2 - BC^2}{2 \cdot AC \cdot AB}$$
$$\cos \angle CAB = \frac{120^2 + 150^2 - 180^2}{2 \cdot 120 \cdot 150}$$
$$\cos \angle CAB = \frac{14400 + 22500 - 32400}{36000} = \frac{4500}{36000} = 0.125$$
$$\angle CAB = \arccos(0.125) \approx 82.819^\circ$$
Since the bearing of $C$ from $A$ is between $090^\circ$ and $180^\circ$, we add this angle to the bearing of $B$ from $A$:
$$\text{Bearing} = 75^\circ + 82.819^\circ = 157.819^\circ \approx 157.8^\circ$$

(b)(i) The top of the tower $T$ is vertically above $C$.
The height of $T$ above $C$ is the height of $D$ above $C$ plus the height of the tower $TD$:
$$TC = 25\text{ m} + 40\text{ m} = 65\text{ m}$$
Triangle $TAC$ is a right-angled triangle with the right angle at $C$:
$$AT = \sqrt{AC^2 + TC^2} = \sqrt{120^2 + 65^2} = \sqrt{14400 + 4225} = \sqrt{18625} \approx 136.47\text{ m}$$
Correct to 3 significant figures, $AT = 136\text{ m}$.

(b)(ii) Triangle $TBC$ is a right-angled triangle with the right angle at $C$.
Let $\theta$ be the angle of elevation of $T$ from $B$:
$$\tan \theta = \frac{TC}{BC} = \frac{65}{180}$$
$$\theta = \arctan\left(\frac{65}{180}\right) \approx 19.854^\circ$$
Correct to 3 significant figures, the angle of elevation is $19.9^\circ$.

(a)(i) [3 marks]
M1 for $\cos \angle ACB = \frac{120^2 + 180^2 - 150^2}{2 \cdot 120 \cdot 180}$
A1 for $\cos \angle ACB = 0.5625$
A1 for $\arccos(0.5625) = 55.77...$ showing rounding to $55.8^\circ$

(a)(ii) [3 marks]
M1 for $\cos \angle CAB = \frac{120^2 + 150^2 - 180^2}{2 \cdot 120 \cdot 150}$
A1 for $\angle CAB = 82.8^\circ$ or $82.81...^\circ$
A1 for bearing $= 157.8^\circ$ or $158^\circ$

(b)(i) [4 marks]
B1 for $TC = 65$ seen or implied
M1 for $AT^2 = 120^2 + TC^2$
M1 for $\sqrt{120^2 + 65^2}$
A1 for $136$ or $136.47...$ (accept 136.5)

(b)(ii) [2 marks]
M1 for $\tan \theta = \frac{65}{180}$
A1 for $19.9^\circ$ or $19.85...^\circ$

(a) Differentiating $y = 2x^3 - 9x^2 - 24x + 15$ with respect to $x$:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x - 24$$

(b) Stationary points occur when $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$$6x^2 - 18x - 24 = 0$$
Divide by 6:
$$x^2 - 3x - 4 = 0$$
$$(x - 4)(x + 1) = 0$$
So, $x = 4$ or $x = -1$.

When $x = 4$:
$$y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97$$
Stationary point: $(4, -97)$.

When $x = -1$:
$$y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28$$
Stationary point: $(-1, 28)$.

(c) Using the second derivative to determine the nature:
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18$$

For $x = 4$:
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(4) - 18 = 30 > 0$$
Since $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$, the point $(4, -97)$ is a local minimum.

For $x = -1$:
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 18 = -30 < 0$$
Since $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$, the point $(-1, 28)$ is a local maximum.

(d) Find the gradient of the tangent at $x = 1$:
$$m = 6(1)^2 - 18(1) - 24 = 6 - 18 - 24 = -36$$
Find the $y$-value at $x = 1$:
$$y = 2(1)^3 - 9(1)^2 - 24(1) + 15 = -16$$
The equation of the tangent is:
$$y - (-16) = -36(x - 1)$$
$$y + 16 = -36x + 36$$
$$y = -36x + 20$$

(a) [2 marks]
B2 for $6x^2 - 18x - 24$ (B1 for two terms correct)

(b) [5 marks]
M1 for setting their derivative equal to 0
M1 for factorising/solving quadratic to find $x = 4$ and $x = -1$
A1 for $x = 4$ and $x = -1$
A1 for $y = -97$
A1 for $y = 28$

(c) [3 marks]
M1 for finding $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18$ (or using first-derivative method)
A1 for $(4, -97)$ is a minimum with explanation (e.g., $30 > 0$)
A1 for $(-1, 28)$ is a maximum with explanation (e.g., $-30 < 0$)

(d) [2 marks]
M1 for substituting $x = 1$ into derivative to get $m = -36$ and finding $y = -16$
A1 for $y = -36x + 20$ or equivalent

(a)(i) The probability that both disks are the same color is the sum of the probabilities of choosing RR, BB, or YY:
$$P(RR) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$$
$$P(BB) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}$$
$$P(YY) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}$$
$$P(\text{same color}) = \frac{20 + 12 + 6}{132} = \frac{38}{132} = \frac{19}{66} \approx 0.288$$

(a)(ii) The probability that they are of different colors is:
$$P(\text{different colors}) = 1 - P(\text{same color}) = 1 - \frac{19}{66} = \frac{47}{66} \approx 0.712$$

(b)(i) The probability of selecting Red first, Blue second, and Yellow third is:
$$P(R_1 \text{ and } B_2 \text{ and } Y_3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22} \approx 0.0455$$

(b)(ii) Exactly two of the three disks are Red. The possible sequences are RRX, RXR, and XRR, where X is a non-Red disk (7 non-Red disks originally):
$$P(RRX) = \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} = \frac{140}{1320}$$
$$P(RXR) = \frac{5}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{140}{1320}$$
$$P(XRR) = \frac{7}{12} \times \frac{5}{11} \times \frac{4}{10} = \frac{140}{1320}$$
$$P(\text{exactly two Red}) = 3 \times \frac{140}{1320} = \frac{420}{1320} = \frac{7}{22} \approx 0.318$$

(a)(i) [3 marks]
M1 for any correct product of two probabilities (e.g. $\frac{5}{12} \times \frac{4}{11}$)
M1 for summing three correct products: $\frac{20}{132} + \frac{12}{132} + \frac{6}{132}$
A1 for $\frac{19}{66}$ or $0.288$ or $0.2878...$

(a)(ii) [2 marks]
M1 for $1 - \text{their } P(\text{same})$
A1 for $\frac{47}{66}$ or $0.712$ or $0.7121...$

(b)(i) [2 marks]
M1 for $\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}$
A1 for $\frac{1}{22}$ or $0.0455$ or $0.04545...$

(b)(ii) [4 marks]
M1 for the probability of one combination, e.g. $\frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} = \frac{140}{1320}$
M1 for identifying there are 3 such combinations (RRX, RXR, XRR)
M1 for $3 \times \text{their product}$
A1 for $\frac{7}{22}$ or $0.318$ or $0.3181...$