2024 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

Multiply the numbers and powers of 10 separately:
\[4.5 \times 6 = 27\]
\[10^7 \times 10^{-4} = 10^{7 + (-4)} = 10^3\]
This gives:
\[27 \times 10^3\]
To write this in standard form, move the decimal point one place to the left and increase the power of 10 by 1:
\[2.7 \times 10^4\]

M1 for 27 \times 10^3 or 27000 seen
A1 for 2.7 \times 10^4 as final answer

Find the highest common factor of the coefficients: the HCF of 15 and 20 is 5.
Find the highest common factor of the algebraic terms: the HCF of \(x^2y\) and \(xy^2\) is \(xy\).
Therefore, the common factor is \(5xy\).
Dividing both terms by this common factor:
\[15x^2y \div 5xy = 3x\]
\[-20xy^2 \div 5xy = -4y\]
This gives:
\[5xy(3x - 4y)\]

B1 for a correct partial factorisation, e.g. 5(3x^2y - 4xy^2), x(15xy - 20y^2), y(15x^2 - 20xy) or xy(15x - 20y)
B2 for 5xy(3x - 4y) as final answer

Find the first differences between consecutive terms:
- \(9 - 2 = 7\)
- \(20 - 9 = 11\)
- \(35 - 20 = 15\)

Now find the second differences (the differences between the first differences):
- \(11 - 7 = 4\)
- \(15 - 11 = 4\)

Since the second difference is constant (+4), the next first difference will be:
\[15 + 4 = 19\]

Therefore, the next term in the sequence is:
\[35 + 19 = 54\]

M1 for finding first differences 7, 11, 15 or second difference of 4
A1 for 54

Apply the power \(\frac{2}{3}\) to each part inside the bracket:
\[(64x^6)^{\frac{2}{3}} = (64)^{\frac{2}{3}} \times (x^6)^{\frac{2}{3}}\]

Calculate the numerical part:
\[(64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = (4)^2 = 16\]

Calculate the variable part:
\[(x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\]

Combine the parts to get the final simplified expression:
\[16x^4\]

B1 for 16 or x^4 in the final answer, or for (4x^2)^2 seen
B2 for 16x^4 as final answer

In a right-angled triangle, the side opposite the smallest angle is always the shortest side. Here, the shortest side is \(5\text{ cm}\).
Let \(\theta\) be the smallest angle.
Using the sine ratio:
\[\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}\]

Solve for \(\theta\):
\[\theta = \sin^{-1}\left(\frac{5}{13}\right) \approx 22.6198^\circ\]

Rounding to 1 decimal place gives:
\[22.6^\circ\]

M1 for sin(\theta) = 5/13 or cos(\theta) = 12/13 or tan(\theta) = 5/12
A1 for 22.6

A reduction of \(8\%\) means the sale price is \(100\% - 8\% = 92\%\) of the original price.
Let the original price be \(P\).
We can write the equation:
\[0.92 \times P = 345\]

Solve for \(P\):
\[P = \frac{345}{0.92} = 375\]

Therefore, the original price of the smartphone was \(\$375\).

M1 for 345 / 0.92 or 345 / 92 * 100
A1 for 375

First, find \(2\mathbf{a}\) and \(3\mathbf{b}\):
\[2\mathbf{a} = 2 \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}\]
\[3\mathbf{b} = 3 \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}\]

Now, subtract the vectors:
\[2\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} -6 \\ 8 \end{pmatrix} - \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} -6 - 6 \\ 8 - (-3) \end{pmatrix} = \begin{pmatrix} -12 \\ 11 \end{pmatrix}\]

B1 for \begin{pmatrix} -6 \\ 8 \end{pmatrix} or \begin{pmatrix} 6 \\ -3 \end{pmatrix} seen, or for one component correct in the final vector
B2 for \begin{pmatrix} -12 \\ 11 \end{pmatrix}

The measurements are given to the nearest millimetre (\(0.1\text{ cm}\)), so the maximum absolute error for each dimension is \(0.05\text{ cm}\).

Find the upper bounds for the length and width:
- Upper bound of the length: \(8.4 + 0.05 = 8.45\text{ cm}\)
- Upper bound of the width: \(5.2 + 0.05 = 5.25\text{ cm}\)

Calculate the upper bound of the perimeter:
\[\text{Perimeter} = 2 \times (\text{Length} + \text{Width})\]
\[\text{Upper bound of perimeter} = 2 \times (8.45 + 5.25) = 2 \times 13.70 = 27.4\text{ cm}\]

M1 for 2 * (their UB_length + their UB_width) where at least one upper bound is correct (8.45 or 5.25)
A1 for 27.4

Multiply the coefficients: \( 3.5 \times 8 = 28 \). Multiply the powers of 10: \( 10^6 \times 10^{-3} = 10^3 \). Combine: \( 28 \times 10^3 = 2.8 \times 10^4 \).

M1 for \( 28 \times 10^3 \) or 28000. A1 for \( 2.8 \times 10^4 \).

Find the highest common factor of \( 6x^2y \) and \( 15xy^2 \), which is \( 3xy \). Divide both terms by \( 3xy \): \( 6x^2y \div 3xy = 2x \) and \( 15xy^2 \div 3xy = 5y \). Thus, we get \( 3xy(2x - 5y) \).

B1 for any correct partial factorisation, e.g., \( 3x(2xy - 5y^2) \) or \( 3xy(2x - 5y) \) seen then spoilt. B1 for \( 3xy(2x - 5y) \) as final answer.

The measurement is correct to the nearest millimetre (\( 0.1\text{ cm} \)). The lower bound of the side length is \( 7.4 - 0.05 = 7.35\text{ cm} \). The perimeter is \( 4 \times \text{side length} \). Lower bound of perimeter = \( 4 \times 7.35 = 29.4\text{ cm} \).

M1 for \( 7.35 \) seen or for \( 4 \times \text{their lower bound} \). A1 for \( 29.4 \).

Use the simple interest formula: \( I = \frac{PRT}{100} \). Here, \( I = \frac{650 \times 2.4 \times 5}{100} = 78 \).

M1 for \( \frac{650 \times 2.4 \times 5}{100} \) oe. A1 for \( 78 \).

To divide by a fraction, multiply by its reciprocal: \( \frac{5}{6} \div \frac{15}{16} = \frac{5}{6} \times \frac{16}{15} \). Simplify by cross-cancelling: \( = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9} \).

M1 for \( \frac{5}{6} \times \frac{16}{15} \) oe. A1 for \( \frac{8}{9} \) cao.

The size of each exterior angle is \( 180^\circ - 140^\circ = 40^\circ \). The sum of the exterior angles of any regular polygon is \( 360^\circ \). Number of sides = \( \frac{360^\circ}{40^\circ} = 9 \).

M1 for \( 180 - 140 \) or for setting up the equation \( \frac{180(n-2)}{n} = 140 \) oe. A1 for \( 9 \).

Apply the power \( \frac{2}{3} \) to both parts of the product: \( (27x^6)^{\frac{2}{3}} = 27^{\frac{2}{3}} \times (x^6)^{\frac{2}{3}} \). First, \( 27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9 \). Next, \( (x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4 \). Combining these gives \( 9x^4 \).

B1 for \( 9 \) or \( x^4 \) in the final answer, or for showing \( \sqrt[3]{27^2}x^4 \). B1 for \( 9x^4 \) as final answer.

The probability that the first pen is red is \( \frac{5}{8} \). Since the pen is not replaced, there are now \(4\) red pens and a total of \(7\) pens left in the box. The probability that the second pen is red is \( \frac{4}{7} \). The combined probability that both are red is \( \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \).

M1 for \( \frac{5}{8} \times \frac{4}{7} \) oe. A1 for \( \frac{5}{14} \) oe (or decimal \(0.357\) or \(0.3571\dots\)).

Apply the power of $-\frac{3}{4}$ to each factor inside the bracket:
$$\left(81a^8b^{-4}\right)^{-\frac{3}{4}} = 81^{-\frac{3}{4}} \times \left(a^8\right)^{-\frac{3}{4}} \times \left(b^{-4}\right)^{-\frac{3}{4}}$$

Evaluate each component:
- $81^{-\frac{3}{4}} = \left(3^4\right)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{27}$
- $\left(a^8\right)^{-\frac{3}{4}} = a^{8 \times -\frac{3}{4}} = a^{-6} = \frac{1}{a^6}$
- $\left(b^{-4}\right)^{-\frac{3}{4}} = b^{-4 \times -\frac{3}{4}} = b^3$

Multiply the simplified components together:
$$\frac{1}{27} \times \frac{1}{a^6} \times b^3 = \frac{b^3}{27a^6}$$

This can also be written in index notation as $\frac{1}{27}a^{-6}b^3$.

M1 for applying the negative fractional power correctly to at least two terms, e.g. obtaining $a^{-6}$, $b^3$, or $81^{-\frac{3}{4}} = \frac{1}{27}$
A1 for $\frac{b^3}{27a^6}$ or $\frac{1}{27}a^{-6}b^3$ or equivalent

Let $x$ be the original price of the smart watch.

A reduction of 15% means the sale price is $100\% - 15\% = 85\%$ of the original price.

Write down the equation:
$$0.85x = 132.60$$

Solve for $x$:
$$x = \frac{132.60}{0.85} = 156$$

Therefore, the original price of the watch was $156.

M1 for $132.60 \div 0.85$ or $132.60 \div 85 \times 100$ or equivalent
A1 for $156$ (accept $156.00$ or $156)

\(2.4 \times 10^7 - 6 \times 10^6 = 2.4 \times 10^7 - 0.6 \times 10^7 = 1.8 \times 10^7\).

Dividing this by \(3 \times 10^{-2}\):

\(\frac{1.8 \times 10^7}{3 \times 10^{-2}} = 0.6 \times 10^{7 - (-2)} = 0.6 \times 10^9 = 6 \times 10^8\).

[1] For finding the numerator as \(1.8 \times 10^7\) (or 18,000,000).
[1] For dividing by \(3 \times 10^{-2}\) to get 600,000,000.
[1] For the final answer in correct standard form: \(6 \times 10^8\).

Using the Cosine Rule:

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

\(AC^2 = 7.5^2 + 6.2^2 - 2(7.5)(6.2)\cos(112^\circ)\)

\(AC^2 = 56.25 + 38.44 - 93\cos(112^\circ)\)

\(AC^2 \approx 94.69 - 93(-0.3746) = 94.69 + 34.838 = 129.528\)

\(AC = \sqrt{129.528} \approx 11.38\text{ cm}\).

To 3 significant figures, the length of \(AC\) is 11.4 cm.

[1] For correct substitution into Cosine Rule, i.e., \(7.5^2 + 6.2^2 - 2(7.5)(6.2)\cos(112^\circ)\).
[1] For finding \(AC^2 \approx 129.5\).
[1] For the final answer 11.4 (accept 11.38 to 11.40).

To find the upper bound of \(\frac{p}{q}\), we use the upper bound of \(p\) and the lower bound of \(q\):

Upper bound of \(p = 4.85\).

Lower bound of \(q = 0.245\).

Upper bound of \(\frac{p}{q} = \frac{4.85}{0.245} \approx 19.7959\).

To 3 significant figures, this is 19.8.

[1] For identifying the upper bound of \(p\) as 4.85 or the lower bound of \(q\) as 0.245.
[1] For the division \(\frac{4.85}{0.245}\).
[1] For the correct final answer of 19.8 (or 19.79 to 19.8).

\(\overrightarrow{PM} = \overrightarrow{PO} + \overrightarrow{OM}\)

Since \(OP : PB = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{b}\), so \(\overrightarrow{PO} = -\frac{2}{3}\mathbf{b}\).

\(M\) is the midpoint of \(AB\), so \(\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

Thus,

\(\overrightarrow{PM} = -\frac{2}{3}\mathbf{b} + \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{1}{2}\mathbf{a} - \frac{1}{6}\mathbf{b}\).

[1] For finding \(\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or \(\overrightarrow{PO} = -\frac{2}{3}\mathbf{b}\).
[1] For a correct vector addition route, e.g., \(\overrightarrow{PM} = \overrightarrow{PO} + \overrightarrow{OM}\).
[1] For the final simplified answer: \(\frac{1}{2}\mathbf{a} - \frac{1}{6}\mathbf{b}\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 3\), \(b = -4\), and \(c = -5\):

\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)}\)

\(x = \frac{4 \pm \sqrt{16 + 60}}{6}\)

\(x = \frac{4 \pm \sqrt{76}}{6}\)

\(x_1 = \frac{4 + 8.7178}{6} \approx 2.12\)

\(x_2 = \frac{4 - 8.7178}{6} \approx -0.79\)

[1] For correct substitution into the quadratic formula (or a completed square form).
[1] For finding \(\sqrt{76}\) or 8.72 seen, or showing \(\frac{4 \pm \sqrt{76}}{6}\).
[1] For both correct solutions to 2 d.p.: 2.12 and -0.79.

Calculate the total score of the first class: \(24 \times 65 = 1560\).

Calculate the total score of the second class: \(16 \times 72 = 1152\).

Add the total scores: \(1560 + 1152 = 2712\).

Divide by the total number of students (40):

Overall mean = \(\frac{2712}{40} = 67.8\%\).

[1] For finding the total score of either class (1560 or 1152).
[1] For the sum of total scores (2712) divided by 40.
[1] For the final correct mean: 67.8 (or 67.8%).

Rearrange the equation to solve for \(\sin x\):

\(5\sin x = -2 \implies \sin x = -0.4\).

Find the principal value: \(x = \arcsin(-0.4) \approx -23.58^\circ\).

In the range \(0^\circ \le x \le 360^\circ\):

First solution: \(x_1 = 180^\circ - (-23.58^\circ) = 203.58^\circ \approx 203.6^\circ\).

Second solution: \(x_2 = 360^\circ + (-23.58^\circ) = 336.42^\circ \approx 336.4^\circ\).

[1] For rearranging to \(\sin x = -0.4\).
[1] For one correct angle (either 203.6 or 336.4) to 1 d.p.
[1] For both correct angles: 203.6 and 336.4.

To find the coordinates of point $C$ which divides the line segment $AB$ in the ratio $3 : 1$:

1. Find the total number of parts in the ratio:
$$3 + 1 = 4$$
So, point $C$ is $\frac{3}{4}$ of the distance from $A$ to $B$.

2. Calculate the change in the $x$-coordinate from $A$ to $B$:
$$\Delta x = x_B - x_A = 5 - (-3) = 8$$
Multiply this by $\frac{3}{4}$:
$$\frac{3}{4} \times 8 = 6$$

3. Calculate the change in the $y$-coordinate from $A$ to $B$:
$$\Delta y = y_B - y_A = -1 - 11 = -12$$
Multiply this by $\frac{3}{4}$:
$$\frac{3}{4} \times (-12) = -9$$

4. Add these changes to the coordinates of $A$:
$$x_C = -3 + 6 = 3$$
$$y_C = 11 + (-9) = 2$$

Therefore, the coordinates of $C$ are $(3, 2)$.

**M1** for finding the difference in $x$ and $y$ coordinates, i.e., $\Delta x = 8$ and $\Delta y = -12$ (or finding vector $\vec{AB} = \begin{pmatrix} 8 \\ -12 \end{pmatrix}$)
**M1** for multiplying the coordinate differences by $\frac{3}{4}$ to get $+6$ and $-9$
**M1** for adding these changes to the coordinates of $A$, i.e., $x = -3 + 6$ and $y = 11 - 9$
**A1** for $(3, 2)$ (accept written as $x = 3, y = 2$)

To find the upper bound for the difference: $\text{Area of square} - \text{Area of circle}$, we need to use the upper bound of the square's side length and the lower bound of the circle's radius.

1. Find the bounds:
- Side length of the square, $s = 12.4\text{ cm}$ (correct to nearest $0.1\text{ cm}$):
$$\text{Upper bound } s_{\text{upper}} = 12.4 + 0.05 = 12.45\text{ cm}$$
- Radius of the circle, $r = 5.6\text{ cm}$ (correct to nearest $0.1\text{ cm}$):
$$\text{Lower bound } r_{\text{lower}} = 5.6 - 0.05 = 5.55\text{ cm}$$

2. Calculate the bounds of the areas:
- Upper bound for the area of the square:
$$A_{\text{square, upper}} = s_{\text{upper}}^2 = 12.45^2 = 155.0025\text{ cm}^2$$
- Lower bound for the area of the circle:
$$A_{\text{circle, lower}} = \pi r_{\text{lower}}^2 = \pi \times 5.55^2 \approx 96.7689\text{ cm}^2$$

3. Calculate the upper bound of the difference:
$$\text{Difference}_{\text{upper}} = 155.0025 - 96.7689 = 58.2336...\text{ cm}^2$$

Rounding to 3 significant figures gives $58.2$.

**B1** for $12.45$ seen as the upper bound of the side length OR $5.55$ seen as the lower bound of the radius
**M1** for $12.45^2$ OR $\pi \times 5.55^2$ (or $3.142 \times 5.55^2$)
**M1** for their $A_{\text{square, upper}} - A_{\text{circle, lower}}$
**A1** for $58.2$

(a) The recorded distance is \(10\text{ km}\) correct to the nearest \(0.1\text{ km}\).
Lower bound for the distance in km is:
\(10 - 0.05 = 9.95\text{ km}\).
Converting to metres:
\(9.95 \times 1000 = 9950\text{ m}\).

(b) To find the upper bound for the average speed, we use the formula:
\(\text{Speed}_{\text{UB}} = \frac{\text{Distance}_{\text{UB}}}{\text{Time}_{\text{LB}}}\).

The upper bound for the distance is:
\(10 + 0.05 = 10.05\text{ km} = 10050\text{ m}\).

The lower bound for the time is:
\(45 - 0.5 = 44.5\text{ minutes}\).
Converting to seconds:
\(44.5 \times 60 = 2670\text{ seconds}\).

Therefore, the upper bound for speed is:
\(\text{Speed}_{\text{UB}} = \frac{10050}{2670} \approx 3.76404\text{ m/s}\).
Correct to 3 significant figures, this is \(3.76\text{ m/s}\).

(a) [2 marks]
B2 for 9950
(B1 for 9.95 seen)

(b) [6 marks]
M1 for 10.05 (km) or 10050 (m) seen
M1 for 44.5 (mins) or 2670 (s) seen
M1 for converting their chosen distance to metres
M1 for converting their chosen time to seconds
M1 for their max distance divided by their min time
A1 for 3.76

(a) Using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 65^2 + 80^2 - 2(65)(80)\cos(72^\circ)\)
\(AC^2 = 4225 + 6400 - 10400\cos(72^\circ)\)
\(AC^2 = 10625 - 3213.78\)
\(AC^2 = 7411.22\)
\(AC = \sqrt{7411.22} \approx 86.088\text{ m}\).
Correct to 3 significant figures, \(AC = 86.1\text{ m}\).

(b) Using the Area Rule for triangles:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \times 65 \times 80 \times \sin(72^\circ)\)
\(\text{Area} = 2600 \times \sin(72^\circ) \approx 2472.75\text{ m}^2\).
Correct to 3 significant figures, \(\text{Area} = 2470\text{ m}^2\).

(c) The area can also be expressed as:
\(\text{Area} = \frac{1}{2} \times AC \times BD\)
\(2472.75 = \frac{1}{2} \times 86.088 \times BD\)
\(BD = \frac{2 \times 2472.75}{86.088} \approx 57.447\text{ m}\).
Correct to 3 significant figures, \(BD = 57.4\text{ m}\).

(a) [3 marks]
M1 for \(65^2 + 80^2 - 2(65)(80)\cos(72^\circ)\)
A1 for \(7411.22\) or \(7410\)
A1 for \(86.1\) (accept \(86.08\) to \(86.1\))

(b) [2 marks]
M1 for \(\frac{1}{2} \times 65 \times 80 \times \sin(72^\circ)\)
A1 for \(2470\) or \(2473\)

(c) [3 marks]
M1 for equating their area to \(\frac{1}{2} \times \text{their } AC \times BD\)
M1 for correct rearrangement to make \(BD\) the subject
A1 for \(57.4\) (accept \(57.4\) to \(57.5\))

(a) Looking at the differences between consecutive terms:
\(12 - 5 = 7\)
\(25 - 12 = 13\)
\(44 - 25 = 19\)
\(69 - 44 = 25\)

The differences are \(7, 13, 19, 25, ...\)
These differences increase by \(6\) each time, so the next difference is \(25 + 6 = 31\).
Thus, the next term of the sequence is \(69 + 31 = 100\).

(b) Since the second differences are constant at \(6\), the sequence is quadratic and has the general form \(an^2 + bn + c\).
Here, \(2a = 6 \implies a = 3\).
Subtracting \(3n^2\) from the terms:
- For \(n=1\): \(5 - 3(1)^2 = 2\)
- For \(n=2\): \(12 - 3(2)^2 = 0\)
- For \(n=3\): \(25 - 3(3)^2 = -2\)
- For \(n=4\): \(44 - 3(4)^2 = -4\)

The remaining sequence is \(2, 0, -2, -4, ...\), which is linear with first term \(2\) and common difference \(-2\).
The expression for this linear sequence is \(2 + (n-1)(-2) = 4 - 2n\).
Combining the quadratic and linear parts gives the \(n\)th term:
\(3n^2 - 2n + 4\).

(c) Substitute \(n = 4\) into the given expression:
\(\text{4th term} = \frac{3(4) - 1}{2^4} = \frac{12 - 1}{16} = \frac{11}{16}\).
This is already in its simplest form.

(a) [1 mark]
B1 for 100

(b) [4 marks]
M1 for finding constant second difference of 6 or setting \(a = 3\)
M1 for substituting values to set up linear equations in \(b\) and \(c\)
M1 for solving equations to find \(b = -2\) or \(c = 4\)
A1 for \(3n^2 - 2n + 4\) (or equivalent)

(c) [3 marks]
M1 for substituting \(n = 4\) into the expression
A1 for \(\frac{12 - 1}{16}\) or \(\frac{11}{16}\)
A1 for \(\frac{11}{16}\) as final simplest fraction

(a) The total volume of the toy consists of the volume of the cone and the volume of the hemisphere.
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3\).
\(V_{\text{hemisphere}} = \frac{2}{3} \pi r^3\).

Total Volume:
\(V = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3\).
This matches the required formula.

(b) (i) Setting \(V = 250\):
\(\frac{4}{3}\pi r^3 = 250\)
\(\pi r^3 = 187.5\)
\(r^3 = \frac{187.5}{\pi} \approx 59.683\)
\(r = \sqrt[3]{59.683} \approx 3.9079\text{ cm}\).
Correct to 3 significant figures, \(r = 3.91\text{ cm}\).

(b) (ii) The total surface area consists of the curved surface area of the hemisphere and the curved surface area of the cone.
For the hemisphere:
\(\text{Curved surface area} = 2\pi r^2 = 2 \times \pi \times (3.9079)^2 \approx 95.95\text{ cm}^2\).

For the cone, first calculate the slant height \(l\):
\(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5}\).
\(l = 3.9079 \times \sqrt{5} \approx 8.7383\text{ cm}\).

\(\text{Cone curved surface area} = \pi r l = \pi \times 3.9079 \times 8.7383 \approx 107.29\text{ cm}^2\).

Total surface area:
\(\text{Total Area} = 95.95 + 107.29 = 203.24\text{ cm}^2\).
Correct to 3 significant figures, this is \(203\text{ cm}^2\).

(a) [3 marks]
M1 for volume of cone formula \(\frac{1}{3}\pi r^2 (2r)\)
M1 for volume of hemisphere formula \(\frac{2}{3}\pi r^3\)
A1 for completing algebra to show \(\frac{4}{3}\pi r^3\) rigorously

(b) (i) [3 marks]
M1 for setting \(\frac{4}{3}\pi r^3 = 250\)
M1 for rearranging to find \(r^3 = 59.7\) or \(\frac{187.5}{\pi}\)
A1 for \(3.91\) (accept \(3.90\) to \(3.91\))

(b) (ii) [2 marks]
M1 for writing or substituting into curved surface area expression \(2\pi r^2 + \pi r (r\sqrt{5})\) oe
A1 for \(203\) (accept \(203\) to \(204\))

(a) To find the missing values \(p\) and \(q\):
- For \(x = -2\):
\(p = f(-2) = \frac{12}{-2} - (-2)^2 = -6 - 4 = -10\).
- For \(x = 2\):
\(q = f(2) = \frac{12}{2} - (2)^2 = 6 - 4 = 2\).

Thus, the table is completed with \(p = -10\) and \(q = 2\).

(b) (i) Looking at the table, when \(y = 11\), we have \(x = 1\).
So, \(x = 1\).

(b) (ii) Set \(f(x) = 3x - 1\):
\(\frac{12}{x} - x^2 = 3x - 1\)

Multiply the entire equation by \(x\) (since \(x \neq 0\)):
\(12 - x^3 = 3x^2 - x\)

Rearrange all terms to one side to get the standard form:
\(x^3 + 3x^2 - x - 12 = 0\).
This matches the required equation.

(a) [2 marks]
B1 for \(p = -10\)
B1 for \(q = 2\)

(b) (i) [2 marks]
M1 for referring to table at \(y=11\) or solving equation \(\frac{12}{x} - x^2 = 11\)
A1 for \(x = 1\)

(b) (ii) [4 marks]
M1 for setting up \(\frac{12}{x} - x^2 = 3x - 1\)
M1 for multiplying all terms correctly by \(x\)
M1 for rearranging terms to one side of the equation
A1 for obtaining \(x^3 + 3x^2 - x - 12 = 0\) with no errors seen

(a) First selection has total balls \(5 + 4 = 9\).
- \(\text{P(Red)} = \frac{5}{9}\)
- \(\text{P(Blue)} = \frac{4}{9}\)

Second selection (without replacement, total balls \(8\)):
- If first is Red (remaining: 4 Red, 4 Blue):
- \(\text{P(Red | Red)} = \frac{4}{8}\)
- \(\text{P(Blue | Red)} = \frac{4}{8}\)
- If first is Blue (remaining: 5 Red, 3 Blue):
- \(\text{P(Red | Blue)} = \frac{5}{8}\)
- \(\text{P(Blue | Blue)} = \frac{3}{8}\)

(b) (i) The outcomes for both balls being the same colour are Red-Red (RR) or Blue-Blue (BB):
\(\text{P(Same)} = \text{P(RR)} + \text{P(BB)}\)
\(\text{P(RR)} = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\)
\(\text{P(BB)} = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}\)
\(\text{P(Same)} = \frac{20}{72} + \frac{12}{72} = \frac{32}{72} = \frac{4}{9}\) (or \(0.444\)).

(b) (ii) The outcome for at least one ball being red is everything except Blue-Blue (BB):
\(\text{P(At least one Red)} = 1 - \text{P(BB)}\)
\(\text{P(At least one Red)} = 1 - \frac{12}{72} = 1 - \frac{1}{6} = \frac{5}{6}\) (or \(0.833\)).

(a) [3 marks]
B1 for first selection branches correct (probabilities \(\frac{5}{9}\) and \(\frac{4}{9}\))
B1 for second selection branches from Red correct (probabilities \(\frac{4}{8}\) and \(\frac{4}{8}\))
B1 for second selection branches from Blue correct (probabilities \(\frac{5}{8}\) and \(\frac{3}{8}\))

(b) (i) [2 marks]
M1 for \(\frac{5}{9} \times \frac{4}{8} + \frac{4}{9} \times \frac{3}{8}\)
A1 for \(\frac{4}{9}\) (or equivalent fraction, or \(0.444\))

(b) (ii) [3 marks]
M1 for identifying the correct scenarios (RR, RB, BR) or using the complement method
M1 for calculation: e.g. \(1 - \frac{4}{9} \times \frac{3}{8}\) or \(\frac{20}{72} + \frac{20}{72} + \frac{20}{72}\)
A1 for \(\frac{5}{6}\) (or equivalent fraction, or \(0.833\))

(a) Total parts = \(5 + 3 = 8\).
Number of Mountain bicycles = \(960 \times \frac{5}{8} = 600\).

(b) (i) Selling price = \(320 \times 1.35 = 432\).

(ii) 2024 manufacturing cost = \(320 \times 1.08 = 345.60\).
Profit = \(453.60 - 345.60 = 108\).
Percentage profit = \(\frac{108}{345.60} \times 100 = 31.25\%\).

(c) Selling price = \(x \times \left(1 + \frac{y}{100}\right) = 480\)
\(x \times \frac{100 + y}{100} = 480\)
\(x = \frac{480 \times 100}{100 + y} = \frac{48000}{100 + y}\).

(d) Profit on 1 Road bicycle = \(480 - x = 480 - \frac{48000}{100 + y}\)
\(= \frac{480(100 + y) - 48000}{100 + y} = \frac{48000 + 480y - 48000}{100 + y} = \frac{480y}{100 + y}\).
Total profit for 360 bicycles = \(360 \times \frac{480y}{100 + y} = \frac{172800y}{100 + y}\).

(a) M1 for \(960 \div 8\) or \(960 \times \frac{5}{8}\)
A1 for 600

(b)(i) M1 for \(320 \times 1.35\) or \(320 + 0.35 \times 320\)
A1 for 432

(b)(ii) M1 for \(320 \times 1.08 = 345.60\)
M1 for \(\frac{453.60 - \text{their } 345.60}{\text{their } 345.60} \times 100\)
A1 for 31.25

(c) M1 for setting up \(x \left(1 + \frac{y}{100}\right) = 480\) or equivalent
A1 for showing correct algebraic steps to arrive at \(x = \frac{48000}{100 + y}\)

(d) M1 for writing profit per bike as \(480 - \frac{48000}{100 + y}\) or total profit as \(360 \left(480 - \frac{48000}{100 + y}\right)\)
M1 for obtaining common denominator \(100 + y\)
A1 for \(\frac{172800y}{100+y}\) as final simplified answer

(a) The angle \(QPR = 115^\circ - 65^\circ = 50^\circ\).
Using the cosine rule:
\(QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos(50^\circ)\)
\(QR^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos(50^\circ)\)
\(QR^2 = 144 + 324 - 432 \cdot \cos(50^\circ)\)
\(QR^2 \approx 468 - 432 \cdot 0.64279 = 190.32\)
\(QR \approx 13.8\text{ km}\) (13.795 km).

(b) Using the sine rule to find angle \(PQR\):
\(\frac{\sin(PQR)}{18} = \frac{\sin(50^\circ)}{13.795}\)
\(\sin(PQR) = \frac{18 \cdot \sin(50^\circ)}{13.795} \approx 0.9995\)
\(PQR \approx \sin^{-1}(0.9995) \approx 88.2^\circ\).
Angle made by line PQ with the South-North line at Q is \(180^\circ - 65^\circ = 115^\circ\) (co-interior).
This means the line QP points South-West at \(245^\circ\), which is \(65^\circ\) West of South.
Since angle \(PQR = 88.2^\circ\), line QR is rotated \(88.2^\circ\) clockwise from QP.
Angle between South at Q and QR is \(88.2^\circ - 65^\circ = 23.2^\circ\) East of South.
Bearing of R from Q = \(180^\circ - 23.2^\circ = 156.8^\circ\).

(c) The shortest distance from P to the line QR is the perpendicular distance, \(d\).
Using right-angled triangle trigonometry in the perpendicular height from P to line QR:
\(d = PQ \cdot \sin(PQR) = 12 \cdot \sin(88.2^\circ) \approx 12.0\text{ km}\).

(d) Distance = 13.795 km, Speed = 15 km/h.
Time taken = \(\frac{13.795}{15} \text{ hours} \approx 0.9197\text{ hours}\).
In minutes: \(0.9197 \times 60 = 55.18 \approx 55\text{ minutes}\).

(a) M1 for angle \(QPR = 50^\circ\) seen or used
M1 for \(QR^2 = 12^2 + 18^2 - 2 \times 12 \times 18 \cos(50^\circ)\)
A1 for 13.8 or 13.79 to 13.80

(b) M1 for \(\frac{\sin(PQR)}{18} = \frac{\sin(50^\circ)}{\text{their } QR}\)
A1 for angle \(PQR = 88.2^\circ\) or 88.15 to 88.25
M1 for bearing method e.g. \(180 - (\text{angle } PQR - 65)\) or equivalent
A1 for 156.8 or 156.75 to 156.85

(c) M1 for recognising shortest distance is the perpendicular height of the triangle
M1 for \(12 \sin(\text{their } 88.2^\circ)\) or \(\frac{1}{2} \times 12 \times 18 \sin(50^\circ) = \frac{1}{2} \times 13.795 \times d\)
A1 for 12.0 or 11.98 to 12.01

(d) M1 for \(\frac{\text{their } QR}{15} \times 60\)
A1 for 55

(a) \(f'(x) = 3x^2 - 6x - 9\).

(b) Setting \(f'(x) = 0\):
\(3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0\)
So, \(x = 3\) or \(x = -1\).
When \(x = 3\), \(y = 3^3 - 3(3^2) - 9(3) + 5 = -22\).
When \(x = -1\), \(y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\).
Turning points are \((-1, 10)\) and \(3, -22\).

(c) First-derivative or second-derivative test:
\(f''(x) = 6x - 6\).
At \(x = -1\), \(f''(-1) = -12 < 0\), so \((-1, 10)\) is a local maximum.
At \(x = 3\), \(f''(3) = 12 > 0\), so \((3, -22)\) is a local minimum.

(d) When \(x = 1\):
\(y = 1^3 - 3(1^2) - 9(1) + 5 = -6\).
Gradient of the tangent at \(x = 1\) is \(f'(1) = 3(1^2) - 6(1) - 9 = -12\).
Using the point-slope form:
\(y - (-6) = -12(x - 1) \implies y + 6 = -12x + 12 \implies y = -12x + 6\).

(a) B2 for \(3x^2 - 6x - 9\) (B1 for 2 terms correct)

(b) M1 for setting their derivative to 0
M1 for factorising or solving to get \(x = 3\) and \(x = -1\)
A1 for y-coordinates \(-22\) and \(10\)
A1 for writing as coordinates \((-1, 10)\) and \((3, -22)\)

(c) M1 for finding second derivative \(6x - 6\) or testing values on either side of turning points
A1 for identifying \((-1, 10)\) as maximum and \((3, -22)\) as minimum with correct justification

(d) M1 for finding \(y = -6\) when \(x = 1\)
M1 for substituting \(x = 1\) into their derivative to find gradient \(-12\)
M1 for substituting their \(y\), gradient, and \(x = 1\) into \(y = mx + c\)
A1 for \(y = -12x + 6\)

(a) The tree diagram should show:
- First branch: Red (R) with probability \(\frac{5}{8}\), Blue (B) with probability \(\frac{3}{8}\).
- Second branch after R: R with probability \(\frac{4}{7}\), B with probability \(\frac{3}{7}\).
- Second branch after B: R with probability \(\frac{5}{7}\), B with probability \(\frac{2}{7}\).

(b) (i) \(P(R_1 \cap R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\).

(ii) \(P(\text{different}) = P(R_1 \cap B_2) + P(B_1 \cap R_2) = \left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(c) Using conditional probability formula:
\(P(R_1 | R_2) = \frac{P(R_1 \cap R_2)}{P(R_2)}\).
We first find \(P(R_2)\):
\(P(R_2) = P(R_1 \cap R_2) + P(B_1 \cap R_2) = \frac{20}{56} + \left(\frac{3}{8} \times \frac{5}{7}\right) = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}\).
Then,
\(P(R_1 | R_2) = \frac{20/56}{35/56} = \frac{20}{35} = \frac{4}{5} = 0.8\).

(a) B1 for first branch probabilities correct (\(5/8\), \(3/8\))
B2 for second branch probabilities correct (\(4/7\), \(3/7\) and \(5/7\), \(2/7\))

(b)(i) M1 for \(\frac{5}{8} \times \frac{4}{7}\)
A1 for \(\frac{5}{14}\) or equivalent decimal \(0.357\)

(b)(ii) M1 for identifying two cases: RB and BR
M1 for \(\left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right)\)
A1 for \(\frac{15}{28}\) or equivalent decimal \(0.536\)

(c) M1 for writing conditional probability formula \(\frac{P(R_1 \cap R_2)}{P(R_2)}\)
M1 for finding \(P(R_2) = \frac{35}{56}\) or \(\frac{5}{8}\)
M1 for \(\frac{20/56}{35/56}\)
A1 for \(\frac{4}{5}\) or \(0.8\)

**(a)**
Using the cosine rule on triangle $ABC$:
$$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)$$
$$AC^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(75^\circ)$$
$$AC^2 = 14400 + 22500 - 36000 \times 0.258819...$$
$$AC^2 = 36900 - 9317.48...$$
$$AC^2 = 27582.51...$$
$$AC = 166.079...\text{ m}$$
$$AC = 166\text{ m} \text{ (correct to 3 s.f.)}$$

**(b)**
Using the area of a triangle formula:
$$\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)$$
$$\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(75^\circ)$$
$$\text{Area} = 9000 \times 0.965925...$$
$$\text{Area} = 8693.33...\text{ m}^2$$
$$\text{Area} = 8690\text{ m}^2 \text{ (correct to 3 s.f.)}$$

**(c)(i)**
Since $D$ is the point on $AC$ closest to $B$, the line $BD$ is perpendicular to $AC$.
Using the formula for the area of a triangle:
$$\text{Area} = \frac{1}{2} \times AC \times BD$$
$$8693.33... = \frac{1}{2} \times 166.079... \times BD$$
$$BD = \frac{2 \times 8693.33...}{166.079...}$$
$$BD = 104.688...\text{ m}$$
This rounds to $105\text{ m}$ to the nearest metre.

**(c)(ii)**
In the right-angled triangle $TDB$:
$$\tan(18^\circ) = \frac{TD}{BD}$$
$$TD = BD \times \tan(18^\circ)$$
Using the more accurate value of $BD = 104.688...$:
$$TD = 104.688... \times \tan(18^\circ)$$
$$TD = 34.015...\text{ m}$$
$$TD = 34.0\text{ m} \text{ (correct to 3 s.f.)}$$
*(Note: If using the rounded value of $BD = 105\text{ m}$, then $TD = 105 \times \tan(18^\circ) = 34.1\text{ m}$)*

**(c)(iii)**
First, calculate the length of $AD$ in the right-angled triangle $ADB$:
$$AD^2 = AB^2 - BD^2$$
$$AD^2 = 120^2 - 104.688...^2$$
$$AD^2 = 14400 - 10959.70...$$
$$AD^2 = 3440.30...$$
$$AD = 58.654...\text{ m}$$

In the right-angled triangle $TDA$, the angle of elevation of $T$ from $A$ is angle $TAD$:
$$\tan(\angle TAD) = \frac{TD}{AD}$$
$$\tan(\angle TAD) = \frac{34.015...}{58.654...} = 0.57993...$$
$$\angle TAD = \arctan(0.57993...) = 30.11^\circ$$
Therefore, the angle of elevation is $30.1^\circ$ (correct to 1 d.p.).
*(Note: If using the rounded values $BD = 105\text{ m}$ and $TD = 34.1\text{ m}$, then $AD = \sqrt{120^2 - 105^2} = 57.879\text{ m}$ and $\tan(\angle TAD) = \frac{34.1}{57.879}$ leading to $30.5^\circ$)*

**(a) [3 Marks]**
* **M1** for $120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(75)$
* **A1** for $[AC^2 = ]$ $27580$ to $27590$
* **A1** for $166$ (accept $166.0$ to $166.1$)

**(b) [2 Marks]**
* **M1** for $\frac{1}{2} \times 120 \times 150 \times \sin(75)$
* **A1** for $8690$ (accept $8693$ to $8694$)

**(c)(i) [3 Marks]**
* **M1** for identifying $BD$ is perpendicular to $AC$ (or finding angle $A$ e.g., $\sin(A) = \frac{150 \sin(75)}{\text{their } AC}$)
* **M1** for $\frac{1}{2} \times \text{their } AC \times BD = \text{their } \text{Area}$ or $120 \sin(\text{their } A)$
* **A1** for completing proof with value $104.68...$ shown before rounding to $105$

**(c)(ii) [3 Marks]**
* **M1** for $\tan(18) = \frac{TD}{\text{their } BD}$
* **A1** for $TD = 104.68... \times \tan(18)$ or $105 \times \tan(18)$
* **A1** for $34.0$ (or $34.1$ from using $105$)

**(c)(iii) [3 Marks]**
* **M1** for $AD^2 = 120^2 - \text{their } BD^2$ or $AD = 120 \cos(\text{their } A)$
* **M1** for $\tan(\text{angle}) = \frac{\text{their } TD}{\text{their } AD}$
* **A1** for $30.1$ to $30.2$ (or $30.5$ if using rounded values $105$ and $34.1$)