2024 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

Total selling price = \(80 \times 2.10 = \$168\)

Profit = \(\$168 - \$120 = \$48\)

Percentage profit = \(\frac{48}{120} \times 100 = 40\%\)

M1 for \(80 \times 2.10\) [= 168] or profit of 48 seen
M1 for \(\frac{\text{their } 168 - 120}{120} \times 100\) or \(\frac{\text{their profit}}{120} \times 100\)
A1 for 40

Using Pythagoras' theorem:

\(\text{Height}^2 + 2.5^2 = 6.5^2\)

\(\text{Height}^2 = 42.25 - 6.25\)

\(\text{Height}^2 = 36\)

\(\text{Height} = \sqrt{36} = 6\)

M1 for \(6.5^2 - 2.5^2\)
M1 for \(\sqrt{6.5^2 - 2.5^2}\) or \(\sqrt{36}\)
A1 for 6

Find the highest common factor of \(15a^2b\) and \(20ab^2\), which is \(5ab\).

Divide both terms by \(5ab\):

\(\frac{15a^2b}{5ab} = 3a\)

\(\frac{20ab^2}{5ab} = 4b\)

So, \(15a^2b - 20ab^2 = 5ab(3a - 4b)\).

B2 for \(5ab(3a - 4b)\) final answer
or B1 for \(5(3a^2b - 4ab^2)\) or \(a(15ab - 20b^2)\) or \(b(15a^2 - 20ab)\) or \(ab(15a - 20b)\) or \(5ab(\text{two-term expression})\)

To write 0.000305 in standard form \(A \times 10^n\), where \(1 \le A < 10\):

Move the decimal point 4 places to the right to get 3.05.

Since the decimal point was moved to the right, the index is negative: \(n = -4\).

So, \(3.05 \times 10^{-4}\).

B1 for \(3.05 \times 10^{-4}\)

Using the laws of indices:

\(\frac{3^8}{3^p} = 3^{8-p}\)

So, \(3^{8-p} = 3^2\)

\(8 - p = 2\)

\(p = 6\)

B1 for 6

The sequence is arithmetic with a first term of 18 and a common difference of \(-4\) (since \(14 - 18 = -4\)).

The formula for the \(n\)th term is:

\(\text{Term} = a + (n - 1)d\)

\(\text{Term} = 18 + (n - 1)(-4) = 18 - 4n + 4 = 22 - 4n\)

B2 for \(-4n + 22\) or \(22 - 4n\) oe final answer
or B1 for \(-4n + c\) or \(kn + 22\) (\(k \neq 0\))

The degree of accuracy is 0.5 kg.

Half of the degree of accuracy is \(0.5 \div 2 = 0.25\) kg.

Lower bound = \(18 - 0.25 = 17.75\) kg.

M1 for \(18 - 0.25\) or \(0.5 \div 2\)
A1 for 17.75

First find the intersection of sets \(P\) and \(Q\):

\(P \cap Q = \{3, 5, 7\}\)

The number of elements in \(P \cap Q\) is 3.

M1 for identifying the set \(P \cap Q = \{3, 5, 7\}\) or listing these elements
A1 for 3

42 098 000 written in words is forty-two million ninety-eight thousand.

B1 for forty-two million ninety-eight thousand (allow minor spelling errors, but place value must be correct).

The reciprocal of 0.8 is:
$$\frac{1}{0.8} = \frac{10}{8} = 1.25$$

B1 for 1.25 or \frac{5}{4} or 1\frac{1}{4}.

Convert each number to a decimal to compare:
- \(\frac{5}{8} = 0.625\)
- \(8.2 \times 10^{-1} = 0.82\)
- \(\frac{9}{11} \approx 0.818\)
- \(81.5\% = 0.815\)

Comparing the decimals: \(0.625 < 0.815 < 0.818 < 0.82\).

Therefore, the correct order is:
$$\frac{5}{8} < 81.5\% < \frac{9}{11} < 8.2 \times 10^{-1}$$

M1 for converting at least two of the numbers correctly to decimals or percentages for comparison.
A1 for the fully correct order as given.

1. Find the amount spent on the ticket:
$$\frac{1}{5} \times 600 = \$120$$

2. Find the total amount left after spending on the ticket and food is $315.

3. Calculate the total spent on ticket and food:
$$600 - 315 = \$285$$

4. Calculate the amount spent on food:
$$285 - 120 = \$165$$

5. Write this as a fraction of the total $600 and simplify:
$$\frac{165}{600} = \frac{33}{120} = \frac{11}{40}$$

M1 for finding the cost of the ticket ($120) or total spent ($285).
M1 for finding the amount spent on food ($165) or for writing \frac{165}{600} oe.
A1 for \frac{11}{40} or equivalent fraction in its simplest form.

$$\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} -3 \\ 8 \end{pmatrix} - 2\begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 8 \end{pmatrix} - \begin{pmatrix} 8 \\ -4 \end{pmatrix} = \begin{pmatrix} -3 - 8 \\ 8 - (-4) \end{pmatrix} = \begin{pmatrix} -11 \\ 12 \end{pmatrix}$$

B1 for either component correct in the final vector.
B1 for both components correct in the final vector.

The first significant figure is the 7 in the thousandths place.
The second significant figure is the 0 in the ten-thousandths place.
Looking at the next digit, which is 8 (since 8 >= 5), we round the 0 up to 1.

Thus, correct to 2 significant figures, the number is 0.0071.

B1 for 0.0071.

Using the cosine ratio in the right-angled triangle:
$$\cos(x) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{6.4}{11}$$

$$x = \arccos\left(\frac{6.4}{11}\right) \approx 54.417^{\circ}$$

Correct to 1 decimal place, \(x = 54.4\).

M1 for \cos(x) = \frac{6.4}{11} or equivalent trig ratio setup.
A1 for 54.4 or 54.41 to 54.42.

The degree of accuracy is to the nearest 5 g. Half of this accuracy is:
$$5 \div 2 = 2.5\text{ g}$$

Lower bound:
$$75 - 2.5 = 72.5\text{ g}$$

Upper bound:
$$75 + 2.5 = 77.5\text{ g}$$

Thus, the completed statement is:
$$72.5 \le m < 77.5$$

B1 for 72.5 as the lower bound.
B1 for 77.5 as the upper bound.

To compare the numbers, convert each to a decimal:
- \(\frac{5}{7} \approx 0.714\)
- \(7.3 \times 10^{-1} = 0.73\)
- \(0.72 = 0.72\)
- \(71.5\% = 0.715\)

Comparing these decimals:
\(0.714 < 0.715 < 0.72 < 0.73\)

Therefore, in ascending order: \(\frac{5}{7}\), \(71.5\%\), \(0.72\), \(7.3 \times 10^{-1}\).

M1 for converting at least 2 numbers correctly to decimals or percentages to enable comparison
A1 for correct order

Using right-angled trigonometry:
\[\cos(y^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{9}{15} = 0.6\]
\[y = \cos^{-1}(0.6) \approx 53.13\]
To 1 decimal place, \(y = 53.1\).

M1 for \(\cos(y) = \frac{9}{15}\) or better
A1 for 53.1

Find the highest common factor of both terms, which is \(8b\).
Factorise this out of the expression:
\[24ab - 16b^2 = 8b(3a - 2b)\]

B1 for partial factorisation: \(2(12ab - 8b^2)\), \(b(24a - 16b)\), or \(4b(6a - 4b)\)
B2 for \(8b(3a - 2b)\) completely correct

The price is reduced by 15%, which means the sale price is 85% of the normal price:
\[84 \times \left(1 - \frac{15}{100}\right) = 84 \times 0.85 = 71.40\]
The sale price is $71.40.

M1 for \(84 \times 0.15\) or \(84 \times 0.85\) or better
A1 for 71.40 (or 71.4)

The sequence is linear (arithmetic) with a first term of \(a = 31\) and a common difference of \(d = 25 - 31 = -6\).
Using the formula for the \(n\)th term:
\[a + (n - 1)d = 31 + (n - 1)(-6) = 31 - 6n + 6 = 37 - 6n\]

B1 for \(-6n + k\) (where \(k\) is any constant) or \(31 - 6(n - 1)\)
B2 for \(37 - 6n\) or equivalent final answer

First, identify the elements of each set within the universal set:
- \(\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)
- \(A = \{2, 3, 5, 7\}\)
- \(B = \{1, 2, 3, 4, 6\}\)

Find the intersection set \(A \cap B\):
\(A \cap B = \{2, 3\}\)

The number of elements is:
\(\text{n}(A \cap B) = 2\).

B1 for listing elements of \(A\) and \(B\) correctly, or for finding \(A \cap B = \{2, 3\}\)
B1 for 2

To write in standard form, represent the number as \(a \times 10^n\), where \(1 \le a < 10\).
Move the decimal point 4 places to the right:
\[0.000305 = 3.05 \times 10^{-4}\]

B1 for \(3.05 \times 10^{-4}\)

The volume \(V\) of a cylinder is given by the formula:
\[V = \pi r^2 h\]
Substitute the given values:
\[V = \pi \times 4^2 \times 11 = 176\pi \approx 552.920\text{ cm}^3\]
Rounding to 1 decimal place gives \(552.9\text{ cm}^3\).

M1 for \(\pi \times 4^2 \times 11\)
A1 for 552.9

Find the highest common factor of the numerical coefficients $24$ and $16$, which is $8$.

Find the highest common factor of the variable terms $y^2$ and $y$, which is $y$.

Therefore, the highest common factor of both terms is $8y$.

Dividing each term by $8y$ gives:
$$\frac{24y^2}{8y} = 3y$$

$$\frac{-16y}{8y} = -2$$

Combining these gives the fully factorised expression:
$$8y(3y - 2)$$

B2 for $8y(3y - 2)$ final answer

or B1 for a correct partial factorisation, such as $8(3y^2 - 2y)$, $y(24y - 16)$, $2y(12y - 8)$, $4y(6y - 4)$, or $8y(3y - 2)$ seen and then spoilt.

Convert the capacity of the tank to litres:
\(15\text{ m}^3 = 15 \times 1000 = 15\,000\text{ litres}\).

Convert the pumping rate to litres per second:
\(1.8 \times 10^4\text{ litres/hour} = 18\,000\text{ litres/hour}\).
\(18\,000 \div 3600 = 5\text{ litres/second}\).

Calculate the time taken in seconds:
\(\text{Time} = 15\,000 \div 5 = 3000\text{ seconds}\).

Write in standard form:
\(3 \times 10^3\).

M1 for converting capacity to litres (\(15\,000\)) or rate to \(\text{m}^3\text{/h}\) (\(18\)).
M1 for dividing their capacity by their rate (e.g., \(15\,000 \div 5\) or equivalent).
A1 for \(3 \times 10^3\) (or \(3.0 \times 10^3\)).

Use the Cosine Rule to find the length of \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(PQR)\)
\(PR^2 = 8.4^2 + 5.5^2 - 2(8.4)(5.5)\cos(42^\circ)\)
\(PR^2 = 70.56 + 30.25 - 92.4 \cdot 0.74314\)
\(PR^2 = 100.81 - 68.666\)
\(PR^2 = 32.144\)
\(PR = \sqrt{32.144} \approx 5.6696\text{ cm}\).

Correct to 3 significant figures, \(PR = 5.67\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(8.4^2 + 5.5^2 - 2 \cdot 8.4 \cdot 5.5 \cdot \cos(42)\).
A1 for \(PR^2 = 32.1\dots\) or better.
A1 for \(5.67\) (accept \(5.669\dots\)).

Analyze the differences between successive terms:
Terms: \(3, 8, 15, 24, 35\)
First differences: \(5, 7, 9, 11\)
Second differences: \(2, 2, 2\)

Since the second differences are constant and equal to \(2\), the quadratic term is \(an^2\) where \(a = 2 \div 2 = 1\).
Subtract \(n^2\) from each term:
\(3 - 1^2 = 2\)
\(8 - 2^2 = 4\)
\(15 - 3^2 = 6\)
\(24 - 4^2 = 8\)
\(35 - 5^2 = 10\)

The remaining sequence is \(2, 4, 6, 8, 10, \dots\), which has a linear term of \(2n\).
Thus, the \(n\)th term is \(n^2 + 2n\).

M1 for identifying the second difference is constant (\(2\)) or setting up the simultaneous equations for \(an^2+bn+c\).
M1 for establishing the quadratic coefficient is \(1\) or finding the remaining linear sequence is \(2n\).
A1 for \(n^2 + 2n\).

First simplify inside the brackets:
\(\frac{64x^6}{y^{-3}} = 64x^6 y^3\).

Now apply the power of \(-\frac{2}{3}\):
\((64x^6 y^3)^{-\frac{2}{3}} = (64)^{-\frac{2}{3}} \cdot (x^6)^{-\frac{2}{3}} \cdot (y^3)^{-\frac{2}{3}}\).

Evaluate each part:
\(64^{-\frac{2}{3}} = \frac{1}{64^{2/3}} = \frac{1}{4^2} = \frac{1}{16}\).
\((x^6)^{-\frac{2}{3}} = x^{-4} = \frac{1}{x^4}\).
\((y^3)^{-\frac{2}{3}} = y^{-2} = \frac{1}{y^2}\).

Combine the terms:
\(\frac{1}{16x^4y^2}\).

M1 for simplifying inside the bracket to \(64x^6y^3\) or evaluating at least two powers incorrectly but applying power index correctly.
M1 for evaluating \(64^{-2/3} = \frac{1}{16}\).
A1 for \(\frac{1}{16x^4y^2}\) or equivalent with negative indices (e.g. \(\frac{1}{16}x^{-4}y^{-2}\)).

Total number of pens in the box = \(5 + 4 = 9\).

Calculate the probability that both pens are blue:
\(P(\text{both blue}) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}\).

Subtract this probability from 1 to find the probability of at least one red pen:
\(P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - \frac{1}{6} = \frac{5}{6}\).

M1 for finding the probability of drawing two blue pens: \(\frac{4}{9} \times \frac{3}{8}\).
M1 for subtracting their \(P(\text{both blue})\) from 1, or for a fully correct sum of three products: \(\frac{5}{9}\times\frac{4}{8} + \frac{4}{9}\times\frac{5}{8} + \frac{5}{9}\times\frac{4}{8}\).
A1 for \(\frac{5}{6}\) (or equivalent fraction, or \(0.833\) or better).

First find vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\).

Now find vector \(\overrightarrow{AM}\):
\(\overrightarrow{AM} = \frac{1}{4}\overrightarrow{AB} = \frac{1}{4}(-\mathbf{a} + \mathbf{b})\).

Find the position vector of \(M\), which is \(\overrightarrow{OM}\):
\(\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{1}{4}(-\mathbf{a} + \mathbf{b}) = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\).

M1 for finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) (seen or implied).
M1 for using a correct route for \(\overrightarrow{OM}\), such as \(\mathbf{a} + \frac{1}{4}\overrightarrow{AB}\) or \(\mathbf{b} - \frac{3}{4}\overrightarrow{AB}\).
A1 for \(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\) or equivalent simplest form.

Find the gradient of the given line:
\(3x - 2y = 8 \implies 2y = 3x - 8 \implies y = \frac{3}{2}x - 4\).
So, the gradient of the given line is \(m_1 = \frac{3}{2}\).

The perpendicular gradient is:
\(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\).

Use the perpendicular gradient and point \((6, -1)\) to find the equation of the line:
\(y - (-1) = -\frac{2}{3}(x - 6)\)
\(y + 1 = -\frac{2}{3}x + 4\)
\(y = -\frac{2}{3}x + 3\).

M1 for finding gradient of given line is \(\frac{3}{2}\) or equivalent.
M1 for finding the perpendicular gradient of \(-\frac{2}{3}\) and substituting \((6, -1)\) into a linear equation.
A1 for \(y = -\frac{2}{3}x + 3\) or equivalent.

First calculate the number of students who study History and/or Geography:
\(n(H \cup G) = n(H) + n(G) - n(H \cap G)\)
\(n(H \cup G) = 28 + 18 - 8 = 38\).

Subtract this from the total number of students to find those who study neither:
\(\text{Neither} = 45 - 38 = 7\).

M1 for writing down \(20\) (History only) and/or \(10\) (Geography only) or drawing a Venn diagram with at least one correct region.
M1 for a complete method: \(45 - (20 + 8 + 10)\) or \(45 - (28 + 18 - 8)\).
A1 for \(7\).

To find how many times larger the mass of the first planet is compared to the second, we divide the first mass by the second mass:
\(\frac{4.8 \times 10^{24}}{1.6 \times 10^{21}}\)

Dividing the coefficients:
\(\frac{4.8}{1.6} = 3\)

Subtracting the exponents of 10:
\(10^{24 - 21} = 10^3\)

Combining these gives:
\(3 \times 10^3\)

M1 for \(\frac{4.8 \times 10^{24}}{1.6 \times 10^{21}}\)
A1 for 3000
A1 for \(3 \times 10^3\) (final answer in standard form)

First, find the upper bound for the length and the width.

The length is measured correct to the nearest \(5\text{ m}\), so the error interval is \(\pm 2.5\text{ m}\).
Upper bound for length, \(L_{\text{upper}} = 85 + 2.5 = 87.5\text{ m}\).

The width is measured correct to the nearest \(1\text{ m}\), so the error interval is \(\pm 0.5\text{ m}\).
Upper bound for width, \(W_{\text{upper}} = 40 + 0.5 = 40.5\text{ m}\).

The perimeter of a rectangle is given by \(P = 2(L + W)\).
The upper bound for the perimeter is:
\(P_{\text{upper}} = 2(L_{\text{upper}} + W_{\text{upper}}) = 2(87.5 + 40.5) = 2 \times 128 = 256\text{ m}\).

B1 for upper bound of length \(87.5\) or upper bound of width \(40.5\) seen
M1 for \(2 \times (\text{their } L_{\text{upper}} + \text{their } W_{\text{upper}})\)
A1 for \(256\)

First, find the common factors of the terms \(15x^2 y\) and \(-60y\).
The common factor is \(15y\).
Factorising out \(15y\):
\(15y(x^2 - 4)\)

Next, factorise the expression inside the bracket. This is a difference of two squares:
\(x^2 - 4 = (x - 2)(x + 2)\)

So, the completely factorised expression is:
\(15y(x - 2)(x + 2)\)

M1 for \(15y(x^2 - 4)\) or \(y(15x^2 - 60)\) or \(15(x^2y - 4y)\)
M1 for recognition of difference of two squares, e.g., \((x-2)(x+2)\)
A1 for \(15y(x-2)(x+2)\) oe

Let \(\theta\) be the angle the ladder makes with the ground.

The ladder represents the hypotenuse of a right-angled triangle, and the distance from the base of the ladder to the wall is the adjacent side to angle \(\theta\).

Using the cosine ratio:
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5}\)
\(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right)\)
\(\theta \approx 67.380...\)

Correct to 1 decimal place, the angle is \(67.4^\circ\).

M1 for \(\cos(\theta) = \frac{2.5}{6.5}\) or \(\sin(\theta) = \frac{\sqrt{6.5^2 - 2.5^2}}{6.5}\)
M1 for \(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right)\)
A1 for \(67.4\) or \(67.38...\)

Multiply each term of the equation by the common denominator \(x(x-4)\):
\(3x + 2(x-4) = 1x(x-4)\)

Expand the brackets:
\(3x + 2x - 8 = x^2 - 4x\)

Combine like terms:
\(5x - 8 = x^2 - 4x\)

Rearrange into a quadratic equation of the form \(ax^2 + bx + c = 0\):
\(x^2 - 9x + 8 = 0\)

Factorise the quadratic:
\((x - 1)(x - 8) = 0\)

Thus, \(x = 1\) or \(x = 8\).

M1 for \(3x + 2(x-4) = x(x-4)\) oe
M1 for \(x^2 - 9x + 8 = 0\)
A1 for \(1\) or \(8\) (both required)

Let the terms be \(u_1 = 2\), \(u_2 = 9\), \(u_3 = 20\), \(u_4 = 35\), \(u_5 = 54\).

Find the first differences:
\(9 - 2 = 7\)
\(20 - 9 = 11\)
\(35 - 20 = 15\)
\(54 - 35 = 19\)
The first differences are \(7, 11, 15, 19\).

Find the second differences:
\(11 - 7 = 4\)
\(15 - 11 = 4\)
\(19 - 15 = 4\)
The second differences are constant and equal to \(4\).

Since the second differences are constant, the sequence is quadratic and of the form \(an^2 + bn + c\), where:
\(2a = 4 \implies a = 2\).

Now, subtract the \(2n^2\) term from the original sequence to find the linear part:
For \(n=1\): \(2 - 2(1)^2 = 0\)
For \(n=2\): \(9 - 2(2)^2 = 1\)
For \(n=3\): \(20 - 2(3)^2 = 2\)
For \(n=4\): \(35 - 2(4)^2 = 3\)
For \(n=5\): \(54 - 2(5)^2 = 4\)

The remaining terms form the sequence \(0, 1, 2, 3, 4\), which is a linear sequence of the form \(dn + e\) with common difference \(1\) and first term \(0\), which simplifies to \(n - 1\).

Combining both parts, the \(n\)th term is:
\(2n^2 + n - 1\).

M1 for finding second difference of 4, leading to \(2n^2\)
M1 for subtracting \(2n^2\) from terms of sequence to get \(0, 1, 2, 3...\)
A1 for \(2n^2 + n - 1\) oe

First, find the vector \(2\mathbf{a} + 3\mathbf{b}\):
\(2\mathbf{a} = 2 \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}\)
\(3\mathbf{b} = 3 \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 15 \end{pmatrix}\)

\(2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \begin{pmatrix} -3 \\ 15 \end{pmatrix} = \begin{pmatrix} 6 + (-3) \\ -8 + 15 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}\).

Next, calculate the magnitude of the resulting vector \(\begin{pmatrix} 3 \\ 7 \end{pmatrix}\):
\(\left| 2\mathbf{a} + 3\mathbf{b} \right| = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58} \approx 7.6157...\)

To 3 significant figures, the magnitude is \(7.62\).

M1 for \(2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}\) (or finding individual components)
M1 for \(\sqrt{3^2 + 7^2}\) (Pythagoras on their components)
A1 for \(7.62\) or \(\sqrt{58}\) (accept \(7.615...\))

The total number of balls in the bag is \(5 + 3 = 8\).
We want the probability of getting two balls of different colours. This can happen in two ways:
1. Red first, then Blue (RB)
2. Blue first, then Red (BR)

Case 1: Red then Blue
Probability of Red on 1st pick: \(\frac{5}{8}\)
Probability of Blue on 2nd pick: \(\frac{3}{7}\) (since there is no replacement, 7 balls remain)
\(P(\text{RB}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\).

Case 2: Blue then Red
Probability of Blue on 1st pick: \(\frac{3}{8}\)
Probability of Red on 2nd pick: \(\frac{5}{7}\)
\(P(\text{BR}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\).

Total probability:
\(P(\text{different}) = P(\text{RB}) + P(\text{BR}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28} \approx 0.536\).

M1 for \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\)
M1 for addition of two different product scenarios, e.g., \(\left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right)\)
A1 for \(\frac{15}{28}\) or equivalent fraction or \(0.536\)

1. Find the area in square millimeters:
\(\text{Area} = (2.5 \times 10^2) \times (1.6 \times 10^2) = 4.0 \times 10^4\text{ mm}^2\).

2. Convert square millimeters to square meters:
\(1\text{ m}^2 = 1,000,000\text{ mm}^2 = 10^6\text{ mm}^2\).
\(\text{Area in m}^2 = \frac{4.0 \times 10^4}{10^6} = 4.0 \times 10^{-2}\text{ m}^2\).

3. Express in standard form:
\(4 \times 10^{-2}\) (or \(4.0 \times 10^{-2}\)).

M1 for calculating the area in \(\text{mm}^2\) as \(40000\) oe
M1 for dividing by \(10^6\) to convert to \(\text{m}^2\)
A1 for \(4 \times 10^{-2}\) or \(4.0 \times 10^{-2}\) in standard form

Use the Cosine Rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(115^\circ)\)
\(AC^2 = 54.76 + 84.64 - 136.16(-0.422618)\)
\(AC^2 = 139.4 + 57.5438\)
\(AC^2 = 196.9438\)
\(AC = \sqrt{196.9438} \approx 14.0336\text{ cm}\)

To 3 significant figures, \(AC = 14.0\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(115^\circ)\)
A1 for \(AC^2 = 196.9...\) or better
A1 for \(14.0\) (accept \(14\))

Find the differences between consecutive terms:
First differences: \(7, 11, 15, 19\)
Second differences: \(4, 4, 4\)

Since the second differences are constant and equal to 4, the sequence is quadratic of the form \(an^2 + bn + c\), where \(2a = 4 \implies a = 2\).

Subtract \(2n^2\) from each term:
- \(n = 1: 4 - 2(1^2) = 2\)
- \(n = 2: 11 - 2(2^2) = 3\)
- \(n = 3: 22 - 2(3^2) = 4\)
- \(n = 4: 37 - 2(4^2) = 5\)
- \(n = 5: 56 - 2(5^2) = 6\)

The resulting linear sequence \(2, 3, 4, 5, 6\) has the \(n\)th term \(n + 1\).

Therefore, the \(n\)th term of the original sequence is \(2n^2 + n + 1\).

M1 for finding constant second difference of 4 to identify a quadratic term of \(2n^2\)
M1 for subtracting \(2n^2\) and finding the linear part \(n + 1\)
A1 for \(2n^2 + n + 1\) (or equivalent)

1. Calculate the final value of the investment after 5 years using the compound interest formula:
\(A = P(1 + r)^n\)
\(A = 4500(1 + 0.028)^5\)
\(A = 4500(1.028)^5\)
\(A \approx 4500 \times 1.1480626 \approx 5166.28\)

2. Calculate the total interest earned by subtracting the principal from the final value:
\(\text{Interest} = A - P = 5166.28 - 4500 = 666.28\)

3. Round to the nearest dollar:
\(\text{Interest} = 666\).

M1 for correct substitution into compound interest formula: \(4500 \times 1.028^5\)
A1 for finding total value \(5166.28\) or interest \(666.28\)
A1 for \(666\) (rounded to nearest integer)

Express both fractions with a common denominator of \((x-2)(x+3)\):

\[\frac{3}{x-2} - \frac{2}{x+3} = \frac{3(x+3) - 2(x-2)}{(x-2)(x+3)}\]

Expand the numerator:

\[3(x+3) - 2(x-2) = 3x + 9 - 2x + 4 = x + 13\]

Thus, the single fraction in its simplest form is:

\[\frac{x+13}{(x-2)(x+3)}\]

M1 for writing with a common denominator \((x-2)(x+3)\) oe
M1 for correct expansion of the numerator: \(3x + 9 - 2x + 4\)
A1 for \(\frac{x+13}{(x-2)(x+3)}\) or \(\frac{x+13}{x^2+x-6}\)

Let \(T\) be the set of students playing tennis and \(B\) be the set of students playing basketball.

1. Find the number of students playing at least one of the sports:
\(n(T \cup B) = 30 - 5 = 25\).

2. Find the number of students playing both sports using the intersection formula:
\(n(T \cap B) = n(T) + n(B) - n(T \cup B)\)
\(n(T \cap B) = 18 + 15 - 25 = 8\).

3. Find the number of students playing tennis but not basketball:
\(n(T \text{ only}) = n(T) - n(T \cap B) = 18 - 8 = 10\).

4. Calculate the probability:
\(P(\text{plays tennis but not basketball}) = \frac{10}{30} = \frac{1}{3}\).

M1 for finding the number of students playing both sports: \(18+15-25 = 8\)
M1 for finding the number of students playing tennis only: \(18 - 8 = 10\)
A1 for \(\frac{1}{3}\) (or equivalent fraction, e.g. \(\frac{10}{30}\), or decimal \(0.333\))

Set \(y = g(x)\) and rearrange to solve for \(x\):

\[y = \frac{2}{x+1}\]

Multiply by \(x+1\):

\[y(x+1) = 2\]

Divide by \(y\):

\[x+1 = \frac{2}{y}\]

Subtract 1:

\[x = \frac{2}{y} - 1\]

Replace \(x\) with \(g^{-1}(x)\) and \(y\) with \(x\):

\[g^{-1}(x) = \frac{2}{x} - 1 \quad \left(\text{or } \frac{2-x}{x}\right)\]

M1 for setting up the equation \(y = \frac{2}{x+1}\) and multiplying by \(x+1\) oe
M1 for rearranging to make \(x\) the subject: \(x = \frac{2}{y} - 1\) oe
A1 for \(\frac{2}{x} - 1\) or \(\frac{2-x}{x}\)

1. Find the angle of the sector, \(\theta\), in degrees:
\(\text{Arc Length} = \frac{\theta}{360} \times 2\pi r\)
\[8\pi = \frac{\theta}{360} \times 2\pi \times 12\]
\[8\pi = \frac{24\pi\theta}{360}\]
\[8 = \frac{\theta}{15} \implies \theta = 120^\circ\]

2. Calculate the area of the sector:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)
\[\text{Area} = \frac{120}{360} \times \pi \times 12^2\]
\[\text{Area} = \frac{1}{3} \times 144\pi = 48\pi\text{ cm}^2\]

Alternative Method:
Using the formula for area of a sector directly from arc length \(L\) and radius \(r\):
\(\text{Area} = \frac{1}{2} r L = \frac{1}{2} \times 12 \times 8\pi = 48\pi\text{ cm}^2\).

M1 for setting up a correct equation for arc length to find \(\theta\): \(8\pi = \frac{\theta}{360} \times 24\pi\) oe, or using direct formula \(\frac{1}{2} r L\)
M1 for calculating \(\theta = 120^\circ\) or substitution into sector area formula
A1 for \(48\pi\)