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

First, expand the brackets:
\(4x - 8 - 3x + 15 = 19\)

Simplify the left-hand side by collecting like terms:
\(x + 7 = 19\)

Subtract 7 from both sides:
\(x = 19 - 7\)
\(x = 12\)

M1 for correct expansion of at least one bracket: \(4x - 8\) or \(-3x + 15\)
A1 for \(12\)

First, find the amount of Euros to be converted after the fee is deducted:
\(€450 - €12 = €438\)

Next, convert this amount into Dollars using the exchange rate €1 = $1.18:
\(438 \times 1.18 = 516.84\)

Liam receives $516.84.

M1 for \(450 - 12\) or \(438 \times 1.18\)
A1 for 516.84

First, find the common difference between consecutive terms:
\(11 - 5 = 6\)
\(17 - 11 = 6\)
\(23 - 17 = 6\)

The sequence increases by 6 each time, so the expression for the \(n\)-th term starts with \(6n\).

When \(n = 1\), \(6(1) = 6\).
To get the first term, 5, we must subtract 1 from 6.
Thus, the \(n\)-th term is \(6n - 1\).

M1 for \(6n + c\) (where \(c\) is a constant) or for finding the common difference is 6
A1 for \(6n - 1\)

To factorise \(8a^2b - 12ab^2\) completely, we look for the highest common factor of both terms. The highest common factor of 8 and 12 is 4. The highest common factor of \(a^2\) and \(a\) is \(a\). The highest common factor of \(b\) and \(b^2\) is \(b\). Thus, the highest common factor is \(4ab\). Dividing each term by \(4ab\) gives \(8a^2b \div 4ab = 2a\) and \(12ab^2 \div 4ab = 3b\). Putting these together, we get \(4ab(2a - 3b)\).

M1 for finding a common factor of the form \(2ab\), \(4a\), \(4b\) or for the expression \(4ab(2a + kb)\) where \(k \neq -3\). A1 for the fully correct factorised expression \(4ab(2a - 3b)\).

The total surface area of a cuboid is calculated using the formula \(2(lw + lh + wh)\). Substituting the given dimensions: Length \(l = 7\) cm, Width \(w = 5\) cm, Height \(h = 4\) cm. Total surface area = \(2 \times (7 \times 5 + 7 \times 4 + 5 \times 4) = 2 \times (35 + 28 + 20) = 2 \times 83 = 166\) cm\(^2\).

M1 for showing a correct method to find the area of at least 3 faces, such as \(7 \times 5 + 7 \times 4 + 5 \times 4\) or \(2 \times 35 + 2 \times 28 + 2 \times 20\). A1 for 166.

First, expand the bracket on the left-hand side: \(6x - 15 = 4x + 7\). Next, subtract \(4x\) from both sides to group the \(x\) terms: \(2x - 15 = 7\). Then, add 15 to both sides: \(2x = 22\). Finally, divide both sides by 2: \(x = 11\).

M1 for a correct first step, such as expanding the bracket to get \(6x - 15\) or correctly isolating the variable terms after an expansion error. A1 for the final answer 11.

Group the terms to factorise by grouping:

1. Group the first two terms and the last two terms:
\((15ax - 10ay) + (12bx - 8by)\)

2. Factorise out the common factor from each group:
\(5a(3x - 2y) + 4b(3x - 2y)\)

3. Factorise out the common binomial factor \((3x - 2y)\):
\((3x - 2y)(5a + 4b)\)

M1 for partial factorisation, e.g. \(5a(3x - 2y) + 4b(3x - 2y)\) or \(3x(5a + 4b) - 2y(5a + 4b)\)
A1 for \((3x - 2y)(5a + 4b)\) or equivalent

1. Convert the price in London from GBP (£) to JPY:
\(480 \times 162 = 77\,760\text{ JPY}\)

2. Find the difference between the two prices:
\(77\,760 - 75\,000 = 2\,760\text{ JPY}\)

M1 for converting the currency, e.g., \(480 \times 162\) or \(75\,000 \div 162\)
A1 for 2760

1. Multiply every term in the equation by 12 (the lowest common multiple of 4 and 3) to clear the fractions:
\(12 \times \left(\frac{2x - 3}{4}\right) - 12 \times \left(\frac{x - 1}{3}\right) = 12 \times 2\)

\(3(2x - 3) - 4(x - 1) = 24\)

2. Expand the brackets:
\(6x - 9 - 4x + 4 = 24\)

3. Simplify the terms:
\(2x - 5 = 24\)

4. Solve for \(x\):
\(2x = 29\)
\(x = 14.5\)

M1 for multiplying by 12 correctly to eliminate fractions: \(3(2x - 3) - 4(x - 1) = 24\) (allow one sign or arithmetic error)
A1 for 14.5 or \(\frac{29}{2}\) or \(14\frac{1}{2}\)

To find the selling price after a 15% loss, we can calculate 85% of the original price: \(240 \times (1 - 0.15) = 240 \times 0.85 = 204\). Alternatively, the loss is \(0.15 \times 240 = 36\) dollars, so the selling price is \(240 - 36 = 204\).

M1 for \(240 \times 0.85\) or for finding the loss as \(240 \times 0.15 = 36\) A1 for 204

First, expand each set of brackets: \(3(2x - 5) = 6x - 15\) and \(-4(x - 3) = -4x + 12\). Next, collect the like terms: \(6x - 4x - 15 + 12 = 2x - 3\).

M1 for correct expansion of at least one bracket, i.e., \(6x - 15\) or \(-4x + 12\) (or \(4x - 12\) if preceded by subtraction) A1 for 2x - 3

Let \(\theta\) be the angle of elevation. The flagpole and its shadow form a right-angled triangle where the opposite side to \(\theta\) is \(8.5\text{ m}\) and the adjacent side is \(5.2\text{ m}\). Using the tangent ratio: \(\tan(\theta) = \frac{8.5}{5.2}\). To find \(\theta\), calculate \(\theta = \tan^{-1}\left(\frac{8.5}{5.2}\right) \approx 58.544^{\circ}\). Rounding to 1 decimal place gives \(58.5^{\circ}\).

M1 for \(\tan(\theta) = \frac{8.5}{5.2}\) or \(\tan^{-1}\left(\frac{8.5}{5.2}\right)\) A1 for 58.5

First, expand the bracket on the right-hand side of the equation: \(7x - 4 = 6x + 15\). Next, subtract \(6x\) from both sides: \(x - 4 = 15\). Finally, add \(4\) to both sides: \(x = 19\).

M1 for expanding the brackets correctly to get \(6x + 15\) (or for isolating terms in \(x\) on one side and constant terms on the other side of their incorrect expansion). A1 for \(19\).

The perimeter of a sector is the sum of the arc length and the two radii. First, calculate the arc length: \(\text{Arc length} = \frac{120}{360} \times 2 \times \pi \times 8.4 = \frac{1}{3} \times 16.8\pi \approx 17.593\text{ cm}\). Now, add the two radii to get the total perimeter: \(\text{Perimeter} = 17.593 + 2 \times 8.4 = 17.593 + 16.8 = 34.393\text{ cm}\). Correct to 1 decimal place, the perimeter is \(34.4\text{ cm}\).

M1 for \(\frac{120}{360} \times 2 \times \pi \times 8.4\) or \(17.59...\) seen. M1 for adding \(2 \times 8.4\) to their arc length. A1 for \(34.4\).

Find the difference between consecutive terms: \(9 - 2 = 7\), \(16 - 9 = 7\), etc. Since the common difference is \(7\), the \(n\)-th term formula will contain \(7n\). Compare the sequence with the \(7n\) times-table (\(7, 14, 21, 28, \dots\)): each term in the sequence is \(5\) less than \(7n\). Therefore, the \(n\)-th term is \(7n - 5\).

M1 for a term of \(7n\) or for an expression of the form \(7n + k\) (where \(k\) is any constant). A1 for \(7n - 5\) (or equivalent).

First, find the common difference between consecutive terms: \(13 - 7 = 6\), \(19 - 13 = 6\), etc. Since the difference is constant, the sequence is linear and its \(n\)-th term has the form \(6n + c\). To find \(c\), substitute \(n = 1\): \(6(1) + c = 7\), which gives \(c = 1\). Therefore, the expression for the \(n\)-th term is \(6n + 1\).

M1 for finding a common difference of 6 (or for an expression of \(6n + k\) where \(k\) is any constant). A1 for the correct expression \(6n + 1\) (or equivalent).

Multiply the decimal parts: \(4 \times 3 = 12\). Multiply the powers of 10: \(10^{5} \times 10^{-8} = 10^{5 + (-8)} = 10^{-3}\). Combine these to get: \(12 \times 10^{-3}\). Convert to standard form by writing 12 as \(1.2 \times 10^{1}\): \(1.2 \times 10^{1} \times 10^{-3} = 1.2 \times 10^{-2}\).

M1 for \(12 \times 10^{-3}\) or \(0.012\) or an answer of the form \(1.2 \times 10^{k}\) where \(k \neq -2\). A1 for the correct standard form answer: \(1.2 \times 10^{-2}\).

The total number of parts in the ratio is \(3 + 7 = 10\). To find the value of one part: \(\frac{140}{10} = 14\). Mia's share is 7 parts: \(7 \times 14 = 98\).

M1 for \(140 \div (3+7)\) or \(140 \div 10\) or 14 seen, or \(\frac{7}{10} \times 140\). A1 for 98.

First, find the gradient \(m\) of the line using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the coordinates of \(A\) and \(B\) gives: \(m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3\). Next, substitute the gradient and the coordinates of point \(A(1, 4)\) into \(y = mx + c\) to find \(c\): \(4 = 3(1) + c\), which simplifies to \(c = 1\). Therefore, the equation of the line is \(y = 3x + 1\).

M1 for finding the gradient as \(3\) or for substituting their gradient and a given point into \(y = mx + c\). A1 for \(y = 3x + 1\).

First, convert both numbers into ordinary numbers: \(1.2 \times 10^3 = 1200\) and \(3.4 \times 10^2 = 340\). Add these two values together: \(1200 + 340 = 1540\). Finally, convert \(1540\) back into standard form: \(1.54 \times 10^3\).

M1 for converting at least one number correctly to an ordinary number (e.g. \(1200\) or \(340\)) or showing a step to align the powers. A1 for \(1.54 \times 10^3\).

Find the common difference between consecutive terms: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\). Since the common difference is \(4\), the formula starts with \(4n\). To find the constant term, subtract \(4\) from the first term: \(7 - 4 = 3\). Alternatively, substitute \(n = 1\) into \(4n + c = 7\) to get \(c = 3\). Therefore, the expression for the \(n\)-th term is \(4n + 3\).

M1 for identifying the common difference is \(4\) or for any expression of the form \(4n + k\), where \(k\) is any constant. A1 for \(4n + 3\) or equivalent.

Find the highest common factor of the coefficients 14 and 21, which is 7. Find the highest common factor of the variables \(x^2\) and \(xy\), which is \(x\). Thus, the overall common factor is \(7x\). Divide each term by this factor to find the terms inside the bracket: \(14x^2 \div 7x = 2x\) and \(-21xy \div 7x = -3y\). This gives the factorised expression \(7x(2x - 3y)\).

M1 for a correct partial factorisation such as \(7(2x^2 - 3xy)\) or \(x(14x - 21y)\), or for the correct common factor \(7x\) outside a bracket with a minor error inside. A1 for \(7x(2x-3y)\).

First, expand the bracket on the left side of the equation: \(4x - 20 = 2x + 6\). Next, subtract \(2x\) from both sides to gather the terms in \(x\) on one side: \(2x - 20 = 6\). Then, add \(20\) to both sides to isolate the term with \(x\): \(2x = 26\). Finally, divide both sides by 2 to find the value of \(x\): \(x = 13\).

M1 for correct expansion of the bracket to \(4x - 20\) or for isolating the \(x\) terms and constant terms correctly on opposite sides of the equation from their previous step. A1 for 13.

The perimeter of a semicircle consists of the curved arc and the straight diameter. The formula for the curved arc length of a semicircle is \(\frac{1}{2} \times \pi \times d\), where \(d\) is the diameter. Substituting \(d = 10\text{ cm}\), the arc length is \(\frac{1}{2} \times \pi \times 10 = 5\pi \approx 15.708\text{ cm}\). The total perimeter is the sum of this arc length and the diameter: \(\text{Perimeter} = 15.708 + 10 = 25.708\text{ cm}\). Rounding to 3 significant figures gives \(25.7\text{ cm}\).

M1 for calculating the arc length of the semicircle: \(\frac{1}{2} \times \pi \times 10\) or \(5\pi\) or \(15.7\) or better. M1 for adding the diameter \(10\) to their arc length. A1 for \(25.7\) (accept answers in the range \([25.7, 25.71]\)).

Expand the brackets on both sides of the equation:
\(6x + 12 = 2x - 4\)

Subtract \(2x\) from both sides:
\(4x + 12 = -4\)

Subtract \(12\) from both sides:
\(4x = -16\)

Divide both sides by \(4\):
\(x = -4\)

M1 for correct expansion of at least one bracket (e.g. \(6x + 12\) or \(2x - 4\)) or for correct algebraic separation of their terms.
A1 for \(-4\) (or \(x = -4\))

First, find the vector \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \begin{pmatrix} 3 \\ k \end{pmatrix} = \begin{pmatrix} 9 \\ k - 8 \end{pmatrix}\). The magnitude of this vector is given by \(\sqrt{9^2 + (k - 8)^2} = 15\). Squaring both sides gives \(81 + (k - 8)^2 = 225\). Subtracting 81 from both sides gives \((k - 8)^2 = 144\). Taking the square root gives \(k - 8 = 12\) or \(k - 8 = -12\). Since \(k > 0\), we have \(k - 8 = 12\), which gives \(k = 20\).

M1 for finding \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 9 \\ k - 8 \end{pmatrix}\). M1 for setting up the equation \(9^2 + (k-8)^2 = 15^2\) or equivalent. A1 for \(k = 20\).

Express both sides of the equation as powers of 3: \(27^{2x - 1} = (3^3)^{2x - 1} = 3^{3(2x - 1)} = 3^{6x - 3}\) and \(\frac{1}{9^{x - 3}} = 9^{-(x - 3)} = (3^2)^{-x + 3} = 3^{-2x + 6}\). Equating the exponents: \(6x - 3 = -2x + 6\). Rearranging terms gives \(8x = 9\), which simplifies to \(x = 1.125\) (or \(\frac{9}{8}\)).

M1 for writing \(27 = 3^3\) and \(9 = 3^2\) to express both sides in base 3. M1 for equating exponents to obtain \(6x - 3 = -2x + 6\) or equivalent. A1 for \(1.125\) (or \(\frac{9}{8}\)).

The area of a sector of a circle with radius \(r\) and angle \(\theta\) is given by \(\text{Area} = \frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(150 = \frac{\theta}{360} \times \pi \times 15^2\), which simplifies to \(\frac{\theta}{360} \times \pi = \frac{150}{225} = \frac{2}{3}\). The arc length \(L\) of the sector is \(L = \frac{\theta}{360} \times 2\pi r = 2 \times 15 \times \left(\frac{\theta}{360} \times \pi\right) = 30 \times \frac{2}{3} = 20\text{ cm}\). The perimeter of the sector is \(L + 2r = 20 + 2(15) = 50\text{ cm}\).

M1 for setting up the area relation to find \(\frac{\theta}{360}\pi = \frac{2}{3}\) (or equivalent angle calculation). M1 for using the sector angle to calculate the arc length of 20 cm. A1 for the final perimeter of 50.

Factorise the numerator: \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\). Factorise the denominator using the difference of two squares: \(9x^2 - 1 = (3x + 1)(3x - 1)\). Divide both the numerator and the denominator by the common factor \((3x + 1)\) to get \(\frac{x - 5}{3x - 1}\).

M1 for factorising the numerator to \((3x + 1)(x - 5)\) or M1 for factorising the denominator to \((3x + 1)(3x - 1)\). A1 for the final answer \(\frac{x - 5}{3x - 1}\).

In the right-angled triangle \(ABC\), \(\cos(BAC) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{7.2}{12.5}\). Therefore, \(\text{angle } BAC = \cos^{-1}\left(\frac{7.2}{12.5}\right) \approx 54.833^\circ\). Rounding to 1 decimal place gives \(54.8^\circ\).

M1 for \(\cos(BAC) = \frac{7.2}{12.5}\). A1 for \(54.8\) or \(54.83...\)

First differences: \(7, 11, 15, 19\). Second differences: \(4, 4, 4\). Since the second differences are constant, the sequence is quadratic with \(an^2 + bn + c\) where \(a = \frac{4}{2} = 2\). Subtracting \(2n^2\) from each term gives: \(4 - 2(1) = 2\), \(11 - 2(4) = 3\), \(22 - 2(9) = 4\). This linear sequence \(2, 3, 4, \dots\) has \(n\)-th term \(n + 1\). Combining these, the \(n\)-th term is \(2n^2 + n + 1\).

M1 for finding second differences are constant (value 4) or for showing the coefficient of \(n^2\) is 2. A1 for the correct expression \(2n^2 + n + 1\).

Let \( V \) be the original value. After 2 years of 12% depreciation, the value of the machinery is given by: \( V \times (1 - 0.12)^2 = 13552 \). This simplifies to: \( V \times 0.88^2 = 13552 \), which gives: \( V \times 0.7744 = 13552 \). Solving for \( V \): \( V = \frac{13552}{0.7744} = 17500 \).

M1 for setting up the equation \( V \times 0.88^2 = 13552 \) or equivalent. M1 for evaluating \( 0.88^2 = 0.7744 \) or finding the intermediate value \( 13552 / 0.88 = 15400 \). A1 for 17500.

The area of a sector is given by \( \frac{\theta}{360} \times \pi r^2 = 75\pi \). Substituting \( r = 15 \), we get: \( \frac{\theta}{360} \times 225\pi = 75\pi \). Dividing both sides by \( 225\pi \) yields: \( \frac{\theta}{360} = \frac{75}{225} = \frac{1}{3} \). The arc length of the sector is: \( \frac{\theta}{360} \times 2\pi r = \frac{1}{3} \times 2\pi \times 15 = 10\pi \) cm. The total perimeter of the sector is: \( \text{Arc length} + 2r = 10\pi + 2(15) = 10\pi + 30 \approx 61.4159 \) cm. Correct to 1 decimal place, this is 61.4 cm.

M1 for setting up the sector area equation to find the fraction \( \frac{\theta}{360} = \frac{1}{3} \) or angle \( \theta = 120^\circ \). M1 for calculating the arc length as \( 10\pi \) or the total perimeter expression \( 10\pi + 30 \). A1 for 61.4.

Express both sides of the equation as powers of 2. Since \( 16 = 2^4 \), the right-hand side can be rewritten as: \( \frac{1}{16^{x + 2}} = 16^{-(x + 2)} = (2^4)^{-(x + 2)} = 2^{-4(x + 2)} = 2^{-4x - 8} \). Equating the exponents from both sides: \( 3x - 1 = -4x - 8 \). Rearranging terms to solve for \( x \): \( 3x + 4x = -8 + 1 \), which gives: \( 7x = -7 \). Dividing by 7, we obtain: \( x = -1 \).

M1 for writing the right-hand side as \( 2^{-4(x+2)} \) or \( 2^{-4x-8} \) (or equivalent base conversion). M1 for equating the exponents to form the linear equation \( 3x - 1 = -4x - 8 \) (or equivalent). A1 for \( x = -1 \).

First, write all parts of the equation with a base of 3:
\( 3^{2x-1} = \frac{(3^3)^{\frac{1}{2}}}{(3^2)^{x-2}} \)

Simplify the powers on both sides:
\( 3^{2x-1} = \frac{3^{\frac{3}{2}}}{3^{2x-4}} \)

Using index laws for division on the right-hand side:
\( 3^{2x-1} = 3^{\frac{3}{2} - (2x - 4)} \)
\( 3^{2x-1} = 3^{1.5 - 2x + 4} \)
\( 3^{2x-1} = 3^{5.5 - 2x} \)

Since the bases are the same, equate the indices:
\( 2x - 1 = 5.5 - 2x \)

Solve for \( x \):
\( 4x = 6.5 \)
\( x = 1.625 \) or \( \frac{13}{8} \)

M1: For expressing at least two terms as powers of 3 (e.g., \( 3^{3/2} \) or \( 3^{2x-4} \))
M1: For equating the indices correctly to form a linear equation (e.g., \( 2x - 1 = 1.5 - 2x + 4 \))
A1: For the correct final answer: \( 1.625 \) or \( \frac{13}{8} \) or \( 1\frac{5}{8} \)

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

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

Step 3: Find the gradient of the perpendicular line.
\( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{2} = -0.5 \)

Step 4: Form the equation using point-slope form with the midpoint \( (4, 1) \) and gradient \( -0.5 \).
\( y - 1 = -0.5(x - 4) \)
\( y - 1 = -0.5x + 2 \)
\( y = -0.5x + 3 \)

M1: For finding the coordinates of the midpoint \( (4, 1) \)
M1: For finding the gradient of the perpendicular bisector \( -0.5 \) (or \( -\frac{1}{2} \)) from their gradient of \( AB \)
A1: For the correct equation \( y = -0.5x + 3 \) (or equivalent form in \( y = mx + c \) such as \( y = -\frac{1}{2}x + 3 \))

Let \( B \) be the event of choosing a blue counter and \( R \) be the event of choosing a red counter.
Total number of counters initially is \( 5 + 3 = 8 \).

The easiest way to find the probability of "at least one blue" is to subtract the probability of "no blue" (which is picking two red counters) from 1.

Calculate the probability of picking two red counters:
\( P(R, R) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \)

Subtract this from 1 to find the probability of at least one blue:
\( P(\text{at least one blue}) = 1 - P(R, R) = 1 - \frac{5}{14} = \frac{9}{14} \)

Alternatively, calculate the sum of the favorable outcomes:
\( P(B, R) + P(R, B) + P(B, B) = \left(\frac{3}{8} \times \frac{5}{7}\right) + \left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{2}{7}\right) \)
\( = \frac{15}{56} + \frac{15}{56} + \frac{6}{56} = \frac{36}{56} = \frac{9}{14} \)

M1: For calculating the probability of picking two red counters: \( \frac{5}{8} \times \frac{4}{7} \) (or equivalent product representing other combinations)
M1: For correctly setting up the calculation for the complement \( 1 - \frac{20}{56} \) (or summing the three individual combinations)
A1: For the final answer \( \frac{9}{14} \) (or equivalent simplified fraction or decimal to 3 sig figs: \( 0.643 \))

To solve the equation:
\(\frac{6}{x+1} - \frac{2}{x} = 1\)

Multiply every term by the common denominator \(x(x+1)\) to clear the fractions:
\(6x - 2(x+1) = x(x+1)\)

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

Rearrange into a standard quadratic equation form \(ax^2 + bx + c = 0\):
\(x^2 - 3x + 2 = 0\)

Factorise the quadratic expression:
\((x-1)(x-2) = 0\)

This gives the solutions:
\(x = 1\) or \(x = 2\).

M1 for correctly eliminating denominators: \(6x - 2(x+1) = x(x+1)\) or better.
M1 for simplifying to a 3-term quadratic equation: \(x^2 - 3x + 2 = 0\).
A1 for both correct answers: \(x = 1\) and \(x = 2\) (or \(1\) and \(2\)).

First, calculate the volume of the sphere using the formula \(V = \frac{4}{3}\pi r^3\):
\(V_{\text{sphere}} = \frac{4}{3} \times \pi \times 6^3\)
\(V_{\text{sphere}} = \frac{4}{3} \times 216\pi = 288\pi\text{ cm}^3\)

Next, write down the formula for the volume of a cone, \(V = \frac{1}{3}\pi r^2 h\), and substitute the known base radius \(r = 3\text{ cm}\):
\(V_{\text{cone}} = \frac{1}{3} \times \pi \times 3^2 \times h = 3\pi h\text{ cm}^3\)

Since the volume of metal remains the same, set the two volumes equal:
\(3\pi h = 288\pi\)

Divide both sides by \(3\pi\) to find the height \(h\):
\(h = \frac{288}{3} = 96\text{ cm}\).

M1 for a correct expression or calculation for the volume of the sphere: \(\frac{4}{3} \times \pi \times 6^3\) (or \(288\pi\) or \(904.8\)).
M1 for equating their sphere volume to the formula for the volume of the cone: \(\frac{1}{3} \times \pi \times 3^2 \times h = 288\pi\).
A1 for \(96\).

The total number of balls in the bag is \(5 + 3 = 8\).

To find the probability that at least one ball is blue, we can use the complement method:
\(\text{P(at least one blue)} = 1 - \text{P(no blue balls)}\)

"No blue balls" means that both balls selected must be red.

The probability of selecting a red ball first is \(\frac{5}{8}\).
Since the selection is without replacement, the probability of selecting a second red ball is \(\frac{4}{7}\).

Therefore, the probability of selecting two red balls is:
\(\text{P(Red, Red)} = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)

Subtract this from 1 to find the probability of selecting at least one blue ball:
\(\text{P(at least one blue)} = 1 - \frac{5}{14} = \frac{9}{14}\) (or approximately \(0.643\)).

M1 for a correct product representing two red balls selected without replacement: \(\frac{5}{8} \times \frac{4}{7}\).
M1 for subtracting their product from 1: \(1 - \text{their P(Red, Red)}\), or for calculating the sum of alternative valid outcomes: \(\frac{3}{8}\times\frac{2}{7} + \frac{5}{8}\times\frac{3}{7} + \frac{3}{8}\times\frac{5}{7}\).
A1 for \(\frac{9}{14}\) or equivalent fraction or decimal to at least 3 significant figures (\(0.643\) or \(0.6428...\)).

Multiply the entire equation by the common denominator \(2x(x - 4)\) to clear fractions: \(3(2x) - 2(2)(x - 4) = 1(x)(x - 4)\). Expand both sides: \(6x - 4x + 16 = x^2 - 4x\). Simplify to get: \(2x + 16 = x^2 - 4x\). Rearrange into standard quadratic form: \(x^2 - 6x - 16 = 0\). Factor the quadratic equation: \((x - 8)(x + 2) = 0\). Solving for \(x\) gives \(x = 8\) or \(x = -2\).

M1 for clearing fractions correctly: \(6x - 4(x-4) = x(x-4)\). A1 for establishing the correct quadratic equation in standard form: \(x^2 - 6x - 16 = 0\). A1 for finding both correct solutions: \(x = 8\) and \(x = -2\).

The perimeter of a sector is given by the sum of its two radii and its arc length. Arc length = \(\frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2\pi r = \frac{2}{3}\pi r\). Therefore, the total perimeter is \(2r + \frac{2}{3}\pi r = 25\). Factor out \(r\): \(r(2 + \frac{2}{3}\pi) = 25\). Solve for \(r\): \(r = \frac{25}{2 + \frac{2}{3}\pi} \approx \frac{25}{4.0944} \approx 6.106\) cm. Rounded to 3 significant figures, \(r = 6.11\).

M1 for setting up the perimeter equation: \(2r + \frac{120}{360} \times 2\pi r = 25\). M1 for factorising or simplifying to solve for \(r\): \(r(2 + \frac{2}{3}\pi) = 25\). A1 for the final answer \(6.11\) (accept answers in the range 6.10 to 6.11).

Find the first differences of the sequence: 7, 11, 15, 19. Find the second differences: 4, 4, 4. Since the second difference is constant, the sequence is quadratic and has the form \(an^2 + bn + c\), where \(a = \frac{4}{2} = 2\). Subtract \(2n^2\) from each term of the sequence: for \(n = 1\): \(5 - 2(1) = 3\); for \(n = 2\): \(12 - 2(4) = 4\); for \(n = 3\): \(23 - 2(9) = 5\); for \(n = 4\): \(38 - 2(16) = 6\); for \(n = 5\): \(57 - 2(25) = 7\). The resulting sequence is linear: 3, 4, 5, 6, 7, ... with an \(n\)-th term of \(n + 2\). Combining these, the overall \(n\)-th term is \(2n^2 + n + 2\).

M1 for finding the constant second difference of 4, indicating a term of \(2n^2\). M1 for subtracting \(2n^2\) from the terms and finding the linear sequence 3, 4, 5... A1 for the correct final expression \(2n^2 + n + 2\).

Multiply both sides of the equation by the common denominator \((x-1)(x+3)\): \(3(x+3) + 2(x-1) = (x-1)(x+3)\). Expand the brackets: \(3x + 9 + 2x - 2 = x^2 + 2x - 3\). Simplify both sides: \(5x + 7 = x^2 + 2x - 3\). Rearrange the equation into a standard quadratic form: \(x^2 - 3x - 10 = 0\). Factorise the quadratic expression: \((x-5)(x+2) = 0\). Solving this gives the final solutions: \(x = 5\) or \(x = -2\).

M1 for multiplying through by the common denominator and expanding correctly to get \(3x + 9 + 2x - 2 = x^2 + 2x - 3\) (or equivalent). M1 for simplifying to a three-term quadratic equation \(x^2 - 3x - 10 = 0\). A1 for both correct solutions \(x = 5\) and \(x = -2\).

Let the sequence be represented by \(T_n\). Find the first and second differences of the sequence. Sequence: 3, 8, 15, 24, 35. First differences: 5, 7, 9, 11. Second differences: 2, 2, 2. Since the second difference is constant, the sequence is quadratic and has the form \(an^2 + bn + c\), where the coefficient \(a = \frac{\text{second difference}}{2} = \frac{2}{2} = 1\). Subtracting \(n^2\) from each term of the sequence: For \(n=1\): \(3 - 1^2 = 2\). For \(n=2\): \(8 - 2^2 = 4\). For \(n=3\): \(15 - 3^2 = 6\). This yields the linear sequence \(2, 4, 6, \dots\), which has the general term \(2n\). Combining the parts, the \(n\)-th term of the original sequence is \(n^2 + 2n\).

M1 for identifying that the second difference is constant (2) or identifying the term \(n^2\). M1 for attempting to find the linear part, e.g. finding the differences of \(T_n - n^2\) or setting up and solving simultaneous equations. A1 for the final correct expression \(n^2 + 2n\) (or equivalent, such as \(n(n+2)\)).

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\). For a sphere of radius \(3\text{ cm}\): \(V_{\text{sphere}} = \frac{4}{3} \times \pi \times 3^3 = 36\pi\text{ cm}^3\). The volume of a cone is given by the formula \(V = \frac{1}{3}\pi r^2 h\). For a cone with base radius \(2\text{ cm}\) and height \(h\): \(V_{\text{cone}} = \frac{1}{3} \times \pi \times 2^2 \times h = \frac{4}{3}\pi h\text{ cm}^3\). Since the metal sphere is recast into the cone, their volumes are equal: \(\frac{4}{3}\pi h = 36\pi\). Dividing both sides by \(\pi\) and multiplying by \(\frac{3}{4}\) gives: \(h = \frac{36 \times 3}{4} = 27\text{ cm}\).

M1 for finding the volume of the sphere as \(36\pi\) (or approx 113.1). M1 for setting up the volume equation \(\frac{1}{3} \times \pi \times 2^2 \times h = 36\pi\) and attempting to solve for \(h\). A1 for the correct height of 27.

To simplify the algebraic fraction, we factorise both the numerator and the denominator.

1. Factorise the quadratic expression in the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

2. Factorise the expression in the denominator using the difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

3. Write the fraction with the factorised forms:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\)

4. Cancel the common factor \((2x + 1)\) from both the numerator and the denominator:
\(\frac{x - 3}{2x - 1}\)

M1 for factorising the numerator to \((2x + 1)(x - 3)\)
M1 for factorising the denominator to \((2x - 1)(2x + 1)\)
A1 for the fully simplified answer \(\frac{x - 3}{2x - 1}\)

1. Use the formula for the area of a sector to find the radius \(r\):
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)
\(27\pi = \frac{120}{360} \times \pi r^2\)
\(27\pi = \frac{1}{3} \pi r^2\)
Divide both sides by \(\pi\):
\(27 = \frac{1}{3} r^2\)
Multiply by 3:
\(r^2 = 81\)
Since the radius must be positive, \(r = 9\) cm.

2. Calculate the arc length of the sector:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)
\(\text{Arc length} = \frac{120}{360} \times 2\pi(9) = \frac{1}{3} \times 18\pi = 6\pi\) cm.

3. Calculate the total perimeter of the sector (the arc length plus two radii):
\(\text{Perimeter} = \text{Arc length} + 2r\)
\(\text{Perimeter} = 6\pi + 2(9) = 6\pi + 18\) cm.

M1 for forming an equation for the radius: \(\frac{120}{360} \pi r^2 = 27\pi\) or showing \(r = 9\)
M1 for calculating the arc length: \(\frac{120}{360} \times 2 \times \pi \times 9\) or \(6\pi\)
A1 for the correct final perimeter: \(6\pi + 18\)

1. Find the first differences between successive terms:
11 - 4 = 7
20 - 11 = 9
31 - 20 = 11
44 - 31 = 13
The first differences are 7, 9, 11, 13.

2. Find the second differences:
9 - 7 = 2
11 - 9 = 2
13 - 11 = 2
Since the second difference is constant and equal to 2, the sequence is quadratic and has the general form \(an^2 + bn + c\), where \(2a = 2 \implies a = 1\).

3. Subtract the \(n^2\) term from each term of the original sequence to find the remaining linear sequence:
For \(n = 1\): \(4 - 1^2 = 3\)
For \(n = 2\): \(11 - 2^2 = 7\)
For \(n = 3\): \(20 - 3^2 = 11\)
For \(n = 4\): \(31 - 4^2 = 15\)
For \(n = 5\): \(44 - 5^2 = 19\)
The remaining sequence is 3, 7, 11, 15, 19, ... which has a first term of 3 and a common difference of 4.

4. Write the linear expression: \(4n - 1\).

5. Combine the quadratic and linear parts:
The \(n\)-th term is \(n^2 + 4n - 1\).

M1 for finding second differences are 2, which gives \(1n^2\) as the leading term of the quadratic expression (or setting up simultaneous equations)
M1 for subtracting \(n^2\) to find the linear sequence 3, 7, 11, 15, ... (or solving simultaneous equations for \(b\) and \(c\))
A1 for the correct final expression: \(n^2 + 4n - 1\) (or equivalent)

(a) Cost Price = $120.
Total Selling Price = \(15 \times 14.50 = 217.50\) dollars.
Profit = \(217.50 - 120 = 97.50\) dollars.
Percentage Profit = \(\frac{97.50}{120} \times 100\% = 81.25\%\).

(b) Simple Interest = \(P \times R \times T = 850 \times 0.024 \times 5 = 102\) dollars.
Total Value = \(850 + 102 = 952\) dollars.

(c) Sale Price = \(480 \times (1 - 0.15) = 480 \times 0.85 = 408\) dollars.
Remaining Balance = \(408 - 150 = 258\) dollars.
Monthly Installment = \(\frac{258}{8} = 32.25\) dollars.

(a) M1 for \(15 \times 14.50\) [1]
M1 for \(\frac{\text{their } 217.50 - 120}{120} \times 100\) [1]
A1 for 81.25% [1]

(b) M1 for \(850 \times 0.024 \times 5\) [1]
M1 for adding their interest to 850 [1]
A1 for 952 [1]

(c) M1 for \(480 \times 0.85\) [1]
M1 for \(\text{their } 408 - 150\) [1]
M1 for \(\text{their } 258 \div 8\) [1]
A1.56 for 32.25 [1.56]

(a) Since \(\triangle ABC\) is right-angled at \(B\), we apply Pythagoras' theorem:
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 2.4^2 + 1.8^2 = 5.76 + 3.24 = 9.0\)
\(AC = \sqrt{9.0} = 3\) m.

(b) In \(\triangle ABC\):
\(\tan(ACB) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{2.4}{1.8}\)
\(\text{Angle } ACB = \tan^{-1}\left(\frac{2.4}{1.8}\right) \approx 53.13^\circ\).

(c) In right-angled triangle \(\triangle ABD\), the hypotenuse is \(AD = 4.5\) m and the height is \(AB = 2.4\) m.
\(BD^2 = AD^2 - AB^2 = 4.5^2 - 2.4^2 = 20.25 - 5.76 = 14.49\)
\(BD = \sqrt{14.49} \approx 3.8066\) m. Correct to 3 significant figures, \(BD = 3.81\) m.

(d) The angle of elevation is \(\angle ADB\).
\(\sin(ADB) = \frac{AB}{AD} = \frac{2.4}{4.5}\)
\(\text{Angle } ADB = \sin^{-1}\left(\frac{2.4}{4.5}\right) \approx 32.23^\circ\). Correct to 1 decimal place, the angle is \(32.2^\circ\).

(a) M1 for \(2.4^2 + 1.8^2\) [1]
M1 for \(\sqrt{2.4^2 + 1.8^2}\) [1]
A1 for 3 [1]

(b) M1 for \(\tan(ACB) = \frac{2.4}{1.8}\) or equivalent trig ratio [1]
M1 for \(\tan^{-1}\left(\frac{2.4}{1.8}\right)\) [1]
A1 for 53.1 or 53.13 [1]

(c) M1 for \(4.5^2 - 2.4^2\) [1]
M1 for \(\sqrt{4.5^2 - 2.4^2}\) [1]
A1 for 3.81 or 3.806 to 3.807 [1]

(d) M1 for \(\sin(ADB) = \frac{2.4}{4.5}\) or equivalent [1]
A1.56 for 32.2 or 32.23 [1.56]

(a) The sequence of circles is 4, 6, 8, 10, ... which increases by 2 each time. For Pattern 5, the number of circles is \(10 + 2 = 12\).

(b) Since the sequence increases by 2 each time, the term-to-term rule is \(2n\).
When \(n = 1\), \(2(1) + c = 4 \implies c = 2\).
Therefore, the expression for the number of circles in Pattern \(n\) is \(2n + 2\).

(c) Set the expression equal to 46:
\(2n + 2 = 46\)
\(2n = 44\)
\(n = 22\).
Since the number of squares in Pattern \(n\) is equal to \(n\), there are 22 squares.

(d)(i) Perimeter of 1 square = \(4 \times 3 = 12\) cm.

(d)(ii) A rectangle made of \(n\) squares joined side-by-side has width 3 cm and length \(3n\) cm.
Perimeter = \(2 \times (\text{width} + \text{length}) = 2(3 + 3n) = 6 + 6n\) (or \(6n + 6\)) cm.

(d)(iii) Substituting \(n = 15\) into the expression \(6n + 6\):
Perimeter = \(6(15) + 6 = 90 + 6 = 96\) cm.

(a) B1 for 12 [1]

(b) B1 for \(2n\) or \(a + 2n\) [1]
B1 for \(2n + 2\) or equivalent [1]

(c) M1 for \(2n + 2 = 46\) or equivalent [1]
A1.56 for 22 [1.56]

(d)(i) B1 for 12 [1]

(d)(ii) M1 for \(2(3 + 3n)\) [1]
A1 for \(6n + 6\) or \(6(n + 1)\) [1]

(d)(iii) M1 for substituting \(n = 15\) into their expression [1]
M1 for \(6 \times 15 + 6\) [1]
A1 for 96 [1]

(a) Interest = \(\frac{P \times R \times T}{100} = \frac{3600 \times 3.5 \times 4}{100} = 504\) dollars.

(b) Total Amount = \(P \times \left(1 + \frac{R}{100}\right)^T = 3600 \times (1.032)^4 \approx 3600 \times 1.1342673 = 4083.362... \approx 4083.36\) dollars.

(c)(i) Amount in DKK = \(500 \times 8.64 = 4320\) DKK.

(c)(ii) Remaining DKK = \(4320 - 3150 = 1170\) DKK.
Amount received back = \(\frac{1170}{8.64} = 135.4166... \approx 135.42\) pounds.

(a) M1 for \(\frac{3600 \times 3.5 \times 4}{100}\)
A1 for 504

(b) M1 for \(3600 \times (1.032)^4\)
A1 for 4083.36 (must be correct to 2 decimal places)

(c)(i) B1 for 4320

(c)(ii) M1 for subtracting \(4320 - 3150\) (or their \(\text{(c)(i)} - 3150\))
M1 for dividing by 8.64
A1 for 135.42 (must be correct to 2 decimal places)

(a) Using Pythagoras' theorem: \(AC^2 = AB^2 + BC^2 = 120^2 + 50^2 = 14400 + 2500 = 16900\).
\(AC = \sqrt{16900} = 130\text{ m}\).

(b) \(\tan(BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{50}{120}\).
\(\angle BAC = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.6198...^\circ \approx 22.6^\circ\).

(c)(i) Let \(P\) be the top of the flagpole. \(PC = 35\text{ m}\).
In the right-angled triangle \(PCB\):
\(PB^2 = PC^2 + BC^2 = 35^2 + 50^2 = 1225 + 2500 = 3725\).
\(PB = \sqrt{3725} \approx 61.032... \approx 61.0\text{ m}\).

(c)(ii) The angle of elevation is \(\angle PBC\).
\(\tan(PBC) = \frac{PC}{BC} = \frac{35}{50} = 0.7\).
\(\angle PBC = \tan^{-1}(0.7) \approx 34.992...^\circ \approx 35.0^\circ\).

(a) M1 for \(120^2 + 50^2\)
A1 for 130

(b) M1 for \(\tan(BAC) = \frac{50}{120}\)
A1 for 22.6 or 22.62

(c)(i) M1 for \(35^2 + 50^2\)
A1 for 61.0 or 61

(c)(ii) M1 for \(\tan(\theta) = \frac{35}{50}\)
A1 for 35.0 or 35

(a) Area of trapezium = \(\frac{1}{2} (a + b) h = \frac{1}{2} (40 + 60) \times 30 = 50 \times 30 = 1500\text{ cm}^2\).

(b) Length of the trough = \(1.2\text{ m} = 120\text{ cm}\).
Volume = \(\text{Area of cross-section} \times \text{length} = 1500 \times 120 = 180\ 000\text{ cm}^3\).

(c) Since \(1000\text{ cm}^3 = 1\text{ litre}\),
Volume in litres = \(\frac{180000}{1000} = 180\text{ litres}\).

(d) Time in minutes = \(\frac{180}{5.4} = 33.333...\text{ minutes}\).
\(0.333...\text{ minutes} = \frac{1}{3} \times 60 = 20\text{ seconds}\).
Total time = 33 minutes and 20 seconds.

(a) M1 for \(\frac{1}{2} (40 + 60) \times 30\)
A1 for 1500

(b) M1 for \(1.2\text{ m} = 120\text{ cm}\)
A1 for showing \(1500 \times 120 = 180\ 000\)

(c) B1 for 180

(d) M1 for \(\frac{180}{5.4}\) (or their \(\text{(c)} / 5.4\))
A1 for 33.33... minutes or 33 mins 20 secs
A1 for 33 minutes 20 seconds

(a) First, find the cost of 1 cover: \(110 / 50 = \$2.20\). Next, find the cost of 1 page: \(25 / 1000 = \$0.025\). Since each notebook uses 60 pages, the cost of the pages is \(60 \times 0.025 = \$1.50\). Total cost for one notebook is \(2.20 + 1.50 = \$3.70\). (b) Selling price is $5.55, and the cost price is $3.70. Profit is \(5.55 - 3.70 = \$1.85\). Percentage profit is \((\text{Profit} / \text{Cost Price}) \times 100 = (1.85 / 3.70) \times 100 = 50\%\). (c) Using the compound interest formula: \(\text{Value} = 1500 \times (1 + 3/100)^4 = 1500 \times 1.03^4 = 1500 \times 1.12550881 = 1688.2632...\). Correct to the nearest cent, the value is $1688.26. (d) First, find the simple interest earned: \(1762.50 - 1500 = \$262.50\). Using the simple interest formula \(I = (P \times r \times t)/100\): \(262.50 = (1500 \times r \times 5)/100\), which simplifies to \(262.50 = 75r\). Solving for \(r\) gives \(r = 262.50 / 75 = 3.5\).

(a) M1 for \(110 / 50\) or \(25 / 1000\). M1 for \(2.20 + 60 \times 0.025\). A1 for 3.70 (or 3.7). (b) M1 for \(5.55 - 3.70\) or 1.85. M1 for \((\text{their 1.85} / \text{their 3.70}) \times 100\). A1 for 50. (c) M1 for \(1500 \times 1.03^4\) or \(1500 \times 1.03^n\) (where \(n \ge 2\)). A1 for 1688.263... B1 for rounding their value to 2 decimal places. (d) M1 for \(1762.50 - 1500\) or 262.50. M1 for \(262.50 = 1500 \times r \times 5 / 100\) or \(262.50 / (1500 \times 5) \times 100\). A1 for 3.5.

(a) Area of the trapezium cross-section is \(\frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (40 + 60) \times 30 = 1500\text{ cm}^2\). (b)(i) Volume of the trough is \(\text{cross-sectional area} \times \text{length} = 1500 \times 120 = 180000\text{ cm}^3\). (b)(ii) Since \(1\text{ litre} = 1000\text{ cm}^3\), the volume in litres is \(180000 / 1000 = 180\text{ litres}\). (c) Time taken in minutes is \(180 / 8 = 22.5\text{ minutes}\). Since \(0.5\text{ minutes} = 30\text{ seconds}\), the time is 22 minutes and 30 seconds. (d) Volume of the cylinder is \(\pi \times r^2 \times h = \pi \times 5^2 \times 40 = 1000\pi \approx 3141.59265...\text{ cm}^3\). Correct to 3 significant figures, this is \(3140\text{ cm}^3\).

(a) M1 for \(\frac{1}{2} \times (40 + 60) \times 30\). A1 for 1500. (b)(i) M1 for \(\text{their 1500} \times 120\). A1 for 180000. (b)(ii) B1 for 180 (or \(\text{their (b)(i)} / 1000\)). (c) M1 for \(\text{their (b)(ii)} / 8\). A1 for 22.5. B1 for converting 22.5 minutes to 22 minutes 30 seconds. (d) M1 for \(\pi \times 5^2 \times 40\). A1 for \(1000\pi\) or 3141.59... B1 for rounding to 3 significant figures (3140). B1 for unit \(\text{cm}^3\).

(a)(i) The number of triangular tiles increases by 2 each time: 4, 6, 8, ... so Pattern 4 has \(8 + 2 = 10\) triangular tiles. (a)(ii) Pattern 5 has \(10 + 2 = 12\) triangular tiles. (b) The sequence of triangular tiles is 4, 6, 8, 10, ... The first term \(a = 4\) and the common difference \(d = 2\). The nth term is \(a + (n-1)d = 4 + 2(n-1) = 2n + 2\). (c) Setting \(2k + 2 = 76\) gives \(2k = 74\), so \(k = 37\). (d) In Pattern 20, there are 20 square tiles and \(2(20) + 2 = 42\) triangular tiles. Cost of square tiles is \(20 \times \$0.80 = \$16.00\). Cost of triangular tiles is \(42 \times \$0.50 = \$21.00\). Total cost is \(16.00 + 21.00 = \$37.00\).

(a)(i) B1 for 10. (a)(ii) B1 for 12. (b) M1 for \(2n + c\) (where \(c\) is any constant) or \(2n + 2\) seen. A1 for \(2n + 2\) or equivalent. (c) M1 for setting their \(2k + 2 = 76\). M1 for \(2k = 74\). A1 for 37. (d) B1 for finding 20 square tiles. M1 for finding \(2(20) + 2 = 42\) triangular tiles. M1 for \(20 \times 0.80\) or 16.00. M1 for \(\text{their 42} \times 0.50\) or 21.00. A1 for 37 or 37.00.

**(a)**
Volume of a pyramid = \(\frac{1}{3} \times \text{base area} \times \text{height}\)
Base area = \(8 \times 8 = 64\text{ cm}^2\)
Volume = \(\frac{1}{3} \times 64 \times 12 = 256\text{ cm}^3\).

**(b)**
To find the total surface area, we need the area of the 4 triangular faces.
First, find the slant height, \(l\), of each triangular face using Pythagoras' theorem:
\(l = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \approx 12.65\text{ cm}\).
Area of one triangular face = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times \sqrt{160} \approx 50.60\text{ cm}^2\).
Total surface area = \(\text{base area} + 4 \times \text{area of one face}\)
Total surface area = \(64 + 4 \times 50.596 = 266.38\text{ cm}^2\).
To 3 significant figures, this is \(266\text{ cm}^2\).

**(c)**
Volume of the sphere = Volume of the pyramid = \(256\text{ cm}^3\).
\(\frac{4}{3} \pi r^3 = 256\)
\(\pi r^3 = 192\)
\(r^3 = \frac{192}{\pi} \approx 61.1155\)
\(r = \sqrt[3]{61.1155} \approx 3.9389\text{ cm}\).
To 3 significant figures, the radius is \(3.94\text{ cm}\).

**(d)**
Surface area of the sphere = \(4\pi r^2 = 4\pi (3.9389)^2 \approx 194.97\text{ cm}^2\).
Original surface area = \(266.38\text{ cm}^2\).
Reduction in surface area = \(266.38 - 194.97 = 71.41\text{ cm}^2\).
Percentage lost = \(\frac{71.41}{266.38} \times 100 \approx 26.81\%\).
To 3 significant figures, this is \(26.8\%\).

(a) M1 for \(\frac{1}{3} \times 8^2 \times 12\), A1 for 256.
(b) M1 for finding slant height \(\sqrt{12^2 + 4^2}\) or 12.6..., M1 for calculating area of 4 triangles e.g. \(4 \times \frac{1}{2} \times 8 \times \sqrt{160}\), M1 for adding base area 64, A1 for 266 (or 266.38...).
(c) M1 for setting up sphere volume formula \(\frac{4}{3} \pi r^3 = 256\), M1 for rearranging to find \(r^3\) or \(r\), A1 for 3.94.
(d) M1 for calculating surface area of sphere \(4\pi r^2\), M1 for percentage change formula \(\frac{\text{Difference}}{\text{Original}} \times 100\), A1 for 26.8.

**(a)**
First, differentiate the equation of the curve to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = x^2 - 4\)
For turning points, set \(\frac{dy}{dx} = 0\):
\(x^2 - 4 = 0 \implies x^2 = 4 \implies x = 2\text{ or } x = -2\).

Substitute these x-values back into the original curve equation:
For \(x = 2\):
\(y = \frac{2^3}{3} - 4(2) + 2 = \frac{8}{3} - 8 + 2 = -3.33\) (or \(-\frac{10}{3}\))
For \(x = -2\):
\(y = \frac{(-2)^3}{3} - 4(-2) + 2 = -\frac{8}{3} + 8 + 2 = 7.33\) (or \(\frac{22}{3}\))

So the turning points are \((-2, 7.33)\) and \((2, -3.33)\).

**(b)**
We can find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = 2x\)
At \(x = -2\):
\(\frac{d^2y}{dx^2} = 2(-2) = -4 < 0\), so \((-2, 7.33)\) is a local maximum.
At \(x = 2\):
\(\frac{d^2y}{dx^2} = 2(2) = 4 > 0\), so \((2, -3.33)\) is a local minimum.

**(c)**
At \(x = 3\):
\(y = \frac{3^3}{3} - 4(3) + 2 = 9 - 12 + 2 = -1\)
The gradient of the tangent at \(x = 3\) is given by \(\frac{dy}{dx}\) at \(x = 3\):
\(m = 3^2 - 4 = 5\)
Using the equation of a straight line, \(y - y_1 = m(x - x_1)\):
\(y - (-1) = 5(x - 3)\)
\(y + 1 = 5x - 15 \implies y = 5x - 16\).

**(d)**
Let \(f(x) = \frac{x^3}{3} - 4x + 2\).
Evaluate at the interval boundaries:
\(f(0.5) = \frac{0.5^3}{3} - 4(0.5) + 2 = \frac{0.125}{3} - 2 + 2 \approx 0.0417\) (which is positive)
\(f(1.0) = \frac{1^3}{3} - 4(1) + 2 = \frac{1}{3} - 4 + 2 \approx -1.667\) (which is negative)
Since \(f(0.5) > 0\) and \(f(1.0) < 0\), there is a sign change. Since the function is continuous, there must be a root between \(x = 0.5\) and \(x = 1.0\).

(a) M1 for differentiation to get \(x^2 - 4\), M1 for setting derivative to 0 and finding \(x = \pm 2\), A1 for \(y\)-value \(7.33\) or \(22/3\), A1 for \(y\)-value \(-3.33\) or \(-10/3\).
(b) M1 for testing the gradient or second derivative, A1 for correctly identifying max at \(x = -2\) and min at \(x = 2\) with valid justification.
(c) M1 for finding \(y = -1\) at \(x = 3\), M1 for finding gradient \(m = 5\), A1 for \(y = 5x - 16\).
(d) M1 for calculating both \(f(0.5)\) and \(f(1.0)\), A1 for stating the sign change implies a root exists.

Total number of balls = \(5 + 4 + 3 = 12\).

**(a)**
To get two balls of the same colour, the outcomes can be Red-Red (RR), Blue-Blue (BB), or Green-Green (GG).
\(P(RR) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)
\(P(BB) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\)
\(P(GG) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}\)
Total probability = \(P(RR) + P(BB) + P(GG) = \frac{20 + 12 + 6}{132} = \frac{38}{132} = \frac{19}{66} \approx 0.288\) (or \(28.8\%\)).

**(b)**
We can find the probability of getting no blue balls and subtract it from 1.
Number of non-blue balls = \(5 + 3 = 8\).
\(P(\text{no blue}) = \frac{8}{12} \times \frac{7}{11} = \frac{56}{132} = \frac{14}{33}\)
\(P(\text{at least one blue}) = 1 - P(\text{no blue}) = 1 - \frac{14}{33} = \frac{19}{33} \approx 0.576\) (or \(57.6\%\)).

**(c)**
For three balls to be of different colours, we must select one Red, one Blue, and one Green in any order.
There are \(3! = 6\) possible arrangements (RBG, RGB, BRG, BGR, GRB, GBR).
For any single arrangement, say Red then Blue then Green:
\(P(RBG) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\)
Since all 6 arrangements have the same probability, the total probability is:
\(6 \times \frac{1}{22} = \frac{6}{22} = \frac{3}{11} \approx 0.273\) (or \(27.3\%\)).

(a) M1 for summing individual same-colour probabilities, M2 for any two correct terms (e.g. \(5/12 \times 4/11\) and \(4/12 \times 3/11\)), A1 for \(19/66\) or \(0.288\).
(b) M1 for finding the total number of non-blue balls is 8, M2 for \(1 - (8/12 \times 7/11)\) or calculating direct probabilities, A1 for \(19/33\) or \(0.576\).
(c) M1 for product of three fractions representing a single arrangement (e.g. \(5/12 \times 4/11 \times 3/10\)), M2 for multiplying the single product by 6, A1 for \(3/11\) or \(0.273\).

(a) (i) Volume of cylinder = \pi \times r^2 \times 2r = 2\pi r^3. (ii) Volume of cone = \frac{1}{3}\pi r^2 h. Since volume of cone is half of cylinder volume: \frac{1}{3}\pi r^2 h = \frac{1}{2} (2\pi r^3) = \pi r^3. Multiplying both sides by 3 and dividing by \pi r^2 gives h = 3r. (b) (i) Total volume = Volume of cylinder + Volume of cone = 2\pi r^3 + \frac{1}{3}\pi r^2 (3r) = 3\pi r^3. Given 3\pi r^3 = 360, so r^3 = 120/\pi \approx 38.197. Thus r = 3.368 cm. (ii) Slant height of the cone l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (3r)^2} = \sqrt{10r^2} = r\sqrt{10} \approx 3.36785 \times 3.16228 = 10.650 cm. Total surface area = Base of cylinder + Curved area of cylinder + Curved area of cone = \pi r^2 + 2\pi r (2r) + \pi r l = 5\pi r^2 + \pi r l = \pi \times (3.36785)^2 \times (5 + \sqrt{10}) \approx 290.85 \approx 291 \text{ cm}^2. (iii) Volume of sphere = \frac{4}{3}\pi R^3 = 360, so R^3 = 270/\pi \approx 85.944. Thus R = \sqrt[3]{85.944} \approx 4.41 cm.

(a)(i) B1: 2\pi r^3 or equivalent. (a)(ii) M1: \frac{1}{3}\pi r^2 h = \pi r^3 or equivalent. A1: Completing the algebra to show h = 3r. (b)(i) M1: 2\pi r^3 + \pi r^3 = 360 or 3\pi r^3 = 360. M1: r = \sqrt[3]{120/\pi}. A1: 3.37 (accept 3.368 to 3.37). (b)(ii) M1: Finding slant height l = r\sqrt{10} or 10.65. M1: Total area formula 5\pi r^2 + \pi r l. M1: Substituting their r. A1: 291 (accept 290.8 to 291.2). (b)(iii) M1: \frac{4}{3}\pi R^3 = 360. A1: 4.41 (accept 4.41 to 4.42).

(a) Time taken outward = 45/x hours. (b) Time taken return = 40/(x - 3) hours. (c)(i) Return time - Outward time = 20 mins = 1/3 hour. So, 40/(x - 3) - 45/x = 1/3. Multiply both sides by 3x(x - 3): 120x - 135(x - 3) = x(x - 3) \implies 120x - 135x + 405 = x^2 - 3x \implies -15x + 405 = x^2 - 3x \implies x^2 + 12x - 405 = 0. (ii) Factoring the equation: (x + 27)(x - 15) = 0, which gives x = 15 or x = -27. (iii) Since speed must be positive, we choose x = 15. Outward time = 45 / 15 = 3 hours. Return time = 40 / (15 - 3) = 40/12 = 3 hours 20 minutes. Total time = 3 hours + 3 hours 20 minutes = 6 hours 20 minutes.

(a) B1: 45/x. (b) B1: 40/(x-3). (c)(i) B1: For converting 20 minutes to 1/3 hour. M1: For setting up equation 40/(x-3) - 45/x = 1/3. M1: For clearing fractions: 120x - 135(x-3) = x(x-3). A1: Correctly completing algebra to get x^2 + 12x - 405 = 0. (c)(ii) M1: (x + 27)(x - 15) = 0 or correct use of the quadratic formula. A1: x = 15. A1: x = -27. (c)(iii) M1: For selecting positive root x = 15. M1: For calculating total time in hours (6.33 hours or 6 and 1/3 hours). A1: 6 hours 20 minutes.

(a) (i) P(R, R) = 6/12 * 5/11 = 30/132 = 5/22. (ii) P(same colour) = P(R,R) + P(B,B) + P(G,G) = 30/132 + (4/12 * 3/11) + (2/12 * 1/11) = 30/132 + 12/132 + 2/132 = 44/132 = 1/3. P(different colours) = 1 - P(same colour) = 1 - 1/3 = 2/3. (iii) P(at least one green) = 1 - P(no green) = 1 - (10/12 * 9/11) = 1 - 90/132 = 1 - 15/22 = 7/22. (b) Three of the same colour can only be R,R,R or B,B,B (as there are only 2 greens). P(R,R,R) = 6/12 * 5/11 * 4/10 = 120/1320 = 1/11. P(B,B,B) = 4/12 * 3/11 * 2/10 = 24/1320 = 1/55. Total probability = 1/11 + 1/55 = 6/55.

(a)(i) M1: 6/12 * 5/11. A1: 5/22 (or 0.227). (a)(ii) M1: P(B,B) = 12/132 or P(G,G) = 2/132. M1: P(same) = 44/132 or 1/3. M1: 1 - P(same) or summing the alternative 6 probabilities. A1: 2/3 (or 0.667). (a)(iii) M1: 1 - P(no green) or list of green combinations: P(G, non-G) + P(non-G, G) + P(G, G). M1: 10/12 * 9/11. A1: 7/22 (or 0.318). (b) M1: P(R,R,R) = 120/1320. M1: P(B,B,B) = 24/1320. A1: 6/55 (or 0.109).

(a) The volume of the box is \(V = x^2 h = 108\text{ cm}^3\), which gives \(h = \frac{108}{x^2}\). The external surface area \(A\) of an open rectangular box with a square base is \(A = x^2 + 4xh\). Substituting \(h\) into this equation gives: \(A = x^2 + 4x \left(\frac{108}{x^2}\right) = x^2 + \frac{432}{x}\). (b) Differentiating \(A\) with respect to \(x\) gives \(\frac{dA}{dx} = 2x - 432x^{-2} = 2x - \frac{432}{x^2}\). (c) For a stationary value, set \(\frac{dA}{dx} = 0\), which gives \(2x - \frac{432}{x^2} = 0\), leading to \(2x^3 = 432\) and \(x^3 = 216\), hence \(x = 6\). (d) When \(x = 6\), the minimum surface area is \(A = 6^2 + \frac{432}{6} = 36 + 72 = 108\text{ cm}^2\). (e) At \(x = 6\), the area of the base is \(x^2 = 36\text{ cm}^2\) and the area of the sides is \(\frac{432}{6} = 72\text{ cm}^2\). The total cost of plating is \(36 \times 0.05 + 72 \times 0.02 = 1.80 + 1.44 = \$3.24\).

(a) M1 for writing volume formula: \(x^2 h = 108\) or \(h = \frac{108}{x^2}\). M1 for writing surface area formula: \(A = x^2 + 4xh\). A1 for completing the substitution to show \(A = x^2 + \frac{432}{x}\) with no errors. (b) M1 for \(2x\) or \(-432x^{-2}\). A1 for \(2x - \frac{432}{x^2}\). (c) M1 for setting their derivative equal to zero. M1 for solving to get \(x^3 = 216\). A1 for \(x = 6\). (d) M1 for substituting \(x = 6\) into the formula for \(A\). A1 for 108. (e) M1 for \(36 \times 0.05\) or \(72 \times 0.02\). A1 for 3.24.

(a) Using the Cosine Rule on triangle \(ABC\): \(AC^2 = 120^2 + 85^2 - 2 \times 120 \times 85 \times \cos(74^\circ)\) which gives \(AC^2 = 14400 + 7225 - 20400 \times 0.275637 = 16002.00\). Thus, \(AC = \sqrt{16002.00} \approx 126.5\text{ m}\) (to 1 d.p.). (b) Using the Sine Rule on triangle \(ADC\): \(\frac{\sin(CAD)}{CD} = \frac{\sin(ADC)}{AC}\) which gives \(\sin(CAD) = \frac{95 \times \sin(58^\circ)}{126.50} \approx 0.63687\). Therefore, \(\text{Angle } CAD = \arcsin(0.63687) \approx 39.6^\circ\). (c) The total area is the sum of the areas of triangles \(ABC\) and \(ADC\). Area of \(\triangle ABC = \frac{1}{2} \times 120 \times 85 \times \sin(74^\circ) \approx 4902.4\text{ m}^2\). In triangle \(ADC\), the third angle is \(ACD = 180^\circ - 58^\circ - 39.56^\circ = 82.44^\circ\). Area of \(\triangle ADC = \frac{1}{2} \times AC \times CD \times \sin(ACD) = \frac{1}{2} \times 126.50 \times 95 \times \sin(82.44^\circ) \approx 5956.5\text{ m}^2\). Total area = \(4902.4 + 5956.5 = 10858.9\text{ m}^2 \approx 10900\text{ m}^2\) (to 3 s.f.). (d) Let the top of the pole be \(T\). In the right-angled triangle \(TAC\), the angle of elevation is \(\theta = \angle TCA\). Then \(\tan(\theta) = \frac{15}{AC} = \frac{15}{126.50} \approx 0.11858\), which gives \(\theta \approx 6.8^\circ\).

(a) M1 for \(120^2 + 85^2 - 2 \times 120 \times 85 \times \cos(74^\circ)\). A1 for \(16002\) or \(16000\). A1 for \(126.5\) (or 126.49...). (b) M1 for \(\frac{\sin(CAD)}{95} = \frac{\sin(58^\circ)}{126.5}\). M1 for \(\sin(CAD) = \frac{95 \times \sin(58^\circ)}{126.5}\). A1 for \(39.6^\circ\) or \(39.56^\circ...\). (c) M1 for \(\text{Area of } ABC = \frac{1}{2} \times 120 \times 85 \times \sin(74^\circ)\). M1 for finding angle \(ACD = 82.4^\circ\). M1 for \(\text{Area of } ADC = \frac{1}{2} \times 126.5 \times 95 \times \sin(82.4^\circ)\). A1 for 10900 or 10860 (accept range 10850 to 10900). (d) M1 for \(\tan(\theta) = \frac{15}{126.5}\). A1 for \(6.8\) or \(6.76...\) (accept 6.7 to 6.8).

(a)(i) The probability that the first marble is red is \(\frac{7}{n}\). Since there is no replacement, the probability that the second is red is \(\frac{6}{n-1}\). The probability both are red is \(P(RR) = \frac{7}{n} \times \frac{6}{n-1} = \frac{42}{n(n-1)}\). (a)(ii) The number of blue marbles is \(n-7\). One of each color can be drawn in two orders: red-blue or blue-red. The probability is \(P(RB) + P(BR) = \frac{7}{n} \times \frac{n-7}{n-1} + \frac{n-7}{n} \times \frac{7}{n-1} = \frac{14(n-7)}{n(n-1)}\). (b)(i) Since \(P(RR) = \frac{1}{5}\), we have \(\frac{42}{n(n-1)} = \frac{1}{5}\). Multiplying both sides by \(5n(n-1)\) yields \(210 = n(n-1)\), which simplifies to \(n^2 - n - 210 = 0\). (b)(ii) Factoring the quadratic: \((n - 15)(n + 14) = 0\), which gives \(n = 15\) or \(n = -14\). Since the number of marbles must be positive, we have \(n = 15\). (b)(iii) The event 'at least one marble is blue' is the complement of 'both marbles are red'. Thus, the probability is \(1 - P(RR) = 1 - \frac{1}{5} = \frac{4}{5}\) (or 0.8).

(a)(i) M1 for \(\frac{7}{n} \times \frac{6}{n-1}\). A1 for \(\frac{42}{n(n-1)}\). (a)(ii) M1 for identifying both \(RB\) and \(BR\) combinations. M1 for \(2 \times \left(\frac{7}{n} \times \frac{n-7}{n-1}\right)\). A1 for \(\frac{14(n-7)}{n(n-1)}\). (b)(i) M1 for setting their expression from (a)(i) equal to \(\frac{1}{5}\). M1 for cross-multiplying to obtain \(210 = n^2 - n\). A1 for fully showing the target equation with no errors. (b)(ii) M1 for factorisation \((n-15)(n+14) = 0\) (or quadratic formula). A1 for \(n = 15\) (with negative root rejected). (b)(iii) M1 for \(1 - \frac{1}{5}\) oe. A1 for \(\frac{4}{5}\) or 0.8.

(a) Using the cosine rule in triangle \(ABD\): \(BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(BAD) = 85^2 + 60^2 - 2 \cdot 85 \cdot 60 \cdot \cos(110^\circ) = 7225 + 3600 - 10200 \cdot (-0.34202) = 10825 + 3488.6 = 14313.6\). Therefore, \(BD = \sqrt{14313.6} \approx 119.64\text{ m}\), which rounds to \(120\text{ m}\) (to 3 significant figures) or \(119.6\text{ m}\) (to 4 significant figures). (b) Area of triangle \(ABD = \frac{1}{2} \cdot AB \cdot AD \cdot \sin(BAD) = \frac{1}{2} \cdot 85 \t\cdot 60 \cdot \sin(110^\circ) = 2550 \cdot 0.93969 \approx 2396.2\text{ m}^2\), which rounds to \(2400\text{ m}^2\) (to 3 significant figures) or \(2396\text{ m}^2\) (to 4 significant figures). (c) Using the sine rule in triangle \(BCD\): \(\frac{CD}{\sin(CBD)} = \frac{BD}{\sin(BCD)}\). Using \(BD = 119.64\text{ m}\): \(CD = \frac{119.64 \cdot \sin(48^\circ)}{\sin(75^\circ)} = \frac{119.64 \cdot 0.74314}{0.96593} \approx 92.0\text{ m}\) (to 3 significant figures). If using \(BD = 120\text{ m}\), \(CD \approx 92.3\text{ m}\). (d) The shortest distance from \(A\) to \(BD\) is the perpendicular height \(h\) of triangle \(ABD\). \(\text{Area} = \frac{1}{2} \cdot BD \cdot h \implies 2396.2 = \frac{1}{2} \cdot 119.64 \cdot h \implies h = \frac{2 \cdot 2396.2}{119.64} \approx 40.1\text{ m}\) (to 3 significant figures). If using \(BD = 120\text{ m}\), \(h \approx 39.9\text{ m}\).

(a) M1 for \(BD^2 = 85^2 + 60^2 - 2 \cdot 85 \cdot 60 \cdot \cos(110^\circ)\), A1 for \(14310\) to \(14320\), A1 for \(119.6\) to \(120\). (b) M1 for \(\frac{1}{2} \cdot 85 \cdot 60 \cdot \sin(110^\circ)\), A1 for \(2396\) to \(2400\). (c) M1 for \(\frac{CD}{\sin(48^\circ)} = \frac{BD}{\sin(75^\circ)}\), M1 for \(CD = \frac{BD \cdot \sin(48^\circ)}{\sin(75^\circ)}\), A1 for \(92.0\) to \(92.3\). (d) M1 for \(\frac{1}{2} \cdot BD \cdot h = \text{Area}\), M1 for \(h = \frac{2 \cdot 2396.2}{119.64}\) or equivalent, A1 for \(39.9\) to \(40.1\).

(a) The total length of the pool and path is \(15 + 2x\) and the total width is \(10 + 2x\). The total area is given by \((15 + 2x)(10 + 2x) = 150 + 30x + 20x + 4x^2 = 4x^2 + 50x + 150\). Since the total area is \(266\text{ m}^2\): \(4x^2 + 50x + 150 = 266 \implies 4x^2 + 50x - 116 = 0\). Dividing the entire equation by 2 gives: \(2x^2 + 25x - 58 = 0\). (b) We solve \(2x^2 + 25x - 58 = 0\) by factorisation: \((2x + 29)(x - 2) = 0\). This gives \(2x + 29 = 0 \implies x = -14.5\) or \(x - 2 = 0 \implies x = 2\). (c) Since the width of the path must be a positive value, we choose \(x = 2\text{ m}\). (d) Area of the path = Total area - Pool area = \(266 - (15 \cdot 10) = 116\text{ m}^2\). The area of one square tile is \(0.5\text{ m} \cdot 0.5\text{ m} = 0.25\text{ m}^2\). The number of tiles needed is \(\frac{116}{0.25} = 464\).

(a) M1 for \((15 + 2x)(10 + 2x)\), A1 for \(4x^2 + 50x + 150 = 266\), A1 for dividing by 2 to reach \(2x^2 + 25x - 58 = 0\). (b) M2 for \((2x + 29)(x - 2)\) [or M1 for correct use of quadratic formula with at most one sign error], A1 for \(x = 2\), A1 for \(x = -14.5\). (c) B1 for \(2\text{ m}\) (accept \(2\)). (d) M1 for \(266 - 150\), M1 for dividing their path area by \(0.25\) (or multiplying by \(4\)), A1 for \(464\).