2025 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

First, divide both sides of the equation by 3:
\(2x - 5 = 7\)

Next, add 5 to both sides:
\(2x = 12\)

Finally, divide by 2:
\(x = 6\)

M1 for correctly dividing by 3 to get \(2x - 5 = 7\) or for expanding the brackets to get \(6x - 15 = 21\).
A0.5 for the correct final answer of 6.

Find the highest common factor (HCF) of the two terms:
- The HCF of 8 and 12 is 4.
- The HCF of \(a^2\) and \(a\) is \(a\).
- The HCF of \(b\) and \(b^2\) is \(b\).

Therefore, the common factor is \(4ab\).
Factoring this out gives:
\(4ab(2a - 3b)\)

M1 for finding any common factor (e.g. \(2ab(4a - 6b)\) or \(ab(8a - 12b)\)).
A0.5 for the fully factorised correct expression \(4ab(2a - 3b)\).

First, calculate the width of the rectangle:
\(\text{Width} = \frac{\text{Area}}{\text{Length}} = \frac{48}{8} = 6\text{ cm}\)

Next, calculate the perimeter of the rectangle:
\(\text{Perimeter} = 2 \times (\text{Length} + \text{Width})\)
\(\text{Perimeter} = 2 \times (8 + 6) = 2 \times 14 = 28\text{ cm}\)

M1 for finding the width is \(6\text{ cm}\) (or showing \(48 \div 8\)).
A0.5 for the correct perimeter of 28.

First, find the total number of parts in the ratio:
\(2 + 3 + 5 = 10\text{ parts}\)

Next, calculate the value of one part:
\(\$120 \div 10 = \$12\)

The largest share has 5 parts:
\(5 \times \$12 = \$60\)

M1 for finding the value of one share (\(120 \div 10\)) or writing the fraction \(\frac{5}{10} \times 120\).
A0.5 for the correct final answer of 60.

First, calculate the number of pages Maya reads on Monday:
\(\frac{3}{8} \times 160 = 3 \times 20 = 60\text{ pages}\)

Next, add the pages read on Tuesday to get the total pages read:
\(60 + 45 = 105\text{ pages}\)

Finally, subtract the total pages read from the book's total pages:
\(160 - 105 = 55\text{ pages}\)

M1 for finding the pages read on Monday is 60, or showing a total of 105 pages have been read.
A0.5 for the correct final answer of 55.

First, find the sum of all five temperatures:
\((-3) + 2 + 5 + (-1) + 7 = 10\)

Next, divide the sum by the number of temperatures (5):
\(\text{Mean} = \frac{10}{5} = 2^\circ\text{C}\)

M1 for correctly calculating the sum as 10, or showing the division \(\frac{-3 + 2 + 5 - 1 + 7}{5}\).
A0.5 for the correct mean of 2.

Substitute \(x = -3\) into the given equation:
\(y = 2(-3)^2 - 5(-3)\)

Calculate each term:
\((-3)^2 = 9\), so \(2(9) = 18\)
\(-5 \times (-3) = 15\)

Add the terms together:
\(y = 18 + 15 = 33\)

M1 for correct substitution of \(x = -3\), showing \(2(9) - 5(-3)\) or \(18 + 15\).
A0.5 for the correct final answer of 33.

Following the order of operations (BIDMAS/BODMAS):

First, calculate inside the parentheses:
\(2 - 5 = -3\)

Next, perform the multiplication:
\(-4 \times (-3) = 12\)

Substitute this back into the expression:
\(12 + 12 = 24\)

M1 for calculating the bracket as \(-3\) and showing multiplication to get \(+12\), or for showing the intermediate expression \(12 - (-12)\).
A0.5 for the correct final answer of 24.

First, expand each bracket separately:
\(3(2x - 5) = 6x - 15\)
\(-4(x - 3) = -4x + 12\)

Now, collect the like terms together:
\(6x - 4x - 15 + 12 = 2x - 3\).

M1 for correct expansion of at least one bracket (e.g., \(6x - 15\) or \(-4x + 12\))
A0.5 for \(2x - 3\)

Let the width of the rectangle be \(w\text{ cm}\).
The formula for perimeter is \(2(\text{length} + \text{width})\).
\(2(10 + w) = 32\)
\(10 + w = 16\)
\(w = 6\text{ cm}\)

Now calculate the area:
\(\text{Area} = \text{length} \times \text{width} = 10 \times 6 = 60\text{ cm}^2\).

M1 for finding the width is \(6\text{ cm}\) or showing a correct method to find the width (e.g., \((32 - 2 \times 10) \div 2\))
A0.5 for \(60\)

Multiply both sides of the equation by \(4\):
\(3x - 1 = 20\)

Add \(1\) to both sides:
\(3x = 21\)

Divide both sides by \(3\):
\(x = 7\).

M1 for multiplying both sides by \(4\) to obtain \(3x - 1 = 20\), or for adding \(1\) to both sides to obtain \(3x = 21\) (following an arithmetic error)
A0.5 for \(7\)

First, find \(15\%\) of \(\$80\):
\(10\%\) of \(\$80 = \$8\)
\(5\%\) of \(\$80 = \$4\)
So, \(15\%\) of \(\$80 = \$8 + \$4 = \$12\).

Subtract the discount from the original price to find the sale price:
\(\$80 - \$12 = \$68\).

M1 for finding the discount of \(\$12\) or showing a correct method to calculate \(85\%\) of \(\$80\) (e.g., \(0.85 \times 80\))
A0.5 for \(68\)

First, use the laws of indices to simplify the multiplication:
\(5^3 \times 5^{-1} = 5^{3 + (-1)} = 5^2 = 25\)

Next, calculate \(2^2\):
\(2^2 = 4\)

Now, subtract the two values:
\(25 - 4 = 21\).

M1 for simplifying \(5^3 \times 5^{-1}\) to \(5^2\) or finding its value as \(25\)
A0.5 for \(21\)

The given list of ages is already in ascending order: \(4, 6, 8, 8, 9, 13\).
Since there is an even number of values (6 values), the median is the mean of the middle two values (the 3rd and 4th values):
Middle values are \(8\) and \(8\).

\(\text{Median} = \frac{8 + 8}{2} = 8\).

M1 for identifying the two middle values as \(8\) and \(8\)
A0.5 for \(8\)

Find the gradient, \(m\):
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 3}{2 - 0} = \frac{8}{2} = 4\)

The line passes through \((0, 3)\), so the y-intercept, \(c\), is \(3\).

Substituting these into the general equation \(y = mx + c\) gives:
\(y = 4x + 3\).

M1 for finding the correct gradient \(m = 4\)
A0.5 for the complete equation \(y = 4x + 3\)

The frequency of Red is \(12\), and the total number of spins is \(50\).
Write this as a fraction:
\(\frac{12}{50}\)

To find the percentage, convert the fraction so that the denominator is 100:
\(\frac{12 \times 2}{50 \times 2} = \frac{24}{100} = 24\%\).

M1 for showing the fraction \(\frac{12}{50}\) or equivalent
A0.5 for \(24\) (or \(24\%\))

First, group the terms into two pairs: \(10ax - 15ay\) and \(6bx - 9by\). Factorise the highest common factor from each pair: \(10ax - 15ay = 5a(2x - 3y)\) and \(6bx - 9by = 3b(2x - 3y)\). Now, take out the common binomial factor \((2x - 3y)\) to obtain the final factorised expression: \((5a + 3b)(2x - 3y)\).

M1 for finding a common factor of a pair of terms, such as \(5a(2x - 3y)\) or \(3b(2x - 3y)\). A0.5 for the fully correct factorised expression: \((5a + 3b)(2x - 3y)\).

The formula for the area of a trapezium is \(\text{Area} = \frac{1}{2}(a + b)h\), where \(a\) and \(b\) are the lengths of the parallel sides, and \(h\) is the perpendicular height. Substituting the given values: \(36 = \frac{1}{2}(7 + 11)h\). Simplify the terms inside the brackets: \(36 = \frac{1}{2}(18)h\), which becomes \(36 = 9h\). Dividing both sides by 9, we get \(h = 4\).

M1 for substituting correctly into the area of a trapezium formula, e.g. \(\frac{1}{2}(7 + 11)h = 36\) or \(9h = 36\). A0.5 for \(4\).

The range is the difference between the maximum and minimum values in the data set. Identify the maximum temperature, which is \(7^{\circ}\text{C}\), and the minimum temperature, which is \(-5^{\circ}\text{C}\). Calculate the range: \(\text{Range} = 7 - (-5) = 7 + 5 = 12^{\circ}\text{C}\).

M1 for identifying both the correct maximum \(7\) and minimum \(-5\), or for writing a correct calculation such as \(7 - (-5)\). A0.5 for \(12\).

First, eliminate the fraction by multiplying both sides of the equation by 2: \(3x - 5 = 16\). Next, isolate the term with \(x\) by adding 5 to both sides: \(3x = 16 + 5\), which simplifies to \(3x = 21\). Finally, divide both sides by 3 to find \(x\): \(x = \frac{21}{3}\), giving \(x = 7\).

M1 for correctly eliminating the fraction, e.g. \(3x - 5 = 16\), or for writing \(1.5x - 2.5 = 8\). A0.5 for \(7\).

(a) Expand the brackets:
\(3(2x - 5) - 2(x - 4) = 6x - 15 - 2x + 8\)
Group the like terms:
\((6x - 2x) + (-15 + 8) = 4x - 7\)

(b) Find the highest common factor (HCF) of the terms \(12y^2\) and \(18y\), which is \(6y\):
\(12y^2 - 18y = 6y(2y - 3)\)

(a) M1 for correct expansion of at least one bracket (e.g., \(6x - 15\) or \(-2x + 8\))
A1 for final answer \(4x - 7\)

(b) M1 for partial factorisation or taking out a common factor (e.g., \(2y(6y - 9)\) or \(6(2y^2 - 3y)\))
A1 for final fully factorised answer \(6y(2y - 3)\)

(a) The formula for the area of a trapezium is:
\(\text{Area} = \frac{1}{2}(a + b)h\)
Substitute the given values:
\(\text{Area} = \frac{1}{2}(7 + 11) \times 6 = \frac{1}{2}(18) \times 6 = 9 \times 6 = 54\text{ cm}^2\)

(b) The area of the rectangle is also \(54\text{ cm}^2\).
\(\text{Area} = \text{length} \times \text{width}\)
\(54 = \text{length} \times 4\)
\(\text{length} = \frac{54}{4} = 13.5\text{ cm}\)
Now calculate the perimeter of the rectangle:
\(\text{Perimeter} = 2(\text{length} + \text{width}) = 2(13.5 + 4) = 2(17.5) = 35\text{ cm}\)

(a) M1 for \(\frac{1}{2}(7 + 11) \times 6\) or equivalent
A1 for \(54\)

(b) M1 for setting up \(\text{length} = 54 / 4\) or finding the length is \(13.5\)
A1 for final answer \(35\)

(a) Write out the goals in order of size:
\(0, 0, 1, 1, 1, 2, 2, 2, 3, 4\)
Since there are 10 matches, the median is the average of the 5th and 6th values:
The 5th value is \(1\) and the 6th value is \(2\).
\(\text{Median} = \frac{1 + 2}{2} = 1.5\)

(b) Find the total number of goals:
\((0 \times 2) + (1 \times 3) + (2 \times 3) + (3 \times 1) + (4 \times 1) = 0 + 3 + 6 + 3 + 4 = 16\)
Divide by the total frequency (10):
\(\text{Mean} = \frac{16}{10} = 1.6\)

(a) M1 for identifying the 5th and 6th values or for listing the data in order
A1 for \(1.5\)

(b) M1 for finding the total number of goals, \(16\) (e.g., \(0 \times 2 + 1 \times 3 + 2 \times 3 + 3 \times 1 + 4 \times 1\))
A1 for \(1.6\)

(a) To find \(p\), substitute \(x = -1\) into the equation:
\(y = (-1)^2 - 3(-1) - 2 = 1 + 3 - 2 = 2\)
So, \(p = 2\).

To find \(q\), substitute \(x = 2\) into the equation:
\(y = (2)^2 - 3(2) - 2 = 4 - 6 - 2 = -4\)
So, \(q = -4\).

(b) The line of symmetry passes through the vertex of the parabola.
The minimum point occurs halfway between the symmetric values. Looking at the table, \(y = -4\) at both \(x = 1\) and \(x = 2\).
The line of symmetry is halfway between these values:
\(x = \frac{1 + 2}{2} = 1.5\) (or using \(x = -\frac{b}{2a} = -\frac{-3}{2(1)} = 1.5\)).
So the equation is \(x = 1.5\).

(a) B1 for \(p = 2\)
B1 for \(q = -4\)

(b) M1 for finding the midpoint of \(1\) and \(2\) or using \(-\frac{b}{2a}\)
A1 for \(x = 1.5\) (accept \(x = \frac{3}{2}\))

Let the equations be:
1) \(3x + 2y = 11\)
2) \(2x - y = 12\)

Multiply equation (2) by 2 to make the coefficients of \(y\) opposites:
\(4x - 2y = 24\) (3)

Add equation (1) and equation (3):
\((3x + 2y) + (4x - 2y) = 11 + 24\)
\(7x = 35\)
\(x = 5\)

Substitute \(x = 5\) into equation (2):
\(2(5) - y = 12\)
\(10 - y = 12\)
\(-y = 2\)
\(y = -2\)

M1 for a valid method to eliminate one variable (e.g., multiplying second equation by 2)
A1 for finding \(x = 5\) or \(y = -2\)
M1 for substituting their found value back into either equation to find the second variable
A1 for both correct values: \(x = 5\) and \(y = -2\)

(a) A reduction of \(20\%\) means the sale price is \(80\%\) of the original price.
Let \(P\) be the original price.
\(0.80 \times P = 240\)
\(P = \frac{240}{0.8} = \frac{2400}{8} = 300\)
So the original price is \(\$300\).

(b) First, find the actual increase in price:
\(\text{Increase} = 472 - 400 = 72\)
Next, calculate this increase as a percentage of the original price:
\(\text{Percentage increase} = \frac{72}{400} \times 100 = \frac{72}{4} = 18\%\)

(a) M1 for \(240 \div 0.8\) or equivalent
A1 for \(300\)

(b) M1 for finding the increase of \(72\) or writing \(\frac{472}{400}\)
A1 for \(18\)

(a) Find the common difference between consecutive terms:
\(11 - 5 = 6\)
\(17 - 11 = 6\)
Since the difference is \(6\), the formula starts with \(6n\).
Now find the constant offset \(c\) in \(6n + c\) by using the first term (where \(n=1\)):
\(6(1) + c = 5 \implies 6 + c = 5 \implies c = -1\)
So the \(n\)-th term is \(6n - 1\).

(b) To find the 5th term, substitute \(n = 5\) into the given formula:
\(3n^2 - 1 = 3(5)^2 - 1\)
\(5^2 = 25\)
\(3 \times 25 - 1 = 75 - 1 = 74\)

(a) M1 for a term of \(6n\) seen in their final answer
A1 for \(6n - 1\) (or equivalent)

(b) M1 for substituting \(5\) into the expression: \(3(5^2) - 1\)
A1 for \(74\)

Factorise the numerator: \(x^2 - 9 = (x - 3)(x + 3)\). Factorise the denominator: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). Cancel the common factor \((x + 3)\) to get the simplified fraction \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator correctly as \((x-3)(x+3)\) or the denominator correctly as \((2x-1)(x+3)\). A1 for \(\frac{x-3}{2x-1}\).

Expand the brackets to get \(6x - 15 - 2x - 8 = 9\). Simplify this to \(4x - 23 = 9\). Add 23 to both sides to get \(4x = 32\), which gives \(x = 8\).

M1 for correct expansion of at least one bracket, e.g. \(6x - 15\) or \(-2x - 8\) (or better, \(4x - 23 = 9\)). A1 for 8.

Use the formula for the area of a trapezium, \(\text{Area} = \frac{1}{2}(a + b)h\). Substitute the given values: \(42 = \frac{1}{2}(x + x + 4) \times 6\). Simplify to \(42 = 3(2x + 4)\). Divide by 3 to get \(14 = 2x + 4\), which gives \(2x = 10\) and \(x = 5\).

M1 for setting up a correct equation, e.g. \(\frac{1}{2}(2x + 4) \times 6 = 42\). A1 for 5.

Since the numbers are in ascending order, the median is the 4th number, which is \(x\), so \(x = 8\). The mean is 8, so the sum of the seven numbers is \(7 \times 8 = 56\). Thus, \(3 + 4 + 6 + 8 + 10 + 11 + y = 56\), which simplifies to \(42 + y = 56\), giving \(y = 14\).

M1 for setting up the equation \(42 + y = 56\) or finding \(x = 8\) and attempting to sum the numbers. A1 for 14.

We can find the turning point by completing the square: \(y = (x - 3)^2 - 9 + 14 = (x - 3)^2 + 5\). The turning point is therefore \((3, 5)\). Alternatively, use the line of symmetry \(x = -\frac{b}{2a} = 3\) and substitute back to find \(y = 3^2 - 6(3) + 14 = 5\).

M1 for writing the equation in completed square form as \((x - 3)^2 + 5\) or finding the x-coordinate of the turning point as \(x = 3\). A1 for \((3, 5)\).

Divide the coefficients and the powers of 10 separately: \(\frac{1.2}{8} \times \frac{10^3}{10^{-4}} = 0.15 \times 10^{3 - (-4)} = 0.15 \times 10^7\). Convert this to standard form to get \(1.5 \times 10^6\).

M1 for \(0.15 \times 10^7\) or \(1.5 \times 10^k\) where \(k\) is an integer. A1 for \(1.5 \times 10^6\).

Apply the index to each term inside the bracket: \((27x^9)^{-\frac{2}{3}} = 27^{-\frac{2}{3}} \times (x^9)^{-\frac{2}{3}}\). This simplifies to \(\frac{1}{9} \times x^{-6} = \frac{1}{9x^6}\).

M1 for \(27^{-\frac{2}{3}} = \frac{1}{9}\) or \((x^9)^{-\frac{2}{3}} = x^{-6}\). A1 for \(\frac{1}{9x^6}\).

The linear scale factor is \(k = \frac{18}{12} = \frac{3}{2}\). The volume scale factor is \(k^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}\). Calculate the volume of the larger bottle by multiplying the smaller volume by the scale factor: \(240 \times \frac{27}{8} = 810\text{ cm}^3\).

M1 for finding the volume scale factor \(\left(\frac{18}{12}\right)^3\) or \(\frac{27}{8}\). A1 for 810.

To factorise completely, find the highest common factor (HCF) of the two terms.
For the numerical coefficients \(12\) and \(18\), the HCF is \(6\).
For the variable parts \(a^2b\) and \(ab^2\), the HCF is \(ab\).
Thus, the overall HCF is \(6ab\).
Divide each term by the HCF to find the remaining terms in the bracket:
\(12a^2b \div 6ab = 2a\)
\(-18ab^2 \div 6ab = -3b\)
So, the factorised expression is \(6ab(2a - 3b)\).

M1 for a correct partial factorisation (e.g. \(2ab(6a - 9b)\) or \(6a(2ab - 3b^2)\) or \(6b(2a^2 - 3ab)\))
A1 for the fully correct factorised expression \(6ab(2a - 3b)\)

Multiply both sides of the equation by \(3\) to eliminate the fraction:
\(2x - 5 = 3(4 - x)\)
Expand the right side of the equation:
\(2x - 5 = 12 - 3x\)
Add \(3x\) to both sides:
\(5x - 5 = 12\)
Add \(5\) to both sides:
\(5x = 17\)
Divide by \(5\):
\(x = \frac{17}{5} = 3.4\)

M1 for correct step to eliminate the fraction: \(2x - 5 = 3(4 - x)\) (or better)
A1 for \(3.4\) or \(\frac{17}{5}\) or \(3 \frac{2}{5}\)

The area of a rectangle is length multiplied by width:
\((x + 4)(x - 1) = 36\)
Expand the brackets:
\(x^2 + 3x - 4 = 36\)
Subtract \(36\) from both sides to form a quadratic equation equal to zero:
\(x^2 + 3x - 40 = 0\)
Factorise the quadratic expression:
\((x + 8)(x - 5) = 0\)
This gives two possible solutions:
\(x = -8\) or \(x = 5\)
Since the width of the rectangle is \((x - 1)\text{ cm}\), \(x\) must be greater than \(1\) for the width to be positive.
Therefore, we reject \(x = -8\) and the only valid solution is \(x = 5\).

M1 for setting up the equation \((x + 4)(x - 1) = 36\) and expanding to obtain \(x^2 + 3x - 40 = 0\) (or for \((x + 8)(x - 5) = 0\))
A1 for \(5\) (do not accept \(-8\) as part of the final answer)

Calculate the sum of the first five numbers using their mean:
\(\text{Sum of 5 numbers} = 5 \times 12 = 60\)
Calculate the sum of the six numbers using the new mean:
\(\text{Sum of 6 numbers} = 6 \times 13.5 = 81\)
The sixth number is the difference between these two sums:
\(\text{Sixth number} = 81 - 60 = 21\)

M1 for finding either the total of the first 5 numbers (\(60\)) or the total of the 6 numbers (\(81\))
A1 for \(21\)

Substitute the coordinates of the point \((3, 7)\) into the equation of the curve, where \(x = 3\) and \(y = 7\):
\(7 = 2(3)^2 - 5(3) + k\)
Evaluate the numerical terms on the right side:
\(7 = 2(9) - 15 + k\)
\(7 = 18 - 15 + k\)
\(7 = 3 + k\)
Subtract \(3\) from both sides to find \(k\):
\(k = 4\)

M1 for substituting \(x = 3\) and \(y = 7\) into the given equation, resulting in \(7 = 2(3)^2 - 5(3) + k\) (or equivalent)
A1 for \(4\)

Apply the laws of indices to both the coefficient and the variable term:
\(\left(64x^{12}\right)^{-\frac{2}{3}} = 64^{-\frac{2}{3}} \times \left(x^{12}\right)^{-\frac{2}{3}}\)
Evaluate the numerical coefficient first:
\(64^{-\frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}\)
Next, apply the power of a power rule to the variable part:
\(\left(x^{12}\right)^{-\frac{2}{3}} = x^{12 \times -\frac{2}{3}} = x^{-8} = \frac{1}{x^8}\)
Combine both parts to obtain the final simplified expression:
\(\frac{1}{16x^8}\)

M1 for finding either the correct simplified power of the coefficient (\(\frac{1}{16}\)) or the correct simplified power of the variable (\(x^{-8}\))
A1 for \(\frac{1}{16x^8}\) or \(\frac{x^{-8}}{16}\) or \(0.0625x^{-8}\)

Let \(x = 0.3\dot{7} = 0.3777\dots\)
Multiply by \(10\) to shift the decimal place:
\(10x = 3.7777\dots\)
Multiply by \(100\) to shift it further:
\(100x = 37.7777\dots\)
Subtract the first equation from the second equation to eliminate the recurring digits:
\(100x - 10x = 37.7777\dots - 3.7777\dots\)
\(90x = 34\)
Divide by \(90\):
\(x = \frac{34}{90}\)
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is \(2\):
\(x = \frac{17}{45}\)

M1 for establishing two equations that subtract to eliminate the recurring decimal part (e.g. \(100x = 37.77...\) and \(10x = 3.77...\) leading to \(90x = 34\) or equivalent)
A1 for \(\frac{17}{45}\)

First, factorise each quadratic expression:
1) \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
2) \(2x^2 + 5x + 2 = (2x + 1)(x + 2)\)
3) \(x^2 - 4 = (x - 2)(x + 2)\)
4) \(x^2 - 6x + 9 = (x - 3)^2\)

Substitute these factorised forms back into the expression:
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)} \times \frac{(x - 2)(x + 2)}{(x - 3)^2}\)

Cancel common terms in the numerator and denominator:
- Cancel \(2x + 1\)
- Cancel \(x + 2\)
- Cancel one factor of \(x - 3\)

This leaves the fully simplified fraction:
\(\frac{x - 2}{x - 3}\)

M1 for factorising \(2x^2 - 5x - 3\) to \((2x+1)(x-3)\)
M1 for factorising \(2x^2 + 5x + 2\) to \((2x+1)(x+2)\)
M1 for factorising \(x^2 - 4\) to \((x-2)(x+2)\)
M1 for factorising \(x^2 - 6x + 9\) to \((x-3)^2\)
A1 for final simplified answer \(\frac{x-2}{x-3}\)

Let the radius of the sector be \(r = 12\text{ cm}\) and the angle subtended at the centre be \(\theta^\circ\).

The perimeter of a sector is given by:
\(\text{Perimeter} = 2r + \text{arc length}\)
\(24 + 5\pi = 2(12) + \text{arc length}\)
\(24 + 5\pi = 24 + \text{arc length}\)
\text{So, the arc length} = 5\pi\text{ cm}.

The formula for arc length is:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)
\(5\pi = \frac{\theta}{360} \times 2\pi(12)\)
\(5\pi = \frac{24\pi \theta}{360}\)
\(5 = \frac{\theta}{15} \implies \theta = 75^\circ\).

The formula for the area of a sector is:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)
\(\text{Area} = \frac{75}{360} \times \pi \times 12^2\)
\(\text{Area} = \frac{75}{360} \times 144\pi\)
Since \(\frac{144}{360} = \frac{2}{5}\):
\(\text{Area} = 75 \times \frac{2}{5}\pi = 15 \times 2\pi = 30\pi\text{ cm}^2\).

M1 for setting up the perimeter equation: \(2(12) + \text{arc length} = 24 + 5\pi\)
A1 for finding \(\text{arc length} = 5\pi\)
M1 for setting up the arc length equation to find \(\theta\): \(\frac{\theta}{360} \times 24\pi = 5\pi\) (or finding \(\theta = 75^\circ\))
M1 for substituting \(\theta\) and \(r\) into the sector area formula: \(\frac{75}{360} \times \pi \times 12^2\)
A1 for final answer \(30\pi\)

1. Calculate the sum of the scores for the first group:
\(\text{Sum}_1 = 30 \times 62 = 1860\)

2. Express the sum of the scores for the second group in terms of \(n\):
\(\text{Sum}_2 = 74n\)

3. Write an equation for the combined mean:
\(\text{Combined Mean} = \frac{\text{Total Sum}}{\text{Total Number of Students}}\)
\(69.5 = \frac{1860 + 74n}{30 + n}\)

4. Solve the equation for \(n\):
\(69.5(30 + n) = 1860 + 74n\)
\(2085 + 69.5n = 1860 + 74n\)
Subtract \(1860\) and \(69.5n\) from both sides:
\(2085 - 1860 = 74n - 69.5n\)
\(225 = 4.5n\)
Multiply by 2 to clear decimals:
\(450 = 9n\)
\(n = 50\)

M1 for calculating sum of the first group: \(30 \times 62 = 1860\)
M1 for setting up the combined mean equation: \(\frac{1860 + 74n}{30 + n} = 69.5\)
M1 for expanding and rearranging: \(2085 + 69.5n = 1860 + 74n\) (allow arithmetic errors in expansion)
M1 for isolating the term in \(n\): \(4.5n = 225\) (or equivalent)
A1 for \(n = 50\)

1. Since \(y = -1\) is the horizontal asymptote as \(x \to -\infty\), the constant \(c\) is given by:
\(c = -1\).

2. Use the point \((0, 2)\) to find \(a\):
Substitute \(x = 0\) and \(y = 2\) into the equation:
\(2 = a(3^{k(0)}) - 1\)
\(2 = a(1) - 1\)
\(a = 3\).

3. Use the point \((1, 8)\) and the values of \(a\) and \(c\) to find \(k\):
Substitute \(x = 1\), \(y = 8\), \(a = 3\), and \(c = -1\):
\(8 = 3(3^{k(1)}) - 1\)
\(8 = 3 \cdot 3^k - 1\)
\(9 = 3^{k+1}\)
Since \(9 = 3^2\), we have:
\(k + 1 = 2 \implies k = 1\).

Therefore, the constants are \(a = 3\), \(k = 1\), and \(c = -1\).

B1 for identifying \(c = -1\)
M1 for substituting \((0, 2)\) and their \(c\) into the equation: \(2 = a(3^0) + c\)
A1 for finding \(a = 3\)
M1 for substituting \((1, 8)\), \(a = 3\), and \(c = -1\) into the equation to set up an equation in \(k\): \(8 = 3(3^k) - 1\)
A1 for finding \(k = 1\)

1. Rearrange the linear equation to express \(y\) in terms of \(x\):
\(y = 5 - x\)

2. Substitute \(y = 5 - x\) into the quadratic equation:
\(2x^2 - (5 - x)^2 = 14\)

3. Expand and simplify:
\(2x^2 - (25 - 10x + x^2) = 14\)
\(2x^2 - 25 + 10x - x^2 = 14\)
\(x^2 + 10x - 25 - 14 = 0\)
\(x^2 + 10x - 39 = 0\)

4. Factorise the quadratic equation:
\((x + 13)(x - 3) = 0\)
This gives \(x = -13\) or \(x = 3\).

5. Find the corresponding \(y\) values:
If \(x = 3\), then \(y = 5 - 3 = 2\).
If \(x = -13\), then \(y = 5 - (-13) = 18\).

So the solutions are \(x = 3, y = 2\) and \(x = -13, y = 18\).

M1 for rearranging the linear equation to get \(y = 5 - x\) (or \(x = 5 - y\))
M1 for substituting their expression into the quadratic equation
A1 for obtaining the simplified quadratic equation \(x^2 + 10x - 39 = 0\) (or \(y^2 - 20y + 36 = 0\))
M1 for solving their quadratic equation to find two values of \(x\) (or \(y\))
A1 for both correct pairs of solutions: \((3, 2)\) and \((-13, 18)\)

1. First, find the length of the diagonal of the horizontal base, \(BD\).
In the right-angled triangle \(BAD\), using Pythagoras' theorem:
\(BD^2 = AB^2 + AD^2\)
Since \(ABCD\) is a rectangle, \(AD = BC = 9\text{ cm}\).
\(BD^2 = 12^2 + 9^2\)
\(BD^2 = 144 + 81 = 225\)
\(BD = \sqrt{225} = 15\text{ cm}\).

2. Next, find the length of the space diagonal \(BH\).
The triangle \(BDH\) is a right-angled triangle at \(D\), where \(DH\) is a vertical edge of length \(8\text{ cm}\).
Using Pythagoras' theorem in triangle \(BDH\):
\(BH^2 = BD^2 + DH^2\)
\(BH^2 = 15^2 + 8^2\)
\(BH^2 = 225 + 64 = 289\)
\(BH = \sqrt{289} = 17\text{ cm}\).

Therefore, the length of the diagonal \(BH\) is \(17\text{ cm}\).

M1 for applying Pythagoras' theorem to find the base diagonal: \(BD^2 = 12^2 + 9^2\)
A1 for finding \(BD = 15\) (or keeping as \(BD^2 = 225\))
M1 for identifying that triangle \(BDH\) is right-angled at \(D\) and writing \(BH^2 = BD^2 + DH^2\)
M1 for substituting values: \(BH^2 = 15^2 + 8^2\) (or \(225 + 64\))
A1 for \(17\)

Let the original price of the television be \(P\).

1. After the first reduction of 20%, the price is:
\(P \times (1 - 0.20) = 0.8P\)

2. After the second reduction of 15%, the price is:
\(0.8P \times (1 - 0.15) = 0.8P \times 0.85 = 0.68P\)

3. Set up the equation with the final price:
\(0.68P = 272\)

4. Solve for \(P\):
\(P = \frac{272}{0.68}\)
Multiply numerator and denominator by 100 to clear the decimal:
\(P = \frac{27200}{68}\)
Since \(68 \times 4 = 272\), we get:
\(P = 400\).

The original price of the television was $400.

M1 for expressing first reduction: \(0.8P\) (or equivalent)
M1 for expressing second reduction: \(0.8P \times 0.85\) (or equivalent)
M1 for setting up the equation: \(0.68P = 272\) (or finding the intermediate price of \(320\) via \(272 \div 0.85\))
M1 for the division step: \(P = \frac{272}{0.68}\) (or \(320 \div 0.8\))
A1 for \(400\)

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

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

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

4. Use the point-slope form with midpoint \(M(4, 1)\) and perpendicular gradient \(m_{\perp} = -\frac{1}{2}\):
\(y - 1 = -\frac{1}{2}(x - 4)\)
Multiply both sides by 2:
\(2(y - 1) = -1(x - 4)\)
\(2y - 2 = -x + 4\)
Rearrange into the form \(ax + by = c\):
\(x + 2y = 6\).

M1 for finding the midpoint of \(AB\) as \((4, 1)\)
M1 for finding the gradient of \(AB\) as \(2\)
M1 for finding the perpendicular gradient as \(-\frac{1}{2}\) (negative reciprocal of their gradient)
M1 for substituting their midpoint and perpendicular gradient into a linear equation form
A1 for final answer \(x + 2y = 6\) (or any equivalent integer form like \(2x + 4y = 12\))

First, eliminate the fractions by multiplying every term by the common denominator \((x-2)(x+3)\):
\(4(x+3) + 2(x-2) = (x-2)(x+3)\)

Next, expand the brackets on both sides of the equation:
\(4x + 12 + 2x - 4 = x^2 + 3x - 2x - 6\)
\(6x + 8 = x^2 + x - 6\)

Rearrange the equation into a standard quadratic form \(ax^2 + bx + c = 0\):
\(x^2 + x - 6 - 6x - 8 = 0\)
\(x^2 - 5x - 14 = 0\)

Factorise the quadratic equation:
\((x-7)(x+2) = 0\)

Solve for \(x\):
\(x = 7\) or \(x = -2\)

M1 for multiplying by \((x-2)(x+3)\) or putting LHS over a common denominator.
M1 for expanding brackets correctly to get \(4x + 12 + 2x - 4\) and \(x^2 + x - 6\) (allow max 1 sign error total).
M1 for simplifying to a 3-term quadratic: \(x^2 - 5x - 14 = 0\) (or equivalent).
M1 for factorising into \((x-7)(x+2) = 0\) or correct use of the quadratic formula for their quadratic.
A1 for both correct solutions: \(x = 7\) and \(x = -2\).

Factorise by grouping the first two terms and the last two terms:
\(6ax - 9ay = 3a(2x - 3y)\)
\(-8bx + 12by = -4b(2x - 3y)\)
Combining these common binomial factors, we get:
\((3a - 4b)(2x - 3y)\)

M1 for \(3a(2x - 3y)\) or \(-4b(2x - 3y)\) or equivalent partial factorisation
A1 for \((3a - 4b)(2x - 3y)\)

Expand the brackets:
\(8x - 12 - 3x + 15 = 18\)
Simplify the left side:
\(5x + 3 = 18\)
Subtract 3 from both sides:
\(5x = 15\)
Divide by 5:
\(x = 3\)

M1 for correct expansion of brackets: \(8x - 12 - 3x + 15 = 18\) (allow one numerical error)
A1 for 3

Find the dimensions of the larger rectangle including the path:
Length = \(12 + 1.5 + 1.5 = 15\text{ m}\)
Width = \(8 + 1.5 + 1.5 = 11\text{ m}\)

Area of the larger rectangle = \(15 \times 11 = 165\text{ m}^2\)
Area of the lawn = \(12 \times 8 = 96\text{ m}^2\)

Area of the path = \(165 - 96 = 69\text{ m}^2\)

M1 for showing outer dimensions are \(15\text{ m}\) and \(11\text{ m}\) or for calculating the outer area \(15 \times 11 = 165\)
A1 for 69

Calculate the total number of goals:
\((0 \times 3) + (1 \times 7) + (2 \times 5) + (3 \times 4) + (4 \times 1) = 0 + 7 + 10 + 12 + 4 = 33\)

Divide by the total number of matches (20):
Mean = \(33 / 20 = 1.65\)

M1 for \(0 \times 3 + 1 \times 7 + 2 \times 5 + 3 \times 4 + 4 \times 1\) (or 33) seen
A1 for 1.65

To find where the curve crosses the \(x\)-axis, set \(y = 0\):
\(x^2 - 4x + 3 = 0\)
Factorise the quadratic expression:
\((x - 1)(x - 3) = 0\)
This gives \(x = 1\) or \(x = 3\).
Therefore, the coordinates of the points are \((1, 0)\) and \((3, 0)\).

M1 for factorising to \((x - 1)(x - 3)\) or correct use of the quadratic formula
A1 for \((1, 0)\) and \((3, 0)\) (accept \(x=1\) and \(x=3\))

Multiply both sides by \(2 - t\):
\(s(2 - t) = 3t + 4\)
Expand the bracket:
\(2s - st = 3t + 4\)
Rearrange to group all terms with \(t\) on one side:
\(2s - 4 = 3t + st\)
Factorise out \(t\):
\(2s - 4 = t(3 + s)\)
Divide by \(3 + s\):
\(t = \frac{2s - 4}{3 + s}\)

M1 for multiplying by \(2 - t\) and expanding: \(2s - st = 3t + 4\)
A1 for \(t = \frac{2s - 4}{s + 3}\) or equivalent

The perimeter of a semicircle consists of the curved arc and the straight diameter.
Curved arc length = \(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 14 = 7\pi \approx 21.99\text{ cm}\)
Total perimeter = \(7\pi + 14 \approx 35.99\text{ cm}\)
To 3 significant figures, this is \(36.0\text{ cm}\) (or \(36\text{ cm}\)).

M1 for \(\frac{1}{2} \times \pi \times 14 + 14\) or \(\pi \times 7 + 14\)
A1 for 36 or 36.0 or 35.99...

Expand both sides of the inequality:
\(5x - 10 \le 6x + 3\)
Subtract \(5x\) from both sides:
\(-10 \le x + 3\)
Subtract 3 from both sides:
\(-13 \le x\)
This is equivalent to:
\(x \ge -13\)

M1 for expansion of brackets: \(5x - 10 \le 6x + 3\) (allow one numerical error)
A1 for \(x \ge -13\) or \(-13 \le x\)

First, find the highest common factor of \(12a^2b\) and \(18ab^2\).

- The highest common factor of 12 and 18 is 6.
- The highest common factor of \(a^2\) and \(a\) is \(a\).
- The highest common factor of \(b\) and \(b^2\) is \(b\).

So, the highest common factor of the two terms is \(6ab\).

Divide each term by \(6ab\):
- \( \frac{12a^2b}{6ab} = 2a \)
- \( \frac{18ab^2}{6ab} = 3b \)

Thus, the fully factorised expression is \( 6ab(2a - 3b) \).

M1 for a correct partial factorisation, e.g. \( 2ab(6a - 9b) \), \( 3ab(4a - 6b) \), or \( 6a(2ab - 3b^2) \).
A1 for \( 6ab(2a - 3b) \) or equivalent fully factorised form.

To solve the equation:
\( \frac{2x - 3}{5} = 7 \)

Multiply both sides by 5 to clear the fraction:
\( 2x - 3 = 35 \)

Add 3 to both sides:
\( 2x = 38 \)

Divide both sides by 2:
\( x = 19 \)

M1 for multiplying by 5 to get \( 2x - 3 = 35 \) or for dividing by 5 correctly to get \( \frac{2x}{5} = \frac{38}{5} \).
A1 for 19.

The formula for the area of a trapezium is:
\( \text{Area} = \frac{1}{2}(a + b)h \)

Substitute the given values into the formula:
\( a = 7 \)
\( b = 11 \)
\( h = 6.5 \)

\( \text{Area} = \frac{1}{2}(7 + 11) \times 6.5 \)
\( \text{Area} = \frac{1}{2}(18) \times 6.5 \)
\( \text{Area} = 9 \times 6.5 = 58.5 \text{ cm}^2 \)

M1 for \( \frac{1}{2}(7 + 11) \times 6.5 \) or equivalent.
A1 for 58.5.

The formula for the mean is:
\( \text{Mean} = \frac{\text{Sum of heights}}{\text{Number of students}} \)

So:
\( 148 = \frac{142 + 150 + 145 + 153 + h}{5} \)

First, find the sum of the four known heights:
\( 142 + 150 + 145 + 153 = 590 \)

Substitute this back into the equation:
\( 148 = \frac{590 + h}{5} \)

Multiply both sides by 5:
\( 590 + h = 740 \)

Subtract 590 from both sides:
\( h = 740 - 590 = 150 \)

M1 for setting up a correct equation, e.g. \( 142 + 150 + 145 + 153 + h = 148 \times 5 \) or \( 590 + h = 740 \).
A1 for 150.

Substitute \(x = -3\) into the given function equation:
\( y = 2(-3)^2 - 5(-3) + 1 \)

First calculate the square:
\( (-3)^2 = 9 \), so \( 2(-3)^2 = 2 \times 9 = 18 \)

Next calculate the linear term:
\( -5(-3) = +15 \)

Add the terms together:
\( y = 18 + 15 + 1 = 34 \)

M1 for substituting \(x = -3\) with at least one correct term evaluated, e.g. \( 2(9) - 5(-3) + 1 \) or \( 18 + 15 + 1 \).
A1 for 34.

First expand each of the brackets:
\( 4(2a - 3b) = 8a - 12b \)
\( -3(a - 5b) = -3a + 15b \)

Now collect the like terms together:
\( 8a - 3a - 12b + 15b \)
\( = 5a + 3b \)

M1 for correct expansion of at least one bracket, e.g. \( 8a - 12b \) or \( -3a + 15b \).
A1 for \( 5a + 3b \).

Multiply both sides of the formula by 3 to clear the fraction:
\( 3w = t - 4 \)

Add 4 to both sides to isolate \(t\):
\( 3w + 4 = t \)

Therefore, \( t = 3w + 4 \).

M1 for multiplying both sides by 3 to get \( 3w = t - 4 \) or for dividing correctly to get \( w + \frac{4}{3} = \frac{t}{3} \).
A1 for \( 3w + 4 \) (or \( t = 3w + 4 \)).

The perimeter of a semicircle consists of the curved arc plus the straight diameter.

- The curved arc length is half of the circumference of a full circle:
\( \text{Arc length} = \frac{1}{2} \pi d = \frac{1}{2} \times \pi \times 8 = 4\pi \approx 12.566 \text{ cm} \)

- The straight side is the diameter of 8 cm.

Now, add the arc length and the diameter to find the total perimeter:
\( \text{Perimeter} = 12.566 + 8 = 20.566 \text{ cm} \)

Rounding to 1 decimal place, we get \( 20.6 \) cm.

M1 for \( \frac{1}{2} \times \pi \times 8 + 8 \) or \( 4\pi + 8 \).
A1 for 20.6 (accept 20.56 to 20.57).

The mean of the five numbers is 8, so their sum is \(5 \times 8 = 40\). This gives the equation: \(3 + a + 9 + b + 13 = 40\). Simplifying the equation: \(25 + a + b = 40\). Subtracting 25 from both sides gives \(a + b = 15\).

M1 for \(3 + a + 9 + b + 13 = 5 \times 8\) or equivalent. A1 for 15.

Expand the brackets: \(8x - 12 = 18\). Add 12 to both sides: \(8x = 30\). Divide by 8: \(x = \frac{30}{8} = 3.75\). Alternatively, divide both sides by 4 first: \(2x - 3 = 4.5\), then add 3: \(2x = 7.5\), and divide by 2: \(x = 3.75\).

M1 for \(8x - 12 = 18\) or \(2x - 3 = 4.5\) or equivalent first step. A1 for 3.75 or equivalent fraction.

(a) Expanding the brackets: \(5(2x - 3) = 10x - 15\) and \(-3(x - 4) = -3x + 12\). Combining like terms: \(10x - 3x - 15 + 12 = 7x - 3\). (b) The highest common factor of \(15y^2\) and \(10y\) is \(5y\). Factorising this out gives: \(5y(3y - 2)\).

(a) M1 for \(10x - 15\) or \(-3x + 12\), A1 for \(7x - 3\). (b) M1 for finding a common factor such as \(y(15y - 10)\) or \(5(3y^2 - 2y)\), A1.4 for the fully factorised expression \(5y(3y - 2)\).

First, find the lengths of the two missing inner edges. The inner horizontal edge is \(14\text{ m} - 6\text{ m} = 8\text{ m}\). The inner vertical edge is \(10\text{ m} - 4\text{ m} = 6\text{ m}\). (a) Perimeter = \(10 + 14 + 4 + 8 + 6 + 6 = 48\text{ m}\). (b) Split the shape into two rectangles: a vertical split gives a left rectangle of size \(6\text{ m} \times 10\text{ m} = 60\text{ m}^2\) and a right rectangle of size \(8\text{ m} \times 4\text{ m} = 32\text{ m}^2\). Total area = \(60 + 32 = 92\text{ m}^2\).

(a) M1 for identifying missing sides of \(8\text{ m}\) and \(6\text{ m}\), A1 for \(48\text{ m}\). (b) M1 for a complete method to split the shape and find the area of both parts, A1.4 for \(92\text{ m}^2\).

(a) The mode is the value with the highest frequency. The highest frequency is 6, which corresponds to 1 goal. Mode = 1. (b) The total number of goals is \((0 \times 3) + (1 \times 6) + (2 \times 5) + (3 \times 4) + (4 \times 2) = 0 + 6 + 10 + 12 + 8 = 36\). The total number of matches (frequency sum) is \(3 + 6 + 5 + 4 + 2 = 20\). Mean = \(36 \div 20 = 1.8\).

(a) B1 for 1. (b) M1 for showing the sum of goals \((0 \times 3) + (1 \times 6) + (2 \times 5) + (3 \times 4) + (4 \times 2)\) or 36, M1 for dividing their sum by 20, A1.4 for 1.8.

(a) Substitute \(x = -2\) into the equation: \(y = 2(-2)^2 - 3(-2) - 5 = 2(4) + 6 - 5 = 8 + 6 - 5 = 9\). So \(a = 9\). Substitute \(x = 1\) into the equation: \(y = 2(1)^2 - 3(1) - 5 = 2 - 3 - 5 = -6\). So \(b = -6\). (b) The y-intercept is where \(x = 0\). From the table, when \(x = 0\), \(y = -5\). The coordinates are \((0, -5)\).

(a) M1 for substituting either \(x = -2\) or \(x = 1\) into the function, A1 for \(a = 9\), A1 for \(b = -6\). (b) B1.4 for \((0, -5)\) or \(x=0, y=-5\).

Multiply all terms by 12 (the lowest common multiple of 4 and 3) to clear the denominators: \(12 \times \left(\frac{3x - 1}{4}\right) - 12 \times \left(\frac{x + 2}{3}\right) = 12 \times 2\), which simplifies to: \(3(3x - 1) - 4(x + 2) = 24\). Expand the brackets: \(9x - 3 - 4x - 8 = 24\). Combine like terms: \(5x - 11 = 24\). Add 11 to both sides: \(5x = 35\). Divide by 5: \(x = 7\).

M1 for multiplying by 12 or putting terms over a common denominator, e.g., \(\frac{3(3x-1) - 4(x+2)}{12} = 2\). M1 for expanding brackets correctly: \(9x - 3 - 4x - 8\) (allow one sign error). M1 for isolating the x terms to get \(5x = 35\). A1.4 for \(x = 7\).

(a) The sale price represents \(100\% - 15\% = 85\%\) of the original price. Let \(P\) be the original price: \(0.85 \times P = 272\), so \(P = 272 \div 0.85 = 320\). (b) The total value after 2 years is \(400 \times (1.03)^2 = 400 \times 1.0609 = 424.36\). The total interest earned is \(424.36 - 400 = 24.36\).

(a) M1 for \(272 \div 0.85\), A1 for \(320\). (b) M1 for \(400 \times 1.03^2\) or \(424.36\), A1.4 for \(24.36\).

(a) Using Pythagoras' theorem: \(h^2 + 2.5^2 = 6.5^2\), so \(h^2 = 6.5^2 - 2.5^2 = 42.25 - 6.25 = 36\). Taking the square root gives \(h = \sqrt{36} = 6\text{ m}\). (b) Let the angle with the ground be \(\theta\). Using the cosine ratio: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5}\). Therefore, \(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right) \approx 67.38^\circ\), which is \(67.4^\circ\) to 1 decimal place.

(a) M1 for \(6.5^2 - 2.5^2\) or \(\sqrt{6.5^2 - 2.5^2}\), A1 for \(6\). (b) M1 for \(\cos(\theta) = \frac{2.5}{6.5}\) or \(\sin(\theta) = \frac{6}{6.5}\) or \(\tan(\theta) = \frac{6}{2.5}\), A1.4 for \(67.4^\circ\) or \(67.38^\circ\) (accept answers in the range \(67.3^\circ\) to \(67.4^\circ\)).

The total number of balls is \(5 + 3 = 8\). (a) The probability of picking a red ball first is \(\frac{5}{8}\) (or \(0.625\)). (b) Because the ball is replaced, the probabilities remain the same for both picks: \(P(\text{Red}) = \frac{5}{8}\) and \(P(\text{Blue}) = \frac{3}{8}\). The combinations for different colours are (Red, Blue) and (Blue, Red). \(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{8} = \frac{15}{64}\). \(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}\). Adding these together: \(\frac{15}{64} + \frac{15}{64} = \frac{30}{64} = \frac{15}{32}\) (or \(0.46875\)).

(a) B1 for \(\frac{5}{8}\) or equivalent decimal. (b) M1 for \(\frac{5}{8} \times \frac{3}{8}\) or \(\frac{15}{64}\), M1 for adding \(\frac{15}{64} + \frac{15}{64}\), A1.4 for \(\frac{15}{32}\) or equivalent decimal \(0.469\) or \(0.46875\).

First, calculate the area of the rectangular section of the garden:
\(\text{Area of rectangle} = \text{length} \times \text{width} = 14 \times 8 = 112\text{ m}^2\).

Next, find the radius of the semi-circle. Since the diameter of the semi-circle is equal to the width of the rectangle (\(8\text{ m}\)):
\(r = \frac{8}{2} = 4\text{ m}\).

Now, calculate the area of the semi-circular section:
\(\text{Area of semi-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times 4^2 = 8\pi \approx 25.133\text{ m}^2\).

Finally, add the two areas to find the total area of the garden:
\(\text{Total Area} = 112 + 25.133 = 137.133\text{ m}^2\).

Rounding to 1 decimal place, we get \(137.1\text{ m}^2\).

M1 for calculating the area of the rectangle: \(14 \times 8 = 112\)
M1 for identifying the radius of the semi-circle is \(4\text{ m}\) and attempting its area: \(\frac{1}{2} \times \pi \times 4^2\) (or \(25.1\) or \(8\pi\))
M1 for adding their two calculated areas: \(112 + 25.13...\)
A1.4 for the correct final answer of \(137.1\)

Let \(a\) be the cost of an adult ticket and \(c\) be the cost of a child ticket.

From the information given, we can write the following system of equations:
1) \(3a + 5c = 49\)
2) \(4a + 2c = 42\)

We can simplify equation (2) by dividing all terms by 2:
\(2a + c = 21 \implies c = 21 - 2a\)

Substitute this expression for \(c\) into equation (1):
\(3a + 5(21 - 2a) = 49\)
\(3a + 105 - 10a = 49\)
\(-7a + 105 = 49\)
\(-7a = 49 - 105\)
\(-7a = -56\)
\(a = 8\)

Now, substitute \(a = 8\) back into the expression for \(c\):
\(c = 21 - 2(8)\)
\(c = 21 - 16 = 5\)

Thus, the cost of a child ticket is \(\$5\).

M1 for setting up two correct simultaneous equations, e.g., \(3a + 5c = 49\) and \(4a + 2c = 42\) (or any other variables)
M1 for a correct method to eliminate one variable (e.g., multiplying to equate coefficients or substituting)
A1 for finding one correct value (either adult ticket price \(a = 8\) or child ticket price \(c = 5\))
A1.4 for the correct final answer of \(5\) (cost of a child ticket)

Factor the numerator: \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\). Factor the denominator: \(9x^2 - 1 = (3x + 1)(3x - 1)\). Divide both by the common factor \(3x + 1\) to get \(\frac{x - 5}{3x - 1}\).

1 mark for factoring numerator: \((3x + 1)(x - 5)\). 1 mark for factoring denominator: \((3x + 1)(3x - 1)\). 0.5 marks for correct simplification.

Square both sides: \(p^2 = \frac{3t - 4r}{t + r}\). Multiply by the denominator: \(p^2(t + r) = 3t - 4r\) which simplifies to \(p^2 t + p^2 r = 3t - 4r\). Rearrange to group \(t\) terms on one side: \(p^2 r + 4r = 3t - p^2 t\). Factor out \(t\): \(r(p^2 + 4) = t(3 - p^2)\). Divide by \(3 - p^2\) to isolate \(t\): \(t = \frac{r(p^2 + 4)}{3 - p^2}\).

1 mark for squaring both sides: \(p^2 = \frac{3t - 4r}{t + r}\). 1 mark for isolating terms with \(t\): \(3t - p^2 t = p^2 r + 4r\). 0.5 marks for factoring and correct division.

The perimeter of a sector is given by \(2r + \text{arc length}\). The arc length is \(\frac{\theta}{360} \times 2 \pi r = \frac{110}{360} \times 2 \pi \times 8 \approx 15.359\text{ cm}\). The perimeter is therefore \(2(8) + 15.359 = 31.359\text{ cm}\). Rounding to 3 significant figures gives \(31.4\text{ cm}\).

1 mark for calculating correct arc length: \(15.4\) (or \(15.359...\)). 1 mark for adding \(2 \times 8\) to their arc length. 0.5 marks for correct final answer of 31.4.

The area of the shape is the sum of the area of the rectangle and the semi-circle: \(15w + \frac{1}{2} \pi \left(\frac{w}{2}\right)^2 = 120 \Rightarrow 15w + \frac{\pi w^2}{8} = 120\). This can be written as the quadratic equation \(\frac{\pi}{8} w^2 + 15w - 120 = 0\). Using the quadratic formula: \(w = \frac{-15 + \sqrt{15^2 - 4(\frac{\pi}{8})(-120)}}{2(\frac{\pi}{8})} = \frac{-15 + \sqrt{225 + 60\pi}}{\frac{\pi}{2}} \approx 6.792\text{ cm}\). Correct to 3 significant figures, \(w = 6.79\).

1 mark for setting up the correct equation: \(15w + \frac{\pi w^2}{8} = 120\). 1 mark for using the quadratic formula with correct substitutions. 0.5 marks for correct final answer of 6.79 (accept 6.79 to 6.80).

Identify the midpoints of the intervals: 5, 15, 25, and 35. The sum of the products of frequencies and midpoints is \(5(5) + 12(15) + x(25) + 8(35) = 25 + 180 + 25x + 280 = 485 + 25x\). The total number of scores is \(5 + 12 + x + 8 = 25 + x\). Set up the mean equation: \(\frac{485 + 25x}{25 + x} = 21.5 \Rightarrow 485 + 25x = 21.5(25 + x) \Rightarrow 485 + 25x = 537.5 + 21.5x \Rightarrow 3.5x = 52.5 \Rightarrow x = 15\).

1 mark for identifying midpoints and getting the expression for total frequency-midpoint products: \(485 + 25x\). 1 mark for the equation \(485 + 25x = 21.5(25 + x)\) and attempting to solve. 0.5 marks for the correct answer 15.

Let the numbers in ascending order be \(a, b, c, d, e\). Since the median is 8, \(c = 8\). The mode is 6, so at least two numbers must be 6. Since \(a \le b \le 8\), we have \(a = 6\) and \(b = 6\). The range is 10, so \(e - a = 10 \Rightarrow e - 6 = 10 \Rightarrow e = 16\). The mean is 9, so the sum of the five numbers is \(5 \times 9 = 45\). Thus, \(6 + 6 + 8 + d + 16 = 45 \Rightarrow 36 + d = 45 \Rightarrow d = 9\). The numbers are \(6, 6, 8, 9, 16\), and the largest number is 16.

1 mark for identifying that the three smallest numbers must be 6, 6, 8. 1 mark for using the range of 10 to establish the largest number is 16, or setting up the sum equation. 0.5 marks for the correct largest number of 16.

Set the equations equal to find the intersection points: \(2x^3 - 5x^2 - 2x + 10 = 4x - 5 \Rightarrow 2x^3 - 5x^2 - 6x + 15 = 0\). Since \(x = 2.5\) (or \(2x - 5 = 0\)) is a root, divide the cubic by \((2x - 5)\) to get \(x^2 - 3 = 0\). Solving this gives \(x = \pm \sqrt{3}\). Since \(x < 0\), we have \(x = -\sqrt{3} \approx -1.73\) (correct to 3 significant figures).

1 mark for setting up the equation \(2x^3 - 5x^2 - 6x + 15 = 0\). 1 mark for factorizing to find \(x^2 - 3 = 0\). 0.5 marks for the correct negative root of -1.73.

From the first equation, make \(y\) the subject: \(y = 2x - 5\). Substitute this into the second equation: \(x^2 + (2x - 5)^2 = 25 \Rightarrow x^2 + 4x^2 - 20x + 25 = 25 \Rightarrow 5x^2 - 20x = 0\). Factorizing gives \(5x(x - 4) = 0\), so \(x = 0\) or \(x = 4\). Since \(x > 0\), the required value is \(x = 4\).

1 mark for substitution to get \(x^2 + (2x - 5)^2 = 25\). 1 mark for simplifying to the quadratic equation \(5x^2 - 20x = 0\). 0.5 marks for identifying the correct positive solution \(x = 4\).

Multiply both sides by \( (2x + 7) \):
\( y(2x + 7) = 3x - 5 \)

Expand the bracket:
\( 2xy + 7y = 3x - 5 \)

Rearrange to group all terms containing \( x \) on one side and other terms on the other side:
\( 7y + 5 = 3x - 2xy \)

Factorise out \( x \):
\( 7y + 5 = x(3 - 2y) \)

Divide by \( (3 - 2y) \) to isolate \( x \):
\( x = \frac{7y + 5}{3 - 2y} \)

M1 for multiplying both sides by \( (2x + 7) \) to eliminate the fraction.
M1 for expanding brackets, grouping all terms with \( x \) on one side, and factorising.
A0.5 for the correct final formula.

Using the sector area formula, \( \text{Area} = \frac{1}{2} r L \), where \( L \) is the arc length:
\( 29.4 = \frac{1}{2} \times 8.4 \times L \)
\( 29.4 = 4.2 \times L \)
\( L = \frac{29.4}{4.2} = 7\text{ cm} \)

The perimeter of the sector includes the arc length and two radii:
\( \text{Perimeter} = L + 2r = 7 + 2(8.4) = 23.8\text{ cm} \)

M1 for setting up a correct equation for the area to find the arc length or sector angle.
M1 for finding the arc length as \( 7\text{ cm} \) (or finding the angle as \( \theta \approx 47.7^{\circ} \)) and adding \( 2 \times 8.4 \).
A0.5 for the correct final answer of \( 23.8 \).

Find the total mass of the boys:
\( 8 \times 52.5 = 420\text{ kg} \)

Find the total mass of the girls:
\( 12 \times 46.5 = 558\text{ kg} \)

Find the combined total mass of the 20 children:
\( 420 + 558 = 978\text{ kg} \)

Calculate the overall mean mass:
\( \text{Mean} = \frac{978}{20} = 48.9\text{ kg} \)

M1 for finding either the total mass of the boys (\( 420 \)) or the total mass of the girls (\( 558 \)).
M1 for summing the two total masses and dividing by 20.
A0.5 for the correct answer of \( 48.9 \).

Set the two equations equal to find the \( x \)-coordinates of the intersection points:
\( 2x^2 - 5x - 3 = 3x + 7 \)

Rearrange to form a quadratic equation equal to zero:
\( 2x^2 - 8x - 10 = 0 \)

Divide by 2 to simplify:
\( x^2 - 4x - 5 = 0 \)

Factorise the quadratic:
\( (x - 5)(x + 1) = 0 \)

This gives the \( x \)-values:
\( x = 5 \) and \( x = -1 \)

Substitute these back into the linear equation to find the corresponding \( y \)-values:
For \( x = -1 \): \( y = 3(-1) + 7 = 4 \implies (-1, 4) \)
For \( x = 5 \): \( y = 3(5) + 7 = 22 \implies (5, 22) \)

M1 for equating both expressions and obtaining a correct 3-term quadratic equation.
M1 for solving the quadratic equation to find both \( x \)-values (\( x = 5 \) and \( x = -1 \)) or finding one full point correctly.
A0.5 for both correct coordinate pairs: \( (-1, 4) \) and \( (5, 22) \).

First, factorise each quadratic expression:
1. $2x^2 - 7x - 15 = (2x + 3)(x - 5)$
2. $4x^2 - 25 = (2x - 5)(2x + 5)$
3. $x^2 - 9 = (x - 3)(x + 3)$
4. $2x^2 + 11x + 15 = (2x + 5)(x + 3)$

Now, rewrite the division as multiplication by the reciprocal:
$$\frac{(2x + 3)(x - 5)}{(2x - 5)(2x + 5)} \times \frac{(2x + 5)(x + 3)}{(x - 3)(x + 3)}$$

Cancel the common factors $(2x + 5)$ and $(x + 3)$ from the numerator and denominator:
$$\frac{(2x + 3)(x - 5)}{(2x - 5)(x - 3)}$$

- [M1] Factorising $2x^2 - 7x - 15$ to $(2x + 3)(x - 5)$
- [M1] Factorising $4x^2 - 25$ to $(2x - 5)(2x + 5)$
- [M1] Factorising $2x^2 + 11x + 15$ to $(2x + 5)(x + 3)$
- [M1] Factorising $x^2 - 9$ to $(x - 3)(x + 3)$
- [M1] Inverting the second fraction and multiplying
- [A1.5] Correct final simplified fraction: $\frac{(2x + 3)(x - 5)}{(2x - 5)(x - 3)}$ (or expanded form $\frac{2x^2 - 7x - 15}{2x^2 - 11x + 15}$)

The perimeter of the track is given by:
$$P = 2L + 2\pi r = 400 \implies L = 200 - \pi r$$

The area enclosed by the track is:
$$A = 2rL + \pi r^2$$

Substituting $L$ into the area equation:
$$A = 2r(200 - \pi r) + \pi r^2 = 400r - 2\pi r^2 + \pi r^2 = 400r - \pi r^2$$

To find the maximum area, differentiate $A$ with respect to $r$ and set to 0:
$$\frac{\text{d}A}{\text{d}r} = 400 - 2\pi r = 0 \implies r = \frac{200}{\pi}$$

Substitute this optimal radius back into the area equation:
$$A_{\text{max}} = 400\left(\frac{200}{\pi}\right) - \pi \left(\frac{200}{\pi}\right)^2 = \frac{80000}{\pi} - \frac{40000}{\pi} = \frac{40000}{\pi} \approx 12732.4\text{ m}^2$$

- [M1] Writing down the perimeter equation: $2L + 2\pi r = 400$
- [M1] Writing down the area equation: $A = 2rL + \pi r^2$
- [M1.5] Substituting $L$ to obtain a quadratic in terms of $r$: $A = 400r - \pi r^2$
- [M1.5] Finding the optimal radius: $r = \frac{200}{\pi}$ or $\approx 63.7$
- [A1.5] Finding the maximum area: $12732.4$ (accept $12730$ to $12740$ if using rounded intermediate values)

(a) Since the total frequency is 80:
$$8 + 15 + k + 22 + 11 = 80 \implies k + 56 = 80 \implies k = 24$$

(b) Using midpoints $x$ of each class interval:
- $0 < s \le 20$: midpoint $x = 10$, product $f x = 8 \times 10 = 80$
- $20 < s \le 40$: midpoint $x = 30$, product $f x = 15 \times 30 = 450$
- $40 < s \le 60$: midpoint $x = 50$, product $f x = 24 \times 50 = 1200$
- $60 < s \le 80$: midpoint $x = 70$, product $f x = 22 \times 70 = 1540$
- $80 < s \le 100$: midpoint $x = 90$, product $f x = 11 \times 90 = 990$

$$\sum f x = 80 + 450 + 1200 + 1540 + 990 = 4260$$
$$\text{Estimated Mean} = \frac{\sum f x}{\sum f} = \frac{4260}{80} = 53.25$$

(c) The number of students scoring more than 60 marks is $22 + 11 = 33$.
$$\text{Probability} = \frac{33}{80} = 0.4125$$

- [M1.5] (a) $k = 80 - (8+15+22+11) = 24$
- [M1] (b) Midpoints identified: 10, 30, 50, 70, 90
- [M1.5] (b) $\sum f x = 4260$, divided by 80
- [A1] (b) Mean = $53.25$
- [A1.5] (c) Number of students with score $>60$ is $22+11=33$, so probability is $\frac{33}{80}$ or $0.4125$

To find the stationary points, find the derivative $\frac{\text{d}y}{\text{d}x}$ and set it to 0:
$$\frac{\text{d}y}{\text{d}x} = 3x^2 - 6x - 9 = 0$$
Divide by 3:
$$x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$$
So, $x = 3$ or $x = -1$.

Find the corresponding $y$-values:
For $x = -1$:
$$y = (-1)^3 - 3(-1)^2 - 9(-1) + 15 = -1 - 3 + 9 + 15 = 20$$
Since this is the higher $y$-value, the local maximum is $P(-1, 20)$.

For $x = 3$:
$$y = (3)^3 - 3(3)^2 - 9(3) + 15 = 27 - 27 - 27 + 15 = -12$$
Since this is the lower $y$-value, the local minimum is $Q(3, -12)$.

- [M2] Differentiating $y$ to get $\frac{\text{d}y}{\text{d}x} = 3x^2 - 6x - 9$
- [M1.5] Setting derivative to 0 and solving to find $x = 3$ and $x = -1$
- [A1.5] Calculating the $y$-coordinates: $y = 20$ for $x = -1$, and $y = -12$ for $x = 3$
- [A1.5] Correctly identifying $P(-1, 20)$ as the maximum and $Q(3, -12)$ as the minimum

(a) Write expressions for the time taken on each leg of the journey:
- Outward time $t_1 = \frac{40}{v}$
- Return time $t_2 = \frac{36}{v+2}$

Since the return journey takes 1 hour less than the outward journey:
$$\frac{40}{v} - \frac{36}{v+2} = 1$$

Multiply by $v(v+2)$ to clear the fractions:
$$40(v+2) - 36v = v(v+2)$$
$$40v + 80 - 36v = v^2 + 2v$$
$$4v + 80 = v^2 + 2v$$
$$v^2 - 2v - 80 = 0$$
This is shown.

(b) Factorise the quadratic equation:
$$(v - 10)(v + 8) = 0$$
This gives $v = 10$ or $v = -8$.
Since speed must be positive, we discard $v = -8$. Thus, the outward speed is $10\text{ km/h}$.

- [M1] (a) Writing expressions for time: $\frac{40}{v}$ and $\frac{36}{v+2}$
- [M1] (a) Setting up the equation: $\frac{40}{v} - \frac{36}{v+2} = 1$
- [M1.5] (a) Multiplying by $v(v+2)$ and correctly expanding to get $v^2 - 2v - 80 = 0$
- [M1.5] (b) Factorising $v^2 - 2v - 80 = 0$ to $(v-10)(v+8) = 0$ (or using quadratic formula)
- [A1.5] (b) Finding $v = 10$ (rejecting negative solution)

(a) Draw the diagram of the three ports. The bearing of $B$ from $A$ is $065^\circ$. The bearing of $C$ from $B$ is $145^\circ$.
Using parallel North lines at $A$ and $B$, the angle $\angle ABC$ is:
$$\angle ABC = 180^\circ - 145^\circ + 65^\circ = 100^\circ$$

Now, use the cosine rule in triangle $ABC$ to find $AC$:
$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$
$$AC^2 = 12^2 + 18^2 - 2(12)(18)\cos(100^\circ)$$
$$AC^2 = 144 + 324 - 432(-0.173648) = 468 + 75.016 = 543.016$$
$$AC = \sqrt{543.016} \approx 23.3\text{ km}$$

(b) Use the sine rule to find the angle $\angle BAC$:
$$\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC} \implies \frac{\sin(\angle BAC)}{18} = \frac{\sin(100^\circ)}{23.3027}$$
$$\sin(\angle BAC) = \frac{18 \sin(100^\circ)}{23.3027} \approx 0.7607$$
$$\angle BAC = \sin^{-1}(0.7607) \approx 49.5^\circ$$

The bearing of $C$ from $A$ is the bearing of $B$ from $A$ plus the angle $\angle BAC$:
$$\text{Bearing} = 065^\circ + 49.5^\circ = 114.5^\circ \approx 115^\circ$$

- [M1] Finding angle $\angle ABC = 100^\circ$
- [M1] (a) Using cosine rule: $AC^2 = 12^2 + 18^2 - 2 \times 12 \times 18 \cos(100^\circ)$
- [A1.5] (a) $AC = 23.3$ (or $23.30$ to $23.31$)
- [M1] (b) Using sine rule to find $\angle BAC$: $\frac{\sin(\angle BAC)}{18} = \frac{\sin(100^\circ)}{23.3}$
- [A1] (b) $\angle BAC \approx 49.5^\circ$
- [A1] (b) Bearing of $C$ from $A = 65^\circ + 49.5^\circ = 114.5^\circ$ (accept $115^\circ$)

(a) Set the volume of the sphere equal to the volume of the cylinder:
$$\frac{4}{3}\pi R^3 = \pi r^2 h$$
Since $h = 3r$:
$$\frac{4}{3}\pi R^3 = \pi r^2 (3r) = 3\pi r^3$$
$$4R^3 = 9r^3 \implies R^3 = \frac{9}{4}r^3 = 2.25r^3$$
$$R = \sqrt[3]{2.25}r$$

(b) The total surface area of a solid cylinder is:
$$A_{\text{cylinder}} = 2\pi r^2 + 2\pi rh$$
Since $h = 3r$:
$$A_{\text{cylinder}} = 2\pi r^2 + 2\pi r(3r) = 2\pi r^2 + 6\pi r^2 = 8\pi r^2$$
This is shown.

(c) The total surface area of the sphere is:
$$A_{\text{sphere}} = 4\pi R^2 = 4\pi \left(2.25^{1/3}r\right)^2 = 4\pi (2.25)^{2/3} r^2$$

The ratio of the total surface area of the sphere to the cylinder is:
$$\text{Ratio} = \frac{4\pi (2.25)^{2/3} r^2}{8\pi r^2} = \frac{4(2.25)^{2/3}}{8} = 0.5 \times 1.717 = 0.8585 \approx 0.859$$

- [M1] (a) Equating volumes: $\frac{4}{3} \pi R^3 = \pi r^2 h$ with $h = 3r$
- [A1] (a) Correctly simplifying to $R = \sqrt[3]{2.25} r$
- [M1] (b) Formula for cylinder surface area: $2\times\pi r^2 + 2\times\pi r h$
- [A1] (b) Substituting $h=3r$ to get $8\pi r^2$
- [M1.5] (c) Expressing sphere surface area in terms of $r$: $4\pi \left(2.25^{1/3}r\right)^2$
- [A1] (c) Calculating the ratio: $\approx 0.859$ (accept $0.858$ to $0.860$)

(a) Let $R$ be a red counter and $B$ be a blue counter. The total number of counters is 12.
$$P(\text{same colour}) = P(RR) + P(BB)$$
$$P(RR) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$$
$$P(BB) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$$
$$P(\text{same colour}) = \frac{42}{132} + \frac{20}{132} = \frac{62}{132} = \frac{31}{66} \approx 0.470$$

(b) The probability of at least one blue is the complement of getting no blue (i.e., both are red):
$$P(\text{at least one blue}) = 1 - P(RR) = 1 - \frac{42}{132} = \frac{90}{132} = \frac{15}{22} \approx 0.682$$

- [M1] (a) Identifying the two cases: $RR$ and $BB$
- [M1.5] (a) Calculating product of probabilities for $RR$ and $BB$: $\frac{42}{132}$ and $\frac{20}{132}$
- [A1] (a) Summing and simplifying: $\frac{31}{66}$ or $0.470$
- [M2] (b) Using the complement method $1 - P(RR)$ or summing $RB + BR + BB$
- [A1] (b) Correct final probability: $\frac{15}{22}$ or $0.682$

(a)
(i) Time = \(\frac{\text{Distance}}{\text{Speed}}\) = \(\frac{45}{x}\) hours.
(ii) The speed for the return journey is \(x - 3\) km/h.
Time taken = \(\frac{45}{x-3}\) hours.

(b) 30 minutes is equivalent to \(0.5\) hours (or \(\frac{1}{2}\) hours).
The return journey takes longer, so:
\(\frac{45}{x-3} - \frac{45}{x} = 0.5\)
Multiply the entire equation by the common denominator \(x(x-3)\):
\(45x - 45(x-3) = 0.5x(x-3)\)
Expand the brackets:
\(45x - 45x + 135 = 0.5(x^2 - 3x)\)
\(135 = 0.5x^2 - 1.5x\)
Multiply the entire equation by 2 to clear the decimal:
\(270 = x^2 - 3x\)
Rearrange to make the equation equal to zero:
\(x^2 - 3x - 270 = 0\) (as required).

(c) Solve \(x^2 - 3x - 270 = 0\) by factorising:
We need two numbers that multiply to \(-270\) and add to \(-3\). These are \(-18\) and \(+15\).
\((x-18)(x+15) = 0\)
This gives \(x = 18\) or \(x = -15\).
Since the average speed must be a positive value, we reject \(x = -15\).
Therefore, the outward average speed is \(18\) km/h.

(a)(i) B1: For \(\frac{45}{x}\)
(a)(ii) B1: For \(\frac{45}{x-3}\)
(b) M1: For setting up the algebraic equation in terms of difference in hours: e.g. \(\frac{45}{x-3} - \frac{45}{x} = 0.5\) (or equivalent with consistent units, e.g. difference in minutes = 30)
A1: For correctly clearing fractions, expanding brackets, and showing full algebraic steps to arrive at the given quadratic equation \(x^2 - 3x - 270 = 0\) without any errors.
(c) M1: For attempting to solve the quadratic equation, either by factorisation \((x-18)(x+15)\) or correct substitution into the quadratic formula.
A1: For \(x = 18\) (must reject the negative solution to obtain final mark).