2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

First, expand each bracket: \(3(2x - 5) = 6x - 15\) and \(-2(x - 4) = -2x + 8\). Next, collect and combine the like terms: \(6x - 2x - 15 + 8 = 4x - 7\).

M1 for a correct expansion of at least one bracket (e.g. \(6x - 15\) or \(-2x + 8\)) [1 mark]. A0.5 for the fully simplified correct answer \(4x - 7\) [0.5 marks].

Let the width of the rectangle be \(w\) cm. The length is twice the width, which is \(2w\) cm. The perimeter is given by \(2(\text{length} + \text{width}) = 24\) cm. Substituting the expressions gives \(2(2w + w) = 24\), which simplifies to \(6w = 24\). Solving this gives \(w = 4\) cm. Therefore, the width is \(4\) cm and the length is \(8\) cm. The area of the rectangle is \(\text{width} \times \text{length} = 4 \times 8 = 32\text{ cm}^2\).

M1 for setting up a correct equation in one variable, e.g. \(6w = 24\), or for identifying both correct dimensions of the rectangle (length = 8 and width = 4) [1 mark]. A0.5 for the correct final area of \(32\) [0.5 marks].

Multiply both sides of the equation by 2: \(3x + 1 = 16\). Subtract 1 from both sides: \(3x = 15\). Divide both sides by 3 to find the value of \(x\): \(x = 5\).

M1 for correctly multiplying both sides by 2 to obtain \(3x + 1 = 16\) [1 mark]. A0.5 for the correct final answer of \(5\) [0.5 marks].

Expand the bracket first:
\(-2(a - 3b) = -2a + 6b\)

Now collect like terms:
\(3a - 4b - 2a + 6b = (3a - 2a) + (-4b + 6b)\)
\(= a + 2b\)

M1 for correct expansion of the bracket, showing \(-2a + 6b\) (or \(2a - 6b\) inside a subtracted bracket)
A0.5 for final answer \(a + 2b\) (accept \(1a + 2b\))

The perimeter of a rectangle is given by \(2(l + w) = P\), where \(l\) is the length and \(w\) is the width.
Substitute the given values:
\(2(l + 5) = 28\)
\(l + 5 = 14\)
\(l = 9\text{ cm}\)

Now calculate the area:
\(\text{Area} = \text{length} \times \text{width} = 9 \times 5 = 45\text{ cm}^2\)

M1 for showing that the length is \(9\text{ cm}\) or for a correct equation such as \(2l + 10 = 28\)
A0.5 for \(45\)

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 a correct first step of multiplying both sides by \(4\) to get \(3x - 1 = 20\) (or equivalent)
A0.5 for \(7\)

Expand the first bracket: \(3 \times 2x - 3 \times 5 = 6x - 15\). Expand the second bracket: \(-2 \times x - 2 \times (-4) = -2x + 8\). Combine the terms: \(6x - 15 - 2x + 8 = (6x - 2x) + (-15 + 8) = 4x - 7\).

M1 for \(6x - 15\) or \(-2x + 8\) seen. A0.5 for \(4x - 7\).

Multiply both sides of the equation by 5: \(2y + 3 = 15\). Subtract 3 from both sides: \(2y = 12\). Divide by 2: \(y = 6\).

M1 for \(2y + 3 = 15\) or \(2y = 12\). A0.5 for \(6\).

Let the width of the rectangle be \(w\text{ cm}\). The perimeter is given by \(2(\text{length} + \text{width}) = 32\), so \(2(10 + w) = 32\). Dividing by 2 gives \(10 + w = 16\), which means \(w = 6\). The area of the rectangle is \(\text{length} \times \text{width} = 10 \times 6 = 60\text{ cm}^2\).

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

First, expand each of the brackets: \(4(3x - 2) = 12x - 8\) and \(-3(2x - 5) = -6x + 15\). Next, collect and combine the like terms: \(12x - 6x - 8 + 15 = 6x + 7\).

M1 for a correct expansion of at least one bracket (e.g. \(12x - 8\) or \(-6x + 15\)). A0.5 for the fully simplified correct expression \(6x + 7\).

Subtract \(2x\) from both sides of the equation: \(5x - 4 = 11\). Add \(4\) to both sides: \(5x = 15\). Divide by \(5\) to find \(x\): \(x = 3\).

M1 for collecting like terms correctly on both sides (e.g. \(7x - 2x = 11 + 4\) or \(5x = 15\)). A0.5 for the correct solution \(3\).

Use the formula for the area of a trapezium: \(A = \frac{1}{2}(a + b)h\). Substitute the given values: \(A = \frac{1}{2}(6 + 10) \times 4.5 = \frac{1}{2}(16) \times 4.5 = 8 \times 4.5 = 36\text{ cm}^2\).

M1 for a correct substitution into the area formula of a trapezium, e.g., \(\frac{1}{2}(6 + 10) \times 4.5\). A0.5 for the correct value \(36\).

First, expand each bracket separately: \(3(2x - 5) = 6x - 15\) and \(-4(x - 2) = -4x + 8\). Next, collect and simplify the like terms: \(6x - 15 - 4x + 8 = (6x - 4x) + (-15 + 8) = 2x - 7\).

M1 for a correct expansion of at least one bracket (either \(6x - 15\) or \(-4x + 8\) seen). A0.5 for the fully simplified final answer \(2x - 7\).

First, find the width of the rectangle using the formula: \(\text{Area} = \text{length} \times \text{width}\). This gives \(28 = 8 \times \text{width}\), so \(\text{width} = \frac{28}{8} = 3.5\text{ cm}\). Now, calculate the perimeter: \(\text{Perimeter} = 2 \times (\text{length} + \text{width}) = 2 \times (8 + 3.5) = 2 \times 11.5 = 23\text{ cm}\).

M1 for correctly finding the width as \(3.5\) or for writing a correct expression for the perimeter, such as \(2 \times (8 + \frac{28}{8})\). A0.5 for the correct final answer \(23\).

Multiply both sides of the equation by \(2\) to clear the fraction: \(3x + 1 = 16\). Subtract \(1\) from both sides to isolate the term with the variable: \(3x = 15\). Finally, divide both sides by \(3\) to find \(x = 5\).

M1 for isolating the numerator, resulting in \(3x + 1 = 16\) or equivalent. A0.5 for the correct final answer \(5\).

First, expand the brackets: \(3(2x - 5) = 6x - 15\) and \(-2(x - 4) = -2x + 8\). Next, combine like terms: \((6x - 2x) + (-15 + 8) = 4x - 7\).

M1 for correct expansion of at least one bracket (e.g., \(6x - 15\) or \(-2x + 8\)) [0.5 marks]. A1 for the correct final simplified expression \(4x - 7\) [1 mark].

Identify the highest and lowest temperatures in the given set. The highest temperature is \(4^\circ\text{C}\) and the lowest temperature is \(-5^\circ\text{C}\). The range is the difference between the highest and lowest values: \(4 - (-5) = 4 + 5 = 9^\circ\text{C}\).

M1 for identifying both the highest value (4) and lowest value (-5) [0.5 marks]. A1 for the correct range of 9 [1 mark].

Let \(w\) be the width of the rectangle. The perimeter is given by \(2(\text{length} + \text{width}) = 26\). Substituting the given length: \(2(8 + w) = 26 \implies 8 + w = 13 \implies w = 5\text{ cm}\). The area is \(\text{length} \times \text{width} = 8 \times 5 = 40\text{ cm}^2\).

M1 for finding the width of the rectangle to be 5, or setting up a correct equation like \(2(8 + w) = 26\) [0.5 marks]. A1 for the correct area of 40 [1 mark].

First, find the highest common factor of the coefficients 12 and 18, which is 6. Next, find the highest common factor of the variable parts \(a^2b\) and \(ab^2\), which is \(ab\). Thus, the highest common factor is \(6ab\). Dividing each term by \(6ab\) gives \(2a\) and \(-3b\) respectively. Combining these, we write the expression in its factorised form as: \(6ab(2a - 3b)\).

M1 for a correct partial factorisation, such as \(2ab(6a - 9b)\) or \(6a(2ab - 3b^2)\). A0.5 for the correct final answer: \(6ab(2a - 3b)\).

First, convert both quantities to the same unit. Since \(1\text{ kg} = 1000\text{ g}\), we have \(1.5\text{ kg} = 1.5 \times 1000 = 1500\text{ g}\). Next, find the percentage by writing 360 as a fraction of 1500 and multiplying by 100: \(\frac{360}{1500} \times 100 = \frac{360}{15} = 24\). Therefore, 360 g is \(24\%\) of 1.5 kg.

M1 for converting to the same units and setting up the fraction, e.g., \(\frac{360}{1500}\) or \(\frac{0.36}{1.5}\). A0.5 for the correct final answer: 24.

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

Combine the like terms:
\(8x - 3x = 5x\)
\(-12 + 15 = 3\)
So, the simplified expression is \(5x + 3\).

(b) Find the highest common factor (HCF) of the two terms \(10a^2b\) and \(15ab^2\):
- The HCF of the coefficients \(10\) and \(15\) is \(5\).
- The HCF of \(a^2b\) and \(ab^2\) is \(ab\).
Thus, the HCF is \(5ab\).

Divide each term by the HCF to find the terms inside the bracket:
\(10a^2b \div 5ab = 2a\)
\(-15ab^2 \div 5ab = -3b\)
So, the factorised expression is \(5ab(2a - 3b)\).

(a)
M1 for expanding at least one bracket correctly to get \(8x - 12\) or \(-3x + 15\)
A1 for the correct simplified expression \(5x + 3\)

(b)
M1 for finding a common factor of at least \(5\), \(a\), or \(b\) (e.g., \(5(2a^2b - 3ab^2)\) or \(ab(10a - 15b)\))
A1 for the completely factorised expression \(5ab(2a - 3b)\)

(a) To calculate the perimeter, we must first find the two missing inner dimensions:
- The missing inner vertical length is \(8\text{ m} - 3\text{ m} = 5\text{ m}\).
- The missing inner horizontal length is \(9\text{ m} - 4\text{ m} = 5\text{ m}\).

Now sum all six outer boundary sides:
\(\text{Perimeter} = 8 + 9 + 3 + 5 + 5 + 4 = 34\text{ m}\).

(b) To calculate the total area, split the compound shape into two simpler rectangles:
- Method 1 (vertical split):
- Left rectangle: width = \(4\text{ m}\), height = \(8\text{ m}\). \(\text{Area} = 4 \times 8 = 32\text{ m}^2\).
- Right rectangle: width = \(5\text{ m}\), height = \(3\text{ m}\). \(\text{Area} = 5 \times 3 = 15\text{ m}^2\).
- \(\text{Total Area} = 32 + 15 = 47\text{ m}^2\).

- Method 2 (horizontal split):
- Top rectangle: width = \(4\text{ m}\), height = \(5\text{ m}\). \(\text{Area} = 4 \times 5 = 20\text{ m}^2\).
- Bottom rectangle: width = \(9\text{ m}\), height = \(3\text{ m}\). \(\text{Area} = 9 \times 3 = 27\text{ m}^2\).
- \(\text{Total Area} = 20 + 27 = 47\text{ m}^2\).

(a)
M1 for finding the missing lengths of \(5\text{ m}\) and \(5\text{ m}\), or for a complete addition expression of 6 sides
A1 for \(34\)

(b)
M1 for a complete and correct method to split the shape and find the sum of two correct areas (e.g., \(4 \times 8 + 5 \times 3\) or \(4 \times 5 + 9 \times 3\))
A1 for \(47\)

We can solve these equations using either substitution or elimination.

Method: Elimination
Multiply the second equation by \(2\) to align the coefficients of \(y\):
\(2 \times (4x - y = 5) \implies 8x - 2y = 10\)

Add this new equation to the first equation to eliminate \(y\):
\((3x + 2y) + (8x - 2y) = 12 + 10\)
\(11x = 22\)
\(x = 2\)

Substitute \(x = 2\) back into the second equation to solve for \(y\):
\(4(2) - y = 5\)
\(8 - y = 5\)
\(y = 3\)

Therefore, the solution is \(x = 2\) and \(y = 3\).

M1 for a correct method to eliminate one variable (e.g. multiplying the second equation by 2 to get \(8x - 2y = 10\), or writing \(y = 4x - 5\))
M1 for adding/subtracting the equations or substituting to obtain a single equation in one variable (e.g., \(11x = 22\) or \(3x + 2(4x - 5) = 12\))
A1 for \(x = 2\)
A1 for \(y = 3\)

Multiply the second equation by 2 to get \( 8x - 2y = 26 \). Add this to the first equation: \( (3x + 2y) + (8x - 2y) = 18 + 26 \), which gives \( 11x = 44 \), so \( x = 4 \). Substitute \( x = 4 \) into the second equation: \( 4(4) - y = 13 \) which gives \( 16 - y = 13 \), so \( y = 3 \).

M1 for multiplying the second equation by 2 (or first by some factor) to align coefficients. M1 for adding the equations to eliminate one variable. A1 for finding one correct coordinate (either \( x = 4 \) or \( y = 3 \)). A1 for finding both correct coordinates.

The original rectangle has two sides of length 8 cm, one side of length 12 cm, and one side of length 12 cm from which a diameter of 6 cm is removed. The remaining straight segments on that side sum to \( 12 - 6 = 6 \) cm. The curved boundary of the semi-circle has length equal to half the circumference of a circle of diameter 6 cm, which is \( \frac{1}{2} \times \pi \times 6 = 3\pi \) cm. Therefore, the total perimeter is \( 8 + 8 + 12 + 6 + 3\pi = 34 + 3\pi \) cm.

M1 for calculating the remaining straight parts of the side with the cut-out: \( 12 - 6 = 6 \) cm. M1 for calculating the semi-circular arc length: \( \frac{1}{2} \times \pi \times 6 = 3\pi \). M1 for adding all straight sides: \( 8 + 8 + 12 + 6 = 34 \). A1 for the correct final expression \( 34 + 3\pi \).

Since there are 5 numbers and they are ordered, the median is the 3rd number in the list. Thus, \( y = 12 \). The mean of the 5 numbers is 11, which means the sum of the numbers is \( 11 \times 5 = 55 \). Therefore, \( 5 + x + y + 14 + 14 = 55 \). Substituting \( y = 12 \) gives \( 5 + x + 12 + 14 + 14 = 55 \), which simplifies to \( x + 45 = 55 \). Solving this gives \( x = 10 \).

B1 for identifying the median \( y = 12 \). M1 for using the definition of the mean to set up the equation: \( (5 + x + y + 14 + 14)/5 = 11 \). M1 for substituting \( y = 12 \) and simplifying the sum to \( x + 45 = 55 \). A1 for \( x = 10 \) and \( y = 12 \).

Multiply the first equation by 3 and the second equation by 2:
\(9x + 6y = 33\)
\(10x - 6y = 62\)
Add the two equations to eliminate \(y\):
\(19x = 95\)
\(x = 5\)
Substitute \(x = 5\) into the first equation:
\(3(5) + 2y = 11\)
\(15 + 2y = 11\)
\(2y = -4\)
\(y = -2\)

M1 for a correct method to eliminate one variable (e.g. multiplying to get equal coefficients of \(x\) or \(y\)). A1 for \(x = 5\). M1 for substituting their found value to find the second variable. A1 for \(y = -2\).

We can factorise by grouping terms. First group: \(12ax - 8ay = 4a(3x - 2y)\). Second group: \(-3bx + 2by = -b(3x - 2y)\). Combining these gives: \(4a(3x - 2y) - b(3x - 2y) = (4a - b)(3x - 2y)\).

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

Find the sum of the original 8 numbers: \(8 \times 12 = 96\). Find the sum of all 9 numbers: \(9 \times 13 = 117\). Subtract the original sum from the new sum to find the 9th number: \(117 - 96 = 21\).

M1 for showing \(9 \times 13\) or \(8 \times 12\) (or values 117 or 96). A1 for 21.

Multiply both sides by \((x+2)(x-1)\) to clear the denominators: \(5(x-1) = 3(x+2)\). Expand the brackets: \(5x - 5 = 3x + 6\). Subtract \(3x\) from both sides: \(2x - 5 = 6\). Add 5 to both sides: \(2x = 11\). Divide by 2: \(x = 5.5\) (or \(\frac{11}{2}\)).

M1 for \(5(x-1) = 3(x+2)\) or better algebraic step. A1 for 5.5 or \(\frac{11}{2}\) or \(5\frac{1}{2}\).

To factorise \( 6x^2 - 7x - 5 \), we need to find two numbers that multiply to \( 6 \times (-5) = -30 \) and add to \( -7 \). These numbers are \( -10 \) and \( 3 \). We rewrite the quadratic expression as \( 6x^2 - 10x + 3x - 5 \). Factoring by grouping, we get \( 2x(3x - 5) + 1(3x - 5) \), which factorises completely to \( (2x + 1)(3x - 5) \).

M1 for finding the correct split of the middle term (e.g. \( 6x^2 - 10x + 3x - 5 \)) or for any equivalent partial factorisation, or for a product of two linear brackets that gives two of the three terms correct (e.g. \( (2x - 1)(3x + 5) \)). A1 for \( (2x + 1)(3x - 5) \) or \( (3x - 5)(2x + 1) \).

The formula for the arc length of a circle is \( L = \frac{\theta}{360} \times 2\pi r \), where \( \theta \) is the sector angle and \( r \) is the radius. Substituting the given values, we get \( 5\pi = \frac{\theta}{360} \times 2\pi(9) \). Simplifying this gives \( 5\pi = \frac{18\pi\theta}{360} \), which reduces to \( 5 = \frac{\theta}{20} \). Solving for \( \theta \), we find \( \theta = 5 \times 20 = 100 \). Thus, the angle subtended is \( 100^\circ \).

M1 for setting up a correct equation, e.g. \( 5\pi = \frac{\theta}{360} \times 2\pi \times 9 \) or \( \theta = \frac{5\pi}{18\pi} \times 360 \) (or equivalent). A1 for \( 100 \).

To solve the simultaneous equations \( 2x - 3y = 13 \) and \( 5x + 2y = 4 \), we can multiply the first equation by 2 and the second equation by 3 to align the coefficients of \( y \). This yields \( 4x - 6y = 26 \) and \( 15x + 6y = 12 \). Adding these two equations gives \( 19x = 38 \), which simplifies to \( x = 2 \). Substituting \( x = 2 \) back into the first equation, we get \( 2(2) - 3y = 13 \), which simplifies to \( 4 - 3y = 13 \), or \( -3y = 9 \), giving \( y = -3 \). Thus, the solution is \( x = 2, y = -3 \).

M1 for a correct method to eliminate one variable (e.g. multiplying equations to align coefficients and adding/subtracting, or a correct substitution step). A1 for both \( x = 2 \) and \( y = -3 \).

We factorise by grouping the terms in pairs: \((15ab - 10a) + (12b - 8)\). Factor out the common term from each group: \(5a(3b - 2) + 4(3b - 2)\). Now factor out the common binomial: \((5a + 4)(3b - 2)\).

M1 for \(5a(3b - 2) + 4(3b - 2)\) or \(3b(5a + 4) - 2(5a + 4)\) A1 for \((5a + 4)(3b - 2)\) or equivalent (e.g. \((3b - 2)(5a + 4)\))

Cross-multiply to clear the denominators: \(2(2x + 1) = 5(x - 3)\). Expand both sides of the equation: \(4x + 2 = 5x - 15\). Subtract \(4x\) from both sides: \(2 = x - 15\). Add \(15\) to both sides: \(x = 17\).

M1 for \(2(2x + 1) = 5(x - 3)\) or better A1 for 17

Use the formula for the area of a sector: \(\text{Area} = \frac{\theta}{360} \times \pi r^2\). Substituting the given values gives \(5\pi = \frac{\theta}{360} \times \pi \times 6^2\). Divide both sides by \(\pi\) to get \(5 = \frac{36\theta}{360}\). Simplifying this gives \(5 = \frac{\theta}{10}\), which results in \(\theta = 50\).

M1 for \(5\pi = \frac{\theta}{360} \times \pi \times 6^2\) or better A1 for 50

First, factorise the numerator as a difference of two squares:
\(x^2 - 9 = (x - 3)(x + 3)\)

Next, factorise the denominator by taking out the common factor of \(2x\):
\(2x^2 + 6x = 2x(x + 3)\)

Now, simplify the fraction by cancelling the common factor \((x + 3)\):
\(\frac{(x - 3)(x + 3)}{2x(x + 3)} = \frac{x - 3}{2x}\)

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

Express both sides of the equation with a common base of 3:
\(27 = 3^3\) and \(9 = 3^2\)

Substitute these bases into the equation:
\((3^3)^{x-1} = (3^2)^{x+2}\)

Apply the index law \((a^m)^n = a^{mn}\):
\(3^{3(x-1)} = 3^{2(x+2)}
3^{3x-3} = 3^{2x+4}\)

Equate the exponents:
\(3x - 3 = 2x + 4\)

Solve for \(x\):
\(3x - 2x = 4 + 3\)
\(x = 7\)

M1 for expressing both sides as powers of 3, e.g. \(3^{3(x-1)} = 3^{2(x+2)}\) or \(3x-3 = 2x+4\)
A1 for 7

Let the five positive integers in ascending order be \(a, b, c, d, e\).

1. Since the mean is 6, the sum of the five integers is:
\(5 \times 6 = 30\)

2. Since the median is 5, the middle number is 5:
\(c = 5\)

3. To maximize the largest number \(e\), we must minimize the other four numbers \(a, b, c, d\).
- The smallest possible positive integer is 1, so we set \(a = 1\).
- Since 5 is the single mode, 5 must appear more than once and more than any other number. To minimize the other values, we can let 5 appear three times, so \(b = 5\) and \(d = 5\).

This gives the set: \(1, 5, 5, 5, e\).

Summing these values:
\(1 + 5 + 5 + 5 + e = 30\)
\(16 + e = 30\)
\(e = 14\)

(If 5 only appeared twice, e.g., \(1, 5, 5, 6, e\), the sum of the first four would be 17, giving \(e = 13\), which is smaller.)

M1 for recognizing the sum of the five numbers is 30, or identifying the minimal set structure as \(1, 5, 5, 5, e\)
A1 for 14

We factorise by grouping the terms in pairs:

\(12ab - 4a + 3b - 1 = 4a(3b - 1) + 1(3b - 1)\)

Now we factor out the common bracket \((3b - 1)\):

\((4a + 1)(3b - 1)\)

M1 for finding a common factor from a pair of terms, e.g. \(4a(3b - 1)\) or \(3b(4a + 1) - 1(4a + 1)\)
A1 for \((4a + 1)(3b - 1)\) or \((3b - 1)(4a + 1)\)

Express both bases as powers of 3:

\(27 = 3^3\) and \(9 = 3^2\)

Substitute these into the equation:

\((3^3)^{p-1} = (3^2)^{p+2}\)

\(3^{3(p-1)} = 3^{2(p+2)}\)

Since the bases are now equal, we can set the indices equal to each other:

\(3(p - 1) = 2(p + 2)\)

\(3p - 3 = 2p + 4\)

\(p = 7\)

M1 for expressing both sides as a power of 3: \(3^{3(p-1)} = 3^{2(p+2)}\) or better
A1 for \(7\)

The perimeter of a sector is equal to the sum of the two radii and the arc length:

\(\text{Perimeter} = 2r + \text{Arc length}\)

Given that the radius \(r = 6\text{ cm}\):

\(\text{Perimeter} = 12 + \text{Arc length}\)

Comparing this to the given perimeter of \(12 + 3\pi\text{ cm}\), the arc length must be:

\(\text{Arc length} = 3\pi\text{ cm}\)

Using the formula for the arc length of a sector with angle \(\theta\):

\(\frac{\theta}{360} \times 2\pi r = 3\pi\)

\(\frac{\theta}{360} \times 12\pi = 3\pi\)

\(\frac{\theta}{30} = 3\)

\(\theta = 90^{\circ}\)

M1 for equating the arc length to \(3\pi\), e.g. \(\frac{\theta}{360} \times 2 \times \pi \times 6 = 3\pi\) or better
A1 for \(90\) (accept \(90^{\circ}\))

To solve the equation:
\(\frac{2}{x} + \frac{3}{x+2} = 1\)

Multiply every term by the common denominator \(x(x+2)\) to clear the fractions:
\(2(x+2) + 3x = x(x+2)\)

Expand both sides of the equation:
\(2x + 4 + 3x = x^2 + 2x\)
\(5x + 4 = x^2 + 2x\)

Rearrange the equation into standard quadratic form \(ax^2 + bx + c = 0\):
\(x^2 + 2x - 5x - 4 = 0\)
\(x^2 - 3x - 4 = 0\)

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

Solve for \(x\):
\(x = 4\) or \(x = -1\)

M1 for writing with a common denominator or multiplying through by \(x(x+2)\)
A1 for obtaining a correct equation without fractions: \(2(x+2) + 3x = x(x+2)\) or \(5x + 4 = x^2 + 2x\)
M1 for rearranging into standard quadratic form: \(x^2 - 3x - 4 = 0\)
M1 for factorising their quadratic into \((x + a)(x + b) = 0\) where \(ab = -4\) or \(a+b = -3\) (or using the quadratic formula correctly with their coefficients)
A1 for both solutions \(x = 4\) and \(x = -1\) correct

(a) The area of a sector with angle \(\theta\) and radius \(r\) is given by:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)

Substitute \(r = 6\) and \(\text{Area} = 15\pi\):
\(15\pi = \frac{\theta}{360} \times \pi \times 6^2\)
\(15\pi = \frac{\theta}{360} \times 36\pi\)

Divide both sides by \(\pi\):
\(15 = \frac{36\theta}{360}\)
\(15 = \frac{\theta}{10}\)
\(\theta = 150\)

So the angle of the sector is \(150^\circ\).

(b) The perimeter of the sector consists of the arc length plus two radii:
\(\text{Perimeter} = \text{Arc length} + 2r\)

The arc length is:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{150}{360} \times 2 \times \pi \times 6\)
\(\text{Arc length} = \frac{5}{12} \times 12\pi = 5\pi\text{ cm}\)

Therefore, the perimeter is:
\(\text{Perimeter} = 5\pi + 2(6) = 12 + 5\pi\text{ cm}\)

(a)
M1 for setting up the area equation: \(\frac{\theta}{360} \times \pi \times 6^2 = 15\pi\)
A1 for fully correct working leading to \(\theta = 150\)

(b)
M1 for a correct method to find the arc length, e.g., \(\frac{150}{360} \times 2 \times \pi \times 6\) or \(\frac{5}{12} \times 12\pi\)
A1 for arc length \(= 5\pi\) (may be implied by final answer of \(12 + 5\pi\))
A1 for \(12 + 5\pi\) (or \(5\pi + 12\))

(a) To factorise completely:
\(6a^2b - 8ab^2\)

Identify the highest common factor of the terms:
The numerical parts \(6\) and \(8\) have HCF \(2\).
The variable parts \(a^2b\) and \(ab^2\) have HCF \(ab\).
Therefore, the HCF is \(2ab\).

Divide each term by the HCF:
\(6a^2b \div 2ab = 3a\)
\(-8ab^2 \div 2ab = -4b\)

So, the factorised expression is:
\(2ab(3a - 4b)\)

(b) To simplify the fraction:
\(\frac{3x^2 - 10x + 3}{x^2 - 9}\)

First, factorise the numerator \(3x^2 - 10x + 3\):
Look for two numbers that multiply to \(3 \times 3 = 9\) and add to \(-10\). These are \(-9\) and \(-1\).
\(3x^2 - 9x - x + 3 = 3x(x - 3) - 1(x - 3) = (3x - 1)(x - 3)\)

Next, factorise the denominator \(x^2 - 9\) as a difference of two squares:
\(x^2 - 9 = (x - 3)(x + 3)\)

Rewrite the fraction and cancel the common factor \((x - 3)\):
\(\frac{(3x - 1)(x - 3)}{(x - 3)(x + 3)} = \frac{3x - 1}{x + 3}\)

(a)
B1 for \(2(3a^2b - 4ab^2)\) or \(ab(6a - 8b)\) or \(2ab(3a - k \cdot b)\) or \(2ab(k \cdot a - 4b)\) where \(k \neq 0\)
B1 for \(2ab(3a - 4b)\)

(b)
M1 for factorising the numerator: \((3x - 1)(x - 3)\)
M1 for factorising the denominator: \((x - 3)(x + 3)\)
A1 for \(\frac{3x - 1}{x + 3}\) (or \(\frac{3x-1}{x+3}\))

First, factorise each quadratic and linear expression completely:

1. Difference of two squares: \(x^2 - 9 = (x - 3)(x + 3)\)
2. Quadratic factorisation: \(2x^2 + 5x - 3 = 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)\)
3. Common factor: \(2x^2 - 6x = 2x(x - 3)\)
4. Difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now rewrite the division as multiplication by the reciprocal of the second fraction:
\(\frac{(x - 3)(x + 3)}{(2x - 1)(x + 3)} \times \frac{(2x - 1)(2x + 1)}{2x(x - 3)}\)

Cancel the common terms in the numerator and denominator:
- Cancel \((x + 3)\) from the first fraction
- Cancel \((x - 3)\) from the numerator and denominator
- Cancel \((2x - 1)\) from the numerator and denominator

This leaves the simplified expression:
\(\frac{2x + 1}{2x}\)

M1 for factorising \(x^2 - 9\) to \((x - 3)(x + 3)\) or \(4x^2 - 1\) to \((2x - 1)(2x + 1)\)
M1 for factorising \(2x^2 + 5x - 3\) to \((2x - 1)(x + 3)\)
M1 for factorising \(2x^2 - 6x\) to \(2x(x - 3)\)
M1 for multiplying by the reciprocal and showing at least two correct cancellations
A1 for \(\frac{2x + 1}{2x}\) (or \(1 + \frac{1}{2x}\))

1. Set up the equation for the arc length:
\(L = \frac{\theta}{360} \times 2\pi R = 8\pi\)
Dividing both sides by \(2\pi\) gives:
\(\frac{\theta}{360} R = 4\) [Equation 1]

2. Set up the equation for the change in area:
The original area is:
\(A_1 = \frac{\theta}{360} \pi R^2\)
The new area with radius \(R - 4\) is:
\(A_2 = \frac{\theta}{360} \pi (R - 4)^2\)
The difference in area is \(24\pi\):
\(\frac{\theta}{360} \pi R^2 - \frac{\theta}{360} \pi (R - 4)^2 = 24\pi\)
Dividing both sides by \(\pi\):
\(\frac{\theta}{360} [ R^2 - (R - 4)^2 ] = 24\)
\(\frac{\theta}{360} [ R^2 - (R^2 - 8R + 16) ] = 24\)
\(\frac{\theta}{360} [ 8R - 16 ] = 24\) [Equation 2]

3. Expand Equation 2 and substitute Equation 1:
\(8 \left( \frac{\theta}{360} R \right) - 16 \left( \frac{\theta}{360} \right) = 24\)
Substitute \(\frac{\theta}{360} R = 4\):
\(8(4) - 16 \left( \frac{\theta}{360} \right) = 24\)
\(32 - 16 \left( \frac{\theta}{360} \right) = 24\)
\(16 \left( \frac{\theta}{360} \right) = 8\)
\(\frac{\theta}{360} = \frac{1}{2}\)
\(\theta = 180\)

4. Find the value of \(R\):
Substitute \(\frac{\theta}{360} = \frac{1}{2}\) into Equation 1:
\(\frac{1}{2} R = 4 \implies R = 8\)

M1 for setting up the arc length equation: \(\frac{\theta}{360} \times 2\pi R = 8\pi\) (or equivalent)
M1 for setting up the area difference equation: \(\frac{\theta}{360} \pi R^2 - \frac{\theta}{360} \pi (R-4)^2 = 24\pi\)
M1 for simplifying the area equation to \(\frac{\theta}{360}(8R - 16) = 24\)
M1 for a valid method to solve the simultaneous system (e.g. substituting \(\frac{\theta R}{360} = 4\))
A1 for both \(R = 8\) and \(\theta = 180\)

1. Set up the first equation using the sum of frequencies:
\(8 + 12 + x + y = 40\)
\(x + y = 20\) [Equation 1]

2. Determine the midpoints (\(m\)) for each interval:
- For \(0 < h \le 10\), midpoint \(m_1 = 5\)
- For \(10 < h \le 20\), midpoint \(m_2 = 15\)
- For \(20 < h \le 30\), midpoint \(m_3 = 25\)
- For \(30 < h \le 50\), midpoint \(m_4 = 40\)

3. Set up the equation for the estimated mean:
\(\text{Estimated Mean} = \frac{\sum f \cdot m}{\sum f}\)
\(21 = \frac{(8 \times 5) + (12 \times 15) + (x \times 25) + (y \times 40)}{40}\)
\(21 = \frac{40 + 180 + 25x + 40y}{40}\)
\(21 \times 40 = 220 + 25x + 40y\)
\(840 = 220 + 25x + 40y\)
\(25x + 40y = 620\)
Dividing both sides by 5:
\(5x + 8y = 124\) [Equation 2]

4. Solve the simultaneous equations:
From Equation 1, multiply by 5:
\(5x + 5y = 100\)
Subtract this from Equation 2:
\((5x + 8y) - (5x + 5y) = 124 - 100\)
\(3y = 24 \implies y = 8\)

Substitute \(y = 8\) back into Equation 1:
\(x + 8 = 20 \implies x = 12\)

M1 for the total frequency equation: \(x + y = 20\) (or equivalent)
M1 for determining at least three correct midpoints: \(5, 15, 25, 40\)
M1 for setting up the mean equation: \(\frac{8(5) + 12(15) + 25x + 40y}{40} = 21\)
M1 for a valid algebraic method to solve their linear simultaneous equations
A1 for both \(x = 12\) and \(y = 8\)

First, change the division into multiplication by the reciprocal: \(\frac{2x^2 - 5x - 3}{4x^2 - 1} \times \frac{2x^2 - 5x - 3}{x^2 - 6x + 9}\). Next, factorise each quadratic term: 1) \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\), 2) \(4x^2 - 1 = (2x - 1)(2x + 1)\) (difference of two squares), 3) \(x^2 - 6x + 9 = (x - 3)^2\). Substitute these factors back into the expression: \(\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} \times \frac{(2x+1)(x-3)}{(x-3)^2}\). Simplify by cancelling common terms: the \(2x+1\) term cancels between the numerator and denominator of the first fraction. The two \(x-3\) terms in the numerators cancel with the \((x-3)^2\) in the denominator of the second fraction. This leaves the final simplified fraction: \(\frac{2x+1}{2x-1}\).

M1 for factorising \(2x^2 - 5x - 3\) as \((2x+1)(x-3)\) (seen at least once). M1 for factorising \(4x^2 - 1\) as \((2x-1)(2x+1)\). M1 for factorising \(x^2 - 6x + 9\) as \((x-3)^2\). M1 for multiplying by the reciprocal and attempting to cancel common terms. A1 for the final simplified fraction \(\frac{2x+1}{2x-1}\).

1) Express the boundary lengths of the shaded region in terms of \(r\): The inner arc length is \(\frac{60}{360} \times 2\pi r = \frac{1}{3}\pi r\). The outer arc length with radius \(R = 3r\) is \(\frac{60}{360} \times 2\pi (3r) = \pi r\). The two straight radial boundaries each have length \(R - r = 3r - r = 2r\), giving a total straight length of \(4r\). 2) Write an expression for the total perimeter: \(\text{Perimeter} = \frac{1}{3}\pi r + \pi r + 4r = r\left(4 + \frac{4}{3}\pi\right)\). 3) Equate this to the given perimeter of \(12 + 4\pi\): \(r\left(4 + \frac{4}{3}\pi\right) = 12 + 4\pi\). Factorising the right-hand side gives \(3\left(4 + \frac{4}{3}\pi\right)\), so \(r = 3\). 4) Find the area of the shaded region: \(\text{Area} = \text{Area of large sector} - \text{Area of small sector} = \frac{60}{360}\pi R^2 - \frac{60}{360}\pi r^2 = \frac{1}{6}\pi (9r^2 - r^2) = \frac{4}{3}\pi r^2\). 5) Substitute \(r = 3\) into the area expression: \(\text{Area} = \frac{4}{3}\pi (3^2) = 12\pi\) cm\(^2\).

M1 for expressing the arc lengths in terms of \(r\) (e.g., \(\frac{1}{3}\pi r\) or \(\pi r\) or their sum \(\frac{4}{3}\pi r\)). M1 for identifying the straight edges as \(4r\) or \(2(R-r)\). M1 for equating the total perimeter to \(12 + 4\pi\) and solving to find \(r = 3\). M1 for writing a correct expression for the shaded area, e.g., \(\frac{1}{6}\pi R^2 - \frac{1}{6}\pi r^2\) or \(\frac{4}{3}\pi r^2\). A1 for the final exact area \(12\pi\).

1) Since the five numbers are in increasing order (\(a \le b \le c \le d \le e\)), the median is the middle value, \(c\). Thus, \(c = 11\). 2) The unique mode is \(6\). Since \(6 < 11\) and it must appear more than once to be the mode, we must have \(a = 6\) and \(b = 6\). 3) The range is the difference between the largest and smallest values: \(e - a = 12 \implies e - 6 = 12 \implies e = 18\). 4) The mean is \(11\), so the sum of all five integers is \(11 \times 5 = 55\). Therefore, \(a + b + c + d + e = 55\). Substituting the known values: \(6 + 6 + 11 + d + 18 = 55 \implies 41 + d = 55 \implies d = 14\). The values are \(a = 6, b = 6, c = 11, d = 14, e = 18\).

B1 for identifying \(c = 11\). B1 for identifying \(a = 6\) and \(b = 6\). M1 for using the range to write \(e - 6 = 12\) to find \(e = 18\). M1 for setting up the equation for the mean: \(6 + 6 + 11 + d + 18 = 55\). A1 for all five correct values: \(a = 6, b = 6, c = 11, d = 14, e = 18\).

First, expand the brackets:
$$4x - 12 = 18 - 2x$$

Next, add \(2x\) to both sides:
$$6x - 12 = 18$$

Then, add \(12\) to both sides:
$$6x = 30$$

Finally, divide by \(6\):
$$x = 5$$

M1 for a correct expansion of the bracket to \(4x - 12\) or for a correct algebraic step to isolate the terms in \(x\) and the constant terms.
A1 for 5.

The perimeter \(P\) of a rectangle is given by the formula:
$$P = 2(\text{length} + \text{width})$$

Substitute the known values into the formula:
$$38 = 2(12 + w)$$

Divide both sides by 2:
$$19 = 12 + w$$

Subtract 12 from both sides:
$$w = 7\text{ cm}$$

M1 for a correct method to find the width, e.g., \((38 - 2 \times 12) \div 2\) or \(2(12 + w) = 38\).
A1 for 7.

First, calculate the total number of goals scored in all 10 matches by summing the products of the goals and their respective frequencies:
$$\text{Total goals} = (0 \times 2) + (1 \times 4) + (2 \times 3) + (3 \times 1)$$
$$\text{Total goals} = 0 + 4 + 6 + 3 = 13$$

Next, divide the total number of goals by the total number of matches (which is the sum of the frequencies, 10):
$$\text{Mean} = \frac{13}{10} = 1.3$$

M1 for a correct method to find the sum of products: \(0 \times 2 + 1 \times 4 + 2 \times 3 + 3 \times 1\) (implied by 13) or division of their sum of products by 10.
A1 for 1.3.

The perimeter of the rectangle is \(2 \times (14 + 8) = 44\text{ cm}\). Since the square has the same perimeter, the side length of the square is \(44 \div 4 = 11\text{ cm}\). The area of the square is \(11^2 = 121\text{ cm}^2\).

M1 for \(2 \times (14 + 8) \div 4\) or \(44 \div 4\) or finding the side length of the square is 11. A1 for 121.

Expand the brackets: \(4(2x - 3) = 8x - 12\) and \(-3(x - 5) = -3x + 15\). Combine like terms: \(8x - 12 - 3x + 15 = (8x - 3x) + (-12 + 15) = 5x + 3\).

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

The sum of all five numbers is \(5 \times 12 = 60\). The sum of the four given numbers is \(8 + 15 + 14 + 9 = 46\). The fifth number is \(60 - 46 = 14\).

M1 for \(5 \times 12\) or 60, or for \(8 + 15 + 14 + 9 + x = 60\) (or equivalent). A1 for 14.

First, expand each bracket carefully:
\(5(2x - 3) = 10x - 15\)
\(-3(x - 4) = -3x + 12\)

Now, write the full expression and group like terms:
\(10x - 15 - 3x + 12\)
\(= (10x - 3x) + (-15 + 12)\)
\(= 7x - 3\)

M1 for correct expansion of at least one bracket (i.e. \(10x - 15\) or \(-3x + 12\))
A1 for \(7x - 3\) as the final simplified answer

First, find the width of the rectangle using the formula:
\(\text{Area} = \text{length} \times \text{width}\)
\(54 = 12 \times \text{width}\)
\(\text{width} = \frac{54}{12} = 4.5\text{ cm}\)

Next, calculate the perimeter of the rectangle:
\(\text{Perimeter} = 2 \times (\text{length} + \text{width})\)
\(\text{Perimeter} = 2 \times (12 + 4.5) = 2 \times 16.5 = 33\text{ cm}\)

M1 for \(\frac{54}{12}\) or for finding the width is \(4.5\text{ cm}\)
A1 for \(33\)

Let the fifth number be \(x\).
The mean of five numbers is calculated by dividing their sum by \(5\):
\(\frac{8 + 15 + 11 + 14 + x}{5} = 12\)

Multiply both sides by \(5\) to find the sum of all five numbers:
\(8 + 15 + 11 + 14 + x = 60\)

Add the four known numbers together:
\(48 + x = 60\)

Subtract \(48\) from \(60\) to find \(x\):
\(x = 12\)

M1 for showing \(5 \times 12\) or \(60\), or a complete correct method showing the sum of the known values subtracted from 60
A1 for \(12\)

First, find the highest common factor (HCF) of the numerical coefficients 12 and 18, which is 6. Next, find the highest common factor of the algebraic parts \( a^2b \) and \( ab^2 \), which is \( ab \). This gives a common factor of \( 6ab \). Divide both terms by \( 6ab \) to find the remaining terms: \( 12a^2b \div 6ab = 2a \) and \( -18ab^2 \div 6ab = -3b \). Therefore, the fully factorised expression is \( 6ab(2a - 3b) \).

M1 for a partial factorisation, such as \( 2ab(6a - 9b) \), \( 3ab(4a - 6b) \), \( 6a(2ab - 3b^2) \), or \( 6b(2a^2 - 3ab) \).
A1 for the fully correct factorised expression: \( 6ab(2a - 3b) \).

First, find the radius of the semicircle by dividing the diameter by 2: \( r = 14 \div 2 = 7 \) cm. The area of a semicircle is half the area of a full circle: \( \text{Area} = \frac{1}{2} \pi r^2 \). Substitute \( r = 7 \) into the formula: \( \text{Area} = \frac{1}{2} \times \pi \times 7^2 = 24.5\pi \approx 76.969 \) cm\(^2\). Rounding to 1 decimal place gives 77.0 cm\(^2\).

M1 for a correct method to find the area of the semicircle, e.g. \( \frac{1}{2} \times \pi \times 7^2 \) or \( \frac{\pi \times 14^2}{8} \).
A1 for 77.0 or 77 (accept answers in the range 76.9 to 77.0).

The mean of the 5 heights is found by summing the heights and dividing by 5: \( \frac{12 + 15 + 18 + 14 + h}{5} = 16 \). Multiply both sides by 5 to find the total sum of the heights: \( 12 + 15 + 18 + 14 + h = 80 \). Simplify the sum of the known values: \( 59 + h = 80 \). Subtract 59 from 80 to find the value of \( h \): \( h = 80 - 59 = 21 \).

M1 for a correct equation showing the sum of heights equal to \( 16 \times 5 \), e.g., \( 59 + h = 80 \).
A1 for 21.

First, expand both brackets carefully, paying attention to the signs: \(5(2x - 3) = 10x - 15\) and \(-2(3x - 4) = -6x + 8\). Next, combine the terms: \(10x - 6x - 15 + 8 = 4x - 7\).

M1 for correct expansion of at least one bracket (giving \(10x - 15\) or \(-6x + 8\)). A1 for the fully simplified correct answer \(4x - 7\).

Use the formula for the area of a trapezium: \(\text{Area} = \frac{1}{2}(a + b)h\). Substituting the given values: \(\text{Area} = \frac{1}{2}(6.4 + 9.6) \times 4.5 = \frac{1}{2}(16) \times 4.5 = 8 \times 4.5 = 36\text{ cm}^2\).

M1 for a correct substitution into the trapezium area formula, e.g., \(\frac{1}{2}(6.4 + 9.6) \times 4.5\). A1 for the correct answer \(36\).

Let the fifth number be \(x\). The sum of all five numbers is \(5 \times 12 = 60\). The sum of the four given numbers is \(7 + 15 + 13 + 8 = 43\). Therefore, \(x = 60 - 43 = 17\).

M1 for showing that the sum of the five numbers is \(5 \times 12\) or \(60\), or for the equation \(7 + 15 + 13 + 8 + x = 60\). A1 for \(17\).

First, expand the brackets: \( 5(3x - 2) = 15x - 10 \) and \( -2(4x - 7) = -8x + 14 \). Next, group the like terms together: \( 15x - 8x - 10 + 14 \). Simplifying this gives \( 7x + 4 \).

M1 for correct expansion of at least one bracket (e.g. \( 15x - 10 \) or \( -8x + 14 \))
A1 for \( 7x + 4 \)

Calculate the total number of books read by multiplying each value by its frequency and summing them up: \( (0 \times 3) + (1 \times 8) + (2 \times 6) + (3 \times 5) + (4 \times 3) = 0 + 8 + 12 + 15 + 12 = 47 \). Divide this total by the number of students: \( \text{Mean} = \frac{47}{25} = 1.88 \).

M1 for a correct method to find the sum of products (e.g. \( 0 \times 3 + 1 \times 8 + 2 \times 6 + 3 \times 5 + 4 \times 3 \))
A1 for 1.88

Use the formula for the area of a trapezium: \( \text{Area} = \frac{1}{2}(a + b)h \). Substituting the given values: \( \text{Area} = \frac{1}{2}(7 + 13) \times 6.4 = \frac{1}{2}(20) \times 6.4 = 10 \times 6.4 = 64\text{ cm}^2 \).

M1 for \( 0.5 \times (7 + 13) \times 6.4 \) or equivalent
A1 for 64

Let the cost of a banana be \(x\) cents. The cost of an apple is then \(x + 15\) cents. The total cost of 5 bananas and 3 apples is given by the equation: \(5x + 3(x + 15) = 405\). Expanding the brackets gives: \(5x + 3x + 45 = 405\). Combining the like terms gives: \(8x + 45 = 405\). Subtracting 45 from both sides gives: \(8x = 360\). Dividing by 8 gives: \(x = 45\). Therefore, the cost of a banana is 45 cents.

M1 for writing the correct expression for 3 apples: \(3(x + 15)\). M1 for setting up the correct equation: \(5x + 3(x + 15) = 405\). M1 for correctly simplifying the equation to \(8x = 360\) (or equivalent). A1.4 for the correct final answer of 45.

The outer perimeter of the garden path is made of three straight sides of the rectangle and the curved edge of the semi-circle. The three straight sides are: \(12 + 12 + 6 = 30\) m. The curved edge of the semi-circle with diameter \(d = 6\) m has a length of: \(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 6 = 3\pi \approx 9.425\) m. Adding these together: \(\text{Total Perimeter} = 30 + 9.425 = 39.425\) m. Rounded to 1 decimal place, this is 39.4 m.

M1 for finding the sum of the three straight sides: \(12 + 12 + 6 = 30\) m. M1 for calculating the arc length of the semi-circle: \(\frac{1}{2} \times \pi \times 6\) or \(3\pi\) (approx 9.42). M1 for adding the straight sides sum to the arc length: \(30 + 3\pi\). A1.4 for the correct final answer of 39.4 (accept 39.42 to 39.43).

First, calculate the total number of goals scored in all 20 matches by multiplying each number of goals by its frequency: \((0 \times 4) + (1 \times 7) + (2 \times 5) + (3 \times 3) + (4 \times 1) = 0 + 7 + 10 + 9 + 4 = 30\) goals. Next, divide the total number of goals by the total number of matches (sum of frequencies = 20): \(\text{Mean} = \frac{30}{20} = 1.5\).

M1 for attempting to find the sum of products (at least 3 correct products shown or added). M1 for finding the correct total goals: 30. M1 for dividing their total goals by 20. A1.4 for the correct final answer of 1.5.

The perimeter of the compound shape consists of three outer sides of the rectangle and the curved boundary of the semicircle.

1. The three outer sides of the rectangle are:
\(\text{Length} + \text{Length} + \text{Width} = 12\text{ cm} + 12\text{ cm} + 8\text{ cm} = 32\text{ cm}\).

2. The curved boundary of the semicircle with diameter \(d = 8\text{ cm}\) is:
\(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 8 = 4\pi \approx 12.566\text{ cm}\).

3. The total perimeter is:
\(32 + 12.566 = 44.566\text{ cm}\).

To 3 significant figures, this is \(44.6\text{ cm}\).

M1 for adding three sides of the rectangle: \(12 + 12 + 8 = 32\)
M1 for calculation of the semi-circular arc: \(\frac{1}{2} \times \pi \times 8\) (or \(4\pi\))
M1 for adding their two components: \(32 + 4\pi\)
A1 for \(44.6\) (accept answers in the range \(44.56\) to \(44.6\))

To eliminate \(y\), multiply the first equation by 5 and the second equation by 2:
\(15x + 10y = 95\)
\(4x + 10y = 40\)

Subtract the second equation from the first:
\((15x - 4x) + (10y - 10y) = 95 - 40\)
\(11x = 55\)
\(x = 5\)

Substitute \(x = 5\) back into the first equation:
\(3(5) + 2y = 19\)
\(15 + 2y = 19\)
\(2y = 4\)
\(y = 2\)

Check using the second equation:
\(2(5) + 5(2) = 10 + 10 = 20\) (Correct).

M1 for a correct method to eliminate one variable (e.g., multiplying equations to equate coefficients and subtracting)
A1 for finding \(x = 5\) (or \(y = 2\))
M1 for substituting their found value back into one of the original equations to find the second variable
A1 for \(y = 2\) (or \(x = 5\))

Let the five numbers in ascending order be \(a, b, c, d, e\).

1. Since the median is 11, the middle (third) number is 11:
\(c = 11\).

2. Since the unique mode is 9, and 9 is less than 11, the first two numbers must be 9:
\(a = 9\) and \(b = 9\).

3. Since the range is 10, the difference between the largest and smallest number is 10:
\(e - a = 10 \implies e - 9 = 10 \implies e = 19\).

The five numbers are now \(9, 9, 11, d, 19\).

4. Since the mean of the five numbers is 12, their sum is:
\(12 \times 5 = 60\).

Therefore:
\(9 + 9 + 11 + d + 19 = 60\)
\(48 + d = 60\)
\(d = 12\).

So, the five numbers in ascending order are 9, 9, 11, 12, 19.

B1 for identifying that the median is the middle number: \(c = 11\)
B1 for identifying that the mode being 9 means the first two numbers are \(a = 9\) and \(b = 9\)
M1 for using the range of 10 to find the largest number: \(e = 19\)
M1 for setting up the equation for the sum of the numbers equal to 60 to find \(d = 12\)
A1 for the correct final list: 9, 9, 11, 12, 19

First, calculate the area of the rectangle: \(15 \times 8 = 120\text{ m}^2\). Next, find the radius of the semi-circle. Since the diameter of the semi-circle is 8 m, the radius is \(r = 4\text{ m}\). Calculate the area of the semi-circle: \(\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 4^2 = 8\pi \approx 25.133\text{ m}^2\). The total area is the sum of the two areas: \(120 + 25.133 = 145.133\text{ m}^2\). Correct to 1 decimal place, this is 145.1.

M1 for \(15 \times 8\) (or 120)
M1 for \(\frac{1}{2} \pi \times 4^2\) (or 25.1 to 25.2)
M1 for adding their two calculated areas
A1 for 145.1

We can solve these equations using the elimination method. Multiply the second equation by 3:
\(3(2x - y) = 3(12)\) which simplifies to \(6x - 3y = 36\).
Now add this to the first equation:
\((5x + 3y) + (6x - 3y) = 41 + 36\)
\(11x = 77\)
\(x = 7\).
Substitute \(x = 7\) back into the second equation:
\(2(7) - y = 12\)
\(14 - y = 12\)
\(y = 2\).
The solution is \(x = 7, y = 2\).

M1 for a correct method to eliminate one variable (e.g., multiplying second equation by 3)
A1 for finding \(x = 7\) (or \(y = 2\))
M1 for substituting their found value back to find the second variable
A1 for both \(x = 7\) and \(y = 2\)

First, find the midpoints of the intervals:
For \(80 < m \le 100\), the midpoint is 90.
For \(100 < m \le 120\), the midpoint is 110.
For \(120 < m \le 140\), the midpoint is 130.
For \(140 < m \le 160\), the midpoint is 150.
Next, calculate the sum of the products of each midpoint and its corresponding frequency:
\(8 \times 90 = 720\)
\(17 \times 110 = 1870\)
\(15 \times 130 = 1950\)
\(10 \times 150 = 1500\)
Sum of products = \(720 + 1870 + 1950 + 1500 = 6040\).
Finally, divide this sum by the total frequency (50):
Estimated mean = \(6040 / 50 = 120.8\) grams.

M1 for finding correct midpoints of the intervals: 90, 110, 130, 150 (at least three correct)
M1 for calculating \(\sum f \times x\) (with their midpoints): \(8(90) + 17(110) + 15(130) + 10(150)\) (or 6040)
M1 for dividing their sum by 50
A1 for 120.8

To find the area of the rectangle, we first need to determine the value of \(x\). 1. Set up the perimeter equation: The perimeter of a rectangle is given by \(P = 2(\text{length} + \text{width})\). Substituting the expressions: \(2((3x + 1) + (2x - 3)) = 56\). Simplifying the terms: \(2(5x - 2) = 56\) which gives \(10x - 4 = 56\). 2. Solve for \(x\): \(10x = 60\) so \(x = 6\). 3. Calculate the dimensions: \(\text{Length} = 3(6) + 1 = 19\) cm and \(\text{Width} = 2(6) - 3 = 9\) cm. 4. Calculate the area: \(\text{Area} = 19 \times 9 = 171\) cm\(^2\).

M1 for setting up a correct equation for the perimeter, e.g., \(2(3x + 1) + 2(2x - 3) = 56\) or \(5x - 2 = 28\). A1 for solving to get \(x = 6\). B1 for calculating length = 19 and width = 9 (or showing product \(19 \times 9\)). A1 for final answer 171.

First, factorise the quadratic expression in the numerator:
\(2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)\)

Next, factorise the denominator using the difference of two squares:
\(x^2 - 25 = (x - 5)(x + 5)\)

Now, substitute these back into the fraction and cancel the common factor of \((x - 5)\):
\(\frac{(2x + 3)(x - 5)}{(x + 5)(x - 5)} = \frac{2x + 3}{x + 5}\)

M1 for factorising the numerator: \((2x + 3)(x - 5)\)
M1 for factorising the denominator: \((x + 5)(x - 5)\)
A0.5 for final simplified fraction: \(\frac{2x + 3}{x + 5}\)

1. Find the radius, \(r\), using the formula for the area of a sector:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)
\(54\pi = \frac{135}{360} \times \pi r^2\)
Divide both sides by \(\pi\):
\(54 = \frac{3}{8} r^2\)
\(r^2 = 54 \times \frac{8}{3} = 144\)
\(r = 12\text{ cm}\) (since \(r > 0\))

2. Find the arc length of the sector:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{135}{360} \times 2\pi \times 12 = 9\pi\text{ cm}\)

3. Calculate the perimeter of the sector (arc length plus two radii):
\(\text{Perimeter} = 9\pi + 2(12) = 9\pi + 24 \approx 52.274\text{ cm}\)

Rounding to 3 significant figures gives \(52.3\text{ cm}\).

M1 for setting up equation to find \(r\): \(\frac{135}{360} \times \pi r^2 = 54\pi\) leading to \(r = 12\)
M1 for calculating the arc length: \(\frac{135}{360} \times 2\pi \times 12 = 9\pi\) (or approx 28.3)
A0.5 for adding \(2r\) to the arc length to obtain \(52.3\)

1. Set up the first equation using the total frequency:
\(8 + x + 15 + y = 40\)
\(x + y = 17\) --- (Equation 1)

2. Find the mid-interval values for each class:
- For \(0 < t \le 10\): Midpoint = 5
- For \(10 < t \le 20\): Midpoint = 15
- For \(20 < t \le 30\): Midpoint = 25
- For \(30 < t \le 40\): Midpoint = 35

3. Set up the equation for the estimated mean:
\(\text{Mean} = \frac{\sum f m}{\sum f} = \frac{5(8) + 15x + 25(15) + 35y}{40} = 21.25\)
\(40 + 15x + 375 + 35y = 21.25 \times 40\)
\(415 + 15x + 35y = 850\)
\(15x + 35y = 435\)
Divide by 5 to simplify:
\(3x + 7y = 87\) --- (Equation 2)

4. Solve the simultaneous equations:
From Equation 1, \(3x + 3y = 51\).
Subtracting this from Equation 2:
\((3x + 7y) - (3x + 3y) = 87 - 51\)
\(4y = 36\)
\(y = 9\)

Substitute \(y = 9\) back into Equation 1:
\(x + 9 = 17\)
\(x = 8\)

Thus, \(x = 8\) and \(y = 9\).

M1 for creating the total frequency equation: \(x + y = 17\)
M1 for creating the mean equation: \(15x + 35y = 435\) (or simplified \(3x + 7y = 87\))
A0.5 for solving both equations correctly to obtain \(x = 8\) and \(y = 9\)

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator, which is a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Write the expression as: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancel the common factor \((2x + 1)\) from the numerator and denominator to get \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x+1)(x-3)\) or equivalent. M1 for factorising the denominator to \((2x-1)(2x+1)\). A0.5 for the fully simplified fraction \(\frac{x-3}{2x-1}\).

Let \(\theta\) be the angle of the sector. The area of the sector is given by: \(\text{Area} = \frac{\theta}{360} \times \pi r^2 = 35.6\). Since \(r = 8.4\), we have: \(\frac{\theta}{360} \times \pi \times 8.4^2 = 35.6 \Rightarrow \frac{\theta}{360} \approx \frac{35.6}{221.67} \approx 0.1606\). The arc length of the sector is: \(l = \frac{\theta}{360} \times 2\pi r \approx 0.1606 \times 2 \times \pi \times 8.4 \approx 8.48\text{ cm}\). (Alternatively, using \(\text{Area} = \frac{1}{2} r l\), we get \(35.6 = \frac{1}{2} \times 8.4 \times l \Rightarrow l = \frac{71.2}{8.4} \approx 8.476\text{ cm}\)). The perimeter of the sector includes the arc length and two radii: \(\text{Perimeter} = l + 2r \approx 8.476 + 2(8.4) = 8.476 + 16.8 = 25.276\text{ cm}\). Correct to 3 significant figures, the perimeter is \(25.3\text{ cm}\).

M1 for a correct method to find the arc length, e.g., \(\frac{2 \times 35.6}{8.4}\) or finding the sector angle \(\theta \approx 57.8^\circ\). M1 for adding \(2 \times 8.4\) to their arc length. A0.5 for the final answer \(25.3\) (accept answers in the range \(25.27\) to \(25.3\)).

First, find the upper bound for each dimension. The length is \(85\text{ m}\) correct to the nearest metre, so its upper bound is \(85 + 0.5 = 85.5\text{ m}\). The width is \(40\text{ m}\) correct to the nearest \(5\text{ m}\), so its upper bound is \(40 + 2.5 = 42.5\text{ m}\). The upper bound for the area is the product of the upper bounds of the length and width: \(\text{Upper Bound of Area} = 85.5 \times 42.5 = 3633.75\text{ m}^2\).

M1 for identifying either upper bound correctly: length upper bound = \(85.5\) or width upper bound = \(42.5\). M1 for multiplying their upper bound of length by their upper bound of width. A0.5 for the correct final answer \(3633.75\) (or \(3630\) to 3 s.f.).

First, set up the quadratic equation for the area of the garden: \(w(w + 4) = 35\). Expanding and rearranging gives: \(w^2 + 4w - 35 = 0\). Using the quadratic formula, \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get: \(w = \frac{-4 \pm \sqrt{4^2 - 4(1)(-35)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 140}}{2} = \frac{-4 \pm \sqrt{156}}{2}\). Calculating the positive root: \(w = \frac{-4 + 12.48999...}{2} \approx 4.24499\text{ m}\). To 3 significant figures, \(w = 4.24\).

M1 for setting up the quadratic equation \(w(w + 4) = 35\) or \(w^2 + 4w - 35 = 0\). M1 for a correct substitution into the quadratic formula (or completing the square to find \((w + 2)^2 = 39\)). A0.5 for the final answer 4.24.

The area of a sector can be expressed as \(A = \frac{1}{2} r L\), where \(r\) is the radius and \(L\) is the arc length. Substituting the given values: \(18.5 = \frac{1}{2} \times 6.4 \times L\), which simplifies to \(18.5 = 3.2 L\). Solving for \(L\), we get \(L = \frac{18.5}{3.2} = 5.78125\text{ cm}\). Alternatively, using the sector area formula: \(\frac{\theta}{360} \times \pi \times 6.4^2 = 18.5\), which gives \(\theta \approx 51.4^\circ\), and the arc length \(L = \frac{51.4}{360} \times 2 \pi \times 6.4 \approx 5.78\text{ cm}\). The perimeter of the sector is the arc length plus two radii: \(P = L + 2r = 5.78125 + 2 \times 6.4 = 18.58125\text{ cm}\). Rounding to 3 significant figures gives \(18.6\text{ cm}\).

M1 for finding the arc length of the sector, \(L \approx 5.78\text{ cm}\). M1 for adding \(2 \times 6.4\) to their arc length. A0.5 for the final answer 18.6.

Let the five positive integers in ascending order be \(a, b, c, d, e\). Since the median is 9, we must have \(c = 9\). The mean is 8, so the sum of the five integers is \(5 \times 8 = 40\). Since the unique mode is 11, the number 11 must appear at least twice. Since \(c = 9\), the only positions for 11 are \(d = 11\) and \(e = 11\). Thus, the sum of the remaining two integers is \(a + b = 40 - 9 - 11 - 11 = 9\). Since the mode is uniquely 11, the value of \(b\) must be strictly less than 9, so \(b \le 8\). To minimize the range \(e - a = 11 - a\), we need to maximize \(a\). Since \(a \le b\) and \(a + b = 9\), the maximum integer value for \(a\) is 4 (with \(b = 5\)). The set of integers is then \(\{4, 5, 9, 11, 11\}\). The range is \(11 - 4 = 7\).

M1 for determining that the three largest numbers are 9, 11, and 11, or that the sum of the remaining two numbers is 9. M1 for identifying that the maximum value of the smallest number \(a\) is 4 (or listing the set \(\{4, 5, 9, 11, 11\}\)). A0.5 for the final answer 7.

Let the radius of the sector be \(r\) and the arc length be \(l\). The perimeter is given by \(2r + l = 32\) and the area is given by \(\frac{1}{2}rl = 60\). From the perimeter equation, we can write \(l = 32 - 2r\). Substituting this into the area equation gives \(\frac{1}{2}r(32 - 2r) = 60\). Simplifying this equation: \(16r - r^2 = 60\), which rearranges to the quadratic equation \(r^2 - 16r + 60 = 0\). Factorizing the quadratic gives \((r - 10)(r - 6) = 0\). Thus, the two possible values for the radius are \(r = 10\) and \(r = 6\). The larger of these two values is 10.

M1 for setting up simultaneous equations: \(2r + l = 32\) and \(\frac{1}{2}rl = 60\) (or using \(\theta\)). M1 for forming the quadratic equation \(r^2 - 16r + 60 = 0\). A0.5 for identifying the larger radius as 10.

Multiply all terms by the common denominator \((x-2)(x+1)\) to clear the fractions: \(4(x+1) - 3(x-2) = (x-2)(x+1)\). Expanding both sides gives \(4x + 4 - 3x + 6 = x^2 - x - 2\). Simplifying this yields \(x + 10 = x^2 - x - 2\). Rearranging into standard quadratic form gives \(x^2 - 2x - 12 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we have \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2} = \frac{2 \pm \sqrt{52}}{2} = 1 \pm \sqrt{13}\). The positive solution is \(1 + \sqrt{13} \approx 4.60555\). Correct to 3 significant figures, this is 4.61.

M1 for clearing fractions to obtain \(4(x+1) - 3(x-2) = (x-2)(x+1)\). M1 for establishing the quadratic equation \(x^2 - 2x - 12 = 0\). A0.5 for the correct positive solution 4.61 (accept 4.61 or 4.605 to 4.61).

First, square both sides to eliminate the square root: \(P^2 = \frac{w+3}{2w-5}\). Next, multiply both sides by \(2w-5\) to clear the fraction: \(P^2(2w-5) = w+3\). Expand the bracket: \(2wP^2 - 5P^2 = w+3\). Collect all terms involving \(w\) on one side and the other terms on the opposite side: \(2wP^2 - w = 5P^2 + 3\). Factorize out \(w\) on the left side: \(w(2P^2 - 1) = 5P^2 + 3\). Finally, divide both sides by \(2P^2 - 1\) to make \(w\) the subject: \(w = \frac{5P^2 + 3}{2P^2 - 1}\).

M1 for squaring both sides: \(P^2 = \frac{w+3}{2w-5}\). M1 for expanding and grouping terms in \(w\) to obtain \(w(2P^2 - 1) = 5P^2 + 3\) or equivalent. A0.5 for the final correct expression \(w = \frac{5P^2 + 3}{2P^2 - 1}\) or equivalent.

(a) The time taken for the first part of the journey is \(\frac{36}{x}\) hours. The time taken for the second part is \(\frac{45}{x-3}\) hours. The total time is 4.5 hours, so:
\(\frac{36}{x} + \frac{45}{x-3} = 4.5\)
Multiply both sides by \(x(x-3)\):
\(36(x-3) + 45x = 4.5x(x-3)\)
\(36x - 108 + 45x = 4.5x^2 - 13.5x\)
\(81x - 108 = 4.5x^2 - 13.5x\)
Divide the entire equation by 4.5:
\(18x - 24 = x^2 - 3x\)
Rearranging gives:
\(x^2 - 21x + 24 = 0\).

(b) Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-21) \pm \sqrt{(-21)^2 - 4(1)(24)}}{2(1)}\)
\(x = \frac{21 \pm \sqrt{441 - 96}}{2}\)
\(x = \frac{21 \pm \sqrt{345}}{2}\)
\(x = \frac{21 \pm 18.574}{2}\)
\(x = 19.787\) or \(x = 1.213\)
To 2 decimal places:
\(x = 19.79\) or \(x = 1.21\).

(c) Since the speed of the second part of the journey is \(x - 3\) km/h, \(x\) must be greater than 3. Thus, \(x = 19.79\) km/h.
Time taken for the first part = \(\frac{36}{19.787} \approx 1.8194\) hours.
\(0.8194 \times 60 \approx 49.16\) minutes.
So, the time taken is 1 hour and 49 minutes (to the nearest minute).

(a) M1: For setting up the initial equation \(\frac{36}{x} + \frac{45}{x-3} = 4.5\)
M1: For multiplying by \(x(x-3)\) correctly to obtain \(36(x-3) + 45x = 4.5x(x-3)\) or equivalent
A1: For simplifying to the required equation \(x^2 - 21x + 24 = 0\) with no errors seen

(b) M1: For correct substitution into the quadratic formula, e.g. \(\frac{21 \pm \sqrt{(-21)^2 - 4(1)(24)}}{2}\) (allow one sign error)
A1: For \(x = 19.79\) (or \(19.78...\))
A0.5: For \(x = 1.21\) (or \(1.21...\))

(c) B1: For choosing \(x = 19.79\) and calculating \(\frac{36}{\text{their } x}\) leading to 1 hour 49 minutes.

(a) Let \(C\) be the center of the sector and \(O\) be the center of the inner circle. The line of symmetry of the sector bisects the sector angle, so the angle is \(110^\circ / 2 = 55^\circ\). The distance from \(C\) to the outer arc is 15 cm. Since the inner circle is tangent to the outer arc, the distance from \(C\) to the center of the inner circle is \(CO = 15 - r\). A right-angled triangle is formed by \(C\), \(O\), and the point of contact with one of the straight edges, where the side opposite to the \(55^\circ\) angle is the radius of the inner circle, \(r\). Using trigonometry:
\(\sin 55^\circ = \frac{r}{15 - r}\)
\(r = (15 - r)\sin 55^\circ\)
\(r = 15\sin 55^\circ - r\sin 55^\circ\)
\(r(1 + \sin 55^\circ) = 15\sin 55^\circ\)
\(r = \frac{15\sin 55^\circ}{1 + \sin 55^\circ}\)
Since \(\sin 55^\circ \approx 0.819152\):
\(r = \frac{15 \times 0.819152}{1 + 0.819152} \approx 6.7544\) cm
So, \(r \approx 6.75\) cm, correct to 3 significant figures.

(b) Area of the sector = \(\frac{110}{360} \times \pi \times 15^2 = \frac{11}{36} \times \pi \times 225 = 215.984\) \(\text{cm}^2\).
Area of the inner circle = \(\pi \times r^2\).
If using the exact value \(r \approx 6.7544\): \(\text{Area} = \pi \times 6.7544^2 \approx 143.328\) \(\text{cm}^2\).
If using the rounded value \(r = 6.75\): \(\text{Area} = \pi \times 6.75^2 \approx 143.139\) \(\text{cm}^2\).
Area not covered = \(215.984 - 143.328 \approx 72.7\) \(\text{cm}^2\) (or \(215.984 - 143.139 \approx 72.8\) \(\text{cm}^2\)).

(a) M1: For identifying the half-angle as \(55^\circ\)
M1: For expressing the distance to the center of the inner circle as \(15 - r\)
M1: For setting up the trigonometric equation \(\sin 55^\circ = \frac{r}{15 - r}\) or equivalent
A1: For solving for \(r\) and obtaining \(6.75...\) and rounding to \(6.75\) with no errors seen

(b) M1: For calculating the area of the sector, e.g. \(\frac{110}{360} \times \pi \times 15^2 \approx 216\)
M1: For calculating the area of the circle, e.g. \(\pi \times 6.75^2 \approx 143\) or \(\pi \times 6.7544^2 \approx 143.3\)
A0.5: For \(72.7\) or \(72.8\) (accept answers in range \(72.6\) to \(72.9\))

(a) Since there are 8 scores and their mean is 67, the sum of all 8 scores is:
\(8 \times 67 = 536\).
The sum of the 6 known scores is:
\(48 + 55 + 61 + 72 + 72 + 80 = 388\).
Therefore, the sum of the two unknown scores \(a\) and \(b\) is:
\(a + b = 536 - 388 = 148\).

(b) Let the 8 scores be arranged in ascending order. The median of 8 scores is the mean of the 4th and 5th scores when sorted. Since the median is 68, the sum of the 4th and 5th scores must be:
\(2 \times 68 = 136\).
The known scores in ascending order are 48, 55, 61, 72, 72, 80. Since 72 occurs twice, and it is given as the unique mode, neither \(a\) nor \(b\) can be equal to any of the other scores to create a second mode (unless \(a\) or \(b\) is 72, which would just make 72 appear three times). If the 5th score is 72, the 4th score must be \(136 - 72 = 64\). Let's assume \(a = 64\). Since \(a + b = 148\), this gives \(b = 148 - 64 = 84\). Let's test the complete sorted list of scores: 48, 55, 61, 64, 72, 72, 80, 84. The 4th score is 64 and the 5th score is 72. The median is \(\frac{64 + 72}{2} = 68\) (correct). The only repeated value is 72, which appears twice, making it the unique mode (correct). The mean is 67 (correct). Therefore, \(a = 64\) and \(b = 84\).

(a) M1: For calculating the total sum of the 8 scores: \(8 \times 67 = 536\)
A1: For subtracting the sum of the known scores (\(388\)) to show \(a + b = 148\)

(b) M1: For stating that the sum of the 4th and 5th scores must be \(2 \times 68 = 136\)
M1: For identifying that the 5th score is 72 (from the known list) or setting up the equation \(a + 72 = 136\)
M1: For finding \(a = 64\)
M1: For using \(a + b = 148\) to find \(b = 84\)
A0.5: For both \(a = 64\) and \(b = 84\) correctly identified.

(a) The perimeter of the sector consists of two radii of length \(12\text{ cm}\) and the arc length \(AB\).

\text{Arc length } AB = \frac{120}{360} \times 2 \pi \times 12 = 8\pi\text{ cm}.

\text{Perimeter} = \text{Arc length } AB + OA + OB = 8\pi + 12 + 12 = 8\pi + 24\text{ cm}.

(b) Let \(C\) be the center of the circular hole of radius \(r\). Because of symmetry, \(C\) lies on the angle bisector of \u2220AOB, which bisects \(120^\circ\) into two angles of \(60^\circ\). Let \(P\) be the point of tangency on \(OA\). In the right-angled triangle \(OPC\):

\sin 60^\circ = \frac{PC}{OC} = \frac{r}{OC} \implies OC = \frac{r}{\sin 60^\circ} = \frac{2r}{\sqrt{3}}

The distance from \(O\) to the boundary of the sector along the line of symmetry is equal to the radius of the sector, which is \(12\text{ cm}\). Therefore, the distance from \(O\) to the furthest point of the circular hole along this line is \(OC + r = 12\).

\frac{2r}{\sqrt{3}} + r = 12 \implies r\left(\frac{2}{\sqrt{3}} + 1\right) = 12 \implies r = \frac{12}{\frac{2}{\sqrt{3}} + 1} = 24\sqrt{3} - 36 \approx 5.569\text{ cm}.

To 3 significant figures, \(r = 5.57\).

(a)
M1: For \(\frac{120}{360} \times 2 \pi \times 12\) or \(8\pi\) seen.
A1: For fully showing the perimeter is \(8\pi + 24\) (must show addition of the two radii of 12 cm).

(b)
M1: For identifying the angle bisector of \(60^\circ\) and setting up \(\sin 60^\circ = \frac{r}{OC}\) (or equivalent).
M1: For writing \(OC\) in terms of \(r\), i.e., \(OC = \frac{2r}{\sqrt{3}}\).
M1: For setting up the equation \(OC + r = 12\).
A1: For finding an expression for \(r\), such as \(r = \frac{12}{1 + 2/\sqrt{3}}\) or \(24\sqrt{3} - 36\).
A0.5: For correct evaluation of \(r\) to \(5.57\) (accept range [5.56, 5.57]).

(a) \text{Time} = \frac{\text{Distance}}{\text{Speed}}

Time for the first part: \(\frac{40}{x}\) hours.

Time for the second part: \(\frac{36}{x - 8}\) hours.

Since total time is \(5\) hours, we have:

\frac{40}{x} + \frac{36}{x - 8} = 5

Multiply every term by \(x(x - 8)\):

40(x - 8) + 36x = 5x(x - 8)

40x - 320 + 36x = 5x^2 - 40x

76x - 320 = 5x^2 - 40x

Rearranging to standard quadratic form gives:

5x^2 - 116x + 320 = 0. (This is shown)

(b) Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

x = \frac{-(-116) \pm \sqrt{(-116)^2 - 4(5)(320)}}{2(5)}

x = \frac{116 \pm \sqrt{13456 - 6400}}{10} = \frac{116 \pm \sqrt{7056}}{10} = \frac{116 \pm 84}{10}

This yields two potential solutions:

x = \frac{200}{10} = 20 \quad \text{or} \quad x = \frac{32}{10} = 3.2

Since the speed for the second part of the journey is \(x - 8\), we must have \(x > 8\). Thus, \(x = 20\).

The average speed of the first part is \(20\text{ km/h}\).

(a)
M1: For writing correct expressions for the time of both parts: \(\frac{40}{x}\) and \(\frac{36}{x-8}\).
M1: For setting up the equation \(\frac{40}{x} + \frac{36}{x-8} = 5\).
M1: For multiplying by \(x(x-8)\) to clear fractions correctly: \(40(x-8) + 36x = 5x(x-8)\).
A0.5: For simplifying to the given quadratic equation \(5x^2 - 116x + 320 = 0\) with no algebraic errors.

(b)
M1: For correct substitution into the quadratic formula, e.g. \(\frac{116 \pm \sqrt{(-116)^2 - 4(5)(320)}}{2(5)}\) or correct factorization \((5x-16)(x-20) = 0\).
M1: For obtaining the two solutions \(x = 20\) and \(x = 3.2\).
A1: For selecting the correct speed \(x = 20\) (with reason \(x > 8\) stated or implied by rejecting \(3.2\)).

(a) Using the cosine rule in triangle \(PQR\):

PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR)

PR^2 = 140^2 + 180^2 - 2 \times 140 \times 180 \times \cos 72^\circ

PR^2 = 19600 + 32400 - 50400 \times 0.309017

PR^2 = 52000 - 15574.46 = 36425.54

PR = \sqrt{36425.54} \approx 190.85\text{ m}.

To 3 significant figures, \(PR = 191\text{ m}\).

(b) \text{Area} = \frac{1}{2} \cdot PQ \cdot QR \cdot \sin(\angle PQR)

\text{Area} = \frac{1}{2} \times 140 \times 180 \times \sin 72^\circ = 12600 \times 0.9510565 \approx 11983.31\text{ m}^2.

To 3 significant figures, \text{Area} = 12000\text{ m}^2$.

(c) The shortest path from \(Q\) to \(PR\) is the perpendicular height \(h\) from \(Q\) to the base \(PR\).

\text{Area} = \frac{1}{2} \times PR \times h

11983.31 = \frac{1}{2} \times 190.85 \times h \implies h = \frac{2 \times 11983.31}{190.85} \approx 125.58\text{ m}.

To 3 significant figures, the length of the path is \(126\text{ m}\).

(a)
M1: For correct substitution into cosine rule: \(140^2 + 180^2 - 2 \times 140 \times 180 \times \cos 72^\circ\).
M1: For correct calculation of \(PR^2 \approx 36425\) or better.
A1: For \(PR \approx 191\) (accept 190.8 to 191).

(b)
M1: For correct formula for area: \(\frac{1}{2} \times 140 \times 180 \times \sin 72^\circ\).
A1: For \(\approx 12000\) or \(11980\) (accept 11980 to 12000).

(c)
M1: For setting up equation for the perpendicular height: \(\frac{1}{2} \times \text{their (a)} \times h = \text{their (b)}\).
A0.5: For \(126\) (accept range [125.5, 126.1]).

(a) The path has a uniform width of 1 m on all sides, so we add 2 m to both the length and the width of the garden to find the combined dimensions:

Combined length = \((x + 5) + 1 + 1 = x + 7\) m

Combined width = \((2x - 3) + 1 + 1 = 2x - 1\) m

The combined area is \(143 \text{ m}^2\):

\((x + 7)(2x - 1) = 143\)

\(2x^2 - x + 14x - 7 = 143\)

\(2x^2 + 13x - 7 - 143 = 0\)

\(2x^2 + 13x - 150 = 0\)

(b) To solve \(2x^2 + 13x - 150 = 0\), we can factorise the quadratic equation:

\((2x + 25)(x - 6) = 0\)

This gives:

\(2x + 25 = 0 \implies x = -12.5\)

\(x - 6 = 0 \implies x = 6\)

Since the dimensions must be positive, we reject the negative value. Therefore, \(x = 6\).

Using \(x = 6\) to find the dimensions of the garden:

Length = \(6 + 5 = 11\) m

Width = \(2(6) - 3 = 9\) m

Perimeter of the garden = \(2 \times (\text{Length} + \text{Width}) = 2(11 + 9) = 40\) m.

Part (a):
M1: For identifying combined length as \(x + 7\) or combined width as \(2x - 1\).
M1: For setting up the area equation \((x + 7)(2x - 1) = 143\).
A1: For fully expanding and simplifying to obtain the correct quadratic equation \(2x^2 + 13x - 150 = 0\) with no errors shown.

Part (b):
M1: For factorising \((2x + 25)(x - 6) = 0\) or correctly applying the quadratic formula.
A1: For finding \(x = 6\) (and rejecting \(x = -12.5\)).
A1: For calculating the correct perimeter of 40 m.

(a) In the triangle \(OAB\), \(OA = OB = 12 \text{ cm}\) and the angle \(AOB = 70^\circ\).

Using the Cosine Rule to find the chord length \(AB\):

\(AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(70^\circ)\)

\(AB^2 = 12^2 + 12^2 - 2(12)(12)\cos(70^\circ)\)

\(AB^2 = 144 + 144 - 288\cos(70^\circ)\)

\(AB^2 = 288 - 288(0.34202)\)

\(AB^2 \approx 189.498\)

\(AB \approx 13.766 \text{ cm}\)

To 3 significant figures, the length of the chord is \(13.8 \text{ cm}\).

(b) First, find the length of the arc \(AB\):

\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)

\(\text{Arc length} = \frac{70}{360} \times 2 \times \pi \times 12\)

\(\text{Arc length} = \frac{14\pi}{3} \approx 14.661 \text{ cm}\)

Now, the perimeter of the segment is the sum of the chord length and the arc length:

\(\text{Perimeter} = \text{Chord } AB + \text{Arc } AB\)

\(\text{Perimeter} \approx 13.766 + 14.661 = 28.427 \text{ cm}\)

To 3 significant figures, the perimeter of the segment is \(28.4 \text{ cm}\).

Part (a):
M1: For using the Cosine Rule correctly: \(12^2 + 12^2 - 2(12)(12)\cos(70)\) or equivalent trigonometric method (e.g. \(2 \times 12 \sin(35)\)).
A1: For \(AB^2 \approx 189.5\) or \(\sin(35) \approx 0.5736\).
A1: For \(13.8\) or \(13.76...\) (accept answers rounding to 13.8).

Part (b):
M1: For a correct expression for arc length: \(\frac{70}{360} \times 2 \times \pi \times 12\).
A1: For arc length \(14.7\) or \(14.66...\).
A1ft: For adding their chord length to their arc length, giving \(28.4\) or \(28.42...\) (accept answers rounding to 28.4).

(a) The class interval with the highest frequency is \(10 < t \le 20\) (with a frequency of 26).

(b) To calculate an estimate of the mean, we first find the midpoint (\(x\)) of each class interval:
- For \(0 < t \le 10\): midpoint \(x = 5\)
- For \(10 < t \le 20\): midpoint \(x = 15\)
- For \(20 < t \le 30\): midpoint \(x = 25\)
- For \(30 < t \le 50\): midpoint \(x = 40\)
- For \(50 < t \le 60\): midpoint \(x = 55\)

Next, find the sum of products of frequencies and midpoints, \(\sum f x\):

\(\sum f x = (12 \times 5) + (26 \times 15) + (22 \times 25) + (15 \times 40) + (5 \times 55)\)

\(\sum f x = 60 + 390 + 550 + 600 + 275 = 1875\)

The total frequency is \(\sum f = 80\).

Estimated mean = \(\frac{\sum f x}{\sum f} = \frac{1875}{80} = 23.4375 \text{ minutes}\).

Correct to 3 significant figures, the estimated mean is \(23.4 \text{ minutes}\).

(c) The median is the value at position \(\frac{80}{2} = 40\).

Let's calculate the cumulative frequencies:
- up to \(t = 10\): 12 students
- up to \(t = 20\): \(12 + 26 = 38\) students
- up to \(t = 30\): \(38 + 22 = 60\) students

Since the 40th student falls after the 38th student but before the 60th student, the median lies in the class interval \(20 < t \le 30\).

Part (a):
B1: For \(10 < t \le 20\).

Part (b):
M1: For obtaining correct midpoints (at least 4 correct) of 5, 15, 25, 40, 55.
M1: For finding the sum of products \(\sum f x\) using their midpoints (at least 4 correct terms shown: \(60 + 390 + 550 + 600 + 275\)).
M1: For dividing their \(\sum f x\) by 80.
A1: For \(23.4\) or \(23.4375\).

Part (c):
M1: For showing cumulative frequencies of 12, 38, 60 or identifying the 40th/40.5th value.
A1: For \(20 < t \le 30\).