2024 OCR GCSE Mathematics - J560 模拟试题及答案详解

Let the weights of flour, sugar, and butter be \(5x\), \(3x\), and \(2x\) respectively. The difference in parts between flour and sugar is \(5 - 3 = 2\) parts. We are given that this difference is equal to \(120\text{ g}\), so: \(2x = 120\text{ g}\) which gives \(x = 60\text{ g}\). The total weight of the ingredients is the sum of all parts: \(5x + 3x + 2x = 10x\) parts. Substituting the value of \(x\): \(10 \times 60\text{ g} = 600\text{ g}\).

M1 for establishing the difference in parts, e.g. \(5 - 3 = 2\) parts, or setting up the equation \(5x - 3x = 120\). M1 for finding the value of one part, e.g. \(120 \div 2 = 60\). A1 for \(600\) (allow \(600\text{ g}\)).

First, find the perpendicular height, \(h\), of the trapezium. This forms a right-angled triangle where the hypotenuse is the slanted side of \(10\text{ cm}\) and the base is the difference between the two parallel sides: \(18\text{ cm} - 12\text{ cm} = 6\text{ cm}\). Applying Pythagoras' theorem: \(h^2 + 6^2 = 10^2\), which simplifies to \(h^2 = 100 - 36 = 64\). Taking the square root gives \(h = 8\text{ cm}\). The area of a trapezium is given by \(\frac{a + b}{2} \times h\). Substituting the values: \(\text{Area} = \frac{12 + 18}{2} \times 8 = 15 \times 8 = 120\text{ cm}^2\).

M1 for applying Pythagoras' theorem to find the height, e.g. \(\sqrt{10^2 - (18-12)^2}\). A1 for a height of \(8\) (or this value correctly substituted in the area formula). A1 for \(120\).

To clear the denominators, multiply every term in the equation by the lowest common multiple of 3 and 2, which is 6: \(2(2x + 5) - 3(x - 1) = 24\). Expand the brackets: \(4x + 10 - 3x + 3 = 24\). Collect and simplify like terms on the left-hand side: \(x + 13 = 24\). Subtract 13 from both sides to solve for \(x\): \(x = 11\).

M1 for a correct method to eliminate the denominators, e.g. multiplying by 6 to obtain \(2(2x + 5) - 3(x - 1) = 24\) (allow one sign error during bracket expansion for this mark). M1 for simplifying the equation to the form \(x + a = b\) or equivalent, e.g. \(x + 13 = 24\). A1 for \(11\).

1. Calculate the mass of sand and cement in the \( 14\text{ kg} \) taken from Mixer A:
Total parts = \( 5 + 2 = 7 \).
Value of 1 part = \( 14\text{ kg} \div 7 = 2\text{ kg} \).
Sand from Mixer A = \( 5 \times 2 = 10\text{ kg} \).
Cement from Mixer A = \( 2 \times 2 = 4\text{ kg} \).

2. Calculate the mass of sand and cement in the \( 12\text{ kg} \) taken from Mixer B:
Total parts = \( 3 + 1 = 4 \).
Value of 1 part = \( 12\text{ kg} \div 4 = 3\text{ kg} \).
Sand from Mixer B = \( 3 \times 3 = 9\text{ kg} \).
Cement from Mixer B = \( 1 \times 3 = 3\text{ kg} \).

3. Combine the mixtures:
Total sand = \( 10 + 9 = 19\text{ kg} \).
Total cement = \( 4 + 3 = 7\text{ kg} \).

4. Write the ratio in its simplest form:
The ratio of sand to cement is \( 19 : 7 \). Since 19 and 7 share no common factors other than 1, this is already in its simplest form.

M1 for calculating the sand and cement masses for Mixer A (10 kg sand, 4 kg cement) OR Mixer B (9 kg sand, 3 kg cement).
M1 for adding the respective sand and cement masses to get 19 kg of sand and 7 kg of cement.
A1 for 19:7 (accept 19 to 7, but reject unsimplified forms like 38:14).

1. Find the area of the larger sector:
\( \text{Area}_{\text{large}} = \frac{45}{360} \times \pi \times 8^2 = \frac{1}{8} \times 64\pi = 8\pi \approx 25.1327\text{ cm}^2 \).

2. Find the area of the smaller sector:
\( \text{Area}_{\text{small}} = \frac{45}{360} \times \pi \times 5^2 = \frac{1}{8} \times 25\pi = 3.125\pi \approx 9.8175\text{ cm}^2 \).

3. Subtract the smaller sector's area from the larger sector's area to find the remaining area:
\( \text{Area}_{\text{remaining}} = 8\pi - 3.125\pi = 4.875\pi \approx 15.3153\text{ cm}^2 \).

4. Rounding to 3 significant figures gives \( 15.3\text{ cm}^2 \).

M1 for a correct expression or calculation for the area of either sector (e.g. \( 8\pi \) or 25.1, or \( 3.125\pi \) or 9.82).
M1 for subtracting the smaller sector area from the larger sector area.
A1 for 15.3 (accept answers in the range 15.3 to 15.32).

Let \( s \) be the price of a Standard coffee and \( d \) be the price of a Deluxe coffee.

From the given information, we can set up two simultaneous equations:
1) \( 8s + 5d = 112 \)
2) \( 12s + 3d = 114 \)

To eliminate \( s \), find a common multiple of 8 and 12, which is 24.
Multiply equation (1) by 3:
\( 24s + 15d = 336 \) (3)

Multiply equation (2) by 2:
\( 24s + 6d = 228 \) (4)

Subtract equation (4) from equation (3):
\( (24s + 15d) - (24s + 6d) = 336 - 228 \)
\( 9d = 108 \)

Divide both sides by 9:
\( d = 12 \)

So, the price of one Deluxe coffee is £12.

M1 for forming two correct algebraic simultaneous equations (e.g. \( 8s + 5d = 112 \) and \( 12s + 3d = 114 \)).
M1 for a correct method to eliminate one variable to find the other (e.g. multiplying to equate coefficients and subtracting, or correct substitution).
A1 for 12 (accept £12 or 12.00).

To solve the equation:
\(\frac{3x - 1}{4} + \frac{2x + 3}{3} = 5\)

Multiply all terms by the lowest common multiple of 4 and 3, which is 12:
\(12 \left( \frac{3x - 1}{4} \right) + 12 \left( \frac{2x + 3}{3} \right) = 12 \times 5\)
\(3(3x - 1) + 4(2x + 3) = 60\)

Expand the brackets:
\(9x - 3 + 8x + 12 = 60\)

Collect like terms:
\(17x + 9 = 60\)

Subtract 9 from both sides:
\(17x = 51\)

Divide by 17:
\(x = 3\)

M1 for multiplying by 12 (or a common denominator) to clear fractions correctly, e.g., \(3(3x - 1) + 4(2x + 3) = 60\) or \(\frac{3(3x-1) + 4(2x+3)}{12} = 5\)
M1 for expanding brackets and collecting terms to get to \(17x + 9 = 60\) or equivalent
A1 for 3 (or \(x = 3\))

Let \(H\), \(G\), and \(F\) represent the number of students studying History, Geography, and French respectively.
We are given:
\(H : G = 5 : 3\)
\(G : F = 4 : 7\)

To combine these into a single ratio \(H : G : F\), find a common value for \(G\) in both ratios. The lowest common multiple of 3 and 4 is 12.

Multiply the first ratio by 4:
\(H : G = 20 : 12\)

Multiply the second ratio by 3:
\(G : F = 12 : 21\)

So, the combined ratio is:
\(H : G : F = 20 : 12 : 21\)

The difference in parts between French and History is:
\(21 - 20 = 1\text{ part}\)

We are told that 35 more students study French than History, so:
\(1\text{ part} = 35\text{ students}\)

The total number of parts is:
\(20 + 12 + 21 = 53\text{ parts}\)

The total number of students is:
\(53 \times 35 = 1855\)

M1 for attempting to find a combined ratio, e.g. finding \(H:G:F = 20:12:21\) (or showing at least two of these numbers in a correct relative ratio)
M1 for equating the difference of 1 part to 35, or setting up an equation such as \(21y - 20y = 35\)
A1 for 1855

The formula for compound interest is:
\(A = P \left(1 + \frac{r}{100}\right)^t\)

Substitute the given values:
\(5624.32 = 5000 \left(1 + \frac{r}{100}\right)^3\)

Divide by 5000:
\(\left(1 + \frac{r}{100}\right)^3 = \frac{5624.32}{5000}\)
\(\left(1 + \frac{r}{100}\right)^3 = 1.124864\)

Take the cube root of both sides:
\(1 + \frac{r}{100} = \sqrt[3]{1.124864}\)
\(1 + \frac{r}{100} = 1.04\)

Subtract 1:
\(\frac{r}{100} = 0.04\)

Multiply by 100:
\(r = 4\)

M1 for setting up a correct equation, e.g. \(5000 \times x^3 = 5624.32\) or \(\left(1 + \frac{r}{100}\right)^3 = 1.124864\)
M1 for taking the cube root, e.g. \(1 + \frac{r}{100} = 1.04\) or \(x = 1.04\)
A1 for 4 (accept 4\% or r = 4)

Let \(P\) be the profit in 2021. An increase of 15% is represented by a multiplier of \(1.15\). A decrease of 8% is represented by a multiplier of \(0.92\). The overall multiplier from 2021 to 2023 is: \(1.15 \times 0.92 = 1.058\). Therefore, the relationship is: \(P \times 1.058 = 454882\). To find \(P\): \(P = \frac{454882}{1.058} = 430000\). The profit in 2021 was £430,000.

M1: For calculating the combined multiplier \(1.15 \times 0.92 = 1.058\) or for finding the profit in 2022: \(454882 \div 0.92 = 494438\). M1: For a complete method to find the profit in 2021: \(454882 \div 1.058\) or \(494438 \div 1.15\). A1: For 430000 (accept £430,000).

Let the number of red, blue, and green pens on Monday be \(5x\), \(3x\), and \(2x\) respectively. After producing 450 extra blue pens on Tuesday, the number of blue pens becomes \(3x + 450\). The new ratio is \(5x : (3x + 450) : 2x = 5 : 8 : 2\). Comparing the parts, we see that the blue part has increased by 5 parts (\(8 - 3 = 5\)). Therefore, \(5x = 450\), which gives \(x = 90\). The number of green pens is \(2x = 2 \times 90 = 180\). The number of red pens is \(5x = 450\), and the number of blue pens is \(8x = 720\). The total number of non-green (red or blue) pens is \(450 + 720 = 1170\). Each presentation box contains exactly 1 green pen and 11 other pens. The 180 green pens can make at most 180 boxes. To make \(n\) boxes, we need \(11n\) non-green pens. Since we have 1170 non-green pens, we set \(11n \le 1170\), which gives \(n \le 106.36\). Thus, the maximum number of boxes that can be completely filled is 106.

M1 for setting up a ratio relationship, e.g., \(8 - 3 = 5\) parts = 450 or \(\frac{3x+450}{5x} = \frac{8}{5}\). A1 for finding 1 part = 90 pens (or calculating 180 green pens and 1170 non-green pens). M1 for identifying that each box requires 1 green pen and 11 non-green pens. M1 for calculating the limiting factor: \(1170 \div 11\). A1 for the correct final answer of 106.

The perimeter of the logo consists of the straight edge \(OA\), the arc \(AB\), and the two straight edges \(BC\) and \(CO\) of the equilateral triangle \(OBC\). Since \(OBC\) is equilateral with side length \(r\), we have \(BC = r\) and \(CO = r\). The arc length \(AB\) is given by \(\frac{135}{360} \times 2\pi r = 0.75\pi r\). The total perimeter is \(OA + \text{arc } AB + BC + CO = r + 0.75\pi r + r + r = (3 + 0.75\pi)r\). Setting this equal to 52, we get \((3 + 0.75\pi)r = 52\). This simplifies to \(5.3562r = 52\), so \(r = \frac{52}{5.3562} \approx 9.7084\text{ cm}\). To 3 significant figures, this is 9.71.

M1 for arc length formula: \(\frac{135}{360} \times 2\pi r\) or \(0.75\pi r\). M1 for identifying the components of the perimeter: Arc \(AB + 3r\). M1 for setting up the equation: \(r(3 + 0.75\pi) = 52\). M1 for rearranging to find \(r = \frac{52}{3 + 0.75\pi}\). A1 for 9.71 (accept answers in the range 9.7 to 9.71).

Let the number of red counters be \(3k\) and the number of blue counters be \(2k\), so the total number of counters is \(5k\). The probability of picking a red counter first is \(\frac{3k}{5k} = \frac{3}{5}\). Since the choice is without replacement, the probability of picking a second red counter is \(\frac{3k-1}{5k-1}\). The probability that both are red is \(\frac{3}{5} \times \frac{3k-1}{5k-1} = \frac{1}{3}\). Multiplying the fractions, we get \(\frac{9k-3}{25k-5} = \frac{1}{3}\). Cross-multiplying gives \(3(9k - 3) = 25k - 5\), which simplifies to \(27k - 9 = 25k - 5\). Rearranging gives \(2k = 4\), so \(k = 2\). The total number of counters is \(5k = 5 \times 2 = 10\).

M1 for writing red as \(3k\), blue as \(2k\), and total as \(5k\) (or equivalent algebraic representation). M1 for expressing the probability of selecting two red counters: \(\frac{3}{5} \times \frac{3k-1}{5k-1}\). M1 for setting up the equation: \(\frac{3(3k-1)}{5(5k-1)} = \frac{1}{3}\). M1 for solving for \(k\) to find \(k = 2\). A1 for the final answer 10.

First, calculate the volume of each ingredient in the initial \(500\)-litre batch. The ratio is \(5 : 3 : 2\), which has \(5 + 3 + 2 = 10\) parts in total. Volume of orange juice = \(\frac{5}{10} \times 500 = 250\) litres. Volume of pineapple juice = \(\frac{3}{10} \times 500 = 150\) litres. Volume of lime juice = \(\frac{2}{10} \times 500 = 100\) litres. In the new ratio of \(4 : 4 : 3\), the volume of orange juice remains unchanged at \(250\) litres, but it now represents \(4\) parts. Therefore, \(1\) part in the new ratio corresponds to: \(250 \div 4 = 62.5\) litres. Next, calculate the required volume of each of the other ingredients in the new mixture: Required pineapple juice = \(4 \times 62.5 = 250\) litres. Required lime juice = \(3 \times 62.5 = 187.5\) litres. Calculate the extra volumes needed: Extra pineapple juice = \(250 - 150 = 100\) litres. Extra lime juice = \(187.5 - 100 = 87.5\) litres. Total extra volume = \(100 + 87.5 = 187.5\) litres.

M1: For calculating the initial volume of orange juice as \(250\) litres (or finding the volumes of all three: \(250\)L, \(150\)L, \(100\)L). M1: For setting up the equation \(4 \text{ parts} = 250\) litres and finding the value of \(1\) part as \(62.5\) litres. M1: For calculating the new required volume of pineapple juice (\(250\)L) and/or lime juice (\(187.5\)L). M1: For finding the individual volumes to be added: \(100\)L of pineapple juice and \(87.5\)L of lime juice. A1: For a final correct total of \(187.5\) (litres).

Let \(O\) be the center of the sector and \(C\) be the center of the small circle. The line of symmetry of the sector bisects the \(60^\circ\) angle into two \(30^\circ\) angles. If we draw a radius of the small circle to the point of tangency on one of the straight edges, we form a right-angled triangle where the angle at \(O\) is \(30^\circ\) and the opposite side is \(r\). Using trigonometry: \(\sin(30^\circ) = \frac{r}{OC}\). Since \(\sin(30^\circ) = 0.5\), we have: \(0.5 = \frac{r}{OC} \implies OC = 2r\). The distance from \(O\) to the outer boundary of the small circle along the line of symmetry is \(OC + r = 2r + r = 3r\). Since the small circle is tangent to the curved boundary arc of the sector, this total distance must equal the radius of the sector, \(R\). Therefore, \(R = 3r\). Now, calculate the area of the sector: \(\text{Area of sector} = \frac{60}{360} \times \pi R^2 = \frac{1}{6} \pi (3r)^2 = \frac{9}{6} \pi r^2 = 1.5 \pi r^2\). The area of the small circle is \(\pi r^2\). The area of the region inside the sector but outside the circle is: \(\text{Area} = 1.5 \pi r^2 - \pi r^2 = 0.5 \pi r^2\). We are given that this area is \(50\pi\text{ cm}^2\): \(0.5 \pi r^2 = 50\pi \implies 0.5 r^2 = 50 \implies r^2 = 100 \implies r = 10\) (since radius must be positive).

M1: Uses trigonometry on the bisected angle of \(30^\circ\) to show \(OC = 2r\). M1: Combines \(OC\) and \(r\) to establish the relationship \(R = 3r\). M1: Expresses the area of the sector in terms of \(r\) as \(\frac{1}{6}\pi(3r)^2\) or \(1.5\pi r^2\). M1: Sets up the equation \(1.5\pi r^2 - \pi r^2 = 50\pi\) and simplifies to find \(r^2 = 100\). A1: For the correct radius of \(10\).

Let the width of the lawn be \(x\) metres. Since the length of the lawn is \(4\text{ m}\) longer than its width, the length of the lawn is \(x + 4\) metres. The gravel path has a uniform width of \(1.5\text{ m}\) around the outside. Therefore, \(1.5\text{ m}\) is added to both ends of the width and both ends of the length. Total width of lawn and path = \(x + 1.5 + 1.5 = x + 3\) metres. Total length of lawn and path = \(x + 4 + 1.5 + 1.5 = x + 7\) metres. The total combined area is given as \(165\text{ m}^2\): \((x + 3)(x + 7) = 165\). Expand the brackets: \(x^2 + 7x + 3x + 21 = 165 \implies x^2 + 10x + 21 = 165\). Subtract \(165\) from both sides to form a quadratic equation equal to zero: \(x^2 + 10x - 144 = 0\). Factorize the quadratic equation: We need two numbers that multiply to \(-144\) and add to \(10\). These are \(18\) and \(-8\). \((x + 18)(x - 8) = 0\). This gives two potential solutions: \(x = -18\) or \(x = 8\). Since the width must be a positive value, we discard \(x = -18\). Thus, the width of the lawn is \(8\) metres.

M1: For expressing the total dimensions including the path as \(x + 3\) and \(x + 7\) (or equivalent using another variable). M1: For setting up the correct area equation: \((x + 3)(x + 7) = 165\). M1: For expanding and rearranging to the standard quadratic form: \(x^2 + 10x - 144 = 0\) (allow one sign or arithmetic error). M1: For attempting to solve their three-term quadratic equation by factorizing to \((x + 18)(x - 8) = 0\) or by using the quadratic formula. A1: For finding the correct width of \(8\) (must explicitly reject \(-18\) or choose the positive value to secure this mark).

To find the value of \(x\), we set up an equation for the area of the rectangle:
\[\text{Area} = \text{length} \times \text{width}\]
\[(3x + 1)(2x - 3) = 115\]

1. Expand the brackets:
\[3x(2x) + 3x(-3) + 1(2x) + 1(-3) = 115\]
\[6x^2 - 9x + 2x - 3 = 115\]
\[6x^2 - 7x - 3 = 115\]

2. Subtract 115 from both sides to form a quadratic equation equal to 0:
\[6x^2 - 7x - 118 = 0\]

3. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -7\), and \(c = -118\):
\[x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-118)}}{2(6)}\]
\[x = \frac{7 \pm \sqrt{49 - (-2832)}}{12}\]
\[x = \frac{7 \pm \sqrt{2881}}{12}\]

4. Calculate the two possible values of \(x\):
\[x = \frac{7 + 53.6749...}{12} \approx 5.0562...\]
\[x = \frac{7 - 53.6749...}{12} \approx -3.8895...\]

Since \(x\) represents physical dimensions, the width \(2x - 3\) must be positive, which requires \(x > 1.5\). Therefore, we reject the negative value.

Rounding to 3 significant figures:
\[x = 5.06\]

**M1**: Sets up a correct equation for the area: \((3x + 1)(2x - 3) = 115\)
**M1**: Correctly expands and simplifies to a three-term quadratic equation equal to zero: \(6x^2 - 7x - 118 = 0\) (allow one sign or arithmetic error)
**M1**: Uses a correct method to solve their three-term quadratic equation, e.g., substitution into the quadratic formula: \(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-118)}}{2(6)}\)
**A1**: Identifies the two solutions to the quadratic equation: \(x \approx 5.06\) and \(x \approx -3.89\) (or values that round to these)
**A1**: Selects the positive root and rounds correctly to 3 significant figures to give \(5.06\) (with or without rejection of the negative solution explicitly shown)

To find the overall percentage of students who walk to school, we first find a combined ratio for the number of students in Years 7, 8, and 9.

1. Combine the ratios:
- \(\text{Year 7} : \text{Year 8} = 3 : 4\)
- \(\text{Year 8} : \text{Year 9} = 3 : 1\)

To combine these, find a common multiple for the Year 8 parts. The lowest common multiple of 4 and 3 is 12.
Multiply the first ratio by 3:
\(\text{Year 7} : \text{Year 8} = 9 : 12\)

Multiply the second ratio by 4:
\(\text{Year 8} : \text{Year 9} = 12 : 4\)

So, the combined ratio is:
\(\text{Year 7} : \text{Year 8} : \text{Year 9} = 9 : 12 : 4\)

2. Assign quantities based on the ratio:
Let the total number of students in each year group be represented by parts.
Total parts = \(9 + 12 + 4 = 25\) parts.

Let us assume a convenient total or use algebra. For example, let 1 part represent \(100\) students:
- Year 7 students = \(900\)
- Year 8 students = \(1200\)
- Year 9 students = \(400\)
- Total students = \(900 + 1200 + 400 = 2500\)

3. Calculate the number of students who walk in each year group:
- Year 7 walkers = \(40\% \text{ of } 900 = 0.40 \times 900 = 360\)
- Year 8 walkers = \(35\% \text{ of } 1200 = 0.35 \times 1200 = 420\)
- Year 9 walkers = \(60\% \text{ of } 400 = 0.60 \times 400 = 240\)

4. Calculate the total number of walkers:
\(\text{Total walkers} = 360 + 420 + 240 = 1020\)

5. Calculate the overall percentage:
\[\text{Percentage} = \frac{\text{Total walkers}}{\text{Total students}} \times 100\]
\[\text{Percentage} = \frac{1020}{2500} \times 100 = 40.8\%\]

**M1**: Attempts to find a common term for Year 8 in both ratios (e.g. by multiplying \(3:4\) by 3 and \(3:1\) by 4)
**A1**: Correctly identifies the combined ratio of Year 7 : Year 8 : Year 9 as \(9 : 12 : 4\) (or any equivalent ratio, e.g., \(18 : 24 : 8\))
**M1**: Correctly calculates the relative proportion or number of walkers in each year group (e.g., finding \(3.6\) parts, \(4.2\) parts, and \(2.4\) parts; or finding 360, 420, and 240 students)
**M1**: Demonstrates a complete method to find the overall percentage by dividing the total number of walkers by the total number of students and multiplying by 100: \(\frac{3.6 + 4.2 + 2.4}{25} \times 100\) (or equivalent)
**A1**: Correctly calculates the final answer as \(40.8\%\) (accept \(40.8\))

First, write an expression for the area of the rectangle: Area of rectangle = \((3x + 4)(2x - 1) = 6x^2 - 3x + 8x - 4 = 6x^2 + 5x - 4\). Next, write an expression for the area of the square: Area of square = \((x + 2)^2 = x^2 + 4x + 4\). Since the area of the rectangle is \(30\text{ cm}^2\) greater than the area of the square, we can write the equation: \(\text{Area of rectangle} - \text{Area of square} = 30\). Substituting the expressions: \((6x^2 + 5x - 4) - (x^2 + 4x + 4) = 30\). Expanding and simplifying the left-hand side: \(6x^2 + 5x - 4 - x^2 - 4x - 4 = 30\), which simplifies to \(5x^2 + x - 8 = 30\). Subtracting 30 from both sides gives: \(5x^2 + x - 38 = 0\).

M1: For expanding \((3x + 4)(2x - 1)\) to get \(6x^2 + 5x - 4\) (allow one arithmetic slip). M1: For expanding \((x + 2)^2\) to get \(x^2 + 4x + 4\). M1: For setting up the difference equation, i.e., \((6x^2 + 5x - 4) - (x^2 + 4x + 4) = 30\). A1: For fully correct algebraic simplification leading to the target equation \(5x^2 + x - 38 = 0\) with all steps shown clearly.

Let the initial value of the investment be \(V\). An increase of \(x\%\) in the first year corresponds to a multiplier of \(1 + \frac{x}{100}\). The value at the end of the first year is: \(V_1 = V \left(1 + \frac{x}{100}\right)\). A decrease of \(x\%\) in the second year corresponds to a multiplier of \(1 - \frac{x}{100}\). The value at the end of the second year is: \(V_2 = V_1 \left(1 - \frac{x}{100}\right) = V \left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right)\). Using the difference of two squares to expand this expression: \(V_2 = V \left(1^2 - \left(\frac{x}{100}\right)^2\right) = V \left(1 - \frac{x^2}{10000}\right)\). The total change in value is: \(V_2 - V = V \left(1 - \frac{x^2}{10000}\right) - V = -V \frac{x^2}{10000}\). To find the percentage change relative to the initial value: \(\text{Percentage change} = \frac{-V \frac{x^2}{10000}}{V} \times 100\% = -\frac{x^2}{100}\%\). The negative sign indicates a decrease, so the overall percentage change is a decrease of \(\frac{x^2}{100}\%\).

M1: For representing the value after the first year as \(V\left(1 + \frac{x}{100}\right)\) or using a trial initial value like \(100\) to get \(100 + x\). M1: For multiplying by the second-year multiplier to get \(V\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right)\) or equivalent with a trial value. M1: For expanding the brackets correctly to obtain \(V\left(1 - \frac{x^2}{10000}\right)\) or equivalent. A1: For a complete and rigorous proof showing the change is \(-\frac{x^2}{100}\%\), concluding with a decrease of \(\frac{x^2}{100}\%\).

The arc length of a sector with radius \(r\) and angle \(120^\circ\) is given by: \(\text{Arc length} = \frac{120}{360} \times 2\pi r = \frac{2}{3}\pi r\). The perimeter \(P\) of the sector consists of the arc length plus the two straight edges (each of length \(r\)): \(P = \frac{2}{3}\pi r + 2r\). Rearranging this to find the arc length in terms of \(P\) and \(r\): \(\frac{2}{3}\pi r = P - 2r\). Solving for \(r\) gives: \(r = \frac{3(P - 2r)}{2\pi}\). The area \(A\) of the sector is: \(A = \frac{120}{360} \times ̀\pi r^2 = \frac{1}{3}\pi r^2\). Substituting the expression for \(r\) into the area formula: \(A = \frac{1}{3}\pi \left(\frac{3(P - 2r)}{2\pi}\right)^2\). Expanding the squared term: \(A = \frac{1}{3}\pi \frac{9(P - 2r)^2}{4\pi^2}\). Simplifying this gives: \(A = \frac{3(P - 2r)^2}{4\pi}\).

M1: For expressing the arc length as \(\frac{2}{3}\pi r\) or the perimeter as \(P = \frac{2}{3}\pi r + 2r\). M1: For rearranging to express \(r\) in terms of \(P\) and \(r\), specifically finding \(r = \frac{3(P - 2r)}{2\pi}\). M1: For substituting their expression for \(r\) into the sector area formula \(A = \frac{1}{3}\pi r^2\). A1: For fully correct algebraic working leading to the exact expression \(A = \frac{3(P - 2r)^2}{4\pi}\).

1. The total surface area of a solid cylinder is the sum of the areas of its two circular ends and its curved surface: \(\text{Total Surface Area} = 2\pi r^2 + 2\pi r h\). 2. We are given that the total surface area is \(120\pi\text{ cm}^2\), so: \(2\pi r^2 + 2\pi r h = 120\pi\). 3. Divide every term by \(2\pi\) to simplify: \(r^2 + r h = 60\). 4. Rearrange this equation to make \(h\) the subject: \(r h = 60 - r^2\) which gives \(h = \frac{60 - r^2}{r}\). 5. The volume of a cylinder is given by: \(V = \pi r^2 h\). 6. Substitute the expression for \(h\) into the volume formula: \(V = \pi r^2 \left(\frac{60 - r^2}{r}\right)\). 7. Simplify by cancelling \(r\): \(V = \pi r (60 - r^2)\). 8. Expand the bracket to obtain the required formula: \(V = 60\pi r - \pi r^3\).

M1: Set up the correct equation for total surface area, i.e., \(2\pi r^2 + 2\pi r h = 120\pi\) (or equivalent). M1: Rearrange the equation to express \(h\) in terms of \(r\), leading to \(h = \frac{120\pi - 2\pi r^2}{2\pi r}\) or \(h = \frac{60 - r^2}{r}\) (allow equivalent expressions or making \(rh\) the subject: \(rh = 60 - r^2\)). M1: Substitute their expression for \(h\) (or \(rh\)) into the volume formula \(V = \pi r^2 h\) (or \(V = \pi r (rh)\)), e.g., \(V = \pi r^2 \left(\frac{60-r^2}{r}\right)\). A1: Correctly simplify to obtain \(V = 60\pi r - \pi r^3\) with all steps of algebraic working shown clearly.

Since \( y \) is inversely proportional to the square root of \( x \), we can write:
\( y = \frac{k}{\sqrt{x}} \)

Substitute the given values \( x = 16 \) and \( y = 3 \) to find \( k \):
\( 3 = \frac{k}{\sqrt{16}} \)
\( 3 = \frac{k}{4} \)
\( k = 12 \)

Now, substitute \( k = 12 \) and \( x = 36 \) into the formula to find \( y \):
\( y = \frac{12}{\sqrt{36}} \)
\( y = \frac{12}{6} \)
\( y = 2 \)

M1 for writing a correct equation of the form \( y = \frac{k}{\sqrt{x}} \) (or equivalent)
M1 for substituting \( x = 16 \) and \( y = 3 \) to find \( k = 12 \)
A1 for \( 2 \)

First, let's find the number of blue counters using the ratio of blue to green counters, which is \( 4 : 7 \).
Let the number of blue counters be \( B \).
\( \frac{B}{35} = \frac{4}{7} \)
\( B = \frac{4}{7} \times 35 = 4 \times 5 = 20 \).
There are 20 blue counters.

Next, use the ratio of red to blue counters, which is \( 3 : 5 \), to find the number of red counters (\( R \)).
\( \frac{R}{20} = \frac{3}{5} \).
\( R = \frac{3}{5} \times 20 = 3 \times 4 = 12 \).

Alternatively, we can find the combined ratio \( R : B : G \):
Multiply \( R : B = 3 : 5 \) by 4 to get \( 12 : 20 \).
Multiply \( B : G = 4 : 7 \) by 5 to get \( 20 : 35 \).
So, the ratio \( R : B : G = 12 : 20 : 35 \).
Since there are 35 green counters, there must be 12 red counters.

M1 for a method to find the number of blue counters as 20 OR for writing a combined ratio \( R:B:G \) as \( 12:20:35 \)
M1 for a complete method to find the number of red counters from their blue counter total or combined ratio
A1 for 12

First, write \( \sqrt{12} \) in its simplest form:
\( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \)

Now substitute this back into the expression:
\( (4 - \sqrt{3})(2 + 2\sqrt{3}) \)

Expand the brackets:
\( = 4 \times 2 + 4 \times 2\sqrt{3} - 2\sqrt{3} - \sqrt{3} \times 2\sqrt{3} \)
\( = 8 + 8\sqrt{3} - 2\sqrt{3} - 2 \times 3 \)
\( = 8 + 6\sqrt{3} - 6 \)

Collect like terms:
\( = 2 + 6\sqrt{3} \)

Alternatively, expanding directly first:
\( (4 - \sqrt{3})(2 + \sqrt{12}) = 8 + 4\sqrt{12} - 2\sqrt{3} - \sqrt{36} \)
Since \( \sqrt{36} = 6 \) and \( 4\sqrt{12} = 4(2\sqrt{3}) = 8\sqrt{3} \):
\( = 8 + 8\sqrt{3} - 2\sqrt{3} - 6 \)
\( = 2 + 6\sqrt{3} \)

M1 for simplifying \( \sqrt{12} = 2\sqrt{3} \) or expanding to get at least 3 correct terms out of \( 8 + 4\sqrt{12} - 2\sqrt{3} - 6 \) (or equivalent)
M1 for a fully correct intermediate expansion, e.g., \( 8 + 8\sqrt{3} - 2\sqrt{3} - 6 \) or \( 8 + 4\sqrt{12} - 2\sqrt{3} - 6 \)
A1 for \( 2 + 6\sqrt{3} \)

To eliminate \(y\), multiply the first equation by 3 and the second equation by 4:

1) \(3(3x + 4y) = 3(3) \implies 9x + 12y = 9\)

2) \(4(2x - 3y) = 4(19) \implies 8x - 12y = 76\)

Add the two equations together to eliminate \(y\):

\((9x + 8x) + (12y - 12y) = 9 + 76\)

\(17x = 85\)

Divide by 17:

\(x = 5\)

Substitute \(x = 5\) back into the first equation:

\(3(5) + 4y = 3\)

\(15 + 4y = 3\)

\(4y = -12\)

\(y = -3\)

So, the solution is \(x = 5\), \(y = -3\).

M1 for a correct process to eliminate one variable (allow one arithmetic error).
M1 for finding the value of one variable (either \(x = 5\) or \(y = -3\)).
A1 for both \(x = 5\) and \(y = -3\).

The ratio of red to blue to yellow counters is \(3 : 5 : 2\).

Let the number of parts for red, blue, and yellow counters be \(3x\), \(5x\), and \(2x\) respectively.

The difference in parts between the blue and yellow counters is:

\(5 - 2 = 3\) parts

We are given that there are 24 more blue counters than yellow counters, so:

\(3\text{ parts} = 24\)

Dividing by 3 gives the value of 1 part:

\(1\text{ part} = 24 \div 3 = 8\)

The total number of parts is:

\(3 + 5 + 2 = 10\) parts

Therefore, the total number of counters in the box is:

\(10 \times 8 = 80\)

M1 for finding the difference in parts: \(5 - 2 = 3\) parts, and equating this to 24.
M1 for finding the value of one part: \(24 \div 3 = 8\), or the total number of parts: \(3 + 5 + 2 = 10\).
A1 for 80.

The perimeter of a semi-circular shape consists of the curved arc of the semi-circle and the straight diameter.

First, find the length of the curved arc:

\(\text{Arc length} = \frac{1}{2} \times \pi \times d = \frac{1;}{2} \times \pi \times 12 = 6\pi\text{ cm}\)

Next, add the straight diameter of \(12\text{ cm}\) to find the total perimeter:

\(\text{Total perimeter} = 6\pi + 12\text{ cm}\)

This is in the required form \(a\pi + b\), where \(a = 6\) and \(b = 12\).

M1 for a method to calculate the curved arc length: \(\frac{1}{2} \times \pi \times 12\) (or \(6\pi\)).
M1 for adding the diameter (12) to their arc length expression.
A1 for \(6\pi + 12\) (or \(12 + 6\pi\)).

To find the combined ratio of red (\(R\)), blue (\(B\)), and green (\(G\)) counters, we make the parts for blue the same in both ratios. The common multiple of 5 and 4 is 20. Multiply \(R : B = 3 : 5\) by 4 to get \(12 : 20\). Multiply \(B : G = 4 : 7\) by 5 to get \(20 : 35\). So, the ratio \(R : B : G = 12 : 20 : 35\). The difference between the number of green counters and red counters in ratio parts is \(35 - 12 = 23\) parts. We are given that there are 46 more green counters than red counters: \(23\) parts = 46 counters, so \(1\) part = \(46 \div 23 = 2\) counters. To find the total number of counters, we sum the parts in the ratio: \(12 + 20 + 35 = 67\) parts. Total counters = \(67 \times 2 = 134\).

M1: Attempts to find a common ratio for \(R : B : G\) (e.g. shows \(12 : 20\) and \(20 : 35\) or writes \(12 : 20 : 35\)) M1: Finds the value of one part by dividing 46 by their difference in parts (e.g. \(46 \div (35 - 12) = 2\)) A1: Correct total of 134

Multiply the entire equation by the common denominator \((x-1)(x+2)\) to clear the fractions: \(4(x+2) + 5(x-1) = 3(x-1)(x+2)\). Expand both sides: \(4x + 8 + 5x - 5 = 3(x^2 + 2x - x - 2)\), which simplifies to \(9x + 3 = 3(x^2 + x - 2)\) and thus \(9x + 3 = 3x^2 + 3x - 6\). Rearrange into a standard quadratic equation form \(ax^2 + bx + c = 0\): \(3x^2 - 6x - 9 = 0\). Divide the entire equation by 3 to simplify: \(x^2 - 2x - 3 = 0\). Factorise the quadratic expression: \((x - 3)(x + 1) = 0\). This gives the solutions: \(x = 3\) or \(x = -1\).

M1: For clearing the fractions correctly, e.g. showing \(4(x+2) + 5(x-1) = 3(x-1)(x+2)\) M1: For simplifying to a correct quadratic equation, e.g. \(3x^2 - 6x - 9 = 0\) or \(x^2 - 2x - 3 = 0\) A1: For both correct solutions: \(x = 3\) and \(x = -1\) (accept \(3\) and \(-1\))

The formula for the area of a trapezium is \(\text{Area} = \frac{1}{2}(a + b)h\). Substitute the given values into the formula: \(34 = \frac{1}{2}((2x - 3) + (x + 5)) \times 4\). Simplify the equation: \(34 = 2((2x - 3) + (x + 5))\) which becomes \(34 = 2(3x + 2)\). Divide both sides by 2: \(17 = 3x + 2\). Subtract 2 from both sides: \(15 = 3x\). Divide by 3: \(x = 5\).

M1: For setting up a correct equation using the trapezium area formula, e.g. \(\frac{1}{2}((2x - 3) + (x + 5)) \times 4 = 34\) M1: For simplifying the equation to a linear form, e.g. \(2(3x + 2) = 34\) or \(3x + 2 = 17\) A1: For \(x = 5\) (accept 5)

To combine the ratios, we need the number of parts for blue counters to be the same in both ratios.

The lowest common multiple of 5 and 2 is 10.

Multiply the first ratio by 2:
\(R : B = (3 \times 2) : (5 \times 2) = 6 : 10\)

Multiply the second ratio by 5:
\(B : Y = (2 \times 5) : (3 \times 5) = 10 : 15\)

Combining these gives the ratio of Red : Blue : Yellow counters:
\(R : B : Y = 6 : 10 : 15\)

The difference in parts between yellow and red counters is:
\(15 - 6 = 9\text{ parts}\)

We are given that there are 45 more yellow counters than red counters:
\(9\text{ parts} = 45\)

Therefore, 1 part is:
\(45 \div 9 = 5\text{ counters}\)

The total number of parts is:
\(6 + 10 + 15 = 31\text{ parts}\)

The total number of counters is:
\(31 \times 5 = 155\)

- **M1** for attempting to find a common ratio, e.g. \(R : B : Y = 6 : 10 : 15\) (or equivalent with one variable in common)
- **M1** for finding the value of one part, e.g. \(45 \div (15 - 6) = 5\)
- **A1** for \(155\)

The perimeter of a semicircle consists of a curved boundary (half the circumference) and a straight edge (the diameter).

For a semicircle with radius \(r\):
- Curved boundary = \(\pi r\)
- Diameter = \(2r\)

Therefore, the total perimeter is:
\(\text{Perimeter} = \pi r + 2r = r(\pi + 2)\)

We are given that the perimeter is \(24 + 12\pi\) cm.
Factorising this expression gives:
\(24 + 12\pi = 12(2 + \pi) = 12(\pi + 2)\)

Comparing the two expressions:
\(r(\pi + 2) = 12(\pi + 2)\)

Therefore, \(r = 12\).

The radius of the semicircle is 12 cm.

- **M1** for writing a correct expression for the perimeter of a semicircle in terms of \(r\), e.g. \(\pi r + 2r\) or \(\frac{1}{2}\pi d + d\)
- **M1** for equating their expression to \(24 + 12\pi\) and attempting to factorise, e.g. \(r(\pi + 2) = 12(\pi + 2)\)
- **A1** for \(12\)

Multiply all terms by the lowest common multiple of 3 and 2, which is 6, to clear the fractions:

\(6 \left(\frac{2x - 5}{3}\right) + 6 \left(\frac{x + 2}{2}\right) = 6 \times 4\)

\(2(2x - 5) + 3(x + 2) = 24\)

Expand the brackets:

\(4x - 10 + 3x + 6 = 24\)

Simplify the left-hand side by collecting like terms:

\(7x - 4 = 24\)

Add 4 to both sides:

\(7x = 28\)

Divide by 7:

\(x = 4\)

- **M1** for attempting to clear fractions by multiplying by 6 (or another common multiple), e.g. \(2(2x - 5) + 3(x + 2) = 24\) (allow one arithmetic slip)
- **M1** for expanding and simplifying to the form \(ax = b\), e.g. \(7x = 28\) (or \(7x - 4 = 24\))
- **A1** for \(4\) (or \(x = 4\))

Let the total number of pens in the box be \(r + 5\). If two pens are selected without replacement, the probability of selecting exactly one red pen is the sum of the probability of selecting a red then a blue pen, and the probability of selecting a blue then a red pen: \(P(\text{one red}) = P(R, B) + P(B, R)\). We calculate each probability: \(P(R, B) = \frac{r}{r+5} \times \frac{5}{r+4} = \frac{5r}{(r+5)(r+4)}\) and \(P(B, R) = \frac{5}{r+5} \times \frac{r}{r+4} = \frac{5r}{(r+5)(r+4)}\). Therefore, the total probability of selecting exactly one red pen is: \(P(\text{one red}) = \frac{10r}{(r+5)(r+4)}\). We are given that this probability is \(\frac{5}{9}\): \(\frac{10r}{(r+5)(r+4)} = \frac{5}{9}\). Divide both sides of the equation by 5: \(\frac{2r}{(r+5)(r+4)} = \frac{1}{9}\). Cross-multiply to clear the denominators: \(18r = (r+5)(r+4)\), which expands to \(18r = r^2 + 9r + 20\). Rearrange to form a quadratic equation set to 0: \(r^2 - 9r + 20 = 0\). Factorise the quadratic expression: \((r - 4)(r - 5) = 0\). This gives two potential solutions for \(r\): \(r = 4\) or \(r = 5\). Since the question states that \(r < 5\), we must have \(r = 4\).

M1: For a correct expression for the probability of one combination, e.g. \(\frac{5r}{(r+5)(r+4)}\) or a correct probability tree diagram. M1: For setting up the equation for the combined probability: \(\frac{10r}{(r+5)(r+4)} = \frac{5}{9}\) (or equivalent). M1: For expanding and rearranging to form a correct quadratic equation, e.g. \(r^2 - 9r + 20 = 0\). M1: For factorising or solving their quadratic equation to find two roots, e.g. \(r = 4\) and \(r = 5\). A1: For concluding \(r = 4\) (with reference to the given condition \(r < 5\)).

First, calculate the masses of Type A and Type B soils in 40 kg of 'Premium Mix'. The ratio of Type A to Type B is \(3 : 2\). Total parts = \(3 + 2 = 5\). Value of 1 part = \(\frac{40}{5} = 8\) kg. Mass of Type A = \(3 \times 8 = 24\) kg. Mass of Type B from Premium Mix = \(2 \times 8 = 16\) kg. Second, calculate the masses of Type B and Type C soils in 45 kg of 'Super Mix'. The ratio of Type B to Type C is \(5 : 4\). Total parts = \(5 + 4 = 9\). Value of 1 part = \(\frac{45}{9} = 5\) kg. Mass of Type B from Super Mix = \(5 \times 5 = 25\) kg. Mass of Type C = \(4 \times 5 = 20\) kg. Third, find the total mass of each type of soil in the 'Ultimate Mix': Total mass of Type A = 24 kg, Total mass of Type B = 16 kg + 25 kg = 41 kg, Total mass of Type C = 20 kg. Write this as a ratio: Type A : Type B : Type C = \(24 : 41 : 20\). Since 41 is a prime number, the ratio cannot be simplified further.

M1: For sharing 40 kg in the ratio \(3 : 2\) to find the mass of Type A (24 kg) or Type B (16 kg). M1: For sharing 45 kg in the ratio \(5 : 4\) to find the mass of Type B (25 kg) or Type C (20 kg). A1: For finding the correct individual masses of A, B (Premium), B (Super), and C: 24 kg, 16 kg, 25 kg, and 20 kg respectively. M1: For adding the two masses of Type B together to find the total mass of B: \(16 + 25 = 41\) kg. A1: For the final ratio written in simplest form: \(24 : 41 : 20\).

First, use the perimeter of the lawn to set up an equation: Perimeter = \(2(x + y) = 24\). Divide by 2: \(x + y = 12 \implies y = 12 - x\) [Equation 1]. Next, set up an equation for the combined area. Since the path has a uniform width of 1 metre on all sides, the overall width is increased by 2 metres, and the overall length is increased by 2 metres. Combined dimensions = \((x + 2)\) by \((y + 2)\). Combined area = \((x + 2)(y + 2) = 60\). Expand the brackets: \(xy + 2x + 2y + 4 = 60 \implies xy + 2(x + y) + 4 = 60\) [Equation 2]. Substitute the known value \(x + y = 12\) into Equation 2: \(xy + 2(12) + 4 = 60 \implies xy + 24 + 4 = 60 \implies xy + 28 = 60 \implies xy = 32\) [Equation 3]. Substitute \(y = 12 - x\) from Equation 1 into Equation 3: \(x(12 - x) = 32 \implies 12x - x^2 = 32\). Rearrange into a standard quadratic equation: \(x^2 - 12x + 32 = 0\). Factorise the quadratic equation: \((x - 8)(x - 4) = 0\). This gives two possible solutions: \(x = 8\) or \(x = 4\). If \(x = 8\), then \(y = 12 - 8 = 4\). If \(x = 4\), then \(y = 12 - 4 = 8\). Since we are given that \(x > y\), the correct dimensions are \(x = 8\) and \(y = 4\).

M1: For a correct perimeter equation: \(2(x+y)=24\) or \(x+y=12\). M1: For a correct combined area equation: \((x+2)(y+2)=60\). M1: For substituting \(x+y=12\) to simplify the area equation to \(xy=32\), or for substituting \(y=12-x\) directly into the expanded area equation to form a quadratic equation. M1: For solving the quadratic equation \(x^2-12x+32=0\) by factorisation to find the values 8 and 4. A1: For identifying the correct pair \(x=8\) and \(y=4\) based on the condition \(x>y\).

Let the number of staff in Office A be \(5x\) and the number of staff in Office B be \(3x\), where \(x\) is a positive constant.

First, express the number of part-time staff in each office in terms of \(x\):
- Part-time staff in Office A: \(\frac{2}{5} \times 5x = 2x\)
- Part-time staff in Office B: \(\frac{1}{4} \times 3x = \frac{3}{4}x = 0.75x\)

We are given that the total number of part-time staff is 44:
\(2x + 0.75x = 44\)
\(2.75x = 44\)

Convert \(2.75\) to a fraction to solve without a calculator:
\(\frac{11}{4}x = 44\)
\(11x = 176\)
\(x = 16\)

Now, find the total number of staff in the company:
\(\text{Total staff} = 5x + 3x = 8x\)
\(\text{Total staff} = 8 \times 16 = 128\)

- **M1**: Sets up representations for the staff in both offices, e.g., \(5x\) and \(3x\), or writes an expression for part-time staff as a fraction of the total (e.g. \(\frac{2}{5} \times \frac{5}{8} + \frac{1}{4} \times \frac{3}{8}\) of the total).
- **M1**: Sets up a correct equation in terms of a single variable, e.g., \(2x + \frac{3}{4}x = 44\).
- **M1**: Simplifies the equation to find the value of their variable, e.g., \(2.75x = 44 \implies x = 16\).
- **M1**: Multiplies their value of \(x\) by \(8\) to find the total staff, or calculates staff in Office A (\(80\)) and Office B (\(48\)).
- **A1**: Correct final answer of \(128\).

Since \(x > 1\) for the side length \((x - 1)\) to be positive, the longest side of the right-angled triangle must be the hypotenuse, which is \((2x + 1)\text{ cm}\).

Applying Pythagoras' Theorem:
\((x - 1)^2 + (2x)^2 = (2x + 1)^2\)

Expand each bracket:
- \((x - 1)^2 = x^2 - 2x + 1\)
- \((2x)^2 = 4x^2\)
- \((2x + 1)^2 = 4x^2 + 4x + 1\)

Substitute these back into Pythagoras' Theorem:
\(x^2 - 2x + 1 + 4x^2 = 4x^2 + 4x + 1\)

Simplify the equation by subtracting \(4x^2\) from both sides:
\(x^2 - 2x + 1 = 4x + 1\)

Rearrange to set the quadratic equal to zero:
\(x^2 - 6x = 0\)

Factorise the quadratic:
\(x(x - 6) = 0\)

Since \(x\) must be greater than 1 for the side lengths to be positive lengths, \(x = 6\).

Now, calculate the three side lengths:
- \((x - 1) = 6 - 1 = 5\text{ cm}\)
- \(2x = 2 \times 6 = 12\text{ cm}\)
- \((2x + 1) = 2 \times 6 + 1 = 13\text{ cm}\)

The two shorter perpendicular sides are \(5\text{ cm}\) and \(12\text{ cm}\).

Calculate the area:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\)

- **M1**: Uses Pythagoras' Theorem correctly with the hypotenuse as \((2x + 1)\), e.g., \((x - 1)^2 + (2x)^2 = (2x + 1)^2\).
- **M1**: Correctly expands at least two of the squared terms (e.g., \(x^2 - 2x + 1\) or \(4x^2 + 4x + 1\)).
- **M1**: Simplifies to form a correct quadratic equation, e.g., \(x^2 - 6x = 0\).
- **M1**: Solves the quadratic to find \(x = 6\) and calculates the perpendicular sides as \(5\) and \(12\).
- **A1**: Correctly calculates the area of the triangle as \(30\) (units not required).

First, let's find the area of the first sector.
Let \(\theta\) be the angle of the first sector. The arc length formula is:
\(\text{Arc Length} = \frac{\theta}{360} \times 2\pi r_1\)
\(8\pi = \frac{\theta}{360} \times 2\pi(12)\)
\(8\pi = \frac{\theta}{360} \times 24\pi\)

So, the fraction of the circle is:
\(\frac{\theta}{360} = \frac{8\pi}{24\pi} = \frac{1}{3}\)

The area of this first sector is:
\(\text{Area}_1 = \frac{1}{3} \times \pi \times r_1^2\)
\(\text{Area}_1 = \frac{1}{3} \times \pi \times 12^2 = \frac{144\pi}{3} = 48\pi\text{ cm}^2\)

(Alternatively, using the formula \(\text{Area} = \frac{1}{2} L r\):
\(\text{Area}_1 = \frac{1}{2} \times 8\pi \times 12 = 48\pi\text{ cm}^2\))

Now, for the second sector, the area is also \(48\pi\text{ cm}^2\) and the radius \(r_2 = 9\text{ cm}\).
Let \(\phi\) be the angle of the second sector:
\(\text{Area}_2 = \frac{\phi}{360} \times \pi r_2^2\)
\(48\pi = \frac{\phi}{360} \times \pi \times 9^2\)
\(48\pi = \frac{\phi}{360} \times 81\pi\)

So, the fraction of the second circle is:
\(\frac{\phi}{360} = \frac{48\pi}{81\pi} = \frac{16}{27}\)

Next, calculate the arc length of the second sector (\(L_2\)):
\(L_2 = \frac{16}{27} \times 2\pi r_2\)
\(L_2 = \frac{16}{27} \times 2\pi(9) = \frac{16}{27} \times 18\pi = \frac{32}{3}\pi\text{ cm}\)

Finally, the perimeter of the second sector is the sum of its arc length and two radii:
\(\text{Perimeter} = L_2 + 2r_2\)
\(\text{Perimeter} = \frac{32}{3}\pi + 2(9) = 18 + \frac{32}{3}\pi\text{ cm}\)

- **M1**: Calculates the fraction of the circle (\(\frac{1}{3}\)) or the angle of the first sector (\(120^\circ\)), OR correctly uses the relation \(\text{Area} = \frac{1}{2} L r\).
- **M1**: Calculates the area of the first sector as \(48\pi\).
- **M1**: Sets up a correct equation for the area of the second sector to find its fraction/angle or directly find its arc length, e.g., \(48\pi = \frac{1}{2} \times L_2 \times 9\) or \(48\pi = \frac{\phi}{360} \times 81\pi\).
- **M1**: Correctly calculates the arc length of the second sector as \(\frac{32}{3}\pi\) (or equivalent exact fraction).
- **A1**: Gives the final perimeter as \(18 + \frac{32}{3}\pi\) or \(\frac{32}{3}\pi + 18\) (or equivalent simplified exact form, e.g., \(\frac{54 + 32\pi}{3}\)).

Since \(y\) is inversely proportional to \(x^2\), we can write:
\(y = \frac{k}{x^2}\) for some constant \(k\).

Substitute \(x = 3\) and \(y = 16\) to find \(k\):
\(16 = \frac{k}{3^2}\)
\(16 = \frac{k}{9}\)
\(k = 16 \times 9 = 144\)

So, the formula is:
\(y = \frac{144}{x^2}\)

Since \(z\) is directly proportional to \(\sqrt{y}\), we can write:
\(z = c\sqrt{y}\) for some constant \(c\).

Substitute \(y = 16\) and \(z = 12\) to find \(c\):
\(12 = c\sqrt{16}\)
\(12 = 4c\)
\(c = 3\)

So, the formula is:
\(z = 3\sqrt{y}\)

We need to find the value of \(x\) when \(z = 18\).
First, use the second formula to find \(y\):
\(18 = 3\sqrt{y}\)
\(\sqrt{y} = 6\)
\(y = 36\)

Next, use the first formula to find \(x\):
\(36 = \frac{144}{x^2}\)
\(36x^2 = 144\)
\(x^2 = 4\)

Since \(x\) must be a positive value:
\(x = 2\)

M1: For establishing \(y = \frac{k}{x^2}\) (or equivalent) and substituting \(x=3\) and \(y=16\) to find \(k = 144\).

M1: For establishing \(z = c\sqrt{y}\) (or equivalent) and substituting \(y=16\) and \(z=12\) to find \(c = 3\).

M1: For setting up the equation \(18 = 3\sqrt{y}\) and solving to find \(y = 36\).

M1: For substituting \(y = 36\) into their first equation and finding \(x^2 = 4\).

A1: For 2 (accept \(x = 2\) only; do not accept \(\pm 2\)).

First, we find the formula for the area of the sector of the circle:
\(\text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{120}{360} \times \pi r^2 = \frac{1}{3} \pi r^2\)

Next, we find the formula for the area of the rectangle:
\(\text{Area of rectangle} = \text{length} \times \text{width} = (r + 5) \times 2\pi = 2\pi(r + 5)\)

Since the two areas are equal, we set them equal to each other:
\(\frac{1}{3} \pi r^2 = 2\pi(r + 5)\)

Divide both sides by \(\pi\):
\(\frac{1}{3} r^2 = 2(r + 5)\)

Multiply both sides by 3 to clear the fraction:
\(r^2 = 6(r + 5)\)

Expand the right side:
\(r^2 = 6r + 30\)

Rearrange into a standard quadratic equation form:
\(r^2 - 6r - 30 = 0\)

To solve this, we can complete the square:
\((r - 3)^2 - 9 - 30 = 0\)
\((r - 3)^2 = 39\)
\(r - 3 = \pm\sqrt{39}\)
\(r = 3 \pm\sqrt{39}\)

Since the radius \(r\) must be positive, and \(\sqrt{39} \approx 6.24\), the value \(3 - \sqrt{39}\) is negative. Therefore, we discard the negative root.

Thus, the positive value is:
\(r = 3 + \sqrt{39}\)

M1: For a correct expression for the area of the sector: \(\frac{120}{360}\pi r^2\) or \(\frac{1}{3}\pi r^2\).

M1: For a correct expression for the area of the rectangle: \(2\pi(r + 5)\) or \(2\pi r + 10\pi\).

M1: For equating the two area expressions and eliminating \(\pi\) to form a quadratic equation, e.g., \(r^2 - 6r - 30 = 0\).

M1: For a correct method to solve their quadratic equation, such as completing the square: \((r - 3)^2 - 39 = 0\) or using the quadratic formula: \(\frac{6 \pm \sqrt{156}}{2}\).

A1: For \(3 + \sqrt{39}\) (or equivalent exact form, but must discard the negative root).

The area of the rectangle is:
\(\text{Area}_{\text{rect}} = (3x + 1)(2x - 3)\)

Multiplying out the brackets:
\((3x + 1)(2x - 3) = 6x^2 - 9x + 2x - 3 = 6x^2 - 7x - 3\)

The area of the square is:
\(\text{Area}_{\text{square}} = (x + 2)^2\)

Multiplying out the brackets:
\((x + 2)^2 = x^2 + 4x + 4\)

Since the area of the rectangle is equal to the area of the square:
\(6x^2 - 7x - 3 = x^2 + 4x + 4\)

Rearranging the equation to make one side equal to zero:
Subtract \(x^2\) from both sides:
\(5x^2 - 7x - 3 = 4x + 4\)

Subtract \(4x\) from both sides:
\(5x^2 - 11x - 3 = 4\)

Subtract \(4\) from both sides:
\(5x^2 - 11x - 7 = 0\)

This completes the proof.

M1: For expanding the area of the rectangle: \((3x + 1)(2x - 3) = 6x^2 - 9x + 2x - 3\) (accept at least 3 correct terms out of 4)
A1: For simplifying the rectangle's area to \(6x^2 - 7x - 3\) AND correctly expanding the square's area to \(x^2 + 4x + 4\)
M1: For equating their two algebraic areas: \(6x^2 - 7x - 3 = x^2 + 4x + 4\) (or equivalent)
A1: For fully correct algebraic rearrangement leading to the given equation \(5x^2 - 11x - 7 = 0\) with no errors seen

Let the original number of red counters be \(3x\) and the original number of blue counters be \(4x\), where \(x\) is a positive integer constant.

After adding 9 red counters, the number of red counters is \(3x + 9\).
After removing 3 blue counters, the number of blue counters is \(4x - 3\).

The new ratio is \(6 : 5\), which gives the equation:
\(\frac{3x + 9}{4x - 3} = \frac{6}{5}\)

Cross-multiply to solve for \(x\):
\(5(3x + 9) = 6(4x - 3)\)
\(15x + 45 = 24x - 18\)

Subtract \(15x\) from both sides:
\(45 = 9x - 18\)

Add 18 to both sides:
\(63 = 9x\)

Divide by 9:
\(x = 7\)

The original number of red counters was:
\(3x = 3 \times 7 = 21\)

Thus, the original number of red counters in the bag was indeed 21.

M1: For expressing the original quantities algebraically as \(3x\) and \(4x\) (or using \(r\) and \(b\) with \(r = \frac{3}{4}b\))
M1: For setting up a correct equation representing the new ratio, e.g. \(\frac{3x+9}{4x-3} = \frac{6}{5}\) (or \(\frac{r+9}{b-3} = \frac{6}{5}\))
M1: For correctly cross-multiplying and solving their linear equation to find the value of their variable (yielding \(x = 7\) or equivalent)
A1: For showing that the original number of red counters is \(3 \times 7 = 21\) with fully correct working

**Method 1:**
1. Angle \(ACB = 90^\circ\) because the angle subtended by a diameter at the circumference (angle in a semicircle) is a right angle.
2. \(OA = OC\) because both are radii of the circle.
3. Therefore, triangle \(AOC\) is an isosceles triangle, which means angle \(ACO = \angle CAB = x\) (base angles of an isosceles triangle are equal).
4. Angle \(BCO = \angle ACB - \angle ACO = 90^\circ - x\).

**Method 2:**
1. Angle \(ACB = 90^\circ\) because the angle in a semicircle is a right angle.
2. In the right-angled triangle \(ABC\), angle \(ABC = 180^\circ - 90^\circ - x = 90^\circ - x\) (angles in a triangle sum to \(180^\circ\)).
3. \(OB = OC\) because both are radii of the circle.
4. Therefore, triangle \(BOC\) is an isosceles triangle, which means angle \(BCO = \angle OBC = 90^\circ - x\) (base angles of an isosceles triangle are equal).

M1: For stating that angle \(ACB = 90^\circ\) with the correct reason (e.g., 'angle in a semicircle is \(90^\circ\)' or 'angle subtended by a diameter is a right angle')
M1: For identifying equal radii (e.g., \(OA = OC\) or \(OB = OC\)) to conclude that either triangle \(AOC\) or triangle \(BOC\) is isosceles
A1: For finding either angle \(ACO = x\) (with reason 'base angles of an isosceles triangle are equal') OR angle \(ABC = 90^\circ - x\) (with reason 'angles in a triangle add up to \(180^\circ\)')
A1: For obtaining angle \(BCO = 90^\circ - x\) with clear, complete, and correct geometric reasons throughout

Let the consecutive odd integers be \(2n - 1\) and \(2n + 1\), where \(n\) is an integer. The sum of their squares is \((2n - 1)^2 + (2n + 1)^2\). Expanding the brackets gives \((4n^2 - 4n + 1) + (4n^2 + 4n + 1)\). Simplifying this expression yields \(8n^2 + 2\). Since \(n\) is an integer, \(n^2\) is also an integer, which means \(8n^2\) is a multiple of 8. Therefore, \(8n^2 + 2\) is 2 more than a multiple of 8.

M1: Writes expressions for two consecutive odd integers, e.g., \(2n - 1\) and \(2n + 1\), where \(n\) is an integer. M1: Sets up the sum of their squares and expands both brackets, allowing for at most one error. A1: Simplifies the algebraic expression to \(8n^2 + 2\) (or equivalent form like \(8n^2 + 16n + 10\) if using \(2n+1\) and \(2n+3\)). A1: Factorises or structures the expression as \(8(n^2) + 2\) (or \(8(n^2+2n+1)+2\)) and concludes with a clear statement that \(8(n^2)\) is a multiple of 8, hence the sum is 2 more.

Since \(y\) is inversely proportional to the square of \(x\), we have \(y = \frac{k}{x^2}\) for some constant \(k\). Substitute the given values \(x = 5\) and \(y = 12\) to find \(k\): \(12 = \frac{k}{5^2}\) which gives \(k = 12 \times 25 = 300\). Thus, the formula is \(y = \frac{300}{x^2}\). Now, substitute \(y = 3\) into the formula: \(3 = \frac{300}{x^2}\). Rearranging gives \(3x^2 = 300\), so \(x^2 = 100\). Since we are looking for the positive value of \(x\), we take the positive square root: \(x = \sqrt{100} = 10\).

M1 for setting up the proportional relationship, e.g., \(y = \frac{k}{x^2}\) or \(y \propto \frac{1}{x^2}\). M1 for finding the constant of proportionality \(k = 300\) or for writing \(3 \times x^2 = 12 \times 5^2\). A1 for the final answer of 10.

Let the original number of part-time workers be \(3x\) and the original number of full-time workers be \(5x\). When \(\frac{1}{3}\) of the part-time workers leave, the number of remaining part-time workers is \(3x - \frac{1}{3}(3x) = 2x\). After 15 new full-time workers are hired, the number of full-time workers is \(5x + 15\). The new ratio of part-time to full-time workers is \(1 : 3\), so we can write the equation: \(\frac{2x}{5x + 15} = \frac{1}{3}\). Cross-multiplying gives: \(3(2x) = 1(5x + 15)\), which simplifies to \(6x = 5x + 15\). Solving for \(x\) gives \(x = 15\). The total number of workers before the restructure was \(3x + 5x = 8x\). Substituting \(x = 15\) gives \(8 \times 15 = 120\) workers.

M1 for expressing the new quantities in terms of \(x\), e.g., \(2x\) and \(5x + 15\), and setting up the ratio equation \(\frac{2x}{5x + 15} = \frac{1}{3}\) (or equivalent). M1 for correctly solving for \(x = 15\) (or finding the original parts). A1 for 120.

The area of a rectangle is found by multiplying its length and width: \(\text{Area} = \text{length} \times \text{width}\). Substitute the given values: \((2x + 1)(x - 2) = 35\). Expand the brackets: \(2x^2 - 4x + x - 2 = 35\), which simplifies to \(2x^2 - 3x - 2 = 35\). Subtract 35 from both sides to form a quadratic equation equal to zero: \(2x^2 - 3x - 37 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = -3\), and \(c = -37\): \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-37)}}{2(2)}
\), which simplifies to \(x = \frac{3 \pm \sqrt{9 + 296}}{4} = \frac{3 \pm \sqrt{305}}{4}\). Calculating the two possible values of \(x\): \(x = \frac{3 + 17.4642...}{4} \approx 5.116\) and \(x = \frac{3 - 17.4642...}{4} \approx -3.616\). Since the width of the rectangle \((x - 2)\) must be a positive length, \(x\) must be greater than 2. Therefore, we reject the negative solution. Rounding \(x = 5.116...\) to 3 significant figures gives \(x = 5.12\).

M1 for setting up the equation \((2x+1)(x-2) = 35\) and expanding to get a quadratic equation of the form \(2x^2 - 3x - 37 = 0\) (or equivalent). M1 for applying the quadratic formula correctly to find the positive root. A1 for 5.12 (accept 5.116 to 5.12, but reject negative root).

1. Find the arc length of the sector by subtracting the two straight edges (radii) from the perimeter: \(\text{Arc length} = 24.2 - 2 \times 8.5\) which gives \(\text{Arc length} = 24.2 - 17 = 7.2\text{ cm}\). 2. Set up the equation for the arc length of a sector: \(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\), so \(7.2 = \frac{\theta}{360} \times 2 \times \pi \times 8.5\), which simplifies to \(7.2 = \frac{17\pi \theta}{360}\). 3. Rearrange to solve for \(\theta\): \(\theta = \frac{7.2 \times 360}{17\pi}\), so \(\theta = \frac{2592}{17\pi} \approx 48.5328...\). Correct to 1 decimal place, \(\theta = 48.5^\circ\).

M1 for calculating the arc length: \(24.2 - 2 \times 8.5 = 7.2\text{ cm}\)
M1 for setting up the equation for arc length and rearranging to find \(\theta\), e.g., \(\theta = \frac{7.2 \times 360}{17\pi}\) (or equivalent)
A1 for \(48.5\) (accept answers in the range \(48.5\) to \(48.6\))

1. Express the inverse proportion relationship as an equation: \(y = \frac{k}{x^2}\) where \(k\) is a constant. 2. Substitute the given values \(x = 2.5\) and \(y = 12.8\) to find \(k\): \(12.8 = \frac{k}{2.5^2}\), which gives \(12.8 = \frac{k}{6.25}\), so \(k = 12.8 \times 6.25 = 80\). The equation connecting \(y\) and \(x\) is \(y = \frac{80}{x^2}\). 3. Substitute \(x = 8\) into the equation to find \(y\): \(y = \frac{80}{8^2}\), which gives \(y = \frac{80}{64} = 1.25\).

M1 for setting up the correct equation \(y = \frac{k}{x^2}\) and substituting \(x = 2.5\), \(y = 12.8\) to find \(k = 80\)
M1 for substituting \(x = 8\) into their equation with their value of \(k\), e.g., \(y = \frac{80}{8^2}\)
A1 for \(1.25\) (or equivalent fraction, e.g., \(\frac{5}{4}\) or \(1\frac{1}{4}\))

1. Find the initial number of administrators and engineers: Let the number of administrators be \(3x\) and the number of engineers be \(7x\). The difference in parts is \(7 - 3 = 4\) parts. Since there are 32 more engineers than administrators, we have \(4x = 32\implies x = 8\). Thus, initially: Number of administrators = \(3 \times 8 = 24\), and Number of engineers = \(7 \times 8 = 56\). 2. Find the new number of administrators after hiring: The number of engineers does not change, so it remains \(56\). The new ratio of administrators to engineers is \(5 : 8\). Let \(A_{\text{new}}\) be the new number of administrators: \(\frac{A_{\text{new}}}{56} = \frac{5}{8}\), which gives \(A_{\text{new}} = \frac{5}{8} \times 56 = 35\). 3. Calculate the number of new administrators hired: \(35 - 24 = 11\).

M1 for finding the initial number of engineers as \(56\) (or administrators as \(24\))
M1 for calculating the new number of administrators as \(35\) (by using their initial number of engineers and the new ratio \(5:8\))
A1 for \(11\)

Since \(t\) is inversely proportional to \(v\), we can write:

\(t = \frac{k}{v}\)

Substitute the known values \(v = 80\) and \(t = 4.5\) to find the constant \(k\):

\(4.5 = \frac{k}{80}\)

\(k = 4.5 \times 80 = 360\)

Now, find \(t\) when \(v = 100\):

\(t = \frac{360}{100} = 3.6\text{ hours}\)

To convert \(3.6\) hours into hours and minutes:

\(3\text{ hours and } 0.6 \times 60\text{ minutes} = 3\text{ hours } 36\text{ minutes}\).

M1 for setting up a correct relationship, e.g. \(t = \frac{k}{v}\) or \(t_1 v_1 = t_2 v_2\), and substituting values to find \(k = 360\) (or showing \(80 \times 4.5 = 100 \times t\))

M1 for finding \(t = 3.6\) hours

A1 for 3 hours 36 minutes (accept 3h 36m)

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

After a \(15\%\) reduction, the price is:
\(P \times (1 - 0.15) = 0.85P\)

After a further \(10\%\) reduction, the price is:
\(0.85P \times (1 - 0.10) = 0.85P \times 0.90 = 0.765P\)

We are given that the final clearance price is \(\pounds 535.50\), so:
\(0.765P = 535.50\)

Solve for \(P\):
\(P = \frac{535.50}{0.765} = 700\)

So the original price of the laptop was \(\pounds 700\).

M1 for calculating the combined multiplier: \(0.85 \times 0.90 = 0.765\) (or equivalent equation setup like \(0.9x = 535.50\))

M1 for \(\frac{535.50}{0.765}\) or finding the price before the second reduction: \(\frac{535.50}{0.9} = 595\), then \(\frac{595}{0.85}\)

A1 for 700 (accept \pounds700)

To solve the equation, we first multiply all terms by the common denominator, which is \(x(x+2)\):

\(6(x+2) + 3x = 2x(x+2)\)

Expand the brackets:

\(6x + 12 + 3x = 2x^2 + 4x\)

Combine like terms:

\(9x + 12 = 2x^2 + 4x\)

Rearrange into a quadratic equation equal to zero:

\(2x^2 - 5x - 12 = 0\)

Factorise the quadratic expression:

\(2x^2 - 8x + 3x - 12 = 0\)

\(2x(x - 4) + 3(x - 4) = 0\)

\((2x + 3)(x - 4) = 0\)

This gives two solutions:

\(2x + 3 = 0 \implies x = -1.5\)

\(x - 4 = 0 \implies x = 4\)

Since we are looking for the positive value of \(x\), we have \(x = 4\).

M1 for multiplying through by \(x(x+2)\) to obtain a correct equation without denominators: \(6(x+2) + 3x = 2x(x+2)\) or equivalent

M1 for rearranging into a standard quadratic form: \(2x^2 - 5x - 12 = 0\) (allow one sign or arithmetic error)

A1 for 4 (accept \(x=4\) and must reject or omit the negative solution for full marks)

Let the original price of the coat be \(x\).

First, we find the overall multiplier for the two changes:
1. A 15% increase corresponds to a multiplier of \(1.15\).
2. A 20% decrease corresponds to a multiplier of \(1 - 0.20 = 0.80\).

The combined multiplier is:
\(1.15 \times 0.80 = 0.92\)

This means the sale price is 92% of the original price.

We set up the equation:
\(0.92x = 82.80\)

To find \(x\), we divide:
\(x = \frac{82.80}{0.92} = 90\)

**Alternative method (working backwards):**
1. Find the price before the 20% reduction:
\(£82.80 \div 0.80 = £103.50\)

2. Find the original price before the 15% increase:
\(£103.50 \div 1.15 = £90\)

The original price of the coat was £90.

- **M1**: For a correct method to reverse the 20% reduction, e.g. \(82.80 \div 0.8\) (implied by 103.50), OR for finding the combined multiplier, e.g. \(1.15 \times 0.8\) (implied by 0.92 or 92%).
- **M1**: For a fully correct method to find the original price, e.g. \(\text{their } 103.50 \div 1.15\) or \(82.80 \div 0.92\).
- **A1**: For 90 (accept £90 or 90.00).

First, find the midpoints of each interval: \(0 < t \le 10\) midpoint is 5; \(10 < t \le 20\) midpoint is 15; \(20 < t \le 40\) midpoint is 30; \(40 < t \le 50\) midpoint is 45. Next, set up the equation for the estimated mean: \(\text{Mean} = \frac{\sum f t}{\sum f}\). The sum of products of frequencies and midpoints is \(\sum f t = (12 \times 5) + (18 \times 15) + (w \times 30) + (10 \times 45) = 60 + 270 + 30w + 450 = 780 + 30w\). The total frequency is \(\sum f = 12 + 18 + w + 10 = 40 + w\). Using the estimated mean of 24.4: \(\frac{780 + 30w}{40 + w} = 24.4\). Multiplying both sides by \(40 + w\) gives: \(780 + 30w = 24.4(40 + w)\) which simplifies to \(780 + 30w = 976 + 24.4w\). Rearranging to solve for \(w\): \(30w - 24.4w = 976 - 780\) so \(5.6w = 196\) which gives \(w = 35\). The total number of students is \(40 + 35 = 75\). To estimate the number of students who took more than 35 minutes: For the interval \(20 < t \le 40\) (width 20, frequency 35), the proportion of the interval above 35 is \(\frac{40 - 35}{20} = \frac{5}{20} = 0.25\). The estimated number of students in this part is \(0.25 \times 35 = 8.75\). All 10 students in the interval \(40 < t \le 50\) also took more than 35 minutes. Total students taking more than 35 minutes = \(8.75 + 10 = 18.75\). The percentage of students is \(\frac{18.75}{75} \times 100 = 25\%\).

M1: For calculating midpoints 5, 15, 30, 45. M1: For setting up the algebraic equation for the mean: \(780 + 30w = 24.4(40 + w)\) or equivalent and solving to find \(w = 35\). M1: For using linear interpolation to estimate the number of students between 35 and 40 minutes: \(\frac{5}{20} \times 35\) or 8.75. M1: For finding the total estimated number of students taking more than 35 minutes (8.75 + 10 = 18.75) and dividing by their total frequency (75). A1: For 25% (or 25).

Let \(L\) be the arc length of the sector. The perimeter of the sector is given by \(P = 2r + L = 25\), which means \(L = 25 - 2r\). The area of a sector can be expressed as \(A = \frac{1}{2} r L = 35\). Substitute the expression for \(L\) into the area formula: \(\frac{1}{2} r (25 - 2r) = 35\). Expand and simplify the equation: \(r(25 - 2r) = 70\) which gives \(25r - 2r^2 = 70\). Rearrange into standard quadratic form: \(2r^2 - 25r + 70 = 0\). Solve this using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -25\), and \(c = 70\). This gives \(r = \frac{25 \pm \sqrt{(-25)^2 - 4(2)(70)}}{2(2)} = \frac{25 \pm \sqrt{625 - 560}}{4} = \frac{25 \pm \sqrt{65}}{4}\). Since \(\sqrt{65} \approx 8.0623\), we have two solutions: \(r = \frac{25 + 8.0623}{4} \approx 8.2656\) and \(r = \frac{25 - 8.0623}{4} \approx 4.2344\). Correct to 3 significant figures, the two possible values of \(r\) are 8.27 m and 4.23 m.

M1: For writing an expression for the perimeter, e.g., \(2r + L = 25\). M1: For writing an expression for the area, e.g., \(\frac{1}{2}r L = 35\) or equivalent. M1: For eliminating \(L\) to form a quadratic equation in \(r\), e.g., \(2r^2 - 25r + 70 = 0\). M1: For a correct method to solve their quadratic equation, such as the quadratic formula. A1: For both correct values of \(r\): 4.23 and 8.27 (accept 4.23 m and 8.27 m, must be 3 significant figures).

Since \(y\) is inversely proportional to the square of \(x\), we can write the formula as \(y = \frac{k}{x^2}\) where \(k\) is a constant. Using the first set of values: \(12 = \frac{k}{d^2}\) which gives \(k = 12d^2\). Using the second set of values: \(3 = \frac{k}{(d + 2)^2}\) which gives \(k = 3(d + 2)^2\). Since \(k\) is constant, we can set the two expressions equal to each other: \(12d^2 = 3(d + 2)^2\). Divide both sides by 3 to simplify: \(4d^2 = (d + 2)^2\). Take the square root of both sides (since \(d > 0\) and \(d + 2 > 0\)): \(2d = d + 2\). Subtract \(d\) from both sides to find \(d\): \(d = 2\). Now, substitute \(d = 2\) back to find the constant \(k\): \(k = 12(2)^2 = 48\). Thus, the formula connecting \(y\) and \(x\) is \(y = \frac{48}{x^2}\). Finally, substitute \(x = 5\) into the formula to find \(y\): \(y = \frac{48}{5^2} = \frac{48}{25} = 1.92\).

M1: For setting up the relationship \(y = \frac{k}{x^2}\) or equivalent. M1: For writing two equations in terms of \(d\) and \(k\), e.g., \(k = 12d^2\) and \(k = 3(d+2)^2\). M1: For equating and solving to find \(d = 2\). M1: For finding \(k = 48\) and substituting \(x = 5\) into their formula. A1: For 1.92 (or equivalent fraction like \(\frac{48}{25}\) or \(1\frac{23}{25}\)).

1. Set up the inverse proportion equation: \(T = \frac{k}{\sqrt{W}}\). 2. Use the given values to find the constant of proportionality, \(k\): \(12 = \frac{k}{\sqrt{9}} \implies 12 = \frac{k}{3} \implies k = 36\). So the formula is: \(T = \frac{36}{\sqrt{W}}\). 3. Calculate the number of workers required to complete the job in 4.5 hours: \(4.5 = \frac{36}{\sqrt{W}} \implies \sqrt{W} = \frac{36}{4.5} \implies \sqrt{W} = 8 \implies W = 8^2 = 64\). 4. Determine the number of additional workers needed: \(64 - 16 = 48\).

M1: Set up the correct proportion equation \(T = \frac{k}{\sqrt{W}}\) (or equivalent). A1: Find \(k = 36\). M1: Substitute \(T = 4.5\) and attempt to rearrange for \(\sqrt{W}\) or \(W\). A1: Find \(W = 64\). A1: Correct final answer of 48 (subtracting 16 from 64).

1. The formula for the perimeter of a sector is: \(\text{Perimeter} = 2r + \frac{\theta}{360} \times 2\pi r\). 2. Substitute the given perimeter and angle into the formula: \(38 = 2r + \frac{120}{360} \times 2\pi r \implies 38 = 2r + \frac{2\pi r}{3}\). 3. Factorise \(r\) to solve for the radius: \(38 = r \left(2 + \frac{2\pi}{3}\right) \implies r = \frac{38}{2 + \frac{2\pi}{3}} \approx 9.281\text{ cm}\). 4. Calculate the area of the sector: \(\text{Area} = \frac{120}{360} \times \pi r^2 = \frac{1}{3} \pi r^2 \implies A = \frac{1}{3} \pi (9.281)^2 \approx 90.203\text{ cm}^2\). 5. Round to 3 significant figures: \(90.2\text{ cm}^2\).

M1: Writes a correct expression for the total perimeter, e.g. \(2r + \frac{120}{360} \times 2\pi r = 38\). M1: Factorises and rearranges to make \(r\) the subject, e.g. \(r = \frac{38}{2 + 2\pi/3}\). A1: Obtains \(r \approx 9.28\) (or better). M1: Uses the sector area formula with their value of \(r\). A1: Correct final answer of 90.2 (accept 90.1 to 90.3).

(a) Probability of first green counter is \(\frac{7}{n}\). Since the counter is not replaced, the probability of the second green counter is \(\frac{6}{n-1}\). The probability of getting two green counters is: \(\frac{7}{n} \times \frac{6}{n-1} = \frac{42}{n(n-1)}\). We are given this probability is \(\frac{7}{15}\): \(\frac{42}{n(n-1)} = \frac{7}{15} \implies 42 \times 15 = 7n(n-1) \implies 630 = 7n^2 - 7n\). Divide every term by 7: \(90 = n^2 - n \implies n^2 - n - 90 = 0\). (b) Solve the quadratic equation: \((n - 10)(n + 9) = 0\). Since \(n\) must be positive, \(n = 10\). The number of blue counters is \(10 - 7 = 3\).

M1: Formulates the probability of two greens without replacement, e.g. \(\frac{7}{n} \times \frac{6}{n-1}\). M1: Equates to \(\frac{7}{15}\) and demonstrates clear algebraic manipulation to remove fractions. A1: Correctly derives the given quadratic equation \(n^2 - n - 90 = 0\) with no errors shown. M1: Solves the quadratic equation to find \(n = 10\). A1: Correctly calculates 3 blue counters.

Let the number of red counters be \(3x\) and the number of blue counters be \(5x\), where \(x\) is a positive integer. The total number of counters is \(3x + 5x = 8x\). The probability of choosing two red counters without replacement is: \(P(R, R) = \frac{3x}{8x} \times \frac{3x-1}{8x-1} = \frac{3(3x-1)}{8(8x-1)}\). The probability of choosing two blue counters without replacement is: \(P(B, B) = \frac{5x}{8x} \times \frac{5x-1}{8x-1} = \frac{5(5x-1)}{8(8x-1)}\). The probability that both counters are the same colour is: \(P(R, R) + P(B, B) = \frac{3(3x-1) + 5(5x-1)}{8(8x-1)} = \frac{9x - 3 + 25x - 5}{8(8x-1)} = \frac{34x - 8}{8(8x-1)} = \frac{17x - 4}{4(8x-1)}\). We are given that this probability is \(\frac{16}{31}\): \(\frac{17x - 4}{32x - 4} = \frac{16}{31}\). Cross-multiplying gives: \(31(17x - 4) = 16(32x - 4)\), which simplifies to \(527x - 124 = 512x - 64\). Rearranging to solve for \(x\): \(15x = 60\), so \(x = 4\). The total number of counters originally in the bag is \(8x = 8 \times 4 = 32\).

M1: For representing the number of red, blue, and total counters as \(3x\), \(5x\), and \(8x\) respectively. M1: For writing a correct expression for the probability of selecting two red counters or two blue counters, e.g., \(\frac{3x}{8x} \times \frac{3x-1}{8x-1}\) or \(\frac{5x}{8x} \times \frac{5x-1}{8x-1}\). M1: For equating the sum of the two probabilities to \(\frac{16}{31}\) to form an equation: \(\frac{34x-8}{8(8x-1)} = \frac{16}{31}\) (or equivalent). M1: For algebraic simplification leading to a linear equation in \(x\), e.g., \(15x = 60\). A1: For the correct total number of counters: 32.

The area of a trapezium is given by \(\frac{a+b}{2} \times h\). Here, the parallel sides are \(AB = x\) and \(CD = x + 6\), and the perpendicular height is \(AD = 2x\). \(\text{Area of trapezium} = \frac{x + (x + 6)}{2} \times 2x = (2x + 6) \times x = 2x^2 + 6x\). The semicircle has diameter \(AD = 2x\), which means its radius is \(r = x\). \(\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi x^2\). The remaining area is the area of the trapezium minus the area of the semicircle: \(\text{Remaining Area} = 2x^2 + 6x - \frac{1}{2}\pi x^2 = 120\). This can be written as: \(x^2(2 - 0.5\pi) + 6x - 120 = 0\). Using \(\pi \approx 3.14159\), the coefficient of \(x^2\) is \(2 - 0.5\pi \approx 0.4292\). This gives the quadratic equation: \(0.4292x^2 + 6x - 120 = 0\). Solving this using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \(x = \frac{-6 \pm \sqrt{6^2 - 4(0.4292)(-120)}}{2(0.4292)}\), which simplifies to \(x = \frac{-6 \pm \sqrt{36 + 206.016}}{0.8584}\), so \(x = \frac{-6 \pm \sqrt{242.016}}{0.8584}\). Taking the positive square root for a positive length: \(x = \frac{-6 + 15.557}{0.8584} \approx 11.133\). Correct to 3 significant figures, \(x = 11.1\).

M1: For a correct expression for the area of the trapezium, e.g., \(\frac{x + x + 6}{2} \times 2x\) or \(2x^2 + 6x\). M1: For a correct expression for the area of the semicircle, e.g., \(\frac{1}{2} \pi x^2\). M1: For setting up the equation: \(2x^2 + 6x - \frac{1}{2}\pi x^2 = 120\) (or equivalent). M1: For rearranging into the standard quadratic form \(ax^2 + bx + c = 0\) and showing a valid method to solve it, e.g., applying the quadratic formula. A1: For \(11.1\) (accept answers in the range 11.1 to 11.13).

The area of a trapezium is given by the formula:

\(\text{Area} = \frac{1}{2}(a + b)h\)

Here, the parallel sides are \(a = x + 3\) and \(b = 3x - 1\), and the perpendicular height is \(h = 2x - 4\).

Substitute these expressions into the formula:

\(\text{Area} = \frac{1}{2}\left((x + 3) + (3x - 1)\right)(2x - 4)\)

Simplify the term inside the first brackets:

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

Substitute this back into the area equation:

\(\text{Area} = \frac{1}{2}(4x + 2)(2x - 4)\)

Multiply \(\frac{1}{2}\) by \((4x + 2)\):

\(\text{Area} = (2x + 1)(2x - 4)\)

Expand the brackets:

\(\text{Area} = 2x(2x) + 2x(-4) + 1(2x) + 1(-4)\)
\(\text{Area} = 4x^2 - 8x + 2x - 4\)
\(\text{Area} = 4x^2 - 6x - 4\)

Thus, the area of the trapezium is \(4x^2 - 6x - 4\).

M1: For writing a correct expression for the area of the trapezium using the formula, e.g., \(\frac{1}{2}((x + 3) + (3x - 1))(2x - 4)\).
M1: For simplifying the sum of the parallel sides to obtain \(4x + 2\).
M1: For expanding \((2x + 1)(2x - 4)\) or \(\frac{1}{2}(8x^2 - 12x - 8)\) with at least 3 terms correct.
A1: For fully correct algebraic simplification leading to \(4x^2 - 6x - 4\) with all steps shown clearly.

Let the initial value of the book be \(V\).

An increase of \(2x\%\) in the first year means the value is multiplied by:

\(1 + \frac{2x}{100}\)

In the second year, a decrease of \(x\%\) of this new value means it is multiplied by:

\(1 - \frac{x}{100}\)

The final value \(V_{\text{final}}\) after two years is:

\(V_{\text{final}} = V \left(1 + \frac{2x}{100}\right)\left(1 - \frac{x}{100}\right)\)

Expand the brackets:

\(\left(1 + \frac{2x}{100}\right)\left(1 - \frac{x}{100}\right) = 1 - \frac{x}{100} + \frac{2x}{100} - \frac{2x^2}{10000}\)

\(= 1 + \frac{x}{100} - \frac{x^2}{5000}\)

This gives the overall multiplier. The fractional increase is:

\(\frac{x}{100} - \frac{x^2}{5000}\)

To find the overall percentage increase, multiply this fractional increase by 100:

\(\text{Percentage Increase} = \left(\frac{x}{100} - \frac{x^2}{5000}\right) \times 100\)

\(= x - \frac{100x^2}{5000}\)

\(= x - \frac{x^2}{50}\)

Therefore, the overall percentage increase is \(\left(x - \frac{x^2}{50}\right)\%\).

M1: For setting up an expression for the multiplier after the first year, e.g., \(1 + \frac{2x}{100}\) or \(1 + 0.02x\).
M1: For setting up the product of the two multipliers: \(\left(1 + \frac{2x}{100}\right)\left(1 - \frac{x}{100}\right)\).
M1: For expanding the product of the brackets correctly to get \(1 + \frac{x}{100} - \frac{2x^2}{10000}\) or equivalent.
A1: For multiplying the change by 100 to show the correct overall percentage of \(\left(x - \frac{x^2}{50}\right)\%\) with all steps clearly presented.

The formula for the area of a sector is:

\(A = \frac{\theta}{360} \pi r^2\)

The perimeter of the sector consists of the arc length plus two radii:

\(P = 2r + \frac{\theta}{360} (2\pi r)\)

We are given that \(A = 3P\). Substituting the expressions into this equation:

\(\frac{\theta}{360} \pi r^2 = 3 \left(2r + \frac{\theta}{360} (2\pi r)\right)\)

Expand the right-hand side:

\(\frac{\theta}{360} \pi r^2 = 6r + \frac{6\theta}{360} \pi r\)

Since \(r > 0\), we can divide the entire equation by \(r\):

\(\frac{\theta}{360} \pi r = 6 + \frac{6\theta}{360} \pi\)

Multiply the entire equation by 360 to clear the denominators:

\(\theta \pi r = 2160 + 6\theta \pi\)

Rearrange to group the terms with \(\theta\) on the left side:

\(\theta \pi r - 6\theta \pi = 2160\)

Factor out \(\theta\pi\) on the left-hand side:

\(\theta \pi (r - 6) = 2160\)

Divide both sides by \(\pi(r - 6)\) to solve for \(\theta\):

\(\theta = \frac{2160}{\pi(r - 6)}\)

This is the required expression.

M1: For writing correct expressions for the area \(A = \frac{\theta}{360}\pi r^2\) and the perimeter \(P = 2r + \frac{\theta}{360}(2\pi r)\).
M1: For equating \(A = 3P\) to set up the equation \(\frac{\theta}{360} \pi r^2 = 3 \left(2r + \frac{\theta}{360} (2\pi r)\right)\).
M1: For dividing by \(r\) and rearranging the terms to group \(\theta\), e.g., \(\theta\pi r - 6\theta\pi = 2160\).
A1: For fully correct algebraic work leading directly to the final expression \(\theta = \frac{2160}{\pi(r - 6)}\).

To find the area of the rectangle, we multiply its length and width: \(\text{Area of rectangle} = (2x + 7)(x + 3)\). Expanding the brackets gives: \((2x + 7)(x + 3) = 2x^2 + 6x + 7x + 21 = 2x^2 + 13x + 21\). To find the area of the square, we square its side length: \(\text{Area of square} = (x + 1)^2 = x^2 + 2x + 1\). We are given that the area of the rectangle is \(46\text{ cm}^2\) greater than the area of the square, which can be written as: \(\text{Area of rectangle} - \text{Area of square} = 46\). Substituting the expanded expressions into this equation: \((2x^2 + 13x + 21) - (x^2 + 2x + 1) = 46\). Simplifying the left-hand side gives: \(x^2 + 11x + 20 = 46\). Subtracting 46 from both sides to set the equation to 0: \(x^2 + 11x + 20 - 46 = 0\), which simplifies to: \(x^2 + 11x - 26 = 0\). This is the required equation.

**M1**: For an attempt to find the area of the rectangle by multiplying \((2x+7)(x+3)\) with at least 3 terms correct in the expansion \(2x^2 + 13x + 21\). **M1**: For an attempt to find the area of the square by expanding \((x+1)^2\) to obtain \(x^2 + 2x + 1\) (allow 1 sign or arithmetic error). **M1**: For setting up the correct equation relating the two areas, e.g., \((2x^2 + 13x + 21) - (x^2 + 2x + 1) = 46\). **A1**: For full, correct algebraic simplification leading to \(x^2 + 11x - 26 = 0\) with no errors in the working.