2024 OCR GCSE Mathematics - J560 Practice Paper with Answers

Let the ratio be Orange : Mango : Water = \(3 : 2 : 7\).

The number of parts for mango juice is \(2\).
We are given that \(2\) parts correspond to \(1.5\) litres.

First, find the volume of \(1\) part:
\(1.5 \div 2 = 0.75\) litres.

Next, find the total number of parts in the mixture:
\(3 + 2 + 7 = 12\) parts.

Finally, calculate the total volume:
\(12 \times 0.75 = 9\) litres.

- **M1**: for calculating the volume of a single part, e.g., \(1.5 \div 2\) (or \(0.75\)), or writing a ratio equivalent to \(3 : 2 : 7\) containing \(1.5\) in the correct position (e.g., \(2.25 : 1.5 : 5.25\)).
- **M1**: for summing the parts (\(3 + 2 + 7 = 12\)) and multiplying by their single-part volume, or summing individual volumes: \(2.25 + 1.5 + 5.25\).
- **A1**: for \(9\) (accept \(9\text{ litres}\)).

Multiply both sides of the equation by \(4\):
\(3x - 5 = 4(2x - 3)\)

Expand the brackets on the right-hand side:
\(3x - 5 = 8x - 12\)

Rearrange to group terms in \(x\) on one side and constant terms on the other:
\(12 - 5 = 8x - 3x\)
\(7 = 5x\)

Divide by \(5\):
\(x = \frac{7}{5} = 1.4\).

- **M1**: for correctly multiplying both sides by \(4\) to clear the fraction, e.g., \(3x - 5 = 4(2x - 3)\) or \(3x - 5 = 8x - 12\).
- **M1**: for isolating the variable terms on one side and numbers on the other in their linear equation, e.g., \(8x - 3x = 12 - 5\) or \(5x = 7\).
- **A1**: for \(1.4\) (or \(\frac{7}{5}\) or \(1\frac{2}{5}\)).

The formula for the circumference \(C\) of a circle is:
\(C = 2\pi r\)

Using the given circumference:
\(2\pi r = 28.5\)

\(r = \frac{28.5}{2\pi} \approx 4.5359\text{ cm}\)

The formula for the area \(A\) of a circle is:
\(A = \pi r^2\)

Using the calculated radius:
\(A = \pi \times (4.5359)^2 \approx 64.636\text{ cm}^2\)

Rounding to 3 significant figures gives:
\(A = 64.6\text{ cm}^2\).

- **M1**: for setting up an equation for the radius or diameter, e.g., \(r = \frac{28.5}{2\pi}\) (approx \(4.54\)) or \(d = \frac{28.5}{\pi}\) (approx \(9.07\)).
- **M1**: for substituting their calculated radius into the area formula \(\pi r^2\).
- **A1**: for \(64.6\) (accept answers in the range \(64.6\) to \(64.65\)).

First, find the area of the trapezium-shaped lawn:
\(\text{Area} = \frac{a + b}{2} \times h\)

\(\text{Area} = \frac{8.4 + 12.6}{2} \times 5.5 = \frac{21}{2} \times 5.5 = 10.5 \times 5.5 = 57.75\text{ m}^2\)

Next, determine the number of bags of fertiliser required:
\(\text{Number of bags} = \frac{57.75}{15} = 3.85\)

Since fertiliser is bought in whole bags, and we must cover the entire lawn, we round up to the next integer:
\(4\) bags.

- **M1**: for a correct calculation to find the area of the lawn, e.g., \(\frac{8.4 + 12.6}{2} \times 5.5\) (implied by \(57.75\)).
- **M1**: for dividing their area by 15, e.g., \(\frac{\text{their area}}{15}\) (approx \(3.85\)).
- **A1**: for \(4\) (must be an integer, rounded up from their quotient).

There are \(5 + 3 = 8\) counters in total initially.

We need the probability that both counters are red or both are blue.

Probability of drawing two red counters:
\(\text{P(Red, Red)} = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\)

Probability of drawing two blue counters:
\(\text{P(Blue, Blue)} = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\)

Probability that they are the same colour:
\(\text{P(same colour)} = \text{P(Red, Red)} + \text{P(Blue, Blue)} = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}\).

Converting to decimal, this is approximately \(0.464\) (to 3 s.f.).

- **M1**: for finding either \(\text{P(Red, Red)}\), e.g., \(\frac{5}{8} \times \frac{4}{7}\), or \(\text{P(Blue, Blue)}\), e.g., \(\frac{3}{8} \times \frac{2}{7}\), without replacement.
- **M1**: for summing their two individual non-replacement probabilities, e.g., \(\text{their P(Red, Red)} + \text{their P(Blue, Blue)}\).
- **A1**: for \(\frac{13}{28}\) or equivalent fraction (e.g., \(\frac{26}{56}\)), or decimal \(0.464\) (or \(0.464\dots\)), or percentage \(46.4\%\).

The sale price represents \(100\% - 15\% = 85\%\) of the original price.

Let the original price be \(P\).
\(0.85 \times P = 391\)

Divide by \(0.85\) to find the original price:
\(P = \frac{391}{0.85} = 460\)

The original price was \(\text{£}460\).

- **M1**: for identifying that \(\text{£}391\) represents \(85\%\) of the original price, e.g., \(85\% = 391\) or \(0.85P = 391\).
- **M1**: for a complete method to find the original price, e.g., \(391 \div 0.85\).
- **A1**: for \(460\) (accept \(\text{£}460\)).

Round each number to 1 significant figure:
- \(48.7 \approx 50\)
- \(98.4 \approx 100\), so \(\sqrt{98.4} \approx \sqrt{100} = 10\)
- \(0.193 \approx 0.2\)

Substitute these estimates into the expression:
\(\frac{50 \times 10}{0.2} = \frac{500}{0.2}\)

Calculate the final estimate:
\(\frac{500}{0.2} = 500 \times 5 = 2500\).

- **M1**: for rounding at least two of the three numbers correctly to 1 significant figure (e.g., \(48.7 \to 50\), \(\sqrt{98.4} \to 10\), or \(0.193 \to 0.2\)).
- **M1**: for a complete correct calculation using their rounded numbers, e.g., \(\frac{50 \times 10}{0.2}\).
- **A1**: for \(2500\) with clear working shown.

Since \(y\) is inversely proportional to the square of \(x\), we can write:
\(y = \frac{k}{x^2}\) where \(k\) is a constant.

Substitute the given values \(x = 4\) and \(y = 9\) into the equation to find \(k\):
\(9 = \frac{k}{4^2}\)
\(9 = \frac{k}{16}\)
\(k = 9 \times 16 = 144\)

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

Now, substitute \(x = 6\) into this equation:
\(y = \frac{144}{6^2}\)
\(y = \frac{144}{36} = 4\).

- **M1**: for setting up the correct proportional equation, e.g., \(y = \frac{k}{x^2}\) or \(y \propto \frac{1}{x^2}\).
- **M1**: for substituting \(x = 4\) and \(y = 9\) to find \(k = 144\), and then substituting \(x = 6\) into their formula.
- **A1**: for \(4\).

First, find the total number of parts in the ratio: \(5 + 3 + 2 = 10\) parts. Next, find the mass of each type of tea leaf in a \(4\text{ kg}\) blend: Assam: \(4 \times \frac{5}{10} = 2\text{ kg}\), Ceylon: \(4 \times \frac{3}{10} = 1.2\text{ kg}\), Darjeeling: \(4 \times \frac{2}{10} = 0.8\text{ kg}\). Now, calculate the cost of each component: Assam cost: \(2 \times 12 = \text{£}24\), Ceylon cost: \(1.2 \times 15 = \text{£}18\), Darjeeling cost: \(0.8 \times 22 = \text{£}17.60\). Sum these to find the total cost: \(24 + 18 + 17.60 = \text{£}59.60\). Alternatively, find the cost of \(1\text{ kg}\) of the blend first: \(\frac{5}{10} \times 12 + \frac{3}{10} \times 15 + \frac{2}{10} \times 22 = 6 + 4.5 + 4.4 = \text{£}14.90\) per kg. For \(4\text{ kg}\): \(14.90 \times 4 = \text{£}59.60\).

M1 for a method to find the quantity of at least one type of tea in a \(4\text{ kg}\) blend (e.g., \(4 \times \frac{5}{10}\) or \(2\text{ kg}\)) OR a method to find the cost of \(1\text{ kg}\) of the blend (e.g., \(0.5 \times 12 + 0.3 \times 15 + 0.2 \times 22\)). M1 for a complete correct method to find the total cost of the \(4\text{ kg}\) blend (e.g., \(2 \times 12 + 1.2 \times 15 + 0.8 \times 22\) or \(14.90 \times 4\)). A1 for £59.60 (accept 59.60 or 59.6).

First, calculate the area of the rectangle: \(\text{Area}_{\text{rectangle}} = 8 \times 5 = 40\text{ m}^2\). Next, calculate the area of the semicircle. The diameter is \(5\text{ m}\), so the radius \(r = \frac{5}{2} = 2.5\text{ m}\). \(\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2.5)^2 = 3.125 \pi \approx 9.817\text{ m}^2\). Add the two areas to find the total area: \(\text{Total Area} = 40 + 9.817 = 49.817\text{ m}^2\). Giving the answer to 3 significant figures results in \(49.8\text{ m}^2\).

M1 for finding the area of the rectangle: \(8 \times 5 = 40\). M1 for a correct method to find the area of the semicircle: \(\frac{1}{2} \times \pi \times 2.5^2\) (implied by \(9.82\) or \(3.125\pi\)). A1 for \(49.8\) (accept answers in the range \(49.8\) to \(49.82\)).

1. Find the frequency density (FD) for the first interval \(0 < t \le 5\): \(\text{FD} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{18}{5} = 3.6\). 2. Determine the scale factor for the height on the histogram: \(\text{Height} = 9\text{ cm}\) represents \(\text{FD} = 3.6\), so the scale factor is \(\frac{9}{3.6} = 2.5\text{ cm per unit of FD}\). 3. Use the height of the bar for the class \(15 < t \le 30\) to find its frequency density: \(\text{Height} = 3.5\text{ cm}\), so \(\text{FD} = \frac{3.5}{2.5} = 1.4\). 4. Calculate the frequency \(b\) for the class \(15 < t \le 30\): \(\text{Class Width} = 30 - 15 = 15\). Therefore, \(b = \text{FD} \times \text{Class Width} = 1.4 \times 15 = 21\). 5. Use the total frequency of 80 to find \(a\): \(18 + a + b + 12 = 80\), which simplifies to \(18 + a + 21 + 12 = 80 \implies 51 + a = 80 \implies a = 29\).

M1: For calculation of frequency density of first class: \(\frac{18}{5} = 3.6\) or scale factor of \(\frac{9}{3.6} = 2.5\). M1: For setting up equation for the third class frequency density: \(\frac{b}{15} \times 2.5 = 3.5\) or equivalent. A1: For \(b = 21\). M1: For using the sum of frequencies to set up the equation for \(a\): \(18 + a + 21 + 12 = 80\). A1: For \(a = 29\).

1. Let the initial numbers of administrators, sales representatives, and managers be \(3k\), \(8k\), and \(4k\) respectively. 2. Calculate the new quantities in terms of \(k\) after the percentage changes: Administrators: \(3k \times 1.20 = 3.6k\); Sales representatives: \(8k \times 0.85 = 6.8k\); Managers: \(4k \times 1.10 = 4.4k\). 3. Use the given number of sales representatives after the changes to find \(k\): \(6.8k = 272 \implies k = \frac{272}{6.8} = 40\). 4. Calculate the total number of employees after the changes: \(\text{Total} = 3.6k + 6.8k + 4.4k = 14.8k\). Substituting \(k = 40\), \(\text{Total} = 14.8 \times 40 = 592\).

M1: For applying percentage multipliers to the ratio parts: \(3 \times 1.2 = 3.6\), \(8 \times 0.85 = 6.8\), \(4 \times 1.1 = 4.4\). M1: For equating the new sales representative part to 272: \(6.8k = 272\). A1: For finding the constant multiplier \(k = 40\) (or equivalent scaling factor). M1: For finding the total number of parts after changes: \(3.6 + 6.8 + 4.4 = 14.8\) and multiplying by \(40\). A1: For 592.

1. Determine the dimensions of the combined area (lawn + path): Since the path is 1 metre wide all around, we add 2 metres to both the length and width of the lawn. Combined length = \((2x + 5) + 2 = 2x + 7\) metres; Combined width = \((x + 3) + 2 = x + 5\) metres. 2. Write down the equation for the combined area: \((2x + 7)(x + 5) = 104\). 3. Expand and simplify to form a quadratic equation: \(2x^2 + 10x + 7x + 35 = 104 \implies 2x^2 + 17x + 35 = 104 \implies 2x^2 + 17x - 69 = 0\). 4. Solve the quadratic equation by factoring: \((2x + 23)(x - 3) = 0\). Since \(x\) must be positive, \(x = 3\). 5. Calculate the area of the lawn: Lawn length = \(2(3) + 5 = 11\) m; Lawn width = \(3 + 3 = 6\) m; Lawn area = \(11 \times 6 = 66\text{ m}^2\).

M1: For expressing the combined dimensions as \((2x+7)\) and \((x+5)\). M1: For setting up the quadratic equation: \((2x+7)(x+5) = 104\) and expanding to \(2x^2 + 17x - 69 = 0\). M1: For solving the quadratic equation to find \(x = 3\) (neglecting the negative root). M1: For substituting \(x=3\) back into the original lawn dimensions to get 11 and 6. A1: For 66.

1. Identify the boundary components of the emblem's perimeter: Radius \(OA = 8\text{ cm}\), Arc \(AB\), Side \(BC = 6\text{ cm}\), and Hypotenuse \(OC\). 2. Calculate the length of the hypotenuse \(OC\) using Pythagoras' Theorem in right-angled triangle \(OBC\): \(OC = \sqrt{OB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\). 3. Calculate the length of the arc \(AB\): \(\text{Arc length} = \frac{130}{360} \times 2\pi \times 8 = \frac{13}{36} \times 16\pi = \frac{52}{9}\pi \approx 18.151\text{ cm}\). 4. Calculate the total perimeter: \(\text{Perimeter} = OA + \text{Arc } AB + BC + OC = 8 + 18.151 + 6 + 10 = 42.151\text{ cm}\). 5. Round to 3 significant figures: \(\text{Perimeter} \approx 42.2\text{ cm}\).

M1: For using Pythagoras' Theorem to find \(OC\): \(\sqrt{8^2+6^2}\). A1: For \(OC = 10\). M1: For using the arc length formula: \(\frac{130}{360} \times 2\pi \times 8\). A1: For arc length \(\approx 18.15\) or \(\frac{52}{9}\pi\). A1: For total perimeter 42.2 (accept 42.1 to 42.2 from correct working).

1. Calculate the area of the right-angled triangle \(ABC\): \(\text{Area of } ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 55 \times 72 = 1980\text{ m}^2\). 2. Use Pythagoras' Theorem to find the shared side \(AC\): \(AC^2 = AB^2 + BC^2 = 55^2 + 72^2 = 3025 + 5184 = 8209 \implies AC = \sqrt{8209} \approx 90.6035\text{ m}\). 3. Calculate the area of triangle \(ACD\) using the formula \(\frac{1}{2}ab\sin C\): \(\text{Area of } ACD = \frac{1}{2} \times AC \times AD \times \sin(\angle CAD) = \frac{1}{2} \times 90.6035 \times 80 \times \sin(42^\circ) \approx 3624.14 \times 0.66913 \approx 2425.02\text{ m}^2\). 4. Calculate the total area of the field: \(\text{Total Area} = \text{Area of } ABC + \text{Area of } ACD \approx 1980 + 2425.02 = 4405.02\text{ m}^2\). 5. Round to 3 significant figures: \(\text{Total Area} \approx 4410\text{ m}^2\).

M1: For calculating the area of triangle \(ABC\): \(\frac{1}{2} \times 55 \times 72 = 1980\). M1: For using Pythagoras to find \(AC\): \(\sqrt{55^2 + 72^2}\). A1: For \(AC \approx 90.6\) (or \(\sqrt{8209}\)). M1: For using the \(\frac{1}{2}ab\sin C\) formula: \(\frac{1}{2} \times 90.6 \times 80 \times \sin(42^\circ)\). A1: For 4410 (accept 4400 to 4420 from correct working).

1. Find the total number of chocolates: 12. 2. Find the number of non-dark chocolates: \(12 - 5 = 7\) (4 milk + 3 white). 3. Use the complement method: \(\text{P(at least one dark)} = 1 - \text{P(no dark chocolates)}\). 4. Calculate \(\text{P(no dark chocolates)}\) where both selected are non-dark: First chocolate is non-dark: \(\frac{7}{12}\); Second chocolate is non-dark: \(\frac{6}{11}\). \(\text{P(no dark chocolates)} = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\). 5. Calculate \(\text{P(at least one dark)}\): \(\text{P(at least one dark)} = 1 - \frac{7}{22} = \frac{15}{22}\).

M1: For identifying that there are 7 non-dark chocolates. M1: For writing the probability of first non-dark as \(\frac{7}{12}\) and second non-dark as \(\frac{6}{11}\). M1: For multiplying these probabilities: \(\frac{7}{12} \times \frac{6}{11}\). M1: For subtracting their product from 1: \(1 - \frac{42}{132}\). A1: For \(\frac{15}{22}\) (or equivalent fraction, e.g., \(\frac{90}{132}\); accept 0.682 or 68.2% with working shown).

1. Set up the proportion equations: \(P = k_1 x^2\) and \(Q = \frac{k_2}{\sqrt[3]{y}}\). 2. Find \(k_1\) using the given values for \(P\) and \(x\): \(18 = k_1 (3^2) \implies 18 = 9k_1 \implies k_1 = 2\). So, \(P = 2x^2\). 3. Find \(k_2\) using the given values for \(Q\) and \(y\): \(5 = \frac{k_2}{\sqrt[3]{8}} \implies 5 = \frac{k_2}{2} \implies k_2 = 10\). So, \(Q = \frac{10}{\sqrt[3]{y}}\). 4. Substitute these into \(P \times Q = 180\): \((2x^2) \times \left(\frac{10}{\sqrt[3]{y}}\right) = 180 \implies \frac{20x^2}{\sqrt[3]{y}} = 180\). 5. Rearrange the formula to make \(y\) the subject: \(\frac{x^2}{\sqrt[3]{y}} = 9 \implies \sqrt[3]{y} = \frac{x^2}{9}\). Cube both sides: \(y = \left(\frac{x^2}{9}\right)^3 = \frac{x^6}{729}\).

M1: For setting up \(P = k_1 x^2\) and finding \(k_1 = 2\) to get \(P = 2x^2\). M1: For setting up \(Q = \frac{k_2}{\sqrt[3]{y}}\) and finding \(k_2 = 10\) to get \(Q = \frac{10}{\sqrt[3]{y}}\). M1: For substituting both expressions into \(P \times Q = 180\) to obtain \(\frac{20x^2}{\sqrt[3]{y}} = 180\). M1: For isolating \(\sqrt[3]{y}\) as \(\frac{x^2}{9}\) or equivalent. A1: For \(y = \frac{x^6}{729}\) or \(y = (\frac{x^2}{9})^3\).

1. Let the original price Liam paid be \(P\). 2. Set up the equation using multipliers for each year's change: Year 1 multiplier is 1.12, Year 2 multiplier is 0.95, Year 3 multiplier is 1.08. Thus, \(P \times 1.12 \times 0.95 \times 1.08 = 34466.40\). 3. Calculate the overall multiplier: \(1.12 \times 0.95 \times 1.08 = 1.14912\). 4. Solve for the original price \(P\): \(P \times 1.14912 = 34466.40 \implies P = \frac{34466.40}{1.14912} = \$30,000\). 5. Calculate the profit made: \(\text{Profit} = \text{Final Value} - \text{Original Price} = \$34,466.40 - \$30,000 = \$4,466.40\).

M1: For identifying the three percentage multipliers: 1.12, 0.95, and 1.08. M1: For multiplying the three multipliers together: \(1.12 \times 0.95 \times 1.08 = 1.14912\). M1: For setting up the reverse percentage equation: \(P \times 1.14912 = 34466.40\). A1: For finding the original price of \(\$30,000\). A1: For the final profit of 4466.40 (or \(\$4,466.40\)).

1. **Express the perimeter in terms of \(x\) and \(y\):**
The boundary of the shape consists of three sides of the rectangle (one of length \(2x\) and two of length \(y\)) and the curved arc of the semicircle.
The length of the semicircular arc is:
\(\text{Arc} = \pi \times x\)
So, the total perimeter is:
\(P = 2y + 2x + \pi x\)
Given that \(P = 40\):
\(2y + 2x + \pi x = 40\)
\(2y = 40 - 2x - \pi x\)
\(y = 20 - x - \frac{\pi}{2}x\)

2. **Find the total area, \(A\), of the shape:**
The total area is the sum of the area of the rectangle and the area of the semicircle.
\(A_{\text{rectangle}} = 2x \times y = 2x\left(20 - x - \frac{\pi}{2}x\right) = 40x - 2x^2 - \pi x^2\)
\(A_{\text{semicircle}} = \frac{1}{2} \pi x^2\)

Combine the two areas:
\(A = A_{\text{rectangle}} + A_{\text{semicircle}}\)
\(A = 40x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2\)
\(A = 40x - 2x^2 - \frac{\pi}{2}x^2\)
\(A = 40x - \left(2 + \frac{\pi}{2}\right)x^2\) (as required)

- **M1**: For setting up a correct expression for the perimeter, e.g., \(2y + 2x + \pi x = 40\).
- **M1**: For rearranging the perimeter equation to make \(y\) (or \(2y\)) the subject, e.g., \(y = 20 - x - \frac{\pi}{2}x\).
- **M1**: For establishing a correct expression for the total area in terms of \(x\) and \(y\), e.g., \(A = 2xy + \frac{1}{2}\pi x^2\).
- **A1**: For substituting \(y\) and correctly simplifying to show \(A = 40x - \left(2 + \frac{\pi}{2}\right)x^2\) with all intermediate algebraic steps clearly shown.

1. **Express the area of the triangle:**
\(A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(3x + 2)(2x + 4)\)
\(A_{\text{triangle}} = (3x + 2)(x + 2) = 3x^2 + 6x + 2x + 4 = 3x^2 + 8x + 4\)

2. **Express the area of the rectangle:**
\(A_{\text{rectangle}} = (x + 3)(x + 1) = x^2 + x + 3x + 3 = x^2 + 4x + 3\)

3. **Set up the relationship:**
The area of the triangle is \(25\text{ cm}^2\) greater than the area of the rectangle:
\(A_{\text{triangle}} - A_{\text{rectangle}} = 25\)
\((3x^2 + 8x + 4) - (x^2 + 4x + 3) = 25\)

4. **Simplify the equation:**
\(2x^2 + 4x + 1 = 25\)
\(2x^2 + 4x - 24 = 0\)

Divide each term by \(2\):
\(x^2 + 2x - 12 = 0\) (as required)

- **M1**: For setting up and expanding a correct expression for the area of the triangle: \(3x^2 + 8x + 4\).
- **M1**: For setting up and expanding a correct expression for the area of the rectangle: \(x^2 + 4x + 3\).
- **M1**: For equating the two expressions with the difference of 25: \((3x^2 + 8x + 4) - (x^2 + 4x + 3) = 25\).
- **A1**: For correctly simplifying to \(2x^2 + 4x - 24 = 0\) and dividing by 2 to reach the final form \(x^2 + 2x - 12 = 0\).

1. **Find the relationship between \(y\) and \(x\):**
Since \(y\) is directly proportional to the square of \(x\):
\(y = k_1 x^2\)
Substitute \(x = 3\) and \(y = 18\):
\(18 = k_1 (3^2) \implies 18 = 9k_1 \implies k_1 = 2\)
So, \(y = 2x^2\).

2. **Find the relationship between \(z\) and \(y\):**
Since \(z\) is inversely proportional to the square root of \(y\):
\(z = \frac{k_2}{\sqrt{y}}\)
Substitute \(y = 16\) and \(z = 5\):
\(5 = \frac{k_2}{\sqrt{16}} \implies 5 = \frac{k_2}{4} \implies k_2 = 20\)
So, \(z = \frac{20}{\sqrt{y}}\).

3. **Express \(z\) in terms of \(x\):**
Substitute \(y = 2x^2\) into the equation for \(z\):
\(z = \frac{20}{\sqrt{2x^2}}\)
Since \(x > 0\), \(\sqrt{2x^2} = x\sqrt{2}\):
\(z = \frac{20}{x\sqrt{2}}\)

4. **Rationalise the denominator:**
\(z = \frac{20 \times \sqrt{2}}{x\sqrt{2} \times \sqrt{2}} = \frac{20\sqrt{2}}{2x} = \frac{10\sqrt{2}}{x}\) (as required)

- **M1**: For finding the first constant of proportionality to show \(y = 2x^2\).
- **M1**: For finding the second constant of proportionality to show \(z = \frac{20}{\sqrt{y}}\).
- **M1**: For substituting \(y = 2x^2\) to get \(z = \frac{20}{\sqrt{2x^2}}\) or \(z = \frac{20}{x\sqrt{2}}\).
- **A1**: For rationalising the denominator and simplifying to reach \(z = \frac{10\sqrt{2}}{x}\) with all steps shown.

1. **Find \(\overrightarrow{AB}\):**
Using the vector triangle:
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\)

2. **Find \(\overrightarrow{AP}\):**
Since \(AP : PB = 3 : 1\), \(P\) splits the line segment \(AB\) into 4 equal parts, so:
\(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}(-\mathbf{a} + \mathbf{b}) = -\frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\)

3. **Construct a vector path for \(\overrightarrow{MP}\):**
We can write:
\(\overrightarrow{MP} = \overrightarrow{MO} + \overrightarrow{OA} + \overrightarrow{AP}\)

Since \(M\) is the midpoint of \(OB\), we have:
\(\overrightarrow{MO} = -\frac{1}{2}\mathbf{b}\)

4. **Substitute and simplify:**
\(\overrightarrow{MP} = -\frac{1}{2}\mathbf{b} + \mathbf{a} + \left(-\frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right)\)
\(\overrightarrow{MP} = \left(1 - \frac{3}{4}\right)\mathbf{a} + \left(\frac{3}{4} - \frac{1}{2}\right)\mathbf{b}\)
\(\overrightarrow{MP} = \frac{1}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\)
\(\overrightarrow{MP} = \frac{1}{4}(\mathbf{a} + \mathbf{b})\) (as required)

- **M1**: For expressing \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) (or equivalent).
- **M1**: For finding a fractionally scaled segment of \(AB\), e.g., \(\overrightarrow{AP} = \frac{3}{4}(\mathbf{b} - \mathbf{a})\) or \(\overrightarrow{PB} = \frac{1}{4}(\mathbf{b} - \mathbf{a})\).
- **M1**: For writing a valid vector route for \(\overrightarrow{MP}\), such as \(\overrightarrow{MP} = \overrightarrow{MO} + \overrightarrow{OA} + \overrightarrow{AP}\) or \(\overrightarrow{MP} = \overrightarrow{MB} + \overrightarrow{BP}\).
- **A1**: For correctly substituting all component vectors and showing complete, logical algebraic simplification to reach \(\frac{1}{4}(\mathbf{a} + \mathbf{b})\).

Let S be salads, H be hot meals, and W be sandwiches. We are given S : H = 3 : 5 and H : W = 4 : 3. To combine these ratios, we find a common multiple for H, which is 20. Multiplying the first ratio by 4 gives S : H = 12 : 20. Multiplying the second ratio by 5 gives H : W = 20 : 15. The combined ratio is S : H : W = 12 : 20 : 15. The total number of parts is 12 + 20 + 15 = 47. Each part represents 188 / 47 = 4 lunches. The number of sandwiches sold is 15 * 4 = 60.

M1 for finding the combined ratio S : H : W as 12 : 20 : 15 (or equivalent ratio). M1 for dividing 188 by their sum of parts (188 / 47). A1 for 60.

Multiply both sides of the equation by 6 to clear the denominators: \(3(3x - 1) - 2(x + 4) = 6x\). Expand the brackets: \(9x - 3 - 2x - 8 = 6x\). Simplify the left hand side: \(7x - 11 = 6x\). Subtract 6x from both sides: \(x - 11 = 0\). Add 11 to both sides: \(x = 11\).

M1 for multiplying through by 6 or writing with a common denominator of 6, e.g. \(3(3x - 1) - 2(x + 4) = 6x\). M1 for expanding and simplifying to a linear equation in the form \(ax = b\) or \(x - b = 0\), e.g. \(7x - 11 = 6x\). A1 for 11.

The area of a sector is given by \(\frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(10\pi = \frac{\theta}{360} \times \pi \times 6^2\). This simplifies to \(10\pi = \frac{36\pi\theta}{360}\, which reduces to \)10 = \frac{\theta}{10}\). Multiplying both sides by 10 gives \(\theta = 100\).

M1 for setting up the correct equation \(\frac{\theta}{360} \times \pi \times 6^2 = 10\pi\). M1 for simplifying the equation to \(\frac{\theta}{10} = 10\) or equivalent. A1 for 100.

The area of a trapezium is given by \(\frac{1}{2}(a+b)h\). Substituting the given values: \(54 = \frac{1}{2}((x + 2) + (x + 8)) \times 6\). Simplify the equation: \(54 = 3(2x + 10)\). Divide by 3: \(18 = 2x + 10\). Subtract 10: \(2x = 8\), so \(x = 4\).

M1 for substituting the terms into the area formula: \(\frac{1}{2}(x + 2 + x + 8) \times 6 = 54\). M1 for simplifying to a linear equation, e.g., \(2x + 10 = 18\) or \(6x + 30 = 54\). A1 for 4.

The probability of drawing the first red counter is \(\frac{n}{8}\). Since it is without replacement, the probability of drawing a second red counter is \(\frac{n-1}{7}\). The probability of both being red is \(\frac{n}{8} \times \frac{n-1}{7} = \frac{5}{14}\). This gives \(\frac{n(n-1)}{56} = \frac{5}{14}\). Multiplying both sides by 56 gives \(n(n-1) = 20\). Solving the equation \(n^2 - n - 20 = 0\) gives \((n - 5)(n + 4) = 0\). Since \(n\) must be positive, \(n = 5\).

M1 for expressing the probability of two red counters as \(\frac{n}{8} \times \frac{n-1}{7}\). M1 for forming the equation \(n(n-1) = 20\) or solving \(n^2 - n - 20 = 0\). A1 for 5.

Since \(y\) is inversely proportional to \(x^2\), we can write \(y = \frac{k}{x^2}\) where \(k\) is a constant. Substitute \(x = 3\) and \(y = 4\) to find \(k\): \(4 = \frac{k}{3^2} = \frac{k}{9}\), which gives \(k = 36\). The equation is \(y = \frac{36}{x^2}\). When \(x = 6\), \(y = \frac{36}{6^2} = \frac{36}{36} = 1\).

M1 for the relationship \(y = \frac{k}{x^2}\) (or identifying that when \(x\) is doubled, \(y\) is divided by 4). M1 for finding \(k = 36\) (or evaluating \(4 \div 4\)). A1 for 1.

The sale price of \(£76.50\) represents \(100\% - 15\% = 85\%\) of the original price. Therefore, \(0.85 \times \text{Original Price} = 76.50\). To find the original price, calculate \(76.50 / 0.85 = 7650 / 85\). Simplifying this: \(7650 / 85 = 90\). The original price was \(£90\).

M1 for equating \(85\%\) to \(76.50\), e.g., \(0.85x = 76.50\). M1 for a complete method to find the original price, e.g., \(76.50 \div 0.85\). A1 for 90.

Let's find the first differences between successive terms: \(10 - 3 = 7\), \(21 - 10 = 11\), \(36 - 21 = 15\). The first differences are \(7, 11, 15\). Now find the second differences: \(11 - 7 = 4\), \(15 - 11 = 4\). The constant second difference is 4, which means the coefficient of \(n^2\) is \(4 \div 2 = 2\). Subtract \(2n^2\) from each term of the sequence: For \(n=1\): \(3 - 2(1^2) = 1\). For \(n=2\): \(10 - 2(2^2) = 2\). For \(n=3\): \(21 - 2(3^2) = 3\). For \(n=4\): \(36 - 2(4^2) = 4\). The resulting sequence is \(1, 2, 3, 4\), which has the \(n\)-th term of \(n\). Combining these, the \(n\)-th term of the original sequence is \(2n^2 + n\).

M1 for finding the second difference (4) and identifying that the \(n^2\) coefficient is 2 (e.g., \(2n^2\)). M1 for subtracting \(2n^2\) from the sequence terms to find the linear sequence \(1, 2, 3, 4\) and finding its \(n\)-th term is \(n\) (or equivalent algebraic method). A1 for \(2n^2 + n\).

To find the overall ratio of Red : Blue : Yellow, we need to make the parts for Blue 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:
\(R : B : Y = 6 : 10 : 15\)

We are given that there are 45 yellow counters.
\(15\text{ parts} = 45\text{ counters}\)
\(1\text{ part} = 45 \div 15 = 3\text{ counters}\)

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

The total number of counters is:
\(31 \times 3 = 93\)

M1 for scaling the ratios to get a common term for blue, e.g. \(6 : 10\) and \(10 : 15\) (or any equivalent ratio \(6 : 10 : 15\))
M1 for dividing 45 by their yellow part (15) to find the value of one share (3), or for finding the number of red (18) and blue (30) counters
A1 for 93

To eliminate \(y\), multiply the first equation by 5 and the second equation by 2:

\(15x - 10y = 95\)
\(4x + 10y = 0\)

Add the two equations together:
\((15x + 4x) + (-10y + 10y) = 95 + 0\)
\(19x = 95\)
\(x = 5\)

Substitute \(x = 5\) into the second equation to find \(y\):
\(2(5) + 5y = 0\)
\(10 + 5y = 0\)
\(5y = -10\)
\(y = -2\)

Thus, the solution is \(x = 5\) and \(y = -2\).

M1 for a correct process to eliminate one variable (e.g. multiplying the equations correctly to get coefficients of \(y\) to be \(-10\) and \(10\), or coefficients of \(x\) to be \(6\) and \(6\))
M1 for substituting their found value of one variable back into either equation to find the other, or for finding one variable correctly (e.g., \(x = 5\) or \(y = -2\))
A1 for both \(x = 5\) and \(y = -2\)

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

Substitute the given values into the formula:
\(54 = \frac{1}{2}\left(x + (2x + 3)\right) \times 6\)

Simplify the equation:
\(54 = 3(3x + 3)\)
\(54 = 9x + 9\)

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

Divide by 9:
\(x = 5\)

M1 for setting up a correct equation using the trapezium area formula, e.g. \(54 = \frac{1}{2}(x + 2x + 3) \times 6\)
M1 for simplifying the equation to a linear form, e.g. \(54 = 9x + 9\) or \(18 = 3x + 3\)
A1 for 5

There are 8 chocolates in total. Since Janice eats the first chocolate, there are 7 chocolates remaining for the second draw.

There are two combined outcomes that result in one milk and one dark chocolate:

1) Milk chocolate first, then Dark chocolate (MD):
\(P(\text{MD}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

2) Dark chocolate first, then Milk chocolate (DM):
\(P(\text{DM}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

To find the overall probability of getting one of each, add these two probabilities together:
\(P(\text{one of each}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56}\)

Simplifying the fraction:
\(\frac{30}{56} = \frac{15}{28}\)

M1 for calculating the probability of a single correct combined outcome, e.g., \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\) (must show division by 7 to indicate non-replacement)
M1 for adding two correct probabilities of the two alternative outcomes, e.g., \(\frac{15}{56} + \frac{15}{56}\)
A1 for \(\frac{15}{28}\) (or equivalent fraction such as \(\frac{30}{56}\))

Since angle \(DAB = 90^\circ\), triangle \(DAB\) is a right-angled triangle with hypotenuse \(BD\).

Using Pythagoras' theorem on triangle \(DAB\):
\(BD^2 = AB^2 + AD^2\)
\(13^2 = 5^2 + AD^2\)
\(169 = 25 + AD^2\)
\(AD^2 = 144\)
\(AD = 12\text{ m}\) (since length must be positive).

Since both \(DAB\) and \(ADC\) are right angles, \(AD\) is perpendicular to both parallel sides \(AB\) and \(CD\). Therefore, \(AD\) is the perpendicular height of the trapezium.

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

Substitute \(a = 5\), \(b = 11\), and \(h = 12\):
\(\text{Area} = \frac{1}{2}(5 + 11) \times 12\)
\(\text{Area} = 8 \times 16 = 96\text{ m}^2\).

M1: For applying Pythagoras' theorem on triangle \(DAB\): \(13^2 = 5^2 + AD^2\).
A1: For calculating \(AD = 12\text{ m}\).
M1: For using the correct formula for the area of a trapezium: \(\frac{1}{2}(5 + 11) \times h\).
M1: For substituting their calculated height into the area formula: \(\frac{1}{2}(5 + 11) \times 12\).
A1: For a final correct answer of \(96\) (allow \(96\text{ m}^2\)).

Let the original number of red counters be \(3x\) and the original number of blue counters be \(5x\).
The original total number of counters is \(3x + 5x = 8x\).

When \(18\) red counters are added, the new number of red counters is \(3x + 18\).
The number of blue counters remains \(5x\).

The new ratio of red to blue is:
\(\frac{3x + 18}{5x} = \frac{3}{2}\)

Cross-multiply to solve for \(x\):
\(2(3x + 18) = 3(5x)\)
\(6x + 36 = 15x\)
\(9x = 36\)
\(x = 4\)

Now, calculate the original total number of counters:
\(\text{Total} = 8x = 8 \times 4 = 32\).

M1: For representing the original counters as \(3x\) and \(5x\) (or an equivalent parts method).
M1: For setting up the ratio equation after adding \(18\) red counters: \(\frac{3x + 18}{5x} = \frac{3}{2}\).
M1: For correct algebraic manipulation to eliminate fractions: \(2(3x + 18) = 15x\).
A1: For finding \(x = 4\).
A1: For finding the correct original total of \(32\) counters.

Substitute the expression for \(y\) from the first equation into the second equation:
\(x^2 + (2x - 3)^2 = 18\)

Expand the brackets:
\(x^2 + (4x^2 - 12x + 9) = 18\)
\(5x^2 - 12x + 9 = 18\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\(5x^2 - 12x - 9 = 0\)

Factorise the quadratic:
We need two numbers that multiply to \(5 \times (-9) = -45\) and add to \(-12\). These are \(-15\) and \(3\).
\(5x^2 - 15x + 3x - 9 = 0\)
\(5x(x - 3) + 3(x - 3) = 0\)
\((5x + 3)(x - 3) = 0\)

This gives:
\(x = 3\) or \(x = -\frac{3}{5} = -0.6\)

Find the corresponding values of \(y\):
If \(x = 3\):
\(y = 2(3) - 3 = 3\)

If \(x = -0.6\):
\(y = 2(-0.6) - 3 = -1.2 - 3 = -4.2\)

So the two pairs of solutions are:
\(x = 3, y = 3\) and \(x = -0.6, y = -4.2\).

M1: For substituting \(y = 2x - 3\) into the quadratic equation: \(x^2 + (2x - 3)^2 = 18\).
M1: For expanding and simplifying to a 3-term quadratic: \(5x^2 - 12x - 9 = 0\).
M1: For a complete method to solve their quadratic equation, e.g., factorising to \((5x + 3)(x - 3) = 0\).
A1: For finding both values of \(x\) correctly: \(x = 3\) and \(x = -0.6\) (or \(-\frac{3}{5}\)).
A1: For finding both corresponding values of \(y\) correctly: \(y = 3\) and \(y = -4.2\) (or \(-\frac{21}{5}\)).

The formula for the area of a sector is:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)

Substitute the given values of radius \(r = 6\text{ cm}\) and area \(10\pi\text{ cm}^2\):
\(10\pi = \frac{\theta}{360} \times \pi \times 6^2\)
\(10\pi = \frac{36\theta}{360} \times \pi\)
\(10\pi = \frac{\theta}{10} \times \pi\)

Divide both sides by \(\pi\):
\(10 = \frac{\theta}{10} \implies \theta = 100^2\)

Now, find the perimeter of the sector. The perimeter consists of two straight edges (radii) and the curved arc:
\(\text{Perimeter} = 2r + \text{Arc length}\)

The formula for arc length is:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)

Substitute \(\theta = 100\) and \(r = 6\):
\(\text{Arc length} = \frac{100}{360} \times 2\pi \times 6 = \frac{100}{360} \times 12\pi = \frac{10}{36} \times 12\pi = \frac{10}{3}\pi\text{ cm}\)

Calculate the total perimeter:
\(\text{Perimeter} = 2(6) + \frac{10}{3}\pi = 12 + \frac{10}{3}\pi\text{ cm}\).

This is in the form \(a + b\pi\) where \(a = 12\) and \(b = \frac{10}{3}\).

M1: For substituting \(r=6\) and \(\text{Area}=10\pi\) into the sector area formula: \(\frac{\theta}{360} \times ̃\pi \times 6^2 = 10\pi\).
A1: For finding \(\theta = 100^\circ\) (or finding that the sector represents \(\frac{5}{18}\) of a full circle).
M1: For a correct formula to find the arc length: \(\frac{100}{360} \times 2 \times \pi \times 6\) (or equivalent).
A1: For finding the arc length as \(\frac{10}{3}\pi\) (or \(3\frac{1}{3}\pi\)).
A1: For the final answer in the correct form: \(12 + \frac{10}{3}\pi\) (or \(12 + 3\frac{1}{3}\pi\)).

Let \(x\) be the total number of counters in the box.
The probability of choosing a green counter on the first draw is \(\frac{4}{x}\).

Since the first green counter is not replaced, there are now \(x - 1\) counters remaining in the box, and \(3\) of them are green.
The probability of choosing a green counter on the second draw is \(\frac{3}{x - 1}\).

The probability that both counters are green is:
\(\text{P}(\text{Green, Green}) = \frac{4}{x} \times \frac{3}{x - 1} = \frac{12}{x(x - 1)}\)

We are given that this probability is \(\frac{2}{15}\):
\(\frac{12}{x(x - 1)} = \frac{2}{15}\)

Cross-multiply to solve for \(x\):
\(12 \times 15 = 2x(x - 1)\)
\(180 = 2x^2 - 2x\)

Divide the entire equation by \(2\):
\(x^2 - x - 90 = 0\)

Factorise the quadratic equation:
\((x - 10)(x + 9) = 0\)

This gives two possible solutions:
\(x = 10\) or \(x = -9\)

Since the total number of counters \(x\) must be a positive integer, we reject \(x = -9\).
Therefore, \(x = 10\).

M1: For writing the product of the two probabilities: \(\frac{4}{x} \times \frac{3}{x-1}\).
M1: For setting up the equation: \(\frac{12}{x(x-1)} = \frac{2}{15}\).
M1: For rearranging into a standard quadratic equation: \(x^2 - x - 90 = 0\) (or \(2x^2 - 2x - 180 = 0\)).
M1: For a complete method to solve their 3-term quadratic equation, e.g., factorising to \((x - 10)(x + 9) = 0\).
A1: For the final correct answer of \(10\) (with the negative solution rejected).

Step 1: Express \(y\) in terms of \(x\).
Since \(y\) is inversely proportional to \(\sqrt{x}\):
\(y =
\frac{k_1}{\sqrt{x}}\) where \(k_1\) is a constant.

Substitute \(x = 16\) and \(y = 5\):
\(5 = \frac{k_1}{\sqrt{16}}\)
\(5 = \frac{k_1}{4} \implies k_1 = 20\)

So, \(y = \frac{20}{\sqrt{x}}\).

Step 2: Express \(z\) in terms of \(y\).
Since \(z\) is directly proportional to \(y^2\):
\(z = k_2 y^2\) where \(k_2\) is a constant.

Substitute \(y = 4\) and \(z = 48\):
\(48 = k_2 (4)^2\)
\(48 = 16 k_2 \implies k_2 = 3\)

So, \(z = 3y^2\).

Step 3: Find \(z\) when \(x = 100\).
First, find \(y\) when \(x = 100\):
\(y = \frac{20}{\sqrt{100}} = \frac{20}{10} = 2\)

Now, substitute \(y = 2\) into the equation for \(z\):
\(z = 3(2)^2 = 3 \times 4 = 12\).

M1: For establishing the equation \(y = \frac{k_1}{\sqrt{x}}
\) and calculating \(k_1 = 20\).
M1: For establishing the equation \(z = k_2 y^2\) and calculating \(k_2 = 3\).
M1: For a valid method to find \(y\) when \(x = 100\) using their formula: \(y = \frac{20}{\sqrt{100}}\).
A1: For finding \(y = 2\).
A1: For substituting \(y = 2\) into their equation for \(z\) to obtain \(12\).

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

A reduction of \(20\%\) means the television is worth \(80\%\) of its original price:
\(\text{Reduced Price} = 0.8P\)

An increase of \(15\%\) on this reduced price means the final price is \(115\%\) of the reduced price:
\(\text{Final Price} = 1.15 \times \text{Reduced Price} = 1.15 \times (0.8P)\)

Calculate the combined multiplier:
\(1.15 \times 0.8 = \frac{115}{100} \times \frac{4}{5} = \frac{23}{20} \times \frac{4}{5} = \frac{92}{100} = 0.92\)

We are given that the final price is \(\text{£}414\):
\(0.92P = 414\)

Solve for \(P\):
\(P = \frac{414}{0.92} = \frac{41400}{92}
\)

Simplify the fraction by dividing numerator and denominator by \(2\):
\(P = \frac{20700}{46}\)

Divide by \(23\) (since \(46 = 2 \times 23\) and \(20700 = 900 \times 23\)):
\(P = \frac{900}{2} = 450\)

So, the original price of the television was \(\text{£}450\).

M1: For representing the two percentage changes as multipliers: \(0.8\) and \(1.15\) (or equivalent fractions).
M1: For calculating the overall combined multiplier: \(0.8 \times 1.15 = 0.92\) (or equivalent fraction: \(\frac{23}{25}\)).
M1: For setting up the equation to solve for the original price: \(0.92P = 414\) (or equivalent).
A1: For simplifying to a form ready to divide, e.g., \(P = \frac{41400}{92}\) or showing correct intermediate steps (such as finding the intermediate price of \(\text{£}360\)).
A1: For the final correct answer of \(450\) (accept \(\text{£}450\)).

Step 1: Find the \(n\)-th term of the sequence.
Write out the sequence and find the first and second differences:
Sequence: \(4, \quad 7, \quad 14, \quad 25, \quad 40\)
First differences: \(3, \quad 4\text{ is not the diff}, \quad 7, \quad 11, \quad 15\)
Second differences: \(4, \quad 4, \quad 4\)

Since the second differences are constant and equal to \(4\), the sequence is quadratic, and the coefficient of the \(n^2\) term is \(\frac{4}{2} = 2\).

So, the \(n\)-th term begins with \(2n^2\).

Now, subtract \(2n^2\) from each term of the original sequence:
Original sequence: \(4, \quad 7, \quad 14, \quad 25, \quad 40\)
\(2n^2\) terms: \(2, \quad 8, \quad 18, \quad 32, \quad 50\)
Difference: \(2, \quad -1, \quad -4, \quad -7, \quad -10\)

This difference sequence is a linear sequence: \(2, -1, -4, -7, -10\).
It has a first term of \(2\) and a common difference of \(-3\).
So, its \(n\)-th term is \(-3n + 5\).

Combining these, the \(n\)-th term of the original sequence is:
\(u_n = 2n^2 - 3n + 5\)

Step 2: Find which term has a value of \(257\).
Set \(u_n = 257\):
\(2n^2 - 3n + 5 = 257\)

Subtract \(257\) from both sides to form a quadratic equation:
\(2n^2 - 3n - 252 = 0\)

Solve this quadratic equation by factorising. We need two numbers that multiply to \(2 \times (-252) = -504\) and add to \(-3\). These numbers are \(-24\) and \(21\).

Rewrite and factorise:
\(2n^2 - 24n + 21n - 252 = 0\)
\(2n(n - 12) + 21(n - 12) = 0\)
\((2n + 21)(n - 12) = 0\)

This gives \(n = 12\) or \(n = -10.5\).
Since the position in a sequence \(n\) must be a positive integer, \(n = 12\).
So, the \(12\)-th term has a value of \(257\).

M1: For finding the first differences (3, 7, 11, 15) and constant second differences of 4, and establishing that the coefficient of \(n^2\) is \(2\).
M1: For subtracting \(2n^2\) from the sequence and finding the linear part \(-3n + 5\).
A1: For the correct \(n\)-th term expression: \(2n^2 - 3n + 5\) (or equivalent).
M1: For setting their \(n\)-th term equal to \(257\), rearranging to standard quadratic form \(2n^2 - 3n - 252 = 0\), and attempting to solve.
A1: For the final answer of \(12\) (or \(12\)-th term) with any negative solution rejected.

Let the two consecutive odd numbers be \(2n - 1\) and \(2n + 1\), where \(n\) is an integer. The difference between their squares is: \((2n + 1)^2 - (2n - 1)^2\). Expanding the squares gives: \((2n + 1)^2 = 4n^2 + 4n + 1\) and \((2n - 1)^2 = 4n^2 - 4n + 1\). Subtracting these expressions, we get: \((4n^2 + 4n + 1) - (4n^2 - 4n + 1) = 4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n\). Since \(n\) is an integer, \(8n\) is always a multiple of 8. Hence, the difference is always a multiple of 8.

M1: Represents two consecutive odd numbers algebraically, e.g., \(2n - 1\) and \(2n + 1\). M1: Sets up the difference of their squares, e.g., \((2n + 1)^2 - (2n - 1)^2\). A1: Correctly expands both quadratic expressions and simplifies to \(8n\). A1: Concludes clearly that since \(n\) is an integer, \(8n\) is a multiple of 8.

The area of a sector is given by \(\frac{\theta}{360} \pi r^2\). Substituting the given values: \(\frac{60}{360} \pi r^2 = 24\pi\). Simplifying gives \(\frac{1}{6} \pi r^2 = 24\pi\). Dividing by \(\pi\) and multiplying by 6 gives \(r^2 = 144\), so \(r = 12\text{ cm}\). The arc length of the sector is given by \(\frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2\pi(12) = 4\pi\text{ cm}\). The perimeter is the sum of two radii and the arc length: \(\text{Perimeter} = 2r + \text{Arc length} = 2(12) + 4\pi = 24 + 4\pi\text{ cm}\).

M1: Sets up an equation for the sector area, e.g., \(\frac{60}{360} \pi r^2 = 24\pi\). A1: Solves to find the radius \(r = 12\). M1: Calculates the arc length as \(\frac{60}{360} \times 2 \pi \times 12\) to get \(4\pi\). A1: Fully correct show that with perimeter calculated as \(2(12) + 4\pi = 24 + 4\pi\).

Since \(y\) is inversely proportional to the square root of \(x\), we can write \(y = \frac{C}{\sqrt{x}}\) for some constant \(C\). Substituting \(x = 9d\) and \(y = k\) gives \(k = \frac{C}{\sqrt{9d}} = \frac{C}{3\sqrt{d}}\), so \(C = 3k\sqrt{d}\). Now substituting \(x = 36d\) into the formula gives \(y = \frac{3k\sqrt{d}}{\sqrt{36d}} = \frac{3k\sqrt{d}}{6\sqrt{d}}\). Cancelling the common factor of \(\sqrt{d}\) yields \(y = \frac{3k}{6} = \frac{1}{2}k\).

M1: Writes the correct proportionality statement, e.g., \(y = \frac{C}{\sqrt{x}}\) or \(y\sqrt{x} = C\). M1: Substitutes \(x = 9d\) and \(y = k\) to express the constant \(C\) in terms of \(k\) and \(d\), e.g., \(C = 3k\sqrt{d}\). M1: Substitutes \(x = 36d\) and their expression for \(C\) into the formula. A1: Simplifies the algebraic fraction correctly to obtain \(y = \frac{1}{2}k\).

First, find \(T_{n+1}\) by substituting \(n+1\) for \(n\): \(T_{n+1} = (n+1)^2 + 5(n+1) + 6 = n^2 + 2n + 1 + 5n + 5 + 6 = n^2 + 7n + 12\). Next, factorise both expressions: \(T_n = n^2 + 5n + 6 = (n+2)(n+3)\) and \(T_{n+1} = n^2 + 7n + 12 = (n+3)(n+4)\). Setting up the ratio gives \(\frac{T_{n+1}}{T_n} = \frac{(n+3)(n+4)}{(n+2)(n+3)}\). Cancelling the common factor of \((n+3)\) results in \(\frac{T_{n+1}}{T_n} = \frac{n+4}{n+2}\).

M1: Substitutes \(n+1\) into the expression to write \(T_{n+1} = (n+1)^2 + 5(n+1) + 6\). A1: Expands and simplifies to obtain \(T_{n+1} = n^2 + 7n + 12\). M1: Factorises both quadratics to get \((n+2)(n+3)\) and \((n+3)(n+4)\). A1: Divides and cancels the common factor \((n+3)\) to show \(\frac{n+4}{n+2}\).

Let the initial revenue be \(100\). After a 12% increase, the revenue is \(100 \times 1.12 = 112\). After a 5% decrease, the revenue is \(112 \times 0.95 = 106.4\). The overall percentage increase is \(106.4 - 100 = 6.4\%\).

M1 for attempting to multiply the multipliers: \(1.12 \times 0.95\) (or showing a step-by-step calculation starting with a value like 100). M1 for finding the final multiplier of \(1.064\) (or final value of \(106.4\)). A1 for \(6.4\%\) (or \(6.4\)).

The ratio is flour : sugar : butter = \(5 : 3 : 2\). The 3 parts corresponding to sugar represent \(450\text{ g}\). Therefore, 1 part represents \(450 \div 3 = 150\text{ g}\). The total ratio parts for flour and butter combined is \(5 + 2 = 7\) parts. The total weight of flour and butter is \(7 \times 150 = 1050\text{ g}\).

M1 for dividing sugar weight by its ratio share: \(450 \div 3\). M1 for multiplying 1 part weight by total parts of flour and butter: \((5 + 2) \times 150\). A1 for \(1050\).

Subtract 6 from both sides of the equation: \(\frac{4x - 3}{5} = 5\). Multiply both sides by 5: \(4x - 3 = 25\). Add 3 to both sides: \(4x = 28\). Divide by 4: \(x = 7\).

M1 for isolating the fraction: \(\frac{4x - 3}{5} = 5\) or multiplying all terms by 5: \(4x - 3 + 30 = 55\). M1 for isolating the term with \(x\): \(4x = 28\). A1 for \(7\).

The radius of the inner pond is \(r = 3.5\text{ m}\). The outer radius, including the path, is \(R = 3.5 + 1.2 = 4.7\text{ m}\). The area of the path is the difference between the areas of the outer circle and inner circle: \(\text{Area} = \pi R^2 - \pi r^2 = \pi (4.7^2) - \pi (3.5^2) = 22.09\pi - 12.25\pi = 9.84\pi \approx 30.913\text{ m}^2\). To 3 significant figures, the area is \(30.9\text{ m}^2\).

M1 for finding the outer radius \(R = 4.7\text{ m}\) or writing a correct expression for the difference in area: \(\pi \times 4.7^2 - \pi \times 3.5^2\). M1 for evaluating \(9.84\pi\) or showing intermediate areas: \(69.397... - 38.484...\) or \(30.913...\). A1 for \(30.9\) (accept \(30.91\) or \(30.92\) depending on pi approximation used).

The formula for the area of a trapezium is \(\text{Area} = \frac{1}{2}(a + b)h\). Substituting the given values: \(54 = \frac{1}{2}(x + (x + 4)) \times 6\). This simplifies to \(54 = 3(2x + 4)\). Expanding the bracket gives \(54 = 6x + 12\). Subtracting 12 gives \(42 = 6x\). Dividing by 6 gives \(x = 7\).

M1 for setting up a correct equation: \(\frac{1}{2}(x + x + 4) \times 6 = 54\) (or equivalent). M1 for simplifying to a linear equation in \(x\), e.g., \(6x + 12 = 54\) or \(2x + 4 = 18\). A1 for \(7\).

There are two scenarios: 1) A red ball is chosen from Bag A, then a red ball is chosen from Bag B. The probability is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). 2) A blue ball is chosen from Bag A, then a red ball is chosen from Bag B. The probability is \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\). Adding these together gives the total probability of choosing a red ball from Bag B: \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\).

M1 for finding either the probability of drawing red then red \(\frac{3}{8} \times \frac{5}{7}\) or blue then red \(\frac{5}{8} \times \frac{4}{7}\). M1 for adding both correct products: \(\frac{15}{56} + \frac{20}{56}\). A1 for \(\frac{5}{8}\) (or equivalent simplified fraction).

Factorise the numerator: \(3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)\). Factorise the denominator: \(x^2 + 5x + 6 = (x + 2)(x + 3)\). The fraction becomes \(\frac{3(x - 2)(x + 2)}{(x + 2)(x + 3)}\). Divide both the numerator and the denominator by the common factor \((x + 2)\) to get the fully simplified form: \(\frac{3(x - 2)}{x + 3}\) or \(\frac{3x - 6}{x + 3}\).

M1 for factorising the numerator to \(3(x - 2)(x + 2)\). M1 for factorising the denominator to \((x + 2)(x + 3)\). A1 for \(\frac{3(x - 2)}{x + 3}\) (or \(\frac{3x - 6}{x + 3}\)).

Since \(y\) is inversely proportional to \(\sqrt{x}\), we can write the formula as \(y = \frac{k}{\sqrt{x}}\) for some constant \(k\). Substitute \(x = 16\) and \(y = 3\) to find \(k\): \(3 = \frac{k}{\sqrt{16}} \implies 3 = \frac{k}{4} \implies k = 12\). So the formula is \(y = \frac{12}{\sqrt{x}}\). When \(x = 36\), we substitute this into the formula: \(y = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\).

M1 for setting up the proportion equation \(y = \frac{k}{\sqrt{x}}\) and substituting \(x = 16\), \(y = 3\) to solve for \(k\). M1 for finding \(k = 12\). A1 for \(2\).

Let the initial number of adults be \(3x\) and the number of children be \(5x\). After 12 adults join, the number of adults becomes \(3x + 12\). The new ratio of adults to children is \(2 : 3\), which gives the equation: \(\frac{3x + 12}{5x} = \frac{2}{3}\). Cross-multiplying to solve for \(x\): \(3(3x + 12) = 2(5x)\) which simplifies to \(9x + 36 = 10x\). Subtracting \(9x\) from both sides gives \(x = 36\). The number of children is \(5x = 5 \times 36 = 180\).

M1 for setting up a correct equation representing the ratio change, e.g. \(\frac{3x+12}{5x} = \frac{2}{3}\) or establishing equivalent ratios. M1 for correctly solving their equation to find the multiplier, e.g. \(x = 36\). A1 for 180.

First find the differences between consecutive terms: \(7, 11, 15, 19\). The second differences are constant at \(4\). This means the sequence is quadratic and the coefficient of the \(n^2\) term is \(4 \div 2 = 2\). Subtracting \(2n^2\) from each term: For \(n=1\): \(4 - 2(1)^2 = 2\). For \(n=2\): \(11 - 2(2)^2 = 3\). For \(n=3\): \(22 - 2(3)^2 = 4\). The remaining linear sequence is \(2, 3, 4, 5, ...\) which has the rule \(n + 1\). Therefore, the general formula for the \(n\)-th term is \(2n^2 + n + 1\). To find the 20th term, substitute \(n = 20\): \(2(20)^2 + 20 + 1 = 2(400) + 21 = 821\).

M1 for finding the second difference of 4 and identifying the quadratic term as \(2n^2\). M1 for finding the linear part \(n+1\) to get the general term \(2n^2 + n + 1\), or for attempting to extend the sequence correctly up to the 20th term. A1 for 821.

To estimate the mean, first find the midpoints of each class interval: \(55\), \(65\), \(75\), \(85\), and \(95\). Next, calculate the sum of the products of the known midpoints and frequencies: \(12 \times 55 + 26 \times 65 + 18 \times 85 + 9 \times 95 = 660 + 1690 + 1530 + 855 = 4735\). The sum of the known frequencies is \(12 + 26 + 18 + 9 = 65\). Using the formula for the estimated mean, we set up the equation \((4735 + 75x) / (65 + x) = 73.25\). Multiplying both sides by \((65 + x)\) gives \(4735 + 75x = 73.25(65 + x)\), which expands to \(4735 + 75x = 4761.25 + 73.25x\). Subtracting \(73.25x\) and \(4735\) from both sides gives \(1.75x = 26.25\). Dividing by \(1.75\) gives \(x = 15\).

M1 for finding the correct midpoints of the class intervals (at least 3 correct midpoints shown or used). M1 for setting up the sum of the products, \(4735 + 75x\). M1 for setting up the equation \((4735 + 75x) / (65 + x) = 73.25\). M1 for expanding and rearranging to a linear equation in \(x\) (e.g., \(1.75x = 26.25\)). A1 for the correct answer of 15.

Let the original number of students in Year 9, Year 10, and Year 11 be represented by \(5k\), \(6k\), and \(4k\) respectively, where \(k\) is a constant. After the changes, the new numbers of students are: Year 9: \(5k + 24\), Year 10: \(6k - 12\), and Year 11: \(4k\). The new ratio is \(7 : 5 : 4\). Since the number of students in Year 11 does not change and is still represented by \(4\) parts in the new ratio, we can set up the equation: \((5k + 24) / (4k) = 7 / 4\). Multiplying both sides by \(4k\) gives \(5k + 24 = 7k\), which simplifies to \(2k = 24\), hence \(k = 12\). The original number of students in each year group is: Year 9: \(5 \times 12 = 60\), Year 10: \(6 \times 12 = 72\), and Year 11: \(4 \times 12 = 48\). The original total is \(60 + 72 + 48 = 180\).

M1 for expressing the initial quantities as \(5k\), \(6k\), \(4k\). M1 for writing expressions for the new quantities: \(5k + 24\), \(6k - 12\), and \(4k\). M1 for setting up a correct equation using the new ratios, e.g., \((5k + 24) / (4k) = 7/4\) or \((6k - 12) / (4k) = 5/4\). M1 for solving the equation to find \(k = 12\). A1 for the correct total of 180.

The length of the garden is \((2x + 5)\) metres. Adding the 1.5-metre pathway to both ends, the combined length is \((2x + 5) + 1.5 + 1.5 = 2x + 8\) metres. The width of the garden is \((x + 3)\) metres. Adding the 1.5-metre pathway to both sides, the combined width is \((x + 3) + 1.5 + 1.5 = x + 6\) metres. The combined area is \((2x + 8)(x + 6) = 135\). Expanding the brackets gives \(2x^2 + 12x + 8x + 48 = 135\), which simplifies to \(2x^2 + 20x - 87 = 0\). Apply the quadratic formula \(x = (-b \pm \sqrt{b^2 - 4ac}) / 2a\) with \(a = 2\), \(b = 20\), and \(c = -87\): \(x = (-20 \pm \sqrt{20^2 - 4(2)(-87)}) / (2 \times 2) = (-20 \pm \sqrt{1096}) / 4\). Since \(x\) must be a positive length, \(x = (-20 + 33.1059) / 4 \approx 3.276\) metres. Rounding to 3 significant figures, we get \(x = 3.28\).

M1 for expressing the combined length as \(2x + 8\) or the combined width as \(x + 6\). M1 for setting up the equation \((2x + 8)(x + 6) = 135\). M1 for expanding and rearranging to standard quadratic form: \(2x^2 + 20x - 87 = 0\). M1 for applying the quadratic formula correctly to their quadratic equation. A1 for the correct value of \(x = 3.28\) to 3 significant figures.

Let the angle of the sector at the centre be \(\theta\). The formula for the arc length \(L\) is: \(L = (\theta / 360) \times 2\pi r = 10\pi \implies (\theta / 360) \times r = 5\). The formula for the area \(A\) of the sector is: \(A = (\theta / 360) \times \pi r^2 = 120\pi \implies (\theta / 360) \times r^2 = 120\). We can divide the area equation by the simplified arc length equation: \([ (\theta / 360) \times r^2 ] / [ (\theta / 360) \times r ] = 120 / 5 \implies r = 24\text{ cm}\). The perimeter of the sector is the sum of the arc length and the two bounding radii: \(\text{Perimeter} = L + 2r = 10\pi + 2(24) = 10\pi + 48\). Evaluating this numerically gives \(\text{Perimeter} \approx 31.4159 + 48 = 79.4159\) cm. To 3 significant figures, the perimeter is \(79.4\) cm.

M1 for writing down a correct equation for arc length, e.g., \(\frac{\theta}{360} \times 2\pi r = 10\pi\). M1 for writing down a correct equation for area, e.g., \(\frac{\theta}{360} \times \pi r^2 = 120\pi\). M1 for a valid method to solve for \(r\) (such as dividing the area equation by the arc length equation). A1 for finding \(r = 24\). M1 for adding \(2r\) to \(10\pi\) and evaluating to 3 significant figures. A1 for 79.4.

First, find the area of the original sector: \(A_{\text{orig}} = (120 / 360) \times \pi \times 8^2 = (64/3)\pi \approx 67.0206\text{ cm}^2\). Next, find the area of the smaller concentric sector that is removed: \(A_{\text{removed}} = (120 / 360) \times \pi \times 5^2 = (25/3)\pi \approx 26.1799\text{ cm}^2\). The radius of the drilled hole is \(0.75\text{ cm}\), so its area is \(A_{\text{hole}} = \pi \times 0.75^2 = 0.5625\pi \approx 1.7671\text{ cm}^2\). Calculate the remaining area of the template: \(A_{\text{remain}} = A_{\text{orig}} - A_{\text{removed}} - A_{\text{hole}} = (64/3)\pi - (25/3)\pi - 0.5625\pi = 13\pi - 0.5625\pi = 12.4375\pi \approx 39.0736\text{ cm}^2\). Now, calculate the percentage of the original area that remains: \(\text{Percentage} = (A_{\text{remain}} / A_{\text{orig}}) \times 100 = (12.4375\pi / (64/3)\pi) \times 100 = (12.4375 \times 3 / 64) \times 100 = (37.3125 / 64) \times 100 \approx 58.3008\%\). To 1 decimal place, the percentage of the original sector's area that remains is \(58.3\%\).

M1 for calculating the area of the original sector: \(\frac{64}{3}\pi \approx 67.02\text{ cm}^2\). M1 for calculating the area of the smaller sector: \(\frac{25}{3}\pi \approx 26.18\text{ cm}^2\). M1 for calculating the area of the circular hole: \(\pi \times 0.75^2 \approx 1.77\text{ cm}^2\). M1 for finding the remaining area: \(12.4375\pi \approx 39.07\text{ cm}^2\). A1 for the correct percentage of 58.3% (accept 58.3).

Let the original number of green counters in the bag be \(x\). The total number of counters in the bag originally is \(7 + 5 + x = 12 + x\). The probability of choosing a green counter first is: \(\text{P}(\text{First is Green}) = x / (12 + x)\). Since the counter is not replaced, the number of green counters remaining is \(x - 1\), and the total number of counters remaining is \(11 + x\). The probability of choosing a green counter second is: \(\text{P}(\text{Second is Green}) = (x - 1) / (11 + x)\). The probability that both counters are green is given by: \([x(x - 1)] / [(12 + x)(11 + x)] = 1 / 20\). Clearing the fractions gives: \(20(x^2 - x) = (12 + x)(11 + x) \implies 20x^2 - 20x = x^2 + 23x + 132\). Rearranging into a standard quadratic equation set to 0: \(19x^2 - 43x - 132 = 0\). Solving using the quadratic formula: \(x = (43 \pm \sqrt{(-43)^2 - 4(19)(-132)}) / (2 \times 19) = (43 \pm \sqrt{1849 + 10032}) / 38 = (43 \pm \sqrt{11881}) / 38 = (43 \pm 109) / 38\). Since \(x\) must be a positive integer: \(x = (43 + 109) / 38 = 152 / 38 = 4\). Originally, there were 4 green counters in the bag.

M1 for expressing the total number of counters as \(12 + x\). M1 for writing the probability of both counters being green: \(\frac{x}{12 + x} \times \frac{x - 1}{11 + x}\). M1 for equating to \(\frac{1}{20}\) and multiplying out brackets. M1 for rearranging to a standard quadratic equation of the form \(19x^2 - 43x - 132 = 0\) and attempting to solve. A1 for the correct answer of 4.

First, write the relationship between \(y\) and \(x\) as \(y = k_1 / x^2\). Substitute \(x = 3\) and \(y = 12\) to find \(k_1\): \(12 = k_1 / 3^2 \implies k_1 = 12 \times 9 = 108\). Thus, \(y = 108 / x^2\). Next, write the relationship between \(z\) and \(y\) as \(z = k_2 \sqrt[3]{y}\). Substitute \(y = 8\) and \(z = 5\) to find \(k_2\): \(5 = k_2 \sqrt[3]{8} \implies 5 = 2k_2 \implies k_2 = 2.5\). Thus, \(z = 2.5 \sqrt[3]{y}\). Now, find \(y\) when \(x = 2\): \(y = 108 / 2^2 = 108 / 4 = 27\). Finally, find \(z\) when \(y = 27\): \(z = 2.5 \sqrt[3]{27} = 2.5 \times 3 = 7.5\).

M1 for setting up \(y = k_1 / x^2\) and finding \(k_1 = 108\). M1 for setting up \(z = k_2 \sqrt[3]{y}\) and finding \(k_2 = 2.5\). M1 for substituting \(x = 2\) into their equation for \(y\) to find \(y = 27\). M1 for substituting their value of \(y\) into their equation for \(z\). A1 for the correct answer of 7.5.

Let \(V\) be the original value of the painting. After 3 years of \(5\%\) annual compound growth, the selling price \(S\) of the painting is: \(S = V \times (1.05)^3\). The seller pays a \(20\%\) tax on this selling price \(S\), meaning they keep \(80\%\) of the selling price. The amount left is: \(\text{Amount Left} = 0.80 \times S = 4630.50\). Substitute \(S\) into the equation: \(0.80 \times V \times (1.05)^3 = 4630.50\). First, calculate the compound multiplier: \((1.05)^3 = 1.157625\). Now multiply by \(0.80\): \(0.80 \times 1.157625 = 0.9261\). So we have: \(0.9261 \times V = 4630.50\). Solve for \(V\): \(V = 4630.50 / 0.9261 = 5000\). The original value of the painting was \(\$5000\).

M1 for representing the value after 3 years as \(V \times 1.05^3\). M1 for recognizing that the amount left represents \(80\%\) of the selling price, i.e., \(0.8 \times \text{selling price}\). M1 for setting up the equation \(0.8 \times V \times 1.05^3 = 4630.50\). M1 for evaluating \(0.8 \times 1.05^3 = 0.9261\) and attempting to divide \(4630.50\) by this multiplier. A1 for the correct answer of 5000.

Let the two consecutive odd integers be \(2n - 1\) and \(2n + 1\), where \(n\) is an integer. Find their squares: \((2n + 1)^2 = 4n^2 + 4n + 1\) and \((2n - 1)^2 = 4n^2 - 4n + 1\). Find the difference between these squares: \((4n^2 + 4n + 1) - (4n^2 - 4n + 1) = 4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n\). Since \(n\) is an integer, \(8n\) is always a multiple of 8. Alternatively, let the odd integers be \(2n + 1\) and \(2n + 3\). Their squares are \((2n+3)^2 = 4n^2 + 12n + 9\) and \((2n+1)^2 = 4n^2 + 4n + 1\). Difference: \((4n^2 + 12n + 9) - (4n^2 + 4n + 1) = 8n + 8 = 8(n + 1)\) since \(n+1\) is an integer, \(8(n+1)\) is a multiple of 8.

M1: Writes expressions for two consecutive odd integers, e.g., \(2n - 1\) and \(2n + 1\). M1: Correctly expands both squared expressions. M1: Subtracts the two expansions to get \(8n\) or \(8n + 8\). A1: Correctly factorises or concludes that the resulting expression is a multiple of 8, with a clear statement that \(n\) is an integer.

The area of the sector is given by: \(\text{Area of sector} = \frac{60}{360} \times \pi r^2 = \frac{1}{6} \pi r^2\). The triangle formed by the two radii and the chord is an equilateral triangle (since the angle is 60 degrees and the two sides are equal to \(r\)). Its area is given by: \(\text{Area of triangle} = \frac{1}{2} a b \sin(C) = \frac{1}{2} r^2 \sin(60^{\circ}) = \frac{1}{2} r^2 \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{4} r^2\). The area of the segment is the area of the sector minus the area of the triangle: \(\text{Area of segment} = \frac{1}{6} \pi r^2 - \frac{\sqrt{3}}{4} r^2 = r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\). This is the required expression.

M1: Calculates the area of the sector as \(\frac{60}{360}\pi r^2\) or \(\frac{1}{6}\pi r^2\). M1: Uses \(\frac{1}{2}ab\sin(C)\) or another valid method to express the area of the triangle. M1: Correctly substitutes \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\) to find the triangle area as \(\frac{\sqrt{3}}{4}r^2\). A1: Correctly subtracts and factorises to obtain \(r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Since \(y\) is inversely proportional to \(x^2\), we can write \(y = \frac{k}{x^2}\) for some constant \(k\). Let the initial value of \(x\) be \(x_1\) and the initial value of \(y\) be \(y_1 = \frac{k}{x_1^2}\). If \(x\) increases by 25%, the new value of \(x\) is \(x_2 = 1.25x_1\). Substitute \(x_2\) into the formula to find the new value of \(y\): \(y_2 = \frac{k}{(1.25x_1)^2} = \frac{k}{1.5625 x_1^2}\). This can be written as \(y_2 = \frac{1}{1.5625} \times \frac{k}{x_1^2} = 0.64 y_1\). The percentage change in \(y\) is: \(\frac{y_2 - y_1}{y_1} \times 100\\% = \frac{0.64y_1 - y_1}{y_1} \times 100\\% = -0.36 \times 100\\% = -36\\%\). This represents a decrease of 36%.

M1: Formulates the proportion equation \(y = \frac{k}{x^2}\) or equivalent. M1: Expresses the new value of \(x\) as \(1.25x\). M1: Substitutes \(1.25x\) to find the new \(y\) expression and evaluates \(\frac{1}{1.25^2}\) as \(0.64\). A1: Shows that \(0.64\) represents a 36% decrease clearly.

First, find \(\overrightarrow{AB}\): \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). Since \(AN:NB = 1:3\), \(\overrightarrow{AN} = \frac{1}{4}\overrightarrow{AB} = \frac{1}{4}(\mathbf{b} - \mathbf{a})\). Find \(\overrightarrow{ON}\): \(\overrightarrow{ON} = \overrightarrow{OA} + \overrightarrow{AN} = \mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a}) = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\). Since \(M\) is the midpoint of \(OB\), \(\overrightarrow{OM} = \frac{1}{2}\mathbf{b}\). Now find \(\overrightarrow{MN}\): \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\frac{1}{2}\mathbf{b} + \left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = \frac{3}{4}\mathbf{a} - \frac{1}{4}\mathbf{b}\). This is the required expression.

M1: Finds \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or \(\overrightarrow{BA} = \mathbf{a} - \mathbf{b}\). M1: Finds a correct expression for \(\overrightarrow{AN}\) or \(\overrightarrow{NB}\), e.g., \(\overrightarrow{AN} = \frac{1}{4}(\mathbf{b} - \mathbf{a})\). M1: Formulates a correct vector path for \(\overrightarrow{MN}\), e.g., \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON}\) or equivalent. A1: Correctly simplifies the vectors to obtain \(\frac{3}{4}\mathbf{a} - \frac{1}{4}\mathbf{b}\).