2023 OCR GCSE Mathematics - J560 模拟试题及答案详解

The ratio of orange juice to lemonade is \( 3 : 5 \). The quantity of lemonade is \( 1200\text{ ml} \), which represents \( 5 \) parts. First, find the volume of \( 1 \) part: \( 1200 \div 5 = 240\text{ ml} \). The total number of parts is \( 3 + 5 = 8 \) parts. Now, find the total volume of punch: \( 8 \times 240 = 1920\text{ ml} \).

M1 for \( 1200 \div 5 \) (or \( 240 \)) or \( 1200 \times \frac{8}{5} \) A1 for \( 1920 \) (accept \( 1920\text{ ml} \))

The sale price represents \( 100\% - 15\% = 85\% \) of the original price. Therefore, \( 85\% \) of the original price is \( £544 \). To find the original price: \( \text{Original price} = 544 \div 0.85 = 640 \). Thus, the original price of the laptop was \( £640 \).

M1 for \( 544 \div 0.85 \) or \( \frac{544}{85} \times 100 \) A1 for \( 640 \) (accept \( £640 \))

The formula for the volume of a cylinder is: \( V = \pi r^2 h \) where \( r \) is the radius of the base and \( h \) is the height. Substitute the given values into the formula: \( V = \pi \times 5^2 \times 8 \) which gives \( V = \pi \times 25 \times 8 = 200\pi \). So, the volume is \( 200\pi\text{ cm}^3 \).

M1 for \( \pi \times 5^2 \times 8 \) A1 for \( 200\pi \) (accept \( 200\pi\text{ cm}^3 \))

The difference in parts between blue and red counters is \( 7 - 3 = 4 \) parts. These \( 4 \) parts represent \( 36 \) counters. Therefore, \( 1 \) part represents \( 36 \div 4 = 9 \) counters. The total number of parts is \( 3 + 7 = 10 \) parts. The total number of counters is \( 10 \times 9 = 90 \).

M1 for finding that 4 parts equal 36, or for \( 36 \div (7 - 3) \). A1 for 90.

Multiply both sides by \( 5 \): \( 4x - 3 = 25 \). Add \( 3 \) to both sides: \( 4x = 28 \). Divide by \( 4 \): \( x = 7 \).

M1 for isolating the numerator, e.g. \( 4x - 3 = 25 \). A1 for 7.

The formula for the total surface area of a cuboid is \( 2(lw + lh + wh) \). Substituting the given values: \( 2(8 \times 5 + 8h + 5h) = 184 \). Divide by 2: \( 40 + 13h = 92 \). Subtract 40: \( 13h = 52 \). Divide by 13: \( h = 4 \). The height is \( 4\text{ cm} \).

M1 for setting up a correct equation for surface area, e.g. \( 2(40 + 13h) = 184 \) or \( 80 + 26h = 184 \). A1 for 4.

The ratio is apples : bananas : oranges = \( 3 : 5 : 4 \). The difference in ratio parts between bananas and apples is \( 5 - 3 = 2 \) parts. We are given that this difference represents 6 bananas. Therefore, 1 part represents \( 6 \div 2 = 3 \) pieces of fruit. The total number of parts is \( 3 + 5 + 4 = 12 \) parts. Thus, the total number of pieces of fruit is \( 12 \times 3 = 36 \).

M1 for finding the value of one share, e.g. \( 6 \div (5 - 3) = 3 \), or for writing a correct equation such as \( 5x - 3x = 6 \). A1 for 36.

A reduction of \( 15\% \) means the sale price is \( 100\% - 15\% = 85\% \) of the original price. Let \( P \) be the original price. We have \( 0.85 \times P = 68 \). Dividing both sides by 0.85 gives \( P = \frac{68}{0.85} = 80 \). So the original price of the coat was £80.

M1 for \( 68 \div 0.85 \) or for equating \( 85\% \) to 68 (e.g. \( 85\% = 68 \)). A1 for 80.

Multiply both sides of the equation by 5 to clear the fraction: \( 2x - 3 = 35 \). Add 3 to both sides: \( 2x = 38 \). Divide by 2: \( x = 19 \).

M1 for multiplying both sides by 5 to get \( 2x - 3 = 35 \) (or equivalent first step). A1 for 19.

First, find the total parts in the ratio: \( 5 + 3 + 2 = 10 \) parts. Next, convert the total volume from litres to millilitres: \( 3.5 \text{ litres} = 3500 \text{ ml} \). Find the volume of one part: \( 3500 \div 10 = 350 \text{ ml} \). Mango juice represents 3 parts of the ratio, so multiply the volume of one part by 3: \( 3 \times 350 = 1050 \text{ ml} \).

M1 for translating the total volume to ml (3500) OR calculating \( 3.5 \div 10 = 0.35 \) litres. A1 for 1050 (accept 1050 ml).

To solve the equation, first multiply both sides by 3 to eliminate the fraction: \( 4x - 1 = 9 \times 3 \) which simplifies to \( 4x - 1 = 27 \). Next, add 1 to both sides: \( 4x = 28 \). Finally, divide both sides by 4: \( x = 7 \).

M1 for showing multiplication by 3, e.g., \( 4x - 1 = 27 \). A1 for 7.

The sale price represents \( 100\% - 10\% = 90\% \) of the original price. To find the original price, divide the sale price by 0.90: \( 468 \div 0.90 = 520 \). Therefore, the original price of the laptop was \( \pounds 520 \).

M1 for setting up the equation \( 0.90 \times P = 468 \) or writing \( 468 \div 0.90 \). A1 for 520 (accept \( \pounds 520 \)).

We are given the ratio of Red : Blue : Green is \(3 : 5 : 4\). We know there are 24 green counters. The ratio part for green is 4. First, find the value of 1 part of the ratio: \(24 \div 4 = 6\) counters. Next, find the total number of parts in the ratio: \(3 + 5 + 4 = 12\) parts. Finally, multiply the total parts by the value of 1 part: \(12 \times 6 = 72\) counters.

M1 for \(24 \div 4\) (or showing that 1 part is 6, or evaluating \(\frac{24}{4} \times 12\)). A1 for 72.

First, expand the bracket on the left side of the equation: \(4x - 12 = 18 - 2x\). Next, add \(2x\) to both sides to collect the \(x\) terms: \(6x - 12 = 18\). Then, add 12 to both sides to isolate the term with \(x\): \(6x = 30\). Finally, divide by 6 to find \(x\): \(x = 5\).

M1 for expanding the bracket correctly to \(4x - 12\) or for a correct rearrangement of their terms (e.g., reaching \(6x = 30\)). A1 for 5 (or \(x = 5\)).

The formula for the volume of a cylinder is \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. Substitute the given values into the formula: \(V = \pi \times 3^2 \times 8 = \pi \times 9 \times 8 = 72\pi\text{ cm}^3\).

M1 for substituting correctly into the volume formula: \(\pi \times 3^2 \times 8\). A1 for \(72\pi\).

To find the number of Chocolate cupcakes, we first need to express the ratios with a common term for Chocolate cupcakes.

We are given:
- \(\text{Vanilla} : \text{Chocolate} = 3 : 5\)
- \(\text{Chocolate} : \text{Strawberry} = 4 : 7\)

The lowest common multiple of 5 and 4 (the parts representing Chocolate) is 20.

Multiply the first ratio by 4:
\(\text{Vanilla} : \text{Chocolate} = 12 : 20\)

Multiply the second ratio by 5:
\(\text{Chocolate} : \text{Strawberry} = 20 : 35\)

Now, we can write the combined ratio:
\(\text{Vanilla} : \text{Chocolate} : \text{Strawberry} = 12 : 20 : 35\)

Next, find the total number of parts in this ratio:
\(12 + 20 + 35 = 67\text{ parts}\)

Since the total number of cupcakes sold is 1005, find the value of one part:
\(\text{Value of 1 part} = \frac{1005}{67} = 15\)

Finally, calculate the number of Chocolate cupcakes sold (which has 20 parts):
\(\text{Chocolate cupcakes} = 20 \times 15 = 300\)

M1: For a process to find a common value for Chocolate in both ratios, e.g., multiplying the first ratio by 4 and/or the second ratio by 5.
M1: For writing the combined ratio \(\text{Vanilla} : \text{Chocolate} : \text{Strawberry} = 12 : 20 : 35\) (or equivalent).
M1: For dividing 1005 by the sum of their parts, i.e., \(1005 \div 67\).
A1: For 300.

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

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

After the second reduction of 10%, the price is:
\(0.8P \times (1 - 0.10) = 0.8P \times 0.9 = 0.72P\)

We are given that the price in the second week is £324, so:
\(0.72P = 324\)

To find the original price \(P\):
\(P = \frac{324}{0.72} = 450\)

Alternatively, working backwards:
- Price before the second reduction = \(\frac{324}{0.90} = £360\)
- Original price before the first reduction = \(\frac{360}{0.80} = £450\)

M1: For a process to find the overall multiplier for both reductions, e.g., \(0.8 \times 0.9 = 0.72\), or for calculating the intermediate price as \(\frac{324}{0.9}\).
M1: For setting up the equation \(0.72P = 324\) or finding the price before the second reduction to be £360.
M1: For a complete method to find the original price, e.g., \(\frac{324}{0.72}\) or \(\frac{360}{0.8}\).
A1: For 450.

The area of a rectangle is found by multiplying its length by its width:
\(\text{Area} = \text{length} \times \text{width}\)

Forming the equation:
\((2x + 3)(x - 1) = 52\)

Expand the brackets:
\(2x^2 - 2x + 3x - 3 = 52\)
\(2x^2 + x - 3 = 52\)

Rearrange the quadratic equation into the form \(ax^2 + bx + c = 0\) by subtracting 52 from both sides:
\(2x^2 + x - 55 = 0\)

Factorise the quadratic expression. We need two numbers that multiply to \(2 \times (-55) = -110\) and add up to 1.
These numbers are \(+11\) and \(-10\).

Rewrite and factorise by grouping:
\(2x^2 - 10x + 11x - 55 = 0\)
\(2x(x - 5) + 11(x - 5) = 0\)
\((2x + 11)(x - 5) = 0\)

This gives two possible solutions:
\(2x + 11 = 0 \implies x = -5.5\)
\(x - 5 = 0 \implies x = 5\)

Since the width of the rectangle is \((x - 1)\) cm, the value of \(x\) must be greater than 1 for the width to be positive.
Therefore, we reject \(x = -5.5\).

Thus, \(x = 5\).

M1: For setting up the equation \((2x + 3)(x - 1) = 52\) and attempting to expand the brackets to get at least three correct terms in the expansion.
M1: For rearranging into the standard form \(2x^2 + x - 55 = 0\) (condone a minor arithmetic slip in the constant term).
M1: For a valid method to solve their quadratic equation, e.g., factorising to \((2x + 11)(x - 5) = 0\) or using the quadratic formula.
A1: For \(x = 5\) only (with the negative value either explicitly rejected or omitted from the final answer).

First, calculate the total number of parts in the ratio:
\(10 + 6 + 1 = 17\) parts.

Next, find the mass of one part by dividing the total mass of one batch by 17:
\(8.5\text{ kg} = 8500\text{ g}\)
\(8500\text{ g} / 17 = 500\text{ g}\) per part.

Since yeast represents 1 part of the ratio, the mass of yeast used in one batch is:
\(1 \times 500\text{ g} = 500\text{ g}\).

For 3 batches, the total mass of yeast needed is:
\(3 \times 500\text{ g} = 1500\text{ g}\).

Finally, calculate the number of packets of yeast needed:
\(1500\text{ g} / 125\text{ g} = 12\) packets.

M1: For calculating the total parts in the ratio: \(10 + 6 + 1 = 17\).
M1: For finding the mass of yeast for one batch: \(8.5 / 17 = 0.5\text{ kg}\) or \(500\text{ g}\).
M1: For finding the total yeast for 3 batches (\(1500\text{ g}\) or \(1.5\text{ kg}\)) and dividing by \(125\text{ g}\) (or \(0.125\text{ kg}\)).
A1: 12 (cao).

First, calculate the volume of the cylinder using \(V = \pi r^2 h\):
\(V_{\text{cylinder}} = \pi \times 3^2 \times 8 = 72\pi \approx 226.195\text{ cm}^3\).

Since the cylinder is melted and recast into a sphere, their volumes are equal:
\(\frac{4}{3}\pi r^3 = 72\pi\).

Divide both sides by \(\pi\):
\(\frac{4}{3}r^3 = 72\).

Multiply both sides by \(\frac{3}{4}\):
\(r^3 = 72 \times \frac{3}{4} = 54\).

Find the cube root of 54:
\(r = \sqrt[3]{54} \approx 3.77976...\text{ cm}\).

Rounding to 3 significant figures gives \(r = 3.78\).

M1: For a correct method to find the cylinder volume: \(\pi \times 3^2 \times 8\) (or \(72\pi\) or \(226.195\)).
M1: For setting up the volume equation: \(\frac{4}{3}\pi r^3 = 72\pi\) (or their cylinder volume).
M1: For rearranging to get \(r^3 = 54\) (or \(r = \sqrt[3]{54}\)).
A1: 3.78 (accept answers in the range 3.77 to 3.80).

The area of a rectangle is length multiplied by width:
\((2x + 3)(x - 2) = 15\)

Expand the brackets:
\(2x^2 - 4x + 3x - 6 = 15\)
\(2x^2 - x - 6 = 15\)

Rearrange into a standard quadratic form \(ax^2 + bx + c = 0\):
\(2x^2 - x - 21 = 0\)

Factorise the quadratic expression:
We look for two numbers that multiply to \(2 \times -21 = -42\) and add to \(-1\). These are \(-7\) and \(6\).
\(2x^2 - 7x + 6x - 21 = 0\)
\(x(2x - 7) + 3(2x - 7) = 0\)
\((2x - 7)(x + 3) = 0\)

This gives two possible solutions:
\(x = 3.5\) or \(x = -3\)

Since the width of the rectangle is \((x - 2)\text{ cm}\), \(x\) must be greater than 2 for the width to be positive.
Therefore, we reject \(x = -3\).

So, the value of \(x\) is \(3.5\).

M1: For establishing the initial equation: \((2x + 3)(x - 2) = 15\).
M1: For expanding and rearranging to form \(2x^2 - x - 21 = 0\).
M1: For factorising their quadratic to \((2x - 7)(x + 3) = 0\) (or using the quadratic formula correctly).
A1: 3.5 (accept \(\frac{7}{2}\), must reject \(x = -3\)).

1. The radius of the hemisphere is \(r = 3\text{ cm}\).
2. The total height of the solid is \(15\text{ cm}\), so the height of the cylinder is \(h = 15 - r = 15 - 3 = 12\text{ cm}\).
3. The volume of the hemisphere is calculated using the formula \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\):
\(V_{\text{hemisphere}} = \frac{2}{3}\pi (3)^3 = \frac{2}{3} \times 27\pi = 18\pi\text{ cm}^3\).
4. The volume of the cylinder is calculated using the formula \(V_{\text{cylinder}} = \pi r^2 h\):
\(V_{\text{cylinder}} = \pi (3)^2 (12) = 9 \times 12\pi = 108\pi\text{ cm}^3\).
5. The total volume is the sum of these two volumes:
\(V_{\text{total}} = 18\pi + 108\pi = 126\pi\text{ cm}^3\).

M1: For finding the height of the cylinder: \(15 - 3 = 12\text{ cm}\).
M1: For a correct method to find the volume of either the hemisphere or the cylinder (e.g. \(\frac{2}{3} \times \pi \times 3^3\) or \(\pi \times 3^2 \times 12\)).
M1: For a complete method to find the total volume by adding both volume formulas with correct dimensions.
A1: For \(126\pi\) (accept exact form with unit, e.g. \(126\pi\text{ cm}^3\)).

1. Let the initial number of French students be \(3x\) and the initial number of Spanish students be \(5x\).
2. After 12 students switch from French to Spanish, the number of French students becomes \(3x - 12\) and the number of Spanish students becomes \(5x + 12\).
3. Since the new ratio is \(1 : 3\), we can write the equation:
\(\frac{3x - 12}{5x + 12} = \frac{1}{3}\)
4. Cross-multiply to solve for \(x\):
\(3(3x - 12) = 1(5x + 12)\)
\(9x - 36 = 5x + 12\)
\(4x = 48\)
\(x = 12\)
5. The total number of students studying these two languages is:
\(3x + 5x = 8x = 8 \times 12 = 96\).

M1: For representing the initial number of students algebraically (e.g. \(3x\) and \(5x\)).
M1: For setting up a correct equation for the new ratio (e.g. \(\frac{3x - 12}{5x + 12} = \frac{1}{3}\)).
M1: For a correct method to solve the equation, leading to \(x = 12\) (or finding initial French = 36 and initial Spanish = 60).
A1: For \(96\).

1. Using Pythagoras' theorem, \(a^2 + b^2 = c^2\), we set up the equation:
\(x^2 + (2x + 2)^2 = (2x + 3)^2\)
2. Expand both squared binomial terms:
\(x^2 + (4x^2 + 8x + 4) = 4x^2 + 12x + 9\)
3. Collect terms and simplify into a standard quadratic equation form:
\(5x^2 + 8x + 4 = 4x^2 + 12x + 9\)
\(x^2 - 4x - 5 = 0\)
4. Factorise the quadratic equation:
\((x - 5)(x + 1) = 0\)
Since lengths must be positive, we reject \(x = -1\) and find \(x = 5\).
5. Substitute \(x = 5\) back to find the perpendicular sides of the triangle:
Base = \(5\text{ cm}\)
Height = \(2(5) + 2 = 12\text{ cm}\)
6. Calculate the area:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\).

M1: For applying Pythagoras' theorem to set up: \(x^2 + (2x + 2)^2 = (2x + 3)^2\).
M1: For correctly expanding the brackets and simplifying to the quadratic equation \(x^2 - 4x - 5 = 0\).
M1: For solving the quadratic equation to find \(x = 5\) (condone inclusion of \(x = -1\) if subsequently rejected) and calculating the two perpendicular sides as \(5\) and \(12\).
A1: For \(30\) (accept \(30\text{ cm}^2\)).

To combine the ratios, we find a common value for the blue counters (the term common to both ratios). The ratio of Red to Blue is \(3 : 5\) and the ratio of Blue to Green is \(4 : 7\). The lowest common multiple of 5 and 4 is 20. Multiplying the Red : Blue ratio by 4 gives \(R : B = 12 : 20\). Multiplying the Blue : Green ratio by 5 gives \(B : G = 20 : 35\). Now we can combine them into a single ratio: \(R : B : G = 12 : 20 : 35\). The difference between the green and blue parts in the ratio is \(35 - 20 = 15\) parts. Since there are 45 more green counters than blue counters, we have: \(15\text{ parts} = 45\), which means \(1\text{ part} = 3\text{ counters}\). The total number of parts is \(12 + 20 + 35 = 67\) parts. Thus, the total number of counters in the bag is \(67 \times 3 = 201\).

M1: scales the ratios to find a common term for blue, e.g., \(R:B = 12:20\) and \(B:G = 20:35\). M1: writes the combined ratio \(R:B:G = 12:20:35\). M1: finds the value of one part by calculating \(45 / (35 - 20) = 3\). A1: for 201.

The area of a rectangle is found by multiplying its length by its width: \(\text{Area} = (2x + 3)(x - 1)\). Since the area is \(52\text{ cm}^2\), we can write the equation: \((2x + 3)(x - 1) = 52\). Expanding the brackets gives: \(2x^2 - 2x + 3x - 3 = 52\), which simplifies to \(2x^2 + x - 3 = 52\). Rearranging to make the quadratic equal to zero: \(2x^2 + x - 55 = 0\). To factorise the quadratic, we seek two numbers that multiply to \(-110\) and add to \(1\). These are \(11\) and \(-10\). Rewriting the middle term: \(2x^2 - 10x + 11x - 55 = 0\) which factorises to \((2x + 11)(x - 5) = 0\). This yields solutions \(x = -5.5\) or \(x = 5\). Since length and width must be positive, \(x - 1 > 0\), so we reject the negative value to get \(x = 5\). Substituting \(x = 5\) into the dimensions: Length = \(2(5) + 3 = 13\text{ cm}\), and Width = \(5 - 1 = 4\text{ cm}\). The perimeter is \(2 \times (\text{Length} + \text{Width}) = 2 \times (13 + 4) = 34\text{ cm}\).

M1: sets up the initial area equation \((2x + 3)(x - 1) = 52\). M1: simplifies to a standard quadratic equation \(2x^2 + x - 55 = 0\). M1: solves the quadratic equation to find \(x = 5\) (accepting omission of the negative root). A1: for 34.

The volume of a sphere is given by \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). The volume of a cylinder is given by \(V_{\text{cylinder}} = \pi R^2 h\). Here, the cylinder has radius \(R = 2r\) and height \(h = 12\), so: \(V_{\text{cylinder}} = \pi (2r)^2 \times 12 = \pi (4r^2) \times 12 = 48\pi r^2\). Since the volume of the cylinder is 3 times the volume of the sphere, we set up the equation: \(V_{\text{cylinder}} = 3 \times V_{\text{sphere}}\) which gives \(48\pi r^2 = 3 \times \left(\frac{4}{3}\pi r^3\right)\). Simplifying the right-hand side gives \(48\pi r^2 = 4\pi r^3\). Dividing both sides by \(4\pi r^2\) (since \(r \neq 0\)) gives \(12 = r\). So, \(r = 12\).

M1: writes a correct formula for the volume of the sphere in terms of \(r\), i.e., \(\frac{4}{3}\pi r^3\). M1: writes a simplified expression for the volume of the cylinder in terms of \(r\), i.e., \(48\pi r^2\). M1: sets up the equation equating the volumes, e.g., \(48\pi r^2 = 4\pi r^3\). A1: for 12.

First, write an expression for the total surface area of the cuboid. The cuboid has six faces with areas in pairs:
- Two faces of size \(x \times (x + 2)\) with total area \(2x(x + 2) = 2x^2 + 4x\)
- Two faces of size \(x \times 4\) with total area \(2 \times 4x = 8x\)
- Two faces of size \((x + 2) \times 4\) with total area \(2 \times 4(x + 2) = 8x + 16\)

Adding these together gives the total surface area:
\text{Total Surface Area} = (2x^2 + 4x) + 8x + (8x + 16) = 2x^2 + 20x + 16

We are given that the total surface area is \(128\text{ cm}^2\), so we set up the equation:
\(2x^2 + 20x + 16 = 128\)

Subtract \(128\) from both sides:
\(2x^2 + 20x - 112 = 0\)

Divide the entire equation by \(2\):
\(x^2 + 10x - 56 = 0\)

Factorise the quadratic equation:
\((x + 14)(x - 4) = 0\)

This gives two possible solutions for \(x\):
\(x = -14\) or \(x = 4\)

Since \(x\) represents a physical length, it must be positive. Therefore, \(x = 4\).

Now, calculate the dimensions of the cuboid:
- Length = \(4\text{ cm}\)
- Width = \(4 + 2 = 6\text{ cm}\)
- Height = \(4\text{ cm}\)

Calculate the volume of the cuboid:
\text{Volume} = \text{length} \times \text{width} \times \text{height}
\text{Volume} = 4 \times 6 \times 4 = 96\text{ cm}^3\).

**M1:** For a correct expression for the total surface area in terms of \(x\), e.g., \(2(x(x+2) + 4x + 4(x+2))\) or \(2x^2 + 20x + 16\).
**M1:** For setting up the equation \(2x^2 + 20x + 16 = 128\) and reducing it to a three-term quadratic equation, e.g., \(x^2 + 10x - 56 = 0\) or \(2x^2 + 20x - 112 = 0\).
**M1:** For solving their quadratic equation to find the positive root, leading to \(x = 4\).
**A1:** For \(96\) (units not required, but reject if incorrect units are given, e.g., \(96\text{ cm}^2\)).

First, we find a combined ratio for Alloy A. The ratio of copper to zinc is \(5 : 3\) and the ratio of zinc to tin is \(4 : 1\). To combine these, we find a common multiplier for zinc, which is the lowest common multiple of 3 and 4, i.e., 12. Multiplying the copper to zinc ratio by 4 gives \(\text{Copper} : \text{Zinc} = 20 : 12\). Multiplying the zinc to tin ratio by 3 gives \(\text{Zinc} : \text{Tin} = 12 : 3\). So, the ratio of \(\text{Copper} : \text{Zinc} : \text{Tin}\) in Alloy A is \(20 : 12 : 3\). The total number of parts in Alloy A is \(20 + 12 + 3 = 35\). Thus, the fraction of Alloy A that is copper is \(\frac{20}{35} = \frac{4}{7}\). Alloy B is made by mixing Alloy A with pure copper in the ratio \(3 : 1\) by mass. Let the total mass of Alloy B be 4 units, consisting of 3 units of Alloy A and 1 unit of pure copper. The mass of copper in 3 units of Alloy A is \(3 \times \frac{4}{7} = \frac{12}{7}\) units. We add 1 unit of pure copper, so the total mass of copper in Alloy B is \(\frac{12}{7} + 1 = \frac{19}{7}\) units. The percentage of Alloy B that is copper is \(\frac{19/7}{4} \times 100\% = \frac{19}{28} \times 100\% \approx 67.857\%\). To 1 decimal place, this is \(67.9\%\).

M1: Method to find a common ratio for Alloy A, e.g., \(\text{Copper} : \text{Zinc} : \text{Tin} = 20 : 12 : 3\). M1: Calculates the fraction of Alloy A that is copper, e.g., \(\frac{20}{35}\) or \(\frac{4}{7}\). M1: Sets up an equation or scenario for Alloy B, e.g., using 3 parts of Alloy A and 1 part of pure copper. M1: Calculates the total mass of copper in Alloy B, e.g., \(\frac{12}{7} + 1 = \frac{19}{7}\). M1: Finds the fraction of Alloy B that is copper, e.g., \(\frac{19}{28}\). A1: Correctly rounds to 1 decimal place to give \(67.9\%\) (accept 67.9).

The total volume \(V\) of the ornament is the sum of the volume of the hemisphere and the volume of the cylinder: \(V = \frac{2}{3}\pi r^3 + \pi r^2 h\). Substituting \(h = 3r\) into the volume equation gives: \(V = \frac{2}{3}\pi r^3 + \pi r^2(3r) = \frac{2}{3}\pi r^3 + 3\pi r^3 = \frac{11}{3}\pi r^3\). We are given that the total volume is \(99\pi\text{ cm}^3\), so: \(\frac{11}{3}\pi r^3 = 99\pi \implies \frac{11}{3}r^3 = 99 \implies r^3 = \frac{99 \times 3}{11} = 27\). Taking the cube root gives \(r = 3\text{ cm}\), and therefore \(h = 3 \times 3 = 9\text{ cm}\). The total surface area \(A\) of the ornament includes the curved surface area of the hemisphere, the curved surface area of the cylinder, and the circular base of the cylinder (not the joined inner faces): \(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\). Substituting \(r = 3\) and \(h = 9\) gives: \(A = 3\pi(3^2) + 2\pi(3)(9) = 27\pi + 54\pi = 81\pi\text{ cm}^2\).

M1: Writes an expression for the total volume of the ornament, i.e., \(V = \frac{2}{3}\pi r^3 + \pi r^2 h\). M1: Substitutes \(h = 3r\) to obtain \(V = \frac{11}{3}\pi r^3\). A1: Solves for \(r\) to find \(r = 3\). M1: Correctly identifies the formula for the outer surface area of the combined shape: \(A = 2\pi r^2 + 2\pi r h + \pi r^2\) or \(3\pi r^2 + 2\pi r h\). M1: Substitutes their values of \(r = 3\) and \(h = 9\) into their surface area expression. A1: Obtains \(81\pi\).

First, find the equation of the line \(L\). The gradient \(m\) is: \(m = \frac{5 - (-7)}{4 - (-2)} = \frac{12}{6} = 2\). Using the point \(B(4, 5)\), the equation of the line is: \(y - 5 = 2(x - 4) \implies y = 2x - 3\). Since the line \(L\) is a tangent to the curve \(C\), they intersect at exactly one point. Set their equations equal to each other: \(2x - 3 = x^2 - kx + 13\). Rearranging into standard quadratic form: \(x^2 - (k + 2)x + 16 = 0\). For the line to be a tangent, the discriminant of this quadratic equation must be zero (\(\Delta = b^2 - 4ac = 0\)): \([-(k+2)]^2 - 4(1)(16) = 0 \implies (k+2)^2 - 64 = 0 \implies (k+2)^2 = 64\). Taking the square root of both sides gives: \(k + 2 = \pm 8\). Solving these two equations: 1) \(k + 2 = 8 \implies k = 6\), 2) \(k + 2 = -8 \implies k = -10\).

M1: Method to find the gradient of the line, e.g., \(\frac{5 - (-7)}{4 - (-2)} = 2\). M1: Finds the correct equation of the line \(L\): \(y = 2x - 3\). M1: Equates the line and curve equations: \(2x - 3 = x^2 - kx + 13\). M1: Rearranges into standard quadratic form, e.g., \(x^2 - (k+2)x + 16 = 0\). M1: Uses the discriminant condition for tangency (\(\Delta = 0\)), e.g., \((k+2)^2 - 64 = 0\). A1: Correctly identifies both values of \(k\): \(k = 6\) and \(k = -10\).