2023 OCR GCSE Mathematics - J560 模擬試題連答案詳解

Number of hot meals = 120, which represents 8 parts of the ratio. One part represents \(120 \div 8 = 15\) meals. The total number of parts is \(3 + 8 = 11\). Therefore, the total number of salads and hot meals sold is \(11 \times 15 = 165\).

M1 for \(120 \div 8\) or \(120 \div 8 \times 3\) or for finding 45 salads. A1 for 165.

The total number of ratio parts is \(1 + 2 + 4 = 7\). A total mass of 350 kg is divided into 7 equal parts: \(350 \text{ kg} \div 7 = 50\text{ kg}\) per part. Since sand represents 2 parts, the mass of sand needed is \(2 \times 50\text{ kg} = 100\text{ kg}\).

M1 for \(350 \div (1 + 2 + 4)\) or showing 50. A1 for 100.

The sale price of %1 represents \(100\% - 15\% = 85\%\) of the normal price. Let the normal price be \(x\). We have \(0.85x = 51\), so \(x = 51 \div 0.85 = 60\). The normal price is &0.

M1 for \(51 \div 0.85\) or setting up \(85\% = 51\). A1 for 60.

The actual increase in the number of members is \(108 - 80 = 28\). The percentage increase is the increase divided by the original number, multiplied by 100: \(\frac{28}{80} \times 100 = 0.35 \times 100 = 35\%\).

M1 for \(\frac{108 - 80}{80}\) or \(\frac{28}{80}\). A1 for 35.

Completing the square for the expression: \(x^2 - 8x + 19 = (x - 4)^2 - 4^2 + 19 = (x - 4)^2 - 16 + 19 = (x - 4)^2 + 3\). The coordinates of the turning point are derived from the form \(y = (x - a)^2 + b\), which gives the turning point \((a, b)\). Thus, the turning point is \((4, 3)\).

M1 for completing the square to get \((x - 4)^2 + k\) where \(k \neq 19\) or for finding the x-coordinate of the turning point is 4. A1 for \((4, 3)\) (accept \(x = 4, y = 3\)).

Since line \(M\) is parallel to line \(L\), it has the same gradient, so the gradient \(m = 3\). The equation of line \(M\) is of the form \(y = 3x + c\). Substituting the point \((2, 10)\) into this equation gives \(10 = 3(2) + c\), which simplifies to \(10 = 6 + c\), so \(c = 4\). Therefore, the equation of the line is \(y = 3x + 4\).

M1 for identifying the gradient is 3 or writing \(y = 3x + c\) or substituting \((2, 10)\) into \(y = mx + c\). A1 for \(y = 3x + 4\) (or equivalent).

Multiply both sides of the equation by 3 to get \(5x - 2 = 18\). Add 2 to both sides to get \(5x = 20\). Divide both sides by 5 to find \(x = 4\).

M1 for multiplying both sides by 3 to get \(5x - 2 = 18\). A1 for \(x = 4\) (accept 4).

The volume of a cylinder is given by the formula \(V = \pi r^2 h\). Substituting the given values: \(V = \pi \times 3^2 \times 10 = \pi \times 9 \times 10 = 90\pi\text{ cm}^3\).

M1 for substituting correctly into the cylinder volume formula \(V = \pi \times 3^2 \times 10\). A1 for \(90\pi\).

The ratio of red to blue marbles is \(3 : 5\). The difference in ratio parts is \(5 - 3 = 2\) parts. Since there are 24 more blue marbles, we have:
\(2 \text{ parts} = 24\)
\(1 \text{ part} = 12\)
The total number of parts is \(3 + 5 = 8\).
Total marbles = \(8 \times 12 = 96\).

M1 for setting up an equation with parts, e.g., \(5 - 3 = 2\) parts represents 24, or finding 1 part = 12. A1 for 96.

To find \(a : c\), make the term for \(b\) the same in both ratios.
Multiply \(a : b\) by 3: \(a : b = 12 : 15\).
Multiply \(b : c\) by 5: \(b : c = 15 : 35\).
Now we have \(a : b : c = 12 : 15 : 35\).
Therefore, \(a : c = 12 : 35\).

M1 for converting both ratios to have a common value for \(b\) (e.g. \(12:15\) and \(15:35\)). A1 for \(12 : 35\).

The sale price of £51 represents \(100\% - 15\% = 85\%\) of the original price.
Let \(P\) be the original price:
\(0.85 \times P = 51\)
\(P = \frac{51}{0.85} = 60\).
The original price of the coat was £60.

M1 for \(51 \div 0.85\) or showing that \(85\% = 51\). A1 for 60.

Let the initial value of the house be \(100\%\) (or 1).
After Year 1, the value is \(1 \times 1.04 = 1.04\).
After Year 2, the value is \(1.04 \times (1 - 0.05) = 1.04 \times 0.95 = 0.988\).
The overall percentage multiplier is \(0.988\), which is equivalent to \(98.8\%\) of the original value.
Overall change = \(98.8\% - 100\% = -1.2\%\) (a 1.2% decrease).

M1 for multiplying the percentage scales together, e.g., \(1.04 \times 0.95\) or showing \(0.988\) / \(98.8\%\). A1 for -1.2% or 1.2% decrease.

The equation is written in completed square form, \(y = (x - h)^2 + k\), where the coordinates of the turning point are \((h, k)\).
For the curve \(y = (x - 3)^2 + 7\), the minimum value of \(y\) occurs when \(x - 3 = 0\), which is \(x = 3\).
When \(x = 3\), \(y = 7\).
So, the coordinates of the turning point are \((3, 7)\).

B2 for \((3, 7)\) (B1 for a coordinate with either 3 as the x-coordinate or 7 as the y-coordinate, e.g., \((-3, 7)\) or \((3, -7)\)).

First, multiply both sides by 3 to eliminate the denominator:
\(2x + 5 = 7 \times 3\)
\(2x + 5 = 21\)
Next, subtract 5 from both sides:
\(2x = 16\)
Finally, divide by 2:
\(x = 8\).

M1 for correctly multiplying both sides by 3, resulting in \(2x + 5 = 21\). A1 for 8.

The formula for the volume of a cylinder is:
\(V = \pi r^2 h\)
Given \(r = 5\) and \(h = 8\):
\(V = \pi \times 5^2 \times 8\)
\(V = \pi \times 25 \times 8\)
\(V = 200\pi\text{ cm}^3\).

M1 for substitute values correctly into the cylinder volume formula, e.g., \(\pi \times 5^2 \times 8\) or \(25 \times 8\). A1 for \(200\pi\).

First, we find a combined ratio for \(A : B : C\). We are given \(A : B = 3 : 4\) and \(B : C = 5 : 2\). To link these, we find a common multiple for \(B\), which is \(20\). Multiplying the terms of \(A : B\) by 5 gives \(15 : 20\). Multiplying the terms of \(B : C\) by 4 gives \(20 : 8\). Thus, the combined ratio is \(A : B : C = 15 : 20 : 8\). Let the shares be \(15x\), \(20x\), and \(8x\). Chris receives £1,400 less than Alice, so we can write the equation: \(15x - 8x = 1400\), which simplifies to \(7x = 1400\). Solving for \(x\) gives \(x = 200\). The total profit is the sum of all shares: \(15x + 20x + 8x = 43x\). Substituting \(x = 200\) gives \(43 \times 200 = 8600\). Therefore, the total profit shared is £8,600.

M1: Attempts to find a combined ratio for \(A : B : C\) (e.g., \(15 : 20 : 8\) or equivalent). M1: Sets up a correct equation representing the difference between Alice and Chris, e.g., \(15x - 8x = 1400\) or \(7 \text{ parts} = 1400\). M1: Solves to find the value of one part or share, e.g., \(x = 200\) or \(1 \text{ part} = 200\). A1: Correct final answer of 8600 (or £8600).

To complete the square for \(3x^2 - 12x + 17\), we first factor out the coefficient of \(x^2\) from the first two terms: \(3(x^2 - 4x) + 17\). Next, we complete the square inside the bracket: \(x^2 - 4x = (x - 2)^2 - 4\). Substituting this back into the expression, we get: \(3[(x - 2)^2 - 4] + 17 = 3(x - 2)^2 - 12 + 17 = 3(x - 2)^2 + 5\). The completed square form is \(y = 3(x - 2)^2 + 5\). The coordinates of the turning point are given by making the squared term zero, which occurs when \(x = 2\), yielding \(y = 5\). Thus, the turning point is \((2, 5)\).

M1: Factorises the first two terms by dividing by 3: \(3(x^2 - 4x) + 17\). M1: Completes the square inside the bracket: \((x - 2)^2 - 4\). M1: Expands and simplifies correctly to obtain the form \(3(x - 2)^2 + 5\). A1: Correctly identifies the turning point coordinates as \((2, 5)\) (accept \(x = 2, y = 5\)).

Let \(V\) be the original value of the watch. An increase of 15% is represented by the multiplier \(1.15\). A decrease of 8% is represented by the multiplier \(0.92\). The combined change over the two years can be written as: \(V \times 1.15 \times 0.92 = 4232\). First, calculate the combined multiplier: \(1.15 \times 0.92 = 1.058\). This gives the equation: \(1.058V = 4232\). To find the original value \(V\), we divide: \(V = \frac{4232}{1.058} = 4000\). Thus, the original value of the watch was £4,000.

M1: Identifies correct multipliers for both changes: \(1.15\) and \(0.92\). A1: Calculates the combined multiplier as \(1.058\) (or equivalent fraction). M1: Sets up the equation \(1.058V = 4232\) and divides \(4232\) by their combined multiplier. A1: Correct final answer of 4000 (or £4000).

The area of a rectangle is found by multiplying its length and width: \(\text{Area} = (2x + 3)(x - 1)\). Since the area is \(35\text{ cm}^2\), we write: \((2x + 3)(x - 1) = 35\). Expanding the brackets gives: \(2x^2 - 2x + 3x - 3 = 35 \implies 2x^2 + x - 3 = 35\). Subtracting 35 from both sides yields the required quadratic equation: \(2x^2 + x - 38 = 0\). Now, we solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -38\). This gives: \(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-38)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 304}}{4} = \frac{-1 \pm \sqrt{305}}{4}\). Since we require the positive value for length, we choose the positive root: \(x = \frac{-1 + \sqrt{305}}{4} \approx \frac{-1 + 17.4642}{4} = \frac{16.4642}{4} \approx 4.116\). Rounding to 2 decimal places, we get \(x = 4.12\).

M1: Formulates the area equation \((2x + 3)(x - 1) = 35\) and expands the brackets to reach \(2x^2 + x - 3 = 35\). M1: Rearranges the expanded equation to show clearly the step leading to \(2x^2 + x - 38 = 0\). M1: Substitutes the correct coefficients into the quadratic formula: \(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-38)}}{4}\). A1: Calculates the positive value of \(x\) to be 4.12 (accept only 4.12; negative root must be discarded or ignored).

The total volume is the sum of the volume of the hemisphere and the volume of the cylinder. The volume of a hemisphere of radius \(r\) is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\). The volume of a cylinder of radius \(r\) and height \(h\) is \(V_{\text{cylinder}} = \pi r^2 h\). Here, the height is \(4r\), so \(V_{\text{cylinder}} = \pi r^2 (4r) = 4\pi r^3\). Adding these together gives the total volume: \(V_{\text{total}} = \frac{2}{3}\pi r^3 + 4\pi r^3 = \left(\frac{2}{3} + \frac{12}{3}\right)\pi r^3 = \frac{14}{3}\pi r^3\). We are given that the total volume is \(1008\pi\text{ cm}^3\), so we write: \(\frac{14}{3}\pi r^3 = 1008\pi\). Dividing both sides by \(\pi\) gives: \(\frac{14}{3}r^3 = 1008\). Multiplying both sides by 3 and dividing by 14 gives: \(r^3 = 1008 \times \frac{3}{14} = 72 \times 3 = 216\). Taking the cube root of both sides gives: \(r = \sqrt[3]{216} = 6\text{ cm}\).

M1: Correctly identifies or uses the formula for the volume of a hemisphere as \(\frac{2}{3}\pi r^3\) or a cylinder as \(\pi r^2 (4r)\). M1: Sums the volumes to write a single algebraic expression in terms of \(\pi r^3\), e.g., \(\frac{14}{3}\pi r^3\) (or equivalent). M1: Equates their expression to \(1008\pi\) and solves for \(r^3\) to find \(r^3 = 216\). A1: Correctly finds the radius \(r = 6\).

Since \(y\) is inversely proportional to the square root of \(x\), we can write the relationship as: \(y = \frac{k}{\sqrt{x}}\) for some constant \(k\). We are given that when \(x = 16\), \(y = 9\). Substituting these values into our equation gives: \(9 = \frac{k}{\sqrt{16}} \implies 9 = \frac{k}{4}\). Solving for \(k\) gives: \(k = 9 \times 4 = 36\). Now we have the complete equation: \(y = \frac{36}{\sqrt{x}}\). To find the value of \(x\) when \(y = 12\), we substitute \(y = 12\) into the equation: \(12 = \frac{36}{\sqrt{x}} \implies \sqrt{x} = \frac{36}{12} \implies \sqrt{x} = 3\). Squaring both sides of the equation gives: \(x = 3^2 = 9\).

M1: Sets up the correct proportional equation: \(y = \frac{k}{\sqrt{x}}\) or equivalent. M1: Substitutes \(x = 16\) and \(y = 9\) to correctly find the constant of proportionality \(k = 36\). M1: Substitutes \(y = 12\) and rearranges to find \(\sqrt{x} = 3\). A1: Correctly calculates \(x = 9\).

First, calculate the value of the machinery after the first year of 10% depreciation: \(25000 \times (1 - 0.10) = 25000 \times 0.90 = 22500\). For the next two years, the value depreciates by \(x\%\) per year. Let \(y = 1 - \frac{x}{100}\) be the multiplier for this depreciation. The value at the end of three years is given by: \(22500 \times y^2 = 16256.25\). Dividing both sides by \(22500\) gives: \(y^2 = \frac{16256.25}{22500} = 0.7225\). Taking the square root of both sides gives: \(y = \sqrt{0.7225} = 0.85\). Since \(y = 1 - \frac{x}{100} = 0.85\), we solve for \(x\): \(\frac{x}{100} = 1 - 0.85 = 0.15 \implies x = 15\). Thus, the depreciation rate for the next two years is 15%.

M1: Calculates the depreciated value after the first year: \(25000 \times 0.90 = 22500\). M1: Sets up a correct compound depreciation equation for the remaining two years, e.g., \(22500 \times y^2 = 16256.25\). M1: Divides and finds the square root of the ratio, yielding \(y = 0.85\) or equivalent. A1: Correctly identifies the value of \(x\) as 15.

First, we find the gradient \(m_1\) of line \(L_1\) using the formula \(m_1 = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the coordinates \((-2, 5)\) and \((4, 8)\) gives: \(m_1 = \frac{8 - 5}{4 - (-2)} = \frac{3}{6} = \frac{1}{2}\). Since line \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is the negative reciprocal of \(m_1\): \(m_2 = -\frac{1}{m_1} = -2\). Now, using the point-slope formula \(y - y_1 = m(x - x_1)\) for \(L_2\) through the point \((3, -1)\), we get: \(y - (-1) = -2(x - 3) \implies y + 1 = -2x + 6\). Rearranging this equation into the form \(ay + bx = c\) gives: \(y + 2x = 5\). Here, \(a = 1\), \(b = 2\), and \(c = 5\), which are all integers.

M1: Calculates the gradient of line \(L_1\) as \(\frac{1}{2}\) (or equivalent). M1: Recognises that perpendicular gradients multiply to \(-1\) and finds the gradient of line \(L_2\) to be \(-2\). M1: Substitutes the gradient \(-2\) and the point \((3, -1)\) into a linear equation form, e.g., \(-1 = -2(3) + c\). A1: Correctly states the final equation in the required form, e.g., \(y + 2x = 5\) (or any equivalent integer multiple like \(2x + y = 5\)).

Let the number of boys be \( 3y \) and the number of girls be \( 5y \). Since there are 60 more girls than boys: \( 5y - 3y = 60 \), which gives \( 2y = 60 \) and thus \( y = 30 \). The number of boys is \( 3 \times 30 = 90 \) and the number of girls is \( 5 \times 30 = 150 \). The total number of children is \( 90 + 150 = 240 \). The ratio of adults to children is \( 5 : 8 \). Let \( A \) be the number of adults. Then \( \frac{A}{240} = \frac{5}{8} \), which gives \( A = 150 \). The total number of people is \( 150 + 240 = 390 \).

M1: For establishing a method to find the value of one part of the child ratio, e.g. \( 60 \div (5 - 3) = 30 \)
M1: For finding the total number of children, e.g. \( 240 \) (or finding boys = 90 and girls = 150)
M1: For using the adult to child ratio to find the number of adults, e.g. \( 240 \div 8 \times 5 = 150 \)
A1: For 390

To find the points of intersection, equate the two equations: \( x^2 - 3x - 4 = 2x + 10 \). Rearranging into standard quadratic form gives: \( x^2 - 5x - 14 = 0 \). Factorising the quadratic equation: \( (x - 7)(x + 2) = 0 \). This gives the x-coordinates: \( x = 7 \) and \( x = -2 \). Substitute these back into the line equation to find the corresponding y-coordinates: For \( x = -2 \), \( y = 2(-2) + 10 = 6 \). For \( x = 7 \), \( y = 2(7) + 10 = 24 \). The points of intersection are \( (-2, 6) \) and \( (7, 24) \).

M1: For equating the equations and rearranging to standard quadratic form, e.g. \( x^2 - 5x - 14 = 0 \)
M1: For solving the quadratic by factorisation or formula to find the two x-values, \( x = 7 \) and \( x = -2 \)
M1: For substituting at least one x-value back into either equation to find a y-value
A1: For both coordinates \( (-2, 6) \) and \( (7, 24) \) (either order)

Let \( P \) be the normal price of the laptop. After a 20% reduction, the price is \( 0.8P \). The additional 10% reduction is applied to this sale price, giving a final price of \( 0.8P \times 0.9 = 0.72P \). We are given that the final price is £576, so: \( 0.72P = 576 \). Solving for \( P \): \( P = \frac{576}{0.72} = 800 \). The normal price of the laptop was £800.

M1: For expressing a 20% reduction as a multiplier of 0.8, or a 10% reduction as 0.9
M1: For combining the two percentage reductions, e.g. \( 0.8 \times 0.9 = 0.72 \) or establishing the intermediate price before the loyalty club discount: \( 576 \div 0.9 = 640 \)
M1: For completing the reverse percentage calculation, e.g. \( 576 \div 0.72 \) or \( 640 \div 0.8 \)
A1: For 800

The area of the rectangle is given by multiplying length and width: \( (2x + 5)(x - 2) = 18 \). Expanding the brackets: \( 2x^2 - 4x + 5x - 10 = 18 \), which simplifies to \( 2x^2 + x - 10 = 18 \). Rearranging to equal zero: \( 2x^2 + x - 28 = 0 \). Factorising the quadratic equation: \( (2x - 7)(x + 4) = 0 \). This gives two possible solutions: \( x = 3.5 \) or \( x = -4 \). Since the width of the rectangle must be positive, \( x - 2 > 0 \), so we choose \( x = 3.5 \). Substituting \( x = 3.5 \) into the dimensions: Length = \( 2(3.5) + 5 = 12\text{ cm} \), Width = \( 3.5 - 2 = 1.5\text{ cm} \). The perimeter is \( 2 \times (\text{length} + \text{width}) = 2 \times (12 + 1.5) = 2 \times 13.5 = 27\text{ cm} \).

M1: For expanding the product and setting up the quadratic equation, e.g. \( 2x^2 + x - 28 = 0 \)
M1: For factorising or using the quadratic formula to find the positive solution \( x = 3.5 \)
M1: For substituting \( x = 3.5 \) back into the expressions to find both the length (12) and the width (1.5)
A1: For 27

The volume of the cylinder is \( V_{\text{cylinder}} = \pi r^2 h = 8\pi r^2 \). The volume of the sphere is \( V_{\text{sphere}} = \frac{4}{3} \pi r^3 \). Since the volumes are equal: \( 8\pi r^2 = \frac{4}{3} \pi r^3 \). Divide both sides by \( \pi r^2 \) (since \( r \neq 0 \)): \( 8 = \frac{4}{3} r \), which simplifies to \( r = 6 \). The total surface area of a cylinder is given by the formula: \( A = 2\pi r^2 + 2\pi r h \). Substituting \( r = 6 \) and \( h = 8 \): \( A = 2\pi (6)^2 + 2\pi (6)(8) = 72\pi + 96\pi = 168\pi \).

M1: For equating the volume formulas: \( 8\pi r^2 = \frac{4}{3} \pi r^3 \) (or equivalent)
M1: For solving to find \( r = 6 \)
M1: For substituting their value of \( r \) and \( h = 8 \) into the total surface area formula \( 2\pi r^2 + 2\pi r h \)
A1: For \( 168\pi \) (accept exact equivalents, do not accept decimal approximations unless 168 is clearly shown)

1. **Find the original number of teachers and guides:**
Since there are 108 students and the initial ratio is \(18 : 2 : 1\):
- Multiplier = \(108 \div 18 = 6\)
- Number of teachers = \(2 \times 6 = 12\)
- Number of guides = \(1 \times 6 = 6\)

2. **Calculate the original total cost:**
- Cost of student tickets = \(108 \times \pounds 12 = \pounds 1296\)
- Cost of teacher tickets = \(12 \times \pounds 15 = \pounds 180\)
- Cost of guide tickets = \(6 \times \pounds 0 = \pounds 0\)
- Original total cost = \(1296 + 180 = \pounds 1476\)

3. **Find the new number of teachers and guides:**
Using the new ratios with the same 108 students:
- Ratio of students to guides is \(12 : 1\), so number of guides = \(108 \div 12 = 9\)
- Ratio of students to teachers is \(6 : 1\), so number of teachers = \(108 \div 6 = 18\)

4. **Calculate the new total cost:**
- Cost of student tickets = \(108 \times \pounds 12 = \pounds 1296\)
- Cost of teacher tickets = \(18 \times \pounds 15 = \pounds 270\)
- Cost of guide tickets = \(9 \times \pounds 0 = \pounds 0\)
- New total cost = \(1296 + 270 = \pounds 1566\)

5. **Calculate the percentage increase:**
- Increase in cost = \(1566 - 1476 = \pounds 90\)
- Percentage increase = \(\frac{90}{1476} \times 100 \approx 6.09756\%\)
- To 1 decimal place, the percentage increase is \(6.1\%\).

- **M1**: For finding the original number of teachers (12) and guides (6) by using the ratio \(18 : 2 : 1\).
- **A1**: For calculating the correct original total cost of \(\pounds 1476\).
- **M1**: For calculating the new number of teachers (18) and/or guides (9) using the new ratios.
- **A1**: For calculating the correct new total cost of \(\pounds 1566\).
- **M1**: For a complete and correct method to find the percentage increase, i.e., \(\frac{1566 - 1476}{1476} \times 100\) or \(\frac{90}{1476} \times 100\).
- **A1**: For the correct final answer of \(6.1\) (accept \(6.1\%\)).

1. **Formulate the volume of the cylinder:**
\(V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (4.5r) = 4.5\pi r^3 = \frac{9}{2}\pi r^3\)

2. **Equate this to the volume of the sphere to find the relation between \(R\) and \(r\):**
\(V_{\text{sphere}} = \frac{4}{3}\pi R^3\)
Since volume is conserved:
\(\frac{4}{3}\pi R^3 = \frac{9}{2}\pi r^3\)
Multiply by \(\frac{3}{4}\) and divide by \(\pi\):
\(R^3 = \frac{27}{8}r^3\)
Taking the cube root of both sides gives:
\(R = \frac{3}{2}r = 1.5r\)

3. **Find the total surface area of the cylinder:**
\(A_{\text{cylinder}} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r (4.5r) = 2\pi r^2 + 9\pi r^2 = 11\pi r^2\)

4. **Find the surface area of the sphere:**
\(A_{\text{sphere}} = 4\pi R^2 = 4\pi (1.5r)^2 = 4\pi (2.25r^2) = 9\pi r^2\)

5. **Find the ratio of the surface area of the sphere to the total surface area of the cylinder:**
\(\text{Ratio} = \frac{A_{\text{sphere}}}{A_{\text{cylinder}}} = \frac{9\pi r^2}{11\pi r^2} = \frac{9}{11}\)
In its simplest integer form, the ratio is \(9 : 11\).

- **M1**: For expressing the volume of the cylinder as \(\pi r^2 (4.5r)\) or \(4.5\pi r^3\).
- **M1**: For equating the two volumes: \(\frac{4}{3}\pi R^3 = 4.5\pi r^3\).
- **A1**: For finding \(R = 1.5r\) or \(R^3 = 3.375r^3\) (or equivalent ratio \(R : r = 3 : 2\)).
- **M1**: For finding the total surface area of the cylinder: \(2\pi r^2 + 2\pi r (4.5r) = 11\pi r^2\).
- **M1**: For finding the surface area of the sphere in terms of \(r\): \(4\pi (1.5r)^2 = 9\pi r^2\).
- **A1**: For simplifying the ratio to \(9 : 11\) (accept \(9:11\)).

1. **Define the variables for the original costs:**
Let the original cost of the sculpture be \(S\).
Since the painting costs 20% more than the sculpture, the original cost of the painting is:
\(P = 1.2S\)

2. **Express the selling prices:**
- The painting's value decreased by 15%, so its selling price is:
\(P_{\text{sell}} = 0.85 \times P = 0.85 \times 1.2S = 1.02S\)
- The sculpture's value increased by 25%, so its selling price is:
\(S_{\text{sell}} = 1.25 \times S = 1.25S\)

3. **Set up the equation for the combined selling price:**
\(P_{\text{sell}} + S_{\text{sell}} = 10215\)
\(1.02S + 1.25S = 10215\)
\(2.27S = 10215\)

4. **Solve for \(S\):**
\(S = \frac{10215}{2.27} = 4500\)
So, the original cost of the sculpture was \(\pounds 4500\).

5. **Calculate the original cost of the painting:**
\(P = 1.2 \times 4500 = 5400\)
Thus, the original cost of the painting was \(\pounds 5400\).

- **M1**: For defining the original cost of the painting in terms of the sculpture, e.g., \(P = 1.2S\) (or vice-versa).
- **M1**: For expressing the selling price of either item algebraically, e.g., \(0.85P\) or \(1.25S\).
- **M1**: For expressing both selling prices in terms of a single variable, e.g., \(1.02S\) and \(1.25S\).
- **M1**: For setting up a correct equation: \(1.02S + 1.25S = 10215\) (or equivalent in terms of \(P\)).
- **A1**: For finding the original cost of the sculpture as \(\pounds 4500\) (or finding \(P\) directly if using a \(P\)-only equation).
- **A1**: For the correct final answer of \(5400\) (accept \(\pounds 5400\)).