2022 OCR GCSE Mathematics - J560 模擬試題連答案詳解

Expand the brackets: \(5x(2x - 3) = 10x^2 - 15x\) and \(-3(x^2 - 4x) = -3x^2 + 12x\). Combine the like terms: \(10x^2 - 3x^2 = 7x^2\) and \(-15x + 12x = -3x\). This gives \(7x^2 - 3x\).

M1 for expanding at least one of the brackets correctly to get \(10x^2 - 15x\) or \(-3x^2 + 12x\). A1 for \(7x^2 - 3x\).

The ratio of cement : sand : gravel is \(1 : 2 : 4\). The total number of parts is \(1 + 2 + 4 = 7\) parts. The sand represents 2 parts, which is \(35\text{ kg}\). The weight of 1 part is \(35 / 2 = 17.5\text{ kg}\). The total weight is \(7 \times 17.5 = 122.5\text{ kg}\).

M1 for a correct method to find the weight of 1 part, e.g. \(35 / 2\), or for using the total parts, 7. A1 for \(122.5\) (accept \(122.5\text{ kg}\)).

To increase \pounds 42 by \(8\%\), multiply by \(1.08\): \(42 \times 1.08 = 45.36\). Alternatively, find \(8\%\) of \pounds 42, which is \(0.08 \times 42 = 3.36\), and add this to the original cost: \(42 + 3.36 = 45.36\).

M1 for a complete method to find the increased amount, e.g. \(42 \times 1.08\) or \(42 + (42 \times 0.08)\). A1 for \(45.36\) (accept \pounds 45.36).

Expand the brackets: \(6x - 15 = 4x + 7\). Subtract \(4x\) from both sides: \(2x - 15 = 7\). Add 15 to both sides: \(2x = 22\). Divide by 2: \(x = 11\).

M1 for expanding the bracket correctly to get \(6x - 15\) or for a correct algebraic step to collect terms on one side. A1 for \(11\).

Multiply the lead numbers: \(3 \times 8 = 24\). Multiply the powers of 10: \(10^5 \times 10^{-2} = 10^3\). Combine these: \(24 \times 10^3\). Convert to standard form: \(2.4 \times 10^4\).

M1 for obtaining \(24 \times 10^3\) or \(24000\) or showing a clear step of multiplying lead numbers and adding powers. A1 for \(2.4 \times 10^4\).

The area of a trapezium is given by \(\text{Area} = \frac{1}{2}(a + b)h\). Substitute the given values: \(66 = \frac{1}{2}(8 + 14)h \implies 66 = 11h \implies h = 6\text{ cm}\).

M1 for setting up a correct equation, e.g. \(\frac{1}{2}(8 + 14)h = 66\). A1 for \(6\) (accept \(6\text{ cm}\)).

First, find the gradient \(m\): \(m = \frac{5 - (-3)}{3 - (-1)} = \frac{8}{4} = 2\). The equation of the line is of the form \(y = 2x + c\). Substitute the coordinates of \((3, 5)\): \(5 = 2(3) + c \implies 5 = 6 + c \implies c = -1\). The equation is \(y = 2x - 1\).

M1 for finding the gradient is \(2\) or for substituting a point into \(y = mx + c\) correctly. A1 for \(y = 2x - 1\).

Write down the prime factors: \(48 = 2^4 \times 3\) and \(72 = 2^3 \times 3^2\). The Highest Common Factor is the product of the lowest powers of the common prime factors: \(\text{HCF} = 2^3 \times 3 = 8 \times 3 = 24\). Alternatively, list the factors of both numbers to find the largest common factor, which is 24.

M1 for prime factorisation of both numbers showing \(2^4 \times 3\) and \(2^3 \times 3^2\) or for listing at least 4 factors of each number including 24. A1 for 24.

To work out the product: 1. Multiply the decimal parts: \(1.5 \times 6 = 9\). 2. Add the indices of the powers of 10: \(10^4 \times 10^{-7} = 10^{4 + (-7)} = 10^{-3}\). 3. Combine to write in standard form: \(9 \times 10^{-3}\). 4. Convert to an ordinary number: \(9 \times 10^{-3} = 0.009\).

M1 for a correct step, e.g. finding the digits 9 with any incorrect decimal place (such as 0.09 or 0.0009), or finding 15,000 and 0.0000006. A1 for 0.009 (or equivalent fraction, e.g. \(\frac{9}{1000}\)).

1. Add the parts of the ratio: \(2 + 5 + 8 = 15\) parts. 2. Find the value of one part: \(120 \div 15 = 8\) sweets. 3. Find the difference in parts between Colin and Anna: \(8 - 2 = 6\) parts. 4. Calculate the difference in sweets: \(6 \times 8 = 48\) sweets. (Alternatively, Anna receives \(2 \times 8 = 16\) sweets, Colin receives \(8 \times 8 = 64\) sweets, and the difference is \(64 - 16 = 48\).)

M1 for \(120 \div (2 + 5 + 8)\) or 8, or for finding Anna's share (16) or Colin's share (64). A1 for 48.

Let the shares of A, B, and C be represented as \(3x\), \(5x\), and \(8x\) respectively. The difference between C's share and A's share is given as \(\pounds 120\). This gives the equation: \(8x - 3x = 120\), which simplifies to \(5x = 120\). Dividing both sides by 5 gives \(x = 24\). The total amount of money shared is the sum of all parts: \(3x + 5x + 8x = 16x\). Substituting \(x = 24\) into this expression: \(16 \times 24 = 384\). Therefore, the total amount of money shared is \(\pounds 384\).

M1 for finding the difference in ratio parts, i.e., \(8 - 3 = 5\) parts.
M1 for setting up the equation to find the value of one part: \(120 \div 5 = 24\).
M1 for calculating the total number of parts, \(3 + 5 + 8 = 16\), and multiplying by the value of one part: \(16 \times 24\).
A1 for 384 (accept \(\pounds 384\)).

First, expand the double brackets: \((3x - 4)(2x + 5) = 6x^2 + 15x - 8x - 20 = 6x^2 + 7x - 20\). Next, subtract the second expression, ensuring to distribute the negative sign to each term inside the bracket: \(-(x^2 - 3x + 8) = -x^2 + 3x - 8\). Now, collect the like terms: \(6x^2 - x^2 = 5x^2\), \(7x + 3x = 10x\), and \(-20 - 8 = -28\). Combining these terms gives the fully simplified expression: \(5x^2 + 10x - 28\).

M1 for expanding the double brackets to get at least 3 correct terms from \(6x^2 + 15x - 8x - 20\).
M1 for simplifying the expansion of the double brackets to \(6x^2 + 7x - 20\).
M1 for correctly subtracting the second bracket terms, leading to \(-x^2 + 3x - 8\).
A1 for \(5x^2 + 10x - 28\).

The formula for the total surface area of a cylinder is \(2\pi r^2 + 2\pi r h\). Substituting \(r = 4\) and the total surface area \(96\pi\) into the formula gives: \(2\pi(4)^2 + 2\pi(4)h = 96\pi\). Simplifying the terms: \(32\pi + 8\pi h = 96\pi\). Dividing the entire equation by \(\pi\) gives: \(32 + 8h = 96\). Subtracting 32 from both sides gives: \(8h = 64\). Dividing by 8 gives: \(h = 8\text{ cm}\).

M1 for writing down or using the correct formula for the total surface area of a cylinder: \(2\pi r^2 + 2\pi r h = 96\pi\).
M1 for substituting \(r = 4\) to get \(32\pi + 8\pi h = 96\pi\) (or equivalent equation).
M1 for simplifying to a linear equation in \(h\), e.g., \(32 + 8h = 96\) or \(8h = 64\).
A1 for 8.

A reduction of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price. Let the original price be \(P\). This gives: \(0.85P = 272\), so \(P = \frac{272}{0.85} = 320\). Therefore, the original price of the bicycle is \(\pounds 320\). After the sale, this original price is increased by \(5\%\): \(\text{New Price} = 1.05 \times 320 = 336\). Thus, the price of the bicycle after the increase is \(\pounds 336\).

M1 for identifying that \(\pounds 272\) is \(85\%\) of the original price (e.g., dividing by \(0.85\) or writing \(100\% - 15\% = 85\%\)).
M1 for calculating the original price: \(272 \div 0.85 = 320\).
M1 for calculating a \(5\%\) increase on \(320\): \(320 \times 1.05\) or \(320 + 16\).
A1 for 336 (accept \(\pounds 336\)).

The total number of counters in the bag is \(7 + 5 = 12\). Since the selection is without replacement, the events are dependent. We can find the probability of getting at least one blue counter by subtracting the probability of getting two red counters from 1: \(P(\text{at least one blue}) = 1 - P(\text{Red, Red})\). The probability of selecting a red counter first is: \(P(\text{first Red}) = \frac{7}{12}\). Since the first counter is not replaced, there are now 11 counters left, of which 6 are red. The probability of selecting a second red counter is: \(P(\text{second Red}) = \frac{6}{11}\). The probability of selecting two red counters is: \(P(\text{Red, Red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\). Thus, the probability of selecting at least one blue counter is: \(P(\text{at least one blue}) = 1 - \frac{7}{22} = \frac{15}{22}\).

M1 for identifying the dependent nature of the events (without replacement) by showing a correct second-draw denominator of 11.
M1 for calculating the probability of selecting two red counters: \(\frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\) (or \(\frac{7}{22}\)) OR for calculating at least one of the individual probabilities involving blue (e.g., \(P(\text{Red, Blue}) = \frac{35}{132}\)).
M1 for a complete and correct method to find the total probability, e.g., \(1 - \frac{42}{132}\) or summing the three favorable outcomes: \(\frac{35}{132} + \frac{35}{132} + \frac{20}{132}\).
A1 for \(\frac{15}{22}\) (or equivalent fraction such as \(\frac{90}{132}\), or decimal \(0.682\) or \(68.2\%\) rounded to at least 3 significant figures).

The total number of counters in the bag is \(x + 5\). The probability of drawing a blue counter first is \(\frac{x}{x+5}\). Since the counter is not replaced, the number of blue counters left is \(x - 1\) and the total number of counters left is \(x + 4\). The probability of drawing a second blue counter is \(\frac{x-1}{x+4}\). The probability that both are blue is: \(\frac{x}{x+5} \times \frac{x-1}{x+4} = \frac{3}{11}\). Expanding and multiplying out: \(\frac{x^2 - x}{x^2 + 9x + 20} = \frac{3}{11}\). Multiplying both sides by the denominators gives: \(11(x^2 - x) = 3(x^2 + 9x + 20) \implies 11x^2 - 11x = 3x^2 + 27x + 60\). Rearranging to form a quadratic equation equal to zero: \(8x^2 - 38x - 60 = 0\). Dividing the entire equation by 2: \(4x^2 - 19x - 30 = 0\). Factoring this quadratic: \((4x + 5)(x - 6) = 0\). This gives two possible solutions: \(x = -1.25\) or \(x = 6\). Since the number of counters must be a positive integer, \(x = 6\).

M1 for writing an expression for the probability of two blue counters: \(\frac{x(x-1)}{(x+5)(x+4)} = \frac{3}{11}\). M1 for expanding and forming a quadratic equation, e.g., \(4x^2 - 19x - 30 = 0\). M1 for factorising or solving the quadratic to find \(x = 6\). A1 for the final answer of 6.

First, find the gradient of the line \(L_1\): \(m_1 = \frac{14 - 5}{4 - (-2)} = \frac{9}{6} = \frac{3}{2}\). Since \(L_2\) is perpendicular to \(L_1\), the gradient of \(L_2\) is the negative reciprocal of \(m_1\): \(m_2 = -\frac{2}{3}\). Using the equation of a straight line with the point \((3, -1)\) and gradient \(-\frac{2}{3}\): \(y - (-1) = -\frac{2}{3}(x - 3) \implies y + 1 = -\frac{2}{3}x + 2\). Multiply both sides by 3 to eliminate the fraction: \(3y + 3 = -2x + 6\). Rearranging to the form \(ay + bx = c\): \(3y + 2x = 3\).

M1 for finding the gradient of \(L_1\) as \(\frac{3}{2}\). M1 for identifying the perpendicular gradient as \(-\frac{2}{3}\). M1 for substituting their gradient and \((3, -1)\) into a straight line equation. A1 for the correct equation in the requested form, e.g., \(3y + 2x = 3\).

First, factorise the quadratic numerator \(2x^2 + 5x - 12\). We look for two numbers that multiply to \(2 \times (-12) = -24\) and add to 5. These numbers are 8 and -3. Rewriting the middle term: \(2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\). Next, factorise the denominator \(3x^2 + 12x\) by factoring out the common term \(3x\): \(3x(x + 4)\). Write the fraction with the factorised expressions: \(\frac{(2x - 3)(x + 4)}{3x(x + 4)}\). Cancelling the common factor \((x + 4)\) from both numerator and denominator gives \(\frac{2x - 3}{3x}\).

M1 for factorising the numerator to find the term \((x+4)\) or \((2x-3)\). M1 for fully factorising the numerator as \((2x - 3)(x + 4)\). M1 for factorising the denominator as \(3x(x + 4)\). A1 for the fully simplified fraction \(\frac{2x - 3}{3x}\).

First, calculate the value of the car after the first year after a \(10\%\) decrease: \(25000 \times (1 - 0.10) = 22500\). In each of the next two years, the value increases by \(x\%\). Let the multiplier for this increase be \(m = 1 + \frac{x}{100}\). The value at the end of the three years is given by: \(22500 \times m^2 = 26244\). Solve for \(m^2\): \(m^2 = \frac{26244}{22500} = 1.1664\). Find \(m\) by taking the square root: \(m = \sqrt{1.1664} = 1.08\). Since \(m = 1 + \frac{x}{100} = 1.08\), we have \(\frac{x}{100} = 0.08 \implies x = 8\).

M1 for calculating the value of the car after 1 year as 22,500. M1 for setting up the equation \(22500 \times (1 + \frac{x}{100})^2 = 26244\). M1 for finding the multiplier: \(1 + \frac{x}{100} = 1.08\). A1 for the final answer of 8.

From the first equation, express \(y\) in terms of \(x\): \(y = 2x + 3\). Substitute this into the second equation: \(x^2 + (2x + 3)^2 = 18\). Expand the bracket: \(x^2 + 4x^2 + 12x + 9 = 18 \implies 5x^2 + 12x + 9 = 18\). Rearrange into standard quadratic form: \(5x^2 + 12x - 9 = 0\). Solve by factorisation: \(5x^2 + 15x - 3x - 9 = 0 \implies 5x(x + 3) - 3(x + 3) = 0 \implies (5x - 3)(x + 3) = 0\). This yields \(x = 0.6\) and \(x = -3\). Substitute these back into \(y = 2x + 3\): For \(x = 0.6\), \(y = 2(0.6) + 3 = 4.2\). For \(x = -3\), \(y = 2(-3) + 3 = -3\). The solutions are \((0.6, 4.2)\) and \((-3, -3)\).

M1 for substituting \(y = 2x + 3\) into the quadratic equation. M1 for expanding and simplifying to a three-term quadratic, e.g., \(5x^2 + 12x - 9 = 0\). M1 for finding both \(x\) values (0.6 and -3) or both \(y\) values (4.2 and -3). A1 for both correct pairs of coordinates: \((0.6, 4.2)\) and \((-3, -3)\).

Let the number of adults be \(3k\) and the number of children be \(5k\), where \(k\) is a constant. The total income is: \((3k \times 8.50) + (5k \times 4.50) = 2400 \implies 25.5k + 22.5k = 2400\). Combine the terms: \(48k = 2400\). Solve for \(k\): \(k = 50\). The total number of people is \(3k + 5k = 8k\). Substituting \(k = 50\) gives: \(8 \times 50 = 400\).

M1 for writing an expression for total cost in terms of a variable, e.g., \(3k \times 8.50 + 5k \times 4.50\). M1 for simplifying to \(48k = 2400\). M1 for finding the multiplier \(k = 50\) or finding the number of adults (150) or children (250). A1 for the correct total number of people: 400.

First, calculate the area of the rectangle: \(\text{Area} = 12 \times 5 = 60\text{ cm}^2\). Next, the area of the sector is given by: \(\text{Area of sector} = \frac{\theta}{360} \times \pi \times 10^2 = 60\). Simplify the equation: \(\frac{100\pi\theta}{360} = 60 \implies \frac{5\pi\theta}{18} = 60\). Multiply by 18 and divide by \(5\pi\) to solve for \(\theta\): \(\theta = \frac{1080}{5\pi} = \frac{216}{\pi} \approx 68.7549\). Rounding to 1 decimal place gives \(\theta = 68.8\).

M1 for finding the area of the rectangle as \(60\text{ cm}^2\). M1 for setting up the sector equation: \(\frac{\theta}{360} \times \pi \times 10^2 = 60\). M1 for rearranging to solve for \(\theta\): \(\theta = \frac{216}{\pi}\). A1 for the correct answer of 68.8 (accept 68.7 to 68.8).

First, calculate the volume of the cylinder: \(V_{\text{cylinder}} = \pi R^2 H = \pi \times 4^2 \times 18 = 288\pi\text{ cm}^3\). The volume of a sphere is given by: \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). Since the volume remains constant: \ \frac{4}{3}\pi r^3 = 288\pi\). Dividing both sides by \(\pi\) gives: \(\frac{4}{3}r^3 = 288\). Solve for \(r^3\): \(r^3 = 288 \times \frac{3}{4} = 216\). Find the cube root of 216: \(r = \sqrt[3]{216} = 6\).

M1 for finding the volume of the cylinder as \(288\pi\). M1 for equating the volume of the sphere to the volume of the cylinder: \(\frac{4}{3}\pi r^3 = 288\pi\). M1 for simplifying to find \(r^3 = 216\). A1 for the correct radius \(r = 6\).

First, find the total number of parts in the ratio:
\(3 + 5 + 2 = 10\) parts.

Next, calculate the volume represented by 1 part:
\(\text{Volume of 1 part} = \frac{20 \text{ litres}}{10} = 2 \text{ litres}\).

Now, calculate the required volume of each ingredient:
- Orange juice needed: \(3 \times 2 = 6\) litres.
- Pineapple juice needed: \(5 \times 2 = 10\) litres.
- Mango juice needed: \(2 \times 2 = 4\) litres.

Compare the required volumes with the quantities currently in stock:
- For Orange juice: \(6\) litres are required, and the café has \(7.5\) litres. Since \(7.5 \ge 6\), no extra orange juice needs to be bought (0 litres).
- For Pineapple juice: \(10\) litres are required, and the café has \(6.2\) litres. Extra needed: \(10 - 6.2 = 3.8\) litres.
- For Mango juice: \(4\) litres are required, and the café has \(1.8\) litres. Extra needed: \(4 - 1.8 = 2.2\) litres.

Finally, calculate the total extra volume of juice to buy:
\(\text{Total extra juice} = 3.8 + 2.2 = 6\) litres.

- **M1**: For a method to find the quantity of one part, e.g., \(20 \div 10\), or finding the required amounts for all three juices (Orange = 6, Pineapple = 10, Mango = 4).
- **M1**: For identifying that no extra orange juice is needed (since \(7.5 > 6\)).
- **M1**: For a correct subtraction method to find the extra amount needed for Pineapple (\(10 - 6.2\)) or Mango (\(4 - 1.8\)).
- **A1**: For the correct total of 6 litres (accept 6).

Let the multiplier for the compound interest be \(M\).

The formula for compound interest over 2 years gives:
\(6000 \times M^2 = 6365.40\)

Divide both sides by \(6000\) to find \(M^2\):
\(M^2 = \frac{6365.40}{6000}\)
\(M^2 = 1.0609\)

Take the square root of both sides to find \(M\):
\(M = \sqrt{1.0609} = 1.03\)

Since the multiplier is \(1.03\), the annual percentage increase is:
\((1.03 - 1) \times 100 = 3\%\)

Therefore, \(r = 3\).

- **M1**: For setting up a correct compound interest equation, e.g., \(6000 \times (1 + \frac{r}{100})^2 = 6365.40\) or \(6000 \times M^2 = 6365.40\).
- **M1**: For a method to isolate the square of the multiplier, e.g., \(M^2 = 1.0609\).
- **M1**: For finding the square root of their multiplier, e.g., \(\sqrt{1.0609} = 1.03\).
- **A1**: For \(r = 3\) (accept 3, do not accept 3% unless 3 is clearly identified as the value of the variable \(r\)).

First, find the number of pens of each colour in the box:
1. The number of green pens is \( 0.2 \times 50 = 10 \).
2. The remaining number of pens is \( 50 - 10 = 40 \), which are red and blue.
3. The ratio of red to blue is \( 2 : 3 \), which means there are \( 5 \) parts in total.
Each part is worth \( 40 \div 5 = 8 \) pens.
- Number of red pens = \( 2 \times 8 = 16 \)
- Number of blue pens = \( 3 \times 8 = 24 \)

We need to find the probability that both pens are the same colour when two are selected without replacement. This can happen in three mutually exclusive ways: both red (RR), both blue (BB), or both green (GG).
- \( P(RR) = \frac{16}{50} \times \frac{15}{49} = \frac{240}{2450} \)
- \( P(BB) = \frac{24}{50} \times \frac{23}{49} = \frac{552}{2450} \)
- \( P(GG) = \frac{10}{50} \times \frac{9}{49} = \frac{90}{2450} \)

Summing these probabilities:
\( P(\text{same colour}) = \frac{240 + 552 + 90}{2450} = \frac{882}{2450} \)

Simplifying the fraction:
Dividing numerator and denominator by 2 gives \( \frac{441}{1225} \).
Dividing by 49 gives \( \frac{9}{25} \) (or \( 0.36 \)).

- **M1**: Correctly calculates the number of green pens as \( 10 \) (or calculates the combined probability of red/blue as \( 0.8 \)).
- **M1**: Correctly determines the count of red pens (\( 16 \)) and blue pens (\( 24 \)).
- **M1**: Identifies the sum of probabilities for the same colour: \( P(RR) + P(BB) + P(GG) \).
- **M1**: Shows at least two correct probability product terms for selection without replacement, e.g., \( \frac{16}{50} \times \frac{15}{49} \) and \( \frac{24}{50} \times \frac{23}{49} \).
- **A1**: Reaches the final simplified probability of \( \frac{9}{25} \) (or \( 0.36 \) / \( 36\% \)).

Let the multiplier for the first 3 years of growth be \( y \), where \( y = 1 + \frac{x}{100} \).
Using the formula for compound growth:
\( 2000 \times y^3 = 2662 \)
\( y^3 = \frac{2662}{2000} \)
\( y^3 = 1.331 \)

Taking the cube root of both sides:
\( y = \sqrt[3]{1.331} = 1.1 \)

Since the multiplier is \( 1.1 \), the growth rate is:
\( x = 10 \)

For the next 2 years, the value decreases by \( (x - 2)\% = (10 - 2)\% = 8\% \) per year.
The multiplier for depreciation is \( 1 - 0.08 = 0.92 \).

Using the value at the end of the 3rd year (£2662) as the starting value for this period:
\( \text{Value after 5 years} = 2662 \times (0.92)^2 \)
\( \text{Value after 5 years} = 2662 \times 0.8464 = 2253.1168 \)

Rounding to the nearest penny gives £2253.12.

- **M1**: Sets up a correct equation for the first 3 years of compound interest, e.g., \( 2000 \times y^3 = 2662 \).
- **M1**: Finds the multiplier \( y = 1.1 \) or growth rate \( x = 10 \) by correctly calculating the cube root of \( 1.331 \).
- **M1**: Determines the depreciation rate is \( 8\% \) (hence multiplier is \( 0.92 \)).
- **M1**: Applies the compound decay formula for 2 years: \( 2662 \times 0.92^2 \).
- **A1**: Correct final value of \( 2253.12 \) (must be rounded to the nearest penny; accept with or without £ sign).

Step 1: Determine individual volumes.
The total volume is \( 102\pi \) and the ratio of cylinder volume to cone volume is \( 15 : 2 \).
Total parts = \( 15 + 2 = 17 \).
- Volume of cone, \( V_{\text{cone}} = \frac{2}{17} \times 102\pi = 12\pi \) \( \text{cm}^3 \).
- Volume of cylinder, \( V_{\text{cyl}} = \frac{15}{17} \times 102\pi = 90\pi \) \( \text{cm}^3 \).

Step 2: Find the radius \( r \) and cone height \( h \).
We are given \( h = \frac{4}{3}r \).
The volume of a cone is given by \( V = \frac{1}{3}\pi r^2 h \). Thus:
\( \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = 12\pi \)
\( \frac{4}{9}r^3 = 12 \)
\( r^3 = 12 \times \frac{9}{4} = 27 \)
\( r = 3 \) cm.
Then, \( h = \frac{4}{3} \times 3 = 4 \) cm.

Step 3: Find the cylinder height \( H \).
The volume of a cylinder is \( V = \pi r^2 H \). Thus:
\( \pi (3)^2 H = 90\pi \)
\( 9H = 90 \implies H = 10 \) cm.

Step 4: Find the slant height \( l \) of the cone.
Using Pythagoras' theorem:
\( l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = 5 \) cm.

Step 5: Calculate the total surface area of the ornament.
The surface consists of:
- The bottom base of the cylinder: \( \pi r^2 = \pi \times 3^2 = 9\pi \) \( \text{cm}^2 \).
- The curved surface of the cylinder: \( 2\pi r H = 2\pi \times 3 \times 10 = 60\pi \) \( \text{cm}^2 \).
- The curved surface of the cone: \( \pi r l = \pi \times 3 \times 5 = 15\pi \) \( \text{cm}^2 \).

Total Surface Area = \( 9\pi + 60\pi + 15\pi = 84\pi \) \( \text{cm}^2 \).

- **M1**: For correctly dividing the total volume in the given ratio to find \( V_{\text{cone}} = 12\pi \) or \( V_{\text{cyl}} = 90\pi \).
- **M1**: For writing an algebraic equation for the cone volume using \( h = \frac{4}{3}r \) and correctly solving to find \( r = 3 \).
- **M1**: For using the cylinder's volume to find its height \( H = 10 \).
- **M1**: For finding the slant height of the cone as \( l = 5 \) using Pythagoras' theorem.
- **A1**: For summing the three parts of the surface area to obtain the final correct answer of \( 84\pi \).

Let her actual average speed be \( v \) km/h.
The time taken for the journey is \( \frac{36}{v} \) hours.

If she increases her speed by 2 km/h, her new speed is \( v + 2 \) km/h.
The time taken for the journey at this faster speed is \( \frac{36}{v+2} \) hours.

The difference in time is 36 minutes. Converting this to hours:
\( 36 \text{ minutes} = \frac{36}{60} \text{ hours} = 0.6 \text{ hours} \) (or \( \frac{3}{5} \) hours).

Set up the equation:
\( \frac{36}{v} - \frac{36}{v+2} = 0.6 \)

Multiply through by the common denominator \( v(v+2) \) to clear fractions:
\( 36(v+2) - 36v = 0.6v(v+2) \)
\( 36v + 72 - 36v = 0.6v^2 + 1.2v \)
\( 72 = 0.6v^2 + 1.2v \)

Divide the entire equation by 0.6 (or multiply by 5 and divide by 3):
\( v^2 + 2v - 120 = 0 \)

Factorise the quadratic equation:
\( (v + 12)(v - 10) = 0 \)

This gives two solutions:
\( v = -12 \) or \( v = 10 \)

Since speed must be a positive value, her actual average speed is \( 10 \) km/h.

- **M1**: Writes correct algebraic expressions for time in terms of \( v \): \( \frac{36}{v} \) and \( \frac{36}{v+2} \).
- **M1**: Sets up a correct difference equation equal to \( 0.6 \) (or \( \frac{36}{60} \) or \( \frac{3}{5} \)).
- **M1**: Correctly clears denominators and expands terms, leading to a quadratic equation of the form \( 0.6v^2 + 1.2v - 72 = 0 \) or \( v^2 + 2v - 120 = 0 \).
- **M1**: Solves the quadratic equation using a valid method (factorisation, completing the square, or formula).
- **A1**: Correctly identifies the positive root to find the actual average speed as \( 10 \) km/h (with negative root discarded).