2024 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

1. Multiply both sides by \(5 - x\):
\(y(5 - x) = 4x + 3\)

2. Expand the bracket:
\(5y - xy = 4x + 3\)

3. Collect terms with \(x\) on one side and other terms on the other side:
\(5y - 3 = 4x + xy\)

4. Factorise \(x\) out of the right-hand side:
\(5y - 3 = x(4 + y)\)

5. Divide both sides by \(4 + y\):
\(x = \frac{5y - 3}{y + 4}\)

M1 for multiplying by \(5-x\) and expanding the bracket, yielding \(5y - xy = 4x + 3\)
M1 for collecting \(x\) terms on one side and factorising to get \(x(4+y)\) or \(x(y+4)\)
A1 for \(x = \frac{5y - 3}{y + 4}\) or equivalent

Let \(V\) be the initial value of the watch at the start of 2021.
An increase of \(12\%\) corresponds to a multiplier of \(1.12\).
A further increase of \(15\%\) corresponds to a multiplier of \(1.15\).

Therefore, we have:
\(V \times 1.12 \times 1.15 = 1932\)
\(V \times 1.288 = 1932\)

Solving for \(V\):
\(V = \frac{1932}{1.288} = 1500\)

M1 for writing \(V \times 1.12 \times 1.15 = 1932\) or finding the combined multiplier \(1.12 \times 1.15 = 1.288\)
M1 for a complete method to find the initial value, e.g. \(1932 \div 1.288\) or \(1932 \div 1.15 \div 1.12\)
A1 for \(1500\)

1. In the right-angled triangle \(DBC\), we can find the side \(BC\) using trigonometry:
\(\tan(35^\circ) = \frac{DB}{BC} = \frac{8}{BC}\)
\(BC = \frac{8}{\tan(35^\circ)} \approx 11.425\text{ cm}\)

2. The total length \(AB = AD + DB = 5 + 8 = 13\text{ cm}\).

3. In the right-angled triangle \(ABC\), use Pythagoras' theorem to find \(AC\):
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 13^2 + 11.425^2 = 169 + 130.534 = 299.534\)
\(AC = \sqrt{299.534} \approx 17.307\text{ cm}\)

To 3 significant figures, \(AC = 17.3\text{ cm}\).

M1 for a correct trigonometric ratio to find \(BC\), e.g., \(\tan(35^\circ) = \frac{8}{BC}\) or \(BC = 11.4\dots\)
M1 for using Pythagoras' theorem in triangle \(ABC\) with their value of \(BC\) and \(AB = 13\), e.g., \(\sqrt{13^2 + 11.425^2}\)
A1 for \(17.3\) (accept answers in range \(17.3\) to \(17.31\))

The total number of counters in the bag is \(7 + 5 + 3 = 15\).

We need the probability of choosing two red counters, two blue counters, or two green counters:

1. Probability of Red and Red:
\(P(\text{R, R}) = \frac{7}{15} \times \frac{6}{14} = \frac{42}{210}\)

2. Probability of Blue and Blue:
\(P(\text{B, B}) = \frac{5}{15} \times \frac{4}{14} = \frac{20}{210}\)

3. Probability of Green and Green:
\(P(\text{G, G}) = \frac{3}{15} \times \frac{2}{14} = \frac{6}{210}\)

Adding these probabilities together:
\(P(\text{Same colour}) = \frac{42}{210} + \frac{20}{210} + \frac{6}{210} = \frac{68}{210} = \frac{34}{105}\) (or approximately \(0.324\))

M1 for any one correct probability product for two identical colours, e.g., \(\frac{7}{15} \times \frac{6}{14}\) or \(\frac{5}{15} \times \frac{4}{14}\) or \(\frac{3}{15} \times \frac{2}{14}\)
M1 for summing the three correct products: \(\left(\frac{7}{15} \times \frac{6}{14}\right) + \left(\frac{5}{15} \times \frac{4}{14}\right) + \left(\frac{3}{15} \times \frac{2}{14}\right)\)
A1 for \(\frac{34}{105}\) or equivalent fraction, or decimal \(0.324\) or better

1. Find the area scale factor between cylinder \(A\) and cylinder \(B\):
\(\text{Area Scale Factor} = \frac{405}{180} = 2.25\)

2. Find the linear scale factor (\(k\)) by taking the square root of the area scale factor:
\(k = \sqrt{2.25} = 1.5\)

3. Find the volume scale factor by cubing the linear scale factor:
\(\text{Volume Scale Factor} = k^3 = 1.5^3 = 3.375\)

4. Calculate the volume of cylinder \(B\):
\(\text{Volume}_B = \text{Volume}_A \times 3.375 = 480 \times 3.375 = 1620\text{ cm}^3\)

M1 for finding the linear scale factor \(k = \sqrt{\frac{405}{180}}\) or \(1.5\) (or reciprocal)
M1 for cubing their linear scale factor and multiplying by 480, e.g. \(480 \times (1.5)^3\)
A1 for \(1620\)

1. The perimeter of a sector consists of the arc length plus two radii:
\(\text{Perimeter} = \text{Arc Length} + 2r\)

2. Calculate the arc length:
\(\text{Arc Length} = \frac{\theta}{360} \times 2\pi r = \frac{135}{360} \times 2 \times \pi \times 12\)
\(\text{Arc Length} = \frac{3}{8} \times 24\pi = 9\pi\text{ cm}\)

3. Add the two radii to get the total perimeter:
\(\text{Perimeter} = 9\pi + 2(12) = 9\pi + 24\text{ cm}\)

M1 for a correct expression for the arc length, e.g., \(\frac{135}{360} \times 2 \times \pi \times 12\) or \(9\pi\)
M1 for adding \(2 \times 12\) (or \(24\)) to their arc length expression
A1 for \(9\pi + 24\) or \(24 + 9\pi\)

To find the mass of one particle, divide the total mass by the number of particles:

\(\text{Mass of one particle} = \frac{2.7 \times 10^4}{4.5 \times 10^7}\)

Separate the numerical and power of 10 parts:
\(\frac{2.7}{4.5} = 0.6\)
\(\frac{10^4}{10^7} = 10^{4 - 7} = 10^{-3}\)

Combine these to get:
\(0.6 \times 10^{-3}\text{ grams}\)

Convert this into standard form:
\(6 \times 10^{-4}\text{ grams}\)

M1 for the division expression \(\frac{2.7 \times 10^4}{4.5 \times 10^7}\)
M1 for reaching \(0.6 \times 10^{-3}\) or \(0.0006\) or equivalent non-standard form
A1 for \(6 \times 10^{-4}\) (accept \(6 \times 10^{-4}\text{ g}\))

1. Find the slant height (\(l\)) of the cone using Pythagoras' theorem:
\(l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\)

2. The formula for the total surface area of a cone is:
\(A = \pi r^2 + \pi r l\)

3. Substitute \(r = 5\) and \(l = 13\) into the formula:
\(A = \pi (5)^2 + \pi (5)(13)\)
\(A = 25\pi + 65\pi = 90\pi\text{ cm}^2\)

4. Calculate the decimal value:
\(90\pi \approx 282.743\text{ cm}^2\)

To 3 significant figures, the surface area is \(283\text{ cm}^2\).

M1 for finding the slant height \(l = 13\) using Pythagoras' theorem
M1 for substituting \(r=5\) and their \(l\) into the formula \(\pi r^2 + \pi r l\), e.g., \(\pi \times 5^2 + \pi \times 5 \times 13\)
A1 for \(283\) (accept answers in the range \(282.7\) to \(283\))

Let \(P\) be the initial value of the investment.\
At the end of the first year, after a decrease of \(8\\%\), the value of the investment is:\
\[P \times (1 - 0.08) = 0.92P\]\
At the end of the second year, after an increase of \(15\\%\), the value is:\
\[0.92P \times (1 + 0.15) = 0.92P \times 1.15 = 1.058P\]\
We are given that the value at the end of the second year is \(£5290\), so:\
\[1.058P = 5290\]\
Solving for \(P\):\
\[P = \frac{5290}{1.058} = 5000\]

M1 for writing an expression for the value after one year, e.g. \(P \times 0.92\) or \(0.92\), or for a correct multiplier for the two-year period, e.g. \(0.92 \times 1.15\) or \(1.058\), or for dividing the final amount by the second year's increase, e.g. \(5290 / 1.15\) (which gives \(4600\)).\
M1 for setting up a complete equation to find \(P\), e.g. \(1.058P = 5290\) or \(0.92P = 4600\).\
A1 for \(5000\).

Multiply both sides of the formula by the denominator \(2w + 5\):\
\[p(2w + 5) = 4 - 3w\]\
Expand the bracket on the left-hand side:\
\[2pw + 5p = 4 - 3w\]\
Rearrange the equation to collect all terms containing \(w\) on one side and terms without \(w\) on the other side:\
\[2pw + 3w = 4 - 5p\]\
Factorise \(w\) from the terms on the left-hand side:\
\[w(2p + 3) = 4 - 5p\]\
Divide both sides by \(2p + 3\) to isolate \(w\):\
\[w = \frac{4 - 5p}{2p + 3}\]

M1 for multiplying both sides by \((2w + 5)\) to clear the fraction, e.g., \(p(2w + 5) = 4 - 3w\).\
M1 for expanding the brackets and isolating terms containing \(w\) on one side of the equation, e.g., \(2pw + 3w = 4 - 5p\) (allow at most one sign error in rearranging).\
A1 for \(w = \frac{4 - 5p}{2p + 3}\) or any equivalent correct fraction, such as \(w = \frac{5p - 4}{-2p - 3}\).

Perimeter of the sector is given by:
\(P = 2r + \frac{\theta}{360} \times 2\pi r = 24 + 4\pi\)

Area of the sector is given by:
\(A = \frac{\theta}{360} \times \pi r^2 = 24\pi\)

From the area equation, we can express:
\(\frac{\theta}{360} = \frac{24}{r^2}\)

Substitute this expression into the perimeter equation:
\(2r + \left(\frac{24}{r^2}\right) \times 2\pi r = 24 + 4\pi\)
\(2r + \frac{48\pi}{r} = 24 + 4\pi\)

Multiply the entire equation by \(r\) to clear the fraction:
\(2r^2 + 48\pi = (24 + 4\pi)r\)
\(2r^2 - (24 + 4\pi)r + 48\pi = 0\)

This quadratic equation in terms of \(r\) can be factorised as:
\((2r - 4\pi)(r - 12) = 0\)

This gives two possible values for \(r\):
1) \(2r - 4\pi = 0 \implies r = 2\pi\) (not an integer)
2) \(r - 12 = 0 \implies r = 12\) (an integer)

Since \(r\) is given as an integer, we must have \(r = 12\).

Substitute \(r = 12\) back into \(\frac{\theta}{360} = \frac{24}{r^2}\):
\(\frac{\theta}{360} = \frac{24}{144} = \frac{1}{6}\)
\(\theta = \frac{360}{6} = 60\).

M1: For writing correct expressions for perimeter and area of a sector in terms of \(r\) and \(\theta\).
M1: For substituting one equation into another to eliminate \(\theta\) and obtain an equation in terms of \(r\) only, e.g., \(2r + \frac{48\pi}{r} = 24 + 4\pi\).
M1: For forming a quadratic equation and attempting to solve for \(r\).
A1: For identifying \(r = 12\) as the correct integer solution.
A1: For \(\theta = 60\).

(a) The area of a triangle is given by \(\frac{1}{2} \times \text{base} \times \text{height}\).
\(\frac{1}{2}(2x + 1)(x + 2) = 7\)
\((2x + 1)(x + 2) = 14\)
\(2x^2 + 4x + x + 2 = 14\)
\(2x^2 + 5x - 12 = 0\)

(b) Solve the quadratic equation by factorising:
\(2x^2 + 5x - 12 = 0\)
\((2x - 3)(x + 4) = 0\)

This gives \(x = 1.5\) or \(x = -4\).
Since a length cannot be negative, we must have \(x = 1.5\).

(c) When \(x = 1.5\):
\(\text{Base} = 2(1.5) + 1 = 4\text{ cm}\)
\(\text{Height} = 1.5 + 2 = 3.5\text{ cm}\)

Using Pythagoras' theorem to find the hypotenuse, \(h\):
\(h = \sqrt{4^2 + 3.5^2} = \sqrt{16 + 12.25} = \sqrt{28.25} \approx 5.315\text{ cm}\)

\(\text{Perimeter} = 4 + 3.5 + 5.315 = 12.815\text{ cm} \approx 12.8\text{ cm}\) (to 3 s.f.).

M1 (Part a): For writing the correct equation for the area of the triangle: \(\frac{1}{2}(2x + 1)(x + 2) = 7\).
A1 (Part a): For expanding and rearranging to show \(2x^2 + 5x - 12 = 0\) clearly.
M1 (Part b): For attempting to factorise or solve the quadratic equation.
A1 (Part b): For selecting \(x = 1.5\) and rejecting the negative root.
M1 (Part c): For applying Pythagoras' theorem to find the hypotenuse \(\sqrt{4^2 + 3.5^2}\).
A1 (Part c): For finding the perimeter as \(12.8\) (accept answers in the range \(12.8\) to \(12.82\)).

(a) There are \(n\) red counters, so there are \(9 - n\) yellow counters.
\(P(\text{both same colour}) = P(\text{Red, Red}) + P(\text{Yellow, Yellow})\)
\(P(\text{Red, Red}) = \frac{n}{9} \times \frac{n-1}{8} = \frac{n(n-1)}{72}\)
\(P(\text{Yellow, Yellow}) = \frac{9-n}{9} \times \frac{8-n}{8} = \frac{(9-n)(8-n)}{72}\)

Set the sum of these probabilities to \(\frac{1}{2}\):
\(\frac{n(n-1) + (9-n)(8-n)}{72} = \frac{1}{2}\)
\(n(n-1) + (9-n)(8-n) = 36\)
\(n^2 - n + 72 - 17n + n^2 = 36\)
\(2n^2 - 18n + 36 = 0\)

Divide by 2:
\(n^2 - 9n + 18 = 0\)

(b) Solve the quadratic equation:
\(n^2 - 9n + 18 = 0 \implies (n-3)(n-6) = 0\)
So, \(n = 3\) or \(n = 6\).

Since there are more yellow than red counters:
\(9-n > n \implies 9 > 2n \implies n < 4.5\).
Therefore, \(n = 3\).
There are 3 red counters and 6 yellow counters in the bag.

We want to find the probability that at least one of the counters is red:
\(P(\text{at least one red}) = 1 - P(\text{Yellow, Yellow})\)
\(P(\text{Yellow, Yellow}) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12}\)
\(P(\text{at least one red}) = 1 - \frac{5}{12} = \frac{7}{12}\).

M1 (Part a): For writing the probability equation in terms of \(n\): \(\frac{n}{9} \times \frac{n-1}{8} + \frac{9-n}{9} \times \frac{8-n}{8} = \frac{1}{2}\).
A1 (Part a): For expanding correctly and simplifying to show \(n^2 - 9n + 18 = 0\).
M1 (Part b): For solving the quadratic equation to get \(n = 3\) (rejecting \(n = 6\) based on the condition).
M1 (Part b): For calculating \(1 - P(\text{Yellow, Yellow})\) or using an alternative tree diagram path method.
A1 (Part b): For \(\frac{7}{12}\) (accept equivalent fractions, or decimal \(0.583\) or better).

The projection of the line \(DF\) on the horizontal base \(ABCD\) is \(DB\).
The angle that the line \(DF\) makes with the base is \(\angle FDB\).

First, calculate the length of \(DB\) using Pythagoras' theorem on the horizontal rectangular base \(ABCD\):
\(DB^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169\)
\(DB = \sqrt{169} = 13\text{ cm}\).

The line \(BF\) is vertical and thus perpendicular to any line in the base, including \(DB\).
In the right-angled triangle \(FBD\):
\(\tan(\angle FDB) = \frac{BF}{DB} = \frac{8}{13}\)

\(\angle FDB = \arctan\left(\frac{8}{13}\right) \approx 31.6075^\circ\).

To 1 decimal place, the angle is \(31.6^\circ\).

M1: For identifying that the required angle is \(\angle FDB\).
M1: For using Pythagoras' theorem to find the diagonal \(DB = \sqrt{12^2 + 5^2}\).
A1: For \(DB = 13\).
M1: For using a correct trigonometric ratio, e.g., \(\tan(\theta) = \frac{8}{13}\).
A1: For \(31.6\) (accept answers in the range \(31.6\) to \(31.61\)).

(a) Let \(x = 1 + \frac{r}{100}\) be the compound interest multiplier.
At the end of year 1, the investment value is:
\(P x = 8320\) [Equation 1]

At the end of year 3, the investment value is:
\(P x^3 = 8998.91\) [Equation 2]

Divide Equation 2 by Equation 1:
\(\frac{P x^3}{P x} = \frac{8998.91}{8320}\)
\(x^2 = 1.08159976\)
\(x = \sqrt{1.08159976} = 1.04\)

Since \(1 + \frac{r}{100} = 1.04\), we have \(r = 4\).

(b) Substitute \(x = 1.04\) into Equation 1:
\(P \times 1.04 = 8320\)
\(P = \frac{8320}{1.04} = 8000\).

M1 (Part a): For expressing the values at year 1 and year 3 in terms of \(P\) and a multiplier, e.g., \(P(1 + r/100) = 8320\) and \(P(1 + r/100)^3 = 8998.91\).
M1 (Part a): For dividing the equations to obtain \((1 + r/100)^2 = \frac{8998.91}{8320}\).
A1 (Part a): For \(r = 4\).
M1 (Part b): For substituting their value of \(r\) into the year 1 equation to find \(P\).
A1 (Part b): For \(P = 8000\).

(a) The ratio of the surface areas of cones \(A\) and \(B\) is \(9 : 16\).
Therefore, the scale factor of length is:
\(\sqrt{\frac{16}{9}} = \frac{4}{3}\)

The scale factor of volume is:
\(\left(\frac{4}{3}\right)^3 = \frac{64}{27}\)

Volume of cone \(B = \text{Volume of cone } A \times \frac{64}{27}\)
\(\text{Volume} = 135 \times \frac{64}{27} = 5 \times 64 = 320\text{ cm}^3\).

(b) Since both cones are made of the same material, they have the same density. Therefore, their masses are in the same ratio as their volumes (\(27 : 64\)).

Mass of cone \(B = 1.28\text{ kg} = 1280\text{ g}\).

Mass of cone \(A = 1280 \times \frac{27}{64} = 20 \times 27 = 540\text{ g}\).

M1 (Part a): For finding the linear scale factor \(\sqrt{16/9} = 4/3\) (or ratio \(3:4\)).
M1 (Part a): For cubing the linear scale factor to get the volume scale factor \(64/27\) (or ratio \(27:64\)).
A1 (Part a): For \(320\).
M1 (Part b): For converting \(1.28\text{ kg}\) to \(1280\text{ g}\) and multiplying by \(\frac{27}{64}\).
A1 (Part b): For \(540\).

(a) Velocity is the derivative of displacement with respect to time:
\(v = \frac{ds}{dt} = 6t^2 - 30t + 24\).

(b) The particle is momentarily at rest when its velocity is zero:
\(6t^2 - 30t + 24 = 0\)
\(t^2 - 5t + 4 = 0\)
\((t - 1)(t - 4) = 0\)

So the particle is at rest at \(t = 1\) and \(t = 4\).

Acceleration is the derivative of velocity with respect to time:
\(a = \frac{dv}{dt} = 12t - 30\).

Substitute \(t = 1\) into the acceleration formula:
\(a = 12(1) - 30 = -18\text{ m/s}^2\).

Substitute \(t = 4\) into the acceleration formula:
\(a = 12(4) - 30 = 18\text{ m/s}^2\).

The values of the acceleration when the particle is at rest are \(-18\text{ m/s}^2\) and \(18\text{ m/s}^2\).

M1 (Part a): For attempting to differentiate the displacement function with at least one term correct.
A1 (Part a): For \(6t^2 - 30t + 24\).
M1 (Part b): For setting their velocity expression to 0 and solving to find \(t = 1\) and \(t = 4\).
M1 (Part b): For differentiating velocity to find acceleration \(12t - 30\) and substituting at least one of their times.
A1 (Part b): For both \(-18\) and \(18\).

Multiply every term in the equation by \((2x-1)(x+3)\) to clear the denominators:
\(3(x+3) - 2(2x-1) = 1(2x-1)(x+3)\)

Expand the terms:
\(3x + 9 - 4x + 2 = 2x^2 + 6x - x - 3\)
\(11 - x = 2x^2 + 5x - 3\)

Rearrange the terms to set the equation to 0:
\(2x^2 + 5x + x - 3 - 11 = 0\)
\(2x^2 + 6x - 14 = 0\)

Divide all terms by 2:
\(x^2 + 3x - 7 = 0\) (as required).

Now solve this equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2(1)}\)
\(x = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}\)

Calculating the two solutions:
\(x = \frac{-3 + 6.08276...}{2} \approx 1.54\) (to 3 s.f.)
\(x = \frac{-3 - 6.08276...}{2} \approx -4.54\) (to 3 s.f.).

M1: For clearing fractions by multiplying correctly: \(3(x+3) - 2(2x-1) = (2x-1)(x+3)\).
A1: For expanding and simplifying to show \(x^2 + 3x - 7 = 0\) clearly.
M1: For substituting correctly into the quadratic formula.
A1: For \(1.54\) (accept \(1.54\) to \(1.542\)).
A1: For \(-4.54\) (accept \(-4.54\) to \(-4.542\)).

The formula for the total surface area of a cylinder is:
\(\text{Total Surface Area} = 2\pi r^2 + 2\pi r h\)

We are given that the total surface area is \(120\pi\text{ cm}^2\):
\(2\pi r^2 + 2\pi r h = 120\pi\)

Divide the entire equation by \(2\pi\):
\(r^2 + r h = 60\)

We are also given that the height \(h\) is \(3\text{ cm}\) more than twice its radius \(r\):
\(h = 2r + 3\)

Substitute this expression for \(h\) into the equation:
\(r^2 + r(2r + 3) = 60\)
\(r^2 + 2r^2 + 3r = 60\)
\(3r^2 + 3r - 60 = 0\)

Divide by 3:
\(r^2 + r - 20 = 0\)

Factorise the quadratic equation:
\((r - 4)(r + 5) = 0\)

Since the radius must be positive, \(r = 4\) cm.

Now find the height \(h\):
\(h = 2(4) + 3 = 11\) cm.

Calculate the volume of the cylinder:
\(V = \pi r^2 h = \pi \times 4^2 \times 11 = 176\pi\text{ cm}^3\).

M1 for setting up the equation \(2\pi r^2 + 2\pi r h = 120\pi\) or \(r^2 + r h = 60\)
M1 for substituting \(h = 2r + 3\) to form a quadratic equation
A1 for obtaining the correct radius \(r = 4\)
M1 for substituting their radius to find the height and substituting into the volume formula \(\pi r^2 h\)
A1 for the correct answer of \(176\pi\)

The probability of selecting two green counters without replacement is:
\(P(\text{Green, Green}) = \frac{7}{n} \times \frac{6}{n-1}\)

We are given that this probability is \(\frac{7}{30}\):
\(\frac{42}{n(n-1)} = \frac{7}{30}\)

Divide both sides by 7:
\(\frac{6}{n(n-1)} = \frac{1}{30}\)

Cross-multiply:
\(n(n-1) = 180\)
\(n^2 - n - 180 = 0\)

Factorise the quadratic:
\((n - 15)(n + 12) = 0\)

Since \(n\) must be a positive integer, the total number of counters in the bag is \(n = 15\).

The number of green counters is 7.
The number of blue counters is \(15 - 7 = 8\).

The probability of choosing two counters of different colours is:
\(P(\text{different}) = P(\text{Green, Blue}) + P(\text{Blue, Green})\)
\(P(\text{different}) = \left(\frac{7}{15} \times \frac{8}{14}\right) + \left(\frac{8}{15} \times \frac{7}{14}\right)\)
\(P(\text{different}) = \frac{56}{210} + \frac{56}{210} = \frac{112}{210}\)

Simplify the fraction:
\(\frac{112}{210} = \frac{8}{15}\).

M1 for writing down an equation for the probability of two green counters: \(\frac{7}{n} \times \frac{6}{n-1} = \frac{7}{30}\)
M1 for forming and solving the quadratic equation to find \(n = 15\)
M1 for calculating the number of blue counters as 8
M1 for a complete method to find the probability of different colours: e.g., \(2 \times \frac{7}{15} \times \frac{8}{14}\)
A1 for \(\frac{8}{15}\) or equivalent fraction (e.g., \(\frac{112}{210}\))

First, find the length of the diagonal \(AC\) of the base:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10\text{ cm}\).

The centre of the base \(O\) is the midpoint of \(AC\), so:
\(OC = \frac{10}{2} = 5\text{ cm}\).

Since \(VO\) is perpendicular to the base, triangle \(VOC\) is a right-angled triangle. Use Pythagoras' theorem to find the height of the pyramid \(VO\):
\(VO = \sqrt{VC^2 - OC^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}\).

Let \(M\) be the midpoint of \(BC\). The angle between the face \(VBC\) and the base is the angle \(\angle VMO\).
In triangle \(VOM\), \(VO\) is perpendicular to \(OM\).
The length \(OM\) is parallel to \(AB\) and is equal to half of \(AB\):
\(OM = \frac{AB}{2} = \frac{8}{2} = 4\text{ cm}\).

In the right-angled triangle \(VOM\):
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{12}{4} = 3\).

Therefore, the angle is:
\(\angle VMO = \tan^{-1}(3) \approx 71.565^\circ\).

To 1 decimal place, this is \(71.6^\circ\).

M1 for finding the diagonal length \(AC = 10\) or half-diagonal \(OC = 5\)
M1 for a correct application of Pythagoras' theorem to find the vertical height \(VO = 12\)
M1 for finding the perpendicular distance from the centre to the edge \(BC\), which is \(OM = 4\)
M1 for setting up the trigonometric ratio \(\tan(\theta) = \frac{12}{4}\)
A1 for an answer in the range \(71.5\) to \(71.6\)

Since spheres \(A\) and \(B\) are made of the same metal, their masses are directly proportional to their volumes:
\(\frac{\text{Volume of } B}{\text{Volume of } A} = \frac{\text{Mass of } B}{\text{Mass of } A} = \frac{320}{135} = \frac{64}{27}\).

For any mathematically similar solids, the ratio of their volumes is the cube of the linear scale factor \(k\):
\(k^3 = \frac{64}{27}\)

Taking the cube root of both sides gives the linear scale factor:
\(k = \sqrt[3]{\frac{64}{27}} = \frac{4}{3}\).

The ratio of the surface areas is the square of the linear scale factor \(k^2\):
\(k^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}\).

Therefore, the surface area of sphere \(B\) is:
\(\text{Surface Area of } B = \text{Surface Area of } A \times \frac{16}{9}\)
\(\text{Surface Area of } B = 54 \times \frac{16}{9} = 6 \times 16 = 96\text{ cm}^2\).

M1 for using the mass ratio to represent the volume ratio: \(\frac{320}{135}\) or \(\frac{64}{27}\) (or inverse)
M1 for finding the linear scale factor by taking the cube root: \(\sqrt[3]{\frac{64}{27}} = \frac{4}{3}\) (or inverse)
M1 for squaring the linear scale factor to get the area scale factor: \(\left(\frac{4}{3}\right)^2 = \frac{16}{9}\) (or inverse)
M1 for multiplying the surface area of \(A\) by the area scale factor: \(54 \times \frac{16}{9}\)
A1 for \(96\)

To solve the equation, first eliminate the denominators by multiplying every term by \((x-3)(x+1)\):
\(5(x+1) - 2(x-3) = 1(x-3)(x+1)\)

Expand both sides:
\(5x + 5 - (2x - 6) = x^2 - 2x - 3\)
\(5x + 5 - 2x + 6 = x^2 - 2x - 3\)
\(3x + 11 = x^2 - 2x - 3\)

Rearrange the terms to form a quadratic equation equal to zero:
\(x^2 - 5x - 14 = 0\)

Factorise the quadratic expression:
\((x - 7)(x + 2) = 0\)

Set each factor to zero to find the solutions:
\(x - 7 = 0 \implies x = 7\)
\(x + 2 = 0 \implies x = -2\)

The solutions are \(x = 7\) and \(x = -2\).

M1 for multiplying by \((x-3)(x+1)\) to clear fractions: \(5(x+1) - 2(x-3) = (x-3)(x+1)\)
M1 for expanding correctly: \(5x + 5 - 2x + 6 = x^2 - 2x - 3\)
M1 for forming a 3-term quadratic equation: \(x^2 - 5x - 14 = 0\)
M1 for factorising the quadratic correctly: \((x-7)(x+2) = 0\) (or correct use of quadratic formula)
A1 for both solutions \(x = 7\) and \(x = -2\)

Let \(P\) be the profit at the start of 2021. An increase of 15% is represented by a multiplier of 1.15. A decrease of 8% is represented by a multiplier of 0.92. Therefore, \(P \times 1.15 \times 0.92 = 158700\). This simplifies to \(P \times 1.058 = 158700\). Solving for \(P\), we get \(P = \frac{158700}{1.058} = 150000\).

M1: for set-up of the equation, e.g., \(P \times 1.15 \times 0.92 = 158700\) or finding the combined multiplier of 1.058. M1: for a complete method to solve for the initial profit, e.g., \(\frac{158700}{1.058}\). A1: for 150000 (accept £150,000).

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Write as a single fraction and cancel the common factor of \((2x + 1)\) from the numerator and denominator to get \(\frac{x - 3}{2x - 1}\).

M1: for factorising the numerator as \((2x + 1)(x - 3)\) or the denominator as \((2x - 1)(2x + 1)\). M1: for fully factorising both numerator and denominator and identifying the common factor of \((2x + 1)\). A1: for \(\frac{x - 3}{2x - 1}\) or equivalent.

The total number of counters in the bag is \(7 + n\). The probability of selecting a red counter first is \(\frac{7}{7+n}\). If a red counter is selected, the probability of selecting another red counter without replacement is \(\frac{6}{6+n}\). The probability of both being red is: \(\frac{7}{7+n} \times \frac{6}{6+n} = \frac{7}{15}\). This simplifies to \(\frac{42}{(7+n)(6+n)} = \frac{7}{15}\). Dividing both sides by 7 gives \(\frac{6}{(7+n)(6+n)} = \frac{1}{15}\), which leads to \((7+n)(6+n) = 90\). Expanding and simplifying: \(n^2 + 13n + 42 = 90\), which is \(n^2 + 13n - 48 = 0\). Factorising this quadratic gives \((n + 16)(n - 3) = 0\). Since the number of counters \(n\) must be positive, \(n = 3\).

M1: for writing a correct probability equation for two red counters, e.g., \(\frac{7}{7+n} \times \frac{6}{6+n} = \frac{7}{15}\). M1: for expanding and forming a quadratic equation, e.g., \(n^2 + 13n - 48 = 0\). A1: for \(n = 3\) (rejecting \(n = -16\)).

Using the Cosine Rule: \(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\). Substituting the given values: \(BC^2 = 8.4^2 + 12.5^2 - 2 \times 8.4 \times 12.5 \times \cos(56^\circ)\). This gives \(BC^2 = 70.56 + 156.25 - 210 \times \cos(56^\circ)\), so \(BC^2 = 226.81 - 210 \times 0.5591929... = 226.81 - 117.4305... = 109.3795...\). Taking the square root: \(BC = \sqrt{109.3795...} \approx 10.458...\text{ cm}\). Correct to 3 significant figures, \(BC = 10.5\text{ cm}\).

M1: for substituting the correct values into the Cosine Rule, e.g., \(8.4^2 + 12.5^2 - 2 \times 8.4 \times 12.5 \times \cos(56^\circ)\). M1: for a correct evaluation of \(BC^2\) (e.g., \(109.38\)) or \(BC = \sqrt{109.38...}\). A1: for 10.5 (accept 10.45 - 10.5).

The area of a sector is given by \(\text{Area} = \frac{\theta}{360} \times \pi r^2\). Substituting the known values: \(48\pi = \frac{\theta}{360} \times \pi \times 12^2\), which simplifies to \(48\pi = \frac{\theta}{360} \times 144\pi\). Dividing by \(\pi\), we get \(48 = \frac{144\theta}{360}\), which gives the fraction of the circle as \(\frac{\theta}{360} = \frac{48}{144} = \frac{1}{3}\). The arc length of the sector is \(\frac{\theta}{360} \times 2\pi r = \frac{1}{3} \times 2\pi \times 12 = 8\pi\text{ cm}\). The perimeter is the sum of the arc length and the two boundary radii: \(\text{Perimeter} = 8\pi + 12 + 12 = 24 + 8\pi\text{ cm}\).

M1: for establishing the relationship to find the angle or fraction of the circle, e.g., \(48\pi = \frac{\theta}{360} \times 144\pi\) or \(\text{fraction} = \frac{1}{3}\). M1: for calculating the arc length as \(8\pi\) or showing a method to find the perimeter, e.g., \(\text{arc length} + 24\). A1: for \(24 + 8\pi\) (accept \(8\pi + 24\)).

First, differentiate the function: \(\frac{dy}{dx} = 6x^2 - 18x - 24\). Set \(\frac{dy}{dx} = 0\) to find the stationary points: \(6x^2 - 18x - 24 = 0 \implies x^2 - 3x - 4 = 0\). Factorising gives \((x - 4)(x + 1) = 0\), so the stationary points are at \(x = 4\) and \(x = -1\). Find the second derivative to test the nature of these points: \(\frac{d^2y}{dx^2} = 12x - 18\). For \(x = -1\), \(\frac{d^2y}{dx^2} = 12(-1) - 18 = -30 < 0\), so \(x = -1\) is the maximum. Substituting \(x = -1\) back into the original curve equation: \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20\). Therefore, the maximum turning point is \((-1, 20)\).

M1: for differentiating to find \(\frac{dy}{dx} = 6x^2 - 18x - 24\) (at least two terms correct). M1: for setting \(\frac{dy}{dx} = 0\) and solving the quadratic equation to get \(x = -1\) and \(x = 4\), and showing which value corresponds to the maximum. A1: for \((-1, 20)\).

We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3, b = -8, c = 2\). This gives: \(x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)}\), which simplifies to \(x = \frac{8 \pm \sqrt{64 - 24}}{6} = \frac{8 \pm \sqrt{40}}{6}\). Evaluating the two cases: \(x = \frac{8 + 6.3245...}{6} \approx 2.387\) and \(x = \frac{8 - 6.3245...}{6} \approx 0.2792\). Correct to 3 significant figures, the solutions are \(x = 2.39\) or \(x = 0.279\).

M1: for substituting the coefficients correctly into the quadratic formula, e.g., \(\frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3}\). M1: for evaluating the discriminant as 40 or showing a correct intermediate step, e.g., \(\frac{8 \pm \sqrt{40}}{6}\). A1: for both correct answers: \(x = 2.39\) and \(x = 0.279\) (accept answers in range 2.38-2.40 and 0.278-0.280).

The ratio of the surface areas is \(9 : 16\). The linear scale factor (L.S.F) is the square root of the area scale factor: \(\text{L.S.F} = \sqrt{9} : \sqrt{16} = 3 : 4\). The volume scale factor (V.S.F) is the cube of the linear scale factor: \(\text{V.S.F} = 3^3 : 4^3 = 27 : 64\). Therefore, the volume of cylinder B is calculated by multiplying the volume of cylinder A by the scale factor \(\frac{64}{27}\): \(\text{Volume of B} = 162 \times \frac{64}{27} = 6 \times 64 = 384\text{ cm}^3\).

M1: for finding the linear scale factor ratio \(3 : 4\) (or scale factor \(\frac{4}{3}\) or \(\frac{3}{4}\)). M1: for finding the volume scale factor ratio \(27 : 64\) (or scale factor \(\frac{64}{27}\)) and multiplying by 162. A1: for 384.

We first rewrite the curve's equation using negative indices:
\(y = 4x^2 - 16x^{-1}\)

Next, we differentiate \(y\) with respect to \(x\) to find the gradient function, \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{d}{dx}(4x^2) - \frac{d}{dx}(16x^{-1})\)
\(\frac{dy}{dx} = 8x - 16(-1)x^{-2}\)
\(\frac{dy}{dx} = 8x + \frac{16}{x^2}\)

Finally, we substitute \(x = 2\) into the gradient function:
\(\frac{dy}{dx} = 8(2) + \frac{16}{2^2} = 16 + \frac{16}{4} = 16 + 4 = 20\).

M1: For a correct attempt to differentiate, with at least one term differentiated correctly (e.g. \(8x\) or \(16x^{-2}\) seen).
M1: For substituting \(x = 2\) into their derivative.
A1: For \(20\) (with working shown).

Using the Cosine Rule to find the angle \(BAC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(BAC)\)

Substitute the given lengths into the formula:
\(10^2 = 7^2 + 8^2 - 2(7)(8) \cos(BAC)\)
\(100 = 49 + 64 - 112 \cos(BAC)\)
\(100 = 113 - 112 \cos(BAC)\)

Rearrange to solve for \(\cos(BAC)\):
\(112 \cos(BAC) = 113 - 100\)
\(112 \cos(BAC) = 13\)
\(\cos(BAC) = \frac{13}{112}\)

Calculate the angle:
\(BAC = \cos^{-1}\left(\frac{13}{112}\right) \approx 83.336...^\circ\)

To 1 decimal place, the angle is \(83.3^\circ\).

M1: For a correct substitution into the Cosine Rule, e.g. \(10^2 = 7^2 + 8^2 - 2(7)(8) \cos(BAC)\).
M1: For a correct rearrangement to isolate the cosine term, e.g. \(\cos(BAC) = \frac{13}{112}\) (or \(0.116...\)).
A1: For \(83.3\) (accept answers in the range \(83.3\) to \(83.4\)).

The perimeter of a sector is the sum of its arc length and the two boundary radii:
\(\text{Perimeter} = \text{Arc length} + 2r\)

First, calculate the arc length of the sector:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)
\(\text{Arc length} = \frac{150}{360} \times 2 \times \pi \times 12\)
\(\text{Arc length} = \frac{5}{12} \times 24\pi = 10\pi\text{ cm}\)

Now, add the two radii to get the total perimeter:
\(\text{Perimeter} = 10\pi + 2(12) = 24 + 10\pi\text{ cm}\).

This is in the form \(a + b\pi\) where \(a = 24\) and \(b = 10\).

M1: For a correct method to find the arc length, e.g. \(\frac{150}{360} \times 2 \times \pi \times 12\) or \(10\pi\).
M1: For adding two radii (i.e., \(24\)) to their arc length.
A1: For \(24 + 10\pi\) (or \(10\pi + 24\)).

We can solve \(3x^2 - 14x + 8 = 0\) by factorising.
We need two numbers that multiply to \(3 \times 8 = 24\) and add to \(-14\).
These numbers are \(-12\) and \(-2\).

Rewrite the middle term:
\(3x^2 - 12x - 2x + 8 = 0\)

Factorise by grouping terms:
\(3x(x - 4) - 2(x - 4) = 0\)
\((3x - 2)(x - 4) = 0\)

Set each factor to zero to solve for \(x\):
1) \(3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3}\)
2) \(x - 4 = 0 \implies x = 4\)

Thus, the solutions are \(x = \frac{2}{3}\) and \(x = 4\).

M1: For a correct method to factorise the quadratic expression, e.g. \((3x - 2)(x - 4)\), or correct substitution into the quadratic formula: \(\frac{-(-14) \pm \sqrt{(-14)^2 - 4(3)(8)}}{2(3)}\).
M1: For finding the two critical values/roots from their factors or formula.
A1: For \(2/3\) and \(4\) (or exact equivalents).

1. Factorise the denominators: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\) and \(2x^2 - x = x(2x - 1)\). 2. Find the lowest common denominator: \(x(2x - 1)(x + 3)\). 3. Express each fraction over the common denominator: \(\frac{3x}{x(2x - 1)(x + 3)} - \frac{2(x + 3)}{x(2x - 1)(x + 3)}\). 4. Combine the numerators and simplify: \(\frac{3x - 2(x + 3)}{x(2x - 1)(x + 3)} = \frac{3x - 2x - 6}{x(2x - 1)(x + 3)} = \frac{x - 6}{x(2x - 1)(x + 3)}\).

M1: For factorising either denominator to \((2x - 1)(x + 3)\) or \(x(2x - 1)\). M1: For finding a common denominator of \(x(2x - 1)(x + 3)\). M1: For writing the correct combined numerator, e.g., \(3x - 2(x + 3)\). A1: For \(\frac{x - 6}{x(2x - 1)(x + 3)}\) or \(\frac{x - 6}{2x^3 + 5x^2 - 3x}\) (or equivalent fully simplified fraction).

1. Find the diagonal of the base AC: \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm}\). 2. Find the distance from the corner A to the center of the base O: \(AO = \frac{1}{2}AC = 6.5\text{ cm}\). 3. In the right-angled triangle VAO, the angle \(VAO = 62^\circ\). Use trigonometry to find VO: \(\tan(62^\circ) = \frac{VO}{AO} \implies VO = 6.5 \times \tan(62^\circ) \approx 12.2247\text{ cm}\). 4. Round to 3 significant figures: \(VO = 12.2\text{ cm}\).

M1: For using Pythagoras' theorem to find \(AC = 13\). M1: For finding the distance to the center \(AO = 6.5\). M1: For setting up the trigonometric ratio \(\tan(62^\circ) = \frac{VO}{6.5}\). A1: For the final answer of 12.2 (accept 12.2 to 12.23).

1. Formulate the dimensions of the combined garden and path: The new length is \(x + 2(2) = x + 4\) metres and the new width is \((x - 4) + 2(2) = x\) metres. 2. Set up the area equation: \(\text{Area} = (x + 4)x = 120 \implies x^2 + 4x - 120 = 0\). 3. Solve using the quadratic formula: \(x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-120)}}{2} = \frac{-4 \pm \sqrt{496}}{2}\). Since length must be positive, we take the positive root: \(x = \frac{-4 + 22.271}{2} \approx 9.1355\). 4. Round to 3 significant figures: \(x \approx 9.14\).

M1: For expressing the total length as \(x + 4\) and total width as \(x\). M1: For setting up the quadratic equation \(x(x + 4) = 120\) or \(x^2 + 4x - 120 = 0\). M1: For applying the quadratic formula or completing the square correctly. A1: For 9.14 (accept 9.13 to 9.14).

Let the center of the circle be O. By symmetry, the vertical line passing through O also passes through the midpoint of AB (let's call it M) and the point of tangency on CD (let's call it T). The length \(MT = 10\text{ cm}\). Since CD is tangent to the circle, \(OT = r\). Therefore, \(OM = MT - OT = 10 - r\). In the right-angled triangle OMA, \(AM = 5\text{ cm}\) and \(OA = r\). By Pythagoras' theorem: \(OA^2 = AM^2 + OM^2 \implies r^2 = 5^2 + (10 - r)^2 \implies r^2 = 25 + 100 - 20r + r^2 \implies 20r = 125 \implies r = 6.25\).

M1: For identifying a right-angled triangle with sides 5, \(10-r\), and hypotenuse \(r\). M1: For applying Pythagoras' theorem: \(r^2 = 5^2 + (10-r)^2\). M1: For expanding and simplifying to \(20r = 125\). A1: For 6.25.

1. Write down the probability of selecting two green counters without replacement: \(P(\text{Green, Green}) = \frac{7}{n} \times \frac{6}{n-1} = \frac{42}{n(n-1)}\). 2. Set this probability equal to \(\frac{7}{15}\): \(\frac{42}{n(n-1)} = \frac{7}{15} \implies \frac{6}{n(n-1)} = \frac{1}{15} \implies n(n-1) = 90\). 3. Solve the quadratic equation: \(n^2 - n - 90 = 0 \implies (n-10)(n+9) = 0\). Since \(n > 0\), \(n = 10\). 4. Find the number of blue counters: \(10 - 7 = 3\).

M1: For writing the product of probabilities \(\frac{7}{n} \times \frac{6}{n-1}\). M1: For setting up the equation \(\frac{42}{n(n-1)} = \frac{7}{15}\). M1: For forming the quadratic equation \(n^2 - n - 90 = 0\) and solving to get \(n = 10\). A1: For 3.

1. Find the gradient of \(\mathbf{L}_1\): \(3x - 2y = 8 \implies y = \frac{3}{2}x - 4\), so gradient \(m_1 = \frac{3}{2}\). 2. Find the gradient of \(\mathbf{L}_2\): \(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\). 3. Find the equation of \(\mathbf{L}_2\) through P(6, -1): \(y - (-1) = -\frac{2}{3}(x - 6) \implies y = -\frac{2}{3}x + 3\). 4. Find the x-intercept by setting \(y = 0\): \(0 = -\frac{2}{3}x + 3 \implies \frac{2}{3}x = 3 \implies x = 4.5\). Thus, the coordinates of Q are (4.5, 0).

M1: For finding the gradient of \(\mathbf{L}_1\) is \(\frac{3}{2}\). M1: For finding the perpendicular gradient of \(\mathbf{L}_2\) is \(-\frac{2}{3}\). M1: For finding the equation of \(\mathbf{L}_2\) as \(y = -\frac{2}{3}x + 3\) or equivalent. A1: For (4.5, 0) or \(x = 4.5, y = 0\).

1. Set up the compound interest formula: \(5000 \times (1 + \frac{r}{100})^3 = 5384.45\). 2. Divide by 5000: \((1 + \frac{r}{100})^3 = 1.07689\). 3. Take the cube root of both sides: \(1 + \frac{r}{100} = \sqrt[3]{1.07689} \approx 1.025\). 4. Solve for \(r\): \(\frac{r}{100} = 0.025 \implies r = 2.5\).

M1: For the equation \(5000 \times (1 + \frac{r}{100})^3 = 5384.45\) or \(5000 \times M^3 = 5384.45\). M1: For \(M^3 = 1.07689\) or \(1 + \frac{r}{100} = 1.025\). M1: For solving to find \(r\). A1: For 2.5.

1. Express the total volume as the sum of the volumes of the hemisphere and cone: \(V_{\text{total}} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\). 2. Substitute \(h = 3r\): \(V_{\text{total}} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2(3r) = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\). 3. Set equal to \(180\pi\): \(\frac{5}{3}\pi r^3 = 180\pi \implies \frac{5}{3}r^3 = 180 \implies r^3 = 108\). 4. Calculate \(r\): \(r = \sqrt[3]{108} \approx 4.7622\). Rounding to 3 significant figures gives \(r \approx 4.76\).

M1: For expressing the volume of the cone with substituted height as \(\frac{1}{3}\pi r^2(3r)\) or \(\pi r^3\). M1: For combining the volumes to get \(\frac{5}{3}\pi r^3\). M1: For setting up the equation \(\frac{5}{3}\pi r^3 = 180\pi\) and solving for \(r^3\) to get 108. A1: For 4.76 (accept 4.76 to 4.763).

To simplify the expression, we first factorise each quadratic and linear expression:

1. Factorise the first numerator:
$$3x^2 - 10x + 8 = (3x - 4)(x - 2)$$

2. Factorise the first denominator:
$$x^2 - 16 = (x - 4)(x + 4)$$

3. Factorise the second numerator:
$$3x^2 + 5x - 12 = (3x - 4)(x + 3)$$

4. Factorise the second denominator:
$$2x^2 + 8x = 2x(x + 4)$$

Now substitute these factorised forms back into the division expression:
$$\frac{(3x - 4)(x - 2)}{(x - 4)(x + 4)} \div \frac{(3x - 4)(x + 3)}{2x(x + 4)}$$

To divide by a fraction, we multiply by its reciprocal:
$$\frac{(3x - 4)(x - 2)}{(x - 4)(x + 4)} \times \frac{2x(x + 4)}{(3x - 4)(x + 3)}$$

Next, cancel the common terms in the numerator and denominator:
- The terms $(3x - 4)$ cancel out.
- The terms $(x + 4)$ cancel out.

This leaves us with:
$$\frac{2x(x - 2)}{(x - 4)(x + 3)}$$

We can write this in its fully simplified form as:
$$\frac{2x(x-2)}{(x-4)(x+3)}\quad\text{or}\quad\frac{2x^2-4x}{x^2-x-12}$$

M1: Factorising $3x^2 - 10x + 8$ to $(3x - 4)(x - 2)$
M1: Factorising at least one of $x^2 - 16$ to $(x - 4)(x + 4)$, $3x^2 + 5x - 12$ to $(3x - 4)(x + 3)$, or $2x^2 + 8x$ to $2x(x + 4)$
M1: Multiplying by the reciprocal and showing the cancellation of common terms $(3x - 4)$ and $(x + 4)$
A1: Correct final simplified fraction $\frac{2x(x-2)}{(x-4)(x+3)}$ or $\frac{2x^2-4x}{x^2-x-12}$

To find the angle between the line $EC$ and the base $ABCD$, we first identify the projection of the line $EC$ on the horizontal base, which is the diagonal $AC$.

The angle we need to calculate is $\angle ECA$.

First, find the length of $AC$ in the right-angled triangle $ABC$ using Pythagoras' theorem:
$$AC^2 = AB^2 + BC^2$$
$$AC^2 = 12^2 + 5^2 = 144 + 25 = 169$$
$$AC = \sqrt{169} = 13\text{ cm}$$

Since the face $ADFE$ is vertical, the line $AE$ is perpendicular to the base $ABCD$. Therefore, triangle $EAC$ is a right-angled triangle at $A$.

We can find the angle $\theta = \angle ECA$ using the tangent ratio:
$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AE}{AC}$$
$$\tan(\theta) = \frac{8}{13}$$
$$\theta = \arctan\left(\frac{8}{13}\right) \approx 31.6075^\circ$$

Correct to 1 decimal place, the angle is $31.6^\circ$.

M1: Applying Pythagoras' theorem to find $AC$, e.g., $\sqrt{12^2 + 5^2}$
A1: Finding $AC = 13$
M1: Using $\tan(\angle ECA) = \frac{8}{13}$ or equivalent trigonometric setup
A1: Correct angle $31.6$ (accept answers in range $31.6$ to $31.61$)

First, we set up an equation to find the total number of sweets $n$.
The probability that the first sweet is lemon is $\frac{6}{n}$.
Since the first sweet is not replaced, the probability that the second sweet is lemon is $\frac{5}{n-1}$.

The probability that both sweets are lemon is:
$$\frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}$$

We are given that this probability is $\frac{1}{3}$:
$$\frac{30}{n(n-1)} = \frac{1}{3}$$
$$n(n-1) = 90$$
$$n^2 - n - 90 = 0$$

Factorise the quadratic equation:
$$(n - 10)(n + 9) = 0$$

Since the number of sweets must be positive, $n = 10$.

So there are 10 sweets in total:
- Lemon-flavoured: $6$
- Strawberry-flavoured: $10 - 6 = 4$

We need to find the probability that the two sweets have different flavours. This can happen in two ways: (Lemon, Strawberry) or (Strawberry, Lemon).

$$P(\text{Lemon, Strawberry}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$$
$$P(\text{Strawberry, Lemon}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$$

$$P(\text{Different}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$$

(Alternatively, $1 - P(\text{Both Lemon}) - P(\text{Both Strawberry}) = 1 - \frac{1}{3} - \left(\frac{4}{10} \times \frac{3}{9}\right) = 1 - \frac{1}{3} - \frac{12}{90} = 1 - \frac{1}{3} - \frac{2}{15} = \frac{8}{15}$)

M1: Formulating the probability equation $\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}$
M1: Rearranging to the quadratic $n^2 - n - 90 = 0$ and finding $n = 10$
M1: Calculating the probability of different flavours, e.g., $2 \times \left(\frac{6}{10} \times \frac{4}{9}\right)$ or $1 - \frac{30}{90} - \frac{12}{90}$
A1: Correct probability of $\frac{8}{15}$ or equivalent fraction / decimal $0.533$ (to 3 sf)

Let $\theta$ be the angle of the sector in degrees.

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

Substitute the given values into the formula:
$$48\pi = \frac{\theta}{360} \times \pi \times 12^2$$
$$48 = \frac{144\theta}{360}$$
$$144\theta = 48 \times 360 = 17280$$
$$\theta = 120^\circ$$

Now, the area of the segment is the area of the sector minus the area of the triangle $OAB$:
$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle } OAB$$

Calculate the area of the triangle $OAB$:
$$\text{Area of triangle } OAB = \frac{1}{2} \times r^2 \times \sin(\theta)$$
$$\text{Area} = \frac{1}{2} \times 12^2 \times \sin(120^\circ)$$
$$\text{Area} = 72 \times \frac{\sqrt{3}}{2} = 36\sqrt{3} \approx 62.3538\text{ cm}^2$$

Now find the area of the segment:
$$\text{Area of segment} = 48\pi - 36\sqrt{3} \approx 150.7964 - 62.3538 = 88.4426\text{ cm}^2$$

To 3 significant figures, the area is $88.4$ cm$^2$.

M1: Setting up an equation to find the sector angle: $48\pi = \frac{\theta}{360} \times \pi \times 12^2$ or equivalent
M1: Correctly determining $\theta = 120^\circ$ (or $\frac{2\pi}{3}$ radians)
M1: Using $\frac{1}{2} \times 12^2 \times \sin(120^\circ)$ to find the area of the triangle
A1: Correct area of the segment $88.4$ (accept answers in range $88.4$ to $88.5$)

The area of a right-angled triangle is given by:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

Substitute the given expressions and area into the equation:
$$18 = \frac{1}{2} (2x - 3)(x + 4)$$

Multiply both sides by 2 to clear the fraction:
$$36 = (2x - 3)(x + 4)$$

Expand the right side:
$$36 = 2x^2 + 8x - 3x - 12$$
$$36 = 2x^2 + 5x - 12$$

Rearrange into a standard quadratic equation format ($ax^2 + bx + c = 0$):
$$2x^2 + 5x - 48 = 0$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 2, b = 5, c = -48$:
$$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-48)}}{2(2)}$$
$$x = \frac{-5 \pm \sqrt{25 + 384}}{4}$$
$$x = \frac{-5 \pm \sqrt{409}}{4}$$

Since $x$ must be positive (lengths $2x - 3 > 0$ and $x + 4 > 0$ require $x > 1.5$):
$$x = \frac{-5 + \sqrt{409}}{4}$$
$$x \approx \frac{-5 + 20.2237}{4} = 3.8059$$

To 3 significant figures, $x = 3.81$.

M1: Setting up the equation $\frac{1}{2} (2x - 3)(x + 4) = 18$
M1: Expanding and rearranging to $2x^2 + 5x - 48 = 0$ (allow one sign or arithmetic error)
M1: Correct substitution of their coefficients into the quadratic formula
A1: Correct positive solution $3.81$ (accept answers in range $3.80$ to $3.81$)

Since the two vases are mathematically similar, we can find the scale factor of their dimensions.

First, find the ratio of their surface areas:
$$\frac{\text{Area of } A}{\text{Area of } B} = \frac{180}{320} = \frac{18}{32} = \frac{9}{16}$$

The ratio of their corresponding linear dimensions (lengths) is the square root of the ratio of their areas:
$$\text{Linear scale factor } k = \sqrt{\frac{9}{16}} = \frac{3}{4}$$

The ratio of their volumes is the cube of the linear scale factor:
$$\text{Volume scale factor } k^3 = \left(\frac{3}{4}\right)^3 = \frac{27}{64}$$

Now, calculate the volume of vase $A$ using this volume scale factor:
$$\text{Volume of } A = \text{Volume of } B \times \frac{27}{64}$$
$$\text{Volume of } A = 1280 \times \frac{27}{64}$$
$$\text{Volume of } A = 20 \times 27 = 540\text{ cm}^3$$

M1: Finding the ratio of surface areas $\frac{180}{320}$ (or equivalent)
M1: Finding the linear scale factor by taking the square root: $\sqrt{\frac{180}{320}} = \frac{3}{4}$ (or $0.75$)
M1: Finding the volume scale factor by cubing the linear scale factor: $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ (or $0.421875$)
A1: Correct final volume of $540$