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To factorise completely, find the highest common factor (HCF) of the two terms.

The HCF of 15 and 20 is 5.
The HCF of \(x^2\) and \(x\) is \(x\).
The HCF of \(y\) and \(y^2\) is \(y\).

Therefore, the overall HCF is \(5xy\).

Divide each term by the HCF:
\(15x^2y \div 5xy = 3x\)
\(-20xy^2 \div 5xy = -4y\)

Combine these to get the factorised expression:
\(5xy(3x - 4y)\)

M1 for finding any common factor (e.g., \(5\), \(x\), \(y\), \(5x\), \(5y\), or \(xy\)) outside the bracket, e.g., \(5(3x^2y - 4xy^2)\) or \(xy(15x - 20y)\).
A1 for the fully correct factorised expression: \(5xy(3x - 4y)\).

A closed cuboid has 6 faces in total, consisting of 3 pairs of identical faces.

The area of the bottom and top faces is:
\(2 \times (8 \times 5) = 2 \times 40 = 80\text{ cm}^2\)

The area of the front and back faces is:
\(2 \times (8 \times 3) = 2 \times 24 = 48\text{ cm}^2\)

The area of the left and right faces is:
\(2 \times (5 \times 3) = 2 \times 15 = 30\text{ cm}^2\)

Adding these together:
Total Surface Area = \(80 + 48 + 30 = 158\text{ cm}^2\).

M1 for a correct sum of the areas of at least 3 different faces (e.g. \(8 \times 5 + 8 \times 3 + 5 \times 3\) or \(40 + 24 + 15\)).
A1 for 158.

First, expand the bracket on the left side:
\(8x - 12 = 5x + 9\)

Next, subtract \(5x\) from both sides to group the \(x\) terms together:
\(3x - 12 = 9\)

Then, add 12 to both sides to isolate the term with \(x\):
\(3x = 21\)

Finally, divide both sides by 3:
\(x = 7\)

M1 for correct expansion of the bracket to \(8x - 12\) or for a correct step in rearranging terms (e.g., \(8x - 5x = 9 + 12\)).
A1 for 7.

First, find the differences between consecutive terms:
\(7 - 3 = 4\)
\(11 - 7 = 4\)
\(15 - 11 = 4\)

Since the difference is constant, this is an arithmetic sequence with a common difference of 4.

The general formula for the \(n\)th term of an arithmetic sequence is \(dn + c\), where \(d\) is the common difference.
Here, \(d = 4\), so the formula is \(4n + c\).

Using the first term where \(n = 1\):
\(4(1) + c = 3\)
\(4 + c = 3\)
\(c = -1\)

Therefore, the \(n\)th term of the sequence is \(4n - 1\).

M1 for any expression of the form \(4n + k\), where \(k\) is any integer constant, or for identifying the common difference of 4.
A1 for \(4n - 1\).

Using Pythagoras' theorem, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.

Let the third side be \(b\).
\(5^2 + b^2 = 13^2\)
\(25 + b^2 = 169\)

Subtract 25 from both sides:
\(b^2 = 144\)

Take the square root of both sides:
\(b = \sqrt{144} = 12\text{ cm}\).

M1 for a correct application of Pythagoras' theorem to find the shorter side, e.g., \(\sqrt{13^2 - 5^2}\) or \(13^2 = 5^2 + x^2\).
A1 for 12.

To convert Euros to Dollars, multiply the amount in Euros by the exchange rate:
\(450 \times 1.16 = 522\).

Therefore, Maya receives $522.

M1 for the calculation \(450 \times 1.16\).
A1 for 522.

The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius.
First, calculate the radius of the circle:
\(r = \frac{12}{2} = 6\text{ cm}\).

Now, calculate the area:
\(A = \pi \times 6^2 = 36\pi \approx 113.0973\text{ cm}^2\).

Rounding to 1 decimal place gives \(113.1\text{ cm}^2\).

M1 for \(\pi \times 6^2\) or \(\pi \times (12/2)^2\).
A1 for 113.1 (accept answers in the range 113.09 to 113.11).

The probability of rolling an even number (2, 4, or 6) on a six-sided die is:
\(P(\text{even}) = \frac{3}{6} = \frac{1}{2}\).

The probability of tossing a Head on a coin is:
\(P(\text{Head}) = \frac{1}{2}\).

Since rolling a die and tossing a coin are independent events, multiply the probabilities:
\(P(\text{even and Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).

M1 for \(\frac{3}{6} \times \frac{1}{2}\) or for listing all 12 possible outcomes and identifying the 3 successful outcomes (e.g. (2,H), (4,H), (6,H)).
A1 for \(1/4\).

Expand the brackets first:\
\(6(2a - 3b) = 12a - 18b\)\
\(-4(a - 5b) = -4a + 20b\)\
\
Combine like terms:\
\(12a - 18b - 4a + 20b = (12a - 4a) + (-18b + 20b) = 8a + 2b\)

M1 for correct expansion of at least one bracket (e.g. \(12a - 18b\) or \(-4a + 20b\))\
A1 for \(8a + 2b\) or \(2(4a + b)\)

First, convert the cost in the USA from dollars to pounds:\
\(\$351 \div 1.30 = £270\)\
\
Next, find the difference in cost between the USA and the UK:\
\(£270 - £240 = £30\)

M1 for converting USA cost to pounds: \(351 \div 1.30\) (or converting UK cost to dollars: \(240 \times 1.30 = 312\))\
A1 for 30

Let the height of the cuboid be \(h\text{ cm}\).\
The formula for the total surface area of a cuboid is:\
\(\text{Total Surface Area} = 2(lw + lh + wh)\)\
\
Substitute the given values into the formula:\
\(2(8 \times 5 + 8 \times h + 5 \times h) = 236\)\
\(2(40 + 13h) = 236\)\
\
Divide both sides by 2:\
\(40 + 13h = 118\)\
\
Subtract 40 from both sides:\
\(13h = 78\)\
\
Divide by 13:\
\(h = 6\)

M1 for setting up the equation \(2(8 \times 5 + 8h + 5h) = 236\) or \(40 + 13h = 118\)\
A1 for 6

Let \(\theta\) be the angle between the ladder and the ground.\
The ladder is the hypotenuse, and the ground distance is the adjacent side to the angle \(\theta\).\
Using the cosine ratio:\
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5}\)\
\
Find \(\theta\):\
\(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right) \approx 67.38^\circ\)\
\
Rounding to 1 decimal place gives \(67.4^\circ\).

M1 for \(\cos(\theta) = \frac{2.5}{6.5}\) or \(\cos^{-1}\left(\frac{2.5}{6.5}\right)\)\
A1 for 67.4 (accept 67.38 or answers in range [67.3, 67.4])

Multiply the first equation by 5 and the second equation by 2 to align the coefficients of \(y\):\
\(15x - 10y = 95\)\
\(4x + 10y = 0\)\
\
Add the two equations:\
\(19x = 95\)\
\(x = 5\)\
\
Substitute \(x = 5\) into the second equation:\
\(2(5) + 5y = 0\)\
\(10 + 5y = 0\)\
\(5y = -10\)\
\(y = -2\)

M1 for correctly multiplying equations to eliminate one variable or substituting \(x = -2.5y\)\
A1 for both correct values: \(x = 5\) and \(y = -2\)

The curve crosses the \(x\)-axis when \(y = 0\):\
\(x^2 - 4x - 5 = 0\)\
\
Factorise the quadratic expression:\
\((x - 5)(x + 1) = 0\)\
\
So, \(x = 5\) or \(x = -1\).\
\
The coordinates of the points are \((5, 0)\) and \((-1, 0)\).

M1 for setting \(x^2 - 4x - 5 = 0\) and attempting to factorise or use quadratic formula\
A1 for \((5, 0)\) and \((-1, 0)\) (or equivalent list of coordinates, in either order)

The terms of the sequence increase by a constant difference:\
\(11 - 5 = 6\)\
\(17 - 11 = 6\)\
\(23 - 17 = 6\)\
\
Since the common difference is 6, the formula is of the form \(6n + c\).\
For the first term (\(n = 1\)):\
\(6(1) + c = 5 \implies 6 + c = 5 \implies c = -1\)\
\
Therefore, the \(n\)th term is \(6n - 1\).

M1 for identifying the common difference is 6 (e.g. finding \(6n\) as part of their expression) or trying to use \(a + (n-1)d\)\
A1 for \(6n - 1\)

The perimeter of a sector consists of the arc length plus two radii:\
\
\\text{Perimeter} = 2r + \\text{Arc length}\
\
Calculate the arc length:\
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2\pi(9) = \frac{1}{3} \times 18\pi = 6\pi\text{ cm}\)\
\
Add the two radii:\
\(2r = 2 \times 9 = 18\text{ cm}\)\
\
Total perimeter:\
\(18 + 6\pi\text{ cm}\)

M1 for calculating the arc length: \(\frac{120}{360} \times 2\pi \times 9\) (or \(6\pi\))\
A1 for \(18 + 6\pi\) (or \(6\pi + 18\))

First, subtract the flat commission fee from the initial amount: $480 - $15 = $465. Next, convert this remaining amount to Euros using the exchange rate: \(465 \times 0.88 = 409.2\). Therefore, Clara receives 409.20 EUR.

M1 for subtracting the commission fee correctly: 480 - 15, or for setting up the multiplication: \(x \times 0.88\). A1 for 409.20 or 409.2.

First, expand both brackets: \(4(3x - 2) = 12x - 8\) and \(-3(x + 5) = -3x - 15\). Next, combine the terms: \(12x - 3x - 8 - 15 = 9x - 23\).

M1 for a correct expansion of at least one bracket (e.g., \(12x - 8\) or \(-3x - 15\)). A1 for the fully simplified expression \(9x - 23\).

In right-angled triangle \(ABC\), using the sine ratio: \(\sin(BAC) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}\). Substituting the given values: \(\sin(28^\circ) = \frac{BC}{14}\). Multiplying both sides by 14: \(BC = 14 \times \sin(28^\circ) \approx 6.5727\text{ cm}\). To 3 significant figures, this is 6.57.

M1 for \(14 \times \sin(28^\circ)\) or \(\sin(28^\circ) = \frac{BC}{14}\). A1 for 6.57 (accept answers in the range 6.57 to 6.573).

Multiply both sides of the equation by 2 to eliminate the fraction: \(3x - 4 = 20\). Add 4 to both sides: \(3x = 24\). Divide both sides by 3 to find \(x\): \(x = 8\).

M1 for multiplying by 2 to get \(3x - 4 = 20\) (or for another correct first step). A1 for \(x = 8\).

The sequence is arithmetic because the difference between consecutive terms is constant: \(5 - 1 = 4\). The formula for the \(n\)-th term of an arithmetic sequence is \(dn + c\), where \(d\) is the common difference. Here, \(d = 4\), so the term is \(4n + c\). Since the first term is 1, when \(n = 1\), we have \(4(1) + c = 1 \implies c = -3\). Therefore, the \(n\)-th term is \(4n - 3\).

M1 for finding the common difference is 4, or for an expression of the form \(4n + k\) (where \(k\) is any constant). A1 for \(4n - 3\).

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

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

First, find the total number of parts in the ratio: \(3 + 5 + 4 = 12\). Next, find the value of one part: \(\frac{240}{12} = 20\). The largest share corresponds to the largest number in the ratio, which is 5. Therefore, the largest share is \(5 \times 20 = 100\).

M1 for dividing 240 by the sum of ratio parts: \(240 / (3+5+4)\). A1 for 100.

To write \(0.000307\) in standard form, move the decimal point 4 places to the right to obtain a number between 1 and 10, which is \(3.07\). Because the original number is less than 1, the exponent is negative: \(3.07 \times 10^{-4}\).

B1 for \(3.07\) or for \(10^{-4}\) seen in the final answer. B1 for the fully correct standard form \(3.07 \times 10^{-4}\).

Expand the bracket on the left-hand side: \(8x - 12 = 3x + 13\). Rearrange the terms to collect those with \(x\) on one side and the constant terms on the other side: \(8x - 3x = 13 + 12\), which simplifies to \(5x = 25\). Divide both sides by \(5\) to find \(x\): \(x = 5\).

M1 for correct expansion of the bracket: \(8x - 12\) (or equivalent progress, e.g. \(5x = 25\) after an expansion error). A1 for \(5\).

First, factorise the numerator:
\(3x^2 - 14x + 8 = 3x^2 - 12x - 2x + 8 = 3x(x - 4) - 2(x - 4) = (3x - 2)(x - 4)\).

Next, factorise the denominator using the difference of two squares:
\(x^2 - 16 = (x - 4)(x + 4)\).

Now, divide the numerator by the denominator and cancel out the common factor \((x - 4)\):
\(\frac{(3x - 2)(x - 4)}{(x - 4)(x + 4)} = \frac{3x - 2}{x + 4}\).

M1 for factorising the numerator correctly: \((3x - 2)(x - 4)\)
M1 for factorising the denominator correctly: \((x - 4)(x + 4)\)
A1 for the final simplified fraction: \(\frac{3x - 2}{x + 4}\)

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

Substitute the given values into the formula:
\(\text{Arc length} = \frac{135}{360} \times 2 \times \pi \times 7.2\)
\(\text{Arc length} = \frac{3}{8} \times 14.4\pi = 5.4\pi \approx 16.965\text{ cm}\)

The perimeter of the sector is the sum of the arc length and two radii:
\(\text{Perimeter} = \text{Arc length} + 2r\)
\(\text{Perimeter} = 16.965 + 2 \times 7.2 = 16.965 + 14.4 = 31.365\text{ cm}\)

Rounding to 1 decimal place gives \(31.4\text{ cm}\).

M1 for \(\frac{135}{360} \times 2 \times \pi \times 7.2\) (or equivalent arc length calculation)
M1 for adding \(2 \times 7.2\) to their arc length
A1 for \(31.4\)

Let's find the first and second differences:
First differences: \(7, 11, 15, 19\)
Second differences: \(4, 4, 4\)

Since the second differences are constant, the sequence is quadratic and has the general form \(an^2 + bn + c\).
The coefficient \(a\) is half of the second difference:
\(a = \frac{4}{2} = 2\)

Now subtract \(2n^2\) from each term in the sequence:
For \(n=1\): \(2 - 2(1)^2 = 0\)
For \(n=2\): \(9 - 2(2)^2 = 1\)
For \(n=3\): \(20 - 2(3)^2 = 2\)
For \(n=4\): \(35 - 2(4)^2 = 3\)

The remaining sequence is \(0, 1, 2, 3, \dots\), which has the linear formula \(n - 1\).

Combining the quadratic and linear parts, the \(n\)th term is \(2n^2 + n - 1\).

M1 for finding second difference is 4 (or establishing coefficient of \(n^2\) is 2)
M1 for subtracting \(2n^2\) from the terms and attempting to find the linear component (e.g. \(n-1\) or solving equations)
A1 for \(2n^2 + n - 1\)

Multiply all terms in the inequality by the lowest common multiple of 3 and 4, which is 12, to clear the fractions:
\(4(2x - 5) - 3(x + 2) > 12\)

Expand the brackets:
\(8x - 20 - 3x - 6 > 12\)

Simplify the left side by grouping like terms:
\(5x - 26 > 12\)

Add 26 to both sides:
\(5x > 38\)

Divide by 5:
\(x > 7.6\) (or \(x > \frac{38}{5}\))

M1 for multiplying by 12 to clear fractions correctly: \(4(2x - 5) - 3(x + 2) > 12\)
M1 for simplifying terms to get \(5x > k\) where \(k\) is a constant (e.g., \(5x > 38\))
A1 for \(x > 7.6\) or \(x > \frac{38}{5}\)

Let \(\theta\) be the angle between the ladder and the horizontal ground.
The side adjacent to the angle is the distance from the ladder to the wall: \(\text{Adjacent} = 2.5\text{ m}\).
The hypotenuse is the length of the ladder: \(\text{Hypotenuse} = 6.5\text{ m}\).

Using the cosine ratio:
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
\(\cos(\theta) = \frac{2.5}{6.5}\)

Find \(\theta\):
\(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right) \approx 67.380^\circ\)

Rounding to 1 decimal place, \(\theta = 67.4^\circ\).

M1 for writing the correct trigonometric ratio: \(\cos(\theta) = \frac{2.5}{6.5}\) (or equivalent)
M1 for evaluating \(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right)\) to obtain \(67.38...\)
A1 for \(67.4\)

Calculate the total amount in the account after 5 years using the compound interest formula:
\(A = P \left(1 + \frac{r}{100}\right)^t\)
\(A = 4500 \times (1 + 0.024)^5 = 4500 \times 1.024^5\)
\(A \approx 4500 \times 1.1258999 = 5066.5496\)

Calculate the interest earned by subtracting the principal from the total amount:
\(\text{Interest} = A - P = 5066.5496 - 4500 = 566.5496\)

To the nearest cent, the interest earned is \(\$566.55\).

M1 for compound interest calculation: \(4500 \times 1.024^5\)
M1 for subtracting the principal \(4500\) from their calculated total amount
A1 for \(566.55\)

First, find the perpendicular height, \(h\), of the cone using Pythagoras' theorem:
\(h^2 + r^2 = l^2\) where \(l\) is the slant height.
\(h^2 + 4.2^2 = 8.5^2\)
\(h^2 = 8.5^2 - 4.2^2\)
\(h^2 = 72.25 - 17.64 = 54.61\)
\(h = \sqrt{54.61} \approx 7.38986\text{ cm}\)

Now, calculate the volume \(V\):
\(V = \frac{1}{3} \pi r^2 h\)
\(V = \frac{1}{3} \times \pi \times 4.2^2 \times 7.38986\)
\(V \approx \frac{1}{3} \times \pi \times 17.64 \times 7.38986 \approx 136.515\text{ cm}^3\)

Rounding to 3 significant figures gives \(137\text{ cm}^3\).

M1 for \(h = \sqrt{8.5^2 - 4.2^2}\) (or finding \(h \approx 7.39\))
M1 for substituting \(r=4.2\) and their \(h\) into \(\frac{1}{3} \pi r^2 h\)
A1 for \(137\) (accept answers in the range \([136.5, 137.0]\))

The total number of balls in the bag is \(5 + 7 = 12\).
We want to find the probability of choosing either two red balls or two blue balls.

Calculate the probability of choosing two red balls:
\(P(\text{Red, Red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)

Calculate the probability of choosing two blue balls:
\(P(\text{Blue, Blue}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\)

Since these events are mutually exclusive, add the probabilities together:
\(P(\text{same colour}) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132}\)

Simplifying the fraction:
\(P(\text{same colour}) = \frac{31}{66}\) (or approximately \(0.470\))

M1 for \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{7}{12} \times \frac{6}{11}\)
M1 for \(\left(\frac{5}{12} \times \frac{4}{11}\right) + \left(\frac{7}{12} \times \frac{6}{11}\right)\)
A1 for \(\frac{31}{66}\) (or \(0.47\) or \(0.470\))

We can solve the simultaneous equations by equating coefficients.

Multiply the first equation by 2:
\(8x - 6y = 35\)

Multiply the second equation by 3:
\(9x + 6y = 7.5\)

Add the two resulting equations to eliminate \(y\):
\((8x - 6y) + (9x + 6y) = 35 + 7.5\)
\(17x = 42.5\)
\(x = 2.5\)

Substitute \(x = 2.5\) into the second equation:
\(3(2.5) + 2y = 2.5\)
\(7.5 + 2y = 2.5\)
\(2y = -5\)
\(y = -2.5\)

M1 for a valid method to eliminate one variable (e.g., multiplying equations to equate coefficients)
A1 for x = 2.5
A1 for y = -2.5 (or equivalent fractions)

Let the angle of the sector at the center be \(\theta\) degrees.

The arc length of a sector is given by:
\(L = \frac{\theta}{360} \times 2\pi r\)

Substitute the given values into the formula:
\(15 = \frac{\theta}{360} \times 2\pi (12)\)
\(15 = \frac{\theta}{360} \times 24\pi\)
\(\frac{\theta}{360} = \frac{15}{24\pi}\)

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

Substitute the expression for \(\frac{\theta}{360}\):
\(A = \frac{15}{24\pi} \times \pi (12)^2\)
\(A = \frac{15}{24} \times 144\)
\(A = 15 \times 6 = 90\text{ cm}^2\)

M1 for a correct expression linking the sector angle to the arc length, e.g., \(\frac{\theta}{360} \times 2\pi (12) = 15\) or using radians \(\theta = 1.2\)
M1 for substituting their angle expression into the area of a sector formula
A1 for 90

First, calculate the volume of the sphere using \(V = \frac{4}{3}\pi r^3\):
\(V_{\text{sphere}} = \frac{4}{3} \times \pi \times 3^3 = 36\pi\text{ cm}^3\)

The volume of a cone is given by \(V = \frac{1}{3}\pi R^2 h\).
Since the volume is conserved:
\(\frac{1}{3}\pi R^2 (3) = 36\pi\)
\(\pi R^2 = 36\pi\)
\(R^2 = 36\)
\(R = 6\text{ cm}\)

M1 for equating the volume of the sphere and the cone: \(\frac{4}{3}\pi (3)^3 = \frac{1}{3}\pi R^2 (3)\)
M1 for simplifying to \(R^2 = 36\)
A1 for 6

Substitute \(n = 2\) and \(n = 4\) into the formula for \(T_n\):

For \(n = 2\):
\(a(2^2) + b(2) - 3 = 13 \implies 4a + 2b - 3 = 13 \implies 4a + 2b = 16 \implies 2a + b = 8\) (Equation 1)

For \(n = 4\):
\(a(4^2) + b(4) - 3 = 53 \implies 16a + 4b - 3 = 53 \implies 16a + 4b = 56 \implies 4a + b = 14\) (Equation 2)

Subtract Equation 1 from Equation 2:
\((4a + b) - (2a + b) = 14 - 8\)
\(2a = 6 \implies a = 3\)

Substitute \(a = 3\) back into Equation 1:
\(2(3) + b = 8 \implies 6 + b = 8 \implies b = 2\)

M1 for creating two correct simultaneous equations in terms of a and b (e.g. \(4a + 2b = 16\) and \(16a + 4b = 56\))
A1 for a = 3
A1 for b = 2

To get different colours, the two possible mutually exclusive outcomes are:
1) Drawing a Red marble first, then a Blue marble.
2) Drawing a Blue marble first, then a Red marble.

Calculate the probability of each outcome:
\(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
\(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Sum these probabilities to find the total probability:
\(P(\text{Different}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)

M1 for calculating one correct product of probabilities (e.g., \(\frac{5}{8} \times \frac{3}{7}\))
M1 for summing the two correct probabilities (\(\frac{15}{56} + \frac{15}{56}\))
A1 for \(\frac{15}{28}\) (or equivalent decimals such as 0.536)

To solve the compound inequality, perform the same operations on all three parts:

First, add 3 to all parts:
\(-5 + 3 < 2x \le 7 + 3\)
\(-2 < 2x \le 10\)

Next, divide all parts by 2:
\(-1 < x \le 5\)

M1 for adding 3 to all parts of the inequality (e.g., to get \(-2 < 2x \le 10\))
A1 for x > -1
A1 for \(x \le 5\) (or A2 for the full statement \(-1 < x \le 5\))

Let the width of the field be \(w\text{ m}\).

The rectangular field, length, and diagonal path form a right-angled triangle. By Pythagoras' theorem:
\[w^2 + 24^2 = 25^2\]
\[w^2 + 576 = 625\]
\[w^2 = 625 - 576\]
\[w^2 = 49\]
\[w = \sqrt{49} = 7\text{ m}\]

M1 for applying Pythagoras' theorem: \(w^2 + 24^2 = 25^2\)
M1 for \(w^2 = 625 - 576\) or \(w = \sqrt{49}\)
A1 for 7

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

A reduction of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price.

Therefore:
\[0.85 \times P = 578\]
\[P = \frac{578}{0.85}\]
\[P = 680\]

The original price of the laptop was \(\$680\).

M1 for recognizing that \(\$578\) represents \(85\%\) of the original price, i.e., \(0.85P = 578\) or \(\frac{578}{85} \times 100\)
A1 for partial division or calculation step (e.g., \(578 \div 85 = 6.8\))
A1 for 680

Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator: \(4x^2 - 1 = (2x + 1)(2x - 1)\). Divide both by the common factor \((2x + 1)\) to get \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\). M1 for factorising the denominator to \((2x + 1)(2x - 1)\). A1 for the correct simplified fraction.

The radius of the hemisphere is \(3\) cm, so its height is also \(3\) cm. The height of the cone is the total height minus the radius of the hemisphere: \(11 - 3 = 8\) cm. The volume of the hemisphere is \(\frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = 18\pi\) cm\(^3\). The volume of the cone is \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (8) = 24\pi\) cm\(^3\). The total volume is the sum of these two volumes: \(18\pi + 24\pi = 42\pi\) cm\(^3\).

M1 for finding the height of the cone is \(8\) cm. M1 for setting up the volume of the hemisphere or cone, e.g., \(\frac{2}{3} \pi (3)^3\) or \(\frac{1}{3} \pi (3)^2 (8)\). A1 for \(42\pi\).

In the right-angled triangle \(ABD\), we can find angle \(ADB\) using trigonometry: \(\tan(\angle ADB) = \frac{AB}{BD} = \frac{7.2}{4.5} = 1.6\). This gives \(\angle ADB = \arctan(1.6) \approx 58.0^\circ\). Since \(D\) lies on the line segment \(BC\), the angle \(ADC\) and angle \(ADB\) lie on a straight line and sum to \(180^\circ\). Thus, \(\angle ADC = 180^\circ - 58.0^\circ = 122.0^\circ\) (to 1 decimal place).

M1 for \(\tan(\angle ADB) = \frac{7.2}{4.5}\) or \(\angle ADB \approx 58.0^\circ\). M1 for subtracting their angle from \(180^\circ\). A1 for \(122\) or \(122.0\).

Substitute the expression for \(y\) into the second equation: \(x^2 + (2x - 3)^2 = 41\). Expand the bracket: \(x^2 + 4x^2 - 12x + 9 = 41\). Simplify to get a quadratic equation: \(5x^2 - 12x - 32 = 0\). Factorising gives \((5x + 8)(x - 4) = 0\), which yields \(x = 4\) or \(x = -1.6\). If \(x = 4\), then \(y = 2(4) - 3 = 5\). If \(x = -1.6\), then \(y = 2(-1.6) - 3 = -6.2\).

M1 for setting up the quadratic equation \(5x^2 - 12x - 32 = 0\). M1 for solving the quadratic equation to find two values of \(x\) (or \(y\)). A1 for both pairs of solutions: \((4, 5)\) and \((-1.6, -6.2)\).

First, find the derivative of the curve: \(\frac{dy}{dx} = 3x^2 - 12\). Set the derivative to \(0\) to find the stationary points: \(3x^2 - 12 = 0 \implies x^2 = 4 \implies x = 2\) or \(x = -2\). Substitute these values back into the original equation to find the corresponding \(y\)-values: For \(x = -2\): \(y = (-2)^3 - 12(-2) + 5 = -8 + 24 + 5 = 21\). For \(x = 2\): \(y = (2)^3 - 12(2) + 5 = 8 - 24 + 5 = -11\). The stationary point in the second quadrant is \((-2, 21)\).

M1 for differentiating to find \(\frac{dy}{dx} = 3x^2 - 12\). M1 for setting their derivative to 0 and solving for \(x = -2\). A1 for \((-2, 21)\).

Find the differences between terms: First differences: \(8, 12, 16, 20\). Second differences: \(4, 4, 4\). Since the second differences are constant, the sequence is quadratic and has the form \(an^2 + bn + c\), where \(a = \frac{4}{2} = 2\). Subtracting \(2n^2\) from each term of the sequence: For \(n=1\): \(3 - 2(1)^2 = 1\). For \(n=2\): \(11 - 2(2)^2 = 3\). For \(n=3\): \(23 - 2(3)^2 = 5\). For \(n=4\): \(39 - 2(4)^2 = 7\). The resulting sequence is \(1, 3, 5, 7, \dots\), which has an \(n\)-th term of \(2n - 1\). Therefore, the overall \(n\)-th term is \(2n^2 + 2n - 1\).

M1 for finding second difference of 4, or \(a = 2\). M1 for finding the linear sequence part \(2n - 1\). A1 for \(2n^2 + 2n - 1\).

Let \(n\) be the number of years. The investment value is given by \(4500 \times (1.032)^n\). We need this value to exceed \(\$6000\): \(4500 \times (1.032)^n > 6000 \implies 1.032^n > \frac{4}{3} \approx 1.3333\). Testing values of \(n\): For \(n = 9\): \(4500 \times 1.032^9 \approx 5968.04\). For \(n = 10\): \(4500 \times 1.032^{10} \approx 6159.02\). Thus, it will take \(10\) complete years for the investment to exceed \(\$6000\).

M1 for setting up the inequality or equation \(4500 \times 1.032^n > 6000\). M1 for evaluating for \(n=9\) or \(n=10\). A1 for \(10\).

Use the Cosine Rule: \(QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos(\angle QPR)\). Substitute the given values: \(13^2 = 8^2 + 7^2 - 2 \cdot 8 \cdot 7 \cdot \cos(\angle QPR)\). Simplify: \(169 = 64 + 49 - 112 \cos(\angle QPR) \implies 169 = 113 - 112 \cos(\angle QPR)\). Subtract 113: \(56 = -112 \cos(\angle QPR) \implies \cos(\angle QPR) = -0.5\). Thus, \(\angle QPR = \arccos(-0.5) = 120^\circ\).

M1 for substituting correctly into the cosine rule: \(13^2 = 8^2 + 7^2 - 2 \times 8 \times 7 \times \cos(P)\). M1 for simplifying to \(\text{cos}(P) = -0.5\). A1 for \(120\).

(a) Total cost = \$42.00 + \$18.00 = \$60.00. Cost price per mug = \$60.00 \\div 15 = \$4.00. (b) Profit per mug = Selling price - Cost price = \$7.50 - \$4.00 = \$3.50. Percentage profit = (\$3.50 / \$4.00) * 100 = 87.5%. (c) Simple Interest = P * R * T / 100 = 1400 * 3.5 * 4 / 100 = 196. Total value = 1400 + 196 = \$1596.

(a) [3 marks total]: M1 for adding 42.00 + 18.00 (= 60.00), M1 for dividing their total by 15, A1 for 4.00. (b) [4 marks total]: M1 for finding profit per mug: 7.50 - 4.00 (= 3.50), M1 for dividing profit by cost price (3.50 / 4.00), M1 for multiplying by 100, A1 for 87.5. (c) [4.56 marks total]: M1 for interest formula setup: 1400 * 3.5 * 4 / 100, A1 for finding interest = 196, M1 for adding interest to 1400, A1.56 for 1596.

(a) Volume of cylinder = \\pi * r^2 * h = \\pi * 6^2 * 15 = 540\\pi cm^3. (b) Total surface area = 2\\pi r^2 + 2\\pi r h = 2\\pi(6)^2 + 2\\pi(6)(15) = 72\\pi + 180\\pi = 252\\pi \\approx 791.68 cm^2. Correct to 1 d.p. is 791.7 cm^2. (c) Volume of one sphere = 540\\pi / 5 = 108\\pi cm^3. Using volume of sphere formula: (4/3)\\pi r^3 = 108\\pi \\implies (4/3)r^3 = 108 \\implies r^3 = 81 \\implies r = \\sqrt[3]{81} \\approx 4.3267 cm. Correct to 2 d.p. is 4.33 cm.

(a) [3 marks total]: M1 for cylinder volume formula, M1 for substituting 6 and 15, A1 for 540\\pi. (b) [4 marks total]: M1 for surface area formula, M1 for substituting values, A1 for 252\\pi or 791.68..., A1 for 791.7. (c) [4.56 marks total]: M1 for dividing volume by 5: 108\\pi, M1 for setting up equation: (4/3)\\pi r^3 = 108\\pi, M1 for finding r^3 = 81, A1.56 for 4.33.

(a) Using Pythagoras' theorem: h^2 + 1.8^2 = 4.5^2 \\implies h^2 = 20.25 - 3.24 = 17.01 \\implies h = \\sqrt{17.01} \\approx 4.1243 m. To 3 s.f., height is 4.12 m. (b) cos(\\theta) = 1.8 / 4.5 = 0.4 \\implies \\theta = cos^{-1}(0.4) \\approx 66.4218 degrees. To 1 d.p., \\theta = 66.4 degrees. (c) d^2 + 4.2^2 = 4.5^2 \\implies d^2 = 20.25 - 17.64 = 2.61 \\implies d = \\sqrt{2.61} \\approx 1.6155 m. To 3 s.f., distance is 1.62 m.

(a) [3.56 marks total]: M1 for Pythagoras setup 4.5^2 - 1.8^2, A1 for 17.01, M1 for square-rooting, A0.56 for 4.12. (b) [4 marks total]: M1 for cos ratio, M1 for cos(\\theta) = 1.8/4.5, M1 for cos^{-1}(0.4), A1 for 66.4. (c) [4 marks total]: M1 for Pythagoras setup d^2 = 4.5^2 - 4.2^2, M1 for d^2 = 2.61, M1 for \\sqrt{2.61}, A1 for 1.62.

(a) 3(2x - 5) - 4(x - 2) = 6x - 15 - 4x + 8 = 2x - 7. (b) Find HCF of 18x^2 y and 12xy^2, which is 6xy. Factorising gives: 6xy(3x - 2y). (c) Factorise numerator: x^2 - 9 = (x - 3)(x + 3). Factorise denominator: 5x + 15 = 5(x + 3). Cancel common factor (x + 3): (x - 3)/5.

(a) [3 marks total]: M1 for 6x - 15, M1 for -4x + 8, A1 for 2x - 7. (b) [3.56 marks total]: M1 for finding any common factor like 2, 3, x or y, M1.56 for full common factor 6xy, A1 for 6xy(3x - 2y). (c) [5 marks total]: M1 for (x-3)(x+3), M1 for 5(x+3), M2 for cancelling (x+3), A1 for (x-3)/5.

(a) Multiply first equation by 2 and second by 3: 8x + 6y = 50, 9x - 6y = 18. Add equations: 17x = 68 \\implies x = 4. Substitute x = 4 into second equation: 3(4) - 2y = 6 \\implies 12 - 2y = 6 \\implies 2y = 6 \\implies y = 3. (b) Multiply both sides by 3: 5x - 2 = 3(x + 4) \\implies 5x - 2 = 3x + 12 \\implies 2x = 14 \\implies x = 7.

(a) [5.56 marks total]: M1 for attempt to eliminate variable, M1 for 17x = 68, A1 for x = 4, M1 for substituting x back, A1.56 for y = 3. (b) [6 marks total]: M2 for 5x - 2 = 3(x + 4), M1 for 3x + 12, M2 for isolating terms to 2x = 14, A1 for x = 7.

(a) If x = -1, y = (-1)^2 - 3(-1) - 4 = 0, so p = 0. If x = 1, y = 1^2 - 3(1) - 4 = -6, so q = -6. If x = 3, y = 3^2 - 3(3) - 4 = -4, so r = -4. (b) The line of symmetry is halfway between x = 1 and x = 2 (where y = -6), which is x = 1.5. (c) Factorise: (x - 4)(x + 1) = 0 \\implies x = 4 or x = -1.

(a) [3 marks total]: B1 for p = 0, B1 for q = -6, B1 for r = -4. (b) [4 marks total]: M2 for identifying midpoint or using formula -b/2a, A2 for x = 1.5. (c) [4.56 marks total]: M2 for factorising to (x-4)(x+1), A1 for x = 4, A1.56 for x = -1.

(a) Common difference is 6. Next term is 25 + 6 = 31. (b) First term is 7, common difference is 6. n-th term = 6n + (7 - 6) = 6n + 1. (c) 50th term = 6(50) + 1 = 301. (d) Set 6n + 1 = 425 \\implies 6n = 424 \\implies n = 70.67. Since n is not a whole number, 425 is not in the sequence.

(a) [2 marks total]: B2 for 31. (b) [4 marks total]: M2 for finding common difference 6, A2 for 6n + 1. (c) [2 marks total]: M1 for substituting 50, A1 for 301. (d) [3.56 marks total]: M1 for 6n + 1 = 425, A1 for n = 70.67, A1.56 for explanation that n must be an integer.

(a) Area = 0.5 * AB * BC * sin(ABC) = 0.5 * 8.5 * 12.4 * sin(58) \\approx 44.692 cm^2. To 3 s.f., Area is 44.7 cm^2. (b) Cosine Rule: AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(ABC) = 8.5^2 + 12.4^2 - 2 * 8.5 * 12.4 * cos(58) = 72.25 + 153.76 - 210.8 * 0.5299 = 226.01 - 111.71 = 114.30 cm^2. AC = \\sqrt{114.30} \\approx 10.69 cm. To 3 s.f., AC is 10.7 cm. (c) sin(58) = height / 8.5 \\implies height = 8.5 * sin(58) \\approx 7.208 cm. To 3 s.f., height is 7.21 cm.

(a) [4 marks total]: M1 for area formula, M2 for correct substitution, A1 for 44.7. (b) [4.56 marks total]: M1 for Cosine Rule formula, M1 for 114.3, M1 for square root, A1.56 for 10.7. (c) [3 marks total]: M1 for trig formula or area-height formula, M1 for solving, A1 for 7.21.

(a) The perimeter consists of two lengths of the rectangle (\(15\text{ m}\) each), one width of the rectangle (\(8\text{ m}\)), and the curved arc of the semicircle.
Curved arc of semicircle = \(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 8 = 4\pi \approx 12.57\text{ m}\).
Perimeter = \(15 + 15 + 8 + 4\pi = 38 + 12.57 = 50.57\text{ m}\).
Rounding to 1 decimal place gives \(50.6\text{ m}\).

(b) The total area consists of the area of the rectangle and the area of the semicircle.
Area of rectangle = \(15 \times 8 = 120\text{ m}^2\).
Area of semicircle = \(\frac{1}{2} \times \pi \times r^2 = \frac{1}{2} \times \pi \times 4^2 = 8\pi \approx 25.13\text{ m}^2\).
Total Area = \(120 + 25.13 = 145.13\text{ m}^2\).
Rounding to 1 decimal place gives \(145.1\text{ m}^2\).

(c) (i) Number of bags required = \(\frac{\text{Total Area}}{18} = \frac{145.13}{18} \approx 8.06\).
Since seed must be bought in whole bags, we round up to the next integer: \(9\) bags.

(ii) Total cost = \(9 \times 12.50 = 112.50\).

(d) Total cost of border = \(\text{Perimeter} \times 8.20\).
Using the more precise perimeter of \(50.57\text{ m}\): \(50.56637... \times 8.20 = 414.64\).
Using the rounded perimeter of \(50.6\text{ m}\): \(50.6 \times 8.20 = 414.92\).

(a) [3 Marks]
- M1 for \(15 + 15 + 8\) (or \(38\))
- M1 for \(\frac{1}{2} \times \pi \times 8\) (or \(4\pi\) or \(12.57\))
- A1 for \(50.6\)

(b) [3 Marks]
- M1 for \(15 \times 8\) (or \(120\))
- M1 for \(\frac{1}{2} \times \pi \times 4^2\) (or \(8\pi\) or \(25.1\))
- A1 for \(145.1\)

(c)(i) [2 Marks]
- M1 for \(\text{their (b)} \div 18\)
- A1 for \(9\) (must be an integer greater than their division result)

(c)(ii) [1 Mark]
- B1 for \(112.50\) (or FT \(\text{their (c)(i)} \times 12.50\))

(d) [2 Marks]
- M1 for \(\text{their (a)} \times 8.20\)
- A1 for \(414.64\) or \(414.92\) (or FT \(\text{their (a)} \times 8.20\))

(a) The perimeter consists of two lengths of the rectangle (\(15\text{ m}\) each), one width of the rectangle (\(8\text{ m}\)), and the curved arc of the semicircle.
Curved arc of semicircle = \(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 8 = 4\pi \approx 12.57\text{ m}\).
Perimeter = \(15 + 15 + 8 + 4\pi = 38 + 12.57 = 50.57\text{ m}\).
Rounding to 1 decimal place gives \(50.6\text{ m}\).

(b) The total area consists of the area of the rectangle and the area of the semicircle.
Area of rectangle = \(15 \times 8 = 120\text{ m}^2\).
Area of semicircle = \(\frac{1}{2} \times \pi \times r^2 = \frac{1}{2} \times \pi \times 4^2 = 8\pi \approx 25.13\text{ m}^2\).
Total Area = \(120 + 25.13 = 145.13\text{ m}^2\).
Rounding to 1 decimal place gives \(145.1\text{ m}^2\).

(c) (i) Number of bags required = \(\frac{\text{Total Area}}{18} = \frac{145.13}{18} \approx 8.06\).
Since seed must be bought in whole bags, we round up to the next integer: \(9\) bags.

(ii) Total cost = \(9 \times 12.50 = 112.50\).

(d) Total cost of border = \(\text{Perimeter} \times 8.20\).
Using the more precise perimeter of \(50.57\text{ m}\): \(50.56637... \times 8.20 = 414.64\).
Using the rounded perimeter of \(50.6\text{ m}\): \(50.6 \times 8.20 = 414.92\).

(a) [3 Marks]
- M1 for \(15 + 15 + 8\) (or \(38\))
- M1 for \(\frac{1}{2} \times \pi \times 8\) (or \(4\pi\) or \(12.57\))
- A1 for \(50.6\)

(b) [3 Marks]
- M1 for \(15 \times 8\) (or \(120\))
- M1 for \(\frac{1}{2} \times \pi \times 4^2\) (or \(8\pi\) or \(25.1\))
- A1 for \(145.1\)

(c)(i) [2 Marks]
- M1 for \(\text{their (b)} \div 18\)
- A1 for \(9\) (must be an integer greater than their division result)

(c)(ii) [1 Mark]
- B1 for \(112.50\) (or FT \(\text{their (c)(i)} \times 12.50\))

(d) [2 Marks]
- M1 for \(\text{their (a)} \times 8.20\)
- A1 for \(414.64\) or \(414.92\) (or FT \(\text{their (a)} \times 8.20\))

(a) Time = Distance / Speed = \(120/v\) hours. (b) Return speed = \(v - 15\) km/h. Time taken = \(120/(v-15)\) hours. (c)(i) Total time = \(120/v + 120/(v-15) = 5.4\). Multiplying both sides by 10 and dividing by 6 simplifies this to: \(20/v + 20/(v-15) = 0.9\) which becomes \(200(v-15) + 200v = 9v(v-15)\). This expands to \(400v - 3000 = 9v^2 - 135v\), which rearranges to \(9v^2 - 535v + 3000 = 0\). (c)(ii) Using the quadratic formula: \(v = \frac{535 \pm \sqrt{(-535)^2 - 4(9)(3000)}}{2(9)} = \frac{535 \pm \sqrt{178225}}{18}\). Thus, \(v \approx 53.18\) or \(v \approx 6.27\).

(a) B1 for 120/v. (b) B1 for 120/(v-15). (c)(i) M1 for setting up the equation 120/v + 120/(v-15) = 5.4. M1 for clearing fractions to get 120(v-15) + 120v = 5.4v(v-15). A1 for showing step-by-step algebraic simplification to the final form. (c)(ii) M1 for correct substitution into the quadratic formula. A1 for 53.18. A1 for 6.27.

(a) Total surface area of a cylinder is \(2\pi r^2 + 2\pi r h\). Since \(h = 2r\), Area = \(2\pi r^2 + 2\pi r(2r) = 2\pi r^2 + 4\pi r^2 = 6\pi r^2\). (b) Volume of cylinder = \(\pi r^2 h = \pi r^2 (2r) = 2\pi r^3\). Volume of sphere = \(\frac{4}{3}\pi R^3\). Since volumes are equal, \(\frac{4}{3}\pi R^3 = 2\pi r^3 \implies 4R^3 = 6r^3 \implies R^3 = \frac{3}{2}r^3 \implies R = \sqrt[3]{\frac{3}{2}} r\). (c) Total surface area of the sphere is \(4\pi R^2 = 4\pi \left(\sqrt[3]{\frac{3}{2}}r\right)^2 = 4\pi (1.5)^{2/3} r^2\). The ratio of cylinder area to sphere area is \(6\pi r^2 : 4\pi (1.5)^{2/3} r^2 = 6 : 4(1.5)^{2/3} \approx 6 : 5.2415 = 1 : 0.87358...\) which rounds to \(1 : 0.874\).

(a) M1 for 2*pi*r^2 + 2*pi*r*(2r). A1 for establishing 6*pi*r^2. (b) M1 for volume of cylinder 2*pi*r^3. M1 for setting 4/3*pi*R^3 = 2*pi*r^3. A1 for simplifying to R^3 = 1.5r^3 and concluding. (c) M1 for expression of sphere surface area 4*pi*R^2. M1 for substituting R in terms of r. M1 for ratio 6 : 4(1.5)^(2/3). A1 for 1 : 0.874.

(a) Factorise numerator and denominator: \(2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)\). \(x^2 + 5x + 6 = (x+2)(x+3)\). Cancelling \(x+2\) gives \(\frac{2(x-2)}{x+3} = \frac{2x-4}{x+3}\). (b) \(\frac{3(x+3) - 2(2x-1)}{(2x-1)(x+3)} = \frac{3x+9-4x+2}{(2x-1)(x+3)} = \frac{11-x}{(2x-1)(x+3)}
\. (c) Grouping terms: \)4a(3x - 2y) - b(3x - 2y) = (4a - b)(3x - 2y)\).

(a) M1 for 2(x-2)(x+2). M1 for (x+2)(x+3). A1 for correct simplified fraction. (b) M1 for 3(x+3) - 2(2x-1). M1 for common denominator (2x-1)(x+3). A1 for (11-x)/((2x-1)(x+3)). (c) M1 for 4a(3x-2y) or -b(3x-2y). A1 for final factorisation.

(a) By the Cosine Rule: \(AC^2 = 110^2 + 85^2 - 2(110)(85)\cos(72^\circ) = 12100 + 7225 - 18700(0.309017) = 13546.38\). Thus, \(AC = \sqrt{13546.38} \approx 116.39\) m, which is \(116\) m to the nearest meter. (b) Area = \(\frac{1}{2} \times AB \times BC \times \sin(ABC) = \frac{1}{2}(110)(85)\sin(72^\circ) \approx 4446.19\) m\(^2\), which is \(4450\) m\(^2\) (or \(4446\) m\(^2\)). (c) In right-angled triangle \(ABF\), \(\tan(14^\circ) = \frac{\text{height}}{110} \implies \text{height} = 110\tan(14^\circ) \approx 27.4\) m. (d) The shortest distance \(d\) is the perpendicular height. Area = \(\frac{1}{2} \times AC \times d \implies 4446.19 = \frac{1}{2}(116.388)d \implies d = \frac{2 \times 4446.19}{116.388} \approx 76.4\) m.

(a) M1 for Cosine Rule formula with correct values. A1 for 116.39. A1 for 116. (b) M1 for Area formula with correct values. A1 for 4450 or 4446. (c) M1 for tan(14) = h/110. A1 for 27.4. (d) M1 for Area = 0.5 * AC * d. A1 for 76.4.

(a) For \(x = -2\), \(y = 12/(-2) + (-2)^2 - 6 = -6 + 4 - 6 = -8\). For \(x = 1\), \(y = 12/1 + 1^2 - 6 = 7\). For \(x = 3\), \(y = 12/3 + 3^2 - 6 = 4 + 9 - 6 = 7\). (b) Multiply both sides of \(\frac{12}{x} + x^2 - 6 = 2x\) by \(x\): \(12 + x^3 - 6x = 2x^2\). Rearranging gives: \(x^3 - 2x^2 - 6x + 12 = 0\). (c)(i) The straight line is \(y = 2x\). (c)(ii) Factorising: \(x^3 - 2x^2 - 6x + 12 = x^2(x - 2) - 6(x - 2) = (x^2 - 6)(x - 2) = 0\). Since \(x=2\) is one solution, \(x^2 - 6 = 0 \implies x = \pm \sqrt{6}\).

(a) B1 for a = -8. B1 for b = 7. B1 for c = 7. (b) M1 for multiplying by x. A1 for correct rearrangement. (c)(i) B1 for y = 2x. (c)(ii) M1 for grouping or polynomial division. A1 for (x^2 - 6)(x-2). A1 for sqrt(6) and -sqrt(6) (or +/-2.45).

(a) The total number of marbles is 12. The probability of drawing two marbles of the same color is \(P(RR) + P(BB) + P(GG)\). \(P(RR) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\). \(P(BB) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\). \(P(GG) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}\). Total probability = \(\frac{20+12+6}{132} = \frac{38}{132} = \frac{19}{66}\) (or approximately 0.288). (b) The probability that at least one marble is green is \(1 - P(\text{no green})\). \(P(\text{no green}) = P(\text{Red or Blue for both}) = \frac{9}{12} \times \frac{8}{11} = \frac{72}{132} = \frac{6}{11}\). Thus, \(1 - \frac{6}{11} = \frac{5}{11}\) (or approximately 0.455).

(a) M1 for listing P(RR), P(BB), P(GG) with without-replacement denominators. M1 for correct individual products. A1 for 19/66 (or 38/132 or 0.288). (b) M1 for 1 - P(no green) or alternative addition of outcomes. M1 for P(no green) = (9/12)*(8/11). A1 for 5/11 (or 0.455).

(a)(i) Sequence A is an arithmetic sequence with a common difference of 3 and first term 5. The \(n\)-th term is \(3n + 2\). (a)(ii) Sequence B has terms that are 1 more than perfect squares: \(1^2+1, 2^2+1, 3^2+1, 4^2+1\). The \(n\)-th term is \(n^2 + 1\). (a)(iii) Sequence C is formed by dividing Sequence A by Sequence B. The \(n\)-th term is \(\frac{3n + 2}{n^2 + 1}\). (b) For \(n = 50\), the term is \(\frac{3(50) + 2}{50^2 + 1} = \frac{152}{2501}\). (c)(i) Sequence D is a geometric sequence with first term \(a = 3\) and common ratio \(r = 2\). The \(n\)-th term is \(3 \times 2^{n-1}\). (c)(ii) Setting \(3 \times 2^{n-1} = 12288 \implies 2^{n-1} = 4096\). Since \(2^{12} = 4096\), we have \(n - 1 = 12 \implies n = 13\).

(a)(i) B1 for 3n + k, B1 for 3n + 2. (a)(ii) M1 for recognising square numbers, A1 for n^2 + 1. (a)(iii) B1 for (3n+2)/(n^2+1). (b) B1 for 152/2501. (c)(i) B1 for 3 * 2^(n-1). (c)(ii) M1 for 3 * 2^(n-1) = 12288, A1 for n = 13.

(a) Simple interest = \(P \times R \times T / 100 = 4500 \times 3.2 \times 6 / 100 = \$864\). (b) Compound interest formula: \(4000 \times (1 + r/100)^6 = 4850 \implies (1 + r/100)^6 = 1.2125 \implies 1 + r/100 = (1.2125)^{1/6} \approx 1.032644 \implies r = 3.2644\%\), which is 3.26% to 2 decimal places. (c) Total amount = \(5000 \times (1 + 0.025)^3 \times (1 + 0.035)^3 = 5000 \times (1.025)^3 \times (1.035)^3 = 5000 \times 1.0768906 \times 1.1087178 \approx \$5969.84\).

(a) M1 for 4500 * 0.032 * 6. A1 for 864. (b) M1 for 4000 * (1 + r/100)^6 = 4850. M1 for (1.2125)^(1/6). A1 for 3.26. (c) M1 for 5000 * (1.025)^3. M1 for multiplying by (1.035)^3. A1 for 5969.84.

(a) Using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 120^2 + 85^2 - 2 \times 120 \times 85 \times \cos(78^\circ)\)
\(AC^2 = 14400 + 7225 - 20400 \times 0.20791\)
\(AC^2 = 21625 - 4241.4 = 17383.6\)
\(AC = \sqrt{17383.6} \approx 131.85\text{ m}\), which is \(132\text{ m}\) correct to 3 significant figures.

(b) Using the Sine Rule:
\(\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(ACB)}{120} = \frac{\sin(78^\circ)}{131.85}\)
\(\sin(ACB) = \frac{120 \times \sin(78^\circ)}{131.85} \approx 0.8902\)
\(ACB = \arcsin(0.8902) \approx 62.9^\circ\).

(c) Area of the triangle:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \times 120 \times 85 \times \sin(78^\circ) \approx 4988.55\text{ m}^2\), which is \(4990\text{ m}^2\) correct to 3 significant figures.

(d) The shortest distance from \(B\) to \(AC\) is the perpendicular height \(h\) of the triangle with base \(AC\).
\(\text{Area} = \frac{1}{2} \times AC \times h\)
\(4988.55 = \frac{1}{2} \times 131.85 \times h\)
\(h = \frac{2 \times 4988.55}{131.85} \approx 75.67\text{ m}\), which is \(75.7\text{ m}\) correct to 3 significant figures.

(a)
- M1: For \(120^2 + 85^2 - 2 \times 120 \times 85 \times \cos(78)\)
- A1: For \(17383.6...\)
- A1: For \(131.8...\) or \(132\)

(b)
- M1: For \(\frac{\sin(ACB)}{120} = \frac{\sin(78)}{131.85}\)
- M1: For rearranging to find \(\sin(ACB) = ...\)
- A1: For \(62.9^\circ\) or \(62.88^\circ\)

(c)
- M1: For \(\frac{1}{2} \times 120 \times 85 \times \sin(78)\)
- A1: For \(4988.5...\)
- A1: For \(4990\)

(d)
- M1: For equating area formula to \(4988.5...\) or using \(85 \times \sin(62.9)\)
- A1: For \(75.7\)

(a) The volume of the cone:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 6^2 \times 8 = 96\pi\text{ cm}^3\).

The volume of the hemisphere:
\(V_{\text{hemi}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \times 6^3 = 144\pi\text{ cm}^3\).

Total volume:
\(V_{\text{total}} = 96\pi + 144\pi = 240\pi\text{ cm}^3\).

(b) First find the slant height \(l\) of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}\).

Curved surface area of the cone:
\(A_{\text{cone}} = \pi r l = \pi \times 6 \times 10 = 60\pi\text{ cm}^2\).

Curved surface area of the hemisphere:
\(A_{\text{hemi}} = 2 \pi r^2 = 2 \pi \times 6^2 = 72\pi\text{ cm}^2\).

Total surface area:
\(A_{\text{total}} = 60\pi + 72\pi = 132\pi \approx 414.69\text{ cm}^2\), which is \(414.7\text{ cm}^2\) correct to 1 decimal place.

(c) First find the total mass in grams:
\(\text{Mass} = \text{Volume} \times \text{Density} = 240\pi \times 7.8 \approx 753.98 \times 7.8 = 5881.06\text{ g}\).

Convert to kilograms:
\(\text{Mass} = \frac{5881.06}{1000} \approx 5.881\text{ kg}\), which is \(5.88\text{ kg}\) correct to 3 significant figures.

(a)
- M1: For \(\frac{1}{3} \pi \times 6^2 \times 8\)
- A1: For \(96\pi\)
- M1: For \(\frac{2}{3} \pi \times 6^3\)
- A1: For showing sum is \(240\pi\)

(b)
- M1: For finding slant height \(l = \sqrt{6^2 + 8^2}\)
- A1: For \(10\)
- M1: For \(\pi \times 6 \times 10\) (or \(60\pi\))
- M1: For \(2 \pi \times 6^2\) (or \(72\pi\))
- A1: For \(414.7\)

(c)
- M1: For multiplying \(240\pi\) (or \(754\)) by \(7.8\)
- M1: For dividing by \(1000\)
- A1: For \(5.88\)

(a) Time taken for the outward journey = \(\frac{36}{x}\) hours.

(b) Speed for the return journey = \(x + 4\text{ km/h}\).
Time taken for the return journey = \(\frac{36}{x + 4}\) hours.

(c) \(45\text{ minutes} = 0.75\text{ hours}\) or \(\frac{3}{4}\) hours.
\(\frac{36}{x} - \frac{36}{x + 4} = \frac{3}{4}\)
Divide the entire equation by 3:
\(\frac{12}{x} - \frac{12}{x + 4} = \frac{1}{4}\)
Multiply both sides by the common denominator \(4x(x + 4)\):
\(12 \times 4(x + 4) - 12 \times 4x = x(x + 4)\)
\(48x + 192 - 48x = x^2 + 4x\)
\(192 = x^2 + 4x\)
\(x^2 + 4x - 192 = 0\).

(d) Factoring the quadratic equation:
\(x^2 + 4x - 192 = 0\)
We need two numbers that multiply to \(-192\) and add to \(4\).
These numbers are \(+16\) and \(-12\).
\((x + 16)(x - 12) = 0\)
So \(x = -16\) or \(x = 12\).
Since speed cannot be negative, \(x = 12\text{ km/h}\).

(e) Outward journey time = \(\frac{36}{12} = 3\text{ hours}\).
Return journey time = \(\frac{36}{16} = 2.25\text{ hours}\).
Total time = \(3 + 2.25 = 5.25\text{ hours}\) (or 5 hours 15 minutes).

(a)
- B1: For \(\frac{36}{x}\)

(b)
- B1: For \(\frac{36}{x + 4}\)

(c)
- M1: For \(\frac{36}{x} - \frac{36}{x + 4} = 0.75\) (or \(\frac{3}{4}\) or \(45\))
- M1: For algebraic fraction common denominator conversion
- M1: For expanding and eliminating fraction
- A1: For showing clearly the final step leading to \(x^2 + 4x - 192 = 0\)

(d)
- M1: For factorisation \((x+16)(x-12)\) or correct use of formula
- A1: For solutions \(x = 12\) and \(x = -16\)
- M1: For discarding negative root
- A1: For final answer \(12\)

(e)
- M1: For \(\frac{36}{12} + \frac{36}{16}\)
- A1: For \(5.25\) hours (or 5h 15m)