2024 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

First, expand the bracket on the left side of the equation to get \(5x - 15 = 2x + 9\). Next, subtract \(2x\) from both sides to obtain \(3x - 15 = 9\). Then, add 15 to both sides to get \(3x = 24\). Finally, divide both sides by 3 to find \(x = 8\).

M1 for correct expansion of the bracket: \(5x - 15\) M1 for isolating the x terms on one side and constant terms on the other (e.g., \(3x = 24\)) A0.5 for 8

First, calculate the increase in the number of members: \(216 - 160 = 56\). Next, calculate this increase as a percentage of the original number of members: \(\frac{56}{160} \times 100 = 35\%\).

M1 for finding the increase of 56 M1 for \(\frac{56}{160} \times 100\) or \(\frac{216}{160} \times 100 - 100\) A0.5 for 35

The sum of the exterior angles of any regular polygon is \(360^\circ\). The formula relating the exterior angle and the number of sides, \(n\), is \(\text{Exterior angle} = \frac{360^\circ}{n}\). Substituting \(40^\circ\) into the formula gives \(40 = \frac{360}{n}\). Solving for \(n\) gives \(n = \frac{360}{40} = 9\).

M2 for \(\frac{360}{40}\) (or M1 for sight of 360) A0.5 for 9

The exterior angle of a triangle is equal to the sum of the two opposite interior angles: \(\angle ACD = \angle BAC + \angle ABC\). Substituting the given values: \(115^\circ = 42^\circ + \angle ABC\), which gives \(\angle ABC = 115^\circ - 42^\circ = 73^\circ\). Alternatively, we can find angle \(\angle ACB = 180^\circ - 115^\circ = 65^\circ\). Then, \(\angle ABC = 180^\circ - (42^\circ + 65^\circ) = 73^\circ\).

M1 for finding \(\angle ACB = 65^\circ\) or for writing the equation \(115 = 42 + x\). A1.5 for the correct final answer of 73.

The sale price represents \(100\% - 10\% = 90\%\) of the original price. Let the original price be \(x\). Then, \(0.90 \times x = 51.30\), which gives \(x = \frac{51.30}{0.90} = 57\). Thus, the original price of the jacket was \(\$57\).

M1 for setting up the equation \(0.90 \times \text{original price} = 51.30\) or for \(51.30 \div 0.90\). A1.5 for the correct final answer of 57.

Cross-multiply to clear the fractions: \(3(2x - 3) = 4(x + 2)\). Expand both sides to get \(6x - 9 = 4x + 8\). Subtract \(4x\) from both sides to get \(2x - 9 = 8\). Add 9 to both sides to get \(2x = 17\). Divide by 2 to obtain the solution: \(x = 8.5\).

M1 for correctly cross-multiplying to obtain \(3(2x - 3) = 4(x + 2)\) or equivalent. A1 for expanding and collecting terms to get \(2x = 17\) or equivalent. A0.5 for the correct final answer of 8.5 (or 17/2).

First, eliminate the denominators by multiplying every term by 4:
\(4 \times \left(\frac{2x - 3}{4}\right) + 4 \times \left(\frac{x}{2}\right) = 4 \times 5\)
\((2x - 3) + 2x = 20\)
Combine like terms:
\(4x - 3 = 20\)
Add 3 to both sides:
\(4x = 23\)
Divide by 4:
\(x = 5.75\) (or \(\frac{23}{4}\) or \(5\frac{3}{4}\)).

M1 for multiplying both sides by 4 to get \((2x - 3) + 2x = 20\) (or equivalent to obtain a common denominator)
M1 for simplifying to \(4x = 23\)
A0.5 for the correct final answer \(5.75\) (or \(\frac{23}{4}\))

First, calculate the profit:
Profit = \(\$116 - \$80 = \$36\)
Next, calculate the percentage profit based on the original cost price:
Percentage Profit = \(\frac{\text{Profit}}{\text{Cost Price}} \times 100 = \frac{36}{80} \times 100\)
Percentage Profit = \(0.45 \times 100 = 45\%\).

M1 for calculating the profit: \(116 - 80 = 36\)
M1 for setting up the percentage fraction: \(\frac{36}{80} \times 100\) (or equivalent)
A0.5 for the final answer \(45\) (accept \(45\%\))

The sum of the angles in a triangle is \(180^\circ\).
The total number of parts in the ratio is:
\(2 + 3 + 7 = 12\text{ parts}\)
Find the value of one part:
\(180^\circ \div 12 = 15^\circ\)
The largest angle corresponds to 7 parts in the ratio:
\(7 \times 15^\circ = 105^\circ\).

M1 for summing the parts of the ratio: \(2 + 3 + 7 = 12\)
M1 for calculating the size of one share: \(180^\circ \div 12\) (or for the expression \(\frac{7}{12} \times 180\))
A0.5 for \(105\) (or \(105^\circ\))

First, expand the brackets:
\(8x - 12 - 3x + 15 = 18\)

Next, simplify and collect like terms:
\(5x + 3 = 18\)

Subtract 3 from both sides:
\(5x = 15\)

Divide by 5:
\(x = 3\)

M1 for expanding the brackets correctly to get at least three terms correct: \(8x - 12 - 3x + 15\)
M1 for correctly isolating the terms in \(x\) on one side, e.g., \(5x = 15\) (or follow through from a single error in expansion)
A0.5 for the correct final answer: \(3\)

The sale price represents \(100\% - 12\% = 88\%\) of the original price.

Let the original price be \(P\).
\(0.88 \times P = 74.80\)

\(P = \frac{74.80}{0.88} = 85\)

Therefore, the original price of the jacket was $85.

M1.5 for a complete and correct method to find the original price, e.g., \(74.80 \div 0.88\) (or \(74.80 \div 88 \times 100\))
A1 for the correct answer: \(85\)

Method 1: Using the exterior angle of a triangle property.
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
\(x + 47 = 112\)
\(x = 112 - 47 = 65\)

Method 2: Using angles on a straight line and angles in a triangle.
Angle \(ACB\) and angle \(ACD\) lie on a straight line, so:
\(ACB = 180^\circ - 112^\circ = 68^\circ\)

The sum of angles in triangle \(ABC\) is \(180^\circ\):
\(x + 47 + 68 = 180\)
\(x + 115 = 180\)
\(x = 65\)

M1.5 for a correct method to find the interior angle \(ACB\) (e.g., \(180 - 112\)) OR for setting up the exterior angle equation \(x + 47 = 112\)
A1 for the correct value: \(65\)

Expand the bracket on the left-hand side to get \(6x - 15 = 4x + 7\). Subtract \(4x\) from both sides to get \(2x - 15 = 7\). Add 15 to both sides to get \(2x = 22\). Finally, divide by 2 to find \(x = 11\).

M1 for correct expansion of the bracket: \(6x - 15\). M1 for correctly isolating the x term on one side: \(2x = 22\) (or equivalent). A0.5 for 11.

First, calculate the reduction in price by subtracting the sale price from the original price: \(\$85 - \$64.60 = \$20.40\). Next, divide this reduction by the original price and multiply by 100 to find the percentage decrease: \(\frac{20.40}{85} \times 100 = 24\%\).

M1 for finding the price reduction: \(20.40\) (or for calculating \(\frac{64.60}{85} = 0.76\)). M1 for \(\frac{20.40}{85} \times 100\) (or \(100 - 76\)). A0.5 for 24.

The sum of the interior angle and the exterior angle at any vertex of a regular polygon is \(180^\circ\). Thus, the exterior angle is \(180^\circ - 140^\circ = 40^\circ\). Since the sum of all exterior angles in any polygon is \(360^\circ\), the number of sides is \(\frac{360}{40} = 9\). Alternatively, using the interior angle formula, solve \(\frac{(n - 2) \times 180}{n} = 140\) which simplifies to \(180n - 360 = 140n\), then \(40n = 360\), leading to \(n = 9\).

M1 for finding the exterior angle: \(180 - 140 = 40\) (or setting up \(\frac{(n-2) \times 180}{n} = 140\)). M1 for dividing 360 by their exterior angle: \(\frac{360}{40}\) (or isolating \(n\) to show \(40n = 360\)). A0.5 for 9.

Multiply both sides of the equation by 3: \(4x - 7 = 15\). Add 7 to both sides: \(4x = 22\). Divide by 4: \(x = 5.5\).

M1 for \(4x - 7 = 15\) or \(4x = 22\). A1.5 for \(5.5\) (or \(\frac{11}{2}\) or \(5\frac{1}{2}\)).

The sale price is \(100\% - 15\% = 85\%\) of the original price. Therefore, \(0.85 \times \text{original price} = 51\). Dividing both sides by 0.85 gives the original price: \(51 \div 0.85 = 60\).

M1 for \(51 \div 0.85\) or for setting up \(0.85x = 51\) or equivalent. A1.5 for \(60\).

The exterior angle of a regular polygon and its interior angle sum to \(180^\circ\). Each exterior angle is \(180^\circ - 140^\circ = 40^\circ\). Since the sum of exterior angles of any polygon is \(360^\circ\), the number of sides is \(360^\circ \div 40^\circ = 9\).

M1 for finding the exterior angle \(180 - 140 = 40\) or setting up the equation \(\frac{(n-2) \times 180}{n} = 140\). A1.5 for \(9\).

First, expand the brackets on both sides of the equation: \(12x - 4 = 2x + 16\). Next, subtract \(2x\) from both sides to get: \(10x - 4 = 16\). Then, add \(4\) to both sides to isolate the x term: \(10x = 20\). Finally, divide both sides by \(10\) to find the value of \(x\): \(x = 2\).

M1 for correctly expanding at least one bracket, e.g., \(12x - 4\) or \(2x + 16\). M1 for isolating the x terms on one side of the equation, e.g., \(10x = 20\). A0.5 for the correct final answer \(2\).

Let the original price be \(x\). A reduction of \(15\%\) means the sale price is \(85\%\) of the original price. Therefore, we can write the equation: \(0.85x = 561\). To find \(x\), divide \(561\) by \(0.85\): \(x = 561 / 0.85 = 660\). The original price of the computer was \(\$660\).

M1 for setting up a correct relationship, e.g., representing the sale price as \(85\%\) of the original price, or writing \(0.85x = 561\). M1 for the calculation \(561 / 0.85\). A0.5 for the correct answer \(660\).

An interior angle and an exterior angle on a straight line add up to \(180^\circ\). First, find the size of one exterior angle: \(180^\circ - 144^\circ = 36^\circ\). The sum of all exterior angles of any polygon is always \(360^\circ\). To find the number of sides, divide \(360^\circ\) by the size of one exterior angle: \(360^\circ / 36^\circ = 10\). Therefore, the polygon has \(10\) sides.

M1 for finding the size of the exterior angle: \(180 - 144 = 36\) (or for setting up the equation \((n - 2) \times 180 / n = 144\)). M1 for dividing the total sum of exterior angles by the exterior angle: \(360 / 36\) (or for solving the equation to \(36n = 360\)). A0.5 for the correct final answer \(10\).

First, find the fraction of students remaining after excluding those who play football: \(1 - \frac{3}{8} = \frac{5}{8}\). Next, find \(\frac{2}{5}\) of this remaining fraction: \(\frac{2}{5} \times \frac{5}{8} = \frac{1}{4}\).

M1 for \(1 - \frac{3}{8}\) or \(\frac{5}{8}\) seen, or for \(\frac{2}{5} \times \text{their remaining fraction}\). A1 for \(\frac{1}{4}\) or equivalent in simplest form.

First, find the fraction of students remaining after excluding those who play football: \(1 - \frac{3}{8} = \frac{5}{8}\). Next, find \(\frac{2}{5}\) of this remaining fraction: \(\frac{2}{5} \times \frac{5}{8} = \frac{1}{4}\).

M1 for \(1 - \frac{3}{8}\) or \(\frac{5}{8}\) seen, or for \(\frac{2}{5} \times \text{their remaining fraction}\). A1 for \(\frac{1}{4}\) or equivalent in simplest form.

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\). Substitute the given values: \(AC^2 = 7.4^2 + 5.8^2 - 2 \times 7.4 \times 5.8 \times \cos(64^\circ)\). Calculate each term: \(7.4^2 = 54.76\), \(5.8^2 = 33.64\), and \(2 \times 7.4 \times 5.8 \times \cos(64^\circ) \approx 85.84 \times 0.43837 = 37.630\). Subtracting the values gives: \(AC^2 \approx 54.76 + 33.64 - 37.630 = 50.77\). Taking the square root: \(AC = \sqrt{50.77} \approx 7.1253\text{ cm}\). Rounding to 3 significant figures gives \(7.13\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7.4^2 + 5.8^2 - 2 \times 7.4 \times 5.8 \times \cos(64)\). A1 for \(AC^2 = 50.77...\) or \(AC = \sqrt{50.77...}\). A1 for \(7.13\) (accept values in the range \(7.12\) to \(7.13\)).

Multiply both sides by the common denominator \((x+1)(x-1)\): \(3(x-1) + 2(x+1) = 1 \times (x+1)(x-1)\). Expand the brackets: \(3x - 3 + 2x + 2 = x^2 - 1\). Simplify the left-hand side: \(5x - 1 = x^2 - 1\). Rearrange the equation to equal zero: \(x^2 - 5x = 0\). Factorise the quadratic equation: \(x(x - 5) = 0\). Solve for \(x\): \(x = 0\) or \(x = 5\).

M1 for multiplying by \((x+1)(x-1)\) to get \(3(x-1) + 2(x+1) = (x+1)(x-1)\) or equivalent. M1 for simplifying to \(x^2 - 5x = 0\) or equivalent 2-term quadratic. A1 for both correct solutions: \(0\) and \(5\).

Let the original price of the jacket be \(x\). After a 15% reduction, the price is \(x \times (1 - 0.15) = 0.85x\). After a further 10% reduction, the price is \(0.85x \times (1 - 0.10) = 0.85x \times 0.90 = 0.765x\). We are given that this final price is $68.85, so we write the equation: \(0.765x = 68.85\). Solving for \(x\) gives \(x = \frac{68.85}{0.765} = 90\). Thus, the original price of the jacket was $90.

M1 for finding the price before the final 10% reduction: \(\frac{68.85}{0.90} = 76.50\), or for showing the combined multiplier \(0.85 \times 0.90 = 0.765\). M1 for \(\frac{76.50}{0.85}\) or \(\frac{68.85}{0.765}\). A1 for \(90\).

Multiply the entire equation by the common denominator \(x(x-1)\) to clear the fractions: \(6x - 4(x-1) = x(x-1)\). Expand both sides: \(6x - 4x + 4 = x^2 - x\). Simplify to get: \(2x + 4 = x^2 - x\). Rearrange the terms to form a standard quadratic equation: \(x^2 - 3x - 4 = 0\). Factor the quadratic expression: \((x - 4)(x + 1) = 0\). Solving this gives: \(x = 4\) or \(x = -1\).

M1 for clearing the fractions correctly: \(6x - 4(x-1) = x(x-1)\); M1 for simplifying to a correct standard quadratic form: \(x^2 - 3x - 4 = 0\); A1 for both correct solutions: \(x = 4\) and \(x = -1\).

Let \(P\) be the original price of the laptop. After the first reduction of 15%, the price is \(0.85P\). After the second reduction of 10%, the final price is \(0.85P \times 0.90 = 0.765P\). We are given that this final price is $535.50, so we can write the equation: \(0.765P = 535.50\). To find \(P\), divide by 0.765: \(P = \frac{535.50}{0.765} = 700\). Thus, the original price of the laptop was $700.

M1 for expressing successive reductions, e.g., \(P \times 0.85 \times 0.90 = 535.50\) or identifying the overall percentage of 76.5%; M1 for a complete method to find the original price, e.g., \(535.50 \div 0.765\) or finding the intermediate price as \(535.50 \div 0.90 = 595\); A1 for 700.

We use the cosine rule on triangle \(ABC\) to find angle \(ABC\) (opposite side \(AC\)): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). Substituting the given side lengths: \(11^2 = 7^2 + 9^2 - 2(7)(9)\cos(ABC)\). Simplify the numbers: \(121 = 49 + 81 - 126\cos(ABC)\), which becomes \(121 = 130 - 126\cos(ABC)\). Rearrange to solve for \(\cos(ABC)\): \(126\cos(ABC) = 130 - 121 = 9\), giving \(\cos(ABC) = \frac{9}{126} = \frac{1}{14}\). Now, find the angle by taking the inverse cosine: \(ABC = \arccos\left(\frac{1}{14}\right) \approx 85.904^\circ\). Correct to 1 decimal place, angle \(ABC\) is 85.9 degrees.

M1 for a correct substitution into the cosine rule: \(11^2 = 7^2 + 9^2 - 2(7)(9)\cos(ABC)\); M1 for a correct rearrangement to isolate the cosine of the angle: \(\cos(ABC) = \frac{9}{126}\) or \(\cos(ABC) = 0.0714...\); A1 for 85.9.

Multiply both sides by the common denominator \((x-1)(x+3)\): \(3(x+3) + 2(x-1) = (x-1)(x+3)\). Expand both sides: \(3x + 9 + 2x - 2 = x^2 + 2x - 3\). Simplify: \(5x + 7 = x^2 + 2x - 3\). Rearrange into a quadratic equation: \(x^2 - 3x - 10 = 0\). Factorise the quadratic equation: \((x-5)(x+2) = 0\). Therefore, \(x = 5\) or \(x = -2\).

M1 for \(3(x+3) + 2(x-1) = (x-1)(x+3)\) or better. M1 for simplifying to a three-term quadratic equation, e.g., \(x^2 - 3x - 10 = 0\). A1 for both solutions: \(x = 5\) and \(x = -2\).

Use the cosine rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substitute the given lengths: \(12^2 = 7^2 + 8^2 - 2(7)(8)\cos(\angle ABC)\). Simplify: \(144 = 49 + 64 - 112\cos(\angle ABC)\). Combine the terms: \(144 = 113 - 112\cos(\angle ABC)\). Rearrange to solve for \(\cos(\angle ABC)\): \(112\cos(\angle ABC) = 113 - 144\), which gives \(112\cos(\angle ABC) = -31\), so \(\cos(\angle ABC) = -\frac{31}{112}\). Find the angle: \(\angle ABC = \cos^{-1}\left(-\frac{31}{112}\right) \approx 116.071^\circ\). Rounding to 1 decimal place gives \(116.1^\circ\).

M1 for correct substitution into the cosine rule, e.g., \(12^2 = 7^2 + 8^2 - 2(7)(8)\cos(B)\). M1 for \(\cos(B) = -\frac{31}{112}\) or \(\cos(B) \approx -0.277\). A1 for \(116.1\) or \(116.07...\) (accept with or without degree symbol).

First, calculate the volume of one cylinder: \(V_{\text{cylinder}} = \pi r^2 h = \pi \times 1.5^2 \times 4 = 9\pi\text{ cm}^3\). Calculate the total volume of 32 cylinders: \(V_{\text{total}} = 32 \times 9\pi = 288\pi\text{ cm}^3\). Set the volume of the sphere equal to this total volume: \(\frac{4}{3}\pi r^3 = 288\pi\). Divide both sides by \(\pi\): \(\frac{4}{3}r^3 = 288\). Multiply by \(\frac{3}{4}\): \(r^3 = 288 \times \frac{3}{4} = 216\). Find the cube root: \(r = \sqrt[3]{216} = 6\).

M1 for finding the volume of one cylinder: \(\pi \times 1.5^2 \times 4\) (or \(9\pi\) or \(28.27...\)). M1 for setting up the equation: \(\frac{4}{3}\pi r^3 = 32 \times 9\pi\) (or \(\frac{4}{3}r^3 = 288\)). A1 for \(6\).

Multiply both sides by the common denominator \((x-1)(x+1)\):
\(6(x+1) - 4(x-1) = 2(x-1)(x+1)\)

Expand both sides:
\(6x + 6 - 4x + 4 = 2(x^2 - 1)\)
\(2x + 10 = 2x^2 - 2\)

Rearrange to form a quadratic equation:
\(2x^2 - 2x - 12 = 0\)

Divide the entire equation by 2:
\(x^2 - x - 6 = 0\)

Factorise the quadratic:
\((x - 3)(x + 2) = 0\)

So, \(x = 3\) or \(x = -2\).

M1 for multiplying by common denominator to obtain \(6(x+1) - 4(x-1) = 2(x^2-1)\) or equivalent.
M1 for expanding and reducing to a three-term quadratic equation, e.g., \(2x^2 - 2x - 12 = 0\) or \(x^2 - x - 6 = 0\).
A1 for both correct solutions: \(x = 3\) and \(x = -2\) (or \(x = 3, -2\)).

Use the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)

Substitute the given values:
\(AC^2 = 8.2^2 + 11.5^2 - 2(8.2)(11.5)\cos(62^\circ)\)
\(AC^2 = 67.24 + 132.25 - 188.6\cos(62^\circ)\)
\(AC^2 \approx 199.49 - 188.6(0.46947)\)
\(AC^2 \approx 110.948\)
\(AC = \sqrt{110.948} \approx 10.533\text{ cm}\)

To 3 significant figures, \(AC = 10.5\).

M1 for correct substitution into the Cosine Rule: \(8.2^2 + 11.5^2 - 2(8.2)(11.5)\cos(62^\circ)\)
M1 for evaluating to \(AC^2 \approx 110.9\) (or \(\sqrt{110.9}\))
A1 for \(10.5\) (accept answers in range \([10.5, 10.54]\)).

Using the compound interest formula:
\(A = P\left(1 + \frac{r}{100}\right)^n\)

Substitute the known values:
\(4396.17 = 4000\left(1 + \frac{r}{100}\right)^3\)

Divide by 4000:
\(\left(1 + \frac{r}{100}\right)^3 = \frac{4396.17}{4000}\)
\(\left(1 + \frac{r}{100}\right)^3 \approx 1.0990425\)

Take the cube root of both sides:
\(1 + \frac{r}{100} = \sqrt[3]{1.0990425} \approx 1.032\)

Subtract 1:
\(\frac{r}{100} = 0.032\)

Multiply by 100:
\(r = 3.2\)

M1 for setting up the equation: \(4000(1 + r/100)^3 = 4396.17\)
M1 for \(\sqrt[3]{\frac{4396.17}{4000}}\) or \(1.032\)
A1 for \(3.2\).

Use the formula for the area of a triangle:
\(\text{Area} = \frac{1}{2} a c \sin B\)

Substitute the given values into the formula:
\(22 = \frac{1}{2} \times 9 \times 7 \times \sin(\angle ABC)\)
\(22 = 31.5 \sin(\angle ABC)\)
\(\sin(\angle ABC) = \frac{22}{31.5} = \frac{44}{63}\)

Find the acute angle:
\(\angle ABC = \sin^{-1}\left(\frac{44}{63}\right) \approx 44.30^{\circ}\)

Find the obtuse angle:
\(\angle ABC = 180^{\circ} - 44.30^{\circ} \approx 135.70^{\circ}\)

To 1 decimal place, the angles are \(44.3^{\circ}\) and \(135.7^{\circ}\).

M1 for \(\frac{1}{2} \times 7 \times 9 \times \sin(\angle ABC) = 22\) or \(\sin(\angle ABC) = \frac{44}{63}\)
A1 for one correct angle (44.3 or 135.7)
A1 for both correct angles (44.3 and 135.7) with no extra angles in the range 0 to 180

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

Next, set this volume equal to the volume of a sphere:
\(V_{\text{sphere}} = \frac{4}{3} \pi R^3 = 72\pi\)

Solve for \(R\):
\(\frac{4}{3} R^3 = 72\)
\(R^3 = 72 \times \frac{3}{4} = 54\)
\(R = \sqrt[3]{54} \approx 3.77976... \text{ cm}\)

Rounding to 3 significant figures gives \(3.78\text{ cm}\).

M1 for cylinder volume \(\pi \times 3^2 \times 8\) or \(72\pi\) (or equivalent value, e.g. 226.2)
M1 for equating their cylinder volume to \(\frac{4}{3}\pi R^3\) (e.g. \(R^3 = 54\))
A1 for 3.78 (or 3.779 to 3.78)

Multiply the entire equation by the common denominator, \(x(x+1)\):
\(12(x+1) - 12x = x(x+1)\)

Expand both sides:
\(12x + 12 - 12x = x^2 + x\)
\(12 = x^2 + x\)

Rearrange into a quadratic equation in standard form:
\(x^2 + x - 12 = 0\)

Factor the quadratic equation:
\((x + 4)(x - 3) = 0\)

Solve for \(x\):
\(x = -4\) or \(x = 3\)

M1 for multiplying through by \(x(x+1)\) to get \(12(x+1) - 12x = x(x+1)\) or equivalent common denominator expression
M1 for rearranging into standard quadratic form \(x^2 + x - 12 = 0\)
A1 for both correct solutions: 3 and -4

Let the initial value of the house be \(x\). After the first year, the value is \(1.12x\). After the second year, the value is \(1.12x \times (1 - 0.05) = 1.064x\). Since the final value is $255 360, we set up the equation \(1.064x = 255360\). Solving for \(x\) gives \(x = \frac{255360}{1.064} = 240000\).

M1 for \(x \times 1.12 \times 0.95\) or \(1.064x\) seen or implied (or equivalent reverse percentage step, e.g. \(255360 \div 0.95\)). M1 for \(255360 \div 1.064\) (or \(268800 \div 1.12\)). A1 for 240000.

Multiply the entire equation by the common denominator \((x-3)(x+1)\) to clear the fractions: \(5(x+1) - 3(x-3) = 4\). Expand the brackets: \(5x + 5 - 3x + 9 = 4\). Combine like terms: \(2x + 14 = 4\). Subtract 14 from both sides: \(2x = -10\). Divide by 2: \(x = -5\).

M1 for multiplying by the common denominator to get \(5(x+1) - 3(x-3) = 4\) or equivalent. M1 for correct expansion and simplification leading to \(2x = -10\) or equivalent. A1 for -5.

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the given values: \(13^2 = 7^2 + 8^2 - 2 \cdot 7 \cdot 8 \cdot \cos(ABC)\), which simplifies to \(169 = 49 + 64 - 112 \cos(ABC)\). This gives \(169 = 113 - 112 \cos(ABC)\). Rearranging terms, we get \(112 \cos(ABC) = 113 - 169 = -56\), so \( \cos(ABC) = -0.5\). Thus, \(ABC = \arccos(-0.5) = 120^\circ\).

M1 for correct substitution into Cosine Rule: \(13^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos(B)\). M1 for rearranging to get \(\cos(B) = -0.5\) or \(\cos(B) = \frac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8}\). A1 for 120.

The largest angle is always opposite the longest side. Let \(a = 7\), \(b = 9\), and \(c = 11\). The largest angle, \(C\), is opposite the side of length \(11\text{ cm}\). Using the cosine rule: \(\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}\) \(\cos(C) = \frac{7^2 + 9^2 - 11^2}{2 \times 7 \times 9}\) \(\cos(C) = \frac{49 + 81 - 121}{126}\) \(\cos(C) = \frac{9}{126} = \frac{1}{14}\) \(C = \arccos\left(\frac{1}{14}\right) \approx 85.904^\circ\). Correct to 1 decimal place, the angle is \(85.9^\circ\).

M1 for correct substitution into cosine rule, e.g. \(\cos(C) = \frac{7^2 + 9^2 - 11^2}{2 \times 7 \times 9}\)
M1 for simplifying to \(\cos(C) = \frac{9}{126}\) or \(\frac{1}{14}\) (or equivalent decimal \(0.0714...\))
A1 for \(85.9\) (accept \(85.9^\circ\))

Let the original value of the car be \(x\). After the first year, the value is \(x \times (1 - 0.12) = 0.88x\). After the second year, the value is \(0.88x \times (1 - 0.08) = 0.88x \times 0.92 = 0.8096x\). We are given that this final value is \(\$12\,144\): \(0.8096x = 12\,144\) \(x = \frac{12\,144}{0.8096} = 15\,000\). The original value of the car was \(\$15\,000\).

M1 for \(x \times 0.88 \times 0.92 = 12\,144\) (or equivalent)
M1 for dividing \(12\,144\) by \(0.8096\) (or by \(0.88\) and then by \(0.92\))
A1 for \(15\,000\) (or \(15000\))

First, calculate the volume of the sphere: \(V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288\pi\text{ cm}^3\). Next, write the volume of the cylinder: \(V_{\text{cylinder}} = \pi R^2 h = \pi (8)^2 h = 64\pi h\text{ cm}^3\). Since the sphere is recast into the cylinder, their volumes are equal: \(64\pi h = 288\pi\). Dividing both sides by \(\pi\) gives: \(64h = 288\) which simplifies to \(h = \frac{288}{64} = 4.5\).

M1 for correct formula and substitution for sphere volume, e.g. \(\frac{4}{3} \times \pi \times 6^3\) (or \(288\pi\) or \(904.8\))
M1 for setting up the equation \(\pi \times 8^2 \times h = \text{their } V_{\text{sphere}}\)
A1 for \(4.5\)

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

Substitute the given values:
\(11^2 = 8^2 + 6^2 - 2 \times 8 \times 6 \times \cos(BAC)\)
\(121 = 64 + 36 - 96 \cos(BAC)\)
\(121 = 100 - 96 \cos(BAC)\)

Rearrange to solve for \(\cos(BAC)\):
\(96 \cos(BAC) = 100 - 121\)
\(96 \cos(BAC) = -21\)
\(\cos(BAC) = -\frac{21}{96} = -0.21875\)

Calculate the angle:
\(BAC = \arccos(-0.21875) \approx 102.6346^\circ\)

Correct to 1 decimal place, \(BAC = 102.6^\circ\).

M1 for correct substitution into the Cosine Rule: \(11^2 = 8^2 + 6^2 - 2(8)(6)\cos(\theta)\)
M1 for rearranging to \(\cos(\theta) = -\frac{21}{96}\) (or \(-0.21875\))
A1 for \(102.6\) (accept \(102.63...\))

Multiply the entire equation by the common denominator \(x(x+1)\):
\(12(x+1) - 12x = 1 \times x(x+1)\)

Expand the brackets:
\(12x + 12 - 12x = x^2 + x\)
\(12 = x^2 + x\)

Rearrange into a quadratic equation set to 0:
\(x^2 + x - 12 = 0\)

Factorise the quadratic expression:
\((x + 4)(x - 3) = 0\)

Solve for \(x\):
\(x = -4\) or \(x = 3\).

M1 for clearing fractions correctly: \(12(x+1) - 12x = x(x+1)\) or equivalent
M1 for reducing to standard quadratic form: \(x^2 + x - 12 = 0\)
A1 for both solutions \(3\) and \(-4\)

**(a)**
- Sells 25 candles at \(\$4.50\) each:
\(25 \times 4.50 = \$112.50\)
- Remaining candles: \(40 - 25 = 15\) candles
- Discounted price: \(4.50 \times (1 - 0.20) = \$3.60\) each
- Sells 15 candles at \(\$3.60\) each:
\(15 \times 3.60 = \$54.00\)
- Total revenue: \(112.50 + 54.00 = \$166.50\)
- Total profit: \(166.50 - 120.00 = \$46.50\)
- Percentage profit: \(\frac{46.50}{120} \times 100 = 38.75\%\)

**(b)(i)**
- Simple Interest = \(P \times R \times T\)
- \(1500 \times 0.032 \times 4 = \$192\)

**(b)(ii)**
- Compound Interest amount: \(P(1 + r)^n\)
- \(1500 \times (1.03)^4 = 1500 \times 1.12550881 = \$1688.2632...\)
- To the nearest cent: \(\$1688.26\)

**(c)**
- New price: \(850 \times 1.08 = \$918\)

**(a) [5 marks]**
- M1 for \(25 \times 4.50 = 112.50\)
- M1 for \(15 \times 4.50 \times 0.80 = 54.00\)
- M1 for finding total revenue \(= 166.50\) or total profit \(= 46.50\)
- M1 for dividing profit by cost price \(\frac{\text{their } 46.50}{120}\)
- A1 for \(38.75\%\) (accept \(38.8\%\))

**(b)(i) [2 marks]**
- M1 for \(1500 \times 0.032 \times 4\)
- A1 for \(192\)

**(b)(ii) [3 marks]**
- M1 for \(1500 \times (1.03)^4\) or \(1500 \times 1.03 \times 1.03 \times 1.03 \times 1.03\)
- A1 for \(1688.26\)
- A1 for rounding to nearest cent (must have 2 decimal places if correct method shown)

**(c) [1 mark]**
- B1 for \(918\)

**(a)**
- Volume of a cylinder: \(V = \pi r^2 h\)
- \(V = \pi \times 4^2 \times 15 = 240\pi\)
- \(V \approx 753.982... \text{ cm}^3\)
- Correct to 1 decimal place: \(754.0 \text{ cm}^3\)

**(b)**
- Total surface area of a cylinder: \(A = 2\pi r^2 + 2\pi r h\)
- \(A = 2\pi(4)^2 + 2\pi(4)(15) = 32\pi + 120\pi = 152\pi\)
- \(A \approx 477.522... \text{ cm}^2\)
- Correct to 1 decimal place: \(477.5 \text{ cm}^2\)

**(c)**
- Volume of cuboid = Volume of cylinder
- \(5 \times 5 \times h = 240\pi\) (or \(754.0\))
- \(25h = 240\pi\)
- \(h = \frac{240\pi}{25} = 9.6\pi \approx 30.159... \text{ cm}\)
- Correct to 1 decimal place: \(30.2 \text{ cm}\)

**(d)**
- Density = \(7.8 \text{ g/cm}^3\)
- Mass = \(753.982 \times 7.8 = 5881.06 \text{ g}\)
- Convert to kg: \(5881.06 / 1000 = 5.881... \text{ kg}\)
- Correct to 2 decimal places: \(5.88 \text{ kg}\)

**(a) [3 marks]**
- M1 for \(\pi \times 4^2 \times 15\)
- A1 for \(240\pi\) or \(753.98...\)
- A1 for \(754.0\) or \(754\)

**(b) [3 marks]**
- M1 for formula \(2\pi r^2 + 2\pi r h\) with correct values substituted
- A1 for \(152\pi\) or \(477.52...\)
- A1 for \(477.5\) (accept \(477\) or \(478\) with correct working)

**(c) [3 marks]**
- M1 for setting volume equal to cuboid formula: \(5^2 \times h = \text{their (a)}\)
- M1 for rearranging: \(h = \frac{\text{their (a)}}{25}\)
- A1 for \(30.2\) (accept \(30.1\) to \(30.2\))

**(d) [2 marks]**
- M1 for multiplying volume by \(7.8\) and dividing by \(1000\)
- A1 for \(5.88\) (accept \(5.87\) to \(5.89\))

**(a)(i)**
- Exterior angle of a regular 9-sided polygon: \(\frac{360^\circ}{9} = 40^\circ\)
- Interior angle = \(180^\circ - 40^\circ = 140^\circ\)
- Alternatively, using the formula: \(\frac{(9-2) \times 180^\circ}{9} = \frac{1260^\circ}{9} = 140^\circ\)

**(a)(ii)**
- Sum of interior angles = \((9-2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ\)

**(b)**
- Total ratio parts = \(3 + 4 + 5 = 12\)
- Total degrees in a triangle = \(180^\circ\)
- Size of one part = \(\frac{180^\circ}{12} = 15^\circ\)
- Largest angle is represented by 5 parts: \(5 \times 15^\circ = 75^\circ\)

**(c)(i)**
- Corresponding angles are equal, so:
\(2x + 15 = 5x - 45\)

**(c)(ii)**
- Rearrange the equation:
\(15 + 45 = 5x - 2x\)
\(60 = 3x\)
\(x = 20\)

**(a)(i) [3 marks]**
- M1 for finding the exterior angle: \(360 / 9\) (or equivalent)
- M1 for subtracting from 180: \(180 - \text{their exterior angle}\)
- A1 for \(140\)

**(a)(ii) [2 marks]**
- M1 for \((9-2) \times 180\) or \(140 \times 9\)
- A1 for \(1260\)

**(b) [3 marks]**
- M1 for adding ratios: \(3 + 4 + 5 = 12\)
- M1 for \(\frac{5}{12} \times 180\) (or equivalent)
- A1 for \(75\)

**(c)(i) [1 mark]**
- B1 for \(2x + 15 = 5x - 45\) (or equivalent)

**(c)(ii) [2 marks]**
- M1 for correct rearrangement isolating \(x\) terms and constant terms (e.g., \(3x = 60\))
- A1 for \(x = 20\)

(a) Using Pythagoras' theorem on the cone's height and radius:
\(l^2 = r^2 + h^2 = 3^2 + 4^2 = 9 + 16 = 25\)
\(l = \sqrt{25} = 5\text{ cm}\).

(b) Volume of the cylinder:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi\text{ cm}^3\).
Volume of the cone:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h_{\text{cone}} = \frac{1}{3}\pi \times 3^2 \times 4 = 12\pi\text{ cm}^3\).
Total Volume:
\(V_{\text{total}} = 72\pi + 12\pi = 84\pi\text{ cm}^3\).

(c) Mass = Volume \(\times\) Density
Using \(V = 84\pi \approx 263.894\text{ cm}^3\):
\(\text{Mass} = 263.894 \times 7.85 = 2071.567...\text{ g}\).
To 1 decimal place, this is \(2071.6\text{ g}\).

(d) The total height of the original toy is \(8 + 4 = 12\text{ cm}\).
Since the larger toy is mathematically similar and has a total height of \(36\text{ cm}\), the linear scale factor is:
\(k = \frac{36}{12} = 3\).
Therefore, the volume (and consequently mass) is multiplied by \(k^3 = 3^3 = 27\).
\(\text{Mass of larger toy} = 2071.567 \times 27 \approx 55932.3\text{ g}\) (or \(2071.6 \times 27 = 55933.2\text{ g}\)).

(a) [2.5 marks total]:
- [M1] for applying Pythagoras' theorem: \(3^2 + 4^2\)
- [A1] for \(\sqrt{25}\) or showing \(9 + 16 = 25\)
- [A0.5] for final conclusion leading to \(5\)

(b) [3 marks total]:
- [M1] for calculating cylinder volume: \(\pi \times 3^2 \times 8\) or \(72\pi\)
- [M1] for calculating cone volume: \(\frac{1}{3} \times \pi \times 3^2 \times 4\) or \(12\pi\)
- [A1] for correct sum of \(84\pi\)

(c) [3 marks total]:
- [M1] for multiplying their volume by \(7.85\)
- [A1] for a correct value in the range \(2071.5\) to \(2071.6\)
- [A1] for final answer correctly rounded to 1 decimal place (or FT their value)

(d) [3 marks total]:
- [M1] for finding scale factor \(36 / 12 = 3\)
- [M1] for multiplying their mass in (c) by \(3^3\) (or \(27\))
- [A1] for an answer in the range \(55930\) to \(55935\)

(a) The fixed charge is \(\$4.50\) and the cost per kilometer is \(\$1.80\).
Therefore, the formula is:
\(C = 1.8x + 4.5\) (or \(C = 1.80x + 4.50\)).

(b) Set \(C = 31.50\):
\(1.8x + 4.5 = 31.5\)
Subtract 4.5 from both sides:
\(1.8x = 27\)
Divide by 1.8:
\(x = \frac{27}{1.8} = 15\text{ km}\).

(c) Start with \(C = 1.8x + 4.5\):
Subtract 4.5 from both sides:
\(C - 4.5 = 1.8x\)
Divide both sides by 1.8:
\(x = \frac{C - 4.5}{1.8}\) (or equivalent form like \(x = \frac{10C - 45}{18}\)).

(d) Eco-Ride's cost formula is \(C = 2.3x\).
Set both costs equal to find the intersection point:
\(1.8x + 4.5 = 2.3x\)
Subtract \(1.8x\) from both sides:
\(4.5 = 0.5x\)
Divide by 0.5:
\(x = 9\text{ km}\).

(a) [2 marks total]:
- [M1] for seeing \(1.8x\) or \(+ 4.5\) in an algebraic expression
- [A1] for the fully correct formula \(C = 1.8x + 4.5\) (accept \(C = 1.80x + 4.50\))

(b) [3.5 marks total]:
- [M1] for setting up the equation: \(1.8x + 4.5 = 31.5\)
- [M1.5] for isolating the \(x\) term: \(1.8x = 27\)
- [A1] for \(15\)

(c) [3 marks total]:
- [M1] for subtracting 4.5 from both sides to get \(C - 4.5 = 1.8x\)
- [M1] for dividing by 1.8
- [A1] for correct final formula: \(x = \frac{C - 4.5}{1.8}\) (or algebraic equivalent)

(d) [3 marks total]:
- [M1] for setting up the equation: \(1.8x + 4.5 = 2.3x\)
- [M1] for simplifying to \(0.5x = 4.5\)
- [A1] for \(9\)

(a) Rent spending:
\(35\% \text{ of } 2400 = 0.35 \times 2400 = \$840\).

Food spending:
\(\frac{1}{6} \text{ of } 2400 = \frac{1}{6} \times 2400 = \$400\).

Total spent so far:
\(\$840 \text{ (rent)} + \$400 \text{ (food)} + \$320 \text{ (utilities)} = \$1560\).

Remaining amount:
\(\$2400 - \$1560 = \$840\).

(b) (i) Total parts in the ratio \(2 : 3\) is \(2 + 3 = 5\).
Value of one part:
\(\frac{4500}{5} = \$900\).
Larger share (3 parts):
\(3 \times 900 = \$2700\).

(b) (ii) Simple interest formula: \(I = \frac{P \times R \times T}{100}\).
Here, \(P = 2700\), \(R = 2.5\), and \(T = 4\).
\(I = \frac{2700 \times 2.5 \times 4}{100} = 27 \times 10 = \$270\).
Total value after 4 years:
\(\text{Total} = \text{Principal} + \text{Interest} = 2700 + 270 = \$2970\).

(c) Let the original price of the laptop be \(P\).
A \(20\%\) reduction means the sale price is \(80\%\) of the original price:
\(0.80 \times P = 512\)
\(P = \frac{512}{0.80} = \$640\).

(a) [3 marks total]:
- [M1] for finding rent spending (\(840\)) OR food spending (\(400\))
- [M1] for summing the expenditures: \(840 + 400 + 320 = 1560\)
- [A1] for \(840\)

(b)(i) [1.5 marks total]:
- [M1] for dividing \(4500\) by \(5\) or finding \(\frac{3}{5}\) of \(4500\)
- [A0.5] for obtaining \(2700\) with clear intermediate steps shown

(b)(ii) [3 marks total]:
- [M1] for applying the simple interest formula: \(\frac{2700 \times 2.5 \times 4}{100}\) or finding interest of \(270\)
- [M1] for adding the interest to the principal: \(2700 + 270\)
- [A1] for \(2970\)

(c) [4 marks total]:
- [M1] for recognizing that \(80\%\) corresponds to \(512\) (e.g., \(0.8P = 512\))
- [M2] for \(\frac{512}{0.8}\) or equivalent calculation
- [A1] for \(640\)

(a) Selling price = \(8.40 \times (1 + 0.65) = 8.40 \times 1.65 = 13.86\).

(b) Total cost of 5 mugs in USD = \(5 \times 13.86 = 69.30\).
Total cost in GBP = \(69.30 \div 1.28 = 54.140625\).
Rounding to the nearest penny gives £54.14.

(c) Simple Interest = \(P \times R \times T = 4500 \times 0.024 \times 3 = 324\).
Total Value = Principal + Interest = \(4500 + 324 = 4824\).

(d) Profit = Selling Price - New Cost Price = \(13.86 - 9.03 = 4.83\).
Percentage Profit = \(\frac{4.83}{9.03} \times 100 \approx 53.488\%\), which rounds to 53.5% (correct to 3 significant figures).

(a) [2 marks]
M1 for \(8.40 \times 1.65\)
A1 for 13.86

(b) [3 marks]
M1 for finding the total cost in USD: \(5 \times 13.86 = 69.30\) (or FT their part a)
M1 for dividing by 1.28: \(69.30 \div 1.28\)
A1 for 54.14

(c) [3 marks]
M1 for \(4500 \times 0.024 \times 3\) (or 324)
M1 for adding interest to 4500: \(4500 + 324\)
A1 for 4824

(d) [3.5 marks]
M1 for calculating the profit: \(13.86 - 9.03 = 4.83\) (or FT their part a)
M1.5 for \(\frac{\text{their profit}}{9.03} \times 100\)
A1 for 53.5% (accept 53.49%)

(a) Area of cross-section = \(\frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (90 + 50) \times 21 = 70 \times 21 = 1470\text{ cm}^2\).

(b) The horizontal distance for the slant height on each side is \(\frac{90 - 50}{2} = 20\text{ cm}\).
Using Pythagoras' theorem, the slant height \(s\) is:
\(s = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29\text{ cm}\).

(c) The external surface area is made of:
- Bottom face: \(50 \times 150 = 7500\text{ cm}^2\)
- Two slanted side faces: \(2 \times (29 \times 150) = 8700\text{ cm}^2\)
- Two trapezoidal end faces: \(2 \times 1470 = 2940\text{ cm}^2\)
Total Surface Area = \(7500 + 8700 + 2940 = 19140\text{ cm}^2\).

(d) Volume = Cross-sectional area \(\times\) length = \(1470 \times 150 = 220,500\text{ cm}^3\).
Since \(1\text{ litre} = 1000\text{ cm}^3\):
Volume in litres = \(220,500 \div 1000 = 220.5\text{ litres}\).

(a) [2 marks]
M1 for \(\frac{1}{2} \times (90 + 50) \times 21\)
A1 for 1470

(b) [3 marks]
M1 for finding the horizontal offset: \(\frac{90 - 50}{2} = 20\)
M1 for substituting into Pythagoras: \(\sqrt{20^2 + 21^2}\)
A1 for showing that \(\sqrt{841} = 29\)

(c) [3.5 marks]
M1 for bottom area: \(50 \times 150 = 7500\)
M1 for two slanted sides: \(2 \times 29 \times 150 = 8700\) (or FT their slant height)
M1 for two end faces: \(2 \times 1470 = 2940\) (or FT their part a)
A0.5 for adding all components together to get 19140

(d) [3 marks]
M1 for \(1470 \times 150\) (or FT their part a)
A1 for \(220,500\text{ cm}^3\)
A1 for converting to litres: 220.5

(a) Substituting \(x\) values:
- For \(x = -1\): \(y = (-1)^2 - 3(-1) - 2 = 1 + 3 - 2 = 2\)
- For \(x = 2\): \(y = (2)^2 - 3(2) - 2 = 4 - 6 - 2 = -4\)
- For \(x = 4\): \(y = (4)^2 - 3(4) - 2 = 16 - 12 - 2 = 2\)

(b) The minimum point lies exactly halfway between the turning points. The line of symmetry is at the average of the x-coordinates of symmetric points, e.g., halfway between 1 and 2, which is \(x = 1.5\).

(c) Solving \(x^2 - 3x - 2 = 0\) is equivalent to finding where the curve crosses the x-axis (where \(y = 0\)). From the graph, these points are approximately \(x \approx -0.56\) and \(x \approx 3.56\) (accept values in the range \([-0.6, -0.5]\) and \([3.5, 3.6]\)).

(d) Drawing the line \(y = x - 2\). At \(x = 0\), \(y = -2\). At \(x = 4\), \(y = 2\).
The points of intersection of this straight line and the curve are where:
\(x^2 - 3x - 2 = x - 2\)
\(x^2 - 4x = 0\)
\(x(x - 4) = 0\)
This yields \(x = 0\) and \(x = 4\).
The corresponding coordinates are \((0, -2)\) and \((4, 2)\).

(a) [3 marks]
B1 for [i] 2
B1 for [ii] -4
B1 for [iii] 2

(b) [1.5 marks]
M1 for finding 1.5
A0.5 for writing the full equation as \(x = 1.5\)

(c) [3 marks]
B1.5 for one correct root (accept range \([-0.6, -0.5]\))
B1.5 for the second correct root (accept range \([3.5, 3.6]\))

(d) [4 marks]
M1 for drawing the line \(y = x - 2\) correctly
A1.5 for first point of intersection \((0, -2)\)
A1.5 for second point of intersection \((4, 2)\)

(a)(i) \text{Time taken by car} = \frac{120}{x} \text{ hours}.
(ii) \text{Time taken by coach} = \frac{120}{x - 10} \text{ hours}.

(b) Since 36 minutes is equivalent to \(\frac{36}{60} = \frac{3}{5}\) hours, we can write the equation:
$$\frac{120}{x - 10} - \frac{120}{x} = \frac{3}{5}$$
Multiply both sides by \(5x(x - 10)\):
$$5 \times 120x - 5 \times 120(x - 10) = 3x(x - 10)$$
$$600x - 600x + 6000 = 3x^2 - 30x$$
$$6000 = 3x^2 - 30x$$
Divide the entire equation by 3:
$$2000 = x^2 - 10x$$
$$x^2 - 10x - 2000 = 0$$

(c) Factoring the quadratic equation:
$$(x - 50)(x + 40) = 0$$
This gives the solutions:
$$x = 50 \quad \text{or} \quad x = -40$$

(d) Since speed must be positive, \(x = 50\) km/h.
The time taken by the car is:
$$\text{Time} = \frac{120}{50} = 2.4 \text{ hours}$$
To convert 2.4 hours into hours and minutes:
$$2.4 \text{ hours} = 2 \text{ hours and } 0.4 \times 60 \text{ minutes} = 2 \text{ hours and } 24 \text{ minutes}$$

(a)(i) B1 for \(\frac{120}{x}\)
(ii) B1 for \(\frac{120}{x - 10}\)

(b) M1 for establishing \(\frac{120}{x - 10} - \frac{120}{x} = \frac{36}{60}\) oe
M1 for combining fractions: \(\frac{120x - 120(x - 10)}{x(x - 10)} = \frac{3}{5}\)
M1 for clearing fractions: \(6000 = 3(x^2 - 10x)\) oe
A1 for completing the algebra to show \(x^2 - 10x - 2000 = 0\) with no errors seen

(c) M2 for \((x - 50)(x + 40)\) (M1 for \((x + a)(x + b)\) where \(ab = -2000\) or \(a+b = -10\)) or for correct application of quadratic formula with at most one arithmetic error
A1 for \(x = 50\) and \(x = -40\)

(d) B1 for choosing \(x = 50\) as the only valid speed
M1 for \(\frac{120}{\text{their } 50}\) (or \(2.4\) hours)
M1 for converting the fractional part of their hours into minutes (e.g., \(0.4 \times 60\))
A1 for 2 hours 24 minutes

(a) The radius of the hemisphere is \(r = 6\text{ cm}\), so its height is also \(6\text{ cm}\).
Therefore, the height of the cone is:
$$h = 14 - 6 = 8\text{ cm}$$

$$\text{Volume of hemisphere} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi (6)^3 = 144\pi\text{ cm}^3$$
$$\text{Volume of cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (8) = 96\pi\text{ cm}^3$$
$$\text{Total Volume} = 144\pi + 96\pi = 240\pi\text{ cm}^3$$

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

The total surface area is the sum of the curved surface area of the cone and the hemisphere:
$$\text{Curved surface area of cone} = \pi r l = \pi (6)(10) = 60\pi\text{ cm}^2$$
$$\text{Curved surface area of hemisphere} = 2\pi r^2 = 2\pi (6)^2 = 72\pi\text{ cm}^2$$
$$\text{Total Surface Area} = 60\pi + 72\pi = 132\pi \approx 414.69... \approx 415\text{ cm}^2$$

(c) Let the volume of the recast sphere be equal to the volume of the ornament:
$$\frac{4}{3} \pi R^3 = 240\pi$$
$$R^3 = 240 \times \frac{3}{4} = 180$$
$$R = \sqrt[3]{180} \approx 5.6462... \approx 5.65\text{ cm}$$

(d) The ratio of the surface areas of the two similar ornaments is:
$$\text{Area scale factor} = \frac{297\pi}{132\pi} = 2.25$$

Therefore, the linear scale factor \(k\) is:
$$k = \sqrt{2.25} = 1.5$$

The volume scale factor is:
$$k^3 = 1.5^3 = 3.375$$

$$\text{Volume of the larger ornament} = 240\pi \times 3.375 = 810\pi\text{ cm}^3$$

(a) B1 for finding the height of the cone is \(8\text{ cm}\)
M1 for \(\frac{2}{3} \times \pi \times 6^3\) (hemisphere volume)
M1 for \(\frac{1}{3} \times \pi \times 6^2 \times 8\) (cone volume)
A1 for completing calculations to show \(144\pi + 96\pi = 240\pi\) with no errors

(b) M1 for finding slant height \(l = \sqrt{6^2 + 8^2}\) (or \(10\))
M1 for \(\pi \times 6 \times (\text{their } 10)\) (curved area of cone)
M1 for \(2 \times \pi \times 6^2\) (curved area of hemisphere)
A1 for \(415\) (or \(414.6\) to \(414.7\))

(c) M1 for equating \(\frac{4}{3}\pi R^3 = 240\pi\)
M1 for \(R = \sqrt[3]{180}\) or \(\sqrt[3]{\frac{3 \times 240}{4}}\)
A1 for \(5.65\) (accept \(5.646...\))

(d) M1 for finding the linear scale factor \(k = \sqrt{\frac{297}{132}} = 1.5\) (or volume scale factor \(1.5^3 = 3.375\))
A1 for \(810\pi\)

(a) Let North at \(P\) be \(N_P\) and North at \(Q\) be \(N_Q\).
Since the North lines \(N_P\) and \(N_Q\) are parallel, the angle between the line \(PQ\) and the South direction at \(Q\) is \(65^\circ\) (alternate angles).
Therefore, the bearing of \(P\) from \(Q\) is:
$$180^\circ + 65^\circ = 245^\circ$$

The bearing of \(R\) from \(Q\) is given as \(140^\circ\).
The angle \(PQR\) is the difference between these two bearings:
$$\text{Angle } PQR = 245^\circ - 140^\circ = 105^\circ$$

(b) Using the Cosine Rule on triangle \(PQR\):
$$PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(PQR)$$
$$PR^2 = 120^2 + 150^2 - 2(120)(150)\cos(105^\circ)$$
$$PR^2 = 14400 + 22500 - 36000\cos(105^\circ)$$
$$PR^2 \approx 36900 - 36000(-0.258819)$$
$$PR^2 \approx 36900 + 9317.49 = 46217.49$$
$$PR = \sqrt{46217.49} \approx 214.98... \approx 215\text{ m}$$

(c) Using the area formula for non-right-angled triangles:
$$\text{Area of triangle } PQR = \frac{1}{2} \cdot PQ \cdot QR \cdot \sin(PQR)$$
$$\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(105^\circ)$$
$$\text{Area} = 9000 \times 0.9659258 \approx 8693.33... \approx 8690\text{ m}^2$$

(d)(i) In the right-angled triangle \(PQT\), where \(\angle PQT = 90^\circ\):
$$\tan(12^\circ) = \frac{QT}{PQ} = \frac{QT}{120}$$
$$QT = 120 \tan(12^\circ) \approx 25.506... \approx 25.5\text{ m}$$

(d)(ii) In the right-angled triangle \(RQT\), where \(\angle RQT = 90^\circ\):
Let \(\theta\) be the angle of elevation of \(T\) from \(R\).
$$\tan(\theta) = \frac{QT}{QR} = \frac{25.5068}{150} \approx 0.170045$$
$$\theta = \tan^{-1}(0.170045) \approx 9.649...^\circ \approx 9.6^\circ$$

(a) M1 for identifying bearing of \(P\) from \(Q\) is \(245^\circ\) (or alternate angle with North at \(Q\) is \(65^\circ\))
M1 for subtracting the bearing of \(R\) (i.e. \(245^\circ - 140^\circ\)) or equivalent angle subtraction
A1 for showing fully correct working leading to \(105^\circ\)

(b) M1 for \(120^2 + 150^2 - 2(120)(150)\cos(105^\circ)\)
A1 for evaluation of \(PR^2\) between \(46200\) and \(46220\)
A1 for \(215\) (accept \(214.98...\))

(c) M1 for \(\frac{1}{2} \times 120 \times 150 \times \sin(105^\circ)\)
A1 for \(8690\) (accept \(8693...\))

(d)(i) M1 for \(\tan(12^\circ) = \frac{QT}{120}\) or \(120 \tan(12^\circ)\)
A1 for \(25.5\) (accept \(25.50\) to \(25.51\))

(d)(ii) M1 for \(\tan(\theta) = \frac{\text{their } QT}{150}\)
M1 for \(\theta = \tan^{-1}\left(\frac{\text{their } QT}{150}\right)\)
A1 for \(9.6\) (accept \(9.64\) to \(9.65\))

(a) Using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(64^\circ)\)
\(AC^2 = 14400 + 22500 - 36000 \times 0.43837 = 21118.6\)
\(AC = \sqrt{21118.6} \approx 145.32\text{ m}\) (or \(145\text{ m}\) to 3 significant figures).

(b) Using the Sine Rule:
\(\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(ACB)}{120} = \frac{\sin(64^\circ)}{145.32}\)
\(\sin(ACB) = \frac{120 \times \sin(64^\circ)}{145.32} \approx 0.7422\)
\(ACB = \sin^{-1}(0.7422) \approx 47.92^\circ\) (or \(47.9^\circ\) to 1 decimal place).

(c) Area of triangle:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(64^\circ) = 9000 \times 0.89879 \approx 8089.1\text{ m}^2\) (or \(8090\text{ m}^2\) to 3 significant figures).

(d) Let the top of the mast be \(T\). In right-angled triangle \(TAC\):
\(\tan(18^\circ) = \frac{h}{AC}\)
\(h = 145.32 \times \tan(18^\circ) \approx 47.218\text{ m}\).
In right-angled triangle \(TAB\), let the angle of elevation from \(B\) be \(\theta\):
\(\tan(\theta) = \frac{h}{AB} = \frac{47.218}{120} \approx 0.39348\)
\(\theta = \tan^{-1}(0.39348) \approx 21.48^\circ\) (or \(21.5^\circ\) to 1 decimal place).

(a) [M1] for \(120^2 + 150^2 - 2(120)(150)\cos(64)\)
[A1] for \(21118.6...\)
[M1] for \(\sqrt{\text{their } 21118.6}\)
[A1] for \(145\text{ m}\) or \(145.3\text{ m}\)

(b) [M1] for \(\frac{\sin(ACB)}{120} = \frac{\sin(64)}{\text{their } AC}\)
[M1] for rearranging to find \(\sin(ACB)\)
[A1] for \(47.9^\circ\) (accept \(47.9\) to \(47.92\))

(c) [M1] for \(0.5 \times 120 \times 150 \times \sin(64)\)
[A1] for \(8090\) (or \(8089\))

(d) [M1] for \(h = AC \times \tan(18)\)
[A1] for \(h = 47.2\) or \(47.22\)
[M1] for \(\tan(\theta) = \frac{\text{their } h}{120}\)
[A1] for \(21.5^\circ\) (accept \(21.48\) to \(21.5\))

(a) Volume of cylinder = \(\pi \times r^2 \times h_2 = \pi \times 6^2 \times 14 = 504\pi\text{ cm}^3\).
Volume of cone = \(\frac{1}{3} \pi \times r^2 \times h_1 = \frac{1}{3} \pi \times 6^2 \times h_1 = 12\pi h_1\text{ cm}^3\).
Total volume = \(504\pi + 12\pi h_1 = 600\pi\).
\(12\pi h_1 = 96\pi \implies h_1 = 8\text{ cm}\).

(b) Total surface area comprises the cylinder flat base, the cylinder curved surface, and the cone curved surface.
Flat base of cylinder = \(\pi \times r^2 = \pi \times 6^2 = 36\pi\).
Curved surface of cylinder = \(2 \pi r h_2 = 2 \times \pi \times 6 \times 14 = 168\pi\).
Slant height of cone, \(l = \sqrt{r^2 + h_1^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}\).
Curved surface of cone = \(\pi r l = \pi \times 6 \times 10 = 60\pi\).
Total surface area = \(36\pi + 168\pi + 60\pi = 264\pi\text{ cm}^2\).

(c) Mass = \(\text{Volume} \times \text{density} = 600\pi \times 0.75 = 450\pi\text{ g}\).
\(450\pi \approx 1413.72\text{ g}\).
To convert to kilograms: \(\frac{1413.72}{1000} \approx 1.41\text{ kg}\) (to 3 significant figures).

(d) Total height of original toy = \(14 + 8 = 22\text{ cm}\).
Volume scale factor = \(\frac{150\pi}{600\pi} = \frac{1}{4} = 0.25\).
Linear scale factor \(k = \sqrt[3]{0.25} \approx 0.62996\).
Total height of the smaller toy = \(22 \times 0.62996 \approx 13.86\text{ cm}\) (or \(13.9\text{ cm}\) to 3 significant figures).

(a) [M1] for cylinder volume \(\pi \times 6^2 \times 14 = 504\pi\)
[M1] for cone volume \(\frac{1}{3} \pi \times 6^2 \times h_1 = 12\pi h_1\)
[A1] for fully correct completion showing \(h_1 = 8\)

(b) [M1] for finding slant height \(l = \sqrt{6^2+8^2}=10\)
[M1] for setting up any of the three surface components correctly (\(36\pi\), \(168\pi\), or \(60\pi\))
[M1] for adding three correct components
[A1] for \(264\pi\)

(c) [M1] for calculating mass in grams: \(600\pi \times 0.75\)
[M1] for division of mass in grams by 1000
[A1] for \(1.41\) (accept \(1.413\) to \(1.414\))

(d) [M1] for scale factor \(k = \sqrt[3]{0.25}\) (or \(0.63\))
[M1] for multiplying \(22\) by their \(k\)
[A1] for \(13.9\) (accept \(13.85\) to \(13.9\))

(a) Cost price of one notebook is \(\frac{240}{x}\).

(b) Number of remaining notebooks is \(x - 6\).
Selling price of one notebook is \(\frac{240}{x} + 3\).
Total amount received = \((x - 6)\left(\frac{240}{x} + 3\right)\).

(c) Profit = Total received - Total cost
\((x - 6)\left(\frac{240}{x} + 3\right) - 240 = 24\)
\((x - 6)\left(\frac{240 + 3x}{x}\right) = 264\)
Multiply both sides by \(x\):
\((x - 6)(240 + 3x) = 264x\)
\(240x + 3x^2 - 1440 - 18x = 264x\)
\(3x^2 + 222x - 1440 = 264x\)
\(3x^2 - 42x - 1440 = 0\)
Divide the entire equation by 3:
\(x^2 - 14x - 480 = 0\) (as required).

(d) To factorise, we need two numbers that multiply to \(-480\) and add to \(-14\). These are \(-30\) and \(+16\).
\((x - 30)(x + 16) = 0\)
\(x = 30\) or \(x = -16\).

(e) Since \(x\) must be positive, \(x = 30\).
Cost price of one notebook = \(\frac{240}{30} = 8\).
Actual selling price = \(8 + 3 = \$11\).

(a) [B1] for \(\frac{240}{x}\)

(b) [B1] for \(x - 6\) seen or \(\frac{240}{x} + 3\) seen
[B1] for \((x - 6)\left(\frac{240}{x} + 3\right)\)

(c) [M1] for setting up the equation \((x - 6)\left(\frac{240}{x} + 3\right) - 240 = 24\)
[M1] for multiplying by \(x\) to clear the fraction correctly
[M1] for expanding brackets to reach \(240x + 3x^2 - 1440 - 18x\)
[M1] for gathering terms to form a quadratic equation equal to zero
[A1] for dividing by 3 with no errors seen to obtain \(x^2 - 14x - 480 = 0\)

(d) [M1] for \((x - 30)(x + 16)\)
[A1] for \(x = 30\)
[A1] for \(x = -16\)

(e) [M1] for choosing positive root \(x = 30\) and finding \(\frac{240}{30} + 3\)
[A1] for \(11\)

**(a)** Time is given by distance divided by speed.
Time taken = \(\frac{40}{v}\) hours.

**(b)** The speed is increased by 4 km/h, so the new speed is \(v+4\) km/h.
Time taken = \(\frac{40}{v+4}\) hours.

**(c)** The difference in time is 30 minutes, which is \(\frac{30}{60} = \frac{1}{2}\) hour.
We set up the equation:
\[\frac{40}{v} - \frac{40}{v+4} = \frac{1}{2}\]
Multiply both sides by the common denominator \(2v(v+4)\):
\[2(40)(v+4) - 2(40)v = v(v+4)\]
\[80v + 320 - 80v = v^2 + 4v\]
\[320 = v^2 + 4v\]
\[v^2 + 4v - 320 = 0\]
This is the required equation.

**(d)** We solve \(v^2 + 4v - 320 = 0\) by factorisation:
We look for two numbers that multiply to \(-320\) and add to \(4\).
These numbers are \(+20\) and \(-16\).
\[(v + 20)(v - 16) = 0\]
So, \(v = -20\) or \(v = 16\).

**(e)** Since speed must be positive, we choose the solution \(v = 16\) km/h.
The outbound speed on the second day is \(v + 4 = 16 + 4 = 20\) km/h.
The distance is 40 km.
Time = \(\frac{40}{20} = 2\) hours.

**(f)** For the entire trip on the second day:
- Outbound distance = 40 km, Outbound time = 2 hours.
- Return distance = 45 km, Return speed = 15 km/h. Return time = \(\frac{45}{15} = 3\) hours.
- Total distance = \(40 + 45 = 85\) km.
- Total time = \(2 + 3 = 5\) hours.
Average speed = \(\frac{\text{Total Distance}}{\text{Total Time}} = \frac{85}{5} = 17\) km/h.

(a) B1 for \(\frac{40}{v}\) or equivalent.
(b) B1 for \(\frac{40}{v+4}\) or equivalent.
(c) M1 for \(\frac{40}{v} - \frac{40}{v+4} = \frac{1}{2}\) (or 0.5 or \(\frac{30}{60}\))
M1 for correct algebraic steps to clear denominators: \(80(v+4) - 80v = v(v+4)\) or \(\frac{40(v+4) - 40v}{v(v+4)} = 0.5\)
A1 for expanding to get \(80v + 320 - 80v = v^2 + 4v\) or equivalent
A1 for final step showing \(v^2 + 4v - 320 = 0\) without any errors.
(d) M2 for \((v+20)(v-16) = 0\) or correct use of quadratic formula: \(v = \frac{-4 \pm \sqrt{4^2 - 4(1)(-320)}}{2}\)
A1 for \(v = 16\) and \(v = -20\).
(e) M1 for substituting \(v = 16\) to find the speed \(20\) km/h or time expression \(\frac{40}{16+4}\).
A1 for 2 hours.
(f) M1 for finding total distance = 85 km or return time = 3 hours.
A1 for 17 km/h.

**(a)**
(i) For \(x = 0.5\):
\[y = 2(0.5) + \frac{8}{0.5^2} - 5 = 1 + \frac{8}{0.25} - 5 = 1 + 32 - 5 = 28\]
(ii) For \(x = 1.5\):
\[y = 2(1.5) + \frac{8}{1.5^2} - 5 = 3 + \frac{8}{2.25} - 5 = 3 + 3.555... - 5 \approx 1.56\]
(iii) For \(x = 4\):
\[y = 2(4) + \frac{8}{4^2} - 5 = 8 + \frac{8}{16} - 5 = 8 + 0.5 - 5 = 3.5\]

**(b)** As \(x\) becomes very large, the term \(\frac{8}{x^2}\) approaches 0. Therefore, \(y\) approaches \(2x - 5\) (or \(y\) increases towards infinity, behaving like the line \(y = 2x - 5\)).

**(c)** Write \(y = 2x + 8x^{-2} - 5\).
Differentiate with respect to \(x\):
\[\frac{dy}{dx} = 2 - 16x^{-3} = 2 - \frac{16}{x^3}\]
At the minimum point \(M\), the gradient is 0:
\[2 - \frac{16}{x^3} = 0\]
\[2 = \frac{16}{x^3} \implies x^3 = 8 \implies x = 2\]
Substitute \(x = 2\) back into the original curve equation:
\[y = 2(2) + \frac{8}{2^2} - 5 = 4 + 2 - 5 = 1\]
So the coordinates of \(M\) are \((2, 1)\).

**(d)**
(i) The equation is of the form \(y_{\text{curve}} = y_{\text{line}}\).
Here, \(y_{\text{curve}} = 2x + \frac{8}{x^2} - 5\), so the line equation is \(y = 8 - x\).
(ii) Start with:
\[2x + \frac{8}{x^2} - 5 = 8 - x\]
Add \(x\) and subtract 8 from both sides:
\[3x - 13 + \frac{8}{x^2} = 0\]
Multiply the entire equation by \(x^2\) (since \(x > 0\)):
\[3x^3 - 13x^2 + 8 = 0\]
This is the required form.

**(e)** At \(x = 1\), the y-coordinate is:
\[y = 2(1) + \frac{8}{1^2} - 5 = 5\]
The gradient of the tangent is given by the derivative at \(x = 1\):
\[m = \left.\frac{dy}{dx}\right|_{x=1} = 2 - \frac{16}{1^3} = -14\]
Using the equation of a straight line \(y - y_1 = m(x - x_1)\):
\[y - 5 = -14(x - 1)\]
\[y = -14x + 14 + 5\]
\[y = -14x + 19\]

(a)(i) B1 for 28
(a)(ii) B1 for 1.56
(a)(iii) B1 for 3.5
(b) B1 for stating that \(y\) increases, or approaches \(2x - 5\), or \(\frac{8}{x^2}\) becomes negligible.
(c) M1 for differentiating \(2x\) to 2 or \(8x^{-2}\) to \(-16x^{-3}\)
A1 for correct derivative \(\frac{dy}{dx} = 2 - \frac{16}{x^3}\)
M1 for setting derivative to 0 and solving to find \(x = 2\)
A1 for \(y = 1\) leading to \((2, 1)\).
(d)(i) B1 for \(y = 8 - x\)
(d)(ii) M1 for combining terms: \(3x - 13 + \frac{8}{x^2} = 0\) or equivalent
A1 for multiplying by \(x^2\) and obtaining the exact form \(3x^3 - 13x^2 + 8 = 0\).
(e) M1 for finding \(y = 5\) or gradient \(m = -14\) at \(x = 1\).
A1 for \(y = -14x + 19\).

**(a)** Draw a North-South line through \(B\).
Let the North line pointing up from \(A\) be \(N_A\) and the North line pointing up from \(B\) be \(N_B\).
The bearing of \(B\) from \(A\) is \(060^\circ\).
The interior angle between the line \(AB\) and the South line at \(B\) is \(060^\circ\) (alternate angles).
Thus, the bearing of \(A\) from \(B\) is \(180^\circ + 60^\circ = 240^\circ\).
The bearing of \(C\) from \(B\) is \(135^\circ\).
The angle \(\angle ABC\) is the difference between these two bearings:
\[\angle ABC = 240^\circ - 135^\circ = 105^\circ\]
Alternatively, the angle from \(BC\) to the South line at \(B\) is \(180^\circ - 135^\circ = 45^\circ\).
The angle from \(AB\) to the South line at \(B\) is \(60^\circ\).
Therefore, \(\angle ABC = 60^\circ + 45^\circ = 105^\circ\).

**(b)** Using the Cosine Rule in triangle \(ABC\) to find \(AC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]
\[AC^2 = 12^2 + 15^2 - 2(12)(15)\cos(105^\circ)\]
\[AC^2 = 144 + 225 - 360\cos(105^\circ)\]
Using \(\cos(105^\circ) \approx -0.258819\):
\[AC^2 = 369 - 360(-0.258819) = 369 + 93.175 = 462.175\]
\[AC = \sqrt{462.175} \approx 21.50 \text{ km}\]
So, the direct distance is \(21.5\) km (to 3 significant figures).

**(c)** To find the bearing of \(C\) from \(A\), we first find the angle \(\angle BAC\).
Using the Sine Rule:
\[\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}\]
\[\frac{\sin(\angle BAC)}{15} = \frac{\sin(105^\circ)}{21.50}\]
\[\sin(\angle BAC) = \frac{15 \sin(105^\circ)}{21.50} \approx \frac{15 \times 0.965926}{21.50} \approx 0.6739\]
\[\angle BAC = \arcsin(0.6739) \approx 42.37^\circ\]
The bearing of \(B\) from \(A\) is \(060^\circ\).
The bearing of \(C\) from \(A\) is:
\[\text{Bearing} = 060^\circ + \angle BAC = 60^\circ + 42.37^\circ = 102.37^\circ\]
So the bearing is \(102.4^\circ\) (to 1 decimal place).

**(d)** The area of triangle \(ABC\) is:
\[\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\]
\[\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin(105^\circ)\]
\[\text{Area} = 90 \times \sin(105^\circ) \approx 86.93 \text{ km}^2\]
So the area is \(86.9\) km\(^2\) (to 3 significant figures).

**(e)** The shortest distance from \(B\) to the path \(AC\) is the perpendicular height, \(h\), of the triangle \(ABC\) with base \(AC\).
Using the area formula:
\[\text{Area} = \frac{1}{2} \times AC \times h\]
\[86.93 = \frac{1}{2} \times 21.50 \times h\]
\[h = \frac{2 \times 86.93}{21.50} \approx 8.087 \text{ km}\]
Alternatively, using right-angled trigonometry:
\[h = AB \times \sin(\angle BAC) = 12 \times \sin(42.37^\circ) \approx 12 \times 0.6739 = 8.09 \text{ km}\]
The shortest distance is \(8.09\) km (to 3 significant figures).

(a) M1 for alternate angles showing the angle with South at B is \(60^\circ\), or for finding the back-bearing of A from B is \(240^\circ\).
A1 for \(60^\circ + 45^\circ = 105^\circ\) or \(240^\circ - 135^\circ = 105^\circ\) with a clear explanation or diagrammatic justification.
(b) M1 for substituting into the Cosine Rule: \(12^2 + 15^2 - 2(12)(15)\cos(105^\circ)\)
A1 for \(AC^2 \approx 462.17\)
A1 for \(AC = 21.5\) km or \(21.50\) km (accept 21.5 to 21.51)
(c) M1 for using the Sine Rule: \(\frac{\sin(\angle BAC)}{15} = \frac{\sin(105^\circ)}{\text{their } AC}\)
A1 for \(\sin(\angle BAC) \approx 0.6739\) or \(\angle BAC \approx 42.4^\circ\)
M1 for adding \(60^\circ\) to their \(\angle BAC\)
A1 for bearing of \(102.4^\circ\) or \(102^\circ\) (accept 102.3 to 102.5)
(d) M1 for \(\frac{1}{2} \times 12 \times 15 \times \text{sin}(105^\circ)\)
A1 for \(86.9\) km\(^2\) or \(86.93\) km\(^2\) (accept 86.9 to 87.0)
(e) M1 for \(\text{their Area} = \frac{1}{2} \times \text{their } AC \times h\) or \(12 \sin(\text{their } \angle BAC)\)
A1 for \(8.09\) km (accept 8.08 to 8.10)

**(a)** Using the Cosine Rule to find \(QR\):
\(QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(QPR)\)
\(QR^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(64^\circ)\)
\(QR^2 = 14400 + 22500 - 36000 \times 0.438371...\)
\(QR^2 = 36900 - 15781.36...\)
\(QR^2 = 21118.63...\)
\(QR = \sqrt{21118.63...} = 145.322...\text{ m}\)
Rounding to the nearest metre gives \(145\text{ m}\).

**(b)** Using the Sine Rule to find angle \(PRQ\):
\(\frac{\sin(PRQ)}{PQ} = \frac{\sin(QPR)}{QR}\)
Using the more accurate value of \(QR = 145.32\):
\(\frac{\sin(PRQ)}{120} = \frac{\sin(64^\circ)}{145.32}\)
\(\sin(PRQ) = \frac{120 \times \sin(64^\circ)}{145.32} \approx 0.74218\)
\(\text{angle } PRQ = \sin^{-1}(0.74218) \approx 47.917^\circ\)
Thus, angle \(PRQ = 47.9^\circ\) (to 1 d.p.).
*(Note: If using \(QR = 145\), \(\sin(PRQ) = \frac{120 \times \sin(64^\circ)}{145} \approx 0.74383\), giving angle \(PRQ = 48.1^\circ\).)*

**(c)** Area of triangle \(PQR\):
\(\text{Area} = \frac{1}{2} \times PQ \times PR \times \sin(QPR)\)
\(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(64^\circ)\)
\(\text{Area} = 9000 \times 0.898794... \approx 8089.15...\text{ m}^2\)
This rounds to \(8090\text{ m}^2\) (to 3 s.f.) or \(8089\text{ m}^2\).

**(d)** The shortest distance from \(P\) to \(QR\) is the perpendicular height, \(h\):
Using \(\text{Area} = \frac{1}{2} \times QR \times h\):
\(8089.15 = \frac{1}{2} \times 145.32 \times h\)
\(h = \frac{2 \times 8089.15}{145.32} \approx 111.33...\text{ m}\)
Alternatively, using right-angled trigonometry:
\(h = PR \times \sin(PRQ) = 150 \times \sin(47.917^\circ) \approx 111.33...\text{ m}\)
This rounds to \(111\text{ m}\) (to 3 s.f.).
*(Note: If using angle \(48.1^\circ\), \(h = 150 \times \sin(48.1^\circ) \approx 111.6\text{ m}\) or \(112\text{ m}\).)*

**(e)** Let the height of the mast be \(PT\). In the right-angled triangle \(TPQ\):
\(\tan(18.5^\circ) = \frac{PT}{PQ}\)
\(PT = 120 \times \tan(18.5^\circ) \approx 40.151...\text{ m}\)
Now, in the right-angled triangle \(TPR\), let \(\theta\) be the angle of elevation of \(T\) from \(R\):
\(\tan(\theta) = \frac{PT}{PR}\)
\(\tan(\theta) = \frac{40.151}{150} \approx 0.26767\)
\(\theta = \tan^{-1}(0.26767) \approx 14.98^\circ\)
Thus, the angle of elevation is \(15.0^\circ\) (to 1 d.p.) or \(15^\circ\).

**(a)** [3 marks]
M1: for correct substitution into the Cosine Rule, i.e., \(120^2 + 150^2 - 2(120)(150)\cos(64)\)
A1: for evaluation of the Cosine Rule terms to \(21118.6...\) or similar
A1: for showing \(\sqrt{21118.6...} = 145.3...\) and concluding \(145\text{ m}\) (must see a more accurate value before rounding to 145)

**(b)** [3 marks]
M1: for correct substitution into the Sine Rule, e.g., \(\frac{\sin(PRQ)}{120} = \frac{\sin(64)}{\text{their } QR}\)
M1: for rearranging to find \(\sin(PRQ) = \frac{120 \sin(64)}{\text{their } QR}\)
A1: for \(47.9^\circ\) (or \(47.91...^\circ\)) from accurate \(QR\), or \(48.1^\circ\) (or \(48.05...^\circ\)) from using \(QR = 145\)

**(c)** [2 marks]
M1: for \(\frac{1}{2} \times 120 \times 150 \times \sin(64)\)
A1: for \(8090\) or \(8089\) or \(8089.15...\)

**(d)** [2 marks]
M1: for a correct method to find the perpendicular height, e.g., \(\text{their Area} = \frac{1}{2} \times \text{their } QR \times d\) OR \(150 \times \sin(\text{their } PRQ)\)
A1: for \(111\) or \(111.3\) (accept answers in the range \([111, 112]\) depending on previous roundings)

**(e)** [3 marks]
M1: for \(PT = 120 \tan(18.5)\) (evaluates to \(40.15...\))
M1: for \(\tan(\theta) = \frac{\text{their } PT}{150}\)
A1: for \(15.0^\circ\) or \(15^\circ\) (accept \(14.98^\circ\))