2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Expand the brackets: \(12x - 20 = 2x + 15\). Subtract \(2x\) from both sides: \(10x - 20 = 15\). Add 20 to both sides: \(10x = 35\). Divide by 10: \(x = 3.5\).

M1 for correct expansion of the bracket: \(12x - 20\). M1 for collecting like terms: \(10x = 35\) (or equivalent). A1 for \(x = 3.5\) (or \(\frac{7}{2}\)).

Convert the mixed number to an improper fraction: \(2\frac{3}{4} = \frac{11}{4}\). To divide by \(\frac{7}{8}\), multiply by its reciprocal: \(\frac{11}{4} \times \frac{8}{7} = \frac{88}{28}\). Simplify the fraction: \(\frac{88}{28} = \frac{22}{7}\). Convert back to a mixed number: \(\frac{22}{7} = 3\frac{1}{7}\).

M1 for converting to \(\frac{11}{4} \times \frac{8}{7}\) or \(\frac{22}{7}\). A1 for \(3\frac{1}{7}\) (accept \(\frac{22}{7}\) for 1.5 marks if mixed number format is omitted).

Since alternate angles are equal, set the two expressions equal to each other: \(3x - 10 = 2x + 30\). Subtract \(2x\) from both sides: \(x - 10 = 30\). Add 10 to both sides: \(x = 40\).

M1 for setting up the equation \(3x - 10 = 2x + 30\). A1 for \(x = 40\).

Use the formula for the volume of a cylinder: \(V = \pi r^2 h\). Substitute \(r = 6\) and \(h = 15\): \(V = \pi \times 6^2 \times 15 = 540\pi\). Calculate the numerical value: \(V \approx 1696.46\text{ cm}^3\). Rounding to 1 decimal place gives \(1696.5\text{ cm}^3\).

M1 for \(\pi \times 6^2 \times 15\). A1 for \(1696.5\) (accept \(1696.46\) to \(1696.5\)).

Calculate the simple interest earned using the formula \(I = \frac{P \times R \times T}{100}\): \(I = \frac{4800 \times 2.5 \times 4}{100} = 480\). Calculate the total value by adding the interest to the principal amount: \(\text{Total Value} = 4800 + 480 = 5280\).

M1 for correct substitution into simple interest formula: \(\frac{4800 \times 2.5 \times 4}{100}\). A1 for finding the interest as \(480\) (or total value \(5280\)).

The x-coordinate of the minimum point of a quadratic curve \(y = ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). Substitute \(a = 1\) and \(b = -4\): \(x = -\frac{-4}{2(1)} = 2\). Now, substitute \(x = 2\) into the original equation to find the y-coordinate: \(y = 2^2 - 4(2) - 5 = -9\). Thus, the coordinates of the minimum point are \((2, -9)\).

M1 for finding \(x = 2\) (using axis of symmetry or completing the square). A1 for \((2, -9)\).

The diagonal of a rectangle forms a right-angled triangle with the length and width as the legs. Use Pythagoras' Theorem: \(d^2 = 24^2 + 10^2 = 576 + 100 = 676\). Thus, \(d = \sqrt{676} = 26\text{ m}\).

M1 for \(24^2 + 10^2\). A1 for \(26\).

The sum of the angles in a triangle is \(180^\circ\). Find the total number of parts in the ratio: \(3 + 4 + 5 = 12\text{ parts}\). The smallest angle corresponds to the part with the value 3: \(\text{Smallest angle} = \frac{3}{12} \times 180^\circ = 45^\circ\).

M1 for \(\frac{3}{12} \times 180\) (or finding 1 part = \(15^\circ\)). A1 for \(45\).

Expand the bracket first: \(6x - 15 = 4x + 7\). Next, subtract \(4x\) from both sides: \(2x - 15 = 7\). Add 15 to both sides: \(2x = 22\). Finally, divide by 2: \(x = 11\).

M1 for expanding the bracket to get \(6x - 15\). M1 for isolating the \(x\) terms and constant terms (e.g., \(2x = 22\)). A0.5 for the final correct answer 11.

Substitute \(x = -3\) into the equation: \(y = 2(-3)^2 - 3(-3) + 1\). Evaluate the powers and multiplications: \(y = 2(9) + 9 + 1 = 18 + 9 + 1\). Add the numbers together to get: \(y = 28\).

M1 for substituting \(x = -3\) into the expression with at most one sign error. M1 for simplifying the terms to \(18 + 9 + 1\). A0.5 for the final answer 28.

Let \(D\) be the midpoint of the base \(BC\). Since the triangle is isosceles, the perpendicular line from \(A\) to \(BC\) bisects the base, so \(BD = 5\text{ cm}\). In the right-angled triangle \(ABD\), use Pythagoras' theorem: \(AD^2 + BD^2 = AB^2\), which gives \(AD^2 + 5^2 = 13^2\). This simplifies to \(AD^2 + 25 = 169\), so \(AD^2 = 144\). Taking the square root gives \(AD = 12\text{ cm}\).

M1 for identifying the half-base length of \(5\text{ cm}\). M1 for setting up the Pythagoras equation: \(h^2 + 5^2 = 13^2\) or \(h = \sqrt{13^2 - 5^2}\). A0.5 for the final correct answer 12.

First, calculate the area of the triangular cross-section: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2\). Then, calculate the volume by multiplying this cross-sectional area by the length of the prism: \(\text{Volume} = 24 \times 15 = 360\text{ cm}^3\).

M1 for finding the area of the triangular cross-section: \(\frac{1}{2} \times 6 \times 8\). M1 for multiplying their cross-sectional area by \(15\). A0.5 for the final answer 360.

First, find the number of students who play tennis: \(120 - 45 - 33 = 42\text{ students}\). Next, calculate this as a percentage of the total student population: \(\frac{42}{120} \times 100\% = 0.35 \times 100\% = 35\%\).

M1 for finding the number of tennis players: \(120 - 45 - 33 = 42\). M1 for writing the correct fraction multiplied by 100: \(\frac{42}{120} \times 100\). A0.5 for the final answer 35.

The sum of the exterior angles of any regular convex polygon is \(360^\circ\). Let \(n\) represent the number of sides. Since each exterior angle is \(40^\circ\), we have: \(n = \frac{360^\circ}{40^\circ} = 9\).

M1.5 for setting up the division \(\frac{360}{40}\) or the equation \(40 \times n = 360\). A1 for the correct answer 9.

First, calculate the actual profit made: \(\text{Profit} = 208 - 160 = 48\). Then, find the percentage profit relative to the original cost price: \(\text{Percentage Profit} = \frac{48}{160} \times 100\% = 0.3 \times 100\% = 30\%\).

M1 for finding the profit: \(208 - 160 = 48\) (or for working out the scale factor: \(\frac{208}{160} \times 100 = 130\%\)). M1 for calculating the fraction \(\frac{48}{160} \times 100\) (or \(130 - 100\)). A0.5 for the correct answer 30.

Multiply the second equation by 2 to align the coefficients of \(y\): \(8x - 2y = 38\). Now add this new equation to the first equation: \((3x + 2y) + (8x - 2y) = 17 + 38\), which simplifies to \(11x = 55\), giving \(x = 5\). Substitute \(x = 5\) into the second equation: \(4(5) - y = 19\), which gives \(20 - y = 19\), so \(y = 1\).

M1 for a valid method to eliminate one variable (e.g., multiplying the second equation by 2 and adding to the first). M1 for finding one correct variable value (\(x = 5\) or \(y = 1\)). A0.5 for both correct values: \(x = 5, y = 1\).

First, multiply both sides of the equation by 4 to get \(3x - 5 = 28\). Next, add 5 to both sides of the equation to get \(3x = 33\). Finally, divide both sides by 3 to get \(x = 11\).

M1 for multiplying both sides by 4 to get \(3x - 5 = 28\). M1 for adding 5 to both sides to get \(3x = 33\). A0.5 for the correct answer 11.

Use the formula for the volume of a cylinder, \(V = \pi r^2 h\). Substitute \(r = 4\) and \(h = 10\): \(V = \pi \times 4^2 \times 10 = 160\pi\). Evaluating this gives \(V \approx 502.6548\text{ cm}^3\). Rounding to 1 decimal place gives \(502.7\text{ cm}^3\).

M1 for substituting values into correct volume formula: \(\pi \times 4^2 \times 10\). M1 for \(160\pi\) or \(502.65...\). A0.5 for correct answer of 502.7.

The sum of the exterior angles of any convex polygon is \(360^\circ\). For a regular polygon with \(n\) sides, each exterior angle is \(\frac{360^\circ}{n}\). Thus, \(\frac{360}{n} = 40\), which gives \(n = \frac{360}{40} = 9\).

M1 for setting up the equation \(\frac{360}{n} = 40\) or showing \(360 \div 40\). A1.5 for the correct answer 9.

The discount amount is \(15\%\) of \(\$640\), which is \(0.15 \times 640 = 96\). The sale price is the original price minus the discount: \(640 - 96 = 544\). Alternatively, calculate \(85\%\) of \(\$640\): \(640 \times 0.85 = 544\).

M1 for calculating the discount as \(96\) or identifying the multiplier \(0.85\). M1 for subtracting the discount from the original price or evaluating \(640 \times 0.85\). A0.5 for the correct answer 544.

Use the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin(C)\). Substituting the given values: \(\text{Area} = \frac{1}{2} \times 8 \times 12 \times \sin(30^\circ)\). Since \(\sin(30^\circ) = 0.5\), this simplifies to \(\text{Area} = 48 \times 0.5 = 24\text{ cm}^2\).

M1 for substituting into the formula: \(\frac{1}{2} \times 8 \times 12 \times \sin(30^\circ)\). M1 for evaluating \(\sin(30^\circ) = 0.5\) or showing \(48 \times \sin(30^\circ)\). A0.5 for the correct area of 24.

Substitute \(x = -3\) into the equation: \(y = 2(-3)^2 - 5(-3) + 1\). Evaluate each term: \(2(-3)^2 = 2 \times 9 = 18\) and \(-5(-3) = 15\). Summing these terms with 1: \(y = 18 + 15 + 1 = 34\).

M1 for substituting \(x = -3\) correctly into the expression: \(2(-3)^2 - 5(-3) + 1\). M1 for dealing with signs to get \(18 + 15 + 1\). A0.5 for correct answer of 34.

We multiply both sides of the equation by the common denominator \(x(x-1)\):
\[6x - 4(x-1) = x(x-1)\]

Expand the brackets:
\[6x - 4x + 4 = x^2 - x\]
\[2x + 4 = x^2 - x\]

Rearrange the terms into standard quadratic form \(ax^2 + bx + c = 0\):
\[x^2 - 3x - 4 = 0\]

Factorise the quadratic expression:
\[(x - 4)(x + 1) = 0\]

This gives the solutions:
\[x = 4 \quad \text{or} \quad x = -1\]

- M1 for eliminating the fractions correctly: \(6x - 4(x-1) = x(x-1)\) or equivalent
- M1 for rearranging into standard quadratic form: \(x^2 - 3x - 4 = 0\)
- A1 for both answers \(4\) and \(-1\)

**Method 1: Completing the square**
We write the quadratic in vertex form:
\[y = 3(x^2 - 4x) + 7\]
\[y = 3\left[(x - 2)^2 - 4\right] + 7\]
\[y = 3(x - 2)^2 - 12 + 7\]
\[y = 3(x - 2)^2 - 5\]
The coordinates of the turning point are \((2, -5)\).

**Method 2: Calculus**
Find the derivative of \(y\) with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 6x - 12\]
Set the derivative equal to \(0\) to find the \(x\)-coordinate of the turning point:
\[6x - 12 = 0 \implies x = 2\]
Substitute \(x = 2\) back into the original equation to find the \(y\)-coordinate:
\[y = 3(2)^2 - 12(2) + 7 = 12 - 24 + 7 = -5\]
So, the turning point is \((2, -5)\).

- M1 for an attempt to complete the square or finding the derivative: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x - 12\)
- A1 for finding the correct \(x\)-coordinate: \(x = 2\) (or one coordinate correct in the turning point)
- A1 for the correct coordinates: \((2, -5)\)

We use the Cosine Rule to find the length of \(AC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]

Substitute the given values:
\[AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(112^\circ)\]
\[AC^2 = 49 + 81 - 126\cos(112^\circ)\]

Calculate the numerical value:
\[AC^2 \approx 130 - 126(-0.374607)\]
\[AC^2 \approx 130 + 47.2005 = 177.2005\]
\[AC = \sqrt{177.2005} \approx 13.312\text{ cm}\]

Rounding to 3 significant figures, we get \(AC = 13.3\text{ cm}\).

- M1 for correct substitution into the Cosine Rule: \(7^2 + 9^2 - 2(7)(9)\cos(112^\circ)\)
- M1 for \(AC^2 \approx 177.2\) (or \(AC = \sqrt{177.2...}\))
- A1 for \(13.3\) (accept \(13.31\))

The volume of a hemisphere is given by:
\[V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\]

The volume of a cone is given by:
\[V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (12) = 4\pi r^2\]

The volume of the recast sphere of radius \(2r\) is:
\[V_{\text{sphere}} = \frac{4}{3}\pi (2r)^3 = \frac{4}{3}\pi (8r^3) = \frac{32}{3}\pi r^3\]

Since the combined volumes of the hemisphere and the cone equal the volume of the sphere:
\[\frac{2}{3}\pi r^3 + 4\pi r^2 = \frac{32}{3}\pi r^3\]

Divide the entire equation by \(\pi r^2\) (since \(r \neq 0\)):
\[\frac{2}{3}r + 4 = \frac{32}{3}r\]

Subtract \(\frac{2}{3}r\) from both sides:
\[4 = \frac{30}{3}r\]
\[4 = 10r\]
\[r = 0.4\]

- M1 for writing a correct equation for the volumes: \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 (12) = \frac{4}{3}\pi (2r)^3\) or equivalent
- M1 for simplifying to a linear equation in \(r\): \(4 = 10r\) or equivalent
- A1 for \(0.4\) or \(\frac{2}{5}\)

We use the formula for compound interest:
\[A = P\left(1 + \frac{r}{100}\right)^n\]

Substitute the known values into the formula:
\[2969.22 = P(1 + 0.035)^5\]
\[2969.22 = P(1.035)^5\]

Solve for \(P\):
\[P = \frac{2969.22}{1.035^5}\]
\[P \approx \frac{2969.22}{1.187686}\]
\[P \approx 2500.00\]

Thus, the initial investment is \(P = 2500\).

- M1 for setting up the equation: \(P(1.035)^5 = 2969.22\)
- M1 for rearranging to make \(P\) the subject: \(P = \frac{2969.22}{1.035^5}\)
- A1 for \(2500\)

The sum of the interior angles of a polygon with \(n\) sides is given by:
\[\text{Sum} = (n - 2) \times 180^\circ\]

For a pentagon (\(n = 5\)):
\[\text{Sum} = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\]

Sum the given angles and set equal to \(540^\circ\):
\[x + (2x - 15) + (x + 25) + (1.5x + 40) + 160 = 540\]

Combine like terms:
\[(x + 2x + x + 1.5x) + (-15 + 25 + 40 + 160) = 540\]
\[5.5x + 210 = 540\]

Subtract \(210\) from both sides:
\[5.5x = 330\]

Divide by \(5.5\):
\[x = 60\]

- M1 for using the sum of interior angles of a pentagon is \(540^\circ\)
- M1 for setting up the simplified equation: \(5.5x + 210 = 540\) or equivalent
- A1 for \(60\)

Express both bases \(27\) and \(9\) as powers of \(3\):
\[27 = 3^3 \quad \text{and} \quad 9 = 3^2\]

Substitute these into the equation:
\[(3^3)^{x-1} = (3^2)^{2x-5}\]

Apply the power of a power index law \((a^m)^n = a^{mn}\):
\[3^{3(x-1)} = 3^{2(2x-5)}\]

Since the bases are now identical, we can equate the exponents:
\[3(x-1) = 2(2x-5)\]

Expand the brackets:
\[3x - 3 = 4x - 10\]

Rearrange to solve for \(x\):
\[-3 + 10 = 4x - 3x\]
\[x = 7\]

- M1 for writing both bases as powers of 3: \(3^{3(x-1)} = 3^{2(2x-5)}\) or equivalent
- M1 for equating the exponents and setting up the linear equation: \(3(x-1) = 2(2x-5)\)
- A1 for \(7\)

The total number of balls in the bag is \(n + 6\).

The probability of selecting the first blue ball is:
\[P(\text{First is Blue}) = \frac{6}{n+6}\]

Since the first ball is not replaced, there are now \(5\) blue balls remaining out of a total of \(n + 5\) balls.
The probability of selecting a second blue ball is:
\[P(\text{Second is Blue}) = \frac{5}{n+5}\]

We are given that the combined probability of both being blue is \(\frac{1}{3}\):
\[\frac{6}{n+6} \times \frac{5}{n+5} = \frac{1}{3}\]
\[\frac{30}{(n+6)(n+5)} = \frac{1}{3}\]

Multiply both sides by \(3(n+6)(n+5)\):
\[90 = (n+6)(n+5)\]
\[90 = n^2 + 11n + 30\]

Rearrange to form a quadratic equation:
\[n^2 + 11n - 60 = 0\]

Factorise the quadratic expression:
\[(n + 15)(n - 4) = 0\]

Since the number of red balls \(n\) must be a positive integer, we discard \(n = -15\).
Thus, \(n = 4\).

- M1 for setting up the probability product equation: \(\frac{6}{n+6} \times \frac{5}{n+5} = \frac{1}{3}\) (or equivalent with total \(N\))
- M1 for forming a correct quadratic equation: \(n^2 + 11n - 60 = 0\) (or \(N^2 - N - 90 = 0\))
- A1 for \(4\)

Multiply both sides of the equation by the common denominator \((x + 2)(x - 1)\):
\(3(x - 1) + 2(x + 2) = 2(x + 2)(x - 1)\)

Expand the terms on both sides:
\(3x - 3 + 2x + 4 = 2(x^2 + x - 2)\)
\(5x + 1 = 2x^2 + 2x - 4\)

Rearrange the terms into standard quadratic form \(ax^2 + bx + c = 0\):
\(2x^2 - 3x - 5 = 0\)

Factorize the quadratic expression:
\((2x - 5)(x + 1) = 0\)

Set each factor to zero to find the solutions:
\(2x - 5 = 0 \Rightarrow x = 2.5\)
\(x + 1 = 0 \Rightarrow x = -1\)

So, the solutions are \(x = -1\) or \(x = 2.5\).

M1 for multiplying by the common denominator to get \(3(x - 1) + 2(x + 2) = 2(x + 2)(x - 1)\) or equivalent.
M1 for rearranging into a standard three-term quadratic form, \(2x^2 - 3x - 5 = 0\) or equivalent.
A1 for both correct solutions: \(x = -1\) and \(x = 2.5\) (or equivalent fractions).

1. Find the \(y\)-coordinate when \(x = 2\):
\(y = 2(2)^3 - 3(2)^2 - 8 = 16 - 12 - 8 = -4\)
So the point of contact is \((2, -4)\).

2. Find the gradient function by differentiating \(y\):
\(\frac{dy}{dx} = 6x^2 - 6x\)

3. Substitute \(x = 2\) to find the gradient of the tangent, \(m\):
\(m = 6(2)^2 - 6(2) = 24 - 12 = 12\)

4. Write down the equation of the tangent using \(y - y_1 = m(x - x_1)\):
\(y - (-4) = 12(x - 2)\)
\(y + 4 = 12x - 24\)
\(y = 12x - 28\)

M1 for finding \(\frac{dy}{dx} = 6x^2 - 6x\) (at least one term correct).
M1 for finding the \(y\)-coordinate of \(-4\) and substituting \(x = 2\) into their derivative to find the gradient \(12\).
A1 for \(y = 12x - 28\) (or any equivalent form).

We can use the Cosine Rule to find the unknown side \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)\)

Substitute the given values into the formula:
\(PR^2 = 8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos(108^\circ)\)
\(PR^2 = 70.56 + 125.44 - 188.16\cos(108^\circ)\)
\(PR^2 = 196 - 188.16(-0.309017)\)
\(PR^2 \approx 196 + 58.1446 = 254.1446\)

Take the square root of both sides:
\(PR = \sqrt{254.1446} \approx 15.9419\text{ cm}\)

Rounding to 3 significant figures gives \(15.9\text{ cm}\).

M1 for correct substitution of values into the Cosine Rule: \(8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos(108^\circ)\).
A1 for evaluating \(PR^2 \approx 254\) or \(254.1\dots\).
A1 for \(15.9\) (accept \(15.94\dots\)).

1. Find the perpendicular height, \(h\), of the cone using Pythagoras' theorem:
\(h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm\rm}

2. Calculate the volume of the cone:
\)V_{\text{cone}} = \frac{1}{3} \pi (5)^2 (12) = 100\pi\text{ cm}^3\)

3. Set the volume of the sphere equal to the volume of the cone:
\(\frac{4}{3} \pi R^3 = 100\pi\)
Multiply both sides by \(3\) and divide by \(4\pi\):
\(R^3 = 75\)

4. Calculate the radius of the sphere:
\(R = \sqrt[3]{75} \approx 4.2172\text{ cm}\)

To 3 significant figures, the radius is \(4.22\text{ cm}\).

M1 for finding the perpendicular height of the cone, \(h = 12\text{ cm}\), using Pythagoras.
M1 for equating the volume of the sphere to the volume of the cone: \(\frac{4}{3}\pi R^3 = \frac{1}{3}\pi (5^2)(12)\) or better.
A1 for \(4.22\) (accept \(4.217\dots\)).

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

After a \(15\%\) reduction, the price is:
\(P \times (1 - 0.15) = 0.85P\)

After a further \(10\%\) reduction, the price is:
\(0.85P \times (1 - 0.10) = 0.85P \times 0.90 = 0.765P\)

We are given that the final price is \(\$306\):
\(0.765P = 306\)

Solve for \(P\):
\(P = \frac{306}{0.765} = 400\)

The original price was \(\$400\).

M1 for expressing the combined reductions as a single decimal multiplier: \(0.85 \times 0.90 = 0.765\) (or finding the intermediate price after the first reduction: \(306 / 0.90 = 340\)).
M1 for setting up the division equation: \(306 / 0.765\) (or \(340 / 0.85\)).
A1 for \(400\).

Let \(i\) be the size of each interior angle and \(e\) be the size of each exterior angle.

We know that the interior and exterior angles of any polygon sum to \(180^\circ\):
\(i + e = 180^\circ\)

We are given:
\(i = e + 140^\circ\)

Substitute the second equation into the first:
\((e + 140^\circ) + e = 180^\circ\)
\(2e + 140^\circ = 180^\circ\)
\(2e = 40^\circ \Rightarrow e = 20^\circ\)

Since the sum of the exterior angles of any regular polygon is \(360^\circ\):
\(n = \frac{360^\circ}{e} = \frac{360^\circ}{20^\circ} = 18\)

The polygon has \(18\) sides.

M1 for forming an equation using \(i + e = 180\) and the given relation, e.g., \(e + 140 + e = 180\).
A1 for finding the exterior angle \(e = 20^\circ\) (or the interior angle \(i = 160^\circ\)).
A1 for \(18\).

1. Express both terms in the numerator with the same power of 10 to add them:
\(4.5 \times 10^{-6} = 0.45 \times 10^{-5}\)

2. Add the terms in the numerator:
\(0.45 \times 10^{-5} + 1.2 \times 10^{-5} = 1.65 \times 10^{-5}\)

3. Perform the division by the denominator:
\(\frac{1.65 \times 10^{-5}}{5.0 \times 10^3} = \left(\frac{1.65}{5.0}\right) \times 10^{-5 - 3} = 0.33 \times 10^{-8}\)

4. Convert the result into standard form:
\(0.33 \times 10^{-8} = 3.3 \times 10^{-9}\)

M1 for adding the numerator correctly to obtain \(1.65 \times 10^{-5}\) (or \(0.0000165\)).
M1 for dividing by \(5.0 \times 10^3\) to obtain \(0.33 \times 10^{-8}\) (or \(3.3 \times 10^{k}\) where \(k\) is an integer).
A1 for \(3.3 \times 10^{-9}\).

1. Find the time taken for the first part of the journey (from A to B):
\(t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{40\text{ km}}{80\text{ km/h}} = 0.5\text{ hours}\)

2. Find the time taken for the second part of the journey (from B to C):
\(t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{120\text{ km}}{60\text{ km/h}} = 2.0\text{ hours}\)

3. Calculate the total distance and total time for the entire journey:
Total Distance = \(40\text{ km} + 120\text{ km} = 160\text{ km}\)
Total Time = \(0.5\text{ hours} + 2.0\text{ hours} = 2.5\text{ hours}\)

4. Calculate the average speed:
Average Speed = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{160\text{ km}}{2.5\text{ hours}} = 64\text{ km/h}\)

M1 for calculating both individual times: \(t_1 = 0.5\text{ hours}\) and \(t_2 = 2.0\text{ hours}\) (or equivalent in minutes).
M1 for dividing total distance \(160\text{ km}\) by total time \(2.5\text{ hours}\) (or equivalent in minutes).
A1 for \(64\).

Multiply both sides of the equation by the common denominator \((x+3)(x-1)\): \(2(x-1) + 3(x+3) = (x+3)(x-1)\). Expand both sides: \(2x - 2 + 3x + 9 = x^2 + 2x - 3\). Simplify to get: \(5x + 7 = x^2 + 2x - 3\). Rearrange the terms into a quadratic equation: \(x^2 - 3x - 10 = 0\). Factorise the quadratic equation: \((x-5)(x+2) = 0\). This gives the solutions: \(x = 5\) or \(x = -2\).

M1 for writing with a common denominator, e.g. \(2(x-1) + 3(x+3) = (x+3)(x-1)\) or better. M1 for reducing to quadratic form \(x^2 - 3x - 10 = 0\). A1 for both correct answers: \(x = 5\) and \(x = -2\).

First, find the derivative of the curve equation to get the gradient function: \(\frac{dy}{dx} = 2x - 4\). Set the gradient equal to 2: \(2x - 4 = 2\). Solve for \(x\): \(2x = 6 \implies x = 3\). Substitute \(x = 3\) back into the original curve equation to find the corresponding y-coordinate: \(y = 3^2 - 4(3) + 7 = 9 - 12 + 7 = 4\). Thus, the coordinates of the point are \((3, 4)\).

M1 for differentiating to obtain \(2x - 4\). M1 for setting their derivative equal to 2 and solving to find \(x = 3\). A1 for correct coordinates \((3, 4)\).

Using the Cosine Rule to find the side \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substitute the given values into the formula: \(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), this simplifies to: \(AC^2 = 49 + 81 - 126(-0.5) = 130 + 63 = 193\). Calculate \(AC\): \(AC = \sqrt{193} \approx 13.892\text{ cm}\). Rounding to 3 significant figures gives \(13.9\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7^2 + 9^2 - 2(7)(9)\cos(120^\circ)\). M1 for \(AC^2 = 193\) or \(AC = \sqrt{193}\). A1 for \(13.9\) or \(13.89\dots\) (accept 13.9).

The total surface area of a cone is the sum of the base area and the curved surface area: \(A = \pi r^2 + \pi r l\), where \(r\) is the radius and \(l\) is the slant height. Substitute \(r = 4\) and \(l = 9\): \(A = \pi(4)^2 + \pi(4)(9) = 16\pi + 36\pi = 52\pi\text{ cm}^2\).

M1 for calculating base area as \(16\pi\) or curved surface area as \(36\pi\). M1 for adding the two areas together, i.e., \(\pi(4)^2 + \pi(4)(9)\). A1 for \(52\pi\).

Let the original price of the laptop be \(P\). After the first reduction of 20%, the price is \(0.80P\). After the second reduction of 10%, the price becomes \(0.90 \times 0.80P = 0.72P\). We are given that \(0.72P = 576\). Solving for \(P\): \(P = \frac{576}{0.72} = 800\). Thus, the original price of the laptop was $800.

M1 for identifying multipliers \(0.80\) and \(0.90\) (or equivalent). M1 for setting up the equation \(0.72P = 576\) or performing the calculation \(576 \div 0.9 \div 0.8\). A1 for 800.

Let the exterior angle of the polygon be \(x^\circ\). The interior angle is therefore \((x + 140)^\circ\). Since the interior angle and exterior angle at any vertex of a polygon sum to \(180^\circ\), we have: \(x + (x + 140) = 180 \implies 2x + 140 = 180 \implies 2x = 40 \implies x = 20\). The exterior angle is \(20^\circ\). The number of sides \(n\) is given by: \(n = \frac{360^\circ}{\text{exterior angle}} = \frac{360}{20} = 18\).

M1 for setting up an equation for the angles, e.g. \(x + (x+140) = 180\). M1 for finding the exterior angle is \(20^\circ\) or interior angle is \(160^\circ\). A1 for 18.

Express both bases as powers of 3: \(27 = 3^3\) and \(9 = 3^2\). Substitute these back into the equation: \((3^3)^{x-1} = (3^2)^{2x-5}\). Apply the index law \((a^m)^n = a^{mn}\): \(3^{3(x-1)} = 3^{2(2x-5)}\). Since the bases are now identical, equate the exponents: \(3(x-1) = 2(2x-5)\). Expand and solve: \(3x - 3 = 4x - 10 \implies 10 - 3 = 4x - 3x \implies x = 7\).

M1 for writing 27 as \(3^3\) and 9 as \(3^2\) (or alternative common base). M1 for equating the exponents: \(3(x-1) = 2(2x-5)\) or better. A1 for 7.

(a) Expand brackets:
$$5x - 15 - 2x - 2 = 7$$
$$3x - 17 = 7$$
$$3x = 24$$
$$x = 8$$

(b) Multiply the second equation by 2:
$$8x - 2y = 10$$
Add this to the first equation:
$$(3x + 2y) + (8x - 2y) = 12 + 10$$
$$11x = 22$$
$$x = 2$$
Substitute $x = 2$ into the second equation:
$$4(2) - y = 5$$
$$8 - y = 5 \Rightarrow y = 3$$
So, $x = 2$ and $y = 3$.

(c) Perimeter of a rectangle is:
$$2(\text{length} + \text{width}) = 30$$
$$2((2x + 3) + x) = 30$$
$$2(3x + 3) = 30$$
$$6x + 6 = 30$$
$$6x = 24$$
$$x = 4$$

(a) M1 for expanding: $5x - 15 - 2x - 2 = 7$
M1 for simplifying to: $3x - 17 = 7$
M1 for rearranging: $3x = 24$
A1 for final answer $x = 8$

(b) M1 for multiplying second equation to get $8x - 2y = 10$ (or equivalent)
M1 for adding equations to eliminate $y$
A1 for $x = 2$
M1 for substituting their $x$ to find $y$
A0.5 for $y = 3$

(c) M1 for setting up the equation $2(2x + 3 + x) = 30$ (or equivalent)
M1 for simplifying to $6x + 6 = 30$
A1 for final answer $x = 4$

(a) Substitute the values of $x$ into $y = x^2 - 2x - 3$:
- For $x = -2$: $P = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5$
- For $x = 1$: $Q = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$
- For $x = 4$: $R = (4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5$

(b) The solutions to $x^2 - 2x - 3 = 0$ are the values of $x$ where $y = 0$. From the table, these are $x = -1$ and $x = 3$.

(c) The minimum point occurs at the vertex. By symmetry, the $x$-coordinate is halfway between the roots: $x = 1$. The corresponding $y$-value is $-4$. So the turning point is $(1, -4)$.

(d) The line of symmetry is the vertical line passing through the turning point, which has the equation $x = 1$.

(a) B1 for $P = 5$, B1 for $Q = -4$, B1.5 for $R = 5$

(b) B1 for $x = -1$, B1 for $x = 3$ (or M1 for identifying where $y = 0$)

(c) B1 for $x$-coordinate of 1, B1 for $y$-coordinate of -4 (Accept $(1, -4)$)

(d) B2 for $x = 1$ (B1 for just '1' or 'x = ...' with wrong value, or drawing the correct vertical line on a sketch)

(a) The area of a triangle is given by:
$$\text{Area} = \frac{1}{2} a c \sin B$$
$$\text{Area} = \frac{1}{2} \times 6.2 \times 8.5 \times \sin(75^\circ)$$
$$\text{Area} = 26.35 \times 0.9659258... \approx 25.45\text{ cm}^2$$
To 3 significant figures, this is $25.5\text{ cm}^2$.

(b) Using the Cosine Rule:
$$b^2 = a^2 + c^2 - 2ac \cos B$$
$$AC^2 = 6.2^2 + 8.5^2 - 2 \times 6.2 \times 8.5 \times \cos(75^\circ)$$
$$AC^2 = 38.44 + 72.25 - 105.4 \times 0.258819...$$
$$AC^2 = 110.69 - 27.2795...$$
$$AC^2 = 83.410...$$
$$AC = \sqrt{83.410...} \approx 9.1329...\text{ cm}$$
To 3 significant figures, $AC = 9.13\text{ cm}$.

(c) Using the Sine Rule:
$$\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}$$
$$\frac{\sin(ACB)}{8.5} = \frac{\sin(75^\circ)}{9.1329}$$
$$\sin(ACB) = \frac{8.5 \times \sin(75^\circ)}{9.1329} \approx \frac{8.21037}{9.1329} \approx 0.8990$$
$$\text{Angle } ACB = \sin^{-1}(0.8990) \approx 64.03^\circ$$
To 1 decimal place, angle $ACB = 64.0^\circ$.

(a) M1 for $\frac{1}{2} \times 6.2 \times 8.5 \times \sin(75^\circ)$
M1 for $26.35 \times \sin(75^\circ)$
A1.5 for $25.5$ (or $25.45$)

(b) M1 for correct substitution into Cosine Rule: $6.2^2 + 8.5^2 - 2(6.2)(8.5)\cos(75^\circ)$
M1 for correct evaluation of terms: $110.69 - 27.28$
A1 for $AC^2 = 83.4$
A1 for $AC = 9.13$

(c) M1 for correct Sine Rule setup: $\frac{\sin(ACB)}{8.5} = \frac{\sin(75^\circ)}{9.13}$
M1 for rearrangement: $\sin(ACB) = \frac{8.5 \sin(75^\circ)}{9.13}$
A1 for $\sin(ACB) \approx 0.899$
A1 for angle $ACB = 64.0^\circ$ (accept $64.0$ to $64.1$ depending on rounding of AC)

(a) The volume of a cylinder is:
$$V = \pi r^2 h = \pi \times 4.5^2 \times 12 = 243\pi \approx 763.407...\text{ cm}^3$$
To 3 significant figures, $V = 763\text{ cm}^3$.

(b) The total surface area of a closed cylinder is:
$$A = 2\pi r^2 + 2\pi r h = 2\pi(4.5^2) + 2\pi(4.5)(12) = 40.5\pi + 108\pi = 148.5\pi \approx 466.526...\text{ cm}^2$$
To 3 significant figures, $A = 467\text{ cm}^2$.

(c) The volume of the sphere is equal to the volume of the cylinder:
$$\frac{4}{3}\pi R^3 = 243\pi$$
Divide both sides by $\pi$:
$$\frac{4}{3} R^3 = 243$$
Multiply by $\frac{3}{4}$:
$$R^3 = 243 \times \frac{3}{4} = 182.25$$
$$R = \sqrt[3]{182.25} \approx 5.6696...\text{ cm}$$
To 3 significant figures, the radius is $5.67\text{ cm}$.

(a) M1 for formula $\pi r^2 h$
M1 for substituting: $\pi \times 4.5^2 \times 12$
A1.5 for $763$ (or $763.4$)

(b) M1 for area of circular ends: $2 \times \pi \times 4.5^2 = 40.5\pi$
M1 for curved surface area: $2 \times \pi \times 4.5 \times 12 = 108\pi$
A2 for $467$ (or $466.5$)

(c) M1 for equating volumes: $\frac{4}{3}\pi R^3 = 243\pi$ (or $763$)
M1 for isolating $R^3$: $R^3 = 182.25$
M1 for finding cube root: $R = \sqrt[3]{182.25}$
A1 for $5.67$ (accept $5.66$ to $5.68$)

(a) Calculation:
$$\frac{3}{8} \times 1440 = 3 \times 180 = 540$$
So, $\frac{3}{8}$ of $1440$ is $540$.

(b) Let the original price be $P$.
A $15\%$ reduction means the sale price is $85\%$ of the original price:
$$0.85 \times P = 306$$
$$P = \frac{306}{0.85} = 360$$
So, the original price was $360$.

(c) First find the increase in price:
$$\text{Increase} = 513 - 450 = 63$$
Now, calculate the percentage increase relative to the original price:
$$\text{Percentage Increase} = \frac{63}{450} \times 100\%$$
$$\text{Percentage Increase} = 0.14 \times 100\% = 14\%$$
So, the price increased by $14\%$.

(a) M1.5 for $\frac{3}{8} \times 1440$
A1 for $540$

(b) M1 for setting up relation: $0.85 \times P = 306$ (or $85\% = 306$)
M2 for $\frac{306}{0.85}$ (or $\frac{306}{85} \times 100$)
A1 for $360$

(c) M1.5 for finding the difference: $513 - 450 = 63$
M2.5 for dividing by original: $\frac{63}{450} \times 100$
A1 for $14$ or $14\%$

(a)(i) The sum of exterior angles of any polygon is $360^\circ$.
$$\text{Number of sides} = \frac{360^\circ}{24^\circ} = 15$$
There are 15 sides.

(a)(ii) The sum of interior angles of an $n$-sided polygon is given by $(n - 2) \times 180^\circ$:
$$\text{Sum} = (15 - 2) \times 180^\circ = 13 \times 180^\circ = 2340^\circ$$

(b)(i) Co-interior angles between parallel lines add up to $180^\circ$:
$$(3x + 15) + (2x + 10) = 180$$
$$5x + 25 = 180$$
$$5x = 155$$
$$x = 31$$

(b)(ii) Find the size of both angles by substituting $x = 31$:
- Angle 1: $3(31) + 15 = 93 + 15 = 108^\circ$
- Angle 2: $2(31) + 10 = 62 + 10 = 72^\circ$
The larger angle is $108^\circ$.

(a)(i) M1 for $\frac{360}{24}$
A1.5 for $15$

(a)(ii) M1.5 for $(15-2) \times 180$ (allow follow through on their sides)
A1.5 for $2340$

(b)(i) M1 for setting up the equation: $(3x+15) + (2x+10) = 180$
M1.5 for simplifying to: $5x + 25 = 180$ or $5x = 155$
A1.5 for $x = 31$

(b)(ii) M1 for substituting $x = 31$ into $3x + 15$ or $2x + 10$
A1 for $108$

(a) Sum of parts of the ratio:
$$2 + 3 + 4 = 9$$
Value of one part:
$$\frac{450}{9} = 50$$
Multiply each part by 50:
- First share: $2 \times 50 = 100$
- Second share: $3 \times 50 = 150$
- Third share: $4 \times 50 = 200$
So the shares are $100, $150, and $200.

(b)(i) Given relation:
$$\text{mass} = k \times \text{length}$$
$$4.8 = k \times 1.2$$
$$k = \frac{4.8}{1.2} = 4$$

(b)(ii) Use the equation $\text{mass} = 4 \times \text{length}$:
$$\text{mass} = 4 \times 3.5 = 14\text{ kg}$$

(c) Use unit proportion:
$$\text{Flour per person} = \frac{600\text{ g}}{8} = 75\text{ g}$$
For 14 people:
$$\text{Flour needed} = 14 \times 75\text{ g} = 1050\text{ g}$$

(a) M1 for finding sum of ratio parts: $2+3+4 = 9$
M1.5 for dividing $450$ by $9$
A1 for final shares: $100, $150, $200 (must have all three correct for full marks)

(b)(i) M1.5 for substituting values into formula: $4.8 = k \times 1.2$
A1 for $k = 4$

(b)(ii) M1 for substituting $3.5$ into formula: $\text{mass} = 4 \times 3.5$
A1 for $14\text{ kg}$

(c) M1.5 for finding flour per person: $\frac{600}{8} = 75$
M1 for multiplying by 14: $14 \times 75$
A1 for $1050\text{ g}$

(a)(i) Let $N$ be the total number of pens. The probability of choosing a blue pen is given by:
$$\frac{\text{Number of blue pens}}{N} = \frac{1}{4}$$
$$\frac{5}{N} = \frac{1}{4}$$
Multiply both sides by $4N$:
$$20 = N$$
Thus, the total number of pens in the box is 20.

(a)(ii) The number of green pens is the total number minus the other colors:
$$\text{Green pens} = 20 - 8\text{ (red)} - 5\text{ (blue)} = 7\text{ green pens}$$

(b)(i) Probability of choosing a red pen:
$$P(\text{Red}) = \frac{\text{Number of red pens}}{\text{Total pens}} = \frac{8}{20} = \frac{2}{5} = 0.4$$

(b)(ii) Probability of choosing a pen that is not blue:
$$P(\text{Not Blue}) = 1 - P(\text{Blue}) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$$
Alternatively, count other colors: $8\text{ (red)} + 7\text{ (green)} = 15$ pens. So $P(\text{Not Blue}) = \frac{15}{20} = \frac{3}{4}$.

(b)(iii) Since there are no yellow pens in the box, the probability is $0$.

(a)(i) M2 for setting up equation: $\frac{5}{N} = \frac{1}{4}$ or stating $5 \times 4 = 20$
A1.5 for showing clearly $N = 20$

(a)(ii) M1 for subtraction: $20 - 8 - 5$
A1 for $7$

(b)(i) M1 for $\frac{8}{20}$ (or equivalent)
A1 for $\frac{2}{5}$ (or $0.4$)

(b)(ii) M1 for $1 - \frac{1}{4}$ or $\frac{15}{20}$
A1 for $\frac{3}{4}$ (or $0.75$)

(b)(iii) B2 for $0$ (or 'impossible')

(a) Area of the triangular cross-section = 0.5 * base * height = 0.5 * 6 * 8 = 24 cm^2. Volume = Area of cross-section * length = 24 * 15 = 360 cm^3. (b) Using Pythagoras' theorem: hypotenuse^2 = 6^2 + 8^2 = 36 + 64 = 100. Taking the square root gives hypotenuse = sqrt(100) = 10 cm. (c) Total surface area is the sum of the areas of the 2 triangular faces and 3 rectangular faces. Area of 2 triangular faces = 2 * 24 = 48 cm^2. Area of rectangular faces = (6 * 15) + (8 * 15) + (10 * 15) = 90 + 120 + 150 = 360 cm^2. Total surface area = 48 + 360 = 408 cm^2. (d) Mass = Density * Volume = 0.65 * 360 = 234 g. (e) First, convert the coverage area of the paint from square meters to square centimeters: 2.5 m^2 = 2.5 * 10000 = 25000 cm^2. Number of blocks that can be painted = 25000 / 408 = 61.27... Since only complete blocks can be painted, the maximum number of complete wooden blocks is 61.

(a) M1 for 0.5 * 6 * 8, M1 for multiplying their cross-section area by 15, A1 for 360. (b) M1 for 6^2 + 8^2 or sqrt(6^2 + 8^2), A1 for showing it equals 10. (c) M1 for calculating the sum of the rectangular areas (e.g., (6+8+10)*15), M1 for adding the area of the two triangular faces (e.g., 2 * 24), A1 for 408. (d) M1 for 360 * 0.65, A1 for 234. (e) M1 for converting 2.5 m^2 to 25000 cm^2 (or converting 408 cm^2 to 0.0408 m^2), A0.5 for 61 (rounded down to the nearest integer).

(a) Using the compound interest formula: \(8000(1 + r/100)^3 = 8741.81\). Dividing by 8000: \((1 + r/100)^3 = 1.092726\). Taking the cube root: \(1 + r/100 = 1.03\), which gives \(r = 3\). (b) Using the depreciation formula: \(P(1 - 0.12)^2 = 13552 \implies P(0.88)^2 = 13552 \implies 0.7744P = 13552 \implies P = 17500\). (c) Let the original price be \(L\). A 28% loss means he sold it for 72% of the original price: \(0.72L = 468 \implies L = 650\). (d) Let the original price be 100%. After a 15% reduction, the price is 85%. After a further 10% reduction, the price becomes \(85\% \times 0.9 = 76.5\%\). The equivalent single reduction is \(100\% - 76.5\% = 23.5\%\).

(a) M1 for \(8000(1 + r/100)^3 = 8741.81\), M1 for \((1 + r/100)^3 = 1.092726\), M1 for \(1 + r/100 = 1.03\), A1 for \(r = 3\). (b) M1 for \(P(0.88)^2 = 13552\), M1 for \(13552 / 0.7744\), A1 for 17500. (c) M1 for \(0.72L = 468\), M1 for \(468 / 0.72\), A1 for 650. (d) M1 for \(0.85 \times 0.90\), M1 for \(1 - 0.765\), A1 for 23.5%.

(a) Time = Distance / Speed = \(\frac{36}{x}\). (b) Time = \(\frac{45}{x-3}\). (c) Total time is given by \(\frac{36}{x} + \frac{45}{x-3} = 4.5\). Divide by 4.5 to simplify: \(\frac{8}{x} + \frac{10}{x-3} = 1\). Multiply both sides by \(x(x-3)\): \(8(x-3) + 10x = x(x-3) \implies 8x - 24 + 10x = x^2 - 3x \implies 18x - 24 = x^2 - 3x \implies x^2 - 21x + 24 = 0\). (d) Using the quadratic formula: \(x = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(24)}}{2} = \frac{21 \pm \sqrt{345}}{2}\). This gives \(x = \frac{21 \pm 18.574}{2}\), so \(x \approx 19.79\) or \(x \approx 1.21\). (e) Since the speed in the second part is \(x-3\), we must have \(x > 3\). Thus, \(x = 19.79\). Time for the first 36 km is \(\frac{36}{19.79} \approx 1.819\) hours. Multiplying the fractional part by 60: \(0.819 \times 60 = 49.1\) minutes. This gives 1 hour and 49 minutes to the nearest minute.

(a) B1 for \(\frac{36}{x}\). (b) B1 for \(\frac{45}{x-3}\). (c) M1 for \(\frac{36}{x} + \frac{45}{x-3} = 4.5\), M1 for algebraic step to remove denominators: \(36(x-3) + 45x = 4.5x(x-3)\), M1 for expanding: \(36x - 108 + 45x = 4.5x^2 - 13.5x\), M1 for rearranging to quadratic form: \(4.5x^2 - 94.5x + 108 = 0\), A1 for dividing by 4.5 to reach \(x^2 - 21x + 24 = 0\). (d) M1 for correct substitution into quadratic formula, M1 for evaluation of discriminant as 345, A1 for \(x = 19.79\), A1 for \(x = 1.21\). (e) M1 for substituting \(x = 19.79\) into \(\frac{36}{x}\), A1 for 1 hour 49 minutes.

(a) Differentiate the function to find the gradient: \(\frac{dy}{dx} = 3x^2 - 6x - 9\). Set \(\frac{dy}{dx} = 0\) for turning points: \(3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x-3)(x+1) = 0\). This gives \(x = 3\) or \(x = -1\). When \(x = 3\), \(y = 3^3 - 3(3)^2 - 9(3) + 5 = -22\). When \(x = -1\), \(y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\). The turning points are \((3, -22)\) and \((-1, 10)\). (b) A horizontal line \(y = k\) intersects a cubic curve exactly twice only when it passes through one of its local extrema (the turning points). Hence, \(k = 10\) or \(k = -22\). (c) When \(x = 1\), the y-coordinate is \(y = 1^3 - 3(1)^2 - 9(1) + 5 = -6\). The gradient of the tangent at \(x = 1\) is \(\frac{dy}{dx}\big|_{x=1} = 3(1)^2 - 6(1) - 9 = -12\). Using the point-slope form of a straight line: \(y - (-6) = -12(x - 1) \implies y + 6 = -12x + 12 \implies y = -12x + 6\).

(a) M1 for \(\frac{dy}{dx} = 3x^2 - 6x - 9\), M1 for setting their derivative to 0, A1 for \(x = 3\) and \(x = -1\), A1 for \(y = -22\), A1 for \(y = 10\). (b) M2 for recognizing that \(k\) must equal the y-coordinates of the turning points (M1 for identifying one correct y-value), A1 for \(k = 10\) and \(k = -22\). (c) M1 for finding \(y = -6\) when \(x = 1\), M1 for substituting \(x = 1\) into their \(\frac{dy}{dx}\), A1 for gradient \(m = -12\), M1 for substituting their point and gradient into a straight line equation, A1 for \(y = -12x + 6\).

(a)(i) Angle \(BAC = 135^\circ - 65^\circ = 70^\circ\). (a)(ii) Using the Cosine rule on triangle \(ABC\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC) = 120^2 + 180^2 - 2(120)(180)\cos(70^\circ) = 14400 + 32400 - 43200(0.34202) = 46800 - 14775.3 = 32024.7\). Hence, \(BC = \sqrt{32024.7} \approx 178.95\) m, which rounds to 179 m to 3 s.f. (a)(iii) First, find angle \(ABC\) using the Cosine rule: \(\cos(ABC) = \frac{120^2 + 178.95^2 - 180^2}{2(120)(178.95)} = \frac{14024.7}{42948} = 0.32655 \implies \text{Angle } ABC \approx 70.94^\circ\). The bearing of \(A\) from \(B\) is \(65^\circ + 180^\circ = 245^\circ\). Since \(C\) lies clockwise to the line \(BA\) looking from \(B\), the bearing of \(C\) from \(B\) is \(245^\circ - 70.94^\circ = 174.06^\circ\), which is \(174.1^\circ\) to 1 d.p. (b) Area of triangle \(ABC = \frac{1}{2} \times AB \times AC \times \sin(BAC) = \frac{1}{2} \times 120 \times 180 \times \sin(70^\circ) = 10800 \times 0.93969 = 10148.6\text{ m}^2\), or 10100 \(\text{m}^2\) to 3 s.f. (c) The shortest distance \(d\) from \(A\) to \(BC\) is the perpendicular height. Area = \(\frac{1}{2} \times BC \times d \implies 10148.6 = \frac{1}{2} \times 178.95 \times d \implies d \approx 113.4\) m, or 113 m to 3 s.f.

(a)(i) B1 for \(135 - 65 = 70\). (a)(ii) M1 for correct substitution into Cosine rule: \(120^2 + 180^2 - 2(120)(180)\cos(70)\), M1 for evaluation to \(32025\), A1 for \(179\) or \(178.95\). (a)(iii) M1 for Cosine rule or Sine rule to find angle \(ABC\), A1 for angle \(ABC = 70.9^\circ\) (or \(71^\circ\)), M1 for establishing bearing relation (e.g. \(245^\circ - 70.9^\circ\)), A1 for \(174.1^\circ\) (or \(174^\circ\)). (b) M1 for \(\frac{1}{2} \times 120 \times 180 \times \sin(70)\), A1 for 10100 or 10150. (c) M1 for equating area to \(\frac{1}{2} \times \text{their } BC \times d\), M1 for rearranging to make \(d\) the subject, A1 for 113 or 113.4.

(a) Volume of hemisphere = \(\frac{2}{3}\pi r^3\). Volume of cone = \(\frac{1}{3}\pi r^2 h\). Total volume = \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\). (b)(i) Since \(h = 3r\), Total Volume = \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2(3r) = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\). Set this equal to \(360\pi\): \(\frac{5}{3}\pi r^3 = 360\pi \implies \frac{5}{3}r^3 = 360 \implies r^3 = 216 \implies r = 6\) cm. (b)(ii) \(h = 3r = 3(6) = 18\) cm. (c) Total surface area = Curved surface area of hemisphere + Curved surface area of cone. CSA of hemisphere = \(2\pi r^2 = 2\pi (6^2) = 72\pi \approx 226.19\text{ cm}^2\). Slant height of cone \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 18^2} = \sqrt{360} \approx 18.974\) cm. CSA of cone = \(\pi r l = \pi (6)(18.974) \approx 357.64\text{ cm}^2\). Total Surface Area = \(226.19 + 357.64 \approx 583.83\text{ cm}^2\), or 584 \(\text{cm}^2\) to 3 s.f. (d) Mass = Volume \(\times\) Density. Volume = \(360\pi \approx 1130.97\text{ cm}^3\). Mass = \(1130.97 \times 7.8 = 8821.59\) g. In kg, Mass = \(8821.59 / 1000 \approx 8.82\) kg.

(a) B1 for hemisphere volume formula, B1 for cone volume formula. (b)(i) M1 for substituting \(h = 3r\) into their volume expression, M1 for setting \(\frac{5}{3}\pi r^3 = 360\pi\), A1 for \(r^3 = 216\) leading to \(r = 6\). (b)(ii) B1 for 18. (c) M1 for CSA of hemisphere = \(2\pi(6)^2\), M1 for finding slant height \(l = \sqrt{6^2 + 18^2}\) (or \(\sqrt{360}\)), M1 for CSA of cone = \(\pi(6)(\sqrt{360})\), M1 for adding the two curved surface areas, A1 for 584 or 583.8. (d) M1 for finding actual volume in \(\text{cm}^3\) (approx 1131), M1 for \(1130.97 \times 7.8 / 1000\), A1 for 8.82.

(a)(i) Tangents \(TA\) and \(TB\) are perpendicular to the radii \(OA\) and \(OB\) respectively, so angle \(OAT = OBT = 90^\circ\). Since the sum of angles in a quadrilateral is \(360^\circ\), angle \(AOB = 360^\circ - 90^\circ - 90^\circ - 48^\circ = 132^\circ\). (a)(ii) The angle subtended by an arc at the centre is twice the angle subtended at the circumference. Thus, angle \(ACB = 132^\circ / 2 = 66^\circ\). (a)(iii) In triangle \(OAB\), \(OA = OB\) (both are radii), so triangle \(OAB\) is isosceles. Therefore, angle \(OAB = (180^\circ - 132^\circ) / 2 = 24^\circ\). (b)(i) Each exterior angle of a regular polygon is \(180^\circ - \text{interior angle} = 180^\circ - 168^\circ = 12^\circ\). The sum of exterior angles of any polygon is \(360^\circ\). Thus, number of sides \(n = 360 / 12 = 30\). (b)(ii) Sum of interior angles = \((n - 2) \times 180^\circ = (30 - 2) \times 180^\circ = 28 \times 180^\circ = 5040^\circ\).

(a)(i) B1 for identifying tangent-radius angle is \(90^\circ\), M1 for \(360 - 90 - 90 - 48\), A1 for 132. (a)(ii) M1 for \(132 / 2\), A1 for 66. (a)(iii) M1 for realizing \(OA=OB\) and setting up equation, A1 for 24. (b)(i) M1 for exterior angle = \(180 - 168\), M1 for \(360 / 12\), A1 for 30. (b)(ii) M1 for \((n-2) \times 180\) or \(30 \times 168\), M1 for substituting \(n = 30\), A1 for 5040.

(a) Placing over a common denominator: \(\frac{4(x + 5) - 3(2x - 3)}{(2x - 3)(x + 5)} = \frac{4x + 20 - 6x + 9}{(2x - 3)(x + 5)} = \frac{29 - 2x}{(2x - 3)(x + 5)}\ready. (b) Factorize the numerator: \)2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorize the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Simplifying the fraction: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\). (c) Combining the fractions on the LHS: \(\frac{5(x + 1) + 2(x - 2)}{(x - 2)(x + 1)} = 3 \implies \frac{5x + 5 + 2x - 4}{x^2 - x - 2} = 3 \implies \frac{7x + 1}{x^2 - x - 2} = 3 \implies 7x + 1 = 3(x^2 - x - 2) \implies 7x + 1 = 3x^2 - 3x - 6 \implies 3x^2 - 10x - 7 = 0\). Solving using the quadratic formula: \(x = \frac{10 \pm \sqrt{(-10)^2 - 4(3)(-7)}}{6} = \frac{10 \pm \sqrt{184}}{6}\). Since \(\sqrt{184} \approx 13.565\): \(x_1 = \frac{23.565}{6} \approx 3.93\), and \(x_2 = \frac{-3.565}{6} \approx -0.59\).

(a) M1 for \(4(x + 5) - 3(2x - 3)\), M1 for common denominator \((2x - 3)(x + 5)\), A1 for numerator \(29 - 2x\), A1 for final correct fraction. (b) M1 for factorizing numerator to \((2x + 1)(x - 3)\), M1 for factorizing denominator to \((2x - 1)(2x + 1)\), M1 for cancelling the common factor \((2x + 1)\), A1 for \(\frac{x - 3}{2x - 1}\). (c) M1 for combining fractions: \(\frac{5(x + 1) + 2(x - 2)}{(x - 2)(x + 1)}\), M1 for multiplying out and getting \(7x + 1 = 3x^2 - 3x - 6\), A1 for \(3x^2 - 10x - 7 = 0\), M1 for correct use of quadratic formula on their quadratic, A1 for both \(x = 3.93\) and \(x = -0.59\).

(a)(i) Probability both blue: \(P(BB) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\). (a)(ii) Probability of different colours: \(P(BR) + P(RB) = \left(\frac{7}{12} \times \frac{5}{11}\right) + \left(\frac{5}{12} \times \frac{7}{11}\right) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\). (b)(i) Total beads in the second bag is \(n + 4\). The probability of selecting two green beads without replacement is \(\frac{n}{n + 4} \times \frac{n - 1}{n + 3} = \frac{5}{14}\). Expanding this equation: \(\frac{n^2 - n}{n^2 + 7n + 12} = \frac{5}{14} \implies 14(n^2 - n) = 5(n^2 + 7n + 12) \implies 14n^2 - 14n = 5n^2 + 35n + 60 \implies 9n^2 - 49n - 60 = 0\). (b)(ii) Solve the quadratic equation by factorizing: \(9n^2 - 49n - 60 = 0 \implies (9n + 10)(n - 6) = 0\). Thus, \(n = -10/9\) or \(n = 6\). Since the number of beads must be a positive integer, there are 6 green beads in the bag.

(a)(i) M1 for \(\frac{7}{12} \times \frac{6}{11}\), A1 for \(\frac{7}{22}\). (a)(ii) M1 for \(\frac{7}{12} \times \frac{5}{11}\) or \(\frac{5}{12} \times \frac{7}{11}\), M1 for adding both cases, A1 for \(\frac{35}{66}\). (b)(i) M1 for total number of beads expressed as \(n + 4\), M1 for product of probabilities \(\frac{n}{n+4} \times \frac{n-1}{n+3}\), M1 for setting product equal to \(\frac{5}{14}\), M1 for clearing denominators: \(14(n^2 - n) = 5(n^2 + 7n + 12)\), A1 for correctly expanding and simplifying to \(9n^2 - 49n - 60 = 0\). (b)(ii) M1 for factorizing or using quadratic formula, A1 for finding solutions \(6\) and \(-10/9\), A1 for choosing positive integer \(n = 6\).

(a) \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{36}{x}\) hours.

(b) The speed on the return journey is \(x - 3\) km/h.
\(\text{Time} = \frac{36}{x - 3}\) hours.

(c)(i) The total time is 5.5 hours:
\(\frac{36}{x} + \frac{36}{x - 3} = 5.5\)
Multiply the entire equation by 2 to clear the decimal:
\(\frac{72}{x} + \frac{72}{x - 3} = 11\)
Multiply by the common denominator \(x(x - 3)\):
\(72(x - 3) + 72x = 11x(x - 3)\)
\(72x - 216 + 72x = 11x^2 - 33x\)
\(144x - 216 = 11x^2 - 33x\)
Rearranging terms to one side:
\(11x^2 - 33x - 144x + 216 = 0\)
\(11x^2 - 177x + 216 = 0\)

(c)(ii) Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
With \(a = 11\), \(b = -177\), \(c = 216\):
\(x = \frac{177 \pm \sqrt{(-177)^2 - 4(11)(216)}}{2(11)}\)
\(x = \frac{177 \pm \sqrt{31329 - 9504}}{22}\)
\(x = \frac{177 \pm \sqrt{21825}}{22}\)
\(x_1 = \frac{177 + 147.732867}{22} \approx 14.76058... \approx 14.76\)
\(x_2 = \frac{177 - 147.732867}{22} \approx 1.3303... \approx 1.33\)

(d) For the return journey, the speed must be positive: \(x - 3 > 0 \implies x > 3\).
Therefore, we use the valid solution \(x \approx 14.76058\).
The speed for the return journey is:
\(14.76058 - 3 = 11.76058\) km/h.
Time taken = \(\frac{36}{11.76058} \approx 3.06107\) hours.
Converting the fractional part to minutes:
\(0.06107 \times 60 \approx 3.66\) minutes.
To the nearest minute, this is 4 minutes.
So, the time taken is 3 hours 4 minutes.

(a) B1 for \(\frac{36}{x}\)
(b) B1 for \(\frac{36}{x-3}\)
(c)(i)
M1 for \(\frac{36}{x} + \frac{36}{x-3} = 5.5\) (or equivalent)
M1 for eliminating denominators correctly, e.g., \(36(x-3) + 36x = 5.5x(x-3)\)
M1 for expanding brackets correctly, e.g., \(36x - 108 + 36x = 5.5x^2 - 16.5x\)
A1 for fully correct working leading to the target equation \(11x^2 - 177x + 216 = 0\)
(c)(ii)
M1 for correct substitution into quadratic formula, e.g., \(\frac{177 \pm \sqrt{(-177)^2 - 4(11)(216)}}{2 \times 11}\) (condone one sign error)
B1 for \(21825\) inside the square root (or \(147.7...\))
A1 for \(14.76\) (accept 14.760 to 14.761)
A1 for \(1.33\) (accept 1.330 to 1.331)
(d)
M1 for choosing \(x \approx 14.76\) and calculating return speed \(x - 3\) (approx. \(11.76\))
M1 for calculating \(\frac{36}{\text{their speed}}\) (approx. \(3.06\) hours) and attempting to convert decimal hours to minutes
A1 for 3 hours 4 minutes

(a) Let us find the angle \(ABC\).
The bearing of \(B\) from \(A\) is \(040^\circ\).
The back-bearing from \(B\) to \(A\) is \(040^\circ + 180^\circ = 220^\circ\).
The bearing of \(C\) from \(B\) is \(115^\circ\).
Therefore, the angle \(ABC\) is the difference between these two bearings:
\(\angle ABC = 220^\circ - 115^\circ = 105^\circ\).

Using the cosine rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 8^2 + 12^2 - 2(8)(12)\cos(105^\circ)\)
\(AC^2 = 64 + 144 - 192\cos(105^\circ)\)
\(AC^2 \approx 208 - 192(-0.258819)\)
\(AC^2 \approx 208 + 49.693\)
\(AC^2 \approx 257.693\)
\(AC = \sqrt{257.693} \approx 16.0528\) km.
To 3 significant figures, \(AC = 16.1\) km.

(b) To find the bearing of \(C\) from \(A\), we first find the angle \(BAC\) (let us call it \(\theta\)).
Using the sine rule:
\(\frac{\sin\theta}{12} = \frac{\sin(105^\circ)}{16.0528}\)
\(\sin\theta = 12 \times \frac{\sin(105^\circ)}{16.0528}\)
\(\sin\theta \approx 12 \times 0.060172 \approx 0.72206\)
\(\theta = \sin^{-1}(0.72206) \approx 46.226^\circ\).

The bearing of \(B\) from \(A\) is \(040^\circ\).
Since point \(C\) lies in a clockwise direction relative to the line \(AB\) from \(A\), the bearing of \(C\) from \(A\) is:
\(\text{Bearing} = 40^\circ + 46.226^\circ = 086.226^\circ\).
To 1 decimal place, the bearing is \(086.2^\circ\).

(c) The distance from \(C\) to \(A\) is \(16.0528\) km.
Speed = \(4.5\) km/h.
\(\text{Time} = \frac{16.0528}{4.5} \approx 3.5673\) hours.
Converting the decimal part to minutes:
\(0.5673 \times 60 \approx 34.04\) minutes.
To the nearest minute, this is 34 minutes.
So, the time taken is 3 hours 34 minutes.

(d) Area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2} \times 8 \times 12 \times \sin(105^\circ)\)
\(\text{Area} = 48 \times \sin(105^\circ) \approx 46.364\) \(\text{km}^2\).
To 3 significant figures, the area is \(46.4\) \(\text{km}^2\).

(a)
M1 for calculating angle \(ABC = 105^\circ\) (may be shown on a sketch)
M1 for substituting correctly into the Cosine Rule: \(8^2 + 12^2 - 2(8)(12)\cos(105)\)
A1 for \(257.7\) or \(16.05...\)
A1 for \(16.1\) (or \(16.05\) to \(16.06\))

(b)
M1 for correct use of Sine Rule or Cosine Rule to find angle \(BAC\), e.g., \(\frac{\sin(BAC)}{12} = \frac{\sin(105)}{\text{their } AC}\)
M1 for correct rearrangement to find \(\sin(BAC)\) or \(\cos(BAC)\)
A1 for angle \(BAC = 46.2^\circ\) (accept \(46.2^\circ\) to \(46.3^\circ\))
M1 for adding \(40^\circ\) to their angle \(BAC\)
A1 for \(086.2^\circ\) or \(086.3^\circ\) (accept \(86.2\) or \(86.3\))

(c)
M1 for \(\frac{\text{their } AC}{4.5}\) (approx. \(3.57\) hours)
A1 for 3 hours 34 minutes (accept 3h 34m)

(d)
M1 for \(\frac{1}{2} \times 8 \times 12 \times \sin(105)\)
A1 for \(46.4\) (accept \(46.36\) to \(46.4\))