2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

First, factorise each quadratic and binomial expression:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
\(x^2 - 9 = (x - 3)(x + 3)\)
\(4x^2 - 1 = (2x - 1)(2x + 1)\)
\(2x^2 + 6x = 2x(x + 3)\)

Now substitute these factorised forms into the expression:
\[\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} \div \frac{(2x - 1)(2x + 1)}{2x(x + 3)}\]

Simplify the first fraction by cancelling the common factor \((x - 3)\):
\[\frac{2x + 1}{x + 3} \div \frac{(2x - 1)(2x + 1)}{2x(x + 3)}\]

To divide by a fraction, multiply by its reciprocal:
\[\frac{2x + 1}{x + 3} \times \frac{2x(x + 3)}{(2x - 1)(2x + 1)}\]

Cancel the common factors \((2x + 1)\) and \((x + 3)\):
\[\frac{2x}{2x - 1}\]

M1 for correct factorisation of at least two of the quadratic/binomial expressions: \((2x+1)(x-3)\), \((x-3)(x+3)\), \((2x-1)(2x+1)\), or \(2x(x+3)\).
M1 for converting the division to multiplication by the reciprocal and cancelling common factors.
A1 for \(\frac{2x}{2x-1}\) or equivalent.

Square both sides to eliminate the square root:
\[u^2 = \frac{5v + w}{v - 2}\]

Multiply both sides by \((v - 2)\):
\[u^2(v - 2) = 5v + w\]
\[u^2v - 2u^2 = 5v + w\]

Group the terms with \(v\) on one side:
\[u^2v - 5v = 2u^2 + w\]

Factorise \(v\) on the left-hand side:
\[v(u^2 - 5) = 2u^2 + w\]

Divide by \((u^2 - 5)\):
\[v = \frac{2u^2 + w}{u^2 - 5}\]

M1 for squaring both sides: \(u^2 = \frac{5v + w}{v - 2}\).
M1 for isolating \(v\) terms on one side: \(u^2v - 5v = 2u^2 + w\).
A1 for \(v = \frac{2u^2 + w}{u^2 - 5}\) or equivalent.

The \(y\)-intercept of the line is \(5\), which corresponds to the point \((0, 5)\).
Thus, the line passes through the points \((4, -3)\) and \((0, 5)\).

Find the gradient, \(m\):
\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-3)}{0 - 4} = \frac{8}{-4} = -2\]

Since the \(y\)-intercept is \(5\), \(c = 5\).
The equation of the line is:
\[y = -2x + 5\]

M1 for finding the gradient \(m = -2\) using the points \((4, -3)\) and \((0, 5)\).
A1 for \(y = -2x + 5\).

First, find the gradient of \(L_1\). Rearrange the equation of \(L_1\) into the form \(y = mx + c\):
\[3x - 2y = 8 \implies 2y = 3x - 8 \implies y = \frac{3}{2}x - 4\]
So, the gradient of \(L_1\) is \(m_1 = \frac{3}{2}\).

Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) satisfies:
\[m_2 = -\frac{1}{m_1} = -\frac{2}{3}\]

The equation of \(L_2\) passing through \((-6, 7)\) is:
\[y - 7 = -\frac{2}{3}(x - (-6))\]
\[y - 7 = -\frac{2}{3}(x + 6)\]

Multiply both sides by \(3\) to clear the fraction:
\[3(y - 7) = -2(x + 6)\]
\[3y - 21 = -2x - 12\s] Rearrange into the form \(ax + by = d\): \[2x + 3y = 9\]

M1 for finding the gradient of the perpendicular line \(m_2 = -\frac{2}{3}\).
M1 for substituting their gradient and the point \((-6, 7)\) into \(y - y_1 = m(x - x_1)\).
A1 for \(2x + 3y = 9\).

Let the sector angle be \(\theta^\circ\) and the radius be \(r = 12\text{ cm}\).
The formula for the area of a sector is:
\[\text{Area} = \frac{\theta}{360} \times \pi r^2\]

Substitute the given values:
\[54\pi = \frac{\theta}{360} \times \pi (12)^2\]
\[54 = \frac{\theta}{360} \times 144\]
\[\frac{\theta}{360} = \frac{54}{144} = \frac{3}{8}\]

Now, calculate the arc length, \(L\):
\[L = \frac{\theta}{360} \times 2 \pi r = \frac{3}{8} \times 2 \pi (12) = 9\pi\text{ cm}\]

The perimeter of the sector includes the arc length and two radii:
\[\text{Perimeter} = L + 2r = 9\pi + 2(12) = 9\pi + 24\text{ cm}\]

M1 for finding the fraction of the circle \(\frac{\theta}{360} = \frac{3}{8}\) (or finding \(\theta = 135^\circ\)).
M1 for calculating the arc length \(9\pi\).
A1 for \(9\pi + 24\).

The formula for the arc length \(L\) of a sector with angle \(\theta^\circ\) and radius \(r\) is:
\[L = \frac{\theta}{360} \times 2 \pi r\]

Substitute \(r = 15\) and \(L = 8\pi\):
\[8\pi = \frac{\theta}{360} \times 2 \pi (15)\]
\[8\pi = \frac{\theta}{360} \times 30\pi\]

Divide both sides by \(\pi\):
\[8 = \frac{30\theta}{360}\]
\[8 = \frac{\theta}{12}\]
\[\theta = 8 \times 12 = 96^\circ\]

M1 for setting up the equation \(8\pi = \frac{\theta}{360} \times 30\pi\) or equivalent.
A1 for \(96\).

First, factorise the coefficient of \(x^2\) out of the \(x\)-terms:
\[y = 2(x^2 + 6x) - 5\]

Complete the square inside the bracket:
\[x^2 + 6x = (x + 3)^2 - 3^2 = (x + 3)^2 - 9\]

Substitute this back into the equation:
\[y = 2[(x + 3)^2 - 9] - 5\]
\[y = 2(x + 3)^2 - 18 - 5\]
\[y = 2(x + 3)^2 - 23\]

The minimum point of the curve occurs when the squared term \((x + 3)^2 = 0\), which gives \(x = -3\).
At \(x = -3\), the value of \(y\) is \(-23\).
Therefore, the coordinates of the minimum point are \((-3, -23)\).

M1 for correctly factorising \(2(x^2 + 6x) - 5\).
M1 for completing the square to get \(2(x+3)^2 - 23\).
A1 for \((-3, -23)\).

Since the line is a tangent to the curve, equate the two equations to find their intersection:
\[x^2 - 4x + 12 = 2x + k\]
\[x^2 - 6x + (12 - k) = 0\]

For the line to be a tangent, this quadratic equation must have exactly one real root, which means its discriminant \(\Delta = 0\):
\[\Delta = b^2 - 4ac = 0\]
\[(-6)^2 - 4(1)(12 - k) = 0\]
\[36 - 48 + 4k = 0\]
\[-12 + 4k = 0\]
\[4k = 12 \implies k = 3\]

M1 for equating the line and the curve and setting the discriminant of the resulting quadratic to \(0\).
A1 for \(3\).

First, factorise the numerator and the denominator:

$$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$

$$x^2 - 9 = (x - 3)(x + 3)$$

Now, substitute these back into the fraction and cancel the common factor of \((x - 3)\):

$$\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}$$

M1 for factorising the numerator: \((2x + 1)(x - 3)\)
M1 for factorising the denominator: \((x - 3)(x + 3)\)
A1 for the final simplified fraction: \(\frac{2x + 1}{x + 3}\)

First, find the gradient of the line \(3x + 2y = 8\) by rewriting it in the form \(y = mx + c\):

$$2y = -3x + 8$$

$$y = -1.5x + 4$$

The gradient, \(m\), of the parallel line is also \(-1.5\).

Substitute the gradient \(m = -1.5\) and the point \((4, -3)\) into the equation \(y = mx + c\) to find \(c\):

$$-3 = -1.5(4) + c$$

$$-3 = -6 + c$$

$$c = 3$$

Thus, the equation is \(y = -1.5x + 3\).

M1 for finding the gradient of \(-1.5\) (or \(-\frac{3}{2}\))
A1 for the final equation \(y = -1.5x + 3\) (or equivalent with fractions)

The perimeter of a sector is given by:

$$\text{Perimeter} = 2r + \text{arc length}$$

Substitute the given values to find the arc length:

$$38 = 2(12) + \text{arc length}$$

$$38 = 24 + \text{arc length}$$

$$\text{Arc length} = 14\text{ cm}$$

The formula for the arc length of a sector is:

$$\text{Arc length} = \frac{\theta}{360} \times 2\pi r$$

Substitute the arc length and the radius into the formula:

$$14 = \frac{\theta}{360} \times 2\pi (12)$$

$$14 = \frac{24\pi\theta}{360}$$

$$14 = \frac{\pi\theta}{15}$$

$$\theta = \frac{14 \times 15}{\pi} \approx 66.845^\circ$$

To 1 decimal place, \(\theta = 66.8^\circ\).

M1 for finding arc length = \(14\text{ cm}\)
M1 for setting up the equation: \(14 = \frac{\theta}{360} \times 2 \times \pi \times 12\) or equivalent
A1 for \(66.8\) or \(66.85\) or \(66.84...\)

Method 1: Complete the square on the quadratic expression:

$$x^2 - 8x + 15 = (x - 4)^2 - 16 + 15$$

$$y = (x - 4)^2 - 1$$

This is in vertex form \(y = (x - h)^2 + k\), where the vertex (turning point) is \((h, k)\).
Therefore, the turning point is \((4, -1)\).

Method 2: Use differentiation:

$$\frac{dy}{dx} = 2x - 8$$

At the turning point, \(\frac{dy}{dx} = 0\):

$$2x - 8 = 0 \implies x = 4$$

Substitute \(x = 4\) back into the original equation:

$$y = 4^2 - 8(4) + 15 = 16 - 32 + 15 = -1$$

Thus, the coordinates are \((4, -1)\).

M1 for completed square form \((x - 4)^2 - 1\) OR for finding \(x = 4\) using differentiation or axis of symmetry formula \(x = -\frac{b}{2a}\)
A1 for the coordinates \((4, -1)\)

Multiply both sides by the denominator to clear the fraction:

$$w(3 - \sqrt{t}) = 3 + \sqrt{t}$$

$$3w - w\sqrt{t} = 3 + \sqrt{t}$$

Rearrange to group all terms containing \(\sqrt{t}\) on one side:

$$3w - 3 = w\sqrt{t} + \sqrt{t}$$

Factorise \(\sqrt{t}\) on the right side:

$$3w - 3 = \sqrt{t}(w + 1)$$

Divide by \((w + 1)\):

$$\sqrt{t} = \frac{3w - 3}{w + 1}$$

Square both sides to solve for \(t\):

$$t = \left(\frac{3w - 3}{w + 1}\right)^2$$

M1 for removing the fraction: \(w(3 - \sqrt{t}) = 3 + \sqrt{t}\)
M1 for isolating terms with \(\sqrt{t}\) and factorising: \(\sqrt{t}(w + 1) = 3w - 3\)
A1 for the final formula: \(t = \left(\frac{3w - 3}{w + 1}\right)^2\) or equivalent correct simplified form

Step 1: Find the coordinates of the midpoint of \(AB\):

$$\text{Midpoint} = \left(\frac{2 + 8}{2}, \frac{5 - 7}{2}\right) = (5, -1)$$

Step 2: Find the gradient of line \(AB\):

$$m_{AB} = \frac{-7 - 5}{8 - 2} = \frac{-12}{6} = -2$$

Step 3: The gradient of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):

$$m = \frac{-1}{-2} = 0.5$$

Step 4: Use the gradient \(m = 0.5\) and the midpoint \((5, -1)\) in the equation of the line:

$$y - y_1 = m(x - x_1)$$

$$y - (-1) = 0.5(x - 5)$$

$$y + 1 = 0.5x - 2.5$$

$$y = 0.5x - 3.5$$

B1 for finding the midpoint of \(AB\) at \((5, -1)\)
M1 for calculating the gradient of \(AB\) as \(-2\) and finding the perpendicular gradient as \(0.5\) (or \(\frac{1}{2}\))
A1 for the final equation \(y = 0.5x - 3.5\) (or equivalent in fraction form)

Rewrite both base numbers as powers of 3:

$$9 = 3^2 \quad \text{and} \quad 27 = 3^3$$

Substitute these into the equation:

$$(3^2)^{2x + 1} = (3^3)^{x - 2}$$

Using the index law \((a^m)^n = a^{mn}\):

$$3^{2(2x + 1)} = 3^{3(x - 2)}$$

$$3^{4x + 2} = 3^{3x - 6}$$

Since the bases are equal, equate the exponents:

$$4x + 2 = 3x - 6$$

$$4x - 3x = -6 - 2$$

$$x = -8$$

M1 for writing both sides with a common base of 3: \(3^{2(2x + 1)} = 3^{3(x - 2)}\) or equivalent
A1 for \(x = -8\)

(a) To factorise \(3x^2 - 14x - 5\), we find two numbers that multiply to \(3 \times -5 = -15\) and add to \(-14\). These are \(-15\) and \(1\).
\(3x^2 - 15x + x - 5 = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5)\).

(b) \(\frac{3x^2 - 14x - 5}{x^2 - 25} = \frac{(3x + 1)(x - 5)}{(x - 5)(x + 5)}\).
Cancelling the common factor \((x - 5)\) gives \(\frac{3x + 1}{x + 5}\).

(a) B2 for \((3x+1)(x-5)\) or B1 for \((3x+a)(x+b)\) where \(ab = -5\) or \(a+3b = -14\).

(b) B1 for \((x-5)(x+5)\) seen as the denominator.
M1 for cancelling the common factor \((x-5)\).
A1 for \(\frac{3x+1}{x+5}\) as final answer.

Subtract 5 from both sides:
\(T - 5 = \sqrt{\frac{u - v}{3}}\)

Square both sides:
\((T - 5)^2 = \frac{u - v}{3}\)

Multiply both sides by 3:
\(3(T - 5)^2 = u - v\)

Rearrange to make \(v\) the subject:
\(v = u - 3(T - 5)^2\)

M1 for isolating the root term: \(T - 5 = \sqrt{\frac{u - v}{3}}\)
M1 for squaring both sides: \((T - 5)^2 = \frac{u - v}{3}\)
M1 for multiplying by 3: \(3(T - 5)^2 = u - v\)
A1 for \(v = u - 3(T - 5)^2\) or \(v = u - 3(T^2 - 10T + 25)\)

(a) Gradient of \(AB\) is \(m = \frac{7 - (-1)}{-5 - 3} = \frac{8}{-8} = -1\).
Using point \(A(3, -1)\):
\(y - (-1) = -1(x - 3) \implies y + 1 = -x + 3 \implies y = -x + 2\).

(b) Midpoint of \(AB\) is \(\left(\frac{3 + (-5)}{2}, \frac{-1 + 7}{2}\right) = (-1, 3)\).
The perpendicular gradient is \(-\frac{1}{-1} = 1\).
Using the midpoint \((-1, 3)\):
\(y - 3 = 1(x - (-1)) \implies y = x + 4\).

(a) M1 for gradient calculation \(\frac{7 - (-1)}{-5 - 3}\) (or -1).
A1 for \(y = -x + 2\) or equivalent.

(b) B1 for midpoint \((-1, 3)\).
M1 for perpendicular gradient being \(-\frac{1}{\text{their } m}\).
A1 for \(y = x + 4\).

(a) Gradient of \(L_1\) is \(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\).

(b) Since \(L_2\) is parallel to \(L_1\), its gradient is also \(3\).
Using point \((-3, 1)\):
\(y - 1 = 3(x - (-3))\)
\(y - 1 = 3(x + 3) \implies y = 3x + 10\).

(a) B1 for \(3\).

(b) M1 for using gradient \(m = 3\) (or their answer from part a).
M1 for substituting \((-3, 1)\) into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\).
A1 for \(y = 3x + 10\).

(a) The arc length \(L = \frac{\theta}{360} \times 2\pi r\) and area \(A = \frac{\theta}{360} \times \pi r^2\).
Dividing the two equations gives:
\(\frac{A}{L} = \frac{r}{2}\)
Substitute the given values:
\(\frac{92.4}{15.4} = \frac{r}{2} \implies 6 = \frac{r}{2} \implies r = 12\).

(b) Substitute \(r = 12\) into the arc length formula:
\(15.4 = \frac{\theta}{360} \times 2\pi \times 12\)
\(15.4 = \frac{\theta}{360} \times 24\pi \implies \theta = \frac{15.4 \times 360}{24\pi} \approx 73.53^{\circ}\).
Rounding to the nearest degree gives \(74\).

(a) M1 for writing formula for arc length or area.
M1 for setting up the division \(\frac{92.4}{15.4} = \frac{r}{2}\) or equivalent substitution.
A1 for \(12\).

(b) M1 for substituting \(r=12\) into either formula to solve for \(\theta\).
A1 for \(74\).

Area of the sector:
\(A_{\text{sector}} = \frac{120}{360} \times \pi \times 8^2 = \frac{64\pi}{3} \approx 67.02 \text{ cm}^2\).

Area of the triangle \(OAB\):
\(A_{\text{triangle}} = \frac{1}{2} \times 8 \times 8 \times \sin(120^{\circ}) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.71 \text{ cm}^2\).

Area of the segment:
\(A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = 67.02 - 27.71 = 39.31 \text{ cm}^2\).
Rounding to 3 significant figures gives \(39.3\) \(\text{cm}^2\).

M1 for sector area: \(\frac{120}{360} \times \pi \times 8^2\) (or \(67.0\) to \(67.1\)).
M1 for triangle area: \(\frac{1}{2} \times 8^2 \times \sin(120^{\circ})\) (or \(27.7\)).
M1 for subtracting triangle area from sector area.
A1 for \(39.3\).

(a) Factorise the 2 from the \(x\)-terms:
\(2(x^2 - 6x) + 11\).
Complete the square inside the bracket:
\(2[(x - 3)^2 - 9] + 11 = 2(x - 3)^2 - 18 + 11 = 2(x - 3)^2 - 7\).

(b) The turning point is \((3, -7)\).
Since the coefficient of the quadratic term is positive (\(2 > 0\)), the turning point is a minimum.

(a) M1 for \(2(x^2 - 6x) + 11\).
M1 for completing the square inside: \((x - 3)^2 - 9\).
A1 for \(2(x - 3)^2 - 7\).

(b) B1 for turning point \((3, -7)\).
B1 for minimum.

(a) \frac{2(x^2 - 4)}{(x + 2)(x + 3)} = \frac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{2(x - 2)}{x + 3}

(b) \frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)} = \frac{3x + 9 - 4x + 2}{(2x - 1)(x + 3)} = \frac{11 - x}{(2x - 1)(x + 3)}

(c) (2y - 3)(y^2 + 2y - 8) = 2y(y^2 + 2y - 8) - 3(y^2 + 2y - 8) = 2y^3 + 4y^2 - 16y - 3y^2 - 6y + 24 = 2y^3 + y^2 - 22y + 24

(d) 3a(4b - 3) - 2(4b - 3) = (3a - 2)(4b - 3)

(a) M1 for factorising numerator: 2(x-2)(x+2)
M1 for factorising denominator: (x+2)(x+3)
A1 for final simplified fraction

(b) M1 for common denominator: (2x-1)(x+3)
M1 for 3(x+3) - 2(2x-1)
A1 for final simplified fraction

(c) M1 for expanding two brackets (e.g. y^2 + 2y - 8)
M1 for multiplying by third bracket
A1 for any correct intermediate expanded 6-term expression
A1 for final simplified 4-term expression

(d) M1 for grouping and partial factorisation: 3a(4b-3) - 2(4b-3)
A1 for final answer

(a) Gradient of AB = \frac{-3 - 5}{5 - 1} = -2

(b) y - 5 = -2(x - 1) => y = -2x + 7

(c) Midpoint of AB = (\frac{1 + 5}{2}, \frac{5 - 3}{2}) = (3, 1)

(d) Gradient of perpendicular bisector = \frac{-1}{-2} = 0.5. Since it passes through (3, 1): y - 1 = 0.5(x - 3) => y = 0.5x - 0.5

(e) Length of AB = \sqrt{(5 - 1)^2 + (-3 - 5)^2} = \sqrt{16 + 64} = \sqrt{80} \approx 8.94

(a) M1 for rise/run formula substitution: \frac{-3-5}{5-1}
A1 for gradient = -2

(b) M1 for using their gradient in y - y1 = m(x - x1) or y = mx + c
A1 for y = -2x + 7

(c) M1 for midpoint formula substitution
A1 for (3, 1)

(d) M1 for perpendicular gradient = -1 / their gradient
M1 for substituting midpoint (3, 1) into y = mx + c with their perp gradient
A1 for y = 0.5x - 0.5

(e) M1 for distance formula substitution
A1 for 8.94

(a) Arc length = \frac{\theta}{360} \times 2\pi r => 35 = \frac{\theta}{360} \times 2\pi \times 75 => \theta = \frac{35 \times 360}{150\pi} \approx 26.738 \approx 26.7^\circ

(b) Area of sector = \frac{1}{2} \times r \times \text{arc length} = \frac{1}{2} \times 75 \times 35 = 1312.5\text{ cm}^2 (or using 26.7: \frac{26.7}{360} \times \pi \times 75^2 \approx 1310.6\text{ cm}^2)

(c) New area = 1312.5 \times 0.85 = 1115.625\text{ cm}^2. Since area is proportional to $r^2$ when the angle is constant: R^2 = 75^2 \times 0.85 => R = 75 \times \sqrt{0.85} \approx 69.146 \approx 69.1\text{ cm}

(d) New arc length = \frac{26.738}{360} \times 2\pi \times 69.146 \approx 32.3\text{ cm}

(a) M1 for arc length formula substitution: \frac{\theta}{360} \times 2\pi \times 75 = 35
M1 for rearranging for \theta
A1 for showing 26.738... which rounds to 26.7

(b) M1 for sector area formula substitution
A1 for 1310 or 1312.5

(c) M1 for calculating 85% of original area: 1115.625
M1 for setting up new area equation: \frac{26.7}{360} \times \pi \times R^2 = 1115.6
M1 for taking the square root
A1 for 69.1

(d) M1 for new arc length formula substitution or using linear scale factor
M1 for substituting their R into formula
A1 for 32.3

(a) For x = -1: y = -1 - 4 + 2 = -3. For x = 2: y = 8 - 16 + 2 = -6. For x = 3.5: y = 42.875 - 49 + 2 = -4.125.

(b) Plotting the points and drawing a smooth curve.

(c) Draw horizontal line y = -3 and find the x-values of intersection: x = -1, x \approx 1.4, x \approx 3.6.

(d) Rearrange $x^3 - 4x^2 - 2x + 4 = 0 \implies x^3 - 4x^2 + 2 = 2x - 2$. The line to draw is $y = 2x - 2$. Plot the line and find the x-coordinates of the intersection points: x \approx -1.15, 0.85, 4.3.

(a) B1 for each correct missing value (-3, -6, -4.125)

(b) B2 for plotting at least 6 points correctly
B2 for drawing a smooth, continuous curve through the plotted points

(c) M1 for drawing the line y = -3 (or identifying intersections)
A1 for finding all three solutions: -1, [1.3 to 1.5], [3.5 to 3.7]

(d) M1 for rearranging equation to find the line equation: y = 2x - 2
M1 for drawing the line y = 2x - 2 on the grid
A1 for finding the x-values of the intersections (typically around -1.15, 0.85, 4.3)

(a) Height of the cone, $h = \sqrt{10^2 - 6^2} = 8\text{ cm}$. Volume of cone = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6^2)(8) = 96\pi \approx 301.6\text{ cm}^3. Volume of cylinder = \pi r^2 H = \pi (6^2)(10) = 360\pi \approx 1131.0\text{ cm}^3. Total volume = 96\pi + 360\pi = 456\pi \approx 1432.6 \approx 1430\text{ cm}^3.

(b) Curved surface area of cone = \pi r l = \pi (6)(10) = 60\pi \approx 188.5\text{ cm}^2. Curved surface area of cylinder = 2\pi r H = 2\pi (6)(10) = 120\pi \approx 377.0\text{ cm}^2. Total surface area (excluding bottom base and joined surfaces) = 60\pi + 120\pi = 180\pi \approx 565\text{ cm}^2.

(c) Mass = Volume \times Density = 456\pi \times 0.75 = 342\pi\text{ grams} \approx 1074.4\text{ grams} = 1.07\text{ kg}.

(a) M1 for finding the height of the cone: \sqrt{10^2 - 6^2} = 8
M1 for volume of cone formula substitution: \frac{1}{3} \pi \times 6^2 \times 8
M1 for volume of cylinder formula substitution: \pi \times 6^2 \times 10
M1 for adding the two volumes
A1 for 1430

(b) M1 for curved surface area of cone formula substitution: \pi \times 6 \times 10
M1 for curved surface area of cylinder formula substitution: 2 \times \pi \times 6 \times 10
M1 for adding both curved surface areas (no flat surfaces)
A1 for 565

(c) M1 for multiplying their volume by 0.75
M1 for dividing by 1000 to convert to kg
A1 for 1.07

(a) PR^2 = 120^2 + 150^2 - 2(120)(150)\cos(74^\circ) = 14400 + 22500 - 36000(0.275637) = 36900 - 9922.95 = 26977.05 => PR = \sqrt{26977.05} \approx 164.25 \approx 164\text{ m}.

(b) Area = \frac{1}{2} \times 120 \times 150 \times \sin(74^\circ) = 9000 \sin(74^\circ) \approx 8651.4 \approx 8650\text{ m}^2.

(c) (i) In right-angled triangle TQP: \tan(18^\circ) = \frac{TQ}{120} => TQ = 120\tan(18^\circ) \approx 38.99 \approx 39.0\text{ m}.

(ii) In right-angled triangle TQR: \tan(\alpha) = \frac{TQ}{QR} = \frac{38.99}{150} \approx 0.2599 => \alpha = \arctan(0.2599) \approx 14.57^\circ \approx 14.6^\circ.

(a) M1 for cosine rule formula substitution: 120^2 + 150^2 - 2(120)(150)\cos(74)
M1 for correct evaluation of terms: 36900 - 9923
A1 for 164

(b) M1 for area formula substitution: \frac{1}{2} \times 120 \times 150 \times \sin(74)
A1 for 8650

(c) (i) M1 for identifying right-angled triangle TQP
M1 for using tangent ratio: TQ = 120 \tan(18)
A1 for 39.0

(ii) M1 for identifying right-angled triangle TQR
M1 for tangent ratio: \tan(\alpha) = \frac{\text{height}}{150}
M1 for substituting their height into tangent ratio
A1 for 14.6

(a) P(BBB) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720} = \frac{1}{30} \approx 0.0333

(b) P(exactly 2 Red) = P(RRD) + P(RDR) + P(DRR), where D is non-red (8 pens). P(RRD) = \frac{2}{10} \times \frac{1}{9} \times \frac{8}{8} = \frac{16}{720}. Total probability = 3 \times \frac{16}{720} = \frac{48}{720} = \frac{1}{15} \approx 0.0667

(c) P(at least 1 Red) = 1 - P(no Red) = 1 - \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} = 1 - \frac{336}{720} = 1 - \frac{7}{15} = \frac{8}{15} \approx 0.533

(d) P(different colours) = 6 \times P(B, Bl, R) = 6 \times (\frac{4}{10} \times \frac{4}{9} \times \frac{2}{8}) = 6 \times \frac{32}{720} = \frac{192}{720} = \frac{4}{15} \approx 0.267

(a) M1 for \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8}
A1 for 1/30 or 0.0333

(b) M1 for any permutation: \frac{2}{10} \times \frac{1}{9} \times \frac{8}{8}
M1 for multiplying by 3
A1 for 1/15 or 0.0667

(c) M1 for P(no Red): \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8}
M1 for subtracting from 1
A1 for 8/15 or 0.533

(d) M1 for one correct product: \frac{4}{10} \times \frac{4}{9} \times \frac{2}{8}
M1 for multiplying by 6 (or 3!)
A1 for 4/15 or 0.267

(a) Next terms are: A: 14 + 3 = 17. B: 5^2 + 1 = 26. C: 162 \times 3 = 486.

(b) (i) Linear sequence with common difference of 3. 3n + c => 3(1) + c = 5 => c = 2. Term is 3n + 2.
(ii) Sequence of square numbers plus 1: n^2 + 1.
(iii) Geometric sequence with ratio 3. 6 \times 3^{n-1} = 2 \times 3^n.

(c) n^2 - 10n + 30 = 105 => n^2 - 10n - 75 = 0 => (n - 15)(n + 5) = 0. Since n must be positive, n = 15.

(a) B1 for each correct next term (17, 26, 486)

(b) (i) M1 for finding common difference 3 (implied by 3n)
A1 for 3n + 2
(ii) M1 for recognizing square relationship (implied by n^2)
A1 for n^2 + 1
(iii) M1 for geometric term: a \times 3^{n-1}
A1 for 2 \times 3^n (or 6 \times 3^{n-1})

(c) M1 for setting up equation: n^2 - 10n + 30 = 105
M1 for factorising their quadratic: (n - 15)(n + 5) = 0
A1 for n = 15

\((a)\) Arc length of the sector: \(L = \frac{135}{360} \times 2 \times \pi \times 8.4 = 6.3\pi \approx 19.792\text{ cm}\).
Perimeter of the sector = \(L + 2R = 19.792 + 2(8.4) = 36.592 \approx 36.6\text{ cm}\).

\((b)(i)\) The arc length of the sector becomes the circumference of the base of the cone:
\(2\pi r = 6.3\pi \implies r = 3.15\text{ cm}\).

\((b)(ii)\) The radius of the sector is the slant height, \(l\), of the cone:
\(l = 8.4\text{ cm}\).
Using Pythagoras' theorem to find the height, \(h\):
\(h = \sqrt{l^2 - r^2} = \sqrt{8.4^2 - 3.15^2} = \sqrt{70.56 - 9.9225} = \sqrt{60.6375} \approx 7.787 \approx 7.79\text{ cm}\).

\((c)\) Volume of the cone:
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3.15)^2 (7.787) \approx 80.917 \approx 80.9\text{ cm}^3\).

\((a)\)
- M1 for \(\frac{135}{360} \times 2 \times \pi \times 8.4\) or better
- M1 for adding \(2 \times 8.4\) to their arc length
- A1 for \(36.6\) or \(36.59\) to \(36.60\)

\((b)(i)\)
- M2 for \(2\pi r = 19.79\) or \(2\pi r = \text{their arc length}\)
- A1 for \(3.15\)

\((b)(ii)\)
- M1 for identifying slant height \(l = 8.4\)
- M1 for \(h^2 = 8.4^2 - \text{their } r^2\)
- A1 for \(7.79\) or \(7.787\) to \(7.788\)

\((c)\)
- M1 for \(\frac{1}{3} \times \pi \times r^2 \times h\) using their \(r\) and \(h\)
- A1 for \(80.9\) or \(80.91\) to \(80.92\)

\((a)\) Gradient of \(AB\), \(m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\).
Using point \((2, 3)\) in \(y - y_1 = m(x - x_1)\):
\(y - 3 = 2(x - 2) \implies y = 2x - 1\).

\((b)\) Midpoint \(M = \left(\frac{2+6}{2}, \frac{3+11}{2}\right) = (4, 7)\).

\((c)\) Gradient of the perpendicular bisector is the negative reciprocal of the gradient of \(AB\):
\(m_{\perp} = -\frac{1}{2} = -0.5\).
Using the midpoint \((4, 7)\):
\(y - 7 = -0.5(x - 4) \implies y = -0.5x + 9\).

\((d)\) Point \(P\) is the y-intercept: when \(x = 0\), \(y = 9 \implies P(0, 9)\).
Point \(Q\) is the x-intercept: when \(y = 0\), \(0 = -0.5x + 9 \implies x = 18 \implies Q(18, 0)\).
Distance \(PQ = \sqrt{(18 - 0)^2 + (0 - 9)^2} = \sqrt{324 + 81} = \sqrt{405} \approx 20.1\text{ units}\).

\((a)\)
- M1 for gradient \(\frac{11 - 3}{6 - 2}\) or better
- M1 for substituting their gradient into \(y = mx + c\) with either point
- A1 for \(y = 2x - 1\)

\((b)\)
- M1 for \(\left(\frac{2+6}{2}, \frac{3+11}{2}\right)\)
- A1 for \((4, 7)\)

\((c)\)
- M1 for gradient of perpendicular bisector \(= -\frac{1}{\text{their } m}\)
- M1 for substituting their perpendicular gradient and their midpoint \((4, 7)\) into \(y - y_1 = m_{\perp}(x - x_1)\)
- A1 for \(y = -0.5x + 9\) or \(y = -\frac{1}{2}x + 9\)

\((d)\)
- B1 for finding coordinates \(P(0, 9)\) and \(Q(18, 0)\) (or values \(9\) and \(18\))
- M1 for \(\sqrt{(18 - 0)^2 + (0 - 9)^2}\) using their values
- A1 for \(20.1\) or \(20.12\) to \(20.13\)

\((a)(i)\) Time = \(\frac{180}{x}\)
\((a)(ii)\) Time = \(\frac{180}{x + 15}\)

\((b)\) Since the return journey is 1 hour quicker:
\(\frac{180}{x} - \frac{180}{x + 15} = 1\)
Multiply both sides by \(x(x + 15)\):
\(180(x + 15) - 180x = x(x + 15)\)
\(180x + 2700 - 180x = x^2 + 15x\)
\(x^2 + 15x - 2700 = 0\) (as required)

\((c)\) Factorizing \(x^2 + 15x - 2700 = 0\):
\((x - 45)(x + 60) = 0\)
So \(x = 45\) or \(x = -60\).
Since speed must be positive, \(x = 45\text{ km/h}\).

\((d)\) Outward time = \(\frac{180}{45} = 4\text{ hours}\).
Return time = \(\frac{180}{60} = 3\text{ hours}\).
Total time = \(4 + 3 = 7\text{ hours}\).

\((a)(i)\)
- B1 for \(\frac{180}{x}\)

\((a)(ii)\)
- B1 for \(\frac{180}{x + 15}\)

\((b)\)
- M1 for \(\frac{180}{x} - \frac{180}{x + 15} = 1\) (or equivalent)
- M1 for algebraic step multiplying by common denominator: \(180(x + 15) - 180x = x(x + 15)\)
- M1 for expanding brackets correctly: \(180x + 2700 - 180x = x^2 + 15x\)
- A1 for establishing the final simplified form \(x^2 + 15x - 2700 = 0\) clearly with no errors.

\((c)\)
- M1 for correct attempt to solve the quadratic (either factorizing, completing the square or quadratic formula)
- A1 for finding factors \((x - 45)(x + 60) = 0\) or roots \(45\) and \(-60\)
- A1 for choosing the positive root \(x = 45\) as the outward speed

\((d)\)
- M1 for calculating \(\frac{180}{\text{their } x}\)
- M1 for calculating \(\frac{180}{\text{their } x + 15}\)
- A1 for \(7\) hours