2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Factor the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
Factor the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\).
Divide both the numerator and the denominator by the common factor \((2x + 1)\):
\(\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} = \frac{x-3}{2x-1}\).

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

Multiply both sides by \(2 - t\) to clear the fraction:
\(w(2 - t) = 3t + 5\)
Expand the left side:
\(2w - wt = 3t + 5\)
Collect terms with \(t\) on one side and other terms on the opposite side:
\(2w - 5 = 3t + wt\)
Factorize the right side to isolate \(t\):
\(2w - 5 = t(3 + w)\)
Divide by \(3 + w\) (or \(w + 3\)) to solve for \(t\):
\(t = \frac{2w - 5}{w + 3}\).

M1 for multiplying to remove the fraction: \(w(2 - t) = 3t + 5\)
M1 for collecting terms containing \(t\) and factorizing: \(t(3 + w) = 2w - 5\)
A1 for \(t = \frac{2w - 5}{w + 3}\) or equivalent correct expression.

The volume of a hemisphere is given by \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\).
The volume of a cylinder is given by \(V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\).
The total volume of the composite solid is the sum of these volumes:
\(V_{\text{total}} = \frac{2}{3}\pi r^3 + 2\pi r^3 = \frac{8}{3}\pi r^3\).
Set this equal to the given total volume:
\(\frac{8}{3}\pi r^3 = 72\pi\)
Divide both sides by \(\pi\):
\(\frac{8}{3}r^3 = 72\)
Multiply by \(3\) and divide by \(8\):
\(r^3 = 72 \times \frac{3}{8} = 27\)
Take the cube root of both sides:
\(r = 3\).

M1 for expressing either volume correctly in terms of \(r\): \(\frac{2}{3}\pi r^3\) or \(2\pi r^3\)
M1 for setting up the total volume equation: \(\frac{8}{3}\pi r^3 = 72\pi\)
A1 for \(r = 3\)

The ratio of the volumes of similar solids is the cube of the scale factor \(k\) of their linear dimensions:
\(k^3 = \frac{V_1}{V_2} = \frac{24\pi}{81\pi} = \frac{24}{81} = \frac{8}{27}\).
Therefore, the linear scale factor is:
\(k = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\).
The ratio of their surface areas is the square of the linear scale factor:
\(k^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
Let \(A\) be the curved surface area of the larger cone. We set up the area ratio:
\(\frac{36}{A} = \frac{4}{9}\)
Solve for \(A\):
\(A = 36 \times \frac{9}{4} = 81\) \(\text{cm}^2\).

M1 for finding the linear scale factor ratio: \(\sqrt[3]{\frac{24}{81}} = \frac{2}{3}\) (or reciprocal)
M1 for finding the area scale factor ratio: \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\) (or reciprocal)
A1 for \(81\)

1. Find the midpoint of \(AB\):
Midpoint \(M = \left(\frac{-3 + 5}{2}, \frac{8 + 2}{2}\right) = (1, 5)\).
2. Find the gradient of line \(AB\):
\(m_{AB} = \frac{2 - 8}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4}\).
3. Find the gradient of the perpendicular line:
\(m_{\perp} = -\frac{1}{m_{AB}} = \frac{4}{3}\).
4. Write the equation of the perpendicular bisector passing through \((1, 5)\):
\(y - 5 = \frac{4}{3}(x - 1)\)
Multiply by 3:
\(3(y - 5) = 4(x - 1)\)
\(3y - 15 = 4x - 4\)
Rearrange into the form \(ay + bx = c\):
\(3y - 4x = 11\).
Here \(a=3\) (which is \(>0\)), \(b=-4\), and \(c=11\). These have no common factors.

M1 for finding the midpoint \((1, 5)\) OR finding the gradient of \(AB\) as \(-\frac{3}{4}\)
M1 for finding the perpendicular gradient as \(\frac{4}{3}\) and substituting \((1, 5)\) into \(y - y_1 = m_{\perp}(x - x_1)\)
A1 for \(3y - 4x = 11\) (or any equivalent integer format with \(a > 0\) having no common factors)

Find the gradient of line \(L_1\):
\(m_1 = \frac{2 - (-1)}{8 - 2} = \frac{3}{6} = \frac{1}{2}\).
Since \(L_2\) is perpendicular to \(L_1\), its gradient is the negative reciprocal:
\(m_2 = -\frac{1}{m_1} = -2\).
Write down the equation of line \(L_2\) in the form \(y = mx + c\) using the point \((4, 5)\):
\(5 = -2(4) + c\)
\(5 = -8 + c\)
\(c = 13\).
Therefore, the equation of the line \(L_2\) is \(y = -2x + 13\).

M1 for gradient of \(L_1\) as \(\frac{1}{2}\)
M1 for perpendicular gradient as \(-2\) and attempting to substitute \((4, 5)\) to find \(c\)
A1 for \(y = -2x + 13\)

Apply the Cosine Rule to find angle \(ABC\) (which is opposite side \(AC\)):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
Substitute the given lengths:
\(15^2 = 7^2 + 12^2 - 2(7)(12)\cos(ABC)\)
\(225 = 49 + 144 - 168\cos(ABC)\)
\(225 = 193 - 168\cos(ABC)\)
Subtract 193 from both sides:
\(32 = -168\cos(ABC)\)
Calculate \(\cos(ABC)\):
\(\cos(ABC) = -\frac{32}{168} = -\frac{4}{21}\)
Solve for the angle:
\(ABC = \arccos\left(-\frac{4}{21}\right) \approx 100.98^\circ\).
Rounding to one decimal place gives \(101.0^\circ\).

M1 for correct substitution into Cosine Rule: \(15^2 = 7^2 + 12^2 - 2(7)(12)\cos(ABC)\)
M1 for rearrangement to isolate cosine term: \(\cos(ABC) = -\frac{32}{168}\) (or \(-0.1905\))
A1 for \(101.0\) or \(101\)

Let \(P\) be the original price of the television.
After a 15% reduction, the price is \(0.85P\).
After a further 10% reduction, the price is \(0.85P \times 0.90 = 0.765P\).
Set up the equation with the final promotional price:
\(0.765P = 382.50\)
Solve for \(P\):
\(P = \frac{382.50}{0.765} = 500\).
The original price of the television was \(\$500\).

M1 for compound multiplier \(0.85 \times 0.90 = 0.765\) or for showing step-by-step reverse percentage calculation (e.g., \(382.50 / 0.90 = 425\))
M1 for calculating \(\frac{382.50}{0.765}\) or \(\frac{425}{0.85}\)
A1 for \(500\)

Multiply both sides by \((5 - 2x)\):
\(y(5 - 2x) = 3x + a\)

Expand the brackets:
\(5y - 2xy = 3x + a\)

Rearrange to collect all terms with \(x\) on one side and the remaining terms on the other side:
\(5y - a = 3x + 2xy\)

Factor out \(x\) on the right-hand side:
\(5y - a = x(3 + 2y)\)

Divide both sides by \((3 + 2y)\) to make \(x\) the subject:
\(x = \frac{5y - a}{3 + 2y}\) or \(x = \frac{5y - a}{2y + 3}\)

M1 for correctly multiplying by the denominator to get \(y(5 - 2x) = 3x + a\)
M1 for isolating terms containing \(x\) on one side, e.g., \(3x + 2xy = 5y - a\)
A1 for the correct final expression: \(x = \frac{5y - a}{3 + 2y}\) (or equivalent)

First, find the gradient of the given line \(3x - 4y = 12\):
\(4y = 3x - 12 \implies y = \frac{3}{4}x - 3\)
The gradient of this line is \(m_1 = \frac{3}{4}\).

The gradient of the perpendicular line is the negative reciprocal of \(m_1\):
\(m_2 = -\frac{1}{m_1} = -\frac{4}{3}\).

Use the point-slope formula with point \((6, -2)\):
\(y - (-2) = -\frac{4}{3}(x - 6)\)
\(y + 2 = -\frac{4}{3}x + 8\)
\(y = -\frac{4}{3}x + 6\)

M1 for finding the gradient of the given line as \(\frac{3}{4}\) or stating the perpendicular gradient as \(-\frac{4}{3}\)
M1 for substituting their perpendicular gradient and the point \((6, -2)\) into a linear equation
A1 for the correct final equation \(y = -\frac{4}{3}x + 6\) (or equivalent)

The volume of a sphere is given by:
\(V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3\)

The volume of a cone is given by:
\(V_{\text{cone}} = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi R^2 (8) = \frac{8}{3}\pi R^2\)

Since the volume remains the same:
\(\frac{8}{3}\pi R^2 = 36\pi\)
\(\frac{8}{3} R^2 = 36\)
\(R^2 = 36 \times \frac{3}{8} = 13.5\)
\(R = \sqrt{13.5} \approx 3.6742...\text{ cm}\)

Correct to 3 significant figures, \(R = 3.67\text{ cm}\).

M1 for equating volume of a sphere to volume of a cone with correct formulae: \(\frac{4}{3}\pi (3)^3 = \frac{1}{3}\pi R^2 (8)\)
M1 for rearranging to find \(R^2 = 13.5\) or \(R = \sqrt{13.5}\)
A1 for the correct answer \(3.67\) (accept \(3.67\) to \(3.68\))

Using the Cosine Rule to find side \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 7.2^2 + 5.4^2 - 2(7.2)(5.4)\cos(112^\circ)\)
\(AC^2 = 51.84 + 29.16 - 77.76 \times (-0.37460659)\)
\(AC^2 = 81.0 + 29.1294 = 110.1294\)
\(AC = \sqrt{110.1294} \approx 10.494\text{ cm}\)

Correct to 3 significant figures, \(AC = 10.5\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7.2^2 + 5.4^2 - 2(7.2)(5.4)\cos(112^\circ)\)
M1 for evaluating to \(AC^2 \approx 110.1...\) or \(AC = \sqrt{110.1...}\)
A1 for \(10.5\) (accept \(10.49\) to \(10.5\))

Let the original price be \(P\).
An increase of \(15\%\) means:
\(1.15P = 41.40\)
\(P = \frac{41.40}{1.15} = 36\)
So the original price is \(\$36\).

In the sale, the new price \(\$41.40\) is reduced by \(10\%\):
\(\text{Sale Price} = 41.40 \times (1 - 0.10) = 41.40 \times 0.90 = 37.26\)

Now, calculate the percentage change from the original price (\(\$36\)) to the sale price (\(\$37.26\)):
\(\text{Percentage Change} = \frac{37.26 - 36}{36} \times 100 = \frac{1.26}{36} \times 100 = 3.5\%\)

M1 for finding the original price: \(41.40 \div 1.15 = 36\)
M1 for finding the sale price: \(41.40 \times 0.90 = 37.26\)
A1 for \(3.5\%\) or \(3.5\) (accept 3.5% increase)

Factor the numerator, which is a quadratic expression:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Factor the denominator, which is a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now substitute back into the fraction and cancel common factors:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

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

First, find the midpoint of the line segment \(AB\):
\(M = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)\)

Next, find the gradient of \(AB\):
\(m_{AB} = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}\)

The gradient of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):
\(m_{\perp} = -\frac{1}{-2/3} = \frac{3}{2} = 1.5\)

Now, use the point-slope formula with the midpoint \((1, 3)\):
\(y - 3 = 1.5(x - 1)\)
\(y - 3 = 1.5x - 1.5\)
\(y = 1.5x + 1.5\)

M1 for finding the midpoint \((1, 3)\)
M1 for finding the perpendicular gradient \(\frac{3}{2}\) or \(1.5\)
A1 for the correct final equation \(y = 1.5x + 1.5\) (or \(y = \frac{3}{2}x + \frac{3}{2}\))

The volume of a hemisphere is given by:
\(V = \frac{2}{3}\pi r^3\)

Equating to the given volume:
\(\frac{2}{3}\pi r^3 = 144\pi\)
\(\frac{2}{3} r^3 = 144\)
\(r^3 = 144 \times \frac{3}{2} = 216\)
\(r = \sqrt[3]{216} = 6\text{ cm}\)

The total surface area of a solid hemisphere consists of the curved surface area plus the flat circular base:
\(A_{\text{total}} = 2\pi r^2 + \pi r^2 = 3\pi r^2\)

Substituting \(r = 6\):
\(A_{\text{total}} = 3\pi (6)^2 = 3\pi (36) = 108\pi\text{ cm}^2\)

M1 for finding the radius \(r = 6\) from the volume formula
M1 for using \(3\pi r^2\) for the total surface area of a solid hemisphere
A1 for the correct final answer \(108\pi\)

Factorise the numerator:
\(2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 = 2x(x-5) + 3(x-5) = (2x+3)(x-5)\).

Factorise the denominator:
\(x^2 - 25 = (x-5)(x+5)\).

Divide the numerator by the denominator to simplify:
\(\frac{(2x+3)(x-5)}{(x-5)(x+5)} = \frac{2x+3}{x+5}\).

M1 for factorising the numerator: \((2x+3)(x-5)\)
M1 for factorising the denominator: \((x-5)(x+5)\)
A1 for final simplified fraction \(\frac{2x+3}{x+5}\) (or equivalent)

The total surface area of a solid hemisphere is composed of its curved surface area and its circular base:
\text{Total Surface Area} = 2\pi r^2 + \pi r^2 = 3\pi r^2\).

Setting this equal to the given area:
\(3\pi r^2 = 108\pi \implies r^2 = 36 \implies r = 6\text{ cm}\).

The volume of a hemisphere is given by:
\(V = \frac{2}{3}\pi r^3\).

Substituting \(r = 6\):
\(V = \frac{2}{3}\pi (6^3) = \frac{2}{3}\pi (216) = 144\pi\text{ cm}^3\).

M1 for setting up the area equation \(3\pi r^2 = 108\pi\) and finding \(r = 6\)
M1 for substituting their \(r\) into the volume formula \(\frac{2}{3}\pi r^3\)
A1 for final answer \(144\pi\)

First, find the gradient of \(L_1\):
\(m_1 = \frac{6 - 9}{8 - 2} = \frac{-3}{6} = -0.5\).

The gradient of \(L_2\) (perpendicular to \(L_1\)) is:
\(m_2 = -\frac{1}{-0.5} = 2\).

Next, find the midpoint \(M\) of the line segment:
\(M = \left(\frac{2+8}{2}, \frac{9+6}{2}\right) = (5, 7.5)\).

Using the point-gradient form for \(L_2\):
\(y - 7.5 = 2(x - 5) \implies y = 2x - 10 + 7.5 \implies y = 2x - 2.5\).

M1 for gradient of perpendicular line \(m_2 = 2\) (or for showing gradient of \(L_1 = -0.5\))
M1 for finding the midpoint \((5, 7.5)\)
A1 for equation \(y = 2x - 2.5\) (or equivalent, such as \(y = 2x - \frac{5}{2}\))

Using the Sine Rule:
\(\frac{\sin(ACB)}{AB} = \frac{\sin(BAC)}{BC}\)

Substituting the given values:
\(\frac{\sin(ACB)}{7} = \frac{\sin(110^\circ)}{11}\)

Solving for \(\sin(ACB)\):
\(\sin(ACB) = \frac{7 \sin(110^\circ)}{11} \approx 0.59798\)

Calculating the angle:
\(ACB = \sin^{-1}(0.59798) \approx 36.72^\circ\).

To 1 decimal place, the angle is \(36.7^\circ\).

M1 for correct substitution into Sine Rule: \(\frac{\sin(ACB)}{7} = \frac{\sin(110^\circ)}{11}\)
M1 for rearranging to find \(\sin(ACB) \approx 0.598\)
A1 for \(36.7\) (accept answers in range \(36.7\) to \(36.8\))

Let the original price be \(P\).

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 the final price is $459:
\(0.765P = 459 \implies P = \frac{459}{0.765} = 600\).

M1 for setting up the equation \(0.85 \times 0.90 \times P = 459\) or finding overall multiplier \(0.765\)
M1 for dividing \(459\) by their multiplier (or finding the intermediate price after first discount, \(510\))
A1 for \(600\)

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

Substituting the given values:
\(2924.65 = 2500\left(1 + \frac{r}{100}\right)^3\)

Dividing both sides by 2500:
\(\left(1 + \frac{r}{100}\right)^3 = \frac{2924.65}{2500} = 1.16986\)

Taking the cube root of both sides:
\(1 + \frac{r}{100} = \sqrt[3]{1.16986} \approx 1.0537\)

Solving for \(r\):
\(\frac{r}{100} = 0.0537 \implies r = 5.37\).

M1 for setting up equation \(2500(1 + r/100)^3 = 2924.65\)
M1 for taking the cube root of \(1.16986\) to get \(1.0537\)
A1 for \(5.37\) (accept answers in the range \(5.36\) to \(5.38\))

(a) Since there is a border of \(2\) cm all around, we add \(2 \times 2 = 4\) cm to both the length and width of the picture.
Length of cardboard = \(x + 5 + 4 = x + 9\) cm.
Width of cardboard = \(2x - 3 + 4 = 2x + 1\) cm.

(b) Area of cardboard \(A = (x + 9)(2x + 1) = 2x^2 + x + 18x + 9 = 2x^2 + 19x + 9\) cm\(^2\).

(c) Set \(A = 84\):
\(2x^2 + 19x + 9 = 84 \Rightarrow 2x^2 + 19x - 75 = 0\).
Factorizing the quadratic equation:
\((2x + 25)(x - 3) = 0\).
This gives \(x = 3\) or \(x = -12.5\). Since length must be positive, we select \(x = 3\).
Dimensions of the picture:
Length = \(3 + 5 = 8\) cm.
Width = \(2(3) - 3 = 3\) cm.

(d) Common denominator is \((2x - 3)(x + 5)\):
\(\frac{3(x + 5) - 2(2x - 3)}{(2x - 3)(x + 5)} = \frac{3x + 15 - 4x + 6}{(2x - 3)(x + 5)} = \frac{21 - x}{(2x - 3)(x + 5)}\).

(a) M1 for adding 4 to either dimension. A1 for \(x+9\) and \(2x+1\).
(b) M1 for expanding \((x+9)(2x+1)\) with at least 3 correct terms. A1 for fully correct simplification to the given formula.
(c) M1 for setting quadratic expression to 84. M1 for correct factorization attempt or use of quadratic formula on \(2x^2 + 19x - 75 = 0\). A1 for \(x=3\). A1 for dimensions 8 cm and 3 cm.
(d) M1 for finding a common denominator. M1 for correct expansion of numerator. A1 for final fraction \(\frac{21-x}{(2x-3)(x+5)}\).

(a) Outward time \(T_1 = \frac{36}{x}\) hours.

(b) Return distance = \(36 + 4 = 40\) km.
Return speed = \(x - 2\) km/h.
Return time \(T_2 = \frac{40}{x - 2}\) hours.

(c) Return journey is \(30\) minutes (\(0.5\) hours) longer:
\(\frac{40}{x - 2} - \frac{36}{x} = 0.5\)
Multiply both sides by \(2x(x - 2)\) to clear denominators:
\(80x - 72(x - 2) = x(x - 2)\)
\(80x - 72x + 144 = x^2 - 2x\)
\(8x + 144 = x^2 - 2x\)
\(x^2 - 10x - 144 = 0\).

(d) Factorize: \((x - 18)(x + 8) = 0\).
Since speed must be positive, \(x = 18\) km/h.

(e) Outward time = \(\frac{36}{18} = 2\) hours.
Return time = \(\frac{40}{16} = 2.5\) hours.
Total time = \(2 + 2.5 = 4.5\) hours.

(a) B1 for \(\frac{36}{x}\).
(b) B1 for \(\frac{40}{x-2}\).
(c) M1 for setting up the equation \(\frac{40}{x-2} - \frac{36}{x} = 0.5\). M1 for multiplying by \(2x(x-2)\) or \(x(x-2)\) correctly. A1 for establishing the given quadratic equation.
(d) M1 for factorizing \((x-18)(x+8)=0\) or correct use of formula. A1 for \(x = 18\).
(e) M1 for calculating both times. A1 for \(4.5\) hours.

(a) Total volume \(V = V_{\text{hemisphere}} + V_{\text{cylinder}} = \frac{2}{3}\pi r^3 + \pi r^2 h = 576\pi\).
Divide by \(\pi\):
\(\frac{2}{3}r^3 + r^2 h = 576 \Rightarrow r^2 h = 576 - \frac{2}{3}r^3\).
Divide by \(r^2\):
\(h = \frac{576}{r^2} - \frac{2}{3}r\).

(b) (i) Substitute \(h = 12\):
\(12 = \frac{576}{r^2} - \frac{2}{3}r\).
Multiply by \(3r^2\):
\(36r^2 = 1728 - 2r^3 \Rightarrow 2r^3 + 36r^2 - 1728 = 0\).
(ii) Substitute \(r = 6\):
\(2(6^3) + 36(6^2) - 1728 = 2(216) + 36(36) - 1728 = 432 + 1296 - 1728 = 1728 - 1728 = 0\). This is verified.

(c) Total surface area of the composite solid includes the curved surface of the hemisphere, the curved surface of the cylinder, and the bottom circular base of the cylinder:
\(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\).
For \(r = 6\) and \(h = 12\):
\(A = 3\pi(6^2) + 2\pi(6)(12) = 3\pi(36) + 144\pi = 108\pi + 144\pi = 252\pi\) cm\(^2\).

(a) M1 for correct formula of total volume. M1 for dividing by \(\pi\) and attempting to isolate \(h\). A1 for correct algebraic steps showing \(h = \frac{576}{r^2} - \frac{2}{3}r\).
(b)(i) M1 for substituting \(h=12\) and multiplying by \(3r^2\). A1 for correct equation. (ii) B1 for showing numerical verification yielding 0.
(c) M1 for sum of three parts of the surface area: \(2\pi r^2 + 2\pi r h + \pi r^2\) (or \(3\pi r^2 + 2\pi r h\)). M1 for substituting \(r=6\) and \(h=12\). A1 for \(252\pi\).

(a) Curved surface area \(= \pi r l = 135\pi\).
With \(l = 15\):
\(\pi r (15) = 135\pi \Rightarrow 15r = 135 \Rightarrow r = 9\) cm.
Using Pythagoras' theorem, vertical height \(h = \sqrt{l^2 - r^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\) cm.

(b) (i) Volume of water in the cone:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (9^2)(12) = 4\pi \times 81 = 324\pi\) cm\(^3\).
(ii) Volume of the sphere:
\(V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (8^3) = \frac{2048}{3}\pi \approx 682.67\pi\) cm\(^3\).
Percentage filled = \(\frac{V_{\text{cone}}}{V_{\text{sphere}}} \times 100\% = \frac{324\pi}{\frac{2048}{3}\pi} \times 100\% = \frac{972}{2048} \times 100\% \approx 47.46\%\), which is \(47.5\%\) (to 3 s.f.).

(c) The second cone is similar to the first.
Area scale factor \(k^2 = 1.44\).
Linear scale factor \(k = \sqrt{1.44} = 1.2\).
Volume scale factor \(k^3 = 1.2^3 = 1.728\).
Volume of second cone = \(324\pi \times 1.728 = 559.872\pi\) cm\(^3\).

(a) M1 for \(\pi r (15) = 135\pi\). A1 for \(r = 9\). M1 for \(h = \sqrt{15^2 - 9^2}\) showing height is 12.
(b)(i) M1 for \(\frac{1}{3}\pi r^2 h\) formula. A1 for \(324\pi\).
(b)(ii) M1 for \(\frac{4}{3}\pi R^3\) formula. M1 for dividing volume of cone by volume of sphere. A1 for \(47.5\%\) (accept \(47.46\%\)).
(c) M1 for finding linear scale factor \(k = 1.2\). M1 for volume scale factor \(k^3 = 1.728\). A1 for \(559.872\pi\) (or \(560\pi\)).

(a) Gradient of \(L_1\):
\(m_1 = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -0.5\).
Using point \(B(6, 1)\):
\(y - 1 = -0.5(x - 6) \Rightarrow y = -0.5x + 3 + 1 \Rightarrow y = -0.5x + 4\).

(b) (i) Midpoint of \(AB = \left(\frac{-2 + 6}{2}, \frac{5 + 1}{2}\right) = (2, 3)\).
(ii) The gradient of the perpendicular line \(L_2\):
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-0.5} = 2\).
Using the midpoint \((2, 3)\):
\(y - 3 = 2(x - 2) \Rightarrow y - 3 = 2x - 4 \Rightarrow 2x - y - 1 = 0\).

(c) For point \(C\) (on \(x\)-axis, \(y=0\)):
\(2x - 0 - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = 0.5\). So \(C(0.5, 0)\).
For point \(D\) (on \(y\)-axis, \(x=0\)):
\(2(0) - y - 1 = 0 \Rightarrow y = -1\). So \(D(0, -1)\).

(d) Triangle \(OCD\) is right-angled at \(O\), with base \(OC = 0.5\) units and height \(OD = 1\) unit.
Area \(= \frac{1}{2} \times 0.5 \times 1 = 0.25\) square units.

(a) M1 for finding gradient of \(L_1\). A1 for \(y = -0.5x + 4\).
(b)(i) B1 for midpoint \((2, 3)\).
(b)(ii) M1 for finding perpendicular gradient \(2\). M1 for substituting gradient and midpoint into line equation. A1 for establishing \(2x - y - 1 = 0\).
(c) B1 for \(C(0.5, 0)\). B1 for \(D(0, -1)\).
(d) M1 for \(\frac{1}{2} \times \text{base} \times \text{height}\). A1 for \(0.25\).

(a) Rearranging \(3x + 4y = 24\):
\(4y = -3x + 24 \Rightarrow y = -0.75x + 6\).
Gradient of \(M_1 = -0.75\).

(b) Gradient of perpendicular line \(M_2\) is \(-\frac{1}{-0.75} = \frac{4}{3}\).
Using point \(P(7, 7)\):
\(y - 7 = \frac{4}{3}(x - 7) \Rightarrow y = \frac{4}{3}x - \frac{28}{3} + 7 \Rightarrow y = \frac{4}{3}x - \frac{7}{3}\).

(c) (i) We solve:
1) \(3x + 4y = 24\)
2) \(4x - 3y = 7\) (from \(3y = 4x - 7\))
Multiply (1) by 3 and (2) by 4:
\(9x + 12y = 72\)
\(16x - 12y = 28\)
Add both equations:
\(25x = 100 \Rightarrow x = 4\).
Substitute \(x = 4\) into (1):
\(3(4) + 4y = 24 \Rightarrow 12 + 4y = 24 \Rightarrow y = 3\).
So, \(Q(4, 3)\).

(ii) Distance \(PQ = \sqrt{(7-4)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\).

(a) B1 for gradient \(-0.75\) (or \(-\frac{3}{4}\)).
(b) M1 for finding perpendicular gradient \(\frac{4}{3}\). M1 for substituting \(P(7, 7)\). A1 for \(y = \frac{4}{3}x - \frac{7}{3}\).
(c)(i) M1 for method of solving simultaneous equations. A1 for \(x=4\), A1 for \(y=3\) (coordinates \((4,3)\)).
(c)(ii) M1 for using distance formula \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). A1 for \(5\).

(a) Using the Cosine Rule in \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(72^\circ)\)
\(AC^2 = 120^2 + 85^2 - 2(120)(85)\cos(72^\circ) = 14400 + 7225 - 20400(0.309017) = 21625 - 6303.95 = 15321.05\)
\(AC = \sqrt{15321.05} \approx 123.78\) m, which is \(123.8\) m (to 1 d.p.).

(b) (i) Using the Sine Rule in \(\triangle ACD\):
\(\frac{\sin(ACD)}{AD} = \frac{\sin(ADC)}{AC} \Rightarrow \frac{\sin(ACD)}{140} = \frac{\sin(48^\circ)}{123.78}\)
\(\sin(ACD) = \frac{140 \times \sin(48^\circ)}{123.78} \approx \frac{140 \times 0.743145}{123.78} \approx 0.84052\)
Since angle \(ACD\) is acute:
\(ACD = \arcsin(0.84052) \approx 57.19^\circ\), which is \(57.2^\circ\) (to 1 d.p.).

(ii) Angle \(CAD = 180^\circ - 48^\circ - 57.19^\circ = 74.81^\circ\).
Area of \(\triangle ACD = \frac{1}{2} \cdot AD \cdot AC \cdot \sin(CAD) = \frac{1}{2} (140)(123.78)\sin(74.81^\circ) \approx 8664.6 \times 0.96507 \approx 8362\) m\(^2\), which is \(8360\) m\(^2\) (to 3 s.f.).

(c) In right-angled \(\triangle AB\text{(top)}\):
\(\tan(8^\circ) = \frac{\text{height}}{AB} = \frac{\text{height}}{120}\)
\(\text{height} = 120 \times \tan(8^\circ) \approx 120 \times 0.14054 \approx 16.86\) m, which is \(16.9\) m (to 3 s.f.).

(a) M1 for correct Cosine Rule formula setup. A1 for \(AC^2 = 15321\). A1 for \(123.8\) m (or \(124\)).
(b)(i) M1 for correct Sine Rule setup. A1 for \(\sin(ACD) = 0.841\). A1 for \(57.2^\circ\).
(b)(ii) M1 for finding angle \(CAD = 74.8^\circ\). M1 for correct area of triangle formula setup. A1 for \(8360\) m\(^2\).
(c) M1 for \(120 \tan(8^\circ)\). A1 for \(16.9\) m.

(a) Overall profit percentage:
\(60\% \text{ of phones sold at } +35\%\)
\(40\% \text{ of phones sold at } -10\%\)
Overall change = \(0.60(35\%) + 0.40(-10\%) = 21\% - 4\% = 17\%\).

(b) Let \(P\) be the original price.
Markup of \(20\%\) makes the price \(1.20P\).
Local sales tax of \(5\%\) makes the final selling price \(1.05 \times 1.20P = 1.26P\).
\(1.26P = 882 \Rightarrow P = \frac{882}{1.26} = 700\).
Original price paid = \(\$700\).

(c) (i) Let the 2020 revenue be \(R\).
2021 revenue = \(1.08R\).
2022 revenue = \(0.95 \times 1.08R = 1.026R\).
This represents \(102.6\%\) of the 2020 revenue.

(ii) Given \(1.026R = 385020\):
\(R = \frac{385020}{1.026} = 375000\).
Revenue in 2020 was \(\$375,000\).

(a) M1 for \(0.6 \times 35\) or \(1.35 \times 14400\). M1 for \(0.4 \times (-10)\) or \(0.90 \times 9600\). A1 for \(17\%\).
(b) M1 for establishing \(1.20 \times 1.05\) or multiplier \(1.26\). M1 for setting up \(1.26P = 882\). A1 for \(\$700\).
(c)(i) M1 for multiplying multipliers \(1.08 \times 0.95\). A1 for \(102.6\%\).
(c)(ii) M1 for dividing \(385020\) by their multiplier. A1 for \(\$375,000\).

(a) The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\). Substituting \(r = 3\text{ cm}\): \(V = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi (27) = 36\pi\text{ cm}^3\). (b) The volume of the hollow cylinder is given by \(V = \pi(R^2 - r_{\text{inner}}^2)h\). Given \(r_{\text{inner}} = 2\text{ cm}\), \(h = 8\text{ cm}\), and volume = \(36\pi\text{ cm}^3\): \(\pi(R^2 - 2^2) \times 8 = 36\pi\) which simplifies to \(8(R^2 - 4) = 36\). This gives \(R^2 - 4 = 4.5\), so \(R^2 = 8.5\), which yields \(R = \sqrt{8.5} \approx 2.9154... \approx 2.92\text{ cm}\) (to 3 s.f.). (c) The total exposed surface area of the hollow cylinder is the sum of: - Outer curved surface area: \(2\pi Rh = 2\pi(\sqrt{8.5})(8) = 16\pi\sqrt{8.5} \approx 146.55\text{ cm}^2\); - Inner curved surface area: \(2\pi rh = 2\pi(2)(8) = 32\pi \approx 100.53\text{ cm}^2\); - Two circular rings (top and bottom): \(2 \times \pi(R^2 - r^2) = 2 \times \pi(8.5 - 4) = 9\pi \approx 28.27\text{ cm}^2\). Total surface area = \(16\pi\sqrt{8.5} + 32\pi + 9\pi = (41 + 16\sqrt{8.5})\pi \approx 275.35\text{ cm}^2\). To 3 significant figures, this is \(275\text{ cm}^2\).

Part (a): M1: For substituting \(r=3\) into the volume of a sphere formula: \(\frac{4}{3}\pi \times 3^3\). A1: For simplifying correctly to show \(36\pi\). Part (b): M1: For writing a correct expression for the volume of the hollow cylinder: \(\pi(R^2 - 2^2) \times 8\). M1: For setting the volume expression equal to \(36\pi\) and simplifying: \(8(R^2 - 4) = 36\). M1: For solving to find \(R^2 = 8.5\) or \(R = \sqrt{8.5}\). A1: For \(2.92\) (accept \(2.915\) to \(2.92\)). Part (c): M1: For calculating the outer curved surface area: \(2 \times \pi \times \sqrt{8.5} \times 8\) (or using their \(R\)). M1: For calculating the inner curved surface area: \(2 \times \pi \times 2 \times 8 = 32\pi\). M1: For calculating the area of the two rings: \(2 \times \pi(R^2 - 4) = 9\pi\). M1: For summing their three surface areas. A1.8: For \(275\) (accept \(275\) to \(275.4\), or in terms of \(\pi\) as \((41 + 16\sqrt{8.5})\pi\)).

(a) Using the cosine rule in triangle \(ABC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the given values: \(AC^2 = 75^2 + 110^2 - 2 \cdot 75 \cdot 110 \cdot \cos(58^\circ) = 5625 + 12100 - 16500 \cdot \cos(58^\circ) \approx 17725 - 8743.67 = 8981.33\). Thus, \(AC = \sqrt{8981.33} \approx 94.77\text{ m}\). Correct to 1 decimal place, \(AC = 94.8\text{ m}\). (b) Using the area formula: \(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC) = \frac{1}{2} \cdot 75 \cdot 110 \cdot \sin(58^\circ) = 4125 \cdot \sin(58^\circ) \approx 3498.2\text{ m}^2\). To the nearest square metre, the area is \(3498\text{ m}^2\). (c) Let the height of the flagpole be \(h = AT\). Since \(T\) is vertically above \(A\), triangle \(TAB\) is right-angled at \(A\). \(\tan(18^\circ) = \frac{AT}{AB} = \frac{h}{75} \implies h = 75 \tan(18^\circ) \approx 24.369\text{ m}\). For the angle of elevation from \(C\), triangle \(TAC\) is right-angled at \(A\). Let the angle of elevation be \(\theta\). \(\tan(\theta) = \frac{AT}{AC} = \frac{24.369}{94.77} \approx 0.25714 \implies \theta = \arctan(0.25714) \approx 14.42^\circ\). Correct to 1 decimal place, the angle of elevation is \(14.4^\circ\).

Part (a): M1: For correct substitution into cosine rule: \(75^2 + 110^2 - 2(75)(110)\cos(58^\circ)\). A1: For correct calculation of \(17725 - 8743.67\) or \(8981.33\). M1: For taking the square root of their positive value. A1: For \(94.8\) (accept \(94.75\) to \(94.85\)). Part (b): M1: For correct formula \(\frac{1}{2} a b \sin(C)\) with correct values: \(\frac{1}{2} \times 75 \times 110 \times \sin(58^\circ)\). M1: For evaluation of the area to \(\approx 3498.2\). A1: For \(3498\). Part (c): M1: For using \(\tan(18^\circ) = \frac{h}{75}\) to find the height \(h\). A1: For \(h \approx 24.4\) or better. M1: For using \(\tan(\theta) = \frac{h}{\text{their } AC}\). M1: For evaluating \(\arctan\left(\frac{24.369}{94.77}\right)\). A0.8: For \(14.4\) (accept \(14.3\) to \(14.5\)).

(a) The gradient of \(AB\) is \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 5}{6 - (-2)} = \frac{6}{8} = 0.75\) (or \(\frac{3}{4}\)). (b) The perpendicular bisector passes through the midpoint of \(AB\). Midpoint \(M = \left(\frac{-2 + 6}{2}, \frac{5 + 11}{2}\right) = (2, 8)\). The gradient of the perpendicular bisector, \(m_{\perp}\), is \(m_{\perp} = -\frac{1}{0.75} = -\frac{4}{3}\). The equation of the line is given by \(y - 8 = -\frac{4}{3}(x - 2)\). Simplifying: \(y - 8 = -\frac{4}{3}x + \frac{8}{3} \implies y = -\frac{4}{3}x + \frac{32}{3}\) (or \(y = -1.33x + 10.7\)). (c) Point \(P\) lies on the \(x\)-axis, so setting \(y = 0\): \(0 = -\frac{4}{3}x + \frac{32}{3} \implies \frac{4}{3}x = \frac{32}{3} \implies x = 8\). Thus, \(P = (8, 0)\). Point \(Q\) lies on the \(y\)-axis, so setting \(x = 0\): \(y = \frac{32}{3}\). Thus, \(Q = \left(0, \frac{32}{3}\right)\). The distance \(PQ\) is \(PQ = \sqrt{(8 - 0)^2 + \left(0 - \frac{32}{3}\right)^2} = \sqrt{64 + \frac{1024}{9}} = \sqrt{\frac{576 + 1024}{9}} = \sqrt{\frac{1600}{9}} = \frac{40}{3} \approx 13.3\) (or \(13\frac{1}{3}\)).

Part (a): M1: For \(\frac{11 - 5}{6 - (-2)}\). A1: For \(0.75\) (or \(\frac{3}{4}\)). Part (b): M1: For midpoint of \(AB\) calculated as \((2, 8)\). M1: For finding the perpendicular gradient: \(m_{\perp} = -\frac{1}{\text{their gradient}}\). M1: For substituting their midpoint and their perpendicular gradient into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\). M1: For solving to find \(c\). A1: For \(y = -\frac{4}{3}x + \frac{32}{3}\) (or equivalent with decimals rounded to 3 s.f.: \(y = -1.33x + 10.7\)). Part (c): M1: For setting \(y = 0\) to find the coordinates of \(P\). M1: For setting \(x = 0\) to find the coordinates of \(Q\). M1: For applying Pythagoras' theorem to find the distance between their \(P\) and \(Q\). A1.8: For \(\frac{40}{3}\) or \(13.3\) (accept \(13.3\) or \(13.33...\)).