2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

First, find the linear scale factor from the ratio of the volumes:
\(\text{Linear scale factor} = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}\).

Next, find the area scale factor by squaring the linear scale factor:
\(\text{Area scale factor} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}\).

Let \(A\) be the surface area of the larger cylinder:
\(\frac{54}{A} = \frac{9}{25}\)
\(A = 54 \times \frac{25}{9} = 6 \times 25 = 150\text{ cm}^2\).

M1 for finding the linear scale factor \(\frac{3}{5}\) or \(3:5\) (or reciprocal)
M1 for area scale factor \(\frac{9}{25}\) or \(9:25\) (or reciprocal)
A1 for 150

Find the position vector \(\overrightarrow{OQ}\):
\(\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = \begin{pmatrix} -3 \\ 4 \end{pmatrix} + \begin{pmatrix} 8 \\ 8 \end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\).

Now, calculate the magnitude \(|\overrightarrow{OQ}|\):
\(|\overrightarrow{OQ}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).

M1 for \(\overrightarrow{OQ} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\)
M1 for \(\sqrt{5^2 + 12^2}\) or \(\sqrt{(\text{their } 5)^2 + (\text{their } 12)^2}\)
A1 for 13

Let \(C\) be the original cost price.
\(1.15 \times C = 207\)
\(C = \frac{207}{1.15} = 180\).

Now, find the profit made:
\(\text{Profit} = \text{Selling Price} - \text{Cost Price} = 207 - 180 = 27\).

M1 for \(207 \div 1.15\) oe
A1 for 180 (cost price)
A1 for 27

Use the cosine rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(120^\circ)\)
\(AC^2 = 64 + 121 - 176(-0.5)\)
\(AC^2 = 185 + 88 = 273\)
\(AC = \sqrt{273} \approx 16.5227\dots\text{ cm}\).
To 3 significant figures, \(AC = 16.5\text{ cm}\).

M1 for \(8^2 + 11^2 - 2(8)(11)\cos(120)\)
A1 for 273
A1 for 16.5 or 16.52 to 16.53

Multiply the entire equation by \(2(x+1)\) to clear the fractions:
\(x(x+1) + 2(5) = 3 \times 2(x+1)\)
\(x^2 + x + 10 = 6x + 6\)

Rearrange into standard quadratic form:
\(x^2 - 5x + 4 = 0\)

Factorise the quadratic:
\((x-4)(x-1) = 0\)

So, \(x = 1\) or \(x = 4\).

M1 for clearing fractions correctly to get \(x(x+1) + 10 = 6(x+1)\) oe
M1 for rearranging to \(x^2 - 5x + 4 = 0\)
A1 for \(x = 1\) and \(x = 4\)

By the alternate segment theorem, the angle subtended by chord \(AB\) in the major segment is \(64^\circ\).

The angle subtended at the centre \(O\) is twice the angle subtended at the circumference:
\(\text{Angle } AOB \text{ (obtuse)} = 2 \times 64^\circ = 128^\circ\).

To find the reflex angle \(AOB\):
\(\text{Reflex angle } AOB = 360^\circ - 128^\circ = 232^\circ\).

M1 for angle at circumference = \(64^\circ\) or obtuse angle \(AOB = 128^\circ\)
M1 for \(360 - 2 \times 64\) oe
A1 for 232

First simplify inside the bracket:
\(\frac{27x^6}{y^{-3}} = 27x^6 y^3\).

Now apply the exponent \(-\frac{2}{3}\):
\((27x^6 y^3)^{-\frac{2}{3}} = (27)^{-\frac{2}{3}} \cdot (x^6)^{-\frac{2}{3}} \cdot (y^3)^{-\frac{2}{3}}\).

Calculate each term:
\(27^{-\frac{2}{3}} = \frac{1}{(\sqrt[3]{27})^2} = \frac{1}{3^2} = \frac{1}{9}\)
\((x^6)^{-\frac{2}{3}} = x^{-4} = \frac{1}{x^4}\)
\((y^3)^{-\frac{2}{3}} = y^{-2} = \frac{1}{y^2}\).

Multiplying these together yields:
\(\frac{1}{9x^4 y^2}\).

M1 for simplifying inside bracket to \(27x^6 y^3\) or applying power of \(2/3\) first
M1 for dealing with fractional power to get coefficient \(\frac{1}{9}\)
A1 for \(\frac{1}{9x^4y^2}\)

First find the midpoint of \(AB\):
\(\text{Midpoint} = \left( \frac{2+6}{2}, \frac{-3+5}{2} \right) = (4, 1)\).

Next, find the gradient of \(AB\):
\(m_{AB} = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\).

The gradient of the perpendicular line is the negative reciprocal:
\(m_{\perp} = -\frac{1}{2}\).

Use the point-slope form with midpoint \((4, 1)\) and gradient \(-\frac{1}{2}\):
\(y - 1 = -\frac{1}{2}(x - 4)\)
\(y = -\frac{1}{2}x + 2 + 1\)
\(y = -\frac{1}{2}x + 3\).

B1 for Midpoint = \((4, 1)\)
M1 for gradient of perpendicular line = \(-1/2\) (implied by gradient of \(AB = 2\))
A1 for \(y = -1/2x + 3\) oe

For \( L = 12.4\text{ m} \) (correct to 1 d.p.):
The upper bound of \( L \) is \( L_{\text{UB}} = 12.4 + 0.05 = 12.45\text{ m} \).

For \( W = 8.35\text{ m} \) (correct to 2 d.p.):
The upper bound of \( W \) is \( W_{\text{UB}} = 8.35 + 0.005 = 8.355\text{ m} \).

The perimeter \( P \) of a rectangle is given by \( P = 2(L + W) \).

The upper bound of the perimeter is:
\( P_{\text{UB}} = 2(L_{\text{UB}} + W_{\text{UB}}) = 2(12.45 + 8.355) = 2(20.805) = 41.61\text{ m} \).

**B1** for \( 12.45 \) or \( 8.355 \) seen.
**M1** for \( 2(\text{their } L_{\text{UB}} + \text{their } W_{\text{UB}}) \), where \( L_{\text{UB}} > 12.4 \) and \( W_{\text{UB}} > 8.35 \).
**A1** for \( 41.61 \)

Volume of cylinder:
\( V_{\text{cylinder}} = \pi \times r^2 \times h = \pi \times 3^2 \times 8 = 72\pi\text{ cm}^3 \).

Since the volume remains the same during recasting, the volume of the sphere is also \( 72\pi\text{ cm}^3 \):
\( \frac{4}{3}\pi R^3 = 72\pi \)

Divide both sides by \( \pi \):
\( \frac{4}{3} R^3 = 72 \)

Multiply both sides by \( \frac{3}{4} \):
\( R^3 = 72 \times \frac{3}{4} = 54 \)

Calculate the cube root:
\( R = \sqrt[3]{54} \approx 3.77976\text{ cm} \).

Rounding to 3 significant figures gives \( 3.78 \).

**M1** for volume of cylinder \( \pi \times 3^2 \times 8 \) or \( 72\pi \) or \( 226.19\dots \)
**M1** for equating sphere volume to their cylinder volume, e.g., \( \frac{4}{3}\pi R^3 = \text{their } V \), leading to \( R^3 = \dots \)
**A1** for \( 3.78 \) or \( 3.779\dots \)

Factorise the numerator:
\( 2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) \).

Factorise the denominator using the difference of two squares:
\( 4x^2 - 1 = (2x - 1)(2x + 1) \).

Simplify the fraction:
\[ \frac{2x^2 - 5x - 3}{4x^2 - 1} = \frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1} \]

**B1** for numerator factorisation: \( (2x + 1)(x - 3) \)
**B1** for denominator factorisation: \( (2x - 1)(2x + 1) \)
**B1** for final answer \( \frac{x - 3}{2x - 1} \) or equivalent

Total number of counters = \( 5 + 3 = 8 \).

Probability of selecting two red counters:
\( P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} \).

Probability of selecting two blue counters:
\( P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} \).

Probability of selecting two counters of the same colour:
\( P(\text{Same Colour}) = P(\text{Red, Red}) + P(\text{Blue, Blue}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28} \approx 0.464 \).

**M1** for \( \frac{5}{8} \times \frac{4}{7} \) or \( \frac{3}{8} \times \frac{2}{7} \)
**M1** for sum of both probabilities: \( \frac{5}{8} \times \frac{4}{7} + \frac{3}{8} \times \frac{2}{7} \)
**A1** for \( \frac{13}{28} \) or equivalent fraction, or \( 0.464 \) or \( 0.4642\dots \)

Use the first term (\( n = 1 \)):
\( a(1)^2 + b(1) - 1 = 1.5 \Rightarrow a + b = 2.5 \) (Equation 1)

Use the second term (\( n = 2 \)):
\( a(2)^2 + b(2) - 1 = 8 \Rightarrow 4a + 2b = 9 \Rightarrow 2a + b = 4.5 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\( (2a + b) - (a + b) = 4.5 - 2.5 \Rightarrow a = 2 \).

Substitute \( a = 2 \) into Equation 1:
\( 2 + b = 2.5 \Rightarrow b = 0.5 \).

Checking for \( n = 3 \):
\( 2(3)^2 + 0.5(3) - 1 = 18 + 1.5 - 1 = 18.5 \) (Correct).

Therefore, \( a = 2 \) and \( b = 0.5 \).

**M1** for setting up at least one correct equation using a term, e.g. \( a + b - 1 = 1.5 \) or \( 4a + 2b - 1 = 8 \)
**M1** for a correct method to solve their simultaneous equations to find \( a \) or \( b \)
**A1** for both \( a = 2 \) and \( b = 0.5 \) (or \( \frac{1}{2} \))

Convert each of the values to a decimal representation to compare them easily:

- \( \frac{13}{18} = 0.72222\dots = 0.72\dot{2} \)
- \( 72\% = 0.72 \)
- \( 0.722 = 0.722 \)
- \( \frac{7}{10} = 0.7 \)

Comparing the decimal values:
\( 0.7 < 0.72 < 0.722 < 0.7222\dots \)

Therefore, the order starting from the smallest is:
\[ \frac{7}{10}, \quad 72\%, \quad 0.722, \quad \frac{13}{18} \]

**M2** for converting at least three values to a common format (e.g., decimals: \( 0.722\dots, 0.72, 0.722, 0.7 \))
(or **M1** for converting at least two values to a common format)
**A1** for correct order: \( \frac{7}{10}, 72\%, 0.722, \frac{13}{18} \)

The base \( ABCD \) is a rectangle with \( AB = 6\text{ cm} \) and \( BC = 4\text{ cm} \).
First, calculate the length of the base diagonal \( AC \) using Pythagoras' theorem:
\( AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}\text{ cm} \approx 7.2111\text{ cm} \).

The vertical height is \( CG = AE = 5\text{ cm} \).
The angle between the diagonal \( AG \) and the base is the angle \( \theta = \angle CAG \) in the right-angled triangle \( ACG \).

Using tangent:
\( \tan\theta = \frac{CG}{AC} = \frac{5}{\sqrt{52}} \)
\( \theta = \tan^{-1}\left(\frac{5}{\sqrt{52}}\right) \approx 34.7356^\circ \).

Rounding to 1 decimal place gives \( 34.7^\circ \).

**M1** for diagonal of base \( AC = \sqrt{6^2 + 4^2} \) or \( \sqrt{52} \) or \( 7.21\dots \)
**M1** for \( \tan\theta = \frac{5}{\text{their } AC} \) or equivalent trigonometric equation to find angle
**A1** for \( 34.7 \) or \( 34.73\dots \)

Let \( B \) be the set of students studying Biology, and \( C \) be the set of students studying Chemistry.
Total students in the group = 30.

Students studying at least one of Biology or Chemistry:
\( n(B \cup C) = 30 - 5 = 25 \).

We use the set formula:
\( n(B \cup C) = n(B) + n(C) - n(B \cap C) \)
\( 25 = 18 + 15 - n(B \cap C) \)
\( n(B \cap C) = 33 - 25 = 8 \).
So, 8 students study both Biology and Chemistry.

We want to find the probability that a student who studies Biology also studies Chemistry:
\( P(C | B) = \frac{n(B \cap C)}{n(B)} = \frac{8}{18} = \frac{4}{9} \approx 0.444 \).

**M1** for finding the number of students who study both subjects: \( 18 + 15 - (30 - 5) = 8 \)
**M1** for division by number of Biology students: \( \frac{\text{their } 8}{18} \)
**A1** for \( \frac{4}{9} \) or \( 0.444 \) or \( 0.4444\dots \)

Using the volume formula for a cylinder:
\[V = \pi r^2 h\]
\[540\pi = \pi \times r^2 \times 15\]
\[r^2 = \frac{540}{15} = 36\]
\[r = 6\text{ cm}\]

Using the total surface area formula for a cylinder:
\[A = 2\pi r^2 + 2\pi r h\]
\[A = 2\pi(6)^2 + 2\pi(6)(15)\]
\[A = 72\pi + 180\pi = 252\pi\text{ cm}^2\]

M1 for \(540\pi = \pi r^2 (15)\) or \(r^2 = 36\) or \(r = 6\)
M1 for \(2\pi(\text{their } r)^2 + 2\pi(\text{their } r)(15)\)
A1 for \(252\pi\)

First, find the upper bounds of both dimensions:
- Length: \(3.4\text{ m}\) to the nearest \(10\text{ cm}\) (\(0.1\text{ m}\)) has an upper bound of:
\[3.4 + 0.05 = 3.45\text{ m}\]
- Width: \(2.15\text{ m}\) to the nearest \(5\text{ cm}\) (\(0.05\text{ m}\)) has an upper bound of:
\[2.15 + 0.025 = 2.175\text{ m}\]

Calculate the upper bound for the area:
\[\text{Area}_{\text{UB}} = 3.45 \times 2.175 = 7.50375\text{ m}^2\]

B1 for \(3.45\) or \(2.175\) seen
M1 for \((\text{their } L_{\text{UB}}) \times (\text{their } W_{\text{UB}})\) where \(3.4 < L_{\text{UB}} \le 3.45\) and \(2.15 < W_{\text{UB}} \le 2.175\)
A1 for \(7.50375\)

First, perform the division:
\[\frac{7.2 \times 10^5}{1.5 \times 10^{-3}} = \frac{7.2}{1.5} \times 10^{5 - (-3)} = 4.8 \times 10^8\]

Next, perform the addition:
\[4.8 \times 10^8 + 1.2 \times 10^8 = (4.8 + 1.2) \times 10^8 = 6 \times 10^8\]

M1 for \(4.8 \times 10^8\) or \(480\ 000\ 000\) as the result of the division
M1 for adding both terms with the same exponent, e.g., \((4.8 + 1.2) \times 10^8\)
A1 for \(6 \times 10^8\) (or \(6.0 \times 10^8\))

Let \(V_0\) be the original value.
Using the compound depreciation formula:
\[V_0 \times (1 - 0.10)^3 = 14\ 580\]
\[V_0 \times 0.9^3 = 14\ 580\]
\[V_0 \times 0.729 = 14\ 580\]
\[V_0 = \frac{14\ 580}{0.729} = 20\ 000\]

M1 for \(V_0 \times 0.9^3 = 14\ 580\) oe
M1 for \(\frac{14\ 580}{0.9^3}\) oe
A1 for \(20\ 000\)

Apply the exponent \(\frac{2}{3}\) to each term inside the bracket:
\[(125)^{\frac{2}{3}} \times (a^9)^{\frac{2}{3}} \times (b^{12})^{\frac{2}{3}}\]
- For the coefficient:
\[125^{\frac{2}{3}} = (5^3)^{\frac{2}{3}} = 5^2 = 25\]
- For the power of \(a\):
\[a^{9 \times \frac{2}{3}} = a^6\]
- For the power of \(b\):
\[b^{12 \times \frac{2}{3}} = b^8\]

This gives \(25a^6b^8\).

B1 for \(25\)
B1 for \(a^6\)
B1 for \(b^8\)
If B0 scored, award SC1 for two out of three terms correct

Using the cosine rule:
\[PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)\]
\[PR^2 = 8^2 + 11^2 - 2(8)(11)\cos(120^\circ)\]
\[PR^2 = 64 + 121 - 176(-0.5)\]
\[PR^2 = 185 + 88 = 273\]
\[PR = \sqrt{273} \approx 16.5227\text{ cm}\]

To 3 significant figures, the length of \(PR\) is \(16.5\text{ cm}\).

M1 for \(8^2 + 11^2 - 2(8)(11)\cos(120^\circ)\)
A1 for \(273\) (or \(PR^2 = 273\))
A1 for \(16.5\) (accept \(16.52\) to \(16.53\))

The total number of socks is \(5 + 7 = 12\).

Probability of picking two black socks:
\[P(\text{Black, Black}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\]

Probability of picking two white socks:
\[P(\text{White, White}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\]

Probability that both are the same colour:
\[P(\text{Same Colour}) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66} \approx 0.470\]

M1 for \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{7}{12} \times \frac{6}{11}\)
M1 for adding their two product probabilities
A1 for \(\frac{31}{66}\) or \(0.470\) (accept \(0.4696\dots\))

(a)
(i) There are 7 red balls and 12 balls in total. After drawing one red ball, there are 6 red balls and 11 balls left.
\(P(\text{both red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\).

(ii) One is red and one is blue means we can draw Red then Blue, or Blue then Red.
\(P(\text{one red, one blue}) = P(R, B) + P(B, R) = \frac{7}{12} \times \frac{5}{11} + \frac{5}{12} \times \frac{7}{11} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\).

(b) If the first two balls were different colors, 1 red and 1 blue have been removed.
Remaining balls in the bag: 6 red and 4 blue (10 balls in total).
\(P(\text{third is blue}) = \frac{4}{10} = \frac{2}{5}\).

(c) Since there is replacement, the probability of drawing red remains \(\frac{7}{12}\) and blue remains \(\frac{5}{12}\) for each draw.
\(P(\text{at least one blue}) = 1 - P(\text{both red}) = 1 - \left(\frac{7}{12} \times \frac{7}{12}\right) = 1 - \frac{49}{144} = \frac{95}{144}\).

(a)(i) M1 for \(\frac{7}{12} \times \frac{6}{11}\), A1 for \(\frac{7}{22}\) or equivalent (e.g. 0.318)
(a)(ii) M1 for \(\frac{7}{12} \times \frac{5}{11}\) or \(\frac{5}{12} \times \frac{7}{11}\), M1 for adding both products, A1 for \(\frac{35}{66}\) or equivalent (e.g. 0.530)
(b) M1 for determining remaining balls as 6 red and 4 blue, A1 for \(\frac{4}{10}\) or \(\frac{2}{5}\)
(c) M1 for \(\frac{7}{12} \times \frac{7}{12}\), M1 for subtracting their product from 1, A1 for \(\frac{95}{144}\) or equivalent (e.g. 0.660)

(a) \(V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} = \pi r^2 (3r) = 3\pi r^3\).
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h_{\text{cone}} = \frac{1}{3}\pi r^2 (4r) = \frac{4}{3}\pi r^3\).
\(V_{\text{total}} = 3\pi r^3 + \frac{4}{3}\pi r^3 = \frac{13}{3}\pi r^3\).

(b) \(\frac{13}{3}\pi r^3 = 560 \implies r^3 = \frac{1680}{13\pi} \approx 41.1354\).
\(r = \sqrt[3]{41.1354} \approx 3.4518 \approx 3.45\text{ cm}\).

(c) The total surface area consists of the base of the cylinder, the curved surface area of the cylinder, and the curved surface area of the cone.
Base of cylinder: \(\pi r^2\).
Curved surface of cylinder: \(2\pi r h = 2\pi r (3r) = 6\pi r^2\).
Slant height of cone: \(l = \sqrt{r^2 + (4r)^2} = \sqrt{17r^2} = r\sqrt{17}\).
Curved surface of cone: \(\pi r l = \pi r^2\sqrt{17}\).
Total surface area: \(A = \pi r^2 (7 + \sqrt{17})\).
When \(r = 4.2\):
\(A = \pi (4.2)^2 (7 + \sqrt{17}) \approx 55.4177 \times 11.1231 \approx 616.42\text{ cm}^2\).

(a) M1 for cylinder volume \(3\pi r^3\), M1 for cone volume \(\frac{4}{3}\pi r^3\), A1 for final sum leading to \(\frac{13}{3}\pi r^3\)
(b) M1 for equating \(\frac{13}{3}\pi r^3 = 560\), M1 for solving for \(r^3\), A1 for \(r \approx 3.45\)
(c) M1 for finding slant height \(l = r\sqrt{17}\), M1 for cylinder curved area \(6\pi r^2\) and base \(\pi r^2\), M1 for cone curved area \(\pi r^2\sqrt{17}\), A1 for substitution of \(r = 4.2\), A1 for 616 (accept 616 to 617)

(a) In triangle \(ABC\), using the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cos(110^\circ)\)
\(AC^2 = 85^2 + 60^2 - 2(85)(60)\cos(110^\circ)\)
\(AC^2 = 7225 + 3600 - 10200(-0.34202)\)
\(AC^2 = 10825 + 3488.6 = 14313.6\)
\(AC = \sqrt{14313.6} \approx 119.64\text{ m} \approx 119.6\text{ m}\).

(b) In triangle \(ACD\), using the cosine rule to find \(AD\) first:
\(AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cos(48^\circ)\)
\(AD^2 = 119.64^2 + 75^2 - 2(119.64)(75)\cos(48^\circ)\)
\(AD^2 = 14313.6 + 5625 - 17946(0.66913)\)
\(AD^2 = 19938.6 - 12008.2 = 7930.4\)
\(AD \approx 89.05\text{ m}\).

Now using the sine rule in triangle \(ACD\) to find angle \(CAD\):
\(\frac{\sin(\angle CAD)}{CD} = \frac{\sin(48^\circ)}{AD}\)
\(\sin(\angle CAD) = \frac{75 \cdot \sin(48^\circ)}{89.05} \approx \frac{75 \cdot 0.74314}{89.05} \approx 0.62586\)
\(\angle CAD = \arcsin(0.62586) \approx 38.75^\circ \approx 38.8^\circ\).

(c) \(\text{Area of } ABC = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC) = \frac{1}{2} \cdot 85 \cdot 60 \cdot \sin(110^\circ) = 2550 \cdot 0.93969 \approx 2396.2\text{ m}^2\).
To 3 significant figures, this is \(2400\text{ m}^2\).

(a) M1 for correct cosine rule equation: \(AC^2 = 85^2 + 60^2 - 2 \cdot 85 \cdot 60 \cos(110^\circ)\), A1 for evaluation of cos term, A1 for \(AC^2 = 14313.6\), A1 for \(AC \approx 119.6\) or \(120\)
(b) M1 for cosine rule to find \(AD\), A1 for \(AD \approx 89.1\), M1 for sine rule setup: \(\frac{\sin(\angle CAD)}{75} = \frac{\sin(48^\circ)}{89.05}\), A1 for angle \(CAD \approx 38.8^\circ\) (accept 38.7 to 38.9)
(c) M1 for area formula: \(\frac{1}{2} \cdot 85 \cdot 60 \cdot \sin(110^\circ)\), A1 for correct calculation, A1 for \(2400\) or \(2396\)

(a) \(m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\).

(b) Using \(y - y_1 = m(x - x_1)\):
\(y - 5 = 2(x - 6) \implies y = 2x - 7\).

(c) \(M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\).

(d) The gradient of the perpendicular bisector is the negative reciprocal of the gradient of \(AB\).
\(m_{\perp} = -\frac{1}{2} = -0.5\).
Since it passes through the midpoint \((4, 1)\):
\(y - 1 = -0.5(x - 4) \implies y = -0.5x + 3\).

(e) Substitute \(x = -2\) into the perpendicular bisector equation:
\(y = -0.5(-2) + 3 = 1 + 3 = 4\).

(a) M1 for correct formula with values, A1 for 2
(b) M1 for standard line equation format using their gradient, A1 for \(y = 2x - 7\)
(c) M1 for formula setup, A1 for \((4, 1)\)
(d) M1 for perpendicular gradient \(-0.5\), M1 for using midpoint \((4,1)\) with perpendicular gradient, A1 for \(y = -0.5x + 3\)
(e) M1 for substituting \(x = -2\) into their line, A1 for 4

(a) Differentiating term by term:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x - 24\).

(b) Turning points occur when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(6x^2 - 18x - 24 = 0 \implies 6(x^2 - 3x - 4) = 0\).
\((x - 4)(x + 1) = 0\).
So \(x = 4\) or \(x = -1\).

Now find the y-coordinates:
When \(x = -1\):
\(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 12 = -2 - 9 + 24 + 12 = 25\).
Turning point: \((-1, 25)\).

When \(x = 4\):
\(y = 2(4)^3 - 9(4)^2 - 24(4) + 12 = 128 - 144 - 96 + 12 = -100\).
Turning point: \((4, -100)\).

(c) The second derivative is \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\).
At \(x = -1\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 18 = -30 < 0\), which indicates a local maximum.
At \(x = 4\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(4) - 18 = 30 > 0\), which indicates a local minimum.

(a) B2 for \(6x^2 - 18x - 24\) (B1 for at least two terms correct)
(b) M1 for setting their derivative to 0, M1 for factoring/solving to find \(x = 4, x = -1\), M1 for substituting \(x = -1\) into curve equation, M1 for substituting \(x = 4\) into curve equation, A1 for both points \((-1, 25)\) and \((4, -100)\) correct
(c) M1 for finding second derivative \(12x - 18\), M1 for substituting \(x = -1\) to get negative result, M1 for substituting \(x = 4\) to get positive result, A1 for correct conclusion (Maximum at \(x=-1\), Minimum at \(x=4\))

(a) \(V = 4500 \times (1 + 0.028)^8 = 4500 \times (1.028)^8\).
\(V \approx 4500 \times 1.24683 = 5610.74\).
To the nearest dollar, this is \(\$5611\).

(b) Let \(r\) be the rate of interest.
\(5000 \times (1 + r)^5 = 5800 \implies (1 + r)^5 = 1.16\).
\(1 + r = 1.16^{0.2} \approx 1.0301\).
\(r \approx 0.0301 \implies 3.01\%\).

(c) \(P = 240\,000 \times (1 - 0.015)^{10} = 240\,000 \times (0.985)^{10}\).
\(P \approx 240\,000 \times 0.85973 = 206335.2\).
Rounding to the nearest hundred: \(206\,300\).

(a) M1 for compound interest formula setup: \(4500 \times 1.028^8\), A1 for \(5610.74\), B1 for rounding to \(5611\)
(b) M1 for compound interest setup: \(5000(1+r)^5 = 5800\), M1 for dividing by \(5000\) to get \(1.16\), M1 for \(\sqrt[5]{1.16}\), A1 for \(3.01\%\) (accept 3.0% to 3.02%)
(c) M1 for population formula: \(240000 \times 0.985^{10}\), A1 for correct evaluation \(206335\), B1 for rounding to nearest hundred (\(206300\))

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

(b) Using the result from (a):
\(\frac{11 - x}{(2x - 1)(x + 3)} = 1 \implies 11 - x = (2x - 1)(x + 3)\).
\(11 - x = 2x^2 + 5x - 3\).
\(2x^2 + 6x - 14 = 0 \implies x^2 + 3x - 7 = 0\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2} = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}\).
\(\sqrt{37} \approx 6.08276\).
\(x_1 = \frac{-3 + 6.08276}{2} \approx 1.54\).
\(x_2 = \frac{-3 - 6.08276}{2} \approx -4.54\).

(a) M1 for finding a common denominator, M1 for expanding the numerator correctly, A1 for showing the final simplified form \(\frac{11-x}{(2x-1)(x+3)}\)
(b) M1 for setting up the equation \(11 - x = (2x - 1)(x + 3)\), M1 for expanding the RHS: \(2x^2 + 5x - 3\), M1 for rearranging to form quadratic \(2x^2 + 6x - 14 = 0\) or \(x^2 + 3x - 7 = 0\), M1 for correct use of quadratic formula with their coefficients, A1 for \(\sqrt{37}\) or \(\sqrt{148}\), A1 for one correct solution, A1 for other correct solution (accept 1.54 and -4.54)

(a) Sequence A is an arithmetic sequence with a common difference of 3.
\(n\text{th term} = 4 + 3(n - 1) = 3n + 1\).

(b) Sequence B values: 2, 6, 12, 20, 30.
First differences: 4, 6, 8, 10.
Second differences: 2, 2, 2.
This is a quadratic sequence of the form \(an^2 + bn + c\) where \(2a = 2 \implies a = 1\).
Comparing terms to \(n^2\):
\(n=1: 2 - 1 = 1\)
\(n=2: 6 - 4 = 2\)
\(n=3: 12 - 9 = 3\)
This residual is equal to \(n\).
Thus, the \(n\)th term is \(n^2 + n\).

(c) Sequence C values: \(3^1, 3^2, 3^3, \dots\).
This is a geometric sequence where the \(n\)th term is \(3^n\).

(d)
(i) Substitute \(n = 5\) into \(n^3 - 2n\):
\(5^3 - 2(5) = 125 - 10 = 115\).

(ii) Substitute \(n = 10\) into \(n^3 - 2n\):
\(10^3 - 2(10) = 1000 - 20 = 980\).

(a) M1 for realizing difference is 3, A1 for \(3n + 1\)
(b) M1 for quadratic method (finding second difference of 2), M1 for adjusting to \(n^2\), A1 for \(n^2 + n\)
(c) M1 for identifying powers of 3, A1 for \(3^n\)
(d)(i) M1 for substituting \(n = 5\) into formula, A1 for 115
(d)(ii) M1 for substituting \(n = 10\) into formula, A1 for leading to 980

(a) Volume of a cone is V = (1/3) * ΓΙ * r^2 * h. Substituting r = 6 and h = 15 gives V = (1/3) * ΓΙ * 6^2 * 15 = (1/3) * ΓΙ * 36 * 15 = 180ΓΙ cm^3. (b) Volume of a sphere is V_s = (4/3) * ΓΙ * R^3. Substituting R = 1.5 gives V_s = (4/3) * ΓΙ * (1.5)^3 = (4/3) * ΓΙ * 3.375 = 4.5ΓΙ cm^3. (c) Number of spheres = (Volume of cone) / (Volume of one sphere) = 180ΓΙ / 4.5ΓΙ = 40. (d)(i) Let k be the scale factor. The ratio of the volumes of similar shapes is equal to the cube of the scale factor: k^3 = (Volume of larger cone) / (Volume of original cone) = 1440ΓΙ / 180ΓΙ = 8. Therefore, k = ∛8 = 2. (ii) The total surface area of a cone is A = ΓΙ * r^2 + ΓΙ * r * l, where l is the slant height. Since l^2 = r^2 + h^2, we have l = √(6^2 + 15^2) = √(36 + 225) = √261 ≈ 16.155 cm. Total surface area = ΓΙ * 6^2 + ΓΙ * 6 * 16.155 = 36ΓΙ + 96.93ΓΙ = 132.93ΓΙ ≈ 417.6 cm^2. Rounding to 3 significant figures gives 418 cm^2.

(a) M1 for (1/3) * ΓΙ * 6^2 * 15, A1 for completing proof to show 180ΓΙ. (b) M1 for (4/3) * ΓΙ * 1.5^3, A1 for 4.5ΓΙ or 9ΓΙ/2. (c) M1 for 180ΓΙ / their (b), A1 for 40. (d)(i) M1 for k^3 = 1440ΓΙ / 180ΓΙ or k^3 = 8, A1 for 2. (d)(ii) M1 for slant height l = √(6^2 + 15^2), M1 for total surface area formula ΓΙ * r^2 + ΓΙ * r * l with their values, A1 for 418 or 417.6.

(a)(i) P(both red) = P(Red on 1st) * P(Red on 2nd) = (5/8) * (4/7) = 20/56 = 5/14 ≈ 0.357. (ii) P(different colours) = P(Red, Blue) + P(Blue, Red) = (5/8)*(3/7) + (3/8)*(5/7) = 15/56 + 15/56 = 30/56 = 15/28 ≈ 0.536. (b)(i) P(same colour) = P(Red, Red) + P(Blue, Blue) = (5/8)*(4/7) + (3/8)*(2/7) = 20/56 + 6/56 = 26/56 = 13/28. (ii) Expected value = (Profit for same colour) * P(same) + (Profit for different) * P(different) = 5 * (13/28) + (-2) * (15/28) = (65 - 30) / 28 = 35/28 = 5/4 = $1.25. Therefore, the expected win is $1.25.

(a)(i) M1 for (5/8) * (4/7), A1 for 5/14 or 20/56 or 0.357. (ii) M1 for (5/8)*(3/7) or (3/8)*(5/7), M1 for summing both products, A1 for 15/28 or 30/56 or 0.536. (b)(i) M1 for P(Blue, Blue) = (3/8)*(2/7), M1 for summing P(Red, Red) + P(Blue, Blue), A1 for obtaining 13/28. (ii) M1 for 5 * (13/28), M1 for -2 * (15/28), A1 for expected win of $1.25.

(a) Differentiating the equation term by term, we get: dy/dx = d/dx(2x^3) - d/dx(3x^2) - d/dx(12x) + d/dx(5) = 6x^2 - 6x - 12. (b) At the turning points, the derivative is zero: 6x^2 - 6x - 12 = 0. Dividing by 6: x^2 - x - 2 = 0. Factoring: (x - 2)(x + 1) = 0, so x = 2 or x = -1. When x = 2, y = 2(2)^3 - 3(2)^2 - 12(2) + 5 = 16 - 12 - 24 + 5 = -15. When x = -1, y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12. The coordinates of the turning points are (2, -15) and (-1, 12). (c) Using the second derivative: d^2y/dx^2 = 12x - 6. For x = 2: d^2y/dx^2 = 12(2) - 6 = 18. Since this is positive, (2, -15) is a minimum turning point. For x = -1: d^2y/dx^2 = 12(-1) - 6 = -18. Since this is negative, (-1, 12) is a maximum turning point. (d) At x = 1, y = 2(1)^3 - 3(1)^2 - 12(1) + 5 = -8. The gradient at x = 1 is dy/dx = 6(1)^2 - 6(1) - 12 = -12. Using the equation of a straight line, y - y1 = m(x - x1): y - (-8) = -12(x - 1) => y + 8 = -12x + 12 => y = -12x + 4.

(a) B1 for two terms correct in differentiation, B1 for all terms correct. (b) M1 for setting their dy/dx = 0, A1 for x = 2 and x = -1, A1 for y = -15 and y = 12, A1 for coordinates paired correctly. (c) M1 for finding second derivative d^2y/dx^2 = 12x - 6, M1 for evaluating at their x-values, A1 for correct conclusion of maximum at (-1, 12) and minimum at (2, -15). (d) M1 for evaluating gradient at x = 1 to find m = -12 and finding y = -8, A1 for y = -12x + 4.

(a) Using the Cosine Rule: AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(ABC) = 85^2 + 110^2 - 2 * 85 * 110 * cos(64°) = 7225 + 12100 - 18700 * 0.43837 = 19325 - 8197.54 = 11127.46. AC = ∑11127.46 = 105.487 m ≈ 105 m (to 3 significant figures). (b) Area of a triangle = (1/2) * AB * BC * sin(ABC) = (1/2) * 85 * 110 * sin(64°) = 4675 * 0.89879 = 4201.86 m^2 ≈ 4200 m^2 (to 3 significant figures). (c) The shortest distance d from B to AC represents the height of the triangle with base AC. Area = (1/2) * base * height => 4201.86 = (1/2) * 105.487 * d => d = (2 * 4201.86) / 105.487 = 79.666 m ≈ 79.7 m (to 3 significant figures). (d) In the right-angled triangle ABP, BP is the opposite side (height of flagpole = 15 m) and AB is the adjacent side (distance = 85 m). Let θ be the angle of elevation. tan(θ) = BP / AB = 15 / 85 = 3 / 17 ≈ 0.17647. θ = arctan(0.17647) ≈ 10.007° ≈ 10.0° (correct to 1 decimal place).

(a) M1 for 85^2 + 110^2 - 2 * 85 * 110 * cos(64°), A1 for AC^2 ≈ 11127, A1 for AC = 105 or 105.5. (b) M1 for (1/2) * 85 * 110 * sin(64°), A1 for 4200 or 4201.9. (c) M1 for using Area = (1/2) * base * height, M1 for substituting their values of Area and AC, A1 for 79.7 or 79.67. (d) M1 for right-angled triangle ABP with BP = 15 and AB = 85, M1 for tan(θ) = 15 / 85, A1 for 10.0°.