2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Factorise the numerator:
$$4x^2 - 9 = (2x - 3)(2x + 3)$$

Factorise the denominator:
$$2x^2 + 5x - 12 = 2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)$$

Simplify by cancelling the common factor $2x - 3$:
$$\frac{(2x - 3)(2x + 3)}{(2x - 3)(x + 4)} = \frac{2x + 3}{x + 4}$$

M1 for $(2x - 3)(2x + 3)$
M1 for $(2x - 3)(x + 4)$
A1 for final answer \(\frac{2x + 3}{x + 4}\)

Let $P$ be the original price.
Since the price is reduced by 15%, the sale price is 85% of the original price.
$$0.85 \times P = 323$$
$$P = \frac{323}{0.85} = 380$$

M1 for $323 \div 0.85$ or $323 \div (1 - 0.15)$
A1 for 380

Find the linear scale factor:
$$\text{scale factor} = \frac{15}{12} = 1.25$$

Calculate the volume of the larger vase using the cube of the scale factor:
$$\text{Volume} = 640 \times 1.25^3 = 640 \times 1.953125 = 1250\text{ cm}^3$$

M1 for linear scale factor $\frac{15}{12}$ oe
M1 for $640 \times (\text{their scale factor})^3$
A1 for 1250

Using the cosine rule:
$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(BAC)$$
$$8^2 = 7^2 + 10^2 - 2(7)(10) \cdot \cos(BAC)$$
$$64 = 49 + 100 - 140 \cdot \cos(BAC)$$
$$64 = 149 - 140 \cdot \cos(BAC)$$
$$140 \cdot \cos(BAC) = 85$$
$$\cos(BAC) = \frac{85}{140} = \frac{17}{28}$$
$$\text{angle } BAC = \cos^{-1}\left(\frac{17}{28}\right) \approx 52.624^\circ$$

Rounding to 1 decimal place gives $52.6^\circ$.

M1 for $8^2 = 7^2 + 10^2 - 2 \cdot 7 \cdot 10 \cos(BAC)$ or better
M1 for $\cos(BAC) = \frac{85}{140}$ oe
A1 for 52.6 or 52.62...

Rearrange the second equation to express $x$ in terms of $y$:
$$x = -3 - 3y$$

Substitute into the first equation:
$$3(-3 - 3y) - 4y = 17$$
$$-9 - 9y - 4y = 17$$
$$-13y = 26$$
$$y = -2$$

Substitute $y = -2$ back into the expression for $x$:
$$x = -3 - 3(-2) = -3 + 6 = 3$$

So the solution is $x = 3, y = -2$.

M1 for a correct method to eliminate one variable
A1 for $x = 3$ or $y = -2$
A1 for both $x = 3, y = -2$

Apply the fractional index to each part of the term:
$$(64x^9)^{\frac{2}{3}} = (64)^{\frac{2}{3}} \cdot (x^9)^{\frac{2}{3}}$$

Simplify the constant factor:
$$(64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16$$

Simplify the variable factor:
$$(x^9)^{\frac{2}{3}} = x^{9 \times \frac{2}{3}} = x^6$$

Combining the parts gives:
$$16x^6$$

B1 for $16$ or $x^6$ in final answer
B1 for $16x^6$ as final answer

Calculate the total amount after 5 years:
$$\text{Amount} = 1500 \times (1 + 0.024)^5 = 1500 \times 1.024^5 \approx 1688.8475...$$

Calculate the interest earned by subtracting the initial principal:
$$\text{Interest} = 1688.8475... - 1500 \approx 188.85$$

M1 for $1500 \times 1.024^5$
A1 for 1688.85
A1 for 188.85

Analyze the differences between the terms:
Terms: 3, 8, 15, 24
First differences: 5, 7, 9
Second differences: 2, 2

Since the second difference is constant at 2, the sequence contains an $n^2$ term with coefficient $\frac{2}{2} = 1$.

Subtract $n^2$ from each term of the sequence:
$$3 - 1^2 = 2$$
$$8 - 2^2 = 4$$
$$15 - 3^2 = 6$$
$$24 - 4^2 = 8$$

The remaining sequence 2, 4, 6, 8, ... corresponds to the linear term $2n$.

Thus, the $n$th term is:
$$n^2 + 2n$$

M1 for finding second differences of 2
A1 for $n^2 + 2n$ oe

First, factorise the numerator and the denominator.

Numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Denominator:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now, rewrite the fraction with these factors:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\)

Cancel the common factor \((2x + 1)\) from the numerator and denominator:
\(\frac{x - 3}{2x - 1}\)

M1 for factorising the numerator: \((2x + 1)(x - 3)\) or denominator: \((2x - 1)(2x + 1)\)
M1 for cancelling the common factor \((2x + 1)\)
A1 for correct final answer: \(\frac{x - 3}{2x - 1}\)

Let the original price be \(P\).
Since the price is reduced by 15%, the sale price is 85% of the original price.
\(0.85P = 646\)
\(P = \frac{646}{0.85} = 760\)

M1 for \(646 \div 0.85\) or \(0.85x = 646\)
A1 for 760

Find the volume of the sphere:
\(V_{\text{sphere}} = \frac{4}{3} \pi (4.5)^3 = \frac{4}{3} \pi (91.125) = 121.5 \pi\)

Since the sphere is recast into a cone, the volume of the cone is equal to the volume of the sphere:
\(V_{\text{cone}} = \frac{1}{3} \pi (3)^2 h = 3 \pi h\)

Equating the volumes:
\(3 \pi h = 121.5 \pi\)
\(3h = 121.5\)
\(h = 40.5\)

M1 for setting up the volume of sphere: \(\frac{4}{3} \pi (4.5)^3\) (or \(121.5\pi\))
M1 for equating their sphere volume to the cone volume formula: \(3\pi h = \text{their } V_{\text{sphere}}\)
A1 for 40.5

Use the Cosine Rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(64^\circ)\)
\(AC^2 = 49 + 81 - 126\cos(64^\circ)\)
\(AC^2 = 130 - 55.235\)
\(AC^2 = 74.765\)
\(AC = \sqrt{74.765} \approx 8.6467\)

To 3 significant figures, \(AC = 8.65\) cm.

M1 for substituting correctly into the cosine rule: \(7^2 + 9^2 - 2(7)(9)\cos(64^\circ)\)
A1 for \(AC^2 = 74.76\dots\) or better
A1 for 8.65

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

Group terms with \(x\) on one side and other terms on the other side:
\(5y - 2 = 3x + xy\)

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

Divide by \((3 + y)\):
\(x = \frac{5y - 2}{y + 3}\)

M1 for clearing the fraction: \(y(5 - x) = 3x + 2\)
M1 for isolating terms with \(x\): \(5y - 2 = 3x + xy\)
A1 for correct final formula: \(x = \frac{5y - 2}{y + 3}\) (or equivalent)

Multiply the first equation by 2 to make the coefficients of \(y\) equal in magnitude:
\(10x + 4y = 32\)

Now add this equation to the second equation:
\((10x + 4y) + (3x - 4y) = 32 + 7\)
\(13x = 39\)
\(x = 3\)

Substitute \(x = 3\) back into the first equation:
\(5(3) + 2y = 16\)
\(15 + 2y = 16\)
\(2y = 1\)
\(y = 0.5\)

So, the solution is \(x = 3\) and \(y = 0.5\).

M1 for a correct method to eliminate one variable (e.g. multiplying the first equation by 2)
A1 for \(x = 3\)
A1 for \(y = 0.5\)

Apply the power of \(\frac{2}{3}\) to each term inside the bracket:
\((64)^{\frac{2}{3}} \cdot (x^6)^{\frac{2}{3}} \cdot (y^{-12})^{\frac{2}{3}}\)

Simplify each term:
\((64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\)
\((x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\)
\((y^{-12})^{\frac{2}{3}} = y^{-12 \times \frac{2}{3}} = y^{-8}\)

Combine the terms:
\(16x^4 y^{-8} = \frac{16x^4}{y^8}\)

B1 for \((64)^{\frac{2}{3}} = 16\)
B1 for \((x^6)^{\frac{2}{3}} = x^4\) or \((y^{-12})^{\frac{2}{3}} = y^{-8}\)
B1 for fully simplified expression: \(16x^4 y^{-8}\) or \(\frac{16x^4}{y^8}\)

Use the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 3\), \(b = -7\), and \(c = -5\):
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-5)}}{2(3)}\)
\(x = \frac{7 \pm \sqrt{49 + 60}}{6}\)
\(x = \frac{7 \pm \sqrt{109}}{6}\)

Calculate the two roots:
\(x = \frac{7 + 10.4403}{6} \approx 2.91\)
\(x = \frac{7 - 10.4403}{6} \approx -0.57\)

M1 for substituting correctly into the formula: \(\frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-5)}}{2(3)}\)
A1 for \(\sqrt{109}\) seen or implied
A1 for 2.91 and -0.57

First, factorise the numerator by grouping:
\(2a - 3ab + 2c - 3bc = a(2 - 3b) + c(2 - 3b) = (2 - 3b)(a + c)\).

Next, factorise the denominator as a difference of two squares:
\(4 - 9b^2 = (2 - 3b)(2 + 3b)\).

Now divide the numerator by the denominator:
\(\frac{(2 - 3b)(a + c)}{(2 - 3b)(2 + 3b)} = \frac{a + c}{2 + 3b}\).

M1 for factorising the numerator to \((2-3b)(a+c)\) oe
M1 for factorising the denominator to \((2-3b)(2+3b)\) oe
A1 for the correct simplified fraction

Multiply the first equation by 2:
\(8x - 6y = 36\)

Multiply the second equation by 3:
\(9x + 6y = 66\)

Add the two equations to eliminate \(y\):
\(17x = 102\)
\(x = 6\)

Substitute \(x = 6\) into the second equation:
\(3(6) + 2y = 22\)
\(18 + 2y = 22\)
\(2y = 4\)
\(y = 2\)

M1 for a correct method to eliminate one variable (e.g. equating coefficients)
A1 for finding one variable correctly (e.g., \(x=6\) or \(y=2\))
A1 for finding both variables correctly

The ratio of the total surface areas of the similar cylinders is:
\(\frac{A_{\text{large}}}{A_{\text{small}}} = \frac{180}{80} = \frac{9}{4}\).

Since the cylinders are mathematically similar, the linear scale factor \(k\) is:
\(k = \sqrt{\frac{9}{4}} = \frac{3}{2}\).

The volume scale factor is:
\(k^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}\).

Therefore, the volume of the smaller cylinder is:
\(V_{\text{small}} = \frac{V_{\text{large}}}{k^3} = \frac{243}{\frac{27}{8}} = 243 \times \frac{8}{27} = 9 \times 8 = 72\text{ cm}^3\).

M1 for finding the linear scale factor \(\sqrt{\frac{180}{80}}\) or \(1.5\) (or reciprocal)
M1 for cubing their linear scale factor to obtain \(3.375\) (or reciprocal)
A1 for 72

Let \(H\) be the hammer price.
The total sale price is the hammer price plus 20% of the hammer price:
\(1.20 \times H = 1536\).

To find the hammer price:
\(H = \frac{1536}{1.20} = 1280\).

M2 for \(1536 \div 1.20\) oe
(or M1 for equating \(120\%\) to \(1536\))
A1 for 1280

Using the cosine rule:
\[BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\]
\[BC^2 = 8.4^2 + 11.5^2 - 2(8.4)(11.5)\cos(52^\circ)\]
\[BC^2 = 70.56 + 132.25 - 193.2\cos(52^\circ)\]
\[BC^2 \approx 202.81 - 193.2(0.61566) \approx 202.81 - 118.946 = 83.864\]
\[BC = \sqrt{83.864} \approx 9.1577\text{ cm}\]
Correct to 3 significant figures, the length is \(9.16\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(8.4^2 + 11.5^2 - 2(8.4)(11.5)\cos(52)\)
M1 for \(BC^2 \approx 83.9\) (or better)
A1 for 9.16 (or 9.157 to 9.158)

Apply the exponent to each factor inside the brackets:
\((64)^{-\frac{2}{3}} \times (x^{12})^{-\frac{2}{3}}\)

Simplify the numerical part:
\(64^{-\frac{2}{3}} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}\)

Simplify the algebraic part:
\((x^{12})^{-\frac{2}{3}} = x^{12 \times \left(-\frac{2}{3}\right)} = x^{-8}\)

Combine the results:
\(\frac{1}{16}x^{-8} = \frac{1}{16x^8}\).

M1 for obtaining \(64^{-\frac{2}{3}} = \frac{1}{16}\) or \((x^{12})^{-\frac{2}{3}} = x^{-8}\)
A1 for the correct final answer of \(\frac{1}{16x^8}\) or \(\frac{1}{16}x^{-8}\)

First, find the scale factor of lengths:
\(k = \frac{18}{15} = 1.2\).

Since the photographs are mathematically similar, the scale factor of areas is:
\(k^2 = 1.2^2 = 1.44\).

Multiply the area of the smaller photograph by the area scale factor:
\(\text{Area of larger photograph} = 225 \times 1.44 = 324\text{ cm}^2\).

M1 for finding the length scale factor \(\frac{18}{15}\) or \(1.2\)
M1 for squaring their length scale factor to get \(1.44\)
A1 for 324

The total number of socks is \(5 + 4 = 9\).

The socks are different colours if we choose (Blue then Red) or (Red then Blue):

\(P(\text{Blue then Red}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\)

\(P(\text{Red then Blue}) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}\)

Add these two probabilities together:
\(P(\text{Different Colours}) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9}\).

M1 for \(\frac{5}{9} \times \frac{4}{8}\) or \(\frac{4}{9} \times \frac{5}{8}\) oe
M1 for adding two correct probabilities
A1 for \(\frac{5}{9}\) or equivalent fraction/decimal (0.556 or 0.5555...)

Multiply both sides by \(3 - t\):
\(w(3 - t) = 4t + 1\)

Expand the brackets:
\(3w - wt = 4t + 1\)

Rearrange to gather terms with \(t\) on one side and the other terms on the other side:
\(3w - 1 = 4t + wt\)

Factorise \(t\) on the right-hand side:
\(3w - 1 = t(4 + w)\)

Divide by \(4 + w\):
\(t = \frac{3w - 1}{w + 4}\)

M1 for multiplying by \(3 - t\) to get \(w(3 - t) = 4t + 1\) or better
M1 for isolating terms containing \(t\) on one side, e.g. \(3w - 1 = t(4 + w)\) oe
A1 for \(t = \frac{3w - 1}{w + 4}\) or equivalent

Using the compound growth formula:
\[150 \times \left(1 + \frac{r}{100}\right)^3 = 199.65\]

Divide both sides by 150:
\[\left(1 + \frac{r}{100}\right)^3 = \frac{199.65}{150}\]
\[\left(1 + \frac{r}{100}\right)^3 = 1.331\]

Take the cube root of both sides:
\[1 + \frac{r}{100} = \sqrt[3]{1.331}\]
\[1 + \frac{r}{100} = 1.1\]

Subtract 1 from both sides:
\[\frac{r}{100} = 0.1\]

Multiply by 100:
\[r = 10\]

M1 for setting up the equation \(150 \times \left(1 + \frac{r}{100}\right)^3 = 199.65\) or \(150 \times k^3 = 199.65\)
M1 for taking the cube root, e.g., \(\sqrt[3]{\frac{199.65}{150}}\) or \(1.1\) seen
A1 for \(10\)

(a) Simple interest formula: \(I = \frac{P \times R \times T}{100}\)
\(I = \frac{6500 \times 3.2 \times 8}{100} = 1664\)

(b) Compound interest formula: \(A = P \left(1 + \frac{R}{100}\right)^t\)
\(A = 6500 \times (1.029)^8 \approx 8168.37\)
To the nearest dollar, the value is \(\$8168\).

(c) \(5000 \times \left(1 + \frac{r}{100}\right)^{12} = 7340\)
\(\left(1 + \frac{r}{100}\right)^{12} = 1.468\)
\(1 + \frac{r}{100} = (1.468)^{\frac{1}{12}} \approx 1.03239\)
\(\frac{r}{100} = 0.03239 \implies r = 3.24\%\)

(a)
M1 for \(6500 \times 0.032 \times 8\) oe
A1 for 1664

(b)
M1 for \(6500 \times (1.029)^8\) oe
A1 for 8168.37 or 8168
A1 for 8168 (nearest dollar)

(c)
M1 for \(5000 \times (1 + r/100)^{12} = 7340\)
M1 for \((1 + r/100)^{12} = 1.468\)
M1 for \(1 + r/100 = 1.468^{1/12}\)
A1 for 1.03239...
A1 for 3.24

(a) Factorise the numerator: \(2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\)
Factorise the denominator: \(x^2 + 5x + 6 = (x + 2)(x + 3)\)
Cancel out common factor \((x + 2)\):
\(\frac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{2(x - 2)}{x + 3} = \frac{2x - 4}{x + 3}\)

(b) Find a common denominator: \((2a - 1)(a + 3)\)
\(\frac{3(a + 3) - 2(2a - 1)}{(2a - 1)(a + 3)} = \frac{3a + 9 - 4a + 2}{(2a - 1)(a + 3)} = \frac{11 - a}{(2a - 1)(a + 3)}\)

(c) Multiply term by term:
\((2y - 3)(y^2 - 4y + 5) = 2y(y^2 - 4y + 5) - 3(y^2 - 4y + 5)\)
\(= 2y^3 - 8y^2 + 10y - 3y^2 + 12y - 15\)
\(= 2y^3 - 11y^2 + 22y - 15\)

(a)
M1 for factorising numerator: \(2(x-2)(x+2)\) oe
M1 for factorising denominator: \((x+2)(x+3)\) oe
M1 for canceling \((x+2)\)
A1 for final answer \(\frac{2x-4}{x+3}\)

(b)
M1 for \(3(a+3) - 2(2a-1)\) oe
M1 for expanding numerator to \(3a + 9 - 4a + 2\)
M1 for common denominator \((2a-1)(a+3)\) seen
A1 for final answer \(\frac{11-a}{(2a-1)(a+3)}\)

(c)
M1 for partial expansion with at least 3 correct terms
M1 for fully expanded unsimplified expression: \(2y^3 - 8y^2 + 10y - 3y^2 + 12y - 15\)
A1 for final simplified expression: \(2y^3 - 11y^2 + 22y - 15\)

(a) Area = Length \(\times\) Width
\((2x + 5)(x - 1) = 42\)
\(2x^2 - 2x + 5x - 5 = 42\)
\(2x^2 + 3x - 5 - 42 = 0\)
\(2x^2 + 3x - 47 = 0\) (Shown)

(b) Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-47)}}{2(2)}\)
\(x = \frac{-3 \pm \sqrt{9 + 376}}{4} = \frac{-3 \pm \sqrt{385}}{4}\)
\(x = \frac{-3 \pm 19.6214}{4}\)
\(x = 4.155 \approx 4.16\) or \(x = -5.655 \approx -5.66\)

(c) Since width cannot be negative, we use \(x = 4.155\).
Length = \(2(4.155) + 5 = 13.31\) m
Width = \(4.155 - 1 = 3.155\) m
Perimeter = \(2 \times (13.31 + 3.155) = 2 \times 16.465 = 32.93\) m, which is \(32.9\) m (to 3 sf).

(a)
M1 for \((2x + 5)(x - 1) = 42\)
M1 for \(2x^2 - 2x + 5x - 5 = 42\)
A1 for final step showing equation in standard form with no errors

(b)
M1 for substituting correct values into quadratic formula
B1 for \(\sqrt{385}\) or \(3^2 - 4(2)(-47)\)
M1 for splitting into two cases
A1 for \(4.16\) (accept \(4.155\) to \(4.16\))
A1 for \(-5.66\) (accept \(-5.655\) to \(-5.66\))

(c)
M1 for selecting positive root \(x \approx 4.16\)
M1 for substituting into \(2(2x+5 + x-1)\) or equivalent
A1 for \(32.9\) or \(32.93\)

(a) Volume of hemisphere = \(\frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6)^3 = 144\pi \approx 452.39\text{ cm}^3\)
Volume of cone = \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (14) = 168\pi \approx 527.79\text{ cm}^3\)
Total volume = \(144\pi + 168\pi = 312\pi \approx 980.18\text{ cm}^3\) (or \(980\text{ cm}^3\))

(b) Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 14^2} = \sqrt{36 + 196} = \sqrt{232} \approx 15.2315\text{ cm}\)
Which rounds to \(15.23\text{ cm}\) (Shown)

(c) Total Surface Area = Curved Surface Area of Cone + Curved Surface Area of Hemisphere
\(\text{Curved surface area of cone} = \pi r l = \pi \times 6 \times 15.2315 \approx 287.10\text{ cm}^2\)
\(\text{Curved surface area of hemisphere} = 2\pi r^2 = 2 \pi (6)^2 = 72\pi \approx 226.19\text{ cm}^2\)
Total Surface Area = \(287.10 + 226.19 = 513.29\text{ cm}^2\) (or \(513\text{ cm}^2\))

(d) Mass = \(\text{Density} \times \text{Volume} = 1.15 \times 980.18 = 1127.2\text{ g}\) (or \(1130\text{ g}\) to 3 sf)

(a)
M1 for Volume of hemisphere = \(\frac{2}{3} \pi \times 6^3\)
M1 for Volume of cone = \(\frac{1}{3} \pi \times 6^2 \times 14\)
A1 for total volume of \(980\) or \(980.18\)

(b)
M1 for \(\sqrt{6^2 + 14^2}\) oe
A1 for showing \(15.2315...\) and rounding to \(15.23\)

(c)
M1 for curved area of cone = \(\pi \times 6 \times 15.23\)
M1 for curved area of hemisphere = \(2 \times \pi \times 6^2\)
M1 for adding the two areas
A1 for \(513\) or \(513.29\)

(d)
M1 for \(1.15 \times \text{their (a)}\)
A1 for \(1130\) (or \(1127.2\))

(a) Using the Cosine Rule:
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\)
\(PR^2 = 75^2 + 110^2 - 2 \times 75 \times 110 \times \cos(58^\circ)\)
\(PR^2 = 5625 + 12100 - 16500 \times 0.52992\)
\(PR^2 = 17725 - 8743.66 = 8981.34\)
\(PR = \sqrt{8981.34} \approx 94.77\text{ m}\)
PR is \(94.8\text{ m}\) (to 3 sf).

(b) Using the Cosine Rule (or Sine Rule):
\(\cos(QPR) = \frac{PQ^2 + PR^2 - QR^2}{2 \times PQ \times PR}\)
\(\cos(QPR) = \frac{75^2 + 94.77^2 - 110^2}{2 \times 75 \times 94.77}\)
\(\cos(QPR) = \frac{5625 + 8981.36 - 12100}{14215.5} = \frac{2506.36}{14215.5} \approx 0.17631\)
\(\text{Angle } QPR = \cos^{-1}(0.17631) \approx 79.84^\circ\)
Angle QPR is \(79.8^\circ\) (to 3 sf).

(c) Area of the triangle = \(\frac{1}{2} \times a \times b \times \sin(C)\)
\(\text{Area} = \frac{1}{2} \times 75 \times 110 \times \sin(58^\circ)\)
\(\text{Area} = 4125 \times 0.84805 \approx 3498.2\text{ m}^2\)
Area of field is \(3500\text{ m}^2\) (to 3 sf).

(a)
M1 for Cosine Rule formula correctly identified
M2 for \(75^2 + 110^2 - 2 \times 75 \times 110 \times \cos(58^\circ)\)
A1 for \(94.8\) or \(94.77\)

(b)
M1 for Sine Rule or Cosine Rule setup: \(\frac{\sin(QPR)}{110} = \frac{\sin(58)}{94.77}\) oe
M2 for isolating \(\sin(QPR)\) or \(\cos(QPR)\)
A1 for \(79.8\) or \(79.84\)

(c)
M1 for Area formula: \(\frac{1}{2} a b \sin(C)\)
M1 for \(0.5 \times 75 \times 110 \times \sin(58)\)
A1 for \(3500\) or \(3498\)

(a) Multiply both sides by \((5 - t)\):
\(w(5 - t) = 3t + 2\)
\(5w - wt = 3t + 2\)
Collect terms with \(t\) on one side:
\(5w - 2 = 3t + wt\)
Factorise \(t\):
\(5w - 2 = t(3 + w)\)
\(t = \frac{5w - 2}{w + 3}\)

(b)(i) Let \(y = \frac{x+2}{2x-3}\)
Swap \(x\) and \(y\):
\(x = \frac{y+2}{2y-3}\)
\(x(2y - 3) = y + 2\)
\(2xy - 3x = y + 2\)
\(2xy - y = 3x + 2\)
\(y(2x - 1) = 3x + 2\)
\(y = \frac{3x+2}{2x-1}\)
Therefore, \(g^{-1}(x) = \frac{3x+2}{2x-1}\).

(b)(ii) First find \(g(4)\):
\(g(4) = \frac{4+2}{2(4)-3} = \frac{6}{5} = 1.2\)
Now find \(f(1.2)\):
\(f(1.2) = 3(1.2) - 1 = 3.6 - 1 = 2.6\)

(a)
M1 for \(w(5 - t) = 3t + 2\)
M1 for expanding: \(5w - wt = 3t + 2\)
M1 for isolating terms with \(t\): \(t(3 + w) = 5w - 2\) oe
A1 for \(t = \frac{5w - 2}{w + 3}\)

(b)(i)
M1 for changing the subject of \(y = g(x)\)
M1 for \(2xy - 3x = y + 2\) oe
M1 for factorising \(y\): \(y(2x - 1) = 3x + 2\)
A1 for \(\frac{3x+2}{2x-1}\)

(b)(ii)
M1 for \(g(4) = \frac{6}{5}\) or \(1.2\)
M1 for substituting their \(g(4)\) into \(f(x)\)
A1 for \(2.6\)

(a) Total parts = \(5 + 3 = 8\)
Number of jackets = \(\frac{3}{8} \times 120 = 45\)

(b) Number of shirts = \(120 - 45 = 75\)
Total cost of shirts = \(75 \times 15 = \$1125\)
Total cost of jackets = \(45 \times 35 = \$1575\)
Total cost = \(1125 + 1575 = \$2700\)

(c) Selling price of each shirt = \(15 \times 1.40 = \$21\)
Total revenue from shirts = \(75 \times 21 = \$1575\)
Total revenue from jackets = \(45 \times 48 = \$2160\)
Total revenue = \(1575 + 2160 = \$3735\)
Total profit = \(3735 - 2700 = \$1035\)
Percentage profit = \(\frac{1035}{2700} \times 100 = 38.33\%\)
Total percentage profit is \(38.3\%\) (to 3 sf).

(a)
M1 for \(\frac{3}{8} \times 120\)
A1 for 45

(b)
M1 for finding shirts = 75
M1 for \((75 \times 15) + (45 \times 35)\)
A1 for 2700

(c)
M1 for shirt selling price: \(15 \times 1.40 = 21\)
M1 for total shirt sales: \(75 \times 21 = 1575\)
M1 for total jacket sales: \(45 \times 48 = 2160\)
M1 for total revenue: 3735 or total profit: 1035
M1 for \(\frac{\text{their } 1035}{2700} \times 100\)
A1 for 38.3

(a) Differentiate term-by-term:
\(\frac{dy}{dx} = 3x^2 - 12x + 9\)

(b) For turning points, set \(\frac{dy}{dx} = 0\):
\(3x^2 - 12x + 9 = 0\)
\(3(x^2 - 4x + 3) = 0\)
\(3(x - 1)(x - 3) = 0\)
So, \(x = 1\) or \(x = 3\)

When \(x = 1\):
\(y = 1^3 - 6(1)^2 + 9(1) + 5 = 1 - 6 + 9 + 5 = 9\)
Turning point: \((1, 9)\)

When \(x = 3\):
\(y = 3^3 - 6(3)^2 + 9(3) + 5 = 27 - 54 + 27 + 5 = 5\)
Turning point: \((3, 5)\)

(c) At \(x = 2\):
Gradient of tangent, \(m = \left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3\)
Y-coordinate, \(y = 2^3 - 6(2)^2 + 9(2) + 5 = 8 - 24 + 18 + 5 = 7\)
So point of tangency is \((2, 7)\).

Using \(y = mx + c\):
\(7 = -3(2) + c \implies 7 = -6 + c \implies c = 13\)
Equation is \(y = -3x + 13\).

(a)
M1 for 2 correct differentiated terms
A1 for \(3x^2 - 12x + 9\)

(b)
M1 for setting \(\text{their } \frac{dy}{dx} = 0\)
M1 for factorising or solving: \(3(x-1)(x-3) = 0\)
A1 for \(x = 1\) and \(x = 3\)
A1 for \(y = 9\) (when \(x = 1\))
A1 for \(y = 5\) (when \(x = 3\))

(c)
M1 for substituting \(x = 2\) into \(\text{their } \frac{dy}{dx}\) to get \(m = -3\)
M1 for substituting \(x = 2\) into original equation to get \(y = 7\)
M1 for substituting \((2, 7)\) and \(m = -3\) into \(y = mx + c\) oe
A1 for \(y = -3x + 13\)

(a)
\(A = P \left(1 + \frac{r}{100}\right)^t\)
\(A = 4500 \left(1 + \frac{3.2}{100}\right)^8 = 4500(1.032)^8 \approx 5790.15\)
Correct to the nearest dollar: $5790

(b)
Total simple interest earned over 6 years = \(3576 - 3000 = 576\)
Simple interest per year = \(\frac{576}{6} = 96\)
Rate of simple interest \(r = \frac{96}{3000} \times 100 = 3.2\%\)

(c)
Let the initial value at the start of 2022 be \(V\).
Value at the end of 2022 = \(1.15V\)
Value at the end of 2023 = \(1.15V \times 0.92 = 1.058V\)
Given: \(1.058V = 264500\)
\(V = \frac{264500}{1.058} = 250000\)

(a) [3 marks]
M1 for \(4500 \times (1.032)^8\)
A1 for 5790.15
A1 for 5790 (correct to the nearest dollar)

(b) [3 marks]
M1 for \(3576 - 3000\)
M1 for \(\frac{\text{their interest}}{6 \times 3000} \times 100\) or \(3000 \times R \times 6 / 100 = 576\)
A1 for 3.2

(c) [5 marks]
M1 for multiplier 1.15 associated with 2022
M1 for multiplier 0.92 associated with 2023
M1 for equation \(1.15 \times 0.92 \times V = 264500\) or \(1.058V = 264500\)
M1 for \(V = \frac{264500}{1.058}\)
A1 for 250000

(a)
Factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
Factorise the denominator:
\(x^2 - 9 = (x - 3)(x + 3)\)
Simplify the fraction:
\(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\)

(b)
Multiply by common denominator \((2x - 1)(x + 2)\):
\(4(x + 2) - 3(2x - 1) = (2x - 1)(x + 2)\)
\(4x + 8 - 6x + 3 = 2x^2 + 3x - 2\)
\(-2x + 11 = 2x^2 + 3x - 2\)
\(2x^2 + 5x - 13 = 0\)
Use the quadratic formula:
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-13)}}{2(2)} = \frac{-5 \pm \sqrt{129}}{4}\)
\(x \approx 1.59\) or \(x \approx -4.09\)

(c)
Square both sides:
\(w^2 = \frac{t + 3}{2t - 1}\)
Multiply by \(2t - 1\):
\(w^2(2t - 1) = t + 3\)
\(2t w^2 - w^2 = t + 3\)
Collect terms with \(t\):
\(2t w^2 - t = w^2 + 3\)
\(t(2w^2 - 1) = w^2 + 3\)
\(t = \frac{w^2 + 3}{2w^2 - 1}\)

(a) [4 marks]
B1 for factorising the numerator \((2x + 1)(x - 3)\)
B1 for factorising the denominator \((x - 3)(x + 3)\)
M1 for dividing out common factor \((x - 3)\)
A1 for \(\frac{2x + 1}{x + 3}\) as final answer

(b) [4 marks]
M1 for common denominator on LHS or multiplying through by denominator
M1 for \(2x^2 + 5x - 13 = 0\) (or equivalent 3-term quadratic)
M1 for substituting their values into quadratic formula
A1 for 1.59 and -4.09

(c) [3 marks]
M1 for squaring both sides \(w^2 = \frac{t+3}{2t-1}\)
M1 for isolating terms with \(t\) on one side \(t(2w^2 - 1) = w^2 + 3\)
A1 for \(t = \frac{w^2 + 3}{2w^2 - 1}\)

(a)
Height of the cylinder = \(2r = 2(3) = 6\) cm.
Height of the cone = \(\text{Total height} - \text{Height of cylinder} = 15 - 6 = 9\) cm.

(b)
\(V_{\text{cylinder}} = \pi r^2 (2r) = 2\pi r^3 = 2\pi (3)^3 = 54\pi\) cm\(^3\).
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (9) = 27\pi\) cm\(^3\).
\(V_{\text{total}} = 54\pi + 27\pi = 81\pi\) cm\(^3\).

(c)
Total surface area consists of:
1. Base of the cylinder: \(\pi r^2 = \pi (3)^2 = 9\pi\) cm\(^2\).
2. Curved surface area of the cylinder: \(2\pi r (2r) = 4\pi r^2 = 4\pi (3)^2 = 36\pi\) cm\(^2\).
3. Curved surface area of the cone: \(\pi r l\).
Find slant height \(l\):
\(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 9^2} = \sqrt{90} = 3\sqrt{10} \approx 9.4868\) cm.
\(\text{CSA}_{\text{cone}} = \pi \times 3 \times 9.4868 \approx 28.46\pi\) cm\(^2\).
\(\text{Total Surface Area} = 9\pi + 36\pi + 28.46\pi = 73.46\pi \approx 230.78 \approx 231\) cm\(^2\).

(a) [2 marks]
M1 for finding height of cylinder = \(2 \times 3 = 6\)
A1 for showing \(15 - 6 = 9\)

(b) [3 marks]
M1 for volume of cylinder \(\pi \times 3^2 \times 6 = 54\pi\)
M1 for volume of cone \(\frac{1}{3} \pi \times 3^2 \times 9 = 27\pi\)
A1 for \(81\pi\)

(c) [6 marks]
M1 for area of circular base = \(9\pi\) or \(28.3\)
M1 for curved area of cylinder = \(36\pi\) or \(113.1\)
M1 for slant height of cone \(l = \sqrt{3^2 + 9^2}\)
A1 for slant height \(\approx 9.49\) or \(3\sqrt{10}\)
M1 for curved area of cone = \(\pi \times 3 \times 9.49\) (\(\approx 89.4\))
A1 for total surface area 231 (accept 230.7 to 231.2)

(a)
Using the Cosine Rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 7.5^2 + 6.2^2 - 2(7.5)(6.2)\cos(112^\circ)\)
\(AC^2 = 56.25 + 38.44 - 93 \cdot (-0.3746)\)
\(AC^2 = 94.69 + 34.838 = 129.528\)
\(AC \approx 11.38 \approx 11.4\) cm.

(b)
Using the Cosine Rule on triangle \(ACD\) to find \(AD\):
\(AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(ACD)\)
\(AD^2 = 129.528 + 14.5^2 - 2(11.381)(14.5)\cos(38^\circ)\)
\(AD^2 = 129.528 + 210.25 - 330.049 \times 0.7880\)
\(AD^2 = 339.778 - 260.079 = 79.699\)
\(AD \approx 8.927\) cm.
Now, find angle \(ADC\) using the Cosine Rule:
\(\cos(ADC) = \frac{AD^2 + CD^2 - AC^2}{2 \cdot AD \cdot CD}\)
\(\cos(ADC) = \frac{79.699 + 210.25 - 129.528}{2 \times 8.927 \times 14.5} = \frac{160.421}{258.883} \approx 0.6197\)
\(\text{Angle } ADC = \cos^{-1}(0.6197) \approx 51.7^\circ\).

(c)
\(\text{Area of triangle } ABC = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \cdot 7.5 \cdot 6.2 \cdot \sin(112^\circ) = 23.25 \times 0.9272 \approx 21.6\) cm\(^2\).

(d)
\(\text{Area of triangle } ABC = \frac{1}{2} \cdot AC \cdot d\)
\(21.557 = \frac{1}{2} \cdot 11.381 \cdot d\)
\(d = \frac{2 \times 21.557}{11.381} \approx 3.79\) cm.

(a) [3 marks]
M1 for implicit cosine rule formula: \(AC^2 = 7.5^2 + 6.2^2 - 2(7.5)(6.2)\cos(112^\circ)\)
A1 for \(AC^2 \approx 129.5\)
A1 for 11.4 (accept 11.38 to 11.4)

(b) [4 marks]
M1 for finding \(AD^2 = 11.38^2 + 14.5^2 - 2(11.38)(14.5)\cos(38^\circ)\)
A1 for \(AD \approx 8.93\)
M1 for \(\cos(ADC) = \frac{AD^2 + 14.5^2 - 11.38^2}{2 \times AD \times 14.5}\) or equivalent sine rule method
A1 for 51.7 (accept 51.68 to 51.75)

(c) [2 marks]
M1 for area formula: \(0.5 \times 7.5 \times 6.2 \times \sin(112^\circ)\)
A1 for 21.6 (accept 21.55 to 21.6)

(d) [2 marks]
M1 for \(0.5 \times 11.38 \times d = \text{their area in (c)}\)
A1 for 3.79 (accept 3.78 to 3.80)