2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

First, factorise the numerator \(2x^2 - 5x - 3\). Find two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\). Splitting the middle term: \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator \(4x^2 - 1\) as a difference of two squares: \(4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)\). Now write the fraction with these factorised forms: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancelling the common factor \((2x + 1)\) from both the numerator and denominator yields the simplified fraction: \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\) or the denominator to \((2x - 1)(2x + 1)\). M1 for factorising both correctly. A0.69 for the final simplified fraction \(\frac{x - 3}{2x - 1}\).

Let \(V\) be the original value of the motorcycle. The depreciation over \(3\) years at a rate of \(5\%\) per year can be modeled by the equation: \(V \times (1 - 0.05)^3 = 6859\). This simplifies to: \(V \times (0.95)^3 = 6859\). Calculate \(0.95^3\): \(0.95^3 = 0.857375\). Solve for \(V\): \(V = \frac{6859}{0.857375} = 8000\). Thus, the initial value was \(\$8000\).

M1 for setting up a correct equation, e.g., \(V \times 0.95^3 = 6859\) or \(\frac{6859}{0.95^3}\). M1 for evaluation of the growth factor \(0.95^3 = 0.857375\) or showing a division step. A0.69 for the correct final value 8000.

Use the cosine rule to find angle \(ABC\) (which we call angle \(B\), opposite to side \(AC = b = 13\)): \(b^2 = a^2 + c^2 - 2ac \cos(B)\). Substitute the given side lengths \(a = 8\), \(c = 7\), and \(b = 13\) into the formula: \(13^2 = 8^2 + 7^2 - 2(8)(7) \cos(B)\). Simplify the terms: \(169 = 64 + 49 - 112 \cos(B)\), which gives \(169 = 113 - 112 \cos(B)\). Rearrange to solve for \(\cos(B)\): \(169 - 113 = -112 \cos(B)\) implies \(56 = -112 \cos(B)\). Hence, \(\cos(B) = -\frac{56}{112} = -0.5\). Find angle \(B\): \(B = \cos^{-1}(-0.5) = 120^\circ\).

M1 for substituting correctly into the cosine rule: \(13^2 = 8^2 + 7^2 - 2(8)(7)\cos(B)\). M1 for rearranging to find \(\cos(B) = -0.5\). A0.69 for the final answer of 120 (degrees).

Multiply both sides by \(2x - a\) to get: \(y(2x - a) = 3x + 5\). Expand the brackets: \(2xy - ay = 3x + 5\). Rearrange the terms to group all terms containing \(x\) on one side: \(2xy - 3x = ay + 5\). Factorise \(x\) out of the left-hand side: \(x(2y - 3) = ay + 5\). Divide by \(2y - 3\) to isolate \(x\): \(x = \frac{ay + 5}{2y - 3}\).

M1 for multiplying by \(2x - a\) to obtain \(2xy - ay = 3x + 5\) or equivalent. M1 for isolating \(x\) terms on one side of the equation. A1 for the correct final formula.

By the cosine rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the given values: \(AC^2 = 8.3^2 + 6.4^2 - 2 \cdot 8.3 \cdot 6.4 \cdot \cos(118^\circ)\). Calculate each term: \(AC^2 = 68.89 + 40.96 - 106.24 \cdot (-0.46947)\). This simplifies to: \(AC^2 = 109.85 + 49.876 = 159.726\). Taking the square root: \(AC = \sqrt{159.726} \approx 12.638\text{ cm}\). To 3 significant figures, this is \(12.6\).

M1 for correct substitution into the cosine rule: \(8.3^2 + 6.4^2 - 2 \cdot 8.3 \cdot 6.4 \cdot \cos(118)\). A1 for evaluating \(AC^2 \approx 159.7\) or \(159.8\). A1 for final answer \(12.6\) (or \(12.6\text{ cm}\)).

Find the radius \(r\) using the volume formula: \(280 = \frac{4}{3}\pi r^3\), which gives \(r^3 = \frac{210}{\pi} \approx 66.845\). Thus, \(r = \sqrt[3]{66.845} \approx 4.0583\text{ cm}\). Now substitute \(r\) into the surface area formula: \(A = 4\pi r^2 = 4\pi (4.0583)^2 \approx 4\pi \times 16.470 = 206.96\text{ cm}^2\). Rounding to 3 significant figures gives \(207\text{ cm}^2\).

M1 for setting up the volume formula to find \(r\): \(\frac{4}{3}\pi r^3 = 280\). M1 for finding \(r \approx 4.06\) or \(r^2 \approx 16.5\). A1 for final answer \(207\).

Use the cosine rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substituting the given values gives \(AC^2 = 7.4^2 + 5.2^2 - 2 \times 7.4 \times 5.2 \times \cos(62^\circ)\). This simplifies to \(AC^2 = 54.76 + 27.04 - 76.96 \times \cos(62^\circ) = 81.8 - 36.1306 = 45.6694\). Therefore, \(AC = \sqrt{45.6694} \approx 6.76\text{ cm}\) (rounded to 3 significant figures).

M1 for correct substitution into the cosine rule: \(7.4^2 + 5.2^2 - 2 \times 7.4 \times 5.2 \times \cos(62^\circ)\). A1 for \(AC^2 = [45.6, 45.7]\) or \(AC = \sqrt{45.669...}\). A1 for \(6.76\) or \(6.758...\).

Let the original price of the tablet be \(x\) dollars. An 18% discount means the selling price is 82% of the original price, so \(0.82x = 118.90\). Solving for \(x\) gives \(x = 118.90 / 0.82 = 145\). Thus, the original price was $145.

M1 for equating \(118.90\) to \(82\%\) of the original price, e.g. \(0.82x = 118.90\). M1 for \(118.90 / 0.82\). A1 for 145.

To eliminate \(y\), multiply the first equation by 5 to get \(15x - 10y = 95\) and the second equation by 2 to get \(8x + 10y = 20\). Adding these two equations gives \(23x = 115\), which simplifies to \(x = 5\). Substitute \(x = 5\) into the first equation: \(3(5) - 2y = 19 \implies 15 - 2y = 19 \implies -2y = 4 \implies y = -2\).

M1 for a correct method to eliminate one variable (e.g. multiplying both equations to obtain equal coefficient magnitudes for one variable). A1 for \(x = 5\). A1 for \(y = -2\).

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

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

Substitute these factors back into the fraction:
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)}\)

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

M1 for factorising the numerator: \((2x + 1)(x - 3)\)
M1 for factorising the denominator: \((2x + 1)(2x - 1)\)
A0.69 for \(\frac{x - 3}{2x - 1}\)

Let \(P\) be the original price of the smart watch.

A discount of 16% means the selling price is \(100\% - 16\% = 84\%\) of the original price.

Therefore, we can write the equation:
\(0.84P = 268.80\)

Solving for \(P\):
\(P = \frac{268.80}{0.84} = 320\)

So, the original price of the smart watch is $320.

M1 for setting up a correct reverse percentage calculation, e.g. \(268.80 \div 0.84\) or \(x \times 0.84 = 268.80\)
A1.69 for 320 (or $320)

Use the Cosine Rule to find angle \(ABC\) (which is angle \(B\)), where the opposite side is \(b = AC = 11\text{ cm}\), and the adjacent sides are \(c = AB = 7\text{ cm}\) and \(a = BC = 9\text{ cm}\):

\(b^2 = a^2 + c^2 - 2ac \cos(B)\)

Substitute the given values into the formula:
\(11^2 = 9^2 + 7^2 - 2(9)(7) \cos(B)\)

Simplify the terms:
\(121 = 81 + 49 - 126 \cos(B)\)

\(121 = 130 - 126 \cos(B)\)

Rearrange to solve for \(\cos(B)\):
\(126 \cos(B) = 130 - 121\)

\(126 \cos(B) = 9\)

\(\cos(B) = \frac{9}{126} = \frac{1}{14}\)

Calculate the angle:
\(B = \cos^{-1}\left(\frac{1}{14}\right) \approx 85.903^\circ\)

Rounding to 1 decimal place, we get \(85.9^\circ\).

M1 for correct substitution into the Cosine Rule: \(11^2 = 7^2 + 9^2 - 2(7)(9)\cos(B)\)
M1 for rearranging the formula correctly to solve for \(\cos(B)\): \(\cos(B) = \frac{7^2 + 9^2 - 11^2}{2 \times 7 \times 9}\)
A0.69 for 85.9 (accept answers in the range 85.9 to 85.91)

First, factorise the numerator: \( 2x^2 - 5x - 3 = (2x + 1)(x - 3) \). Next, factorise the denominator using the difference of two squares: \( 4x^2 - 1 = (2x - 1)(2x + 1) \). Write the expression with the factored terms: \( \frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} \). Cancel the common factor of \( 2x + 1 \) to get the simplified fraction: \( \frac{x - 3}{2x - 1} \).

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

Let the original price of the item be \( x \). An increase of 15% is represented by multiplying by 1.15, giving \( 1.15x \). A subsequent decrease of 20% is represented by multiplying by 0.80. The final sale price can be written as: \( 0.80 \times 1.15x = 73.60 \). Simplifying this gives: \( 0.92x = 73.60 \). Solving for \( x \): \( x = \frac{73.60}{0.92} = 80 \). Therefore, the original price was $80.

M1 for expressing the final price in terms of \( x \), such as \( 1.15 \times 0.80x \) or \( 0.92x \). M1 for setting up the equation \( 0.92x = 73.60 \) or equivalent. A1 for the correct answer 80.

Use the cosine rule to find the angle \( B \) (which is angle \( ABC \)): \( \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} \). Here, \( a = BC = 9 \), \( c = AB = 7 \), and \( b = AC = 12 \). Substitute these values into the formula: \( \cos(B) = \frac{9^2 + 7^2 - 12^2}{2 \times 9 \times 7} \). Simplify the numerator and denominator: \( \cos(B) = \frac{81 + 49 - 144}{126} = \frac{-14}{126} = -\frac{1}{9} \). Calculate the angle: \( B = \arccos\left(-\frac{1}{9}\right) \approx 96.379^\circ \). Rounding to 1 decimal place gives 96.4.

M1 for correct substitution into the cosine rule, e.g., \( 12^2 = 9^2 + 7^2 - 2(9)(7)\cos(B) \). M1 for simplifying to \( \cos(B) = -\frac{1}{9} \) or \( \cos(B) \approx -0.111 \). A1 for the correct angle 96.4.

Let the original price of the laptop be \(x\). After a 15% reduction, the price is \(0.85x\). After a further 10% reduction, the price is \(0.90 \times 0.85x = 0.765x\). We are given that this final price is $612, so \(0.765x = 612\). Solving for \(x\) gives \(x = \frac{612}{0.765} = 800\). Thus, the original price was $800.

M1 for expressing the final price in terms of the original price, e.g., \(0.85 \times 0.90\) or \(0.765x = 612\).
M1 for \(612 \div 0.765\) or finding intermediate sale price \(612 \div 0.90 = 680\).
A1 for 800.

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the given values: \(AC^2 = 8.5^2 + 12.4^2 - 2 \times 8.5 \times 12.4 \times \cos(62^\circ)\). This simplifies to: \(AC^2 = 72.25 + 153.76 - 210.8 \times \cos(62^\circ)\). Since \(\cos(62^\circ) \approx 0.46947\), we get \(AC^2 \approx 226.01 - 98.9646 = 127.0454\). Taking the square root: \(AC \approx 11.271\text{ cm}\). Correct to 3 significant figures, this is \(11.3\text{ cm}\).

M1 for correct substitution into the Cosine Rule formula, e.g., \(8.5^2 + 12.4^2 - 2 \times 8.5 \times 12.4 \times \cos(62^\circ)\).
M1 for \(AC^2 \approx 127\) or \(AC = \sqrt{127.0...}\).
A1 for 11.3 (accept answers in range 11.27 to 11.3).

First, factorise the numerator: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). Next, factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Now, write the fraction with the factorised expressions and cancel the common term \((2x - 1)\): \(\frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)} = \frac{x + 3}{2x + 1}\).

M1 for factorising the numerator to \((2x - 1)(x + 3)\) or equivalent.
M1 for factorising the denominator to \((2x - 1)(2x + 1)\).
A1 for \(\frac{x + 3}{2x + 1}\) or equivalent fully simplified fraction.

First, factorise the numerator: \(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Now, simplify the fraction by cancelling the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\) or equivalent. M1 for factorising the denominator to \((2x - 1)(2x + 1)\). A1 for the final simplified fraction.

Using the cosine rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substitute the given values: \(AC^2 = 7^2 + 9^2 - 2 \cdot 7 \cdot 9 \cdot \cos(124^\circ)\). \(AC^2 = 49 + 81 - 126 \cdot (-0.55919...)\). \(AC^2 = 130 + 70.458... = 200.458...\). \(AC = \sqrt{200.458...} \approx 14.158\text{ cm}\). To 3 significant figures, the length is \(14.2\text{ cm}\).

M1 for correct substitution into the cosine rule, e.g. \(7^2 + 9^2 - 2 \cdot 7 \cdot 9 \cdot \cos(124)\). A1 for \(200.4...\) or \(200\). A1 for \(14.2\) or \(14.15\dots\).

A reduction of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price. Let \(x\) be the original price: \(0.85x = 646\). Solving for \(x\): \(x = \frac{646}{0.85} = 760\).

M1 for equating \(85\%\) to \(646\), e.g., \(0.85x = 646\) or \(646 \div 0.85\). A1 for \(760\).

First, factorise the numerator as a difference of two squares:
\(4x^2 - 9 = (2x - 3)(2x + 3)\)

Next, factorise the quadratic expression in the denominator:
\(2x^2 + 5x - 12 = 2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\)

Now, substitute these factorised forms back into the fraction:
\(\frac{(2x - 3)(2x + 3)}{(2x - 3)(x + 4)}\)

Cancel out the common factor of \(2x - 3\) from the numerator and the denominator:
\(\frac{2x + 3}{x + 4}\)

This is in its simplest form.

M1 for factorising the numerator: \((2x-3)(2x+3)\)
M1 for factorising the denominator: \((2x-3)(x+4)\)
A1 for the final simplified fraction: \(\frac{2x+3}{x+4}\)

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

An increase of 15% multiplies the price by \(1.15\).
A subsequent decrease of 8% multiplies the price by \(0.92\).

Therefore, we have the equation:
\(P \times 1.15 \times 0.92 = 634.80\)
\(P \times 1.058 = 634.80\)

Solving for \(P\):
\(P = \frac{634.80}{1.058} = 600\)

The original price of the laptop was $600.

M1 for writing \(1.15 \times 0.92\) or recognizing the combined multiplier is \(1.058\)
M1 for setting up the equation \(1.058P = 634.80\) or evaluating \(634.80 \div (1.15 \times 0.92)\)
A1 for the correct answer 600

By the Cosine Rule, for any triangle with sides \(a\), \(b\), and \(c\) opposite to angles \(A\), \(B\), and \(C\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)

Substitute the given values:
\(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(112^\circ)\)
\(AC^2 = 64 + 121 - 176\cos(112^\circ)\)
\(AC^2 = 185 - 176(-0.3746...)\)
\(AC^2 = 185 + 65.930...\)
\(AC^2 = 250.930...\)
\(AC = \sqrt{250.930...} \approx 15.84...\text{ cm}\)

Rounding to 3 significant figures gives \(15.8\text{ cm}\).

M1 for substituting correctly into the cosine rule: \(8^2 + 11^2 - 2(8)(11)\cos(112^\circ)\)
A1 for \(AC^2 = 250.9...\)
A1 for \(15.8\) (accept 15.84...)

To simplify the expression, we factorise both the numerator and the denominator:

1. Factorise the quadratic expression in the numerator, \(2x^2 - 5x - 3\):
We look for two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These numbers are \(-6\) and \(1\).
$$2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$$

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

3. Substitute the factorised forms back into the fraction and divide out the common factor \((2x + 1)\):
$$\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}$$

Thus, the simplified fraction is \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\) or equivalent.
M1 for factorising the denominator to \((2x - 1)(2x + 1)\).
A1 for \(\frac{x - 3}{2x - 1}\) or equivalent fraction (e.g., \(\frac{3 - x}{1 - 2x}\)).

We can find angle \(ABC\) (which is angle \(B\)) using the cosine rule:
$$b^2 = a^2 + c^2 - 2ac \cos(B)$$

Here, the side opposite to angle \(B\) is \(b = AC = 13\text{ cm}\), and the adjacent sides are \(c = AB = 7\text{ cm}\) and \(a = BC = 9\text{ cm}\).

Rearranging the formula for \
\cos(B)\:
$$\cos(B) = \frac{a^2 + c^2 - b^2}{2ac}$$

Substitute the given values into the formula:
$$\cos(B) = \frac{9^2 + 7^2 - 13^2}{2 \times 9 \times 7}$$
$$\cos(B) = \frac{81 + 49 - 169}{126}$$
$$\cos(B) = \frac{130 - 169}{126}$$
$$\cos(B) = -\frac{39}{126} = -\frac{13}{42}$$

Now, calculate angle \(B\):
$$B = \cos^{-1}\left(-\frac{13}{42}\right) \approx 108.0311^{\circ}$$

Rounding to 1 decimal place, we get \(108.0^{\circ}\).

M1 for correct substitution into the cosine rule, e.g., \(13^2 = 9^2 + 7^2 - 2 \times 9 \times 7 \times \cos(ABC)\) or \(\cos(ABC) = \frac{9^2 + 7^2 - 13^2}{2 \times 9 \times 7}\).
A1 for \(\cos(ABC) = -\frac{39}{126}\) or \(\cos(ABC) \approx -0.31\).
A1 for \(108.0\) or \(108\) (accept \(108.03...\)).

\(a\) Angle \(QPR = 130^\circ - 75^\circ = 55^\circ\). Using the cosine rule: \(QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos(QPR)\) which gives \(QR^2 = 12^2 + 15^2 - 2 \cdot 12 \cdot 15 \cdot \cos(55^\circ)\). Thus \(QR^2 = 144 + 225 - 360 \cdot \cos(55^\circ) \approx 162.51\), so \(QR = \sqrt{162.51} \approx 12.75\) km. Correct to 3 significant figures, this is 12.7 km. \(b\) Using the sine rule: \(\frac{\sin(PQR)}{15} = \frac{\sin(55^\circ)}{12.75}\) which gives \(\sin(PQR) = \frac{15 \cdot \sin(55^\circ)}{12.75} \approx 0.9638\). Thus, angle \(PQR \approx 74.54^\circ\). The bearing of \(P\) from \(Q\) is \(180^\circ + 75^\circ = 255^\circ\). Therefore, the bearing of \(R\) from \(Q\) is \(255^\circ - 74.54^\circ = 180.46^\circ\), which is 180.5° (to 1 decimal place). \(c\) Area of triangle \(PQR = \frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(55^\circ) = 90 \cdot \sin(55^\circ) \approx 73.7\) km\(^2\).

(a) M1 for cosine rule with correct values: 12^2 + 15^2 - 2*12*15*cos(55). A1 for 162.5... A1 for 12.7 or 12.748... (b) M1 for sine rule: sin(PQR)/15 = sin(55)/12.75. A1 for angle PQR = 74.5. M1 for 255 - angle PQR. A1 for 180.5 (or 180.4 to 180.5). (c) M1 for 0.5 * 12 * 15 * sin(55). A1 for 73.7 (or 73.72...)

\(a\) Curved surface area of a hemisphere of radius \(x\) is \(2\pi x^2\). Curved surface area of a cone of radius \(x\) and slant height \(l\) is \(\pi x l\). Total surface area of the toy = \(2\pi x^2 + \pi x l\). Surface area of a sphere of radius \(2x\) = \(4\pi (2x)^2 = 16\pi x^2\). Setting them equal: \(2\pi x^2 + \pi x l = 16\pi x^2\) which simplifies to \(\pi x l = 14\pi x^2\), hence \(l = 14x\). \(b\)(i) By Pythagoras' theorem in the cone: \(h = \sqrt{l^2 - x^2} = \sqrt{(14x)^2 - x^2} = \sqrt{195x^2} = x\sqrt{195}\). Given \(x = 5.4\), \(h = 5.4 \cdot \sqrt{195} \approx 75.407 \approx 75.4\) cm. \(b\)(ii) Volume of the hemisphere = \(\frac{2}{3}\pi x^3 = \frac{2}{3}\pi (5.4)^3 \approx 329.79\) cm\(^3\). Volume of the cone = \(\frac{1}{3}\pi x^2 h = \frac{1}{3}\pi (5.4)^2 (75.407) \approx 2302.60\) cm\(^3\). Total volume = \(329.79 + 2302.60 = 2632.39\) cm\(^3\), which is 2630 cm\(^3\) (3 s.f.).

(a) M1 for 2*pi*x^2 + pi*x*l. M1 for 4*pi*(2x)^2 or 16*pi*x^2. M1 for equating and simplifying to at least pi*x*l = 14*pi*x^2. A1 for fully correct working leading to l = 14x. (b)(i) M1 for h^2 + x^2 = (14x)^2. M1 for h = 5.4 * sqrt(195) or h = sqrt(75.6^2 - 5.4^2). A1 for 75.4. (b)(ii) M1 for (2/3)*pi*(5.4)^3. M1 for (1/3)*pi*(5.4)^2 * (their 75.4). A1 for 2630 or 2632 to 2633.

\(a\) Original price of one t-shirt = \(\frac{1200}{n}\). \(b\) The number of t-shirts in the new order is \(n + 20\). The price per t-shirt is reduced by \(\$2\), so the new price is \(\frac{1200}{n} - 2\). The total cost is \(1440\), so \((n + 20)\left( \frac{1200}{n} - 2 \right) = 1440\). Expanding the left side: \(1200 - 2n + \frac{24000}{n} - 40 = 1440\). This simplifies to \(-2n + 1160 + \frac{24000}{n} = 1440\), which is \(-2n + \frac{24000}{n} = 280\). Multiplying both sides by \(n\) gives \(-2n^2 + 24000 = 280n\), or \(2n^2 + 280n - 24000 = 0\). Dividing by 2 gives \(n^2 + 140n - 12000 = 0\). \(c\) Factoring the equation: \((n - 60)(n + 200) = 0\). This gives \(n = 60\) or \(n = -200\). Since the number of t-shirts must be positive, \(n = 60\). \(d\) The original price is \(\frac{1200}{60} = \$20\). The reduced price is \(20 - 2 = \$18\).

(a) B1 for 1200/n. (b) M1 for (n+20)(1200/n - 2) = 1440. M1 for expanding: 1200 - 2n + 24000/n - 40 = 1440. M1 for multiplying by n to get -2n^2 + 24000 = 280n (or equivalent quadratic). A1 for fully showing the progression to n^2 + 140n - 12000 = 0. (c) M1 for (n-60)(n+200) [or use of quadratic formula with correct substitution]. A1 for n = 60 and n = -200. A1 for choosing positive root n = 60. (d) M1 for 1200 / (their 60). M1 for (their 20) - 2 or 1440 / ((their 60) + 20). A1 for 18 (or $18).

\( \textbf{Part (a)} \)
First, calculate the angle \(BAC\):
\(\angle BAC = 145^\circ - 75^\circ = 70^\circ\).

Using the cosine rule to find the distance \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)\)
\(BC^2 = 12^2 + 15^2 - 2(12)(15)\cos(70^\circ)\)
\(BC^2 = 144 + 225 - 360\cos(70^\circ)\)
\(BC^2 \approx 369 - 123.127 = 245.873\)
\(BC = \sqrt{245.873} \approx 15.68\) km.
To 3 significant figures, this is \(15.7\) km.

\( \textbf{Part (b)} \)
First, use the sine rule to find \(\angle ABC\):
\(\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{BC}\)
\(\frac{\sin(\angle ABC)}{15} = \frac{\sin(70^\circ)}{15.68}\)
\(\sin(\angle ABC) \approx \frac{15 \sin(70^\circ)}{15.68} \approx 0.8989\)
\(\angle ABC \approx \arcsin(0.8989) \approx 64.0^\circ\).

Now, find the bearing of \(C\) from \(B\):
The bearing of \(B\) from \(A\) is \(075^\circ\), so the back-bearing (direction from \(B\) to \(A\)) is:
\(075^\circ + 180^\circ = 255^\circ\).
Since \(C\) lies to the south of the line \(AB\), the bearing of \(C\) from \(B\) is:
\(255^\circ - 64.0^\circ = 191.0^\circ\) (or \(191^\circ\)).

\( \textbf{Part (c)} \)
Calculate the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC)\)
\(\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin(70^\circ)\)
\(\text{Area} = 90 \times \sin(70^\circ) \approx 84.57 \text{ km}^2\).
To 3 significant figures, this is \(84.6 \text{ km}^2\).

(a) [4 marks]
M1 for identifying \(\angle BAC = 70^\circ\)
M1 for substituting correctly into the Cosine Rule: \(12^2 + 15^2 - 2(12)(15)\cos(70^\circ)\)
A1 for \(BC^2 \approx 245.87... \)
A1 for \(BC = 15.7\) (or 15.68)

(b) [4 marks]
M1 for substituting correctly into the Sine Rule: \(\frac{\sin(\angle ABC)}{15} = \frac{\sin(70^\circ)}{\text{their } BC}\)
A1 for \(\angle ABC = 64.0^\circ\) or \(64.02^\circ\)
M1 for a correct method to find the bearing using back-bearing: \(255^\circ - \text{their } 64.0^\circ\)
A1 for \(191^\circ\) or \(191.0^\circ\)

(c) [3 marks]
M1 for area formula: \(\frac{1}{2} \times 12 \times 15 \times \sin(70^\circ)\)
A1 for \(90 \sin(70^\circ)\)
A1 for \(84.6\) (or 84.57)

\( \textbf{Part (a)} \)
The total volume \(V\) is the sum of the volume of the cylinder and the volume of the cone:
\(V = V_{\text{cylinder}} + V_{\text{cone}}\)
\(V = \pi r^2 (2r) + \frac{1}{3} \pi r^2 h\)
Factor out \(\pi r^2\):
\(V = \pi r^2 \left( 2r + \frac{1}{3}h \right)\). [Proved]

\( \textbf{Part (b)(i)} \)
Substitute \(h = 3r\) into the volume formula:
\(V = \pi r^2 \left( 2r + \frac{1}{3}(3r) \right) = \pi r^2 (2r + r) = 3\pi r^3\).
We are given \(V = 360\pi\), so:
\(3\pi r^3 = 360\pi\)
\(3r^3 = 360\)
\(r^3 = 120\)
\(r = \sqrt[3]{120} \approx 4.9324\) cm.
To 3 s.f., \(r = 4.93\) cm.

\( \textbf{Part (b)(ii)} \)
The total surface area, \(A\), of the toy consists of:
1. The circular base of the cylinder: \(\pi r^2\)
2. The curved surface of the cylinder: \(2\pi r (2r) = 4\pi r^2\)
3. The curved surface of the cone: \(\pi r l\), where \(l\) is the slant height of the cone.

Find the slant height \(l\):
\(l = \sqrt{r^2 + h^2}\)
Since \(h = 3r\), we have:
\(l = \sqrt{r^2 + (3r)^2} = \sqrt{10r^2} = r\sqrt{10}\).

Thus, the curved surface area of the cone is:
\(\pi r (r\sqrt{10}) = \sqrt{10}\pi r^2\).

The total surface area \(A\) is:
\(A = \pi r^2 + 4\pi r^2 + \sqrt{10}\pi r^2 = (5 + \sqrt{10})\pi r^2\).

Using \(r = 120^{1/3} \approx 4.9324\):
\(r^2 = 120^{2/3} \approx 24.3288\).
\(A = (5 + \sqrt{10})\pi (24.3288)\)
\(A \approx (8.162277)\pi (24.3288) \approx 623.895 \text{ cm}^2\).
To 3 s.f., the total surface area is \(624 \text{ cm}^2\).

(a) [2 marks]
M1 for writing \(V = \pi r^2 (2r) + \frac{1}{3}\pi r^2 h\)
A1 for factoring out \(\pi r^2\) to get the required expression.

(b)(i) [3 marks]
M1 for substituting \(h = 3r\) to get \(V = 3\pi r^3\)
M1 for setting \(3\pi r^3 = 360\pi\) and solving for \(r^3\)
A1 for \(r \approx 4.93\) (or 4.932...)

(b)(ii) [6 marks]
B1 for area of base = \(\pi r^2\)
B1 for curved area of cylinder = \(4\pi r^2\)
M1 for finding slant height \(l = r\sqrt{10}\) (or \(l \approx 15.6\))
M1 for curved area of cone = \(\pi r^2 \sqrt{10}\)
M1 for summing three surfaces: \(A = (5 + \sqrt{10})\pi r^2\)
A1 for \(624\) (accept answers in the range [623.8, 624.2])

\( \textbf{Part (a)} \)
(i) Time taken by the first train is \(\frac{\text{Distance}}{\text{Speed}} = \frac{200}{x}\) hours.
(ii) Time taken by the express train is \(\frac{\text{Distance}}{\text{Speed}} = \frac{200}{x+10}\) hours.

\( \textbf{Part (b)} \)
The express train takes 1 hour less than the first train, so:
\(\frac{200}{x} - \frac{200}{x+10} = 1\)

Multiply the entire equation by the common denominator \(x(x+10)\):
\(200(x+10) - 200x = x(x+10)\)
\(200x + 2000 - 200x = x^2 + 10x\)
\(2000 = x^2 + 10x\)
\(x^2 + 10x - 2000 = 0\). [Proved]

\( \textbf{Part (c)} \)
To solve \(x^2 + 10x - 2000 = 0\), we factorise the quadratic equation:
\((x + 50)(x - 40) = 0\)
This yields:
\(x = -50\) or \(x = 40\).

\( \textbf{Part (d)} \)
Since speed must be positive, we take \(x = 40\) km/h.
The average speed of the express train is \(x + 10 = 40 + 10 = 50\) km/h.
The time taken by the express train is:
\(\text{Time} = \frac{200}{50} = 4\) hours.

(a)(i) [1 mark]
B1 for \(\frac{200}{x}\)

(a)(ii) [1 mark]
B1 for \(\frac{200}{x+10}\)

(b) [4 marks]
M1 for \(\frac{200}{x} - \frac{200}{x+10} = 1\) (or equivalent)
M1 for multiplying by \(x(x+10)\) to clear fractions: \(200(x+10) - 200x = x(x+10)\)
A1 for expanding brackets correctly: \(200x + 2000 - 200x = x^2 + 10x\)
A1 for simplifying to \(x^2 + 10x - 2000 = 0\) with no errors.

(c) [3 marks]
M2 for \((x + 50)(x - 40) = 0\) (M1 for \((x \pm 50)(x \pm 40)\))
A1 for \(x = 40\) and \(x = -50\)

(d) [2 marks]
M1 for substituting \(x = 40\) into \(\frac{200}{x+10}\) or \(200 / 50\)
A1 for 4 hours.

(a) Using the cosine rule on \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 120^2 + 150^2 - 2(120)(150)\cos(78^\circ)\)
\(AC^2 = 14400 + 22500 - 36000 \times 0.207912\)
\(AC^2 = 36900 - 7484.82 = 29415.18\)
\(AC = \sqrt{29415.18} \approx 171.51\text{ m}\)
Correct to 3 significant figures: \(172\text{ m}\).

(b) Area of a triangle:
\(\text{Area} = \frac{1}{2} a b \sin(C)\)
\(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(78^\circ) = 9000 \times 0.978148 \approx 8803.3\text{ m}^2\)
Correct to 3 significant figures: \(8800\text{ m}^2\).

(c) In right-angled \(\triangle PAC\) (where \(\angle PAC = 90^\circ\)):
\(\tan(8.5^\circ) = \frac{PT}{AC}\)
\(PT = AC \times \tan(8.5^\circ)\)
Using the more accurate value \(AC = 171.51\):
\(PT = 171.51 \times \tan(8.5^\circ) \approx 25.63\text{ m}\)
Correct to 3 significant figures: \(25.6\text{ m}\).

(d) The shortest distance from \(B\) to \(AC\) is the perpendicular height, \(h\), of \(\triangle ABC\) from base \(AC\):
\(\text{Area} = \frac{1}{2} \times AC \times h\)
\(8803.3 = \frac{1}{2} \times 171.51 \times h\)
\(h = \frac{2 \times 8803.3}{171.51} \approx 102.66\text{ m}\)
Correct to 3 significant figures: \(103\text{ m}\).

(a) [4 marks]
- M1 for substituting correctly into the Cosine Rule: \(120^2 + 150^2 - 2(120)(150)\cos(78^\circ)\)
- A1 for \(7480\) to \(7485\) or \(29415\) to \(29420\)
- A1 for \(171.5...\)
- A1 for \(172\) or \(171.5\)

(b) [2 marks]
- M1 for \(\frac{1}{2} \times 120 \times 150 \times \sin(78^\circ)\)
- A1 for \(8800\) or \(8803\)

(c) [3 marks]
- M1 for \(\tan(8.5) = \frac{PT}{\text{their } AC}\)
- M1 for \(PT = \text{their } AC \times \tan(8.5)\)
- A1 for \(25.6\) or \(25.63\)

(d) [2 marks]
- M1 for \(\frac{1}{2} \times \text{their } AC \times h = \text{their Area}\)
- A1 for \(103\) or \(102.7\)

(a) Volume of a cone: \(V = \frac{1}{3} \pi r^2 h\)
\(V = \frac{1}{3} \pi \times 6^2 \times 15 = 180\pi \approx 565.49\text{ cm}^3\)
Correct to 3 significant figures: \(565\text{ cm}^3\).

(b) Volume of the sphere: \(V = \frac{4}{3} \pi R^3 = 180\pi\)
\(\frac{4}{3} R^3 = 180 \implies R^3 = 135\)
\(R = \sqrt[3]{135} \approx 5.1299\text{ cm}\)
Correct to 3 significant figures: \(5.13\text{ cm}\).

(c) The rise in water level, \(d\), is caused by the volume of the submerged cone:
\(\text{Volume of cylinder of rise} = \pi \times \text{radius}^2 \times d = 180\pi\)
\(\pi \times 8^2 \times d = 180\pi\)
\(64d = 180\)
\(d = \frac{180}{64} = 2.8125\text{ cm}\)
Correct to 3 significant figures: \(2.81\text{ cm}\).

(d) First, find the slant height, \(l\), of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 15^2} = \sqrt{36 + 225} = \sqrt{261} \approx 16.155\text{ cm}\)
Total surface area of a cone:
\(A = \pi r^2 + \pi r l = \pi \times 6^2 + \pi \times 6 \times \sqrt{261}\)
\(A = 36\pi + 6\pi \sqrt{261} \approx 113.10 + 304.52 = 417.62\text{ cm}^2\)
Correct to 3 significant figures: \(418\text{ cm}^2\).

(a) [2 marks]
- M1 for \(\frac{1}{3} \pi \times 6^2 \times 15\)
- A1 for \(565\) or \(565.48...\) or \(180\pi\)

(b) [3 marks]
- M1 for equating sphere volume formula to their (a): \(\frac{4}{3} \pi R^3 = \text{their } V\)
- M1 for rearranging to make \(R^3\) or \(R\) the subject: \(R = \sqrt[3]{\frac{3 \times \text{their } V}{4\pi}}\)
- A1 for \(5.13\) or \(5.129...\)

(c) [3 marks]
- M1 for equating volume of a cylinder with radius 8 to their (a): \(\pi \times 8^2 \times d = \text{their } V\)
- M1 for rearranging to solve for \(d\): \(d = \frac{\text{their } V}{64\pi}\)
- A1 for \(2.81\) or \(2.8125\)

(d) [3 marks]
- M1 for finding slant height \(l = \sqrt{6^2 + 15^2}\)
- M1 for substituting \(r=6\) and their \(l\) into \(\pi r^2 + \pi r l\)
- A1 for \(418\) or \(417.6...\)

(a) The dimensions of the garden including the path are \((12 + 2x)\text{ m}\) by \((8 + 2x)\text{ m}\).
\(\text{Total Area} = (12 + 2x)(8 + 2x) = 165\)
\(96 + 24x + 16x + 4x^2 = 165\)
\(4x^2 + 40x + 96 - 165 = 0\)
\(4x^2 + 40x - 69 = 0\) (as required).

(b) Solve \(4x^2 + 40x - 69 = 0\):
Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-40 \pm \sqrt{40^2 - 4(4)(-69)}}{2(4)}\)
\(x = \frac{-40 \pm \sqrt{1600 + 1104}}{8}\)
\(x = \frac{-40 \pm \sqrt{2704}}{8}\)
\(x = \frac{-40 \pm 52}{8}\)

This gives two solutions:
\(x = \frac{-40 + 52}{8} = \frac{12}{8} = 1.5\)
\(x = \frac{-40 - 52}{8} = -\frac{92}{8} = -11.5\)

(c) Since width must be positive, \(x = 1.5\text{ m}\).

(d) Area of the path = \(\text{Total Area} - \text{Garden Area}\)
\(\text{Area of path} = 165 - (12 \times 8) = 165 - 96 = 69\text{ m}^2\).

Area of one square tile = \(0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2\).

Number of tiles = \(\frac{\text{Area of path}}{\text{Area of one tile}} = \frac{69}{0.25} = 276\).

(a) [4 marks]
- M1 for writing the expression for outer dimensions: \((12 + 2x)\) and \((8 + 2x)\)
- M1 for expanding \((12 + 2x)(8 + 2x)\) to get \(96 + 24x + 16x + 4x^2\)
- M1 for setting their expansion equal to \(165\)
- A1 for complete and correct algebraic steps leading to \(4x^2 + 40x - 69 = 0\)

(b) [4 marks]
- M1 for substituting correctly into the quadratic formula: \(\frac{-40 \pm \sqrt{40^2 - 4(4)(-69)}}{2(4)}\) (or completing square)
- A1 for \(\sqrt{2704}\) or \(52\)
- A1 for \(1.5\)
- A1 for \(-11.5\)

(c) [1 mark]
- B1 for \(1.5\) (or choice of positive solution from their b)

(d) [2 marks]
- M1 for finding the path area (\(69\text{ m}^2\)) and dividing by the area of one tile (\(0.25\text{ m}^2\))
- A1 for \(276\)

a) The volume of a hemisphere is \(V_{\text{hemi}} = \frac{2}{3}\pi r^3\).
The volume of a cylinder is \(V_{\text{cyl}} = \pi r^2 h\).
Since \(h = 3r\), \(V_{\text{cyl}} = \pi r^2 (3r) = 3\pi r^3\).
Total Volume \(V = \frac{2}{3}\pi r^3 + 3\pi r^3 = \frac{11}{3}\pi r^3\).
We are given \(V = 792\pi\), so:
\(\frac{11}{3}\pi r^3 = 792\pi\)
\(\frac{11}{3}r^3 = 792\)
\(r^3 = 792 \times \frac{3}{11} = 72 \times 3 = 216\)
\(r = \sqrt[3]{216} = 6\).

b) The total surface area consists of the curved surface of the hemisphere, the curved surface of the cylinder, and the flat base.
\(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r(3r) = 3\pi r^2 + 6\pi r^2 = 9\pi r^2\).
Substituting \(r = 6\):
\(A = 9\pi (6^2) = 324\pi\text{ cm}^2\).

c) The radius of the cone is \(R = 1.5r = 1.5 \times 6 = 9\text{ cm}\).
Volume of the cone is \(\frac{1}{3}\pi R^2 H = 792\pi\).
\(\frac{1}{3}\pi (9^2) H = 792\pi\)
\(27\pi H = 792\pi\)
\(H = \frac{792}{27} = 29.333... \approx 29.3\text{ cm}\).

a)
- M1 for writing volume of hemisphere as \(\frac{2}{3}\pi r^3\) or cylinder as \(3\pi r^3\)
- M1 for forming the equation \(\frac{11}{3}\pi r^3 = 792\pi\)
- M1 for solving to get \(r^3 = 216\)
- A1 for \(r = 6\)

b)
- M1 for identifying the three surface area components: \(2\pi r^2\), \(2\pi rh\), and \(\pi r^2\)
- M1 for substituting \(h=3r\) to simplify to \(9\pi r^2\) or substituting \(r=6\) and \(h=18\) into individual components
- A1 for \(324\pi\)

c)
- M1 for setting up cone volume equation: \(\frac{1}{3}\pi (1.5r)^2 H = 792\pi\)
- M1 for simplifying to \(27 H = 792\) or equivalent
- A1 for \(29.3\) (accept 29.33...)

a) Using the compound interest formula:
\(4500 \left(1 + \frac{r}{100}\right)^6 = 5510\)
\(\left(1 + \frac{r}{100}\right)^6 = \frac{5510}{4500} = 1.22444...\)
\(1 + \frac{r}{100} = (1.22444...)^{1/6} \approx 1.034346\)
\(\frac{r}{100} = 0.034346\)
\(r \approx 3.43\)

b) Let \(x\) be the original price.
\(0.85x = 612\)
\(x = \frac{612}{0.85} = 720\)
The original price was \(\$720\).

c) (i) Let the initial number of employees be \(E\).
At the end of 2021, number of employees = \(1.12E\).
At the end of 2022, number of employees = \(1.12E \times 0.95 = 1.064E\).
Percentage change = \((1.064 - 1) \times 100\% = 6.4\%\) increase.

(ii) Let the start of 2021 number of employees be \(E\).
\(1.064E = 280\)
\(E = \frac{280}{1.064} \approx 263.158\)
To the nearest whole number, there were 263 employees.

a)
- M1 for compound interest setup: \(4500(1+r/100)^6 = 5510\)
- M1 for isolating brackets: \(1+r/100 = (5510/4500)^{1/6}\)
- A1 for \(r = 3.43\)

b)
- M1 for set up \(0.85x = 612\) or \(612 / 0.85\)
- A1 for \(720\)

c)
- (i) M1 for multiplying \(1.12 \times 0.95\)
- A1 for \(6.4\%\) increase (accept 6.4)
- (ii) M1 for dividing \(280\) by their multiplier from (i) (e.g. \(1.064\))
- A1 for \(263\)

a) Factorise the numerator by grouping:
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).
Factorise the denominator as a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\).
Simplify by cancelling common factor \(2x + 1\):
\(\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} = \frac{x-3}{2x-1}\).

b) Find a common denominator:
\(\frac{3(x + 4) - 2(2x - 1)}{(2x - 1)(x + 4)}\)
\(= \frac{3x + 12 - 4x + 2}{(2x - 1)(x + 4)}\)
\(= \frac{14 - x}{(2x - 1)(x + 4)}\).

c) Multiply both sides by \(t + 2\):
\(w(t + 2) = 3t - 5\)
\(wt + 2w = 3t - 5\)
Gather terms in \(t\) on one side:
\(2w + 5 = 3t - wt\)
\(2w + 5 = t(3 - w)\)
Divide both sides by \(3 - w\):
\(t = \frac{2w + 5}{3 - w}\).

a)
- M1 for factorising the numerator into \((2x+1)(x-3)\)
- M1 for factorising the denominator into \((2x-1)(2x+1)\)
- A1 for final answer \(\frac{x-3}{2x-1}\)

b)
- M1 for finding common denominator \((2x-1)(x+4)\)
- M1 for expansion \(3x + 12 - 4x + 2\)
- A1 for final answer \(\frac{14-x}{(2x-1)(x+4)}\) or equivalent

c)
- M1 for multiplying by \(t + 2\): \(w(t + 2) = 3t - 5\)
- M1 for collecting terms and factorising: \(t(3 - w) = 2w + 5\) or equivalent
- A1 for final subject formula \(t = \frac{2w+5}{3-w}\)