2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

1. Find the midpoint, \(M\), of \(AB\):
\(M = \left(\frac{-2 + 4}{2}, \frac{5 + 13}{2}\right) = (1, 9)\)

2. Find the gradient of \(AB\):
\(m = \frac{13 - 5}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}\)

3. The gradient of the perpendicular line is:
\(m_{\perp} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4} = -0.75\)

4. Use the point-slope form with \(M(1, 9)\) and \(m_{\perp} = -0.75\):
\(y - 9 = -0.75(x - 1)\)
\(y = -0.75x + 0.75 + 9\)
\(y = -0.75x + 9.75\)

M1 for finding the correct midpoint \((1, 9)\)
M1 for finding the gradient of \(AB\) as \(\frac{4}{3}\)
M1 for using the perpendicular gradient \(m_{\perp} = -\frac{3}{4}\)
A1 for the correct equation \(y = -0.75x + 9.75\) (or equivalent fraction form \(y = -\frac{3}{4}x + \frac{39}{4}\))

1. Factor out the highest common factor, which is 2:
\(18x^2 - 50y^2 = 2(9x^2 - 25y^2)\)

2. Recognise that \(9x^2 - 25y^2\) is a difference of two squares:
\(9x^2 - 25y^2 = (3x)^2 - (5y)^2 = (3x - 5y)(3x + 5y)\)

3. Combine to write the final factorised expression:
\(2(3x - 5y)(3x + 5y)\)

B1 for extracting common factor 2: \(2(9x^2 - 25y^2)\)
M1 for factorising the difference of two squares: \((3x - 5y)(3x + 5y)\)
A1 for the final correct answer: \(2(3x - 5y)(3x + 5y)\)

1. Calculate the volume of the sphere:
\(V_{\text{sphere}} = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi\text{ cm}^3\)

2. Set up the formula for the volume of the cone:
\(V_{\text{cone}} = \frac{1}{3} \pi (2.5)^2 h = \frac{6.25}{3} \pi h\text{ cm}^3\)

3. Since the volume is conserved, equate the two volumes:
\(36\pi = \frac{6.25}{3} \pi h\)

4. Solve for the height \(h\):
\(36 = \frac{6.25}{3} h\)
\(108 = 6.25 h\)
\(h = \frac{108}{6.25} = 17.28\text{ cm}\)

M1 for finding the volume of the sphere as \(36\pi\) (or approximately \(113.1\))
M1 for setting up the cone volume equation with \(r = 2.5\): \(\frac{1}{3} \pi (2.5)^2 h\)
M1 for equating both volumes and attempting to isolate \(h\)
A1 for the correct height of \(17.28\)

1. Use the formula for the area of a non-right-angled triangle:
\(\text{Area} = \frac{1}{2} a b \sin C\)

2. Substitute the given values into the formula:
\(\text{Area} = \frac{1}{2} \times 8.4 \times 11.2 \times \sin(108^\circ)\)

3. Calculate the numerical value:
\(\text{Area} = 4.2 \times 11.2 \times 0.9510565 \approx 44.7377\text{ cm}^2\)

4. Rounding to 3 significant figures gives \(44.7\text{ cm}^2\).

M1 for using the formula \(\frac{1}{2} a b \sin C\)
M1 for correctly substituting values: \(\frac{1}{2} \times 8.4 \times 11.2 \times \sin(108^\circ)\)
A1 for the correct answer of \(44.7\) (or in range \(44.7\) to \(44.74\))

1. Understand that a reduction of \(15\%\) means the sale price is \(85\%\) of the original price.
Let the original price be \(P\).

2. Write the equation:
\(0.85 \times P = 561\)

3. Solve for \(P\):
\(P = \frac{561}{0.85} = 660\)

Thus, the original price of the television was \(\$660\).

M1 for expressing that \(85\%\) corresponds to \(561\) (e.g., \(0.85 P = 561\) or \(\frac{561}{85}\))
M1 for attempting to divide by the decimal or percentage ratio: \(561 \div 0.85\)
A1 for \(660\)

1. Find a common denominator, which is \((x-3)(x+4)\):
\(\frac{5(x+4) - 2(x-3)}{(x-3)(x+4)}\)

2. Expand the numerator:
\(5(x+4) - 2(x-3) = 5x + 20 - 2x + 6\)

3. Simplify the numerator by combining like terms:
\(3x + 26\)

4. Combine into the final fraction:
\(\frac{3x+26}{(x-3)(x+4)}\)

B1 for finding the common denominator as \((x-3)(x+4)\) (or expanded to \(x^2+x-12\))
M1 for expansion of numerator: \(5(x+4) - 2(x-3)\)
A1 for correct final simplified fraction: \(\frac{3x+26}{(x-3)(x+4)}\) or \(\frac{3x+26}{x^2+x-12}\)

1. Rearrange \(L_1\) to find its gradient:
\(3x - 2y = 8 \implies -2y = -3x + 8 \implies y = \frac{3}{2}x - 4\)
So, gradient \(m_1 = \frac{3}{2}\).

2. Find the perpendicular gradient \(m_2\):
\(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\)

3. Use the point-slope formula with point \((6, -1)\):
\(y - (-1) = -\frac{2}{3}(x - 6)\)
\(y + 1 = -\frac{2}{3}x + 4\)
\(y = -\frac{2}{3}x + 3\)

M1 for finding the gradient of the first line as \(\frac{3}{2}\) (or \(1.5\))
M1 for using the relationship for perpendicular lines to find the gradient of the second line: \(m = -\frac{2}{3}\)
A1 for the correct final equation \(y = -\frac{2}{3}x + 3\) (accept decimal equivalent with 3 significant figures: \(y = -0.667x + 3\))

1. Recall the formula for the total surface area of a cylinder:
\(A = 2\pi r^2 + 2\pi r h\)

2. Substitute the given values \(r = 4.5\) and \(h = 12\):
\(A = 2\pi (4.5)^2 + 2\pi (4.5)(12)\)
\(A = 2\pi (20.25) + 2\pi (54)\)
\(A = 40.5\pi + 108\pi = 148.5\pi\)

3. Calculate the numerical value:
\(A \approx 148.5 \times 3.14159265 \approx 466.5265\text{ cm}^2\)

4. Rounding to 3 significant figures gives \(467\text{ cm}^2\).

M1 for the correct formula for the total surface area of a cylinder: \(2\pi r^2 + 2\pi r h\)
M1 for substituting values into the formula correctly: \(2\pi(4.5)^2 + 2\pi(4.5)(12)\)
A1 for finding \(148.5\pi\) (or approximately \(466.5\))
A1 for the correct value rounded to 3 significant figures: \(467\)

We use the cosine rule to find the length of the side \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(PQR)\)
\(PR^2 = 7.4^2 + 5.2^2 - 2 \cdot 7.4 \cdot 5.2 \cdot \cos(112^\circ)\)
\(PR^2 = 54.76 + 27.04 - 76.96 \cdot (-0.3746)\)
\(PR^2 = 81.80 + 28.83 = 110.63\)
\(PR = \sqrt{110.63} \approx 10.518\text{ cm}\)
Rounding to 3 significant figures gives \(10.5\text{ cm}\).

M1 for a correct substitution into the cosine rule: \(7.4^2 + 5.2^2 - 2 \cdot 7.4 \cdot 5.2 \cdot \cos(112^\circ)\)
A1 for \(PR^2 = 110.63\) or better
A1 for \(10.5\) (or a value in the range \(10.51\) to \(10.52\))

First, factor out the common term \(2x\):
\(18x^3 - 50xy^2 = 2x(9x^2 - 25y^2)\)
Then, apply the difference of two squares to factorise the quadratic part inside the parentheses:
\(9x^2 - 25y^2 = (3x - 5y)(3x + 5y)\)
Thus, the completely factorised expression is:
\(2x(3x - 5y)(3x + 5y)\)

B1 for a partial factorisation such as \(2(9x^3 - 25xy^2)\) or \(x(18x^2 - 50y^2)\)
M1 for recognizing the difference of two squares: \(9x^2 - 25y^2 = (3x - 5y)(3x + 5y)\)
A1 for the completely factorised correct expression \(2x(3x - 5y)(3x + 5y)\)

1. Find the midpoint, \(M\), of the line segment \(AB\):
\(M = \left(\frac{2 + 8}{2}, \frac{-3 + 1}{2}\right) = (5, -1)\)
2. Calculate the gradient, \(m_{AB}\), of the line segment \(AB\):
\(m_{AB} = \frac{1 - (-3)}{8 - 2} = \frac{4}{6} = \frac{2}{3}\)
3. Find the gradient of the perpendicular line, \(m_{\perp}\):
\(m_{\perp} = -\frac{1}{m_{AB}} = -\frac{3}{2} = -1.5\)
4. Use the point-slope form with the midpoint \((5, -1)\) and gradient \(-1.5\):
\(y - (-1) = -1.5(x - 5)\)
\(y + 1 = -1.5x + 7.5\)
\(y = -1.5x + 6.5\)

B1 for finding the correct midpoint \((5, -1)\)
M1 for finding the gradient of \(AB\) as \(\frac{2}{3}\) and identifying the perpendicular gradient as \(-\frac{3}{2}\) (or \(-1.5\))
A1 for the final equation \(y = -1.5x + 6.5\) (or any equivalent fraction form such as \(y = -\frac{3}{2}x + \frac{13}{2}\))

We set up the compound interest inequality:
\(4500 \cdot (1.024)^n > 5000\)
\((1.024)^n > \frac{5000}{4500}\)
\((1.024)^n > 1.1111...\)
Now, let's compute values for different whole numbers of \(n\):
For \(n = 4\): \(1.024^4 \approx 1.0995\)
For \(n = 5\): \(1.024^5 \approx 1.1259\)
Since \(1.1259 > 1.1111...\), the smallest whole number of years is \(5\).

M1 for setting up the equation or inequality: \(4500 \cdot (1.024)^n > 5000\) or \(4500 \cdot (1.024)^n = 5000\)
M1 for systematic trial and error or logarithmic evaluation showing values for both \(n=4\) and \(n=5\)
A1 for \(5\)

The total surface area of a closed cylinder is given by the formula:
\(A = 2\pi r^2 + 2\pi r h\)
Given \(r = 5\) and \(A = 130\pi\):
\(130\pi = 2\pi(5)^2 + 2\pi(5)h\)
Divide the entire equation by \(\pi\):
\(130 = 2(25) + 10h\)
\(130 = 50 + 10h\)
\(10h = 80\)
\(h = 8\text{ cm}\)

M1 for a correct total surface area formula used: \(2\pi r^2 + 2\pi r h = 130\pi\)
A1 for a correct substitution and simplification: \(50 + 10h = 130\) (or equivalent)
A1 for \(8\)

The formula for the area of a triangle with two sides and an included angle is:
\(\text{Area} = \frac{1}{2} a b \sin(C)\)
Substitute the given values:
\(\text{Area} = \frac{1}{2} \cdot 8.5 \cdot 6.4 \cdot \sin(58^\circ)\)
\(\text{Area} = 27.2 \cdot \sin(58^\circ)\)
Using a calculator:
\(\sin(58^\circ) \approx 0.8480\)
\(\text{Area} \approx 27.2 \cdot 0.8480 = 23.067\text{ cm}^2\)
Rounding to 3 significant figures gives \(23.1\text{ cm}^2\).

M1 for a correct substitution into the area formula: \(\frac{1}{2} \cdot 8.5 \cdot 6.4 \cdot \sin(58^\circ)\)
A1 for an intermediate step of \(27.2 \cdot \sin(58^\circ)\) or \(23.067...\)
A1 for \(23.1\)

To subtract these fractions, we need a common denominator, which is \((x - 2)(2x + 1)\):
\(\frac{5(2x + 1) - 3(x - 2)}{(x - 2)(2x + 1)}\)
Expand the terms in the numerator:
\(\frac{10x + 5 - 3x + 6}{(x - 2)(2x + 1)}\)
Combine like terms:
\(\frac{7x + 11}{(x - 2)(2x + 1)}\)

M1 for expressing both fractions with a common denominator: \(\frac{5(2x + 1) - 3(x - 2)}{(x - 2)(2x + 1)}\)
M1 for correctly expanding the numerator to \(10x + 5 - 3x + 6\) (allow one sign error)
A1 for \(\frac{7x + 11}{(x - 2)(2x + 1)}\) or \(\frac{7x + 11}{2x^2 - 3x - 2}\)

1. Find the gradient of the given line:
\(3x - 4y = 12 \implies 4y = 3x - 12 \implies y = \frac{3}{4}x - 3\)
So, the gradient of the given line is \(m = \frac{3}{4}\).
2. Find the gradient of the perpendicular line:
\(m_{\perp} = -\frac{4}{3}\)
3. Write the equation of the perpendicular line using the point \((6, -1)\):
\(y - (-1) = -\frac{4}{3}(x - 6)\)
\(y + 1 = -\frac{4}{3}x + 8\)
\(y = -\frac{4}{3}x + 7\)
4. Convert the equation to the form \(ax + by = d\) with integer coefficients:
Multiply by 3:
\(3y = -4x + 21\)
\(4x + 3y = 21\)

B1 for finding the gradient of the original line as \(\frac{3}{4}\) or perpendicular gradient as \(-\frac{4}{3}\)
M1 for substituting the perpendicular gradient and point \((6, -1)\) into a line equation formula
A1 for \(4x + 3y = 21\) (or any non-simplified but correct integer form like \(8x + 6y = 42\))

First, find a common denominator, which is \((x-3)(2x+1)\).

Express each fraction with the common denominator:
\(\frac{5(2x+1)}{(x-3)(2x+1)} - \frac{2(x-3)}{(x-3)(2x+1)}\)

Combine the numerators:
\(\frac{5(2x+1) - 2(x-3)}{(x-3)(2x+1)}\)

Expand the terms in the numerator:
\(\frac{10x + 5 - 2x + 6}{(x-3)(2x+1)}\)

Simplify the numerator by collecting like terms:
\(\frac{8x + 11}{(x-3)(2x+1)}\)

M1 for \(5(2x+1) - 2(x-3)\) as numerator of a single fraction
M1 for common denominator \((x-3)(2x+1)\) seen
A1 for \(\frac{8x+11}{(x-3)(2x+1)}\) or \(\frac{8x+11}{2x^2-5x-3}\) as final answer

First, calculate the radius of the hemisphere using the volume formula:
\(V = \frac{2}{3}\pi r^3\)

\(\frac{2}{3}\pi r^3 = 450\)

\(r^3 = \frac{450 \times 3}{2\pi} = \frac{675}{\pi}\)

\(r = \sqrt[3]{214.859} \approx 5.9896\text{ cm}\)

Next, calculate the total surface area of the hemisphere, which consists of the curved surface area (\(2\pi r^2\)) plus the flat circular base (\(\pi r^2\)):
\(A = 3\pi r^2\)

\(A = 3\pi \times (5.9896)^2 \approx 338.12\text{ cm}^2\)

To 3 significant figures, the total surface area is \(338\text{ cm}^2\).

M1 for setting up \(\frac{2}{3}\pi r^3 = 450\) oe
M1 for using \(3\pi r^2\) with their calculated radius
A1 for \(338\) (or \(338.1\) to \(338.2\))

First, find the midpoint of the line segment \(AB\):
Midpoint \(M = \left(\frac{2+8}{2}, \frac{-3+1}{2}\right) = (5, -1)\)

Next, find the gradient of \(AB\):
Gradient \(m_{AB} = \frac{1 - (-3)}{8 - 2} = \frac{4}{6} = \frac{2}{3}\)

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

Use the point-slope form with midpoint \(M(5, -1)\) and gradient \(m_{\perp} = -1.5\):
\(y - (-1) = -1.5(x - 5)\)
\(y + 1 = -1.5x + 7.5\)
\(y = -1.5x + 6.5\)

M1 for finding the midpoint \((5, -1)\) or the gradient of \(AB\) as \(\frac{2}{3}\)
M1 for substituting their perpendicular gradient \(-\frac{3}{2}\) and their midpoint into a line equation
A1 for \(y = -1.5x + 6.5\) or any equivalent equation in the form \(y = mx + c\)

Use the Cosine Rule to find angle \(PQR\) (let's call it angle \(Q\)):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(Q)\)

Substitute the given values:
\(12.6^2 = 7.5^2 + 10.2^2 - 2(7.5)(10.2)\cos(Q)\)

\(158.76 = 56.25 + 104.04 - 153\cos(Q)\)

\(158.76 = 160.29 - 153\cos(Q)\)

Rearrange to solve for \(\cos(Q)\):
\(153\cos(Q) = 160.29 - 158.76\)
\(153\cos(Q) = 1.53\)
\(\cos(Q) = 0.01\)

Now, calculate the angle \(Q\):
\(Q = \arccos(0.01) \approx 89.427^\circ\)

To 1 decimal place, the angle is \(89.4^\circ\).

M1 for correct substitution into the Cosine Rule: \(12.6^2 = 7.5^2 + 10.2^2 - 2(7.5)(10.2)\cos(Q)\)
M1 for rearrangement to isolate \(\cos(Q)\): \(\cos(Q) = \frac{7.5^2 + 10.2^2 - 12.6^2}{2 \times 7.5 \times 10.2}\) (value of 0.01 oe seen)
A1 for \(89.4\) or \(89.42...\) to \(89.43...\)

A reduction of \(15\%\) means the sale price represents \(100\% - 15\% = 85\%\) of the original price.

Let \(x\) be the original price:
\(0.85x = 612\)

Solve for \(x\):
\(x = \frac{612}{0.85} = 720\)

The original price of the laptop was \(\$720\).

M1 for recognising that \(85\%\) represents \(612\) oe
M1 for \(\frac{612}{0.85}\) or \(612 \div 85 \times 100\) oe
A1 for \(720\) (or \(\$720\))

First, identify the common numerical factor, which is 2, and factorise it out:
\(18x^2 - 50y^2 = 2(9x^2 - 25y^2)\)

Now observe that the terms inside the parentheses represent a difference of two squares:
\(9x^2 - 25y^2 = (3x)^2 - (5y)^2\)

Apply the difference of squares identity \(a^2 - b^2 = (a-b)(a+b)\):
\(2(9x^2 - 25y^2) = 2(3x - 5y)(3x + 5y)\)

M1 for extracting the common factor 2: \(2(9x^2 - 25y^2)\)
M1 for factorising the difference of two squares \(9x^2 - 25y^2\) into \((3x-5y)(3x+5y)\)
A1 for final answer \(2(3x-5y)(3x+5y)\) or \(2(3x+5y)(3x-5y)\) oe

(a)
\(4500 \times \left(1 + \frac{r}{100}\right)^6 = 5188.38\)
\(\left(1 + \frac{r}{100}\right)^6 = 1.152973\)
\(1 + \frac{r}{100} = \sqrt[6]{1.152973} = 1.024\)
\(r = 2.4\)

(b)
\(V = 24000 \times (1 - 0.15)^4\)
\(V = 24000 \times (0.85)^4 = 12528.15\)
To the nearest dollar: $12528.

(c)
\(\text{Original Price} \times 0.80 = 784\)
\(\text{Original Price} = \frac{784}{0.80} = 980\)

(d)
\(x \times 1.08 \times 0.95 = 6156\)
\(x \times 1.026 = 6156\)
\(x = 6000\)

(a) M1 for \(4500 \times k^6 = 5188.38\)
M1 for \(k = 1.024\) or \(\sqrt[6]{\frac{5188.38}{4500}}\)
A1 for 2.4

(b) M1 for \(24000 \times (0.85)^4\)
A1 for 12528.15
A1 for 12528 (nearest dollar)

(c) M1 for \(\frac{784}{0.8}\)
A1 for 980

(d) M1 for \(1.08 \times 0.95\) or 1.026 seen
M1 for \(x \times \text{their } 1.026 = 6156\)
M1 for \(\frac{6156}{1.026}\)
A1 for 6000

(a) \(\angle QPR = 135^\circ - 65^\circ = 70^\circ\).

(b) Using the cosine rule:
\(QR^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(70^\circ)\)
\(QR^2 = 324 + 625 - 900 \times 0.34202\)
\(QR^2 = 949 - 307.818 = 641.182\)
\(QR = \sqrt{641.182} \approx 25.32\text{ km}\)
To 1 decimal place: 25.3 km.

(c) Using the sine rule to find angle \(PQR\):
\(\frac{\sin(PQR)}{25} = \frac{\sin(70^\circ)}{25.32}\)
\(\sin(PQR) = \frac{25 \times \sin(70^\circ)}{25.32} = 0.9278\)
\(\angle PQR = 68.1^\circ\)
Bearing of \(P\) from \(Q = 65^\circ + 180^\circ = 245^\circ\).
Bearing of \(R\) from \(Q = 245^\circ - 68.1^\circ = 176.9^\circ \approx 177^\circ\).

(d) \(\text{Time} = \frac{25.32\text{ km}}{12\text{ km/h}} = 2.11\text{ hours}\)
\(0.11\text{ hours} = 0.11 \times 60 = 6.6\text{ minutes} \approx 7\text{ minutes}\).
Total time: 2 hours 7 minutes.

(a) B1 for \(135 - 65 = 70\)

(b) M1 for \(18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(70)\)
A1 for 641.18...
A1 for 25.3

(c) M1 for \(\frac{\sin(PQR)}{25} = \frac{\sin(70)}{\text{their } 25.3}\)
A1 for angle \(PQR = 68.1^\circ\) (or 68.2)
M1 for bearing of \(P\) from \(Q = 245^\circ\)
M1 for \(245 - \text{their } 68.1\)
A1 for 177

(d) M1 for \(\frac{\text{their } 25.3}{12}\)
M1 for converting fractional part of hour to minutes
A1 for 2 hours 7 minutes (accept 2h 6m to 2h 7m)

(a) Gradient of \(AB\), \(m = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3}\).
Using point \((8, 5)\):
\(y - 5 = \frac{4}{3}(x - 8)\)
\(y = \frac{4}{3}x - \frac{32}{3} + 5\)
\(y = \frac{4}{3}x - \frac{17}{3}\).

(b) Midpoint \(M = \left(\frac{2 + 8}{2}, \frac{-3 + 5}{2}\right) = (5, 1)\).

(c) Gradient of the perpendicular bisector, \(m_{\perp} = -\frac{3}{4}\).
Equation through \(M(5, 1)\):
\(y - 1 = -\frac{3}{4}(x - 5)\)
Multiply by 4:
\(4y - 4 = -3(x - 5)\)
\(4y - 4 = -3x + 15\)
\(4y + 3x = 19\).

(d) The line crosses the \(x\)-axis when \(y = 0\).
\(4(0) + 3x = 19\)
\(3x = 19 \implies x = \frac{19}{3} \approx 6.33\).
Coordinates of \(C\) are \(\left(\frac{19}{3}, 0\right)\) or \((6.33, 0)\).

(a) M1 for \(\frac{5 - (-3)}{8 - 2}\)
M1 for substituting \((2, -3)\) or \((8, 5)\) into \(y = mx + c\)
A1 for \(y = \frac{4}{3}x - \frac{17}{3}\) (accept equivalent fractions)

(b) M1 for \(\left(\frac{2+8}{2}, \frac{-3+5}{2}\right)\)
A1 for \((5, 1)\)

(c) M1 for gradient \(= -\frac{1}{\text{their } m}\)
M1 for \(y - 1 = \text{their } m_{\perp}(x - 5)\)
M1 for rearranging to integer coefficients form
A1 for \(4y + 3x = 19\) (or any integer multiple like \(8y + 6x = 38\))

(d) M1 for substituting \(y = 0\) into their part (c) equation
M1 for solving for \(x\)
A1 for \(\left(\frac{19}{3}, 0\right)\) or \((6.33, 0)\)

(a) \(V_{\text{cone}} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3\).

(b) (i) \(\frac{2}{3} \pi r^3 = 75\pi \implies r^3 = 112.5 \implies r = \sqrt[3]{112.5} \approx 4.827\text{ cm}\).
Correct to 2 decimal places: \(r = 4.83\text{ cm}\).
(ii) \(\text{Slant height } l = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5}\)
\(l = 4.827 \times \sqrt{5} \approx 10.79\text{ cm} \approx 10.8\text{ cm}\).

(c) \(V_{\text{toy}} = V_{\text{cone}} + V_{\text{cylinder}} = 75\pi + \pi r^2 h = 250\pi\)
\(\pi r^2 h = 175\pi \implies r^2 h = 175\)
\(h = \frac{175}{4.827^2} = 7.51\text{ cm}\).

(d) \(\text{Exposed Surface Area} = \pi r^2 + 2\pi r h\)
\(A = \pi (4.827)^2 + 2\pi (4.827)(7.51)\)
\(A = 73.20 + 227.91 \approx 301.1\text{ cm}^2\).
To 3 significant figures: \(301\text{ cm}^2\).

(a) M1 for \(\frac{1}{3} \pi r^2 (2r)\)
A1 for \(\frac{2}{3} \pi r^3\)

(b) (i) M1 for equating their (a) to \(75\pi\)
M1 for \(r^3 = 112.5\)
A1 for \(4.827\) and concluding 4.83
(ii) M1 for \(\sqrt{r^2 + (2r)^2}\) or substituting 4.83
A1 for 10.8 (accept 10.79 to 10.80)

(c) M1 for \(75\pi + \pi r^2 h = 250\pi\)
M1 for \(h = \frac{175}{r^2}\)
A1 for 7.51 (accept 7.50 to 7.51)

(d) M1 for \(\pi r^2 + 2\pi r h\) (with their values of \(r\) and \(h\))
A1 for 301 (accept 300 to 302)

(a)
\(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)} = \frac{3x + 9 - 4x + 2}{2x^2 + 6x - x - 3} = \frac{11 - x}{2x^2 + 5x - 3}\).

(b)
\(\frac{11 - x}{2x^2 + 5x - 3} = 1\)
\(11 - x = 2x^2 + 5x - 3\)
\(2x^2 + 6x - 14 = 0\)
\(x^2 + 3x - 7 = 0\)
Using quadratic formula:
\(x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2}\)
\(x = \frac{-3 \pm \sqrt{37}}{2}\)
\(x \approx 1.54\) or \(x \approx -4.54\).

(c)
\(18a^2b - 50b^3 = 2b(9a^2 - 25b^2)\)
\(= 2b(3a - 5b)(3a + 5b)\).

(a) M1 for common denominator \((2x - 1)(x + 3)\)
M1 for numerator \(3(x+3) - 2(2x-1)\)
A1 for \(\frac{11 - x}{2x^2 + 5x - 3}\) (or denominator left factored)

(b) M1 for their numerator \(=\) their denominator
M1 for rearranging to a quadratic \(2x^2 + 6x - 14 = 0\) (or equivalent)
M1 for correct quadratic formula substitution
A1 for 1.54
A1 for -4.54

(c) B1 for factorising out \(2b\)
M2 for difference of two squares \((3a - 5b)(3a + 5b)\)
A1 for final answer \(2b(3a-5b)(3a+5b)\)

(a) Gradient of \(L_1, m = \frac{-2 - 6}{3 - (-1)} = \frac{-8}{4} = -2\).

(b) Gradient of \(L_2, m_2 = -\frac{1}{-2} = \frac{1}{2}\).
Equation of \(L_2\):
\(y - 5 = \frac{1}{2}(x - 4)\)
\(y = \frac{1}{2}x - 2 + 5\)
\(y = \frac{1}{2}x + 3\).

(c) Equation of \(L_1\):
\(y - 6 = -2(x + 1) \implies y = -2x + 4\).
To find the intersection, equate the two line equations:
\(\frac{1}{2}x + 3 = -2x + 4\)
\(2.5x = 1 \implies x = 0.4\).
Substitute \(x = 0.4\) into \(y = -2x + 4\):
\(y = -2(0.4) + 4 = 3.2\).
Point of intersection is \((0.4, 3.2)\).

(a) M1 for \(\frac{-2 - 6}{3 - (-1)}\)
A1 for -2

(b) M1 for perpendicular gradient \(= \frac{1}{2}\)
M1 for substituting \((4, 5)\) into \(y = \text{their } m_2 x + c\)
A1 for \(y = \frac{1}{2}x + 3\)

(c) M1 for finding the equation of \(L_1\) (e.g., \(y = -2x + 4\))
M1 for equating their two line equations
M1 for finding \(x = 0.4\)
M1 for finding \(y = 3.2\)
A1 for \((0.4, 3.2)\)

(a) Using the sine rule:
\(\frac{\sin(ACB)}{15} = \frac{\sin(40^\circ)}{12}\)
\(\sin(ACB) = \frac{15 \times \sin(40^\circ)}{12} = 1.25 \times 0.642787 = 0.80348\)
\(\angle ACB_1 = \sin^{-1}(0.80348) = 53.46^\circ \approx 53.5^\circ\).
\(\angle ACB_2 = 180^\circ - 53.46^\circ = 126.54^\circ \approx 126.5^\circ\).

(b) Case 1 (acute angle):
\(\angle ACB = 53.46^\circ\)
\(\angle ABC = 180^\circ - 40^\circ - 53.46^\circ = 86.54^\circ\)
\(AC_1 = \frac{12 \times \sin(86.54^\circ)}{\sin(40^\circ)} = \frac{12 \times 0.99815}{0.642787} \approx 18.63\text{ cm} \approx 18.6\text{ cm}\).

Case 2 (obtuse angle):
\(\angle ACB = 126.54^\circ\)
\(\angle ABC = 180^\circ - 40^\circ - 126.54^\circ = 13.46^\circ\)
\(AC_2 = \frac{12 \times \sin(13.46^\circ)}{\sin(40^\circ)} = \frac{12 \times 0.23277}{0.642787} \approx 4.346\text{ cm} \approx 4.35\text{ cm}\) (using precise values, \(4.35\text{ cm} \approx 4.36\text{ cm}\)).

(c) For Case 1:
\(\text{Area} = \frac{1}{2} \times 15 \times 12 \times \sin(86.54^\circ) = 90 \times 0.99815 = 89.83\text{ cm}^2 \approx 89.8\text{ cm}^2\).

(a) M1 for \(\frac{\sin(ACB)}{15} = \frac{\sin(40)}{12}\)
M1 for \(\sin(ACB) = 0.8035\)
A1 for 53.5
A1 for 126.5

(b) M1 for angle \(ABC_1 = 86.5\) or angle \(ABC_2 = 13.5\)
M1 for \(\frac{AC}{\sin(ABC)} = \frac{12}{\sin(40)}\)
A1 for 18.6
A1 for 4.35 or 4.36

(c) M1 for \(\frac{1}{2} \times 15 \times 12 \times \sin(\text{their } 86.5)\)
A1 for 89.8 (accept 89.8 to 89.9)

(a) \(\text{Area} = (2x + 3)(x - 1) = 2x^2 - 2x + 3x - 3 = 2x^2 + x - 3\).

(b) (i) \(2x^2 + x - 3 = 32\)
Subtract 32 from both sides:
\(2x^2 + x - 35 = 0\).
(ii) Factorise \(2x^2 + x - 35 = 0\):
\((2x - 7)(x + 5) = 0\)
\(x = 3.5\) or \(x = -5\).
Since physical dimensions must be positive, \(x\) must be greater than 1, so \(x = 3.5\).

(c) \(\text{Perimeter} = 2 \times [ (2x + 3) + (x - 1) ]\)
\(\text{Perimeter} = 2 \times [ 3x + 2 ] = 6x + 4\)
Substituting \(x = 3.5\):
\(\text{Perimeter} = 6(3.5) + 4 = 21 + 4 = 25\text{ cm}\).

(a) B1 for complete algebraic expansion showing terms \(2x^2 - 2x + 3x - 3\)

(b) (i) M1 for \(2x^2 + x - 3 = 32\)
A1 for correctly deriving \(2x^2 + x - 35 = 0\)
(ii) M1 for correct attempt to factorise or use quadratic formula
M1 for \((2x - 7)(x + 5) = 0\) or roots \(3.5\) and \(-5\)
A1 for rejecting \(x = -5\)
A1 for \(x = 3.5\)

(c) M1 for perimeter expression \(2[(2x+3) + (x-1)]\)
M1 for substituting \(x = 3.5\) into their perimeter expression
A1 for 25

(a) Volume of cylinder: $V_{\text{cyl}} = \pi r^2 h = \pi x^2 (2x) = 2\pi x^3$
Volume of hemisphere: $V_{\text{hemi}} = \frac{1}{2} \left(\frac{4}{3} \pi r^3\right) = \frac{2}{3} \pi x^3$
Total volume: $V = 2\pi x^3 + \frac{2}{3} \pi x^3 = \frac{8}{3} \pi x^3$

(b) Set the volume expression equal to $576\pi$:
$\frac{8}{3} \pi x^3 = 576\pi$
$\frac{8}{3} x^3 = 576$
$x^3 = 576 \times \frac{3}{8}$
$x^3 = 216$
$x = \sqrt[3]{216} = 6$

(c) Curved surface area of cylinder: $2\pi r h = 2\pi x (2x) = 4\pi x^2$
Curved surface area of hemisphere: $2\pi r^2 = 2\pi x^2$
Total outer surface area: $A = 4\pi x^2 + 2\pi x^2 = 6\pi x^2$
Since $x = 6$, $A = 6\pi (6)^2 = 216\pi\text{ cm}^2$

(d) Area to paint: $216\pi \approx 678.58\text{ cm}^2$
Number of tins: $\frac{678.58}{500} \approx 1.36$
Since paint must be purchased in whole tins, the minimum number of tins required is 2.

(a) M1 for $2\pi x^3$ or $\frac{2}{3}\pi x^3$ seen
A1 for $\frac{8}{3}\pi x^3$ (or equivalent fraction, e.g., $\frac{8\pi x^3}{3}$)

(b) M1 for setting their (a) equal to $576\pi$
M1 for $x^3 = 216$ or $x = \sqrt[3]{216}$
A1 for $x = 6$ fully shown with no errors

(c) M1 for curved surface area of cylinder ($4\pi x^2$) or hemisphere ($2\pi x^2$) seen
M1 for $6\pi x^2$ or substituting $x=6$ into their area expression
A1 for $216\pi$

(d) M1 for evaluating $216\pi$ to at least 3 s.f. (678 or 679 or 678.5...)
M1 for dividing their area by 500 (e.g., $1.35...$ or $1.36$)
A1 for 2

(a) Gradient of $AB$: $m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2$
Using $y - y_1 = m(x - x_1)$ with $(6, 5)$:
$y - 5 = 2(x - 6)$
$y = 2x - 7$

(b) (i) Midpoint of $AB = \left(\frac{2 + 6}{2}, \frac{-3 + 5}{2}\right) = (4, 1)$
(ii) Gradient of perpendicular line $L = -\frac{1}{\text{gradient of } AB} = -\frac{1}{2}$
Using midpoint $(4, 1)$:
$y - 1 = -\frac{1}{2}(x - 4)$
$2(y - 1) = -(x - 4)$
$2y - 2 = -x + 4$
$x + 2y = 6$

(c) $P$ lies on the $x$-axis ($y = 0$): $x + 2(0) = 6 \implies P(6, 0)$
$Q$ lies on the $y$-axis ($x = 0$): $0 + 2y = 6 \implies y = 3 \implies Q(0, 3)$
Length of $PQ = \sqrt{(6 - 0)^2 + (0 - 3)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71$ (to 3 s.f.)

(a) M1 for gradient of $AB = 2$
M1 for substituting gradient and a point into linear equation formula
A1 for $y = 2x - 7$

(b) (i) M1 for $\frac{2+6}{2}$ or $\frac{-3+5}{2}$ seen
A1 for $(4, 1)$
(ii) M1 for perpendicular gradient $= -\frac{1}{2}$ (FT their gradient of $AB$)
M1 for substituting their midpoint and perpendicular gradient into linear equation
M1 for rearranging to the given form
A1 for $x + 2y = 6$ fully shown with no errors

(c) M1 for finding coordinates $P(6, 0)$ and $Q(0, 3)$
M1 for using Pythagoras' theorem to find length of $PQ$
A1 for $\sqrt{45}$ or $6.71$ or $6.708...$

(a) Using the Cosine Rule:
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)$
$AC^2 = 80^2 + 110^2 - 2(80)(110)\cos(75^\circ)$
$AC^2 = 6400 + 12100 - 17600\cos(75^\circ)$
$AC^2 = 18500 - 17600(0.258819)$
$AC^2 = 18500 - 4555.22 = 13944.78$
$AC = \sqrt{13944.78} \approx 118.09\text{ m}$ (or $118$ m)

(b) Using the Area of a Triangle formula:
$\text{Area} = \frac{1}{2} a c \sin(B) = \frac{1}{2} (80)(110)\sin(75^\circ) = 4400 \sin(75^\circ) \approx 4250.07\text{ m}^2$ (or $4250\text{ m}^2$)

(c) $\text{Cost} = 4250.07 \times 45 = $191253.15 \approx $191,253$ (nearest dollar)
(If using rounded area $4250$: $4250 \times 45 = $191,250$)

(d) Perimeter of $ABC = AB + BC + AC = 80 + 110 + 118.09 = 308.09\text{ m}$
Total cost before discount $= 308.09 \times 18.50 = $5699.67$
Discounted price ($92\%$) $= 5699.67 \times 0.92 = $5243.69$
(If using rounded $AC = 118$ m:
Perimeter $= 80 + 110 + 118 = 308\text{ m}$
Cost before discount $= 308 \times 18.50 = $5698$
Discounted price $= 5698 \times 0.92 = $5242.16$)

(a) M1 for correct substitution into the Cosine Rule: $80^2 + 110^2 - 2(80)(110)\cos(75^\circ)$
M1 for $AC^2 = 13940$ to $13950$
A1 for $118$ or $118.1$ or $118.08...$

(b) M1 for correct substitution into the area formula: $\frac{1}{2} \times 80 \times 110 \times \sin(75^\circ)$
A1 for $4250$ or $4250.07...$

(c) M1 for multiplying their area by 45
A1 for $191253$ or $191250$ (FT their area)

(d) M1 for finding the perimeter (using their $AC$)
M1 for multiplying their perimeter by 18.50
M1 for multiplying their cost by $0.92$ (or finding $8\%$ and subtracting)
A1 for $5243.69$ (or $5244$) or $5242.16$ (or $5242$) if using $AC = 118$