2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

To find the volume of the prism, we first find the area of the triangular cross-section. The base and height of the right-angled triangle are the two shorter sides, which are \(5\text{ cm}\) and \(12\text{ cm}\) (since \(13\text{ cm}\) is the hypotenuse). \text{Area of cross-section} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2. Now, we multiply the cross-sectional area by the length of the prism: \text{Volume} = \text{Area} \times \text{length} = 30 \times 20 = 600\text{ cm}^3.

M1 for \frac{1}{2} \times 5 \times 12 (or equivalent correct method for the cross-sectional area) M1 for (their 30) \times 20 A1 for 600

We can solve these equations using elimination. Multiply the second equation by 2: \(4x - 2y = 2\). Now, add this equation to the first equation: \((3x + 2y) + (4x - 2y) = 19 + 2\) which simplifies to \(7x = 21\), so \(x = 3\). Substitute \(x = 3\) back into the second original equation: \(2(3) - y = 1\) which simplifies to \(6 - y = 1\), so \(y = 5\).

M1 for a correct method to eliminate one variable (e.g. multiplying the second equation by 2 to get \(4x - 2y = 2\) or rearranging to make \(y\) the subject, e.g. \(y = 2x - 1\)) A1 for \(x = 3\) or \(y = 5\) A1 for \(x = 3\) and \(y = 5\)

The sale price represents \(100\% - 15\% = 85\%\) of the original price. Let \(x\) be the original price: \(0.85x = 204\), which means \(x = \frac{204}{0.85}\). Multiplying the numerator and the denominator by 100 gives \(x = \frac{20400}{85}\). Since both 20400 and 85 are divisible by 17 (where \(17 \times 5 = 85\) and \(17 \times 1200 = 20400\)), we have \(x = \frac{1200}{5} = 240\). Thus, the original price is $240.

M1 for equating 85% to 204 (e.g. \(0.85x = 204\)) M1 for a complete correct method (e.g. \(204 \div 0.85\) or \(\frac{204}{85} \times 100\)) A1 for 240

The sale price represents \(100\% - 15\% = 85\%\) of the original price. Let the original price be \(x\). We have: \(0.85x = 68\). To solve for \(x\), divide \(68\) by \(0.85\): \(x = \frac{68}{0.85} = \frac{6800}{85}\). Simplifying this fraction by dividing both numerator and denominator by 17 gives: \(x = \frac{400}{5} = 80\). The original price of the coat was \(\$80\).

M1 for associating \(\$68\) with \(85\%\) (e.g., \(0.85 \times x = 68\) or \(\frac{68}{85}\))
M1 for an attempt to divide \(68\) by \(0.85\) (e.g., \(\frac{6800}{85}\))
A1 for \(80\)

First, expand both sets of brackets carefully, paying attention to the signs: \(4(3x - 2) = 12x - 8\) and \(-3(2x - 5) = -6x + 15\). Next, combine the terms: \(12x - 8 - 6x + 15 = (12x - 6x) + (-8 + 15) = 6x + 7\).

M1 for \(12x - 8\) or better
M1 for \(-6x + 15\) or better (with particular attention to the \(+15\))
A1 for \(6x + 7\) (or equivalent)

Multiply both sides of the equation by 4 to eliminate the denominators: \(3x - 1 = 2(x + 2)\). Expand the brackets on the right-hand side: \(3x - 1 = 2x + 4\). Subtract \(2x\) from both sides of the equation: \(x - 1 = 4\). Add 1 to both sides of the equation: \(x = 5\).

M1 for removing the denominators correctly (e.g., \(3x - 1 = 2(x + 2)\) or \(2(3x - 1) = 4(x + 2)\))
M1 for isolating terms in \(x\) on one side and constant terms on the other side (e.g., \(3x - 2x = 4 + 1\))
A1 for \(5\)

Multiply the terms of the first bracket by each term of the second bracket: \((3x - 4)(2x + 5) = 3x(2x) + 3x(5) - 4(2x) - 4(5)\) which simplifies to \(6x^2 + 15x - 8x - 20\). Grouping the like terms gives \(15x - 8x = 7x\). So, the simplified expression is \(6x^2 + 7x - 20\).

M1 for 3 out of 4 terms correct of \(6x^2\), \(+15x\), \(-8x\), \(-20\) in expansion. A1 for middle terms simplified to \(+7x\). A1 for fully correct expression \(6x^2 + 7x - 20\).

The sale price represents 100% - 15% = 85% of the original price. Let P be the original price: 0.85P = 68. Solving for P gives P = \frac{68}{0.85} = \frac{6800}{85}. Simplifying the fraction by dividing numerator and denominator by 17 gives P = \frac{400}{5} = 80. The original price is $80.

M1 for equating $68 to 85% (e.g. 85% = 68 or 0.85P = 68). M1 for a complete division/multiplication method to find the original price (e.g. 68 \div 0.85 or 68 \times \frac{100}{85}). A1 for 80.

First, multiply every term by the common denominator 12 to eliminate the fractions: \(12 \times \frac{2x - 3}{4} + 12 \times \frac{x + 1}{3} = 12 \times 5\). This simplifies to \(3(2x - 3) + 4(x + 1) = 60\). Expanding the brackets: \(6x - 9 + 4x + 4 = 60\). Combining like terms: \(10x - 5 = 60\). Adding 5 to both sides: \(10x = 65\). Dividing by 10: \(x = 6.5\) (or \(\frac{13}{2}\)).

M1 for multiplying by 12 to clear fractions correctly, leading to \(3(2x - 3) + 4(x + 1) = 60\) (allow one sign error). M1 for collecting terms to \(ax = b\) (e.g. \(10x = 65\)). A1 for 6.5 or \(\frac{13}{2}\) or \(6\frac{1}{2}\).

Group the terms in pairs and factorise the common terms from each pair:

\(3ax - 6bx + 2ay - 4by = 3x(a - 2b) + 2y(a - 2b)\)

Now, factor out the common bracket \((a - 2b)\):

\((3x + 2y)(a - 2b)\)

M1 for \(3x(a - 2b)\) or \(2y(a - 2b)\) (or grouping as \(a(3x + 2y) - 2b(3x + 2y)\))
M1 for \(3x(a - 2b) + 2y(a - 2b)\) or \(a(3x + 2y) - 2b(3x + 2y)\)
A1 for \((3x + 2y)(a - 2b)\) or \((a - 2b)(3x + 2y)\)

Let the original price be \(x\).

Since the price is reduced by \(15\%\), the sale price is \(85\%\) of the original price:
\(0.85x = 204\)

Solve for \(x\):
\(x = \frac{204}{0.85} = \frac{20400}{85}\)

Simplify the fraction by dividing the numerator and the denominator by 5:
\(\frac{20400}{85} = \frac{4080}{17}\)

Now, divide 4080 by 17:
\(4080 \div 17 = 240\)

Thus, the original price was \(\$240\).

M1 for equating \(85\%\) to 204 or writing \(0.85x = 204\)
M1 for \(\frac{204}{0.85}\) or \(\frac{204}{85} \times 100\)
A1 for 240

We have the system of equations:
1) \(3x + 2y = 11\)
2) \(4x - y = 11\)

Multiply equation (2) by 2:
\(8x - 2y = 22\)

Add this result to equation (1) to eliminate \(y\):
\((3x + 2y) + (8x - 2y) = 11 + 22\)
\(11x = 33\)
\(x = 3\)

Substitute \(x = 3\) back into equation (2):
\(4(3) - y = 11\)
\(12 - y = 11\)
\(y = 1\)

Therefore, the solution is \(x = 3\) and \(y = 1\).

M1 for a correct method to eliminate one variable (e.g. multiplying equation (2) by 2 and adding, or expressing \(y = 4x - 11\) and substituting)
A1 for \(x = 3\)
A1 for \(y = 1\)

First, expand the brackets: \( 4(2x - 3) - 3(x + 1) = 10 \) becomes \( 8x - 12 - 3x - 3 = 10 \). Next, group the like terms on the left side: \( 5x - 15 = 10 \). Add 15 to both sides of the equation: \( 5x = 25 \). Finally, divide both sides by 5 to find \( x \): \( x = 5 \).

M1 for correct expansion of at least one bracket to get \( 8x - 12 \) or \( -3x - 3 \). M1 for fully simplifying to the form \( ax = b \), e.g., \( 5x = 25 \). A1 for 5.

The sale price of \( \$240 \) represents \( 80\% \) of the original price (\( 100\% - 20\% = 80\% \)). To find the original price \( x \), we can write the equation: \( 0.8x = 240 \). Dividing both sides by \( 0.8 \) gives: \( x = \frac{240}{0.8} = 300 \).

M1 for equating \( 80\% \) to \( 240 \) (e.g., \( 0.8x = 240 \)). M1 for a complete method to calculate the original price, e.g., \( \frac{240}{0.8} \) or \( \frac{240}{80} \times 100 \). A1 for 300.

The volume of a prism is calculated by multiplying the area of its cross-section by its length. The cross-section is a right-angled triangle with perpendicular sides \( 5\text{ cm} \) and \( 8\text{ cm} \). Its area is: \( \frac{1}{2} \times 5 \times 8 = 20\text{ cm}^2 \). The volume is: \( \text{area} \times \text{length} = 20 \times 12 = 240\text{ cm}^3 \).

M1 for finding the area of the triangular cross-section: \( \frac{1}{2} \times 5 \times 8 \) (or 20). M1 for multiplying their cross-sectional area by the length 12. A1 for 240.

Let \( x \) be the original price of the bicycle. The sale price represents a 15% reduction, which is \( 100\% - 15\% = 85\% \) of the original price. Therefore, \( 0.85x = 170 \). To find \( x \), calculate \( x = \frac{170}{0.85} = \frac{17000}{85} = 200 \). The original price is $200.

M1 for translating the word problem into an equation, e.g., \( 0.85x = 170 \) or showing \( 85\% = 170 \)
M1 for a complete method to find the original value, e.g., \( \frac{170}{85} \times 100 \)
A1 for 200

Multiply both sides of the equation by 4 to eliminate the fraction: \( 3x - 1 = 4(5 - x) \). Expand the bracket on the right side: \( 3x - 1 = 20 - 4x \). Rearrange the equation by adding \( 4x \) to both sides: \( 7x - 1 = 20 \). Add 1 to both sides: \( 7x = 21 \). Divide both sides by 7 to find \( x \): \( x = 3 \).

M1 for multiplying both sides by 4 to get \( 3x - 1 = 4(5 - x) \) or \( 3x - 1 = 20 - 4x \)
M1 for isolating the x terms on one side, e.g., \( 7x = 21 \)
A1 for 3

The prism has 5 faces: 2 triangular ends and 3 rectangular sides. First, find the total area of the 2 triangular faces: \( 2 \times \left(\frac{1}{2} \times 3 \text{ cm} \times 4 \text{ cm}\right) = 12 \text{ cm}^2 \). Next, find the area of the 3 rectangular faces: \( 3 \text{ cm} \times 10 \text{ cm} = 30 \text{ cm}^2 \), \( 4 \text{ cm} \times 10 \text{ cm} = 40 \text{ cm}^2 \), and \( 5 \text{ cm} \times 10 \text{ cm} = 50 \text{ cm}^2 \). The total area of the rectangular faces is \( 30 + 40 + 50 = 120 \text{ cm}^2 \). The total surface area of the prism is \( 12 \text{ cm}^2 + 120 \text{ cm}^2 = 132 \text{ cm}^2 \).

M1 for finding the total area of both triangular ends, e.g., \( 2 \times (\frac{1}{2} \times 3 \times 4) = 12 \) (or 6 for one end)
M1 for finding the sum of the areas of the three rectangular faces, e.g., \( 10 \times (3 + 4 + 5) = 120 \) (or at least two correct rectangular areas shown)
A1 for 132

Expand the brackets first:
\(3(2x - 5) = 6x - 15\)
\(-2(x + 1) = -2x - 2\)

Substitute these back into the equation:
\(6x - 15 - 2x - 2 = 7\)

Collect like terms:
\(4x - 17 = 7\)

Add 17 to both sides of the equation:
\(4x = 24\)

Divide both sides by 4:
\(x = 6\)

M1 for correct expansion of at least one bracket (e.g. \(6x - 15\) or \(-2x - 2\))
M1 for collecting like terms to the form \(ax = b\) (e.g. \(4x = 24\) or \(4x - 17 = 7\))
A1 for \(6\) (or \(x = 6\))

A 15% reduction means the sale price is 85% of the original price.
Let the original price be \(P\).
\(0.85 \times P = 221\)

To find \(P\), divide 221 by 0.85:
\(P = \frac{221}{0.85} = \frac{22100}{85}\)

Simplify the fraction by dividing both the numerator and denominator by 5:
\(P = \frac{4420}{17}\)

Divide 4420 by 17:
\(4420 \div 17 = 260\)

So, the original price of the bicycle is $260.

M1 for equating 85% to 221, e.g., \(0.85 \times \text{original} = 221\) or equivalent
M1 for \(221 \div 0.85\) or \(221 \times \frac{100}{85}\) or equivalent fraction calculation
A1 for 260

The formula for the total surface area \(A\) of a closed cuboid is:
\(A = 2(lw + lh + wh)\)

Substitute the given dimensions: length \(l = 5\), width \(w = 4\), and total surface area \(A = 148\):
\(148 = 2(5 \times 4 + 5h + 4h)\)
\(148 = 2(20 + 9h)\)

Divide both sides by 2:
\(74 = 20 + 9h\)

Subtract 20 from both sides:
\(54 = 9h\)

Divide by 9:
\(h = 6\)

So, the height of the cuboid is 6 cm.

M1 for correct substitution into total surface area formula, e.g., \(2(5 \times 4 + 5h + 4h) = 148\) or equivalent
M1 for isolating the variable term, e.g., \(18h = 108\) or \(9h = 54\)
A1 for 6

To make \( u \) the subject, we perform inverse operations on both sides of the formula. First, multiply both sides by 3: \( 3v = 2u + 5 \). Next, subtract 5 from both sides: \( 3v - 5 = 2u \). Finally, divide both sides by 2 to isolate \( u \): \( u = \frac{3v - 5}{2} \).

M1 for multiplying both sides by 3 to get \( 3v = 2u + 5 \)
M1 for subtracting 5 to get \( 3v - 5 = 2u \) (or for dividing by 2 as a correct step)
A1 for \( u = \frac{3v - 5}{2} \) or equivalent

The sale price of $51 represents \( 100\% - 15\% = 85\% \) of the original price. Let \( x \) be the original price. This gives the equation: \( 0.85x = 51 \). To find \( x \), we calculate \( x = \frac{51}{0.85} = \frac{5100}{85} \). Dividing both numerator and denominator by 17 gives \( \frac{300}{5} = 60 \). Thus, the original price was $60.

M1 for equating \( 85\% \) to $51 (e.g. \( 0.85x = 51 \))
M1 for a complete correct division method: \( 51 \div 0.85 \) or \( \frac{51}{85} \times 100 \)
A1 for 60

The total surface area of a closed cylinder is calculated by adding the areas of the two circular ends to the curved surface area. 1. Area of the two ends: \( 2 \times \pi \times r^2 = 2 \times \pi \times 3^2 = 18\pi \text{ cm}^2 \). 2. Curved surface area: \( 2 \times \pi \times r \times h = 2 \times \pi \times 3 \times 7 = 42\pi \text{ cm}^2 \). 3. Total surface area: \( 18\pi + 42\pi = 60\pi \text{ cm}^2 \).

M1 for finding the area of the two circular ends: \( 18\pi \) (or one end as \( 9\pi \))
M1 for finding the curved surface area: \( 42\pi \)
A1 for \( 60\pi \)

Multiply both sides of the equation by 3 to eliminate the fraction: \(2x + 5 = 3(x - 1)\). Expand the bracket on the right-hand side: \(2x + 5 = 3x - 3\). Subtract \(2x\) from both sides to get: \(5 = x - 3\). Add 3 to both sides to find the value of \(x\): \(x = 8\).

M1 for multiplying both sides by 3: \(2x + 5 = 3(x - 1)\). M1 for expanding the brackets: \(3x - 3\) (or for isolating terms in \(x\) on one side and constant terms on the other). A1 for \(8\) (or \(x = 8\)).

First, find the area of the triangular cross-section using the formula \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\): \(\text{Area} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2\). Next, calculate the volume of the prism by multiplying this area by the length of the prism: \(\text{Volume} = 24 \times 15 = 360\text{ cm}^3\).

M1 for finding the area of the cross-section: \(\frac{1}{2} \times 6 \times 8\) or 24. M1 for multiplying their cross-sectional area by 15: \(\text{their } 24 \times 15\). A1 for 360.

To make \(x\) the subject, we want to isolate \(x\) on one side of the equation.

1. Multiply both sides of the equation by \(2x + 7\) to clear the fraction:
\[y(2x + 7) = 3 - 5x\]

2. Expand the left side:
\[2xy + 7y = 3 - 5x\]

3. Collect all terms containing \(x\) on one side and terms without \(x\) on the other side:
\[2xy + 5x = 3 - 7y\]

4. Factorise \(x\) from the terms on the left side:
\[x(2y + 5) = 3 - 7y\]

5. Divide both sides by \(2y + 5\) to solve for \(x\):
\[x = \frac{3 - 7y}{2y + 5}\]

M1 for multiplying both sides by \(2x + 7\) to get \(y(2x + 7) = 3 - 5x\)
M1 for expanding and collecting terms with \(x\) on one side: \(2xy + 5x = 3 - 7y\) (allow one sign error)
M1 for factorising \(x\) out: \(x(2y + 5) = 3 - 7y\)
A1 for the correct final expression \(x = \frac{3 - 7y}{2y + 5}\) or equivalent, e.g., \(x = \frac{7y - 3}{-2y - 5}\)

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

1. After a 15% reduction, the price is:
\[0.85P\]

2. After a further 10% reduction on this new price, the final price is:
\[0.90 \times 0.85P = 0.765P\]

3. We are given that this final price is $153:
\[0.765P = 153\]

4. Solve for \(P\):
\[P = \frac{153}{0.765} = \frac{153000}{765}\]

Simplify the fraction by dividing the numerator and denominator by 9:
\[153000 \div 9 = 17000\]
\[765 \div 9 = 85\]
\[P = \frac{17000}{85}\]

Now, divide by 17 (since \(17 \times 5 = 85\)):
\[P = \frac{1000}{5} = 200\]

Thus, the original price of the coat was $200.

M1 for identifying the multiplier for a 15% reduction as \(0.85\) (or \(85\%\))
M1 for calculating the combined multiplier \(0.85 \times 0.90 = 0.765\) (or \(76.5\%\))
M1 for setting up the equation \(0.765P = 153\) or equivalent calculation \(\frac{153}{0.90} \div 0.85\)
A1 for the correct answer 200

1. Find the volume of the cone, \(V_{\text{cone}}\), with base radius \(R = 3r\) and height \(H = 4r\):
\[V_{\text{cone}} = \frac{1}{3}\pi (3r)^2 (4r) = \frac{1}{3}\pi (9r^2)(4r) = 12\pi r^3\]

2. Find the volume of the hemisphere, \(V_{\text{hemi}}\), with radius \(R = 3r\):
\[V_{\text{hemi}} = \frac{1}{2} \times \left(\frac{4}{3}\pi (3r)^3\right) = \frac{2}{3}\pi (27r^3) = 18\pi r^3\]

3. Calculate the total volume by adding both volumes:
\[V_{\text{total}} = 12\pi r^3 + 18\pi r^3 = 30\pi r^3\]

4. Set the total volume equal to the given value of \(240\pi\text{ cm}^3\):
\[30\pi r^3 = 240\pi\]

5. Divide both sides by \(30\pi\):
\[r^3 = 8\]

6. Solve for \(r\):
\[r = \sqrt[3]{8} = 2\]

M1 for writing a correct expression for the volume of the cone in terms of \(r\): \(\frac{1}{3}\pi (3r)^2(4r)\) or \(12\pi r^3\)
M1 for writing a correct expression for the volume of the hemisphere in terms of \(r\): \(\frac{2}{3}\pi (3r)^3\) or \(18\pi r^3\)
M1 for equating the total volume expression to the given total: \(30\pi r^3 = 240\pi\) (or equivalent)
A1 for solving to get \(r = 2\)

First, factorise each expression completely: 1) \(2x^2 - 7x + 3 = (2x - 1)(x - 3)\), 2) \(x^2 - 9 = (x - 3)(x + 3)\), 3) \(x^2 + 3x = x(x + 3)\), and 4) \(4x^2 - 1 = (2x - 1)(2x + 1)\). Substituting these back into the original expression, we get \(\frac{(2x - 1)(x - 3)}{(x - 3)(x + 3)} \times \frac{x(x + 3)}{(2x - 1)(2x + 1)}\). Cancelling the common factors \((x - 3)\), \((x + 3)\), and \((2x - 1)\) from the numerator and the denominator yields the simplified fraction \(\frac{x}{2x + 1}\).

M1 for factorising \(2x^2 - 7x + 3\) to \((2x-1)(x-3)\). M1 for factorising \(x^2-9\) to \((x-3)(x+3)\) or \(4x^2-1\) to \((2x-1)(2x+1)\). M1 for factorising \(x^2+3x\) to \(x(x+3)\) and showing multiplication. A1 for correct final answer \(\frac{x}{2x+1}\).

Let \(F\) be the original number of fiction books. Since it increases by 15% to 575, we have \(1.15 \times F = 575\), which gives \(F = \frac{575}{1.15} = 500\). The new number of fiction books is 575. Using the ratio of fiction to non-fiction books of 5 : 4, the number of non-fiction books is \(\frac{575}{5} \times 4 = 115 \times 4 = 460\). Since the non-fiction book count is unchanged, the total number of books before the increase is \(500 + 460 = 960\).

M1 for setting up reverse percentage: \(575 \div 1.15\) or equivalent. A1 for finding original fiction books is 500. M1 for \(\frac{575}{5} \times 4\) to find non-fiction books (460). A1 for final answer of 960.

The volume of a sphere is given by \(V_{\text{sphere}} = \frac{4}{3}\pi R^3\). The volume of a cone is given by \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\). Substituting the height of the cone \(h = 9R\) gives \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (9R) = 3\pi r^2 R\). Since the volumes are equal, \(\frac{4}{3}\pi R^3 = 3\pi r^2 R\). Dividing both sides by \(\pi R\) gives \(\frac{4}{3}R^2 = 3r^2\). Dividing by 3 gives \(r^2 = \frac{4}{9}R^2\). Taking the positive square root of both sides gives \(r = \frac{2}{3}R\).

M1 for sphere volume formula \(\frac{4}{3}\pi R^3\). M1 for cone volume with height substituted: \(\frac{1}{3}\pi r^2 (9R)\). M1 for equating volumes and solving for \(r^2\) to get \(r^2 = \frac{4}{9}R^2\). A1 for correct final answer \(r = \frac{2}{3}R\) or equivalent.

Let \(V_s\) be the volume of the sphere: \(V_s = \frac{4}{3}\pi r^3\). Let \(V_c\) be the volume of the cone with base radius \(R = 2r\) and height \(h\): \(V_c = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (2r)^2 h = \frac{4}{3}\pi r^2 h\). Since the sphere is recast into the cone, their volumes are equal: \(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^2 h\), which simplifies to \(h = r\). The slant height \(l\) of the cone is given by \(l = \sqrt{R^2 + h^2} = \sqrt{(2r)^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}\). The total surface area \(A\) of the cone is the sum of the base area and the curved surface area: \(A = \pi R^2 + \pi R l = \pi (2r)^2 + \pi (2r)(r\sqrt{5}) = 4\pi r^2 + 2\sqrt{5}\pi r^2 = \pi r^2(4 + 2\sqrt{5})\). Comparing this to \(\pi r^2 (a + b\sqrt{5})\), we get \(a = 4\) and \(b = 2\).

M1 for equating volume of sphere and cone to find \(h = r\). M1 for finding slant height \(l = r\sqrt{5}\) using Pythagoras' theorem. M1 for total surface area formula of cone: \(\pi R^2 + \pi R l\) with substitution of \(R = 2r\). A1 for final correct values of \(a = 4\) and \(b = 2\).

First, factorise each expression: 1) \(2x^2 - 11x + 12 = 2x^2 - 8x - 3x + 12 = 2x(x - 4) - 3(x - 4) = (2x - 3)(x - 4)\). 2) \(3x^2 - 12x = 3x(x - 4)\). 3) \(4x^2 - 9 = (2x - 3)(2x + 3)\). 4) \(6x + 9 = 3(2x + 3)\). Now, write the division as multiplication by the reciprocal: \(\frac{(2x - 3)(x - 4)}{3x(x - 4)} \times \frac{3(2x + 3)}{(2x - 3)(2x + 3)}\). Simplify by cancelling common terms: the \((x - 4)\) terms cancel, the \((2x - 3)\) terms cancel, the \((2x + 3)\) terms cancel, and the factor of 3 cancels. This leaves \(\frac{1}{x}\).

M1 for factorising \(2x^2 - 11x + 12\) to \((2x - 3)(x - 4)\). M1 for factorising the other terms: \(3x(x - 4)\), \((2x - 3)(2x + 3)\), and \(3(2x + 3)\). M1 for inverting the second fraction and multiplying. A1 for correct simplified final answer \(\frac{1}{x}\).

Let the original price of the coat be \(P\). After a \(25\%\) increase, the price becomes \(1.25P\) or \(\frac{5}{4}P\). After decreasing this new price by \(x\%\), the final price is \(1.25P \times (1 - \frac{x}{100})\). We are given that this final price is \(10\%\) less than the original price \(P\), which is \(0.9P\) or \(\frac{9}{10}P\). So we set up the equation: \(\frac{5}{4}P \left(1 - \frac{x}{100}\right) = \frac{9}{10}P\). Divide both sides by \(P\): \(\frac{5}{4} \left(1 - \frac{x}{100}\right) = \frac{9}{10}\). Multiply both sides by \(\frac{4}{5}\): \(1 - \frac{x}{100} = \frac{9}{10} \times \frac{4}{5} = \frac{36}{50} = \frac{72}{100}\). Thus, \(1 - \frac{x}{100} = 0.72\), which gives \(\frac{x}{100} = 0.28\), so \(x = 28\).

M1 for representing the price after a 25% increase as \(1.25P\) or equivalent. M1 for setting up the equation \(1.25P \times (1 - \frac{x}{100}) = 0.9P\) or equivalent. M1 for solving the equation to find \(1 - \frac{x}{100} = 0.72\) or equivalent. A1 for \(x = 28\).

Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator: \(4x^2 - 1 = (2x + 1)(2x - 1)\). Divide both numerator and denominator by the common factor \((2x + 1)\): \(\frac{(2x+1)(x-3)}{(2x+1)(2x-1)} = \frac{x-3}{2x-1}\).

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

Let the original price paid by the shopkeeper be \(x\). The marked price is \(1.30x\). The sale price is a \(20\\%\) reduction on the marked price, which is \(0.80 \times 1.30x = 1.04x\). Since the sale price is \(\\$156\), we have \(1.04x = 156\). Solving this equation: \(x = \frac{156}{1.04} = \frac{15600}{104} = 150\). Therefore, the original price paid was \(\\$150\).

M1 for expressing the marked price as \(1.3x\) or similar; M1 for setting up the equation \(0.8 \times 1.3x = 156\) or \(1.04x = 156\); M1 for a correct non-calculator method to solve for \(x\), e.g., \(x = \frac{156}{1.04}\); A1 for 150.

The volume of a hemisphere of radius \(R\) is \(\frac{2}{3}\pi R^3\). Substituting \(R = 3r\): Volume of hemisphere = \(\frac{2}{3}\pi (3r)^3 = \)\\frac{2}{3}\\pi (27r^3) = 18\\pi r^3\). The volume of a cone of radius \(R\) and height \(h\) is \(\frac{1}{3}\pi R^2 h\). Substituting \(R = 3r\): Volume of cone = \(\frac{1}{3}\pi (3r)^2 h = \)\\frac{1}{3}\\pi (9r^2) h = 3\\pi r^2 h\). Since the volumes are equal: \(18\pi r^3 = 3\pi r^2 h\). Dividing both sides by \(3\pi r^2\) gives \(h = 6r\).

M1 for volume of hemisphere: \(\frac{2}{3}\pi (3r)^3\) or \(18\pi r^3\); M1 for volume of cone: \(\frac{1}{3}\pi (3r)^2 h\) or \(3\pi r^2 h\); M1 for equating their volumes and attempting to isolate \(h\); A1 for \(6r\) or \(h = 6r\).

The volume of a cone is given by: \(V_{\text{cone}} = \frac{1}{3}\pi (\text{radius})^2 \times \text{height}\). Substituting the given values: \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (6r) = 2\pi r^3\). The volume of a hemisphere of radius \(R\) is: \(V_{\text{hemisphere}} = \frac{2}{3}\pi R^3\). Since the volume remains the same: \(\frac{2}{3}\pi R^3 = 2\pi r^3\). Divide both sides by \(2\pi\): \(\frac{1}{3} R^3 = r^3\). Multiply both sides by 3: \(R^3 = 3r^3\). Take the cube root of both sides: \(R = \sqrt[3]{3} r\) or \(R = r\sqrt[3]{3}\).

M1 for writing the correct volume of the cone: \(\frac{1}{3}\pi r^2 (6r)\) or \(2\pi r^3\). M1 for equating the volume of the hemisphere to their volume of the cone: \(\frac{2}{3}\pi R^3 = 2\pi r^3\). M1 for simplifying the equation to \(R^3 = 3r^3\). A1 for the correct final simplified expression: \(r\sqrt[3]{3}\) or \(\sqrt[3]{3}r\).

First, multiply both sides by \((2x + 7)\): \(y(2x + 7) = 5 - 3x\). Expand the brackets: \(2xy + 7y = 5 - 3x\). Rearrange to get all terms involving \(x\) on one side and other terms on the other side: \(2xy + 3x = 5 - 7y\). Factorise \(x\) from the left-hand side: \(x(2y + 3) = 5 - 7y\). Divide both sides by \((2y + 3)\) to isolate \(x\): \(x = \frac{5 - 7y}{2y + 3}\).

M1 for correctly multiplying both sides by \(2x + 7\): \(y(2x + 7) = 5 - 3x\). M1 for expanding and collecting terms with \(x\) on one side: \(2xy + 3x = 5 - 7y\). M1 for factorising out \(x\): \(x(2y + 3) = 5 - 7y\). A1 for the final answer: \(x = \frac{5 - 7y}{2y + 3}\) or \(x = \frac{7y - 5}{-2y - 3}\).

Let \(P\) be the original price of the laptop. After a 20% reduction, the price is: \(P \times (1 - 0.20) = 0.8P\). After a further 10% reduction, the price is: \(0.8P \times (1 - 0.10) = 0.8P \times 0.9 = 0.72P\). We are given that the final price is $324, so: \(0.72P = 324\). To find \(P\), divide 324 by 0.72: \(P = \frac{32400}{72}\). Simplifying the fraction: \(P = \frac{32400 \div 9}{72 \div 9} = \frac{3600}{8} = 450\). Thus, the original price was $450. Alternative Method: Working backwards: Before the second 10% reduction, the price was: \(324 \div 0.9 = 360\). Before the first 20% reduction, the price was: \(360 \div 0.8 = 450\).

M1 for calculating the price before the second reduction: \(324 \div 0.9 = 360\) (or expressing \(0.8 \times 0.9 = 0.72\) as the combined multiplier). M1 for setting up the equation for the first reduction: \(P \times 0.8 = 360\) (or \(0.72P = 324\)). M1 for calculating \(360 \div 0.8\) or \(324 \div 0.72\). A1 for 450.

The volume of a sphere of radius \(r\) is \(V = \frac{4}{3}\pi r^3\). The volume of a cylinder of radius \(R\) and height \(h = 3r\) is \(V = \pi R^2 (3r) = 3\pi R^2 r\). Equating the volumes: \(\frac{4}{3}\pi r^3 = 3\pi R^2 r\). Dividing both sides by \(\pi r\) (where \(r > 0\)) gives \(\frac{4}{3}r^2 = 3R^2\). Dividing both sides by \(3\) gives \(R^2 = \frac{4}{9}r^2\). Taking the square root of both sides gives \(R = \frac{2}{3}r\).

M1 for \(\frac{4}{3}\pi r^3\) seen. M1 for \(\pi R^2(3r)\) or \(3\pi R^2 r\) seen. M1 for simplifying to \(R^2 = \frac{4}{9}r^2\) or \(R = \sqrt{\frac{4}{9}r^2}\). A1 for \(R = \frac{2}{3}r\) or equivalent.

Let \(x\) be the original price of the coat. A \(20\%\) reduction followed by a \(15\%\) reduction can be represented as: \(x \times 0.80 \times 0.85 = 136\). First, calculate the combined multiplier: \(0.80 \times 0.85 = 0.68\). This gives the equation: \(0.68x = 136\). Solving for \(x\): \(x = \frac{136}{0.68} = \frac{13600}{68} = 200\). Thus, the original price of the coat was \(\$200\).

M1 for setting up the equation \(x \times 0.80 \times 0.85 = 136\) or showing two-step reverse calculation: \(136 \div 0.85\). M1 for finding intermediate price of \(\$160\) or combined multiplier of \(0.68\). M1 for \(136 \div 0.68\) or \(160 \div 0.80\). A1 for correct answer 200.

First, factorize the denominator of the second term: \(x^2 - 3x = x(x-3)\). The expression becomes: \(\frac{x}{x-3} - \frac{9}{x(x-3)} - 1\). Express the first two terms over the common denominator \(x(x-3)\): \(\frac{x^2 - 9}{x(x-3)} - 1\). Factorize the numerator using the difference of two squares: \(x^2 - 9 = (x-3)(x+3)\). Substitute this back to get: \(\frac{(x-3)(x+3)}{x(x-3)} - 1\). Simplify the fraction by cancelling the common factor \((x-3)\): \(\frac{x+3}{x} - 1\). Finally, subtract 1: \(\frac{x+3}{x} - \frac{x}{x} = \frac{3}{x}\).

M1 for factorizing denominator to \(x(x-3)\). M1 for combining first two terms to \(\frac{x^2 - 9}{x(x-3)}\) or equivalent. M1 for simplifying the fraction to \(\frac{x+3}{x}\). A1 for final answer \(\frac{3}{x}\).

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 \(1\). This gives \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator, \(4x^2 - 1\), using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Now, write the fraction with the factorised terms and cancel the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

M1.5 for factorising the numerator correctly to \((2x + 1)(x - 3)\). M1.5 for factorising the denominator correctly to \((2x - 1)(2x + 1)\). A1.34 for the final simplified fraction \(\frac{x - 3}{2x - 1}\).

Let the original price be \(x\). A 15% reduction means the sale price is 85% of the original price. This gives the equation \(0.85x = 357\), which can be written as \(\frac{85}{100}x = 357\). Solving for \(x\): \(x = \frac{357 \times 100}{85}\). Simplifying by dividing both 357 and 85 by 17: \(357 \div 17 = 21\) and \(85 \div 17 = 5\). This simplifies the calculation to \(x = \frac{21 \times 100}{5} = 21 \times 20 = 420\). Therefore, the original price was $420.

M2 for setting up a correct equation or method: \(357 \div 0.85\) or \(\frac{357}{85} \times 100\) (or M1 for recognizing that 85% corresponds to $357). A2.34 for the correct final answer of 420.

First, calculate the volume of the cylinder using the formula \(V = \pi r^2 h\): \(V = \pi \times 4^2 \times 18 = 288\pi\text{ cm}^3\). Since the metal is recast into a sphere, the volume of the sphere is equal to the volume of the cylinder: \(\frac{4}{3}\pi R^3 = 288\pi\). Divide both sides by \(\pi\) to get \(\frac{4}{3}R^3 = 288\). Multiply both sides by 3 to get \(4R^3 = 864\). Divide by 4 to get \(R^3 = 216\). Taking the cube root of both sides gives \(R = \sqrt[3]{216} = 6\).

M1.5 for a correct expression for the cylinder's volume: \(\pi \times 4^2 \times 18\) or \(288\pi\). M1.5 for equating the sphere's volume formula to their cylinder's volume and simplifying to \(R^3 = 216\). A1.34 for the final correct value of 6.

First, factorise 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 \(1\). Writing the middle term as \(-6x + x\), we get \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Second, factorise the denominator, \(4x^2 - 1\). This is a difference of two squares: \((2x)^2 - 1^2 = (2x - 1)(2x + 1)\). Finally, substitute these back into the fraction and cancel the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

M2 for factorising the numerator to \((2x + 1)(x - 3)\) (or M1 for partial factorisation). M1.5 for factorising the denominator to \((2x - 1)(2x + 1)\). A0.84 for the final simplified fraction \(\frac{x - 3}{2x - 1}\).

Let \(P\) be the original price of the bicycle. The price after a 15% increase is \(1.15P\). This price is then reduced by 20%, which means multiplying it by \(0.80\). The final sale price is \(1.15P \times 0.80 = 0.92P\). We are given that this final price is $184, so we write the equation: \(0.92P = 184\). Solving for \(P\) gives \(P = \frac{184}{0.92} = \frac{18400}{92} = 200\). Thus, the original price was $200.

M2 for setting up the relation \(P \times 1.15 \times 0.80 = 184\) or finding the combined multiplier of \(0.92\). M1.5 for the calculation \(184 \div 0.92\) or equivalent. A0.84 for the correct final answer of 200.

First, expand each bracket separately: \(5(3x - 2) = 15x - 10\) and \(-2(4x - 7) = -8x + 14\). Next, group and combine the like terms: \(15x - 8x - 10 + 14 = 7x + 4\).

M1 for correct expansion of at least one bracket (e.g. \(15x - 10\) or \(-8x + 14\)). A1.66 for the final simplified expression \(7x + 4\).

Multiply both sides of the equation by 5 to clear the fraction: \(2x + 7 = 3 \times 5\) which simplifies to \(2x + 7 = 15\). Next, subtract 7 from both sides: \(2x = 15 - 7\) which simplifies to \(2x = 8\). Finally, divide both sides by 2: \(x = \frac{8}{2} = 4\).

M1 for multiplying both sides by 5 to get \(2x + 7 = 15\). M1 for isolating the x term to get \(2x = 8\). A0.66 for the final answer \(4\).

A reduction of 18% means the sale price is 100% - 18% = 82% of the original price. We can set up the equation: \(0.82 \times \text{Original Price} = 246\). Dividing both sides by 0.82 gives: \(\text{Original Price} = \frac{246}{0.82} = 300\).

M1 for recognizing that $246 corresponds to 82% of the original price (e.g. dividing 246 by 0.82 or setting up a correct ratio). A1.66 for the correct final answer of 300 (accept $300).

Expand the first bracket: \(3 \times 2x = 6x\) and \(3 \times -5 = -15\). Expand the second bracket (be careful with the negative sign): \(-2 \times x = -2x\) and \(-2 \times -4 = +8\). Combine the terms: \(6x - 15 - 2x + 8 = 6x - 2x - 15 + 8 = 4x - 7\).

M1 for expanding at least one bracket correctly to get \(6x - 15\) or \(-2x + 8\). A1 for correct final simplified expression \(4x - 7\).

Calculate the sales tax: \(15\%\) of \(\$480 = 0.15 \times 480 = 72\). Add the tax to the original price: \(480 + 72 = 552\). Alternatively, calculate \(480 \times 1.15 = 552\).

M1 for \(480 \times 1.15\) or \(480 + 0.15 \times 480\). A1 for \(552\).

Multiply both sides of the equation by 4: \(3x - 5 = 7 \times 4\), which gives \(3x - 5 = 28\). Add 5 to both sides: \(3x = 28 + 5\), which gives \(3x = 33\). Divide both sides by 3: \(x = \frac{33}{3} = 11\).

M1 for multiplying both sides by 4 to get \(3x - 5 = 28\). A1 for \(11\).

First, expand each bracket individually:
\(5(2x - 3) = 10x - 15\)
\(-3(x - 4) = -3x + 12\)

Now, collect and simplify the like terms:
\(10x - 3x - 15 + 12 = 7x - 3\)

M1 for \(10x - 15\) or \(-3x + 12\) seen
A1 for \(7x - 3\)

Use the formula for the volume of a cylinder:
\(V = \pi r^2 h\)

Substitute the given values, \(r = 4\text{ cm}\) and \(h = 9\text{ cm}\):
\(V = \pi \times 4^2 \times 9\)
\(V = 144\pi \approx 452.3893...\text{ cm}^3\)

Rounding to 1 decimal place gives \(452.4\text{ cm}^3\).

M1 for substituting correctly into the volume formula: \(\pi \times 4^2 \times 9\)
A1 for \(452.4\) (or answers in range \(452.3\) to \(452.4\))

To find the selling price, calculate the multiplier for a \(28\%\) increase, which is \(1 + 0.28 = 1.28\).

Selling Price = \(150 \times 1.28 = 192\).

Alternatively, calculate the profit first:
Profit = \(0.28 \times 150 = 42\).
Selling Price = \(150 + 42 = 192\).

M1 for \(150 \times 1.28\) or \(150 + (0.28 \times 150)\) or finding profit of \(42\)
A1 for \(192\)

We use the compound interest formula: \( A = P \left(1 + \frac{r}{100}\right)^n \). Here, \( P = 800 \), \( r = 2.5 \), and \( n = 3 \). This gives: \( A = 800 \times (1 + 0.025)^3 = 800 \times 1.025^3 \). Calculating the value: \( A = 800 \times 1.076890625 = 861.5125 \). Rounding to the nearest cent, we get $861.51.

M1 for \( 800 \times 1.025^3 \) or for finding the interest year-by-year. M1 for \( 861.5125 \) seen. A1 for \( 861.51 \).

First, expand the brackets: \( 12x - 4 - 2x - 10 = 14 \). Next, group like terms together: \( 10x - 14 = 14 \). Add 14 to both sides of the equation: \( 10x = 28 \). Divide both sides by 10: \( x = 2.8 \).

M1 for correct expansion of at least one bracket (e.g., \( 12x - 4 \) or \( -2x - 10 \)). M1 for simplifying to a form \( ax = b \) (e.g., \( 10x = 28 \)). A1 for \( 2.8 \) or equivalent fraction.

Let \( h \) represent the height of the cuboid. The total surface area of a cuboid is calculated using: \( A = 2(lw + lh + wh) \). Substituting the given values: \( 184 = 2(8 \times 5 + 8 \times h + 5 \times h) \). Divide both sides by 2: \( 92 = 40 + 8h + 5h \). Simplify: \( 92 = 40 + 13h \). Subtract 40 from both sides: \( 52 = 13h \). Divide by 13: \( h = 4 \).

M1 for setting up a correct equation, e.g., \( 2(8 \times 5 + 8h + 5h) = 184 \). M1 for simplifying to \( 13h = 52 \) or equivalent. A1 for \( 4 \).

First, expand the brackets by multiplying each term in the first bracket by each term in the second bracket: \((2x \times x) + (2x \times 5) + (-3 \times x) + (-3 \times 5)\). This gives: \(2x^2 + 10x - 3x - 15\). Grouping the like terms together, we get: \(2x^2 + 7x - 15\).

M1 for at least 3 correct terms of the expansion (e.g., \(2x^2\), \(10x\), \(-3x\), \(-15\)). A1.66 for the fully simplified correct final answer \(2x^2 + 7x - 15\).

Let the original price of the coat be \(P\). The sale price represents 100% - 15% = 85% of the original price. This can be written as: \(0.85 \times P = 221\). To find the original price \(P\), we divide the sale price by 0.85: \(P = 221 / 0.85 = 260\).

M1 for \(221 / 0.85\) or equivalent (e.g., \(221 / 85 \times 100\)). A1.66 for the correct final answer 260.

First, calculate the cross-sectional area of the trapezium using the formula Area = \(\frac{a + b}{2} \times h\). Area = \(\frac{18 + 25}{2} \times 12 = 21.5 \times 12 = 258\ \text{cm}^2\). Next, multiply the cross-sectional area by the length of the prism to find the volume: Volume = Area \(\times\) Length = \(258 \times 80 = 20640\ \text{cm}^3\).

M1 for the correct method to find the area of the trapezium cross-section: \(\frac{18 + 25}{2} \times 12\). M1 for multiplying their cross-sectional area by 80. A0.66 for the correct final answer 20640.

Let the original price be \(x\).
Since the price is reduced by \(15\%\), the sale price is \(85\%\) of the original price.
\(0.85x = 272\)
\(x = \frac{272}{0.85}\)
\(x = 320\)

Therefore, the original price was \(\$320\).

M1 for establishing the relationship \(0.85x = 272\) or doing the division \(272 \div 0.85\)
A1 for 320

Find the highest common factor (HCF) of the two terms \(15a^2b\) and \(10ab^2\).
- The HCF of the numerical coefficients \(15\) and \(10\) is \(5\).
- The HCF of the variable parts \(a^2b\) and \(ab^2\) is \(ab\).
Therefore, the HCF is \(5ab\).
Divide both terms by the HCF:
\(15a^2b \div 5ab = 3a\)
\(-10ab^2 \div 5ab = -2b\)
So, the fully factorised expression is:
\(5ab(3a - 2b)\)

M1 for a partial factorisation, such as \(5(3a^2b - 2ab^2)\), \(ab(15a - 10b)\), or \(a(15ab - 10b^2)\)
A1 for the fully correct factorisation: \(5ab(3a - 2b)\)

First, expand the brackets:
\(12x - 8 - 2x - 10 = 12\)

Next, simplify the left side by combining like terms:
\(10x - 18 = 12\)

Add \(18\) to both sides of the equation:
\(10x = 12 + 18\)
\(10x = 30\)

Divide both sides by \(10\):
\(x = 3\)

M1 for correct expansion of at least one bracket, e.g., \(12x - 8\) or \(-2x - 10\)
M1 for isolating the \(x\) terms and constant terms, leading to \(10x = 30\) (or equivalent)
A1 for 3

Let the price of the cake before tax be \(x\). The price including tax is 108% of \(x\). Therefore, \(1.08 \times x = 378\). Solving for \(x\): \(x = \frac{378}{1.08} = 350\). Thus, the price of the cake before tax is $350.

M1 for setting up the relation \(1.08 \times \text{price} = 378\) or \(378 \div 1.08\). A1 for 350.

First, expand each bracket: \(4(2x - 3) = 8x - 12\) and \(-3(x - 5) = -3x + 15\). Next, combine the like terms: \(8x - 3x - 12 + 15 = 5x + 3\).

M1 for expansion yielding \(8x - 12\) or \(-3x + 15\). M1 for collecting like terms. A1 for \(5x + 3\).

First, expand the bracket on the left side of the equation: \(5x - 15 = 2x + 9\). Next, isolate the terms with \(x\) on one side by subtracting \(2x\) from both sides: \(3x - 15 = 9\). Then, add 15 to both sides: \(3x = 24\). Finally, divide by 3: \(x = 8\).

M1 for correct expansion of the bracket to get \(5x - 15\). M1 for isolating terms to get \(3x = 24\) or equivalent. A1 for 8.

First, expand each bracket:
\(4(2a - 3b) = 8a - 12b\)
\(-3(a - 5b) = -3a + 15b\)

Next, collect and combine like terms:
\(8a - 3a - 12b + 15b = 5a + 3b\).

M1 for expansion of one bracket correctly: \(8a - 12b\) or \(-3a + 15b\) (or \(3a - 15b\) if subtracted correctly later).
A1 for \(5a + 3b\) final answer.

Find the profit made:
\(168 - 120 = 48\)

Calculate the percentage profit on the cost price:
\(\frac{48}{120} \times 100 = 40\\%\).

M1 for finding the profit \(168 - 120 = 48\) or for writing \(\frac{168}{120} \times 100\) or \(\frac{168 - 120}{120}\).
A1 for \(40\) or \(40\\%\).

The total surface area \(A\) of a closed cylinder is given by:
\(A = 2\pi r^2 + 2\pi r h\)

Substitute \(r = 3\) and \(h = 8\):
\(A = 2\pi(3)^2 + 2\pi(3)(8)\)
\(A = 18\pi + 48\pi\)
\(A = 66\pi \approx 207.345\text{ cm}^2\)

To 1 decimal place, the area is \(207.3\text{ cm}^2\).

M1 for a correct substitution into a valid formula, e.g., \(2\pi \times 3^2 + 2\pi \times 3 \times 8\) or \(18\pi\) or \(48\pi\) seen.
A1 for \(207.3\) (allow \(207.34...\) to \(207.4\) if a less accurate value of \(\pi\) is used, but \(207.3\) is the standard correct value).

First, expand each bracket: \(4(2x - 3) = 8x - 12\) and \(-3(x - 5) = -3x + 15\). Next, combine the terms by grouping: \(8x - 3x - 12 + 15 = 5x + 3\).

M1 for correct expansion of either term: \(8x - 12\) or \(-3x + 15\) (or \(3x - 15\) after subtraction sign).
M1 for collecting like terms: \(8x - 3x\) or \(-12 + 15\) based on their expansion.
A1 for final answer: \(5x + 3\).

First, calculate the profit made on the sale: \(243 - 180 = 63\) dollars. Next, calculate this profit as a percentage of the original cost price: \(\frac{63}{180} \times 100 = 35\%\).

M1 for finding the profit: \(243 - 180 = 63\).
M1 for \(\frac{\text{their profit}}{180} \times 100\).
A1 for \(35\).

First, find the area of the triangular cross-section: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2\). Then, calculate the volume of the prism by multiplying the cross-sectional area by the length of the prism: \(\text{Volume} = 24 \times 15 = 360\text{ cm}^3\).

M1 for finding the area of the triangular cross-section: \(\frac{1}{2} \times 6 \times 8\) or \(24\).
M1 for \(\text{their area} \times 15\).
A1 for \(360\).

Expand the first bracket: \(6x(3x - 5) = 18x^2 - 30x\). Expand the second bracket: \(-4(2x^2 - 7x) = -8x^2 + 28x\). Combine like terms: \(18x^2 - 8x^2 - 30x + 28x = 10x^2 - 2x\).

M1 for \(18x^2 - 30x\) or better. M1 for \(-8x^2 + 28x\) or better. A1 for \(10x^2 - 2x\) or equivalent fully simplified form.

Let the original price be \(x\). A reduction of 12% means the sale price is 88% of the original price. So, \(0.88x = 264\). Solving for \(x\) gives \(x = 264 / 0.88 = 300\).

M1 for representing the relationship, e.g. 88% = 264. M1 for \(264 / 0.88\). A1 for 300.

The total surface area of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi r h\). Substituting \(r = 4\) and \(h = 15\) gives: \(A = 2\pi(4)^2 + 2\pi(4)(15) = 32\pi + 120\pi = 152\pi\). Using the calculator, \(152\pi \approx 477.522\). Correct to 1 decimal place, this is 477.5.

M1 for substituting correctly into curved surface area or area of two circular ends, e.g. \(2 \times \pi \times 4 \times 15\) or \(2 \times \pi \times 4^2\). M1 for adding both parts: \(2\pi(4)^2 + 2\pi(4)(15)\). A1 for 477.5 (accept answers in range 477.5 to 477.6).

1. Calculate the volume of the cylinder: \(V_{\text{cylinder}} = \pi r^2 h = \pi \times 6^2 \times 15 = 540\pi \approx 1696.46\text{ cm}^3\). 2. Find the remaining volume after \(8\%\) loss: \(V_{\text{remaining}} = 540\pi \times 0.92 = 496.8\pi \approx 1560.74\text{ cm}^3\). 3. Calculate the volume of one sphere: \(V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times 1.5^3 = 4.5\pi \approx 14.137\text{ cm}^3\). 4. Divide the remaining volume by the volume of one sphere: \(N = \frac{496.8\pi}{4.5\pi} = 110.4\). Since we want the number of complete spheres, we round down to 110.

M1 for volume of cylinder: pi * 6^2 * 15. M1 for multiplying by 0.92. M1 for volume of sphere: 4/3 * pi * 1.5^3. M1 for dividing their remaining volume by their sphere volume. A1 for 110.

For Account A: The interest earned is \(I = 3500 \times 0.042 \times 5 = \$735\). The total amount is \(3500 + 735 = \$4235\). For Account B: The total amount is \(3500 \times (1 + 0.04)^5 = 3500 \times 1.04^5 = \$4258.28512\). The difference is \(4258.28512 - 4235 = 23.28512\). To the nearest cent, this is \(\$23.29\).

M1 for simple interest calculation of 735 or total 4235. M1 for compound interest formula 3500 * (1.04)^5. A1 for 4258.29 or 4258.28. M1 for subtracting their two totals. A1 for 23.29.

First, set up the area equation: \((2x + 3)(x + 4) = 117\). Expanding the brackets gives \(2x^2 + 8x + 3x + 12 = 117\), which simplifies to \(2x^2 + 11x - 105 = 0\). Factoring the quadratic expression gives \((2x + 21)(x - 5) = 0\). Since \(x\) must be positive, we find \(x = 5\). Using this value, the length is \(2(5) + 3 = 13\text{ m}\) and the width is \(5 + 4 = 9\text{ m}\). The perimeter is \(2 \times (13 + 9) = 44\text{ m}\).

M1 for setting up the equation (2x + 3)(x + 4) = 117. M1 for expanding and simplifying to 2x^2 + 11x - 105 = 0. M1 for solving the quadratic equation to get x = 5. M1 for substituting x = 5 to find the perimeter. A1 for 44.

(a) The volume of a hemisphere of radius \(r = 6\text{ cm}\) is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (6)^3 = 144\pi\text{ cm}^3\). The volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h = \pi (6)^2 h = 36\pi h\text{ cm}^3\). Since the total volume is \(504\pi\text{ cm}^3\), we have \(144\pi + 36\pi h = 504\pi\). Dividing both sides by \(\pi\) gives \(144 + 36h = 504\), which simplifies to \(36h = 360\), hence \(h = 10\text{ cm}\). (b) The total surface area consists of the curved surface area of the hemisphere \(2\pi r^2 = 2\pi (6)^2 = 72\pi\text{ cm}^2\), the curved surface area of the cylinder \(2\pi r h = 2\pi (6)(10) = 120\pi\text{ cm}^2\), and the circular base of the cylinder \(\pi r^2 = \pi (6)^2 = 36\pi\text{ cm}^2\). Total surface area = \(72\pi + 120\pi + 36\pi = 228\pi \approx 716.28\text{ cm}^2\). Correct to 3 significant figures, this is \(716\text{ cm}^2\).

Part (a) [2 Marks]: M1 for a correct expression for the volume of the hemisphere (e.g., \(\frac{2}{3}\pi \times 6^3\)). A1 for establishing the equation \(144\pi + 36\pi h = 504\pi\) and solving to show \(h = 10\). Part (b) [2.54 Marks]: M1 for curved surface area of hemisphere \(2\pi \times 6^2\) or cylinder \(2\pi \times 6 \times 10\). M1 for a complete sum of the three surface areas: \(2\pi(6)^2 + 2\pi(6)(10) + \pi(6)^2\). A1 for final answer 716 (accept 716.3 or 228\(\pi\)).

(a) The total length of the garden including the path is \(L = x + 1.5 + 1.5 = x + 3\text{ m}\). The total width including the path is \(W = (x - 4) + 1.5 + 1.5 = x - 1\text{ m}\). The combined area is \((x + 3)(x - 1) = 165\). Expanding this gives \(x^2 - x + 3x - 3 = 165\), which simplifies to \(x^2 + 2x - 3 = 165\). Subtracting 165 from both sides gives the required equation: \(x^2 + 2x - 168 = 0\). (b) To factorise \(x^2 + 2x - 168 = 0\), we find two numbers that multiply to \(-168\) and add to \(2\), which are \(14\) and \(-12\). So \((x + 14)(x - 12) = 0\), giving \(x = -14\) or \(x = 12\). (c) Since the length must be positive, \(x = 12\). The dimensions of the garden are: length = \(12\text{ m}\) and width = \(12 - 4 = 8\text{ m}\). The perimeter is \(2 \times (\text{length} + \text{width}) = 2(12 + 8) = 40\text{ m}\).

Part (a) [2 Marks]: M1 for expressing the combined length as \(x + 3\) or combined width as \(x - 1\). A1 for multiplying \((x+3)(x-1) = 165\) and correctly simplifying to show \(x^2 + 2x - 168 = 0\). Part (b) [1.54 Marks]: M1 for factorising into \((x+14)(x-12)\) or using the quadratic formula. A1 for finding both solutions \(x = -14\) and \(x = 12\). Part (c) [1 Mark]: B1 for identifying \(x=12\) and calculating the perimeter of the garden as 40.

(a) After a \(15\%\) reduction, the price is \(85\%\) of the original price. Let \(P\) be the original price: \(0.85 \times P = 646\), so \(P = 646 \div 0.85 = 760\). The original price was \(\$760\). (b) After an \(8\%\) increase, the tablet price is \(450 \times 1.08 = 486\). After a further \(5\%\) increase, the price is \(486 \times 1.05 = 510.30\). The final price is \(\$510.30\). (c) The price increase is \(312 - 240 = 72\). The percentage increase is \(\frac{72}{240} \times 100\% = 30\%\).

Part (a) [1.54 Marks]: M1 for \(646 \div 0.85\) or equating \(85\% = 646\). A1 for 760. Part (b) [1.5 Marks]: M1 for multiplying by \(1.08\) or \(1.05\), or doing \(450 \times 1.134\). A1 for 510.30 (or 510.3). Part (c) [1.5 Marks]: M1 for finding the increase of 72 and dividing by 240, i.e., \(\frac{72}{240}\). A1 for 30.

(a) Time taken by coach = \(180/x\) hours. Time taken by train = \(180/(x + 15)\) hours. Since the coach takes 1 hour longer: \(180/x - 180/(x + 15) = 1\). Multiply both sides by \(x(x + 15)\): \(180(x + 15) - 180x = x(x + 15)\). Expand: \(180x + 2700 - 180x = x^2 + 15x\). Simplifying gives: \(x^2 + 15x - 2700 = 0\). (b) Factoring the quadratic equation: \((x - 45)(x + 60) = 0\). Therefore, \(x = 45\) or \(x = -60\). Since speed must be a positive value, the average speed of the coach is \(45\) km/h.

M1 for writing \(180/x - 180/(x + 15) = 1\). M1 for algebraic manipulation leading to \(180(x + 15) - 180x = x(x + 15)\). A1 for correct simplification to the given quadratic. M1 for factorising or using the quadratic formula: \((x - 45)(x + 60) = 0\). A1 for selecting the positive root \(x = 45\).

(a) Volume of cylinder = \(\pi r^2 (3r) = 3\pi r^3\). Volume of cone = \(\frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\). Total Volume = \(3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\). (b) Given Total Volume = 450: \(\frac{11}{3}\pi r^3 = 450\) implies \(r^3 = \frac{1350}{11\pi}\). Calculating this gives \(r^3 \approx 39.065\). Taking the cube root, \(r \approx 3.393\). To 3 significant figures, \(r = 3.39\) cm.

M1 for cylinder volume formula and cone volume formula in terms of \(r\). A1 for showing the total volume is \(\frac{11}{3}\pi r^3\). M1 for setting up the equation \(\frac{11}{3}\pi r^3 = 450\) and solving for \(r^3\). A1 for \(r = 3.39\) (accept 3.393).

(a) Let \(P\) be the profit in 2020. An 8% increase means \(1.08 \times P = 189000\). Thus, \(P = 189000 / 1.08 = 175000\). The profit in 2020 was $175,000. (b) The profit increases by 10% compound interest each year over 3 years (from 2021 to 2024). Expected profit in 2024 = \(189000 \times (1.10)^3 = 189000 \times 1.331 = 251559\). The expected profit is $251,559.

M1 for \(189000 / 1.08\). A1 for 175000. M1 for \(189000 \times 1.10^3\). A1 for 251559.

(\text{Time} = \frac{\text{Distance}}{\text{Speed}})

(a) The time for the first part of the journey is \( \frac{36}{x} \) hours.
The time for the second part of the journey is \( \frac{45}{x - 3} \) hours.
The total time is 8 hours, so:
\[ \frac{36}{x} + \frac{45}{x - 3} = 8 \]

Multiply all terms by \( x(x - 3) \) to clear the denominators:
\[ 36(x - 3) + 45x = 8x(x - 3) \]

Expand the brackets:
\[ 36x - 108 + 45x = 8x^2 - 24x \]

Collect like terms:
\[ 81x - 108 = 8x^2 - 24x \]

Rearrange into the form \( ax^2 + bx + c = 0 \):
\[ 8x^2 - 105x + 108 = 0 \]

(b) Solve \( 8x^2 - 105x + 108 = 0 \) by factorisation or the quadratic formula:
\[ (x - 12)(8x - 9) = 0 \]
Therefore:
\[ x = 12 \quad \text{or} \quad x = \frac{9}{8} = 1.125 \]

(c) Since the speed in the second part is \( x - 3 \), \( x \) must be greater than 3.
Thus, we choose \( x = 12 \).
The speed for the second part is \( 12 - 3 = 9 \) km/h.
The time taken for the second part is:
\[ \frac{45}{9} = 5 \text{ hours} \]

(a)
M1: For writing the initial equation \( \frac{36}{x} + \frac{45}{x - 3} = 8 \).
M1: For multiplying by the common denominator to get \( 36(x - 3) + 45x = 8x(x - 3) \).
A1: For correct expansion and rearrangement to \( 8x^2 - 105x + 108 = 0 \).

(b)
M1: For factorising to \( (x - 12)(8x - 9) = 0 \) or correct substitution into the quadratic formula.
A1: Both correct values: \( x = 12 \) and \( x = 1.125 \).

(c)
B1: For identifying \( x = 12 \) as the only valid speed (since \( x > 3 \)).
B1: For calculating the time as \( 5 \) hours.

(a) The total height of the cylinder is \(15\text{ cm}\) and the water depth is \(10\text{ cm}\).
The height of the empty space is \(15 - 10 = 5\text{ cm}\).
The volume of the empty space is:
\[ V = \pi r^2 h = \pi \times 6^2 \times 5 = 180\pi\text{ cm}^3 \]

(b) The volume of the sphere is equal to the volume of the empty space, so:
\[ \frac{4}{3}\pi r^3 = 180\pi \]
Divide both sides by \(\pi\):
\[ \frac{4}{3} r^3 = 180 \]
\[ r^3 = 180 \times \frac{3}{4} = 135 \]
\[ r = \sqrt[3]{135} \approx 5.1299\text{ cm} \]
To 3 significant figures, \(r = 5.13\text{ cm}\).

(c) The volume of the sphere is \(180\pi \approx 565.487\text{ cm}^3\).
\[ \text{Mass} = \text{Volume} \times \text{Density} \]
\[ \text{Mass} = 565.487\text{ cm}^3 \times 7.85\text{ g/cm}^3 \approx 4439.07\text{ g} \]
Convert to kilograms:
\[ 4439.07\text{ g} \div 1000 = 4.43907\text{ kg} \]
To 3 significant figures, the mass is \(4.44\text{ kg}\).

(a)
M1: For calculating the remaining height \(15 - 10 = 5\) or subtracting water volume from total volume.
A1: For \(180\pi\).

(b)
M1: For equating sphere volume formula to their part (a) answer: \( \frac{4}{3}\pi r^3 = 180\pi \).
A1: For \(r = \sqrt[3]{135}\) or \(5.13\) (accept \(5.129\dots\)).

(c)
M1: For multiplying their volume of the sphere by \(7.85\).
M1: For dividing by 1000 to convert grams to kilograms.
A1: For \(4.44\) (accept \(4.43\) to \(4.45\)).

(a) The standard selling price is:
\[ 150 \times \left(1 + \frac{40}{100}\right) = 150 \times 1.40 = \$210 \]

(b) The standard selling price is discounted by \(x\%\) to give \(\$178.50\):
\[ 210 \times \left(1 - \frac{x}{100}\right) = 178.50 \]

\[ 1 - \frac{x}{100} = \frac{178.50}{210} = 0.85 \]

\[ \frac{x}{100} = 0.15 \implies x = 15 \]

(c) The new cost price is:
\[ 150 \times 1.08 = \$162 \]
To make a \(25\%\) profit on this new cost price, the new selling price is:
\[ 162 \times 1.25 = \$202.50 \]

(a)
M1: For \(150 \times 1.40\) or \(150 + 60\).
A1: For \(210\).

(b)
M1: For setting up the equation \(210 \times (1 - x/100) = 178.50\) or finding the discount amount \(210 - 178.50 = 31.50\).
A1: For \(15\).

(c)
M1: For finding the new cost price \(150 \times 1.08 = 162\).
M1: For calculating the selling price with \(25\%\) profit: \( \text{their } 162 \times 1.25 \).
A1: For \(202.50\) (or \(202.5\)).

Part (a):
We need to factorise \(3x^2 - 11x - 4\).
Find two numbers that multiply to \(3 \times (-4) = -12\) and add to \(-11\). These are \(-12\) and \(1\).
Split the middle term:
\(3x^2 - 12x + x - 4 = 3x(x - 4) + 1(x - 4) = (3x + 1)(x - 4)\).

Part (b):
We simplify the expression:
\(\frac{3x^2 - 11x - 4}{2x^2 - 32} \div \frac{3x^2 + x}{x^2 - 4x}\)
First, factorise each component:
- \(3x^2 - 11x - 4 = (3x + 1)(x - 4)\) (from part a)
- \(2x^2 - 32 = 2(x^2 - 16) = 2(x - 4)(x + 4)\)
- \(3x^2 + x = x(3x + 1)\)
- \(x^2 - 4x = x(x - 4)\)

Now substitute these back into the expression, changing the division to multiplication by the reciprocal:
\(\frac{(3x + 1)(x - 4)}{2(x - 4)(x + 4)} \times \frac{x(x - 4)}{x(3x + 1)}\)

Cancel out the common factors \((3x + 1)\), \((x - 4)\), and \(x\):
- Cancel \((3x + 1)\) from the first numerator and second denominator.
- Cancel \((x - 4)\) from the first numerator and first denominator.
- Cancel \(x\) from the second numerator and second denominator.

This leaves:
\(\frac{x - 4}{2(x + 4)}\) or \(\frac{x - 4}{2x + 8}\).

Total: 4.54 marks
Part (a): 1.54 marks
- M1 for correct attempt to split the middle term or factorise.
- A1 for \((3x + 1)(x - 4)\).
Part (b): 3 marks
- M1 for factorising denominators and numerators.
- M1 for changing division to multiplication and attempting to cancel.
- A1 for \(\frac{x-4}{2(x+4)}\) or equivalent.

Part (a):
Profit in 2022 = \(126000 \times 1.12 = 141120\).

Part (b):
Let \(P\) be the profit in 2020.
A 16% decrease means the profit in 2021 is \(100\% - 16\% = 84\%\) of \(P\).
\(0.84 \times P = 126000\)
\(P = \frac{126000}{0.84} = 150000\).

Part (c):
Let \(n\) be the number of years after 2022.
We want \(141120 \times (1.12)^n > 250000\).
\((1.12)^n > \frac{250000}{141120} \approx 1.7715\)
Using trial and error with a calculator:
For \(n = 5\): \(141120 \times (1.12)^5 \approx 248701.66\) (does not exceed $250 000)
For \(n = 6\): \(141120 \times (1.12)^6 \approx 278545.86\) (exceeds $250 000)
Thus, the minimum number of complete years is 6.

Total: 4.54 marks
Part (a): 1 mark
- B1 for 141120.
Part (b): 1.54 marks
- M1 for \(\frac{126000}{0.84}\) or equivalent.
- A1 for 150000.
Part (c): 2 marks
- M1 for set up of inequality or equation \(141120 \times (1.12)^n > 250000\) or trial of \(n=5\) or \(n=6\).
- A1 for 6.

Part (a):
Volume of the cylinder: \(V_{cyl} = \pi r^2 (2r) = 2\pi r^3\)
Volume of the cone: \(V_{cone} = \frac{1}{3}\pi r^2 h\)
Total Volume \(V = 2\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{6}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (6r + h)\). This is shown.

Part (b)(i):
Substitute \(V = 90\pi\) and \(h = 4r\) into the formula:
\(90\pi = \frac{1}{3}\pi r^2 (6r + 4r)\)
\(90\pi = \frac{1}{3}\pi r^2 (10r)\)
\(90 = \frac{10}{3}r^3\)
\(r^3 = 90 \times \frac{3}{10} = 27\)
\(r = \sqrt[3]{27} = 3\).

Part (b)(ii):
With \(r = 3\), the height of the cone is \(h = 4(3) = 12\) cm.
The total surface area of the solid consists of:
1. The circular base of the cylinder: \(\text{Area}_1 = \pi r^2 = \pi (3^2) = 9\pi\)
2. The curved surface area of the cylinder: \(\text{Area}_2 = 2\pi r (2r) = 4\pi r^2 = 4\pi (3^2) = 36\pi\)
3. The curved surface area of the cone: \(\text{Area}_3 = \pi r l\), where \(l\) is the slant height of the cone.
\(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 12^2} = \sqrt{9 + 144} = \sqrt{153}\)
\(\text{Area}_3 = 3\pi \sqrt{153}\)

Total surface area \(A = 9\pi + 36\pi + 3\pi \sqrt{153} = 45\pi + 3\pi \sqrt{153} \approx 141.37 + 116.58 = 257.95\) cm\(^2\).
Correct to 3 significant figures, the total surface area is 258.

Total: 4.54 marks
Part (a): 1 mark
- M1 for writing the sum of the volumes of cylinder and cone: \(2\pi r^3 + \frac{1}{3}\pi r^2 h\) and factorising out \(\frac{1}{3}\pi r^2\).

Part (b)(i): 1.54 marks
- M1 for setting up the equation \(\frac{10}{3}r^3 = 90\) or equivalent.
- A1 for \(r = 3\).

Part (b)(ii): 2 marks
- M1 for finding slant height \(l = \sqrt{3^2 + 12^2} = \sqrt{153}\) (or \(\approx 12.4\)) or for the sum formula \(5\pi r^2 + \pi r l\).
- A1 for \(258\) (accept answers in range [257.9, 258.1]).

(a) Volume of a hemisphere = \(\frac{2}{3}\pi r^3\). Volume of a cone = \(\frac{1}{3}\pi r^2 h\). Total Volume \(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r + h)\). (b) Substitute \(V = 250\) and \(r = 4.2\): \(250 = \frac{1}{3}\pi (4.2)^2 (2(4.2) + h)\) \Rightarrow \(750 = 17.64\pi(8.4 + h)\) \Rightarrow \(8.4 + h = \frac{750}{17.64\pi} \approx 13.5336\) \Rightarrow \(h = 13.5336 - 8.4 = 5.13\) cm (to 3 s.f.).

Part (a): [2 marks]
- M1 for writing the sum of the volumes of hemisphere and cone: \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\)
- A1 for showing correct algebraic factorization to reach \(\frac{1}{3}\pi r^2 (2r + h)\)

Part (b): [3 marks]
- M1 for correct substitution of values: \(250 = \frac{1}{3}\pi (4.2)^2 (8.4 + h)\)
- M1 for correct rearrangement to find \(h\): \(h = \frac{750}{17.64\pi} - 8.4\)
- A1 for 5.13 (accept 5.13 to 5.14)

(a) Speed for return is \(x-3\) km/h, so time taken is \(\frac{40}{x-3}\) hours. (b) Outward time is \(\frac{40}{x}\) hours. Difference is 40 minutes = \(\frac{2}{3}\) hours. So, \(\frac{40}{x-3} - \frac{40}{x} = \frac{2}{3}\). Divide by 2: \(\frac{20}{x-3} - \frac{20}{x} = \frac{1}{3}\). Multiply by \(3x(x-3)\): \(60x - 60(x-3) = x(x-3)\) \Rightarrow \(180 = x^2 - 3x\) \Rightarrow \(x^2 - 3x - 180 = 0\). (c) \((x-15)(x+12) = 0\). Since speed must be positive, \(x = 15\) km/h.

Part (a): [1 mark]
- B1 for \(\frac{40}{x-3}\)

Part (b): [2 marks]
- M1 for \(\frac{40}{x-3} - \frac{40}{x} = \frac{40}{60}\) (or equivalent)
- A1 for fully correct algebraic simplification to reach \(x^2 - 3x - 180 = 0\)

Part (c): [2 marks]
- M1 for \((x-15)(x+12) = 0\) or correct use of quadratic formula
- A1 for \(x = 15\) (ignore \(x = -12\))

(a) Total value = \(4500 \times (1 + \frac{3.5}{100})^3 = 4500 \times 1.035^3 = 4989.20625\). Correct to the nearest cent, this is $4989.21. (b) Value after 5 years: \(4500 \times (1 + \frac{r}{100})^5 = 5320\). \((1 + \frac{r}{100})^5 = \frac{5320}{4500} = 1.18222\). \(1 + \frac{r}{100} = (1.18222)^{0.2} \approx 1.03403\). \(\frac{r}{100} = 0.03403\) \Rightarrow \(r = 3.40\) (to 3 s.f.).

Part (a): [2 marks]
- M1 for \(4500 \times 1.035^3\)
- A1 for 4989.21

Part (b): [3 marks]
- M1 for \(4500 \times (1 + \frac{r}{100})^5 = 5320\)
- M1 for \(1 + \frac{r}{100} = \sqrt[5]{\frac{5320}{4500}}\) or equivalent rearrangement
- A1 for 3.40 or 3.4

(a) The time to produce 300 sheets of Type A at a rate of \(x\) sheets per hour is \(\frac{300}{x}\) hours.

(b) The time taken for Type B is \(\frac{300}{x+5}\) hours. The difference is 5 hours:
\(\frac{300}{x} - \frac{300}{x+5} = 5\)
Divide the entire equation by 5:
\(\frac{60}{x} - \frac{60}{x+5} = 1\)
Multiply by \(x(x+5)\):
\(60(x+5) - 60x = x(x+5)\)
\(60x + 300 - 60x = x^2 + 5x\)
\(300 = x^2 + 5x\)
\(x^2 + 5x - 300 = 0\).

(c) Factorising the quadratic equation:
\((x + 20)(x - 15) = 0\)
Since production rate \(x\) must be positive, \(x = 15\).

(d) Time for Type A: \(\frac{300}{15} = 20\) hours.
Time for Type B: \(\frac{300}{20} = 15\) hours.
Total time: \(20 + 15 = 35\) hours.

M1 for expression 300/x
M1 for setting up equation 300/x - 300/(x+5) = 5
M1 for algebraic expansion leading to x^2 + 5x - 300 = 0
M1 for factorisation (x + 20)(x - 15) and selecting x = 15
M1 for calculating total time of 35 hours

(a) Volume of cylinder: \(V_{\text{cylinder}} = \pi r^2 (2r) = 2\pi r^3\).
Volume of cone: \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\).
Since their volumes are equal:
\(\frac{1}{3}\pi r^2 h = 2\pi r^3 \implies h = 6r\).

(b)(i) Total volume = \(V_{\text{cylinder}} + \dots = 2\pi r^3 + 2\pi r^3 = 4\pi r^3\).
Given total volume = \(256\pi\):
\(4\pi r^3 = 256\pi \implies r^3 = 64 \implies r = 4\) cm.

(b)(ii) To find the total external surface area, we sum the area of the flat base, the curved surface of the cylinder, and the curved surface of the cone:
1. Base of cylinder: \(\pi r^2 = \pi (4)^2 = 16\pi\).
2. Curved cylinder area: \(2\pi r (2r) = 4\pi r^2 = 64\pi\).
3. Curved cone area: \(\pi r l\), where slant height \(l = \sqrt{r^2 + h^2}\).
With \(r = 4\) and \(h = 6(4) = 24\):
\(l = \sqrt{4^2 + 24^2} = \sqrt{16 + 576} = \sqrt{592} \approx 24.331\) cm.
Cone area = \(\pi \times 4 \times 24.331 \approx 97.324\pi\).
Total surface area = \(16\pi + 64\pi + 97.324\pi = 177.324\pi \approx 557.076\) \(\text{cm}^2\).
Correct to 3 significant figures, this is \(557\) \(\text{cm}^2\).

M1 for equating cone and cylinder volumes: 1/3 * pi * r^2 * h = 2 * pi * r^3
M1 for total volume expression 4 * pi * r^3 and setting it to 256 * pi to find r = 4
M1 for calculating slant height l = sqrt(4^2 + 24^2) = sqrt(592)
M1 for summing area terms: pi * r^2 + 4 * pi * r^2 + pi * r * l
A1 for 557 (accept 557 to 558)

(a) \(8000 \times \left(1 + \frac{R}{100}\right)^3 = 8741.81\)
\(\left(1 + \frac{R}{100}\right)^3 = \frac{8741.81}{8000} \approx 1.092726\)
\(1 + \frac{R}{100} = \sqrt[3]{1.092726} \approx 1.0300\)
\(\frac{R}{100} = 0.03 \implies R = 3.0\), correct to 1 decimal place.

(b) We want \(8741.81 \times (1.03)^n \ge 10000\)
\((1.03)^n \ge 1.14393\)
\(n \ge \frac{\ln(1.14393)}{\ln(1.03)} \approx 4.55\)
Since \(n\) must be a complete number of years, the minimum value of \(n\) is 5.

(c) Simple interest per year = \(7500 \times 0.032 = \$240\).
Total increase needed = \(10000 - 7500 = \$2500\).
Number of years required = \(\frac{2500}{240} \approx 10.42\).
Since it must be complete years, Maya needs 11 years.

M1 for setting up 8000 * (1 + R/100)^3 = 8741.81 and finding R
M1 for compound interest inequality 8741.81 * 1.03^n >= 10000 or trials
A1 for n = 5
M1 for simple interest calculation: interest per year = 240
M1 for 2500 / 240
A1 for 11

1. Let the base radius of the cone be \(r\text{ cm}\). The height of the cone is given as \(h = 5r\text{ cm}\).

2. Use the formula for the volume of a cone:
\(V = \frac{1}{3}\pi r^2 h\)

Substitute the given values into the formula:
\(\frac{1}{3}\pi r^2 (5r) = 360\pi\)

\(\frac{5}{3}\pi r^3 = 360\pi\)

3. Solve for \(r\):
\(\frac{5}{3}r^3 = 360\)

\(r^3 = 360 \times \frac{3}{5}\)

\(r^3 = 216\)

\(r = \sqrt[3]{216} = 6\text{ cm}\)

4. Find the height \(h\):
\(h = 5 \times 6 = 30\text{ cm}\)

5. Calculate the slant height \(l\) of the cone using Pythagoras' theorem:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 30^2} = \sqrt{36 + 900} = \sqrt{936} \approx 30.594\text{ cm}\)

6. Calculate the total surface area \(A\) of the cone (which includes the circular base and the curved surface area):
\(A = \pi r^2 + \pi r l\)

\(A = \pi (6^2) + \pi (6)(30.594)\)

\(A = 36\pi + 183.564\pi \approx 219.564\pi \approx 689.78\text{ cm}^2\)

Rounding to 3 significant figures, we get \(690\text{ cm}^2\).

M1: For setting up the volume equation, e.g., \(\frac{1}{3}\pi r^2 (5r) = 360\pi\) or equivalent
A1: For finding the radius \(r = 6\)
M1: For finding the slant height \(l = \sqrt{6^2 + 30^2}\) or \(l \approx 30.6\)
M1: For substituting their \(r\) and \(l\) into the total surface area formula \(\pi r^2 + \pi r l\)
A1: For final answer \(690\) or in the range \([689.7, 690.1]\)