2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Expand the brackets first:
\((3a - 2)(2a + 5) = 6a^2 + 15a - 4a - 10\)

Simplify the terms:
\(= 6a^2 + 11a - 10\)

Now subtract \(6a^2\):
\(6a^2 + 11a - 10 - 6a^2 = 11a - 10\)

M1 for expanding \((3a - 2)(2a + 5)\) to obtain 4 terms with at least 3 correct (e.g. \(6a^2 + 15a - 4a - 10\))
M1 for simplifying to \(6a^2 + 11a - 10\)
A1 for \(11a - 10\) (final answer)

First, find the decrease in price:
\(\text{Decrease} = 160 - 128 = 32\)

Next, calculate the percentage decrease relative to the original price:
\(\text{Percentage decrease} = \frac{32}{160} \times 100\)

Simplify the fraction:
\(\frac{32}{160} = \frac{2}{10} = \frac{1}{5}\)

Convert to percentage:
\(\frac{1}{5} \times 100 = 20\%\)

M1 for finding the decrease: \(160 - 128 = 32\)
M1 for setting up percentage calculation: \(\frac{\text{their } 32}{160} \times 100\)
A1 for \(20\) (or \(20\%\))

The formula for the total surface area of a cuboid is:
\(\text{Total Surface Area} = 2(lw + lh + wh)\)

Substitute the given dimensions (length \(l = 8\) cm, width \(w = 5\) cm, height \(h = 4\) cm):
\(\text{Area} = 2(8 \times 5 + 8 \times 4 + 5 \times 4)\)
\(\text{Area} = 2(40 + 32 + 20)\)
\(\text{Area} = 2(92) = 184\text{ cm}^2\)

M1 for substituting dimensions into a correct surface area expression, e.g., \(2(8 \times 5) + 2(8 \times 4) + 2(5 \times 4)\) (at least two pairs of products correct)
M1 for evaluating inside the brackets to get \(40 + 32 + 20\) or showing \(80 + 64 + 40\)
A1 for \(184\)

Multiply both sides of the equation by 5 to eliminate the fraction:
\(4x - 3 = 5(x - 2)\)

Expand the bracket on the right-hand side:
\(4x - 3 = 5x - 10\)

Rearrange to solve for \(x\):
Subtract \(4x\) from both sides:
\(-3 = x - 10\)

Add 10 to both sides:
\(x = 7\)

M1 for multiplying by 5 to get \(4x - 3 = 5(x - 2)\) or better
M1 for isolating terms in \(x\) on one side and constant terms on the other, e.g., \(5x - 4x = 10 - 3\) (allow one sign error)
A1 for \(7\)

Identify the difference in ratio parts between Chloe and Alice:
\(7 \text{ parts} - 2 \text{ parts} = 5 \text{ parts}\)

Since Chloe receives $120 more than Alice:
\(5 \text{ parts} = \$120\)

Find the value of 1 part:
\(1 \text{ part} = \frac{120}{5} = \$24\)

Find the total number of parts shared:
\(2 + 3 + 7 = 12 \text{ parts}\)

Calculate the total sum of money shared:
\(\text{Total sum} = 12 \times 24 = \$288\)

M1 for finding that \(7 - 2 = 5\) parts correspond to $120
M1 for finding the value of one part (\(24\)) or expressing the total as \(\frac{12}{5} \times 120\)
A1 for \(288\)

First, convert the probability of choosing a blue counter to a decimal:
\(\frac{2}{5} = 0.4\)

Find the combined probability of choosing a red or blue counter:
\(0.35 + 0.4 = 0.75\)

Since the sum of all probabilities is 1, find the probability of choosing a yellow counter:
\(1 - 0.75 = 0.25\)

M1 for converting \(\frac{2}{5}\) to \(0.4\) or both to fractions with common denominators
M1 for subtraction from 1: \(1 - (0.35 + \text{their } 0.4)\)
A1 for \(0.25\) (decimal only)

Convert 2 hours into seconds:
\(2 \text{ hours} = 2 \times 60 \times 60 = 7200 \text{ seconds}\)

Calculate the distance in metres:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
\(\text{Distance} = 18 \text{ m/s} \times 7200 \text{ s} = 129\,600 \text{ metres}\)

Convert the distance to kilometres:
\(\text{Distance in km} = \frac{129\,600}{1000} = 129.6 \text{ km}\)

M1 for converting 2 hours to \(7200\) seconds or converting \(18\text{ m/s}\) to \(64.8\text{ km/h}\)
M1 for a correct product of speed and time, e.g., \(18 \times 7200\) or \(64.8 \times 2\)
A1 for \(129.6\)

Simplify each part of the expression separately:

For \(8^{-\frac{2}{3}}\):
\(8^{\frac{1}{3}} = 2\)
So, \(8^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4}\)

For \(16^{\frac{3}{4}}\):
\(16^{\frac{1}{4}} = 2\)
So, \(16^{\frac{3}{4}} = 2^3 = 8\)

Multiply the two simplified results:
\(\frac{1}{4} \times 8 = 2\)

M1 for simplifying \(8^{-\frac{2}{3}}\) to \(\frac{1}{4}\) or \(2^{-2}\)
M1 for simplifying \(16^{\frac{3}{4}}\) to \(8\) or \(2^3\)
A1 for \(2\)

Step 1: Convert both mixed numbers into improper fractions:
\( 1\frac{3}{5} = \frac{1 \times 5 + 3}{5} = \frac{8}{5} \)
\( 2\frac{2}{3} = \frac{2 \times 3 + 2}{3} = \frac{8}{3} \)

Step 2: Divide the fractions by multiplying by the reciprocal:
\( \frac{8}{5} \div \frac{8}{3} = \frac{8}{5} \times \frac{3}{8} \)

Step 3: Simplify the calculation:
\( \frac{8 \times 3}{5 \times 8} = \frac{3}{5} \)

M1 for converting both mixed numbers to improper fractions correctly (\(\frac{8}{5}\) and \(\frac{8}{3}\))
M1 for multiplying by the reciprocal (\(\frac{8}{5} \times \frac{3}{8}\))
A1 for final answer of \(\frac{3}{5}\) (or equivalent simplified fraction)

Simple Interest formula is:
\( I = \frac{P \times R \times T}{100} \)

Where:
- \( P = 600 \)
- \( R = 4 \)
- \( T = 3 \)

Step 1: Substitute the values into the formula:
\( I = \frac{600 \times 4 \times 3}{100} \)

Step 2: Simplify the expression:
\( I = 6 \times 4 \times 3 \)
\( I = 24 \times 3 = 72 \)

Liam earns $72 of interest.

M1 for finding 4% of 600 (e.g. \(600 \times 0.04 = 24\))
M1 for multiplying interest of one year by 3 (e.g. \(24 \times 3\)) or for full correct formula substitution \(\frac{600 \times 4 \times 3}{100}\)
A1 for 72

Step 1: Find the highest common factor (HCF) of the numerical coefficients 12 and 18, which is 6.

Step 2: Find the common algebraic factors. Both terms share \( x \) and \( y \), so the highest common variable factor is \( xy \).

Step 3: Extract \( 6xy \) from both terms:
\( 12x^2 y = 6xy(2x) \)
\( 18xy^2 = 6xy(3y) \)

Step 4: Write in factorised form:
\( 6xy(2x - 3y) \)

B1 for any correct partial factorisation (e.g., \(2xy(6x - 9y)\) or \(6(2x^2y - 3xy^2)\))
B1 for another correct factor extracted (e.g., \(6x(2xy - 3y^2)\))
B1 for fully correct factorised final answer: \(6xy(2x - 3y)\)

Step 1: Multiply both sides of the equation by 4 to clear the fraction:
\( 3x - 5 = 7 \times 4 \)
\( 3x - 5 = 28 \)

Step 2: Add 5 to both sides:
\( 3x = 28 + 5 \)
\( 3x = 33 \)

Step 3: Divide both sides by 3:
\( x = \frac{33}{3} \)
\( x = 11 \)

M1 for multiplying by 4: \(3x - 5 = 28\)
M1 for adding 5 to their RHS: \(3x = 33\)
A1 for 11 (or \(x = 11\))

Step 1: Find the total frequency (number of students):
\( \text{Total Frequency} = 4 + 3 + 2 + 1 = 10 \)

Step 2: Find the sum of all scores (Score \(\times\) Frequency):
\( \text{Total Score} = (1 \times 4) + (2 \times 3) + (3 \times 2) + (4 \times 1) \)
\( \text{Total Score} = 4 + 6 + 6 + 4 = 20 \)

Step 3: Calculate the mean score:
\( \text{Mean} = \frac{\text{Total Score}}{\text{Total Frequency}} = \frac{20}{10} = 2 \)

M1 for sum of products: \(1\times4 + 2\times3 + 3\times2 + 4\times1\) (at least 3 terms correct)
M1 for dividing their sum of products by their total frequency (10)
A1 for 2

An isosceles triangle has two equal angles.

Case 1: The given angle of \( 40^\circ \) is the vertex angle (not one of the equal angles).
The remaining two equal angles sum to: \( 180^\circ - 40^\circ = 140^\circ \).
Each of the equal angles is: \( 140^\circ \div 2 = 70^\circ \).
In this case, the angles are \( 40^\circ, 70^\circ, 70^\circ \), so the largest angle is \( 70^\circ \).

Case 2: The given angle of \( 40^\circ \) is one of the equal angles.
The angles are \( 40^\circ, 40^\circ \), and the third angle.
The third angle is: \( 180^\circ - (40^\circ + 40^\circ) = 100^\circ \).
In this case, the angles are \( 40^\circ, 40^\circ, 100^\circ \), so the largest angle is \( 100^\circ \).

The two possible sizes for the largest angle are \( 70^\circ \) and \( 100^\circ \).

M1 for \(180 - 2 \times 40 = 100\)
M1 for \((180 - 40) \div 2 = 70\)
A1 for both answers 70 and 100 (order does not matter)

Step 1: Find the cross-sectional area of the triangular face:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( \text{Area} = \frac{1}{2} \times 6 \times 4 = 12\text{ cm}^2 \)

Step 2: Calculate the volume of the prism by multiplying the cross-sectional area by the length:
\( \text{Volume} = \text{Area} \times \text{length} \)
\( \text{Volume} = 12 \times 10 = 120\text{ cm}^3 \)

M1 for finding the area of the triangular cross-section: \(\frac{1}{2} \times 6 \times 4 = 12\)
M1 for multiplying their cross-sectional area by the length (10)
A1 for 120

Step 1: Find the total number of marbles:
\( 4 + 6 = 10 \) marbles.

Step 2: Find the probability of picking a red marble on the first turn:
\( P(\text{Red}) = \frac{4}{10} = \frac{2}{5} \)

Step 3: Since the marble is replaced, the probability of picking a red marble on the second turn remains the same:
\( P(\text{Red}) = \frac{2}{5} \)

Step 4: Multiply the probabilities to find the combined probability of both being red:
\( P(\text{Red and Red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25} \) (or \( 0.16 \))

M1 for finding the probability of picking one red marble: \(\frac{4}{10}\) (or \(\frac{2}{5}\) or 0.4)
M1 for multiplying two independent probabilities with replacement: \(\frac{4}{10} \times \frac{4}{10}\) (or \(\frac{2}{5} \times \frac{2}{5}\))
A1 for \(\frac{4}{25}\) (or \(\frac{16}{100}\) or 0.16)

Group the terms to factorise:
\(3ax - 6ay + 2bx - 4by = 3a(x - 2y) + 2b(x - 2y)\)

Factorise out the common bracket \((x - 2y)\):
\((3a + 2b)(x - 2y)\).

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

The sale price represents 80% of the original price.
Let the original price be \(P\).
\(0.80 \times P = 144\)

\(P = \frac{144}{0.8} = \frac{1440}{8} = 180\).

So, the original price of the coat is $180.

M1 for associating $144 with 80%
M1 for \(144 \div 0.8\) or \(144 \times \frac{100}{80}\) oe
A1 for 180

The area of the triangular cross-section is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\)

The volume of the prism is:
\(\text{Volume} = \text{Area of cross-section} \times \text{length} = 30 \times 8 = 240\text{ cm}^3\).

M1 for \(\frac{1}{2} \times 5 \times 12\)
M1 for \(\text{their Area} \times 8\)
A1 for 240

Multiply both sides by 15 (or cross-multiply):
\(3(2x - 3) = 5(x + 1)\)

\(6x - 9 = 5x + 5\)

Subtract \(5x\) from both sides:
\(x - 9 = 5\)

Add 9 to both sides:
\(x = 14\).

M1 for \(3(2x - 3) = 5(x + 1)\) oe
M1 for \(6x - 9 = 5x + 5\) oe (correct expansion of their brackets)
A1 for 14

By order of operations, perform the multiplication first:
\(\frac{2}{3} \times \frac{3}{4} = \frac{2 \times 3}{3 \times 4} = \frac{6}{12} = \frac{1}{2}\)

Now, perform the subtraction:
\(1\frac{5}{6} - \frac{1}{2} = \frac{11}{6} - \frac{3}{6} = \frac{8}{6} = \frac{4}{3}\) (or \(1\frac{1}{3}\)).

M1 for \(\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}\) oe
M1 for \(\frac{11}{6} - \frac{3}{6}\) oe (common denominator for subtraction)
A1 for \(\frac{4}{3}\) or \(1\frac{1}{3}\)

Write both numbers with the same power of 10:
\(4.2 \times 10^4 = 4.2 \times 10^4\)
\(3.8 \times 10^3 = 0.38 \times 10^4\)

Adding them:
\((4.2 + 0.38) \times 10^4 = 4.58 \times 10^4\).

Alternatively, convert to ordinary numbers:
\(42000 + 3800 = 45800 = 4.58 \times 10^4\).

M1 for converting to ordinary numbers: \(42000\) and \(3800\) or showing a common index: \(4.2 \times 10^4 + 0.38 \times 10^4\) oe
M1 for \(45800\) or \(4.58 \times 10^k\) (where \(k \neq 4\))
A1 for \(4.58 \times 10^4\)

The difference in ratio parts between Ben and Alice is:
\(5 - 2 = 3\) parts.

These 3 parts represent $45.
Value of 1 part = \(\frac{45}{3} = \$15\).

The total number of parts is:
\(2 + 5 + 7 = 14\) parts.

The total amount of money shared is:
\(14 \times 15 = 210\).

M1 for \(5 - 2 = 3\) parts represented by $45
M1 for finding 1 part = $15 or total parts = 14
A1 for 210

The formula for the mean is:
\(\frac{12 + 15 + 18 + 14 + h}{5} = 16\)

Multiply both sides by 5:
\(12 + 15 + 18 + 14 + h = 80\)
\(59 + h = 80\)

Subtract 59 from both sides:
\(h = 80 - 59 = 21\).

M1 for \(12 + 15 + 18 + 14 + h = 16 \times 5\) oe
M1 for \(59 + h = 80\) oe
A1 for 21

To solve the equation, first expand the brackets:
\(3(2x) - 3(5) - 4(x) - 4(1) = 9\)
\(6x - 15 - 4x - 4 = 9\)

Next, collect like terms:
\((6x - 4x) + (-15 - 4) = 9\)
\(2x - 19 = 9\)

Add 19 to both sides:
\(2x = 9 + 19\)
\(2x = 28\)

Divide both sides by 2:
\(x = 14\)

M1 for correct expansion of at least one bracket (e.g. \(6x - 15\) or \(-4x - 4\))
M1 for collecting terms correctly to the form \(ax = b\) (e.g. \(2x = 28\) or \(2x - 19 = 9\))
A1 for 14

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

Let the original price be \(P\).
\(85\% \text{ of } P = \$68\)
\(0.85 \times P = 68\)
\(P = \frac{68}{0.85} = \frac{6800}{85}\)

Simplifying the fraction by dividing the numerator and denominator by 17 (since \(17 \times 4 = 68\) and \(17 \times 5 = 85\)):
\(P = \frac{400}{5} = 80\)

The original price of the jacket is \(\$80\).

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

First, factorise the numerator and the denominator:
Numerator: \(3(x^2 - 4) = 3(x - 2)(x + 2)\)
Denominator: \((x - 2)(x + 1)\)

Now, simplify the fraction by cancelling the common factor \((x - 2)\):
\(\frac{3(x - 2)(x + 2)}{(x - 2)(x + 1)} = \frac{3(x + 2)}{x + 1}\).

M1 for factorising the numerator to \(3(x-2)(x+2)\) or \((3x-6)(x+2)\)
M1 for factorising the denominator to \((x-2)(x+1)\)
A1 for \(\frac{3(x+2)}{x+1}\) or \(\frac{3x+6}{x+1}\) as the final answer

Let \(x\) be the original price.
A 15% reduction means the sale price is 85% of the original price.
\(0.85x = 20.40\)
\(x = \frac{20.40}{0.85} = \frac{2040}{85} = 24\).
Thus, the original price was $24.

M1 for recognizing that 85% corresponds to $20.40 (e.g., write \(0.85x = 20.40\))
M1 for a correct division step, e.g., \(\frac{20.40}{0.85}\) or \(\frac{2040}{85}\)
A1 for 24 (or 24.00)

First, calculate the volume of the cone:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6^2)(3) = 36\pi\text{ cm}^3\).

Since the cone is recast into a sphere, their volumes are equal:
\(V_{\text{sphere}} = \frac{4}{3} \pi R^3 = 36\pi\)

Divide both sides by \(\pi\):
\ \frac{4}{3} R^3 = 36\\
\tR^3 = 36 \times \frac{3}{4}\\
\tR^3 = 27\\
\tR = 3\text{ cm}\).

M1 for correct expression for the volume of the cone, e.g., \(\frac{1}{3} \pi \times 6^2 \times 3\) (implied by \(36\pi\))
M1 for equating their cone volume to the sphere volume formula, e.g., \(\frac{4}{3} \pi R^3 = 36\pi\) and solving for \(R^3\)
A1 for 3

Substitute \(y = x - 3\) into the quadratic equation:
\(x^2 + (x - 3)^2 = 29\)
\(x^2 + x^2 - 6x + 9 = 29\)
\(2x^2 - 6x - 20 = 0\)

Divide the entire equation by 2:
\(x^2 - 3x - 10 = 0\)

Factorise the quadratic equation:
\((x - 5)(x + 2) = 0\)
This gives \(x = 5\) or \(x = -2\).

Now, find the corresponding \(y\)-values:
If \(x = 5\), then \(y = 5 - 3 = 2\).
If \(x = -2\), then \(y = -2 - 3 = -5\).

So the solutions are \(x = 5, y = 2\) and \(x = -2, y = -5\).

M1 for substituting \(y = x - 3\) correctly into the quadratic equation
M1 for expanding and simplifying to a 3-term quadratic, e.g., \(2x^2 - 6x - 20 = 0\) or \(x^2 - 3x - 10 = 0\)
M1 for factorising their quadratic, e.g., \((x-5)(x+2) = 0\), or using the quadratic formula correctly
A1 for \(x = 5\) and \(x = -2\)
A1 for \(y = 2\) and \(y = -5\) correctly paired with their \(x\)-values

We can rewrite the expression using index laws:
\(64^{-\frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{64})^2}\).

Since \(\sqrt[3]{64} = 4\), we have:
\(\frac{1}{4^2} = \frac{1}{16}\).

M1 for evaluating the cube root of 64 as 4, or for writing the expression as \(\frac{1}{64^{2/3}}\)
A1 for \(\frac{1}{16}\) or 0.0625

First, factorise the quadratic expression:
\(x^2 - 5x - 14 = (x - 7)(x + 2)\).

The critical values where the expression is equal to 0 are \(x = 7\) and \(x = -2\).

Since we want the expression to be greater than 0, the solution lies outside the interval between the critical values:
\(x < -2\) or \(x > 7\).

M1 for factorising to \((x - 7)(x + 2)\) or finding the critical values 7 and -2
M1 for establishing the correct regions, e.g., checking test points or sketching a quadratic curve
A1 for \(x < -2\) or \(x > 7\) (accept equivalent notation, but reject \(-2 > x > 7\))

Let's find the first and second differences:
Terms: \(3, 10, 21, 36\)
First differences: \(10 - 3 = 7\), \(21 - 10 = 11\), \(36 - 21 = 15\)
Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\)

Since the second difference is constant and equals 4, the sequence is quadratic with leading coefficient \(\frac{4}{2} = 2\), so it starts with \(2n^2\).

Let's subtract \(2n^2\) from the original terms:
For \(n=1\): \(3 - 2(1)^2 = 1\)
For \(n=2\): \(10 - 2(2)^2 = 2\)
For \(n=3\): \(21 - 2(3)^2 = 3\)
For \(n=4\): \(36 - 2(4)^2 = 4\)

The remaining sequence is \(1, 2, 3, 4\), which is simply \(n\).

Therefore, the \(n\)-th term is \(2n^2 + n\).

M1 for finding the second difference of 4
M1 for attempting a quadratic term \(2n^2\)
A1 for the final expression \(2n^2 + n\) (or equivalent)

The magnitude of a vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) is given by \(\sqrt{x^2 + y^2}\).
So:
\(\sqrt{12^2 + k^2} = 13\)
\(144 + k^2 = 13^2 = 169\)
\(k^2 = 169 - 144 = 25\)
\(k = \pm 5\)

Since we are given that \(k < 0\), we must have \(k = -5\).

M1 for setting up the equation \(12^2 + k^2 = 13^2\) (or equivalent)
A1 for -5 (reject 5 or \(\pm 5\))

Factorise the numerator: $$3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)$$ Factorise the denominator: $$x^2 + x - 6 = (x + 3)(x - 2)$$ Divide the numerator and the denominator by the common factor $(x - 2)$: $$\frac{3(x - 2)(x + 2)}{(x + 3)(x - 2)} = \frac{3(x + 2)}{x + 3}$$

M1 for factorising the numerator to $3(x^2 - 4)$ or $3(x - 2)(x + 2)$. M1 for factorising the denominator to $(x + 3)(x - 2)$. M1 for cancelling the common factor $(x - 2)$. A1 for final answer $\frac{3(x+2)}{x+3}$ or $\frac{3x+6}{x+3}$.

Let the original cost be $C$. Since $1440 is a 20% profit on the original cost: $$1.20 \times C = 1440$$ $$C = \frac{1440}{1.2} = 1200$$ The second owner bought it for $1440 and sold it at a 15% loss: $$\text{Loss} = 1440 \times 0.15 = 216$$ $$\text{Final Selling Price} = 1440 - 216 = 1224$$

M1 for setting up the equation for original cost: $1.20 \times C = 1440$. A1 for finding the original cost: $1200. M1 for calculating the 15% reduction on $1440: $1440 \times 0.85$. A1 for the final selling price: $1224.

The volume of a sphere is given by: $$V_{\text{sphere}} = \frac{4}{3}\pi r^3$$ With $r = 3\text{ cm}$: $$V_{\text{sphere}} = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi \times 27 = 36\pi\text{ cm}^3$$ The volume of a cone is given by: $$V_{\text{cone}} = \frac{1}{3}\pi R^2 h$$ With $R = 2\text{ cm}$: $$V_{\text{cone}} = \frac{1}{3}\pi (2)^2 h = \frac{4}{3}\pi h$$ Since the sphere is melted down and recast into the cone, their volumes are equal: $$36\pi = \frac{4}{3}\pi h \Rightarrow 36 = \frac{4}{3}h \Rightarrow h = 27\text{ cm}$$

M1 for substituting $r = 3$ into the sphere volume formula: $\frac{4}{3}\pi (3)^3$. A1 for volume of sphere $= 36\pi$. M1 for equating their sphere volume to the cone volume formula: $36\pi = \frac{1}{3}\pi (2)^2 h$. A1 for height of cone $= 27$.

We factorise the quadratic equation: $$2x^2 - 10x + x - 5 = 0$$ $$2x(x - 5) + 1(x - 5) = 0$$ $$(2x + 1)(x - 5) = 0$$ This gives two solutions: $$2x + 1 = 0 \Rightarrow x = -0.5$$ $$x - 5 = 0 \Rightarrow x = 5$$

M2 for factorisation $(2x+1)(x-5)$ (or M1 for finding two numbers that multiply to $-10$ and add to $-9$, e.g., $-10$ and $1$). A1 for $x = -0.5$. A1 for $x = 5$.

Multiply both sides by $(2 - 5y)$: $$x(2 - 5y) = 3y + 4$$ $$2x - 5xy = 3y + 4$$ Rearrange to get all terms with $y$ on one side: $$2x - 4 = 3y + 5xy$$ Factorise $y$ out of the right-hand side: $$2x - 4 = y(3 + 5x)$$ Divide by $(3 + 5x)$: $$y = \frac{2x - 4}{5x + 3}$$

M1 for clearing the fraction: $x(2 - 5y) = 3y + 4$. M1 for expanding and grouping $y$ terms on one side: $2x - 4 = 3y + 5xy$. M1 for factorising $y$: $2x - 4 = y(3 + 5x)$. A1 for final answer $y = \frac{2x - 4}{5x + 3}$ or equivalent.

At the end of Year 1: $$\text{Value} = 8000 \times 1.05 = 8400$$ At the end of Year 2: $$\text{Value} = 8400 \times 1.05 = 8400 + 420 = 8820$$ At the end of Year 3: $$\text{Value} = 8820 \times 1.05 = 8820 + 441 = 9261$$

M1 for calculation of value at Year 1: $8000 \times 1.05 = 8400$. M1 for calculation of value at Year 2: $8400 \times 1.05 = 8820$. M1 for calculation of value at Year 3: $8820 \times 1.05$. A2 for final answer $9261$ (A1 for $9261$ seen in working but not as final answer).

First, find the radius $r$ of the cylinder using the volume formula: $$V = \pi r^2 h$$ $$72\pi = \pi r^2 (8)$$ Divide by $8\pi$: $$r^2 = 9 \Rightarrow r = 3\text{ cm}$$ Now, calculate the total surface area ($A$): $$A = 2\pi r^2 + 2\pi r h$$ $$A = 2\pi (3)^2 + 2\pi (3)(8)$$ $$A = 18\pi + 48\pi = 66\pi\text{ cm}^2$$

M1 for setting up volume equation: $72\pi = \pi r^2 (8)$. A1 for finding radius $r = 3$. M1 for substituting $r = 3$ and $h = 8$ into total surface area formula: $2\pi(3)^2 + 2\pi(3)(8)$. M1 for expanding terms to get $18\pi + 48\pi$. A1 for final answer $66\pi$.

First, expand the first two brackets: $$(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$$ Next, multiply this result by the third bracket $(3x - 1)$: $$(2x^2 + 5x - 12)(3x - 1)$$ $$= 3x(2x^2 + 5x - 12) - 1(2x^2 + 5x - 12)$$ $$= 6x^3 + 15x^2 - 36x - 2x^2 - 5x + 12$$ Combine like terms: $$= 6x^3 + 13x^2 - 41x + 12$$

M2 for expanding first two brackets to $2x^2 + 5x - 12$ (M1 for 3 of 4 terms correct: $2x^2$, $8x$, $-3x$, $-12$). M1 for multiplying their quadratic by $(3x - 1)$. M1 for collecting like terms. A1 for final answer $6x^3 + 13x^2 - 41x + 12$.

First, factorise the numerator:
$$3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)$$

Next, factorise the quadratic denominator:
$$2x^2 - x - 6 = (2x + 3)(x - 2)$$

Now, simplify by cancelling the common factor $(x - 2)$:
$$\frac{3(x - 2)(x + 2)}{(2x + 3)(x - 2)} = \frac{3(x + 2)}{2x + 3} = \frac{3x + 6}{2x + 3}$$

Therefore, the fully simplified expression is $\frac{3x + 6}{2x + 3}$.

M1 for $3(x^2 - 4)$ or $3(x - 2)(x + 2)$ seen
M2 for $(2x + 3)(x - 2)$ (or M1 for $(2x + a)(x + b)$ where $ab = -6$ or $2b + a = -1$)
A1 for $\frac{3x+6}{2x+3}$ or $\frac{3(x+2)}{2x+3}$ as final answer

Let $V$ be the initial value of the painting at the start of the first year.

After the first year, the value increases by $20\%$:
$$V_1 = 1.20 \times V$$

After the second year, the value decreases by $15\%$:
$$V_2 = V_1 \times (1 - 0.15) = 1.20V \times 0.85$$

Calculate the combined multiplier:
$$1.20 \times 0.85 = 1.02$$

So,
$$1.02V = 10\,200$$
$$V = \frac{10\,200}{1.02} = 10\,000$$

The initial value was $$10\,000$.

M1 for $V \times 1.2 \times 0.85 = 10\,200$ or equivalent
M1 for $1.2 \times 0.85 = 1.02$ seen or implied
M1 for $V = \frac{10\,200}{1.02}$
A1 for $10\,000$ or $$10\,000$

First, calculate the volume of the cylinder:
$$\text{Volume of cylinder} = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi \text{ cm}^3$$

Next, set this equal to the volume of the sphere to find its radius, $R$:
$$\frac{4}{3}\pi R^3 = 72\pi$$

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

Multiply by $\frac{3}{4}$:
$$R^3 = 72 \times \frac{3}{4} = 18 \times 3 = 54$$

Therefore, the radius of the sphere is:
$$R = \sqrt[3]{54}\text{ cm}$$

M1 for $\pi \times 3^2 \times 8$ or $72\pi$ seen
M1 for equating sphere volume to their cylinder volume: $\frac{4}{3}\pi R^3 = 72\pi$
M1 for $R^3 = 54$
A1 for $\sqrt[3]{54}$ (accept $3\sqrt[3]{2}$)

Multiply the entire equation by the common denominator, $x(x + 1)$:
$$2(x + 1) + 3x = 2x(x + 1)$$

Expand both sides:
$$2x + 2 + 3x = 2x^2 + 2x$$
$$5x + 2 = 2x^2 + 2x$$

Rearrange into a standard quadratic equation form:
$$2x^2 - 3x - 2 = 0$$

Factorise the quadratic expression:
$$(2x + 1)(x - 2) = 0$$

Solve for $x$:
$$2x + 1 = 0 \implies x = -0.5$$
$$x - 2 = 0 \implies x = 2$$

M1 for $2(x+1) + 3x = 2x(x+1)$ or equivalent correct step to eliminate denominators
M1 for rearranging to $2x^2 - 3x - 2 = 0$
M1 for factorising to $(2x + 1)(x - 2) = 0$ or correct use of the quadratic formula on their quadratic
A1 for $x = 2$ and $x = -0.5$ (or $-1/2$)

Multiply both sides by $(5 - 2t)$ to clear the fraction:
$$p(5 - 2t) = 3t + 2$$

Expand the left-hand side:
$$5p - 2pt = 3t + 2$$

Group all terms containing $t$ on one side and constant terms on the other side:
$$5p - 2 = 3t + 2pt$$

Factorise $t$ on the right-hand side:
$$5p - 2 = t(3 + 2p)$$

Divide by $(3 + 2p)$ to solve for $t$:
$$t = \frac{5p - 2}{3 + 2p}$$

M1 for $p(5 - 2t) = 3t + 2$
M1 for isolating terms in $t$ on one side: $5p - 2 = 3t + 2pt$ or $2pt + 3t = 5p - 2$
M1 for factorising out $t$: $t(3 + 2p) = 5p - 2$
A1 for $t = \frac{5p - 2}{3 + 2p}$ or $t = \frac{2 - 5p}{-3 - 2p}$

Find the mass of dry wood (non-water component) in the log, which remains constant:
$$\text{Dry wood mass} = 120\text{ kg} \times (1 - 0.25) = 120 \times 0.75 = 90\text{ kg}$$

After drying, the water content is $10\%$, which means the dry wood represents $90\%$ of the new mass, $M$:
$$0.90 \times M = 90\text{ kg}$$

Solve for the new mass, $M$:
$$M = \frac{90}{0.90} = 100\text{ kg}$$

M1 for calculating the dry wood mass: $120 \times 0.75$
A1 for $90$ [kg]
M1 for setting up the equation for the new mass: $90 \div 0.90$ or $0.90M = 90$
A1 for $100$ [kg]

First, find the slant height, $L$, of the cone using Pythagoras' theorem:
$$L = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$

The curved surface area of the cone is unfolded to form a sector. The radius of this sector is equal to the slant height, $L = 10\text{ cm}$.

The arc length of the sector is equal to the base circumference of the cone:
$$\text{Arc length} = 2 \times \pi \times r_{\text{base}} = 2\pi \times 6 = 12\pi\text{ cm}$$

The formula for the arc length of a sector is:
$$\text{Arc length} = 2\pi L \times \frac{\theta}{360}$$

Substitute the known values:
$$12\pi = 2\pi \times 10 \times \frac{\theta}{360}$$
$$12 = \frac{20\theta}{360}$$
$$12 = \frac{\theta}{18}$$
$$\theta = 12 \times 18 = 216$$

M1 for slant height $L = \sqrt{6^2 + 8^2} = 10$
M1 for finding the base circumference of the cone: $2\pi \times 6 = 12\pi$ (or finding the curved surface area: $\pi \times 6 \times 10 = 60\pi$)
M1 for equating arc length (or sector area) to find $\theta$: $12\pi = 20\pi \times \frac{\theta}{360}$ or $60\pi = 100\pi \times \frac{\theta}{360}$
A1 for $216$ (or $216^\circ$)

Find the actual increase: \(282 - 240 = 42\). Calculate the percentage increase: \(\frac{42}{240} \times 100 = 17.5\%\).

M1 for \(282 - 240\) or \(\frac{282}{240}\) seen oe
A1 for 17.5

Using trigonometry, \(\cos(64^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{d}{4.5}\), where \(d\) is the distance from the wall to the ladder base.
\(d = 4.5 \times \cos(64^\circ) \approx 1.97\) m (to 3 significant figures).

M1 for identifying the correct trigonometric ratio: \(\cos(64^\circ) = \frac{d}{4.5}\)
M1 for \(d = 4.5 \times \cos(64^\circ)\)
A1 for 1.97 or 1.972 to 1.973

Find the highest common factor of \(12ab\) and \(18b^2\), which is \(6b\).
Divide both terms by \(6b\):
\(12ab \div 6b = 2a\)
\(-18b^2 \div 6b = -3b\).
This gives the factorised expression \(6b(2a - 3b)\).

B1 for \(6(2ab - 3b^2)\) or \(b(12a - 18b)\) or \(2b(6a-9b)\) or \(3b(4a-6b)\)
B2 for \(6b(2a - 3b)\) final answer

Add 2 hours to 19:45 to get 21:45.
Add 18 minutes to 21:45: \(45 + 18 = 63\) minutes, which is 1 hour and 3 minutes.
Therefore, the film finishes at 22:03.

M1 for a correct time calculation method, e.g., adding 2 hours or adding 18 minutes, or showing 21:45 or 21:63
A1 for 22:03

The formula for the volume of a cylinder is \(V = \pi r^2 h\).
\(V = \pi \times 3.5^2 \times 8.2\)
\(V = \pi \times 12.25 \times 8.2 \approx 315.57\) \(\text{cm}^3\).
To 3 significant figures, this is 316 \(\text{cm}^3\).

M1 for substituting correctly into the volume formula, \(\pi \times 3.5^2 \times 8.2\)
A1 for 315.5 to 316.0

Expand the brackets: \(8x - 12 = 14\).
Add 12 to both sides: \(8x = 26\).
Divide by 8: \(x = \frac{26}{8} = 3.25\) (or \(\frac{13}{4}\)).

M1 for \(8x - 12 = 14\) or \(2x - 3 = 3.5\) oe
A1 for 3.25 or \(3 \frac{1}{4}\) or \(\frac{13}{4}\)

Total number of parts is \(3 + 5 + 4 = 12\).
Value of one part is \(\frac{360}{12} = 30\).
The largest ratio share corresponds to 5 parts.
Value of the largest share: \(5 \times 30 = 150\).

M1 for finding total parts: \(3 + 5 + 4 = 12\)
M1 for \(360 \div 12 \times 5\)
A1 for 150

Add the temperatures: \((-2) + 3 + (-1) + 5 + 1 = 6\).
Divide by the number of days (5): \(6 \div 5 = 1.2^\circ\text{C}\).

M1 for sum of temperatures divided by 5, e.g. \(\frac{-2+3-1+5+1}{5}\) or showing sum is 6
A1 for 1.2

Let the original price be \( x \). The price is reduced by 15%, so the sale price is 85% of the original price.
\( 0.85x = 238 \)
\( x = \frac{238}{0.85} \)
\( x = 280 \)

M1 for translating the reduction into an equation: \( 0.85x = 238 \) or equivalent
M1 for solving: \( x = \frac{238}{0.85} \)
A1 for 280

The volume \( V \) of a cylinder is given by \( V = \pi r^2 h \).
Substituting the given values:
\( V = \pi \times 5^2 \times 12 \)
\( V = 300\pi \approx 942.47779... \)
Rounding to 1 decimal place gives 942.5 cm\(^3\).

M1 for substituting values into volume formula: \( \pi \times 5^2 \times 12 \)
A1 for 942.5

Expand the bracket first:
\( 8x - 12 = 5x + 9 \)
Subtract \( 5x \) from both sides:
\( 3x - 12 = 9 \)
Add 12 to both sides:
\( 3x = 21 \)
Divide by 3:
\( x = 7 \)

M1 for expansion of bracket: \( 8x - 12 \)
M1 for isolating terms: \( 3x = 21 \) or equivalent
A1 for 7

Find the highest common factor of \( 12a^2b \) and \( 18ab^2 \).
The highest common factor of 12 and 18 is 6.
The common variables are \( a \) and \( b \).
Thus, the HCF is \( 6ab \).
Divide both terms by \( 6ab \):
\( 12a^2b \div 6ab = 2a \)
\( 18ab^2 \div 6ab = 3b \)
Factored form: \( 6ab(2a - 3b) \)

B1 for partial factorisation, e.g. \( 3ab(4a - 6b) \) or \( 6(2a^2b - 3ab^2) \)
B2 for fully correct answer: \( 6ab(2a - 3b) \)

Add all the temperatures together:
\( -3 + 1 + (-2) + 4 + 5 = 5 \)
Divide the sum by the number of days (5):
\( \text{Mean} = \frac{5}{5} = 1^\circ\text{C} \)

M1 for the sum of the temperatures divided by 5: \( \frac{-3 + 1 - 2 + 4 + 5}{5} \)
A1 for 1

Using the cosine ratio:
\( \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)
\( \cos(\theta) = \frac{12}{13} \)
\( \theta = \arccos\left(\frac{12}{13}\right) \approx 22.61986^\circ \)
To the nearest degree, \( \theta = 23^\circ \).

M1 for setting up the correct trig ratio: \( \cos(\theta) = \frac{12}{13} \) or equivalent
A1 for 23

Move the decimal point 4 places to the right to get a number between 1 and 10:
\( 0.00045 = 4.5 \times 10^{-4} \)

B1 for \( 4.5 \times 10^{-4} \) (accept equivalent standard form notation)

Total number of balls = \( 5 + 3 + 2 = 10 \).
Number of balls that are not blue (red or green) = \( 5 + 2 = 7 \).
Probability of not choosing a blue ball = \( \frac{7}{10} = 0.7 \).

M1 for finding the number of non-blue balls over the total number of balls: \( \frac{7}{10} \) or equivalent fraction
A1 for 0.7 or \( \frac{7}{10} \) or 70%

Let the original price be \(x\). The price after a 15% reduction is 85% of the original price. Thus, \(0.85x = 561\), which gives \(x = 561 / 0.85 = 660\).

M1 for \(100 - 15 = 85\)% or equivalent. M1 for \(\frac{561}{0.85}\). A1 for 660.

The volume of a cylinder is given by \(V = \pi r^2 h\). Substituting the values, we get \(V = \pi \times 4.5^2 \times 12 = 243\pi \approx 763.407\) cm\(^3\). Correct to 3 significant figures, this is 763.

M1 for \(\pi \times 4.5^2 \times 12\). A1 for \(763.4...\) or \(243\pi\). A1 for 763.

Expand the brackets: \(5x - 15 = 2x + 9\). Rearrange the equation to collect like terms: \(5x - 2x = 9 + 15\), which simplifies to \(3x = 24\). Dividing both sides by 3 gives \(x = 8\).

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

Expanding the expression gives \(2x \times x + 2x \times 5 - 3 \times x - 3 \times 5 = 2x^2 + 10x - 3x - 15\). Combining the like terms yields \(2x^2 + 7x - 15\).

M1 for any 3 correct terms out of 4 from \(2x^2 + 10x - 3x - 15\). A1 for \(2x^2 + 7x - 15\).

From 08 45 to 12 45 is 4 hours. From 12 45 to 13 12 is 27 minutes. Therefore, the total journey duration is 4 hours and 27 minutes.

B1 for 4 hours. B1 for 27 minutes.

The total number of parts is \(3 + 5 + 4 = 12\). The value of each part is \(360 / 12 = 30\). The largest share corresponds to 5 parts, so the value is \(5 \times 30 = 150\).

M1 for adding parts to get 12. M1 for \(\frac{360}{12} \times 5\) or equivalent. A1 for 150.

Using Pythagoras' theorem: \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse. We have \(9^2 + b^2 = 15^2\), which simplifies to \(81 + b^2 = 225\). Thus, \(b^2 = 225 - 81 = 144\), giving \(b = \sqrt{144} = 12\).

M1 for \(15^2 - 9^2\) or \(225 - 81\). A1 for 12.

The numerator is \(18.45 + \sqrt{27.8} \approx 18.45 + 5.27257 = 23.72257\). The denominator is \(3.1^2 - 1.4 = 9.61 - 1.4 = 8.21\). Calculating the fraction gives \(\frac{23.72257}{8.21} \approx 2.88947\). Correct to 2 decimal places, this is 2.89.

M1 for numerator \(23.7...\) or denominator \(8.21\) seen. A1 for \(2.889...\). A1 for 2.89.

Increase = $376 - $320 = $56. The percentage increase is calculated as: \\frac{56}{320} \\times 100 = 17.5\%.

M1 for \\frac{376 - 320}{320} \\times 100 or \\frac{56}{320}
A1 for 17.5

Find the highest common factor of both terms: \(6xy\). Divide each term by the common factor: \(12x^2y \div 6xy = 2x\) and \(18xy^2 \div 6xy = 3y\). Thus, the factorised expression is \(6xy(2x - 3y)\).

B1 for any correct partial factorisation (e.g. \(6x(2xy - 3y^2)\) or \(xy(12x - 18y)\))
B1 for \(6xy(2x - 3y)\)

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substituting the values: \(V = \pi \times 4.5^2 \times 14 = 283.5\pi \approx 890.64\text{ cm}^3\). Rounding to 3 significant figures gives 891.

M1 for \\pi \\times 4.5^2 \\times 14
A1 for 891 or 890.6 to 891

Multiply both sides of the equation by 4: \(3x - 5 = 28\). Add 5 to both sides: \(3x = 33\). Divide both sides by 3: \(x = 11\).

M1 for multiplying by 4: \(3x - 5 = 28\)
M1 for isolating the x term: \(3x = 33\)
A1 for 11

By Pythagoras' theorem, \(a^2 + b^2 = c^2\). Substituting the values: \(7.2^2 + 9.6^2 = c^2\), which gives \(51.84 + 92.16 = c^2\), so \(144 = c^2\). Taking the square root gives \(c = 12\text{ cm}\).

M1 for \(7.2^2 + 9.6^2\)
M1 for \(\sqrt{51.84 + 92.16}\)
A1 for 12

The sum of all probabilities in a single event is 1. Therefore, \(P(\text{yellow}) = 1 - (0.35 + 0.20 + 0.15) = 1 - 0.70 = 0.30\).

M1 for \(1 - (0.35 + 0.20 + 0.15)\
A1 for 0.3 or 0.30