2025 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

Find the highest common factor of \(18\) and \(12\), which is \(6\).
Find the highest common factor of \(x^2\) and \(x\), which is \(x\).
Find the highest common factor of \(y\) and \(y^2\), which is \(y\).

The highest common factor of the terms is \(6xy\).
Divide both terms by \(6xy\):
\(18x^2y \div 6xy = 3x\)
\(-12xy^2 \div 6xy = -2y\)

So, the factorised expression is \(6xy(3x - 2y)\).

M1 for finding any correct common factor outside the bracket, e.g., \(2y(9x^2 - 6xy)\)
M1 for a further step towards full factorisation, e.g., \(6xy(3x - 2y)\) with one minor error
A1 for \(6xy(3x-2y)\)

Find the profit:
\(\text{Profit} = \$116 - \$80 = \$36\)

Calculate the percentage profit:
\(\text{Percentage Profit} = \frac{36}{80} \times 100\)
\(\frac{36}{80} = \frac{9}{20}\)
\(\frac{9}{20} \times 100 = 9 \times 5 = 45\%\)

M1 for finding the profit: \(116 - 80 = 36\)
M1 for \(\frac{\text{their } 36}{80} \times 100\) oe
A1 for \(45\)

First, find the size of one exterior angle of the regular polygon:
\(\text{Exterior angle} = 180^\circ - 144^\circ = 36^\circ\)

The sum of the exterior angles of any convex polygon is \(360^\circ\).
So, the number of sides, \(n\), is:
\(n = \frac{360^\circ}{36^\circ} = 10\)

M1 for \(180 - 144\) or for showing that \(\text{exterior angle} = 36\)
M1 for \(\frac{360}{\text{their exterior angle}}\) or for \((n-2) \times 180 = 144n\) oe
A1 for \(10\)

Multiply each term of the equation by the lowest common multiple of \(4\) and \(3\), which is \(12\):
\(12 \times \frac{3x - 1}{4} - 12 \times \frac{x + 2}{3} = 12 \times 2\)
\(3(3x - 1) - 4(x + 2) = 24\)

Expand the brackets:
\(9x - 3 - 4x - 8 = 24\)

Simplify the left side:
\(5x - 11 = 24\)

Add \(11\) to both sides:
\(5x = 35\)

Divide by \(5\):
\(x = 7\)

M1 for multiplying by a common multiple (usually 12) correctly to clear fractions, e.g., \(3(3x - 1) - 4(x + 2) = 24\)
M1 for correct expansion of brackets and simplification to the form \(ax = b\), e.g., \(5x = 35\) or \(5x - 11 = 24\)
A1 for \(7\)

Find the difference between consecutive terms:
\(11 - 4 = 7\)
\(18 - 11 = 7\)
The sequence is increasing by \(7\) each time, so the formula starts with \(7n\).

Compare \(7n\) to the terms in the sequence:
For \(n = 1\): \(7(1) = 7\), but the first term is \(4\). We need to subtract \(3\).
Let's check for \(n = 2\): \(7(2) - 3 = 14 - 3 = 11\), which is correct.
So the \(nth\) term of the sequence is \(7n - 3\).

B1 for \(7n + k\) where \(k\) is any constant (including 0)
B1 for \(kn - 3\) where \(k \neq 0\)
A1 for \(7n - 3\) as final answer

Let the total number of marbles in the bag be \(T\).
We are given that the probability of choosing a red marble is \(\frac{3}{8}\).
Since there are \(24\) red marbles:
\(\frac{24}{T} = \frac{3}{8}\)

Cross-multiply to solve for \(T\):
\(3T = 24 \times 8\)
\(3T = 192\)
\(T = 64\)

So the total number of marbles is \(64\).
The number of green marbles is the total minus the red and blue marbles:
\(\text{Green marbles} = 64 - 24 - 16 = 24\).

M1 for setting up the equation \(\frac{24}{T} = \frac{3}{8}\) or showing that total is \(24 \div \frac{3}{8}\) oe
M1 for finding the total number of marbles is \(64\)
A1 for \(24\)

The formula for the area of a trapezium is:
\(\text{Area} = \frac{1}{2}(a + b)h\)
where \(a\) and \(b\) are the lengths of the parallel sides, and \(h\) is the perpendicular height.

Substitute the given values into the formula:
\(70 = \frac{1}{2}(8 + 12)h\)
\(70 = \frac{1}{2}(20)h\)
\(70 = 10h\)

Solve for \(h\):
\(h = 7\)

M1 for correct substitution into area of trapezium formula, e.g., \(70 = \frac{1}{2}(8+12)h\) oe
M1 for simplifying to a linear equation in \(h\), e.g., \(10h = 70\) or \(20h = 140\)
A1 for \(7\)

Identify the difference in ratio parts between Chloe and Alex:
\(\text{Chloe's parts} - \text{Alex's parts} = 7 - 3 = 4 \text{ parts}\)

We are given that Chloe receives $36 more than Alex, so:
\(4 \text{ parts} = \$36\)
\(1 \text{ part} = \frac{36}{4} = \$9\)

Find the total number of parts shared:
\(\text{Total parts} = 3 + 5 + 7 = 15 \text{ parts}\)

Calculate the total money:
\(\text{Total money} = 15 \times 9 = \$135\).

M1 for subtracting parts: \(7 - 3 = 4\) parts associated with \(36\) or setting up equivalent equations, e.g., \(7x - 3x = 36\)
M1 for finding the value of one part is \(9\) (or total parts = 15)
A1 for \(135\)

To buy 8 notebooks under the offer, we can group them into two sets of 4 notebooks.

For each set of 4 notebooks:
- 3 notebooks are at full price: \(3 \times \$2.40 = \$7.20\)
- 1 notebook is at half price: \(\$2.40 \div 2 = \$1.20\)
- Total for 4 notebooks: \(\$7.20 + \$1.20 = \$8.40\)

For 8 notebooks, we buy 2 of these sets:
\(2 \times \$8.40 = \$16.80\).

M1 for finding the cost of 4 notebooks under the offer: \(3 \times 2.40 + 1.20\) or showing \(7.20 + 1.20\)
M1 for multiplying their cost of 4 notebooks by 2, or writing a complete expression: \(2 \times (3 \times 2.40 + 1.20)\)
A1 for 16.80 (or 16.8)

Expand the brackets first:
\(5(2x - 3y) = 10x - 15y\)
\(-3(x - 4y) = -3x + 12y\)

Now combine like terms:
\((10x - 3x) + (-15y + 12y)\)
\(= 7x - 3y\).

M1 for correct expansion of the first bracket: \(10x - 15y\)
M1 for correct expansion of the second bracket: \(-3x + 12y\) (watch for the sign)
A1 for \(7x - 3y\)

There are two possible cases for the angles in the isosceles triangle:

Case 1: The \(40^\circ\) angle is the unique angle. The other two angles are equal, say \(x\).
\(2x + 40 = 180\)
\(2x = 140\)
\(x = 70^\circ\)
The angles are \(40^\circ\), \(70^\circ\), and \(70^\circ\). The largest angle is \(70^\circ\).

Case 2: The \(40^\circ\) angle is one of the equal angles. The other equal angle is also \(40^\circ\). Let the third angle be \(y\).
\(40 + 40 + y = 180\)
\(80 + y = 180\)
\(y = 100^\circ\)
The angles are \(40^\circ\), \(40^\circ\), and \(100^\circ\). The largest angle is \(100^\circ\).

Thus, the two possible values for the largest angle are \(70\) and \(100\).

M1 for finding the equal angles in Case 1: \((180 - 40) \div 2\) or showing 70
M1 for finding the third angle in Case 2: \(180 - 2 \times 40\) or showing 100
A1 for both 70 and 100

The ratio of Milk : Dark : White is \(5 : 3 : 2\).

The number of parts for dark chocolate is 3.
If 3 parts correspond to 12 chocolates:
\(1 \text{ part} = 12 \div 3 = 4 \text{ chocolates}\).

The total number of parts is:
\(5 + 3 + 2 = 10 \text{ parts}\).

The total number of chocolates is:
\(10 \times 4 = 40\).

M1 for dividing 12 by 3 to find the value of 1 part
M1 for multiplying the total parts \((5 + 3 + 2)\) by their value of 1 part
A1 for 40

To find the duration from 08:45 to 11:12:

1. From 08:45 to 09:00 is 15 minutes.
2. From 09:00 to 11:00 is 2 hours.
3. From 11:00 to 11:12 is 12 minutes.

Total duration:
\(2 \text{ hours} + 15 \text{ minutes} + 12 \text{ minutes} = 2 \text{ hours } 27 \text{ minutes}\).

M1 for a correct method to calculate the minutes to the next hour (e.g., 15 mins) or expressing the subtraction as \(10:72 - 08:45\)
M1 for summing the hours and minutes correctly
A1 for 2 hours 27 minutes (accept 2h 27m or 147 minutes with M2 scored)

Multiply both sides by 2:
\(3x - 5 = 16\)

Add 5 to both sides:
\(3x = 21\)

Divide by 3:
\(x = 7\).

M1 for multiplying both sides by 2 to get \(3x - 5 = 16\)
M1 for isolating the term with \(x\) to get \(3x = 21\)
A1 for 7

Find the highest common factor (HCF) of the numerical coefficients:
\(\text{HCF of } 12 \text{ and } 18 \text{ is } 6\).

Find the common variables with their lowest power:
- For \(a^2\) and \(a\), the common factor is \(a\).
- For \(b\) and \(b^2\), the common factor is \(b\).

Thus, the HCF is \(6ab\).

Factor out \(6ab\):
\(12a^2b - 18ab^2 = 6ab(2a - 3b)\).

M2 for \(6ab(2a - 3b)\)
or B1 for any correct partial factorisation, e.g., \(2ab(6a - 9b)\) or \(3ab(4a - 6b)\) or \(6a(2ab - 3b^2)\)
A1 for \(6ab(2a - 3b)\) as final answer

The perimeter of a rectangle is given by:
\(P = 2(\text{length} + \text{width})\)

Substitute the given values:
\(38 = 2(\text{length} + 7)\)

Divide by 2:
\(19 = \text{length} + 7\)

Subtract 7:
\(\text{length} = 12 \text{ m}\).

Now, calculate the area:
\(\text{Area} = \text{length} \times \text{width}\)
\(\text{Area} = 12 \times 7 = 84 \text{ m}^2\).

M1 for setting up the equation to find the length: \(2 \times 7 + 2L = 38\) or \(38 \div 2 - 7\)
M1 for finding the length = 12
A1 for 84 (accept 84 m\(^2\))

Expand the brackets: \( 4(2x - 3y) = 8x - 12y \) and \( -3(x - 5y) = -3x + 15y \). Combining like terms: \( 8x - 3x - 12y + 15y = 5x + 3y \).

M1 for \( 8x - 12y \) or \( -3x + 15y \) M1 for combining like terms correctly for their expansion A1 for \( 5x + 3y \)

The machine prints \( 150 \div 4 = 37.5 \) flyers per minute. In 10 minutes, it prints \( 37.5 \times 10 = 375 \) flyers.

M1 for \( 150 \div 4 \) oe M1 for \( \text{their } 37.5 \times 10 \) oe A1 for 375

Expand the bracket: \( 7x - 4 = 2x + 16 \). Subtract \( 2x \) from both sides: \( 5x - 4 = 16 \). Add 4 to both sides: \( 5x = 20 \). Divide by 5: \( x = 4 \).

M1 for \( 2x + 16 \) M1 for isolating the \( x \) terms on one side and constant terms on the other A1 for 4

The common difference is \( -4 \). The term before the first term (for \( n=0 \)) is \( 19 + 4 = 23 \). Therefore, the \(n\)th term is \( 23 - 4n \).

B1 for \( -4n + k \) or \( jn + 23 \) (\(j \neq 0\)) M1 for a complete correct method to find the term A1 for \( 23 - 4n \) or equivalent

The sum of the interior angles of a triangle is \( 180^\circ \). So, \( x + (2x - 10) + 70 = 180 \). Simplifying this gives \( 3x + 60 = 180 \), which means \( 3x = 120 \) and \( x = 40 \).

M1 for \( x + 2x - 10 + 70 = 180 \) oe M1 for simplifying to \( 3x = k \) oe A1 for 40

Using a common denominator of 24: \( \frac{5}{6} = \frac{20}{24} \) and \( \frac{3}{8} = \frac{9}{24} \). Then, \( \frac{20}{24} - \frac{9}{24} = \frac{11}{24} \).

M1 for finding a common denominator (e.g., 24) M1 for converting at least one fraction correctly A1 for \( \frac{11}{24} \)

Total parts = \( 2 + 5 + 7 = 14 \). The value of one part is \( 420 \div 14 = 30 \). The largest share corresponds to 7 parts: \( 7 \times 30 = 210 \).

M1 for finding the total parts: \( 2 + 5 + 7 = 14 \) M1 for \( 420 \div 14 \times 7 \) oe A1 for 210

Perimeter of the rectangle is \( 2 \times (12 + 8) = 40 \) cm. The square has perimeter 40 cm, so each side is \( 40 \div 4 = 10 \) cm. The area of the square is \( 10 \times 10 = 100 \) cm\(^2\).

M1 for perimeter of rectangle = 40 M1 for side of square = 10 A1 for 100

Let the original price of the coat be \(x\).
An increase of 15% means the new price is 115% of the original price.
\(1.15x = 92\)
\(x = \frac{92}{1.15}\)
\(x = \frac{9200}{115}\)
Dividing both the numerator and the denominator by 5:
\(x = \frac{1840}{23}\)
Since \(23 \times 8 = 184\), we have:
\(x = 80\)

The price of the coat before the increase was $80.

M2 for \(92 \div 1.15\) oe
or M1 for \(115\% = 92\) oe
A1 for 80

To find the first term greater than 100, set up an inequality:
\(5n - 3 > 100\)
\(5n > 103\)
\(n > 20.6\)

Since the term position \(n\) must be an integer, the first term greater than 100 occurs at \(n = 21\).

Substitute \(n = 21\) into the formula for the \(n\)th term:
\(5(21) - 3 = 105 - 3 = 102\).

M1 for setting up the inequality \(5n - 3 > 100\) oe
M1 for determining that \(n = 21\) or listing terms showing the 20th term is 97 and 21st term is 102
A1 for 102

Method 1:
Find the exterior angle first.
The exterior angle of a regular polygon with \(n\) sides is \(\frac{360^\circ}{n}\).
For a 12-sided regular polygon:
Exterior angle = \(\frac{360^\circ}{12} = 30^\circ\).

Since the interior and exterior angles on a straight line sum to \(180^\circ\):
Interior angle = \(180^\circ - 30^\circ = 150^\circ\).

Method 2:
Use the sum of the interior angles formula, which is \((n - 2) \times 180^\circ\).
For a 12-sided regular polygon:
Sum of interior angles = \((12 - 2) \times 180^\circ = 10 \times 180^\circ = 1800^\circ\).

Since all interior angles in a regular polygon are equal, divide by the number of sides:
Interior angle = \(\frac{1800^\circ}{12} = 150^\circ\).

M1 for \(360 \div 12\) or \((12 - 2) \times 180\) oe
M1 for \(180 - \text{their exterior angle}\) or \(\text{their sum} \div 12\) oe
A1 for 150

First, factorise the quadratic numerator:
$$3x^2 - 14x - 5 = 3x^2 - 15x + x - 5 = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5)$$

Next, factorise the denominator using the difference of two squares:
$$9x^2 - 1 = (3x - 1)(3x + 1)$$

Substitute these back into the fraction:
$$\frac{(3x + 1)(x - 5)}{(3x - 1)(3x + 1)}$$

Cancel the common factor of $(3x + 1)$ from the numerator and the denominator:
$$\frac{x - 5}{3x - 1}$$

M1 for $(3x + 1)(x - 5)$
M1 for $(3x - 1)(3x + 1)$
A2 for $\frac{x - 5}{3x - 1}$ (A1 if one sign error)

Substitute the coordinates of the first point $(1, 5)$ into the equation:
$$5 = a(1)^2 + \frac{b}{1} \implies a + b = 5 \quad [1]$$

Substitute the coordinates of the second point $(2, 13)$ into the equation:
$$13 = a(2)^2 + \frac{b}{2} \implies 4a + \frac{b}{2} = 13 \implies 8a + b = 26 \quad [2]$$

Subtract equation [1] from equation [2]:
$$(8a + b) - (a + b) = 26 - 5$$
$$7a = 21 \implies a = 3$$

Substitute $a = 3$ back into equation [1]:
$$3 + b = 5 \implies b = 2$$

Thus, $a = 3$ and $b = 2$.

M1 for substitute $(1, 5)$ to get $a + b = 5$ or substitute $(2, 13)$ to get $4a + 0.5b = 13$
M1 for a complete correct method to solve their simultaneous equations to find $a$ or $b$
A1 for $a = 3, b = 2$

The sum of the interior angle and exterior angle of any polygon is $180^\circ$.
$$\text{Exterior angle} = 180^\circ - 162^\circ = 18^\circ$$

The sum of the exterior angles of any regular polygon is $360^\circ$.
$$\text{Number of sides, } n = \frac{360^\circ}{18^\circ} = 20$$

M1 for $180 - 162$
M1 for $\frac{360}{\text{their } 18}$
A1 for 20

Multiply the first equation by $2$ and the second equation by $3$ to align the coefficients of $y$:
$$8x - 6y = 34$$
$$9x + 6y = 51$$

Add the two equations together:
$$(8x + 9x) + (-6y + 6y) = 34 + 51$$
$$17x = 85$$
$$x = 5$$

Substitute $x = 5$ back into the second equation:
$$3(5) + 2y = 17$$
$$15 + 2y = 17$$
$$2y = 2$$
$$y = 1$$

So, the solutions are $x = 5$ and $y = 1$.

M1 for correct method to equate coefficients of $x$ or $y$
M1 for adding or subtracting equations to eliminate one variable
A1 for $x = 5$
A1 for $y = 1$

Convert the sequence into decimal values:
$$-0.5, \quad 1, \quad 3.5, \quad 7, \quad 11.5$$

Find the first differences between consecutive terms:
$$1.5, \quad 2.5, \quad 3.5, \quad 4.5$$

Find the second differences:
$$1, \quad 1, \quad 1$$

Since the second differences are constant and equal to $1$, the sequence is quadratic with leading term $an^2$ where:
$$2a = 1 \implies a = 0.5$$

Subtract $0.5n^2$ from each term in the sequence to find the remaining linear/constant part:
- For $n = 1$: $-0.5 - 0.5(1)^2 = -1$
- For $n = 2$: $1 - 0.5(2)^2 = -1$
- For $n = 3$: $3.5 - 0.5(3)^2 = -1$
- For $n = 4$: $7 - 0.5(4)^2 = -1$
- For $n = 5$: $11.5 - 0.5(5)^2 = -1$

The remaining part is constant and equal to $-1$. Therefore, the $n\text{th}$ term is:
$$\frac{1}{2}n^2 - 1$$

M1 for finding first and second differences
M1 for identifying the $an^2$ term coefficient is $0.5$ (or $\frac{1}{2}$)
A1 for $\frac{1}{2}n^2 - 1$ or $\frac{n^2 - 2}{2}$

Multiply both sides of the equation by $5$:
$$5y = 2\sqrt{x} - 3$$

Add $3$ to both sides:
$$5y + 3 = 2\sqrt{x}$$

Divide both sides by $2$:
$$\sqrt{x} = \frac{5y + 3}{2}$$

Square both sides of the equation to make $x$ the subject:
$$x = \left(\frac{5y + 3}{2}\right)^2 \quad \text{or} \quad x = \frac{(5y + 3)^2}{4}$$

M1 for multiplying by $5$ and adding $3$ ($5y + 3 = 2\sqrt{x}$)
M1 for isolating $\sqrt{x}$ ($\sqrt{x} = \frac{5y + 3}{2}$)
A1 for $x = \frac{(5y+3)^2}{4}$ or $x = \left(\frac{5y+3}{2}\right)^2$

Add $\frac{2}{x + 4}$ to both sides of the equation:
$$\frac{3}{2x - 1} = \frac{2}{x + 4}$$

Cross-multiply to clear the denominators:
$$3(x + 4) = 2(2x - 1)$$

Expand the brackets on both sides:
$$3x + 12 = 4x - 2$$

Rearrange to solve for $x$:
$$12 + 2 = 4x - 3x$$
$$x = 14$$

M1 for writing as $\frac{3}{2x-1} = \frac{2}{x+4}$ or putting over a common denominator
M1 for cross-multiplying: $3(x + 4) = 2(2x - 1)$
M1 for correct expansion of brackets: $3x + 12 = 4x - 2$
A1 for $x = 14$

Opposite angles in a cyclic quadrilateral sum to $180^\circ$.

For opposite angles $A$ and $C$:
$$\text{Angle } A + \text{Angle } C = 180^\circ$$
$$(2x + 15) + (3x - 10) = 180$$
$$5x + 5 = 180$$
$$5x = 175 \implies x = 35$$

For opposite angles $B$ and $D$:
$$\text{Angle } B + \text{Angle } D = 180^\circ$$
$$(y + 20) + (2y - 5) = 180$$
$$3y + 15 = 180$$
$$3y = 165 \implies y = 55$$

Therefore, $x = 35$ and $y = 55$.

M1 for stating that opposite angles sum to $180^\circ$ (e.g., $(2x + 15) + (3x - 10) = 180$)
M1 for solving for $x$: $5x + 5 = 180 \implies x = 35$
M1 for setting up the equation for $y$: $(y + 20) + (2y - 5) = 180$
A1 for $x = 35, y = 55$

To simplify the algebraic fraction, factorise the numerator and the denominator:

Numerator:
$$3x^2 - 14x - 5 = 3x^2 - 15x + x - 5 = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5)$$

Denominator:
$$x^2 - 25 = (x + 5)(x - 5)$$

Now rewrite the fraction with factorised forms and cancel the common factor $x - 5$:
$$\frac{(3x + 1)(x - 5)}{(x + 5)(x - 5)} = \frac{3x + 1}{x + 5}$$

M1 for factorising the numerator: $(3x + 1)(x - 5)$
M1 for factorising the denominator: $(x + 5)(x - 5)$
A1 for final answer: $\frac{3x + 1}{x + 5}$

1. Find the gradient of the given line $3y + 2x = 12$:
$$3y = -2x + 12 \implies y = -\frac{2}{3}x + 4$$
The gradient of the given line is $-\frac{2}{3}$.

2. The gradient $m$ of the perpendicular line is:
$$m = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}$$

3. Find the equation of the line using the point $(4, -3)$:
$$y - (-3) = \frac{3}{2}(x - 4)$$
$$y + 3 = \frac{3}{2}x - 6$$
$$y = \frac{3}{2}x - 9$$

M1 for perpendicular gradient $= \frac{3}{2}$ oe
M1 for substituting $(4, -3)$ into $y = (\text{their } m)x + c$ or $y - y_1 = m(x - x_1)$
A1 for $y = \frac{3}{2}x - 9$ (or $y = 1.5x - 9$)

Each exterior angle of the regular polygon is:
$$180^\circ - 162^\circ = 18^\circ$$

The sum of the exterior angles of any polygon is $360^\circ$. Therefore, the number of sides $n$ is:
$$n = \frac{360^\circ}{18^\circ} = 20$$

Alternatively, using the formula for the interior angle:
$$\frac{(n-2) \times 180}{n} = 162$$
$$180n - 360 = 162n$$
$$18n = 360$$
$$n = 20$$

M1 for finding the exterior angle: $180 - 162 = 18$
M1 for calculating the number of sides: $\frac{360}{18}$
A1 for 20

Equate the two expressions for $y$:
$$x^2 - 3x - 10 = 2x + 4$$

Rearrange to form a quadratic equation:
$$x^2 - 5x - 14 = 0$$

Factorise the quadratic:
$$(x - 7)(x + 2) = 0$$

This gives $x = 7$ or $x = -2$.

Find the corresponding $y$-values using $y = 2x + 4$:
- For $x = 7$, $y = 2(7) + 4 = 18$
- For $x = -2$, $y = 2(-2) + 4 = 0$

So the solutions are $x = 7, y = 18$ and $x = -2, y = 0$.

M1 for setting up the quadratic equation $x^2 - 3x - 10 = 2x + 4$ oe
M1 for correctly factorising or solving their three-term quadratic: $(x-7)(x+2) = 0$
A1 for both $x = 7$ and $x = -2$
A1 for both $y = 18$ and $y = 0$ corresponding to their correct $x$-values

Find the differences between terms:
First differences: $11 - 3 = 8$, $23 - 11 = 12$, $39 - 23 = 16$
Second differences: $12 - 8 = 4$, $16 - 12 = 4$

Since the second difference is constant and equals $4$, the sequence is quadratic and the coefficient of $n^2$ is $\frac{4}{2} = 2$. Therefore, the quadratic term is $2n^2$.

Subtract $2n^2$ from the original sequence to find the linear remainder:
- For $n = 1$: $3 - 2(1)^2 = 1$
- For $n = 2$: $11 - 2(2)^2 = 3$
- For $n = 3$: $23 - 2(3)^2 = 5$
- For $n = 4$: $39 - 2(4)^2 = 7$

The linear sequence is $1, 3, 5, 7, \dots$, which has a common difference of $2$ and a first term of $1$.
Its $n$th term is $2n - 1$.

Combine the quadratic term and linear expression to get the final $n$th term:
$$2n^2 + 2n - 1$$

M1 for finding second difference of 4 and setting the first term coefficient of $n^2$ to 2 (i.e. $2n^2$)
M1 for subtracting $2n^2$ from sequence terms to obtain the linear sequence $1, 3, 5, 7, \dots$ or finding its $n$th term as $2n - 1$
A1 for $2n^2 + 2n - 1$ as the final answer

The ratio of the volumes of the similar containers is:
$$\frac{V_{\text{large}}}{V_{\text{small}}} = \frac{810}{240} = \frac{27}{8}$$

Since the containers are mathematically similar, the linear scale factor $k$ is:
$$k = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}$$

The area scale factor is $k^2$:
$$k^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$

To find the surface area of the larger container, multiply the surface area of the smaller container by the area scale factor:
$$\text{Surface Area}_{\text{large}} = 160 \times \frac{9}{4} = 40 \times 9 = 360\text{ cm}^2$$

M1 for finding the volume scale factor ratio $\frac{27}{8}$ or linear scale factor ratio $\frac{3}{2}$ oe
M1 for using the area scale factor ratio $\frac{9}{4}$ or equivalent: $160 \times \left(\frac{3}{2}\right)^2$
A1 for 360

The total surface area of a solid hemisphere consists of the curved surface area plus the flat base circular area:
$$\text{Curved Surface Area} = 2\pi r^2$$
$$\text{Base Area} = \pi r^2$$
$$\text{Total Surface Area} = 2\pi r^2 + \pi r^2 = 3\pi r^2$$

Given the radius $r = 6\text{ cm}$:
$$\text{Total Surface Area} = 3\pi(6)^2 = 3\pi \times 36 = 108\pi\text{ cm}^2$$

M1 for curved surface area of hemisphere $= 2\pi(6)^2$ or flat circular base area $= \pi(6)^2$
M1 for calculating $3\pi(6)^2$ oe
A1 for $108\pi$ as final answer

Method 1: By completing the square:
$$y = x^2 - 8x + 15$$
$$y = (x - 4)^2 - 4^2 + 15$$
$$y = (x - 4)^2 - 16 + 15$$
$$y = (x - 4)^2 - 1$$

Since the minimum value of $(x-4)^2$ is $0$ at $x = 4$, the minimum value of $y$ is $-1$.
Thus, the coordinates of the turning point (minimum point) are $(4, -1)$.

Method 2: By finding the line of symmetry:
The $x$-coordinate of the turning point is given by:
$$x = -\frac{b}{2a} = -\frac{-8}{2(1)} = 4$$

Substitute $x = 4$ into the quadratic equation to find the $y$-coordinate:
$$y = (4)^2 - 8(4) + 15 = 16 - 32 + 15 = -1$$

So, the turning point is at $(4, -1)$.

M1 for completing the square to get $(x - 4)^2 + c$ or for finding the $x$-coordinate of the turning point as $x = 4$ using $x = -\frac{b}{2a}$
M1 for substituting their $x$-value back into $y = x^2 - 8x + 15$ to solve for $y$ or identifying $y = -1$ from completed square form
A1 for $(4, -1)$ as a coordinate pair

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

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

The sum of the interior angles of a pentagon is:\
\( (5 - 2) \times 180^\circ = 540^\circ \)\
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Sum of the given angles:\
\( x + (2x - 15) + (x + 30) + (2x + 25) + 140 = 540 \)\
\( 6x + 180 = 540 \)\
\( 6x = 360 \)\
\( x = 60 \)

M1 for sum of interior angles is \( 540^\circ \)\
M1 for setting up the equation \( 6x + 180 = 540 \) (or equivalent)\
A1 for \( x = 60 \)

Find the first differences:\
6, 8, 10, 12...\
Find the second differences:\
2, 2, 2...\
Since the second difference is constant, the sequence is quadratic with an \( n^2 \) term (since \( \frac{2}{2} = 1 \)).\
Subtract \( n^2 \) from the sequence:\
\( 3 - 1 = 2 \)\
\( 9 - 4 = 5 \)\
\( 17 - 9 = 8 \)\
\( 27 - 16 = 11 \)\
The remaining sequence is 2, 5, 8, 11... which is a linear sequence with the \( n \)th term \( 3n - 1 \).\
Therefore, the overall \( n \)th term is \( n^2 + 3n - 1 \).

M1 for finding second differences and identifying the term \( n^2 \)\
M1 for subtracting \( n^2 \) and finding the linear sequence term \( 3n - 1 \)\
A1 for \( n^2 + 3n - 1 \)

Equate the two expressions for \( y \):\
\( x^2 - x - 1 = 2x + 3 \)\
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Rearrange into a quadratic equation:\
\( x^2 - 3x - 4 = 0 \)\
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Factorise the quadratic:\
\( (x - 4)(x + 1) = 0 \)\
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So, \( x = 4 \) or \( x = -1 \).\
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Substitute \( x \) back into the linear equation to find \( y \):\
For \( x = 4 \): \( y = 2(4) + 3 = 11 \)\
For \( x = -1 \): \( y = 2(-1) + 3 = 1 \)

M1 for equating expressions to get \( x^2 - x - 1 = 2x + 3 \)\
M1 for getting a quadratic equation and solving it (e.g. \( x = 4, x = -1 \))\
A1 for finding one correct pair of coordinates (e.g. \( x = 4, y = 11 \) or \( x = -1, y = 1 \))\
A1 for finding the other correct pair of coordinates

Complete the square for the quadratic expression:\
\( y = x^2 - 6x + 14 \)\
\( y = (x - 3)^2 - 3^2 + 14 \)\
\( y = (x - 3)^2 - 9 + 14 \)\
\( y = (x - 3)^2 + 5 \)\
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Therefore, the turning point has coordinates \( (3, 5) \).

M1 for completing the square to get \( (x - 3)^2 + c \)\
M1 for finding \( c = 5 \)\
A1 for coordinates \( (3, 5) \)

The formula for the area of a sector is:\
\( A = \frac{\theta}{360} \times \pi r^2 \)\
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Substitute the given values:\
\( 20\pi = \frac{72}{360} \times \pi r^2 \)\
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Simplify the fraction:\
\( 20\pi = \frac{1}{5} \times \pi r^2 \)\
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Divide both sides by \( \pi \):\
\( 20 = \frac{1}{5} r^2 \)\
\( r^2 = 100 \)\
\( r = 10 \) (since radius must be positive)

M1 for setting up the equation \( 20\pi = \frac{72}{360} \times \pi r^2 \)\
M1 for simplifying to \( r^2 = 100 \)\
A1 for \( r = 10 \)

Express all terms with base 3:\
\( 27 = 3^3 \)\
\( 27^2 = (3^3)^2 = 3^6 \)\
\( 9^p = (3^2)^p = 3^{2p} \)\
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Substitute these into the equation:\
\( \frac{3^6 \times 3^{-4}}{3^{2p}} = 3^3 \)\
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Simplify the numerator:\
\( \frac{3^2}{3^{2p}} = 3^3 \)\
\( 3^{2 - 2p} = 3^3 \)\
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Equate the exponents:\
\( 2 - 2p = 3 \)\
\( -2p = 1 \)\
\( p = -0.5 \)

M1 for expressing terms in base 3: \( 3^6 \) or \( 3^{2p} \)\
M1 for equating exponents to get \( 2 - 2p = 3 \) (or equivalent)\
A1 for \( p = -0.5 \) (or equivalent fraction)

The formula for the total surface area of a cylinder is:\
\( A = 2\pi r^2 + 2\pi r h \)\
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Substitute the given values (\( r = 3 \) and \( A = 48\pi \)):\
\( 48\pi = 2\pi(3^2) + 2\pi(3)h \)\
\( 48\pi = 18\pi + 6\pi h \)\
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Divide both sides by \( \pi \):\
\( 48 = 18 + 6h \)\
\( 30 = 6h \)\
\( h = 5 \)

M1 for substituting values into the formula: \( 48\pi = 2\pi(3^2) + 2\pi(3)h \)\
M1 for simplifying to \( 30 = 6h \) (or equivalent)\
A1 for \( h = 5 \)

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

Factorise the denominator using the difference of two squares:
$$4x^2 - 1 = (2x + 1)(2x - 1)$$

Substitute the factorised expressions back into the fraction and cancel the common binomial factor \((2x + 1)\):
$$\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}$$

M1 for factorising the numerator: \((2x + 1)(x - 3)\)
M1 for factorising the denominator: \((2x + 1)(2x - 1)\)
A1.5 for final simplified answer: \(\frac{x - 3}{2x - 1}\)

Find the first and second differences:
Sequence: \(2, \quad 9, \quad 20, \quad 35, \quad 54\)
First differences: \(7, \quad 11, \quad 15, \quad 19\)
Second differences: \(4, \quad 4, \quad 4\)

Since the second difference is constant, the sequence is quadratic with the general term \(an^2 + bn + c\).
Here, \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\).

Subtract \(2n^2\) from each term of the original sequence:
For \(n = 1\): \(2 - 2(1)^2 = 0\)
For \(n = 2\): \(9 - 2(2)^2 = 1\)
For \(n = 3\): \(20 - 2(3)^2 = 2\)
For \(n = 4\): \(35 - 2(4)^2 = 3\)

The resulting linear sequence \(0, \quad 1, \quad 2, \quad 3, \dots\) has the general term \(n - 1\).

Combining the quadratic and linear parts, the overall \(nth\) term is:
$$2n^2 + n - 1$$

M1 for finding second differences are constant at 4
M1 for subtracting \(2n^2\) to yield linear sequence \(0, 1, 2, 3\) or setting up simultaneous equations
A1.5 for the final correct answer: \(2n^2 + n - 1\)

Equate the two expressions for \(y\):
$$2x + 1 = x^2 - x - 3$$

Rearrange into a quadratic equation equal to zero:
$$x^2 - 3x - 4 = 0$$

Factorise the quadratic equation:
$$(x - 4)(x + 1) = 0$$

This gives two values for \(x\):
$$x = 4 \quad \text{or} \quad x = -1$$

Substitute each \(x\) value back into the linear equation \(y = 2x + 1\) to find the corresponding \(y\) values:
For \(x = 4\): \(y = 2(4) + 1 = 9\)
For \(x = -1\): \(y = 2(-1) + 1 = -1\)

So the solutions are \(x = 4, y = 9\) and \(x = -1, y = -1\).

M1 for setting up the equation \(2x + 1 = x^2 - x - 3\)
M1 for factorising/solving to find \(x = 4\) and \(x = -1\)
A1.5 for both correct pairs: \(x = 4, y = 9\) and \(x = -1, y = -1\) (A0.5 for only one correct pair)

Use the Cosine Rule to find \(BC\):
$$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)$$

Substitute the given values into the formula:
$$BC^2 = 4^2 + 6^2 - 2(4)(6)\cos(120^\circ)$$

Recall that \(\cos(120^\circ) = -0.5\):
$$BC^2 = 16 + 36 - 48(-0.5)$$
$$BC^2 = 52 + 24 = 76$$

Solve for the exact value of \(BC\):
$$BC = \sqrt{76}$$

M1 for substituting correctly into the Cosine Rule: \(4^2 + 6^2 - 2(4)(6)\cos(120^\circ)\)
M1 for recalling \(\cos(120^\circ) = -0.5\) (or simplified addition: \(52 + 24\))
A1.5 for the final answer in the correct form: \(\sqrt{76}\)

First, find the gradient of the given line by rewriting it in slope-intercept form:
$$4y = 3x - 8 \implies y = \frac{3}{4}x - 2$$
The gradient of the original line is \(\frac{3}{4}\).

The perpendicular line will have a gradient \(m\) which is the negative reciprocal:
$$m = -\frac{4}{3}$$

Use the point-slope equation with the point \((6, -1)\):
$$y - (-1) = -\frac{4}{3}(x - 6)$$
$$y + 1 = -\frac{4}{3}x + 8$$

Multiply the entire equation by 3 to eliminate the fraction:
$$3(y + 1) = -4(x - 6)$$
$$3y + 3 = -4x + 24$$

Rearrange the terms into the form \(ay + bx = c\):
$$3y + 4x = 21$$

M1 for identifying the perpendicular gradient as \(-\frac{4}{3}\)
M1 for substituting their gradient and the point \((6, -1)\) into a line formula
A1.5 for the final answer in the correct integer form: \(3y + 4x = 21\) (or any equivalent integer format, e.g., \(4x + 3y = 21\))