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

The sequence is: 17, 11, 5, -1, ... First, find the common difference by subtracting consecutive terms: 11 - 17 = -6, and 5 - 11 = -6. Since the common difference is -6, the n-th term contains the term -6n. Now, find the constant term c in the expression -6n + c. For n = 1, the term is 17: -6(1) + c = 17, which gives -6 + c = 17, so c = 23. Therefore, the n-th term is 23 - 6n (or -6n + 23).

M1 for finding the common difference is -6 (or writing -6n). M1 for writing or using -6n + c and substituting n=1 to find c (or equivalent method). A1 for 23 - 6n or -6n + 23 as the final answer.

To factorise the expression 12x^2y - 18xy^2 completely, find the highest common factor (HCF) of the two terms. The HCF of the numbers 12 and 18 is 6. The HCF of the x terms (x^2 and x) is x. The HCF of the y terms (y and y^2) is y. Thus, the HCF of the entire expression is 6xy. Divide each term by 6xy to find the terms inside the brackets: 12x^2y / 6xy = 2x, and -18xy^2 / 6xy = -3y. Combining these gives the completely factorised expression: 6xy(2x - 3y).

M1 for finding a partial common factor out of the bracket, e.g. 6(2x^2y - 3xy^2) or xy(12x - 18y). M1 for a nearly complete factorisation with only one factor missing, e.g. 3xy(4x - 6y) or 6x(2xy - 3y^2). A1 for the completely factorised correct expression: 6xy(2x - 3y).

The sale price represents a 15% reduction from the original price. Therefore, the sale price of $187 is 100% - 15% = 85% of the original price. Let the original price be P. We can write 0.85 * P = 187, which means P = 187 / 0.85 = 18700 / 85. We can simplify this division by dividing both the numerator and the denominator by 17: 187 / 17 = 11, and 85 / 17 = 5. This simplifies the fraction to 1100 / 5 = 220. Thus, the original price of the bicycle was $220.

M1 for equating 85% to 187 or writing 187 / (1 - 0.15) or equivalent. M1 for 187 / 85 * 100 (correct division method without calculator). A1 for 220.

First, add 5 to both sides of the equation: \(T + 5 = 3\sqrt{w}\). Next, divide both sides by 3 to isolate the square root: \(\frac{T + 5}{3} = \sqrt{w}\). Finally, square both sides of the equation to solve for \(w\): \(w = \left(\frac{T + 5}{3}\right)^2\), which can also be written as \(w = \frac{(T + 5)^2}{9}\).

M1 for correctly isolating the term with \(w\) on one side (e.g. \(T + 5 = 3\sqrt{w}\)). M1 for isolating the square root of \(w\) (e.g. \(\sqrt{w} = \frac{T + 5}{3}\)). A1 for \(w = \left(\frac{T + 5}{3}\right)^2\) or \(w = \frac{(T + 5)^2}{9}\).

Multiply all terms in the equation by 6 (the lowest common multiple of 3 and 2) to eliminate the denominators: \(2(2x + 1) - 3(x - 3) = 24\). Expand the brackets carefully: \(4x + 2 - 3x + 9 = 24\). Simplify the left-hand side by collecting like terms: \(x + 11 = 24\). Subtract 11 from both sides to find the value of \(x\): \(x = 13\).

M1 for correctly eliminating the denominators, e.g. \(2(2x + 1) - 3(x - 3) = 24\). M1 for correct expansion of the brackets, e.g. \(4x + 2 - 3x + 9 = 24\) (allow one arithmetic or sign error). A1 for \(x = 13\).

Let the exterior angle of the regular polygon be \(x^\circ\). Since the interior angle is 5 times the exterior angle, the interior angle is \(5x^\circ\). The interior and exterior angles at any vertex lie on a straight line, so they sum to \(180^\circ\): \(x + 5x = 180\), which simplifies to \(6x = 180\), giving \(x = 30\). Therefore, each exterior angle is \(30^\circ\). Since the sum of the exterior angles of any polygon is \(360^\circ\), the number of sides \(n\) is \(n = \frac{360}{30} = 12\).

M1 for setting up a correct equation, e.g. \(x + 5x = 180\), or finding that the exterior angle is \(30^\circ\) or the interior angle is \(150^\circ\). M1 for attempting to find the number of sides using \(\frac{360}{\text{exterior angle}}\). A1 for 12.

First, expand the brackets on both sides of the equation. This gives: \(6x - 15 = 4 - x + 2\). Next, simplify the right-hand side by combining the numerical terms: \(6x - 15 = 6 - x\). Now, collect the variable terms on one side by adding \(x\) to both sides: \(7x - 15 = 6\). Next, collect the constant terms on the other side by adding 15 to both sides: \(7x = 21\). Finally, divide both sides by 7 to solve for \(x\): \(x = 3\).

M1 for correct expansion of brackets on either side: \(6x - 15\) or \(4 - x + 2\). M1 for correctly simplifying and collecting like terms to form a linear equation: \(7x = 21\) or equivalent. A1 for 3.

By the exterior angle theorem of a triangle, the exterior angle is equal to the sum of the two opposite interior angles. This gives the equation: \(4x + (2x + 10) = 130\). Simplify the left-hand side: \(6x + 10 = 130\). Subtract 10 from both sides: \(6x = 120\). Divide by 6 to find \(x\): \(x = 20\). Alternatively, find the third interior angle \(\angle ACB = 180^\circ - 130^\circ = 50^\circ\). Then use the sum of angles in a triangle: \(4x + 2x + 10 + 50 = 180\), which simplifies to \(6x + 60 = 180\), leading to \(6x = 120\) and \(x = 20\).

M1 for setting up a correct geometric relationship, e.g., \(4x + (2x + 10) = 130\) or identifying \(\angle ACB = 50^\circ\). M1 for simplifying to a correct linear equation in \(x\), e.g., \(6x = 120\) or \(6x + 60 = 180\). A1 for 20.

First, identify the pattern of the sequence. The terms increase by 4 each time, so the common difference is \(d = 4\). To find the general formula for the \(n\)-th term of this arithmetic sequence, use the formula \(T_n = a + (n - 1)d\), where \(a = 7\) and \(d = 4\). This gives \(T_n = 7 + (n - 1)4 = 4n + 3\). To find the 40th term, substitute \(n = 40\) into the formula: \(T_{40} = 4(40) + 3 = 160 + 3 = 163\).

M1 for identifying the common difference is 4, or finding the general term of the form \(4n + k\). M1 for a correct expression or calculation to find the 40th term, e.g., \(4(40) + 3\) or \(7 + 39 \times 4\). A1 for 163.

First, expand the brackets on the left-hand side of the equation: \(3(2x - 1) = 6x - 3\) and \(-4(x - 3) = -4x + 12\). Substitute these expansions back into the equation: \(6x - 3 - 4x + 12 = 19\). Group the like terms together: \((6x - 4x) + (-3 + 12) = 19\), which simplifies to \(2x + 9 = 19\). Subtract 9 from both sides of the equation: \(2x = 10\). Finally, divide both sides by 2 to find the value of \(x\): \(x = 5\).

M1 for correct expansion of at least one bracket to obtain \(6x - 3\) or \(-4x + 12\). M1 for simplifying the equation to the form \(ax = b\), such as \(2x = 10\) or equivalent. A1 for the correct answer of 5.

An interior angle and its corresponding exterior angle lie on a straight line, meaning they add up to \(180^\circ\). First, calculate the size of one exterior angle: \(180^\circ - 162^\circ = 18^\circ\). The sum of all exterior angles in any convex polygon is \(360^\circ\). Since this is a regular polygon, all exterior angles are equal. To find the number of sides, divide the total sum of exterior angles by the size of one exterior angle: \(360^\circ \div 18^\circ = 20\). Therefore, the regular polygon has 20 sides.

M1 for calculating the size of an exterior angle: \(180 - 162\) (or showing \(18\)). M1 for dividing 360 by their exterior angle: \(360 \div 18\) (or for setting up the equation \(\frac{(n - 2) \times 180}{n} = 162\)). A1 for the correct answer of 20.

Find the difference between consecutive terms of the sequence: \(13 - 17 = -4\), \(9 - 13 = -4\), and \(5 - 9 = -4\). Since the sequence decreases by 4 each time, the common difference is \(-4\). This means the formula for the \(n\)-th term contains the term \(-4n\). To find the constant part of the expression, use the first term where \(n = 1\): \(-4(1) + c = 17\), which gives \(-4 + c = 17\), so \(c = 21\). Combining these terms gives the expression for the \(n\)-th term: \(21 - 4n\) (or \(-4n + 21\)).

M1 for finding a common difference of \(-4\) (or writing \(-4n\) as part of the formula). M1 for a complete method to find the constant term, such as finding the 'zeroth' term \(17 - (-4) = 21\) or solving \(-4(1) + c = 17\). A1 for the correct expression \(21 - 4n\) or any equivalent expression such as \(-4n + 21\).

To solve the equation \(\frac{2x + 5}{3} - \frac{x - 1}{2} = 3\), we first multiply all terms in the equation by 6 (the lowest common multiple of 3 and 2) to eliminate the denominators:

\[2(2x + 5) - 3(x - 1) = 3 \times 6\]

Expand the brackets carefully, noting the negative sign before the second bracket:

\[4x + 10 - 3x + 3 = 18\]

Collect like terms on the left-hand side:

\[x + 13 = 18\]

Subtract 13 from both sides to find the value of \(x\):

\[x = 5\]

M1 for attempting to multiply all terms by 6 (or equivalent method to clear fractions), e.g., \(2(2x + 5) - 3(x - 1) = 18\)
M1 for correct expansion of brackets, leading to \(4x + 10 - 3x + 3 = 18\) (allow one arithmetic or sign error)
A1 for 5

First, find the formula for the \(n\)-th term of the sequence.
The first term is \(a = 5\) and the common difference is \(d = 6\).
Using the formula for an arithmetic sequence:
\[\text{Term} = a + (n-1)d = 5 + (n-1)6 = 6n - 1\]
Next, set up an equation or inequality to find the term closest to 150:
\[6n - 1 \approx 150\]
\[6n \approx 151\]
\[n \approx 25.17\]
Since \(n\) must be an integer, evaluate the terms for \(n = 25\) and \(n = 26\):
For \(n = 25\): \(6(25) - 1 = 149\) (distance of 1 from 150)
For \(n = 26\): \(6(26) - 1 = 155\) (distance of 5 from 150)
Therefore, the term closest to 150 is 149.

M1 for finding the expression for the \(n\)-th term, \(6n - 1\) (or identifying the common difference is 6 and extending the sequence correctly near 150)
M1 for calculating terms near 150, showing \(6(25) - 1 = 149\) and \(6(26) - 1 = 155\) (or demonstrating 149 and 155 are consecutive terms around 150)
A1 for 149

The sum of the interior angles of a pentagon (a 5-sided polygon) is given by:
\[(5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\]
Sum the given interior angles and set the total equal to \(540^\circ\):
\[x + (2x - 10) + (x + 30) + 115 + 125 = 540\]
Simplify the equation by combining like terms:
\[4x + 260 = 540\]
Subtract 260 from both sides:
\[4x = 280\]
Divide by 4:
\[x = 70\]

M1 for calculating the sum of the interior angles of a pentagon as \(540^\circ\)
M1 for setting up the equation \(x + 2x - 10 + x + 30 + 115 + 125 = 540\) (or using their angle sum)
A1 for 70

First, expand each bracket carefully: \(5(2x - 3) = 10x - 15\) and \(-2(3x - 7) = -6x + 14\). Next, collect the like terms: \(10x - 6x - 15 + 14 = 4x - 1\).

M1 for correct expansion of the first bracket to \(10x - 15\) (or one error). M1 for correct expansion of the second bracket to \(-6x + 14\) (or one error, particularly checking the sign of \(+14\)). A1 for the final simplified expression \(4x - 1\).

Find the difference between consecutive terms: \(11 - 3 = 8\), \(19 - 11 = 8\). Since the difference is constant, the sequence is linear and has the form \(8n + c\). To find \(c\), substitute \(n = 1\): \(8(1) + c = 3\), which gives \(c = -5\). Thus, the \(n\)-th term is \(8n - 5\).

M1 for identifying the common difference is 8 (or writing \(8n\)). M1 for attempting to find the constant term (e.g. \(3 - 8\) or setting up \(8(1) + c = 3\)). A1 for \(8n - 5\) (or any equivalent expression).

First, find the exterior angle of the regular polygon. The interior and exterior angles on a straight line add up to \(180^\circ\). Exterior angle = \(180^\circ - 144^\circ = 36^\circ\). Since the sum of the exterior angles of any polygon is \(360^\circ\), the number of sides \(n\) is given by: \(n = \frac{360^\circ}{36^\circ} = 10\).

M1 for finding the exterior angle \(180 - 144\) (or showing \(36\)), or for setting up the interior angle formula \(\frac{(n-2) \times 180}{n} = 144\). M1 for calculating \(360 \div 36\) or solving the equation to find \(n\). A1 for \(10\).

First, find the common difference between consecutive terms: \(17 - 23 = -6\) and \(11 - 17 = -6\). The common difference is \(-6\), so the \(n\)-th term includes \(-6n\). Now, find the constant term \(c\) such that \(-6n + c\) represents the sequence. For \(n = 1\): \(-6(1) + c = 23 \implies -6 + c = 23 \implies c = 29\). Thus, the \(n\)-th term is \(29 - 6n\) (or \(-6n + 29\)).

M1 for finding the common difference is \(-6\) (or writing \(-6n\) as part of their expression). M1 for setting up a correct method to find the constant, e.g. \(-6(1) + c = 23\). A1 for \(29 - 6n\) or \(-6n + 29\).

First, expand the brackets: \(3(2x - 5) = 6x - 15\) and \(-2(x + 1) = -2x - 2\). Substitute these back into the equation: \(6x - 15 - 2x - 2 = 7\). Combine like terms: \(4x - 17 = 7\). Add 17 to both sides: \(4x = 24\). Divide by 4: \(x = 6\).

M1 for correct expansion of at least one bracket, i.e., \(6x - 15\) or \(-2x - 2\). M1 for simplifying to a linear equation of the form \(ax = b\), e.g., \(4x = 24\). A1 for \(6\).

First, convert both mixed numbers into improper fractions: \(1\frac{3}{5} = \frac{8}{5}\) and \(2\frac{2}{3} = \frac{8}{3}\). Now, divide the first fraction by the second by multiplying by the reciprocal: \(\frac{8}{5} \div \frac{8}{3} = \frac{8}{5} \times \frac{3}{8}\). Multiply and simplify: \(\frac{8 \times 3}{5 \times 8} = \frac{3}{5}\).

M1 for converting both mixed numbers correctly to improper fractions: \(\frac{8}{5}\) and \(\frac{8}{3}\). M1 for multiplying by the reciprocal of the divisor: \(\frac{8}{5} \times \frac{3}{8}\). A1 for \(\frac{3}{5}\).

To make \( x \) the subject, we rearrange the equation step-by-step:

1. Multiply both sides by 3:
\( 3A = 5x - 7 \)

2. Add 7 to both sides:
\( 3A + 7 = 5x \)

3. Divide both sides by 5:
\( x = \frac{3A + 7}{5} \)

M1 for multiplying both sides by 3 to get \( 3A = 5x - 7 \) (or equivalent)
M1 for adding 7 to both sides to get \( 3A + 7 = 5x \) (or equivalent)
A1 for the correct final answer: \( x = \frac{3A + 7}{5} \) (or equivalent, e.g., \( x = \frac{3A}{5} + \frac{7}{5} \))

To find the \( n \)-th term of an arithmetic sequence, we first find the common difference between consecutive terms:
\( 14 - 19 = -5 \)
\( 9 - 14 = -5 \)

Since the difference is constant and equals \( -5 \), the term has the form:
\( -5n + c \)

We find the constant \( c \) by substituting \( n = 1 \) for the first term:
\( -5(1) + c = 19 \)
\( -5 + c = 19 \)
\( c = 24 \)

Thus, the expression for the \( n \)-th term is \( 24 - 5n \) (or \( -5n + 24 \)).

B1 for finding the common difference of \( -5 \) (or seeing \( -5n \) in the final answer)
M1 for setting up a correct equation or method to find the constant term, e.g., \( 19 - (-5) \) or \( -5(1) + c = 19 \)
A1 for \( 24 - 5n \) or \( -5n + 24 \) (or equivalent)

1. Since \( AB = AC \), triangle \( ABC \) is an isosceles triangle, meaning the base angles opposite those sides are equal:
\( \angle ABC = \angle ACB \)

2. The sum of the interior angles of a triangle is \( 180^\circ \):
\( \angle BAC + \angle ABC + \angle ACB = 180^\circ \)
\( 40^\circ + 2 \times \angle ACB = 180^\circ \)
\( 2 \times \angle ACB = 140^\circ \)
\( \angle ACB = 70^\circ \)

3. Since \( BCD \) is a straight line, the angles on a straight line sum to \( 180^\circ \):
\( \angle ACB + \angle ACD = 180^\circ \)
\( 70^\circ + \angle ACD = 180^\circ \)
\( \angle ACD = 180^\circ - 70^\circ = 110^\circ \)

M1 for recognizing the base angles are equal and setting up \( \frac{180 - 40}{2} \)
A1 for finding \( \angle ACB = 70^\circ \)
A1 for the correct final answer of \( 110 \) (or \( 110^\circ \))

The terms decrease by 6 each time, so the common difference is \(-6\). This means the expression for the \(n\)-th term contains \(-6n\). Comparing \(-6n\) with the sequence: for \(n = 1\), \(-6(1) = -6\), but we need the first term to be 17. The difference is \(17 - (-6) = 23\). Therefore, the \(n\)-th term is \(23 - 6n\).

M1 for identifying the common difference of \(-6\) (or writing a term of \(-6n\)). M1 for writing an expression of the form \(a - 6n\) where \(a\) is a constant. A1 for the correct expression \(23 - 6n\) or equivalent.

The interior angle and the exterior angle at any vertex of a polygon lie on a straight line, so their sum is \(180^\circ\). Therefore, the exterior angle is \(180^\circ - 156^\circ = 24^\circ\). The sum of the exterior angles of any regular polygon is always \(360^\circ\). To find the number of sides, divide \(360^\circ\) by the exterior angle: \(360 \div 24 = 15\).

M1 for finding the exterior angle: \(180 - 156\) (or \(24\)) or writing the formula \(((n-2) \times 180)/n = 156\). M1 for \(360 \div \text{their } 24\) or solving the equation to get \(24n = 360\). A1 for \(15\).

Convert \(2\frac{1}{4}\) to an improper fraction: \(2\frac{1}{4} = \frac{9}{4}\). Find a common denominator for \(\frac{9}{4}\) and \(\frac{5}{6}\), which is 12. Write each fraction with denominator 12: \(\frac{9}{4} = \frac{27}{12}\) and \(\frac{5}{6} = \frac{10}{12}\). Subtract the fractions: \(\frac{27}{12} - \frac{10}{12} = \frac{17}{12}\). Convert the improper fraction back to a mixed number: \(\frac{17}{12} = 1\frac{5}{12}\).

M1 for converting to improper fraction: \(\frac{9}{4}\) or equivalent. M1 for converting both fractions to have a common denominator (e.g., \(\frac{27}{12}\) and \(\frac{10}{12}\)). A1 for \(1\frac{5}{12}\) (accept \(1\ 5/12\)).

First, factorise the numerator \(2x^2 - 5x - 3\):
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Next, factorise the denominator \(4x^2 - 1\) using the difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now, substitute these factorised forms back into the fraction:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\)

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

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

Let the sequence be represented by \(u_n\):
\(u_1 = 4\), \(u_2 = 9\), \(u_3 = 18\), \(u_4 = 31\), \(u_5 = 48\)

Find the first differences between consecutive terms:
\(9 - 4 = 5\)
\(18 - 9 = 9\)
\(31 - 18 = 13\)
\(48 - 31 = 17\)

Find the second differences:
\(9 - 5 = 4\)
\(13 - 9 = 4\)
\(17 - 13 = 4\)

Since the second differences are constant and equal to \(4\), the sequence is quadratic of the form \(an^2 + bn + c\), where \(a = \frac{4}{2} = 2\).

Now, subtract \(2n^2\) from each term of the sequence:
For \(n=1\): \(4 - 2(1)^2 = 2\)
For \(n=2\): \(9 - 2(2)^2 = 1\)
For \(n=3\): \(18 - 2(3)^2 = 0\)
For \(n=4\): \(31 - 2(4)^2 = -1\)

The remaining sequence is \(2, \ 1, \ 0, \ -1, \ \dots\), which is a linear sequence of the form \(bn + c\) with a common difference of \(-1\).
Thus, \(b = -1\).
Using \(n=1\): \(-1(1) + c = 2 \implies c = 3\).
So, the linear part is \(-n + 3\).

Combining these parts, the \(n\)-th term of the sequence is:
\(2n^2 - n + 3\)

M1 for finding the constant second difference of 4, or for a term of \(2n^2\)
M1 for attempting to find the linear component by subtracting \(2n^2\) from the sequence terms (obtaining \(2, 1, 0, \dots\))
A1 for the correct final expression \(2n^2 - n + 3\)

Multiply the entire equation by the common denominator \(x(x-3)\) to clear the fractions:
\(2x + 3(x-3) = 2x(x-3)\)

Expand both sides of the equation:
\(2x + 3x - 9 = 2x^2 - 6x\)
\(5x - 9 = 2x^2 - 6x\)

Rearrange into a quadratic equation in standard form:
\(2x^2 - 11x + 9 = 0\)

Factorise the quadratic expression:
\((2x - 9)(x - 1) = 0\)

Solve for \(x\):
\(2x - 9 = 0 \implies x = 4.5\) (or \(\frac{9}{2}\))
\(x - 1 = 0 \implies x = 1\)

M1 for multiplying by \(x(x-3)\) to obtain a correct algebraic equation without fractions
M1 for rearranging and simplifying to a correct quadratic equation, e.g., \(2x^2 - 11x + 9 = 0\)
A1 for both correct solutions: \(x = 1\) and \(x = 4.5\) (or \(\frac{9}{2}\))

First, factorize the numerator and the denominator. The numerator is \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\). The denominator is \(2x^2 - 9x - 5 = (2x + 1)(x - 5)\). Dividing the numerator by the denominator, the common factor of \(x - 5\) cancels out. This leaves \(\frac{3x + 1}{2x + 1}\).

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

We can set up a system of linear equations using the given terms. For \(n = 1\), we have \(a(1)^2 + b(1) + 5 = 12\), which simplifies to \(a + b = 7\). For \(n = 2\), we have \(a(2)^2 + b(2) + 5 = 25\), which simplifies to \(4a + 2b = 20\), or equivalently \(2a + b = 10\). Subtracting the first equation from the second gives \((2a + b) - (a + b) = 10 - 7\), which simplifies to \(a = 3\). Substituting \(a = 3\) back into the first equation gives \(3 + b = 7\), which simplifies to \(b = 4\).

M1 for setting up two simultaneous equations: \(a + b = 7\) and \(4a + 2b = 20\) (or equivalent). M1 for a correct algebraic method to solve their simultaneous equations. A1.5 for \(a = 3\) and \(b = 4\) (accept correct values without working for full marks).

The sum of the interior angles of a polygon with \(n\) sides is given by \((n - 2) \times 180^\circ\). For a pentagon, \(n = 5\), so the sum of the interior angles is \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\). The total number of ratio parts is \(2 + 3 + 4 + 4 + 5 = 18\). Each part represents \(540^\circ / 18 = 30^\circ\). The largest interior angle corresponds to the largest ratio part, which is 5. Therefore, the size of the largest interior angle is \(5 \times 30^\circ = 150^\circ\).

M1 for finding the sum of the interior angles of a pentagon is \(540^\circ\). M1 for dividing \(540\) by the sum of the ratio parts (18) to find the value of one part (\(30^\circ\)). A1.5 for \(150\) (or \(150^\circ\)).

First, factorise the numerator by finding two numbers that multiply to \(-6\) and add to \(-5\), which are \(-6\) and \(1\): \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Substitute these back into the fraction to get \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancelling the common factor of \((2x + 1)\) from the numerator and denominator leaves the simplified fraction: \(\frac{x - 3}{2x - 1}\).

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

Find the first differences of the sequence: 10 - 3 = 7, 21 - 10 = 11, 36 - 21 = 15, 55 - 36 = 19. Find the second differences: 11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4. Since the second difference is a constant 4, the sequence is quadratic with the form \(an^2 + bn + c\), where the coefficient \(a = 4 / 2 = 2\). Subtract \(2n^2\) from each term: for \(n=1\), \(3 - 2(1) = 1\); for \(n=2\), \(10 - 8 = 2\); for \(n=3\), \(21 - 18 = 3\). The resulting sequence is 1, 2, 3, ..., which has the \(n\)-th term of \(n\). Therefore, the overall \(n\)-th term is \(2n^2 + n\).

M1 for finding the second difference is 4 or showing that the coefficient of \(n^2\) is 2. M1 for subtracting \(2n^2\) from the terms to find the linear sequence 1, 2, 3, ... (or attempting to solve simultaneous equations to find coefficients). A1 for the correct expression: \(2n^2 + n\).

Multiply all terms by the common denominator \((x - 1)(x + 1)\) to eliminate the fractions: \(3(x + 1) - 2(x - 1) = (x - 1)(x + 1)\). Expand both sides: \(3x + 3 - 2x + 2 = x^2 - 1\). Simplify: \(x + 5 = x^2 - 1\). Rearrange into standard quadratic form: \(x^2 - x - 6 = 0\). Factorise the quadratic expression: \((x - 3)(x + 2) = 0\). Solving for \(x\) gives \(x = 3\) or \(x = -2\).

M1 for clearing fractions correctly: \(3(x + 1) - 2(x - 1) = (x - 1)(x + 1)\) or equivalent. M1 for simplifying and rearranging into a standard three-term quadratic: \(x^2 - x - 6 = 0\). A1 for both correct solutions: \(x = 3\) and \(x = -2\).

1. Factorise the numerator:
\( 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2) \)

2. Factorise the denominator:
\( 2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (2x + 3)(x + 2) \)

3. Cancel the common factor \( (x + 2) \):
\( \frac{3(x - 2)(x + 2)}{(2x + 3)(x + 2)} = \frac{3(x - 2)}{2x + 3} \) (or \( \frac{3x - 6}{2x + 3} \))

M1 for factorising numerator: \( 3(x - 2)(x + 2) \) or \( 3(x^2 - 4) \)
M1 for factorising denominator: \( (2x + 3)(x + 2) \)
A1.5 for final answer: \( \frac{3(x - 2)}{2x + 3} \) or \( \frac{3x - 6}{2x + 3} \)

1. Express both sides with base 3:
\( 27^{2x - 1} = (3^3)^{2x - 1} = 3^{6x - 3} \)
\( \frac{1}{9^{x - 3}} = 9^{-(x - 3)} = (3^2)^{3 - x} = 3^{6 - 2x} \)

2. Equate the indices:
\( 6x - 3 = 6 - 2x \)

3. Solve for \( x \):
\( 8x = 9 \) which gives \( x = \frac{9}{8} \) (or \( 1.125 \))

M1 for expressing terms in base 3: \( 3^{6x - 3} \) or \( 3^{6 - 2x} \)
M1 for equating indices: \( 6x - 3 = 6 - 2x \) (or equivalent)
A1.5 for \( \frac{9}{8} \) or \( 1.125 \)

1. Let \( r_1 = 4 \) and \( r_2 \) be the radii, and \( V_1 = 24\pi \) and \( V_2 = 81\pi \) be the volumes.

2. The ratio of volumes is the cube of the ratio of linear dimensions:
\( \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 \)

3. Substitute values:
\( \frac{24\pi}{81\pi} = \left(\frac{4}{r_2}\right)^3 \)
\( \frac{8}{27} = \left(\frac{4}{r_2}\right)^3 \)

4. Take the cube root:
\( \frac{2}{3} = \frac{4}{r_2} \)

5. Solve for \( r_2 \):
\( r_2 = \frac{4 \times 3}{2} = 6 \text{ cm} \)

M1 for volume ratio \( \frac{24\pi}{81\pi} \) simplified to \( \frac{8}{27} \)
M1 for taking the cube root to find linear scale factor of \( \frac{2}{3} \) (or \( 1.5 \))
A1.5 for \( 6 \)

1. Factorise the numerator: \(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). 2. Factorise the denominator using the difference of two squares: \(4x^2 - 1 = (2x + 1)(2x - 1)\). 3. Cancel the common factor \((2x + 1)\) from the numerator and denominator: \(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}\).

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

1. Find the \(n\)-th term of the numerators: \(3, 8, 15, 24, 35, \dots\). The second differences are constant and equal to \(2\), showing it is quadratic in the form \(n^2 + bn + c\). Comparing with \(n^2\) (\(1, 4, 9, 16, 25\)), we see each term is \(2n\) more than \(n^2\). Thus, the numerator formula is \(n^2 + 2n\). 2. Find the \(n\)-th term of the denominators: \(4, 7, 10, 13, 16, \dots\). This is an arithmetic sequence with first term \(4\) and common difference \(3\), so the formula is \(4 + 3(n-1) = 3n + 1\). 3. Combine both parts to obtain the general term: \(\frac{n^2+2n}{3n+1}\).

M1 for finding the numerator's \(n\)-th term: \(n^2 + 2n\) (or equivalent). M1 for finding the denominator's \(n\)-th term: \(3n + 1\) (or equivalent). A1 for correct combined fraction \(\frac{n^2+2n}{3n+1}\).

1. Find the midpoint \(M\) of \(AB\): \(M = \left(\frac{2+6}{2}, \frac{3+11}{2}\right) = (4, 7)\). 2. Find the gradient \(m\) of \(AB\): \(m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\). 3. Find the perpendicular gradient: \(m_{\perp} = -\frac{1}{m} = -\frac{1}{2}\). 4. Use the gradient \(-\frac{1}{2}\) and point \((4, 7)\) to write the equation: \(y - 7 = -\frac{1}{2}(x - 4) \implies y = -\frac{1}{2}x + 2 + 7 \implies y = -\frac{1}{2}x + 9\).

M1 for finding the midpoint \((4, 7)\). M1 for finding the perpendicular gradient \(-\frac{1}{2}\) (or gradient of \(AB = 2\)). A1 for final answer \(y = -\frac{1}{2}x + 9\) (or \(y = -0.5x + 9\)).

First, factorise the denominator of the second fraction: \(x^2 - 4 = (x - 2)(x + 2)\). Express both fractions with the common denominator \((x - 2)(x + 2)\): \(\frac{3(x + 2)}{(x - 2)(x + 2)} - \frac{2x - 1}{(x - 2)(x + 2)}\). Subtract the numerators: \(\frac{3x + 6 - (2x - 1)}{(x - 2)(x + 2)} = \frac{3x + 6 - 2x + 1}{(x - 2)(x + 2)} = \frac{x + 7}{(x - 2)(x + 2)}\). This can also be written as \(\frac{x + 7}{x^2 - 4}\).

M1 for a common denominator of \((x-2)(x+2)\) or \(x^2-4\) with at least one correct numerator, e.g. \(3(x+2)\). M1 for correct subtraction of numerators: \(3x + 6 - 2x + 1\). A1 for \(\frac{x+7}{x^2-4}\) or \(\frac{x+7}{(x-2)(x+2)}\).

Let \(I\) be the interior angle and \(E\) be the exterior angle of the regular polygon. We know that the sum of an interior angle and an exterior angle on a straight line is \(180^\circ\), so \(I + E = 180^\circ\). We are given that \(I = E + 140^\circ\). Substituting this into the first equation: \((E + 140^\circ) + E = 180^\circ \implies 2E + 140^\circ = 180^\circ \implies 2E = 40^\circ \implies E = 20^\circ\). The formula for each exterior angle of a regular polygon is \(E = \frac{360^\circ}{n}\). Thus, \(20^\circ = \frac{360^\circ}{n} \implies n = \frac{360}{20} = 18\).

M1 for setting up the equation \(I + E = 180\) or expressing \(I\) and \(E\) in terms of \(n\): \(\frac{180(n-2)}{n} - \frac{360}{n} = 140\). M1 for finding the exterior angle \(E = 20^\circ\) or obtaining a simplified equation in \(n\), e.g. \(180n - 360 - 360 = 140n\). A1 for \(n = 18\).

Multiply all terms by the common denominator \(x(x+1)\) to clear the fractions: \(12(x+1) - 12x = x(x+1)\). Expand the brackets: \(12x + 12 - 12x = x^2 + x\). Simplify: \(12 = x^2 + x\). Rearrange into standard quadratic form: \(x^2 + x - 12 = 0\). Factorise the quadratic expression: \((x + 4)(x - 3) = 0\). Therefore, the solutions are \(x = -4\) and \(x = 3\).

M1 for multiplying by \(x(x+1)\) to get \(12(x+1) - 12x = x(x+1)\) or better. M1 for a correct quadratic equation in standard form, e.g. \(x^2 + x - 12 = 0\) (or equivalent). A1 for \(x = 3\) and \(x = -4\) (both required for full marks).

Factorise the numerator \(6x^2 - 7x - 3\): Find two numbers that multiply to \(6 \times (-3) = -18\) and add to \(-7\). These are \(-9\) and \(2\). This gives \(6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)\). Factorise the denominator \(4x^2 - 9\) using the difference of two squares: \((2x - 3)(2x + 3)\). Substitute these back into the fraction to obtain \(\frac{(3x + 1)(2x - 3)}{(2x - 3)(2x + 3)}\). Cancelling the common factor \((2x - 3)\) yields the simplified fraction \(\frac{3x + 1}{2x + 3}\).

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

Find the first and second differences of the sequence. Terms: \(5, 12, 23, 38, 57\). First differences: \(7, 11, 15, 19\). Second differences: \(4, 4, 4\). Since the second difference is constant at \(4\), the sequence is quadratic and the coefficient of \(n^2\) is \(4 / 2 = 2\). Now subtract \(2n^2\) from each term of the original sequence: for \(n = 1\), \(5 - 2(1) = 3\); for \(n = 2\), \(12 - 2(4) = 4\); for \(n = 3\), \(23 - 2(9) = 5\); for \(n = 4\), \(38 - 2(16) = 6\). The resulting sequence is \(3, 4, 5, 6, \dots\), which is an arithmetic sequence with the \(n\)-th term \(n + 2\). Combining both parts, the \(n\)-th term is \(2n^2 + n + 2\).

M1 for finding the second difference is 4, which implies a term of \(2n^2\) M1 for subtracting \(2n^2\) from terms to find the linear component sequence \(3, 4, 5, \dots\) or attempting to solve simultaneous equations A1 for \(2n^2 + n + 2\)

To solve the equation, combine the left side over a common denominator: \(\frac{3(x+1) - 2(x-1)}{(x-1)(x+1)} = 1\). Multiply both sides by the denominator: \(3(x+1) - 2(x-1) = (x-1)(x+1)\). Expand the brackets: \(3x + 3 - 2x + 2 = x^2 - 1\). Simplify the linear equation on the left: \(x + 5 = x^2 - 1\). Rearrange into standard quadratic form: \(x^2 - x - 6 = 0\). Factorise the quadratic expression: \((x - 3)(x + 2) = 0\). This gives the solutions \(x = 3\) or \(x = -2\).

M1 for clearing the fractions correctly: \(3(x+1) - 2(x-1) = (x-1)(x+1)\) M1 for simplifying to a three-term quadratic equation: \(x^2 - x - 6 = 0\) M1 for factorising their quadratic expression: \((x - 3)(x + 2) = 0\) A1 for both correct solutions: \(x = 3\) and \(x = -2\)

First, factorise the quadratic expression in the numerator: \(2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 = 2x(x-5) + 3(x-5) = (2x+3)(x-5)\). Next, factorise the expression in the denominator using the difference of two squares: \(4x^2 - 9 = (2x-3)(2x+3)\). Substitute these factored forms back into the fraction to get \(\frac{(2x+3)(x-5)}{(2x-3)(2x+3)}\). Cancelling the common factor of \((2x+3)\) from both the numerator and the denominator yields the simplified expression \(\frac{x-5}{2x-3}\).

M1 for factorising the numerator to get \((2x+3)(x-5)\). M1 for factorising the denominator to get \((2x-3)(2x+3)\). A1 for the correct simplified expression \(\frac{x-5}{2x-3}\) or equivalent.

First, express both sides of the equation as powers of 3. The right-hand side can be written as \(27\sqrt{3} = 3^3 \times 3^{0.5} = 3^{3.5}\). Therefore, \(\frac{1}{27\sqrt{3}} = 3^{-3.5}\). Now equate the exponents since the bases are equal: \(2x - 1 = -3.5\). Solving for \(x\) gives \(2x = -2.5\), which simplifies to \(x = -1.25\) (or \(-\frac{5}{4}\)).

M1 for writing \(27\sqrt{3}\) as \(3^{3.5}\) or \(3^{\frac{7}{2}}\). M1 for equating exponents to obtain the linear equation \(2x - 1 = -3.5\). A1 for \(x = -1.25\) or \(-\frac{5}{4}\).

Calculate the first differences between terms: \(12-5=7\), \(23-12=11\), \(38-23=15\). Calculate the second differences: \(11-7=4\), \(15-11=4\). Since the second difference is constant and equal to 4, the coefficient of the quadratic term \(n^2\) is half of 4, which is 2. Subtracting \(2n^2\) from each term of the sequence: for \(n=1\), \(5-2=3\); for \(n=2\), \(12-8=4\); for \(n=3\), \(23-18=5\); for \(n=4\), \(38-32=6\). This leaves a linear sequence 3, 4, 5, 6, ..., which has the general term \(n+2\). Combining these terms gives the overall \(n\)-th term of the sequence: \(2n^2 + n + 2\).

M1 for finding the constant second difference of 4 to determine the \(2n^2\) term. M1 for subtracting \(2n^2\) from the terms to get the linear sequence 3, 4, 5, 6, ... (or setting up equivalent simultaneous equations). A1 for the correct final expression \(2n^2 + n + 2\).

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

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

3. Collect all terms containing \(x\) on one side and the other terms on the opposite side:
\[5y - 2 = 3x + 2xy\]

4. Factorise \(x\) from the right-hand side:
\[5y - 2 = x(3 + 2y)\]

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

M1 for correctly multiplying by the denominator: \(y(5 - 2x) = 3x + 2\)
M1 for isolating terms with \(x\) on one side and factorising: \(x(3 + 2y) = 5y - 2\) (or equivalent)
A1 for the correct final formula: \(x = \frac{5y - 2}{2y + 3}\) or any equivalent representation such as \(x = \frac{2 - 5y}{-2y - 3}\)

1. Since \(TA\) is a tangent to the circle at the point of contact \(A\), the radius \(OA\) is perpendicular to the tangent:
\[\angle OAT = 90^\circ\]

2. In the right-angled triangle \(OAT\), the sum of angles is \(180^\circ\):
\[\angle AOT = 180^\circ - 90^\circ - 36^\circ = 54^\circ\]

3. Since \(O\), \(B\), and \(T\) are collinear, angle \(AOB\) is equal to angle \(AOT\):
\[\angle AOB = 54^\circ\]

4. The angle subtended by an arc at the center is twice the angle subtended by the same arc at the circumference. Therefore:
\[\angle ACB = \frac{1}{2} \angle AOB = \frac{54^\circ}{2} = 27^\circ\]

M1 for identifying that the radius meets the tangent at a right angle: \(\angle OAT = 90^\circ\) (may be implied by calculations)
M1 for calculating the angle at the center: \(\angle AOB = 54^\circ\)
A1 for the correct answer: \(27\)

1. Rewrite both sides of the equation using a base of \(3\):
\[27 = 3^3 \quad \text{and} \quad 9 = 3^2\]

2. Substitute these bases into the original equation:
\[(3^3)^{2x - 1} = \frac{1}{(3^2)^{x + 2}}\]

3. Apply the laws of indices \((a^m)^n = a^{mn}\) and \(\frac{1}{a^n} = a^{-n}\):
\[3^{3(2x - 1)} = 3^{-2(x + 2)}\]
\[3^{6x - 3} = 3^{-2x - 4}\]

4. Equate the exponents since the bases are identical:
\[6x - 3 = -2x - 4\]

5. Solve for \(x\):
\[6x + 2x = -4 + 3\]
\[8x = -1\]
\[x = -\frac{1}{8} \quad \text{or} \quad -0.125\]

M1 for expressing both sides correctly as powers of \(3\): \(3^{3(2x - 1)} = 3^{-2(x + 2)}\) (or equivalent)
M1 for equating exponents to form a linear equation: \(6x - 3 = -2x - 4\) (or equivalent from their index representation)
A1 for the correct answer: \(-\frac{1}{8}\) or \(-0.125\)

To find the \(n\)-th term of the quadratic sequence, we analyze the differences:

1st differences: \(7, \quad 11, \quad 15, \quad 19\)

2nd differences: \(4, \quad 4, \quad 4\)

Since the second difference is constant and equals \(4\), the coefficient of \(n^2\) is \(\frac{4}{2} = 2\).

Now, subtract \(2n^2\) from each term of the original sequence:

- For \(n=1\): \(5 - 2(1)^2 = 3\)
- For \(n=2\): \(12 - 2(2)^2 = 4\)
- For \(n=3\): \(23 - 2(3)^2 = 5\)
- For \(n=4\): \(38 - 2(4)^2 = 6\)

The sequence of differences \(3, 4, 5, 6, \dots\) is linear and has the general form \(n + 2\).

Combining these two parts, the \(n\)-th term is \(2n^2 + n + 2\).

M1 for finding second differences are constant at 4 (or coefficient of \(n^2\) is 2)
M1 for subtracting their \(an^2\) from the terms of the sequence to get a linear sequence
A1.5 for the fully correct final answer: \(2n^2 + n + 2\)

First, simplify the fraction inside the parentheses by moving \(y^{-3}\) to the numerator:

\[\left( \frac{64x^6 y^3}{27} \right)^{-\frac{2}{3}}\]

Next, resolve the negative exponent by taking the reciprocal of the fraction:

\[\left( \frac{27}{64x^6 y^3} \right)^{\frac{2}{3}}\]

Apply the cube root (power of \(\frac{1}{3}\)) to each component inside the parentheses:

\[\left( \frac{\sqrt[3]{27}}{\sqrt[3]{64}\sqrt[3]{x^6}\sqrt[3]{y^3}} \right)^2 = \left( \frac{3}{4x^2 y} \right)^2\]

Finally, square each term inside the parentheses:

\[\frac{3^2}{4^2 (x^2)^2 y^2} = \frac{9}{16x^4 y^2}\]

M1 for resolving the negative exponent or simplifying inside to obtain \(\left(\frac{27}{64x^6y^3}\right)^{\frac{2}{3}}\) or equivalent
M1 for correctly finding the cube root of numerical coefficients to get \(\frac{3}{4}\)
A1.5 for the fully correct simplified expression: \(\frac{9}{16x^4 y^2}\) (or \(\frac{9}{16}x^{-4}y^{-2}\))

First, expand the brackets to get \(8x - 12 - 3x + 6 = 19\). Simplify the left-hand side by combining like terms to get \(5x - 6 = 19\). Add 6 to both sides of the equation to get \(5x = 25\). Divide by 5 to find \(x = 5\).

M1 for expanding at least one bracket correctly, e.g. \(8x - 12\) or \(-3x + 6\). M1 for simplifying to the form \(ax = b\), e.g. \(5x = 25\). A0.75 for the correct answer of 5.

The exterior angle of the regular polygon is \(180^\circ - 156^\circ = 24^\circ\). Since the sum of the exterior angles of any polygon is \(360^\circ\), the number of sides \(n\) is calculated as \(360^\circ \div 24^\circ = 15\).

M1 for finding the exterior angle \(180 - 156 = 24\) or for setting up the equation \((n-2) \times 180 = 156n\). M1 for \(360 \div 24\) or for simplifying the equation to \(24n = 360\). A0.75 for the correct answer of 15.

The difference in ratio parts between sopranos and altos is \(5 - 3 = 2\) parts. Since 2 parts represent 12 members, 1 part represents \(12 \div 2 = 6\) members. The total number of parts in the choir is \(5 + 3 + 2 = 10\) parts. Therefore, the total number of members in the choir is \(10 \times 6 = 60\).

M1 for finding the difference in parts: \(5 - 3 = 2\) parts, or setting up an equation such as \(5x - 3x = 12\). M1 for finding the value of one part: \(12 \div 2 = 6\), or solving to find \(x = 6\). A0.75 for the correct answer of 60.

The first term of the arithmetic sequence is and the common difference is . The general formula for the n-th term of an arithmetic sequence is . Substituting the values into the formula gives the 40th term: .

M1 for finding the common difference of 6 or writing a correct expression for the n-th term. M1 for substituting 40 into their expression. A0.75 for the correct final answer 239.

The sum of the angles in any triangle is 180 degrees. First, find the total number of parts in the ratio: 2 + 3 + 5 = 10 parts. Next, calculate the value of one part: 180 / 10 = 18 degrees. The largest angle corresponds to the largest part of the ratio, which is 5: 5 times 18 = 90 degrees.

M1 for summing the ratio parts to get 10. M1 for setting up the calculation (5/10) times 180. A0.75 for the correct final answer 90.

First, expand the brackets: 8x - 12 - 3x + 15 = 18. Next, simplify the left-hand side by combining like terms: 5x + 3 = 18. Subtract 3 from both sides: 5x = 15. Finally, divide both sides by 5 to find x = 3.

M1 for expanding the brackets correctly to get 8x - 12 - 3x + 15 = 18. M1 for simplifying to 5x = 15. A0.75 for the correct final answer 3.

First, expand the brackets: \(4(2x - 3y) = 8x - 12y\) and \(-3(x - 5y) = -3x + 15y\). Next, group the like terms together: \((8x - 3x) + (-12y + 15y)\). Simplifying these groups gives: \(5x + 3y\).

M1 for expanding one bracket correctly (e.g. \(8x - 12y\) or \(-3x + 15y\)). M1 for completely expanding both brackets correctly: \(8x - 12y - 3x + 15y\). A0.75 for the final simplified expression \(5x + 3y\).

The sum of the angles in a triangle is \(180^\circ\). Set up the equation: \(3x + (2x + 15) + (x + 45) = 180\). Simplify the left-hand side by combining like terms: \(6x + 60 = 180\). Subtract 60 from both sides: \(6x = 120\). Divide by 6 to find \(x\): \(x = 20\).

M1 for setting up the angle-sum equation: \(3x + 2x + 15 + x + 45 = 180\). M1 for simplifying to \(6x = 120\). A0.75 for the correct final value \(20\).

Add the two equations to eliminate \(y\): \((3x + 2y) + (x - 2y) = 19 + 1\), which simplifies to \(4x = 20\). Divide by 4 to find \(x = 5\). Substitute \(x = 5\) into the second equation: \(5 - 2y = 1\). Rearranging to solve for \(y\) gives: \(2y = 4\), which results in \(y = 2\).

M1 for attempting to eliminate one variable, e.g. adding the equations to get \(4x = 20\). M1 for finding \(x = 5\) (or finding \(y\) directly via substitution, e.g. \(3(1 + 2y) + 2y = 19\)). A0.75 for the correct final value \(y = 2\).

1. Find the base angles of the isosceles triangle \(ABC\). Since \(AB = AC\), we have:
\(\text{angle } ACB = \frac{180^\circ - 44^\circ}{2} = 68^\circ\).

2. In the right-angled triangle \(BDC\), the angle \(BDC = 90^\circ\) and the angle \(BCD = 68^\circ\) (since \(D\) lies on the line \(AC\)).

3. Calculate the remaining angle \(DBC\):
\(\text{angle } DBC = 180^\circ - (90^\circ + 68^\circ) = 22^\circ\).

M1 for \(\frac{180 - 44}{2}\) or for finding \(\text{angle } ACB = 68^\circ\)
M1 for \(180 - 90 - \text{their } 68\)
A0.75 for the correct final answer 22

1. Find the difference between consecutive terms:
\(13 - 7 = 6\)
\(19 - 13 = 6\)
\(25 - 19 = 6\)
Since the first difference is constant, the sequence is linear and has the form \(6n + c\).

2. Determine the constant \(c\) by substituting \(n = 1\):
\(6(1) + c = 7 \implies c = 1\).

3. Write down the expression:
\(6n + 1\).

M1 for identifying the common difference is 6 (or for writing an expression of the form \(6n + k\))
M1 for substituting \(n = 1\) (or another term) to find \(c = 1\)
A0.75 for \(6n + 1\) (or equivalent, e.g., \(1 + 6n\))

1. Multiply both sides by 5 to clear the fraction:
\(4x - 3 = 5(2x + 3)\)
\(4x - 3 = 10x + 15\)

2. Rearrange the terms to group the \(x\) terms on one side and the constant terms on the other:
\(4x - 10x = 15 + 3\)
\(-6x = 18\)

3. Divide by \(-6\):
\(x = \frac{18}{-6} = -3\).

M1 for \(4x - 3 = 5(2x + 3)\) or better
M1 for isolating the \(x\) terms and constant terms (e.g., \(-6x = 18\) or \(6x = -18\))
A0.75 for the correct final answer \(-3\)

To solve the equation:
1. Multiply both sides of the equation by 2 to eliminate the fraction:
\( 3(x - 5) = 18 \)

2. Divide both sides by 3:
\( x - 5 = 6 \)

3. Add 5 to both sides to find the value of \( x \):
\( x = 11 \)

M1 for multiplying both sides by 2 to get \( 3(x - 5) = 18 \) or dividing both sides by 3 to get \( \frac{x-5}{2} = 3 \)
M1 for isolating the \( x \) term, e.g., \( x - 5 = 6 \) or \( 3x = 33 \)
A0.75 for the final correct answer 11

The sum of the exterior angles of any convex polygon is always \( 360^\circ \).
For a regular polygon with \( n \) sides, each exterior angle is equal to \( \frac{360^\circ}{n} \).

Set up the equation:
\( \frac{360}{n} = 45 \)

Rearrange to solve for \( n \):
\( n = \frac{360}{45} \)
\( n = 8 \)

M1 for recalling the formula for the exterior angle of a regular polygon, e.g. setting up \( \frac{360}{n} = 45 \) or \( n \times 45 = 360 \)
M1 for rearranging the equation to solve for \( n \), e.g. \( n = \frac{360}{45} \)
A0.75 for the correct final answer of 8

The ratio of their shares is Ava : Benjamin : Chloe = \( 3 : 5 : 4 \).
Chloe's share represents 4 parts of the total ratio.
Since Chloe receives \( \$36 \):
1 part = \( \frac{\$36}{4} = \$9 \).

The total number of parts in the ratio is:
\( 3 + 5 + 4 = 12 \) parts.

The total amount of money shared is:
\( 12 \times \$9 = \$108 \).

M1 for finding the value of one share/part: \( 36 \div 4 = 9 \)
M1 for finding the total number of parts, \( 3 + 5 + 4 = 12 \), or for writing the complete expression \( \frac{36}{4} \times 12 \)
A0.75 for the correct total amount of 108

The sequence is \(11, 18, 25, 32, \dots\). The common difference between consecutive terms is \(18 - 11 = 7\). This is a linear sequence of the form \(7n + c\). Since the first term (\(n = 1\)) is 11, we have \(7(1) + c = 11\), which gives \(c = 4\). Therefore, the \(n\)-th term is \(7n + 4\).

M1 for a term of \(7n\) seen or for identifying a common difference of 7. A1.75 for the correct expression \(7n + 4\).

First, expand each bracket separately: \(5(2x - 3) = 10x - 15\) and \(-3(x - 4) = -3x + 12\). Next, collect and simplify like terms: \(10x - 3x - 15 + 12 = 7x - 3\).

M1 for expanding one bracket correctly (either \(10x - 15\) or \(-3x + 12\) seen). A1.75 for the fully simplified expression \(7x - 3\).

The sum of the angles in a triangle is \(180^\circ\). Set up the equation: \(44 + 2x + (3x - 14) = 180\). Simplifying this gives \(5x + 30 = 180\). Subtracting 30 from both sides gives \(5x = 150\). Dividing by 5 yields \(x = 30\).

M1 for setting up a correct equation: \(44 + 2x + 3x - 14 = 180\) or \(5x + 30 = 180\). A1.75 for \(x = 30\).

To find the number of hours, \(h\), we substitute the total cost \(C = 61\) into the given formula: \(61 = 8h + 25\). Subtracting 25 from both sides gives \(36 = 8h\). Dividing both sides by 8 gives \(h = 36 / 8 = 4.5\). Therefore, Marcus rented the bicycle for 4.5 hours.

M1 for setting up the equation: \(8h + 25 = 61\) M1 for rearranging to isolate the term with \(h\): \(8h = 36\) A0.75 for the final answer: \(4.5\)

The sum of the angles in any triangle is always \(180^\circ\). Therefore, we can write the equation: \(2x + (3x - 10) + (x + 40) = 180\). Combining like terms, we get \((2x + 3x + x) + (-10 + 40) = 180\), which simplifies to \(6x + 30 = 180\). Subtracting 30 from both sides gives \(6x = 150\). Finally, dividing both sides by 6 gives \(x = 150 / 6 = 25\).

M1 for writing the correct angle sum equation: \(2x + (3x - 10) + (x + 40) = 180\) M1 for simplifying to \(6x + 30 = 180\) or \(6x = 150\) A0.75 for the final correct answer: \(25\)

We first find the difference between consecutive terms: \(11 - 5 = 6\), \(17 - 11 = 6\), \(23 - 17 = 6\). Since the first difference is constant and equal to 6, this is an arithmetic sequence, so the \(n\)-th term is of the form \(6n + c\), where \(c\) is a constant. Using the first term where \(n = 1\), we substitute: \(6(1) + c = 5\) which gives \(6 + c = 5\), so \(c = -1\). Thus, the expression for the \(n\)-th term is \(6n - 1\).

M1 for finding the common difference of 6 or writing an expression of the form \(6n + c\) M1 for substituting a term index to find \(c\): \(6(1) + c = 5 \implies c = -1\) A0.75 for the correct final expression: \(6n - 1\)

The perimeter of a rectangle is given by \(2 \times (\text{length} + \text{width})\).

Substitute the given expressions into the perimeter formula:
\(2((2x + 5) + (x - 3)) = 52\)

Simplify the expression inside the brackets:
\(2(3x + 2) = 52\)

Divide both sides by 2:
\(3x + 2 = 26\)

Subtract 2 from both sides:
\(3x = 24\)

Divide by 3:
\(x = 8\)

M1 for setting up the equation \(2((2x + 5) + (x - 3)) = 52\) or \(2x + 5 + x - 3 = 26\)
M1 for simplifying to \(6x + 4 = 52\) or \(3x + 2 = 26\)
A0.75 for \(x = 8\)

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

A reduction of \(15\%\) means the sale price is \(85\%\) of the original price.
\(0.85 \times P = 272\)

Solve for \(P\):
\(P = \frac{272}{0.85}\)
\(P = 320\)

So, the original price of the bicycle was \(\$320\).

M1 for realizing that \(\$272\) represents \(85\%\) of the original price (e.g., writing \(85\% = 272\))
M1 for \(\frac{272}{0.85}\) or \(\frac{272}{85} \times 100\)
A0.75 for \(320\)

First, calculate the arc length of the sector using the formula:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r\)

Substitute the given values:
\(\text{Arc length} = \frac{120}{360} \times 2 \times \pi \times 9 = \frac{1}{3} \times 18\pi = 6\pi \approx 18.85\text{ cm}\)

The perimeter of the sector is the arc length plus two radii:
\(\text{Perimeter} = \text{Arc length} + 2r\)

\(\text{Perimeter} = 6\pi + 2(9) = 18.85 + 18 = 36.85\text{ cm}\)

Rounding to 1 decimal place gives \(36.8\text{ cm}\).

M1 for calculating the arc length: \
\frac{120}{360} \times 2 \times \pi \times 9\ (or \(18.85\))
M1 for adding two radii: \(\text{their arc length} + 18\)
A0.75 for \(36.8\) (accept \(36.84\) to \(36.85\))

Multiply both sides of the formula by 3 to get: \(3w = 7t - 4\). Add 4 to both sides of the equation to get: \(3w + 4 = 7t\). Finally, divide both sides by 7 to make \(t\) the subject: \(t = \frac{3w + 4}{7}\).

M1 for multiplying both sides by 3 to obtain \(3w = 7t - 4\). M1 for adding 4 to both sides to obtain \(3w + 4 = 7t\). A0.75 for the final correct subject formula \(t = \frac{3w + 4}{7}\) or equivalent.

The exterior angle and interior angle of a regular polygon sum to \(180^\circ\). Thus, the exterior angle is \(180^\circ - 156^\circ = 24^\circ\). The sum of all exterior angles in any convex polygon is \(360^\circ\). Therefore, the number of sides is \(360^\circ \div 24^\circ = 15\).

M1 for calculating the exterior angle: \(180 - 156 = 24\). M1 for dividing 360 by their exterior angle: \(360 \div 24\). A0.75 for the correct final answer of 15.

The selling price of $253 represents 115% of the original cost price. Let the cost price be \(C\). We have the equation: \(1.15 \times C = 253\). Solving for \(C\), we find: \(C = \frac{253}{1.15} = 220\).

M1 for setting up the percentage relation: \(115\% = 253\) or \(1.15 \times \text{cost price} = 253\). M1 for the division: \(253 \div 1.15\) or equivalent. A0.75 for the final answer 220.

First, multiply every term by the common denominator, 12, to eliminate the fractions: \(12 \times \frac{2x - 3}{4} - 12 \times \frac{x + 1}{3} = 12 \times 2\). This simplifies to: \(3(2x - 3) - 4(x + 1) = 24\). Next, expand the brackets: \(6x - 9 - 4x - 4 = 24\). Combine like terms: \(2x - 13 = 24\). Add 13 to both sides: \(2x = 37\). Divide by 2: \(x = 18.5\).

M1 for correctly multiplying by a common denominator of 12: \(3(2x - 3) - 4(x + 1) = 24\) (or equivalent). M1 for correct expansion of brackets: \(6x - 9 - 4x - 4 = 24\) (allow one sign error). A0.75 for correct final answer 18.5 or \(\frac{37}{2}\).

The perimeter of the lawn consists of three straight sides of the rectangle and the curved boundary of the semi-circle. 1. Sum of the three straight sides: \(12 + 12 + 8 = 32\text{ m}\). 2. Curved boundary of the semi-circle (half of the circumference of a circle with diameter 8 m): \(\frac{1}{2} \times \pi \times 8 = 4\pi \approx 12.57\text{ m}\). 3. Total perimeter: \(32 + 12.57 = 44.57\text{ m}\). Rounding to 1 decimal place gives 44.6 m.

M1 for finding the curved arc length: \(\frac{1}{2} \times \pi \times 8\) (or \(4\pi\) or 12.57 or better). M1 for summing the three straight sides: \(12 + 12 + 8 = 32\). A0.75 for correct final answer 44.6 (accept 44.56 to 44.6).

We find the differences between consecutive terms. First differences: \(12 - 5 = 7\), \(23 - 12 = 11\), \(38 - 23 = 15\), \(57 - 38 = 19\). Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\). Since the second differences are constant and equal to \(4\), the sequence is quadratic of the form \(an^2 + bn + c\), where \(2a = 4\) which gives \(a = 2\). Subtracting \(2n^2\) from each term of the sequence: for \(n=1\): \(5 - 2(1)^2 = 3\); for \(n=2\): \(12 - 2(2)^2 = 4\); for \(n=3\): \(23 - 2(3)^2 = 5\); for \(n=4\): \(38 - 2(4)^2 = 6\). The remaining linear sequence is \(3, 4, 5, 6, \dots\), which has \(n\)-th term \(n + 2\). Thus, the \(n\)-th term is \(2n^2 + n + 2\).

M1 for finding second differences are constant and equal to 4 (or setting up equations like \(a+b+c=5\), \(4a+2b+c=12\)). M1 for finding \(a = 2\). M1 for finding \(b = 1\) or \(c = 2\). A1 for \(2n^2 + n + 2\) or equivalent.

Multiply both sides by the common denominator \((x - 3)(x + 2)\) to get \(4(x + 2) + 3(x - 3) = 2(x - 3)(x + 2)\). Expanding both sides gives \(4x + 8 + 3x - 9 = 2(x^2 - x - 6)\). Simplifying this yields \(7x - 1 = 2x^2 - 2x - 12\). Rearranging into standard quadratic form gives \(2x^2 - 9x - 11 = 0\). Factorising the quadratic expression results in \((2x - 11)(x + 1) = 0\). Solving for \(x\) gives \(2x - 11 = 0 \implies x = 5.5\) and \(x + 1 = 0 \implies x = -1\).

M1 for clearing the fractions to get \(4(x + 2) + 3(x - 3) = 2(x - 3)(x + 2)\) or equivalent. M1 for expanding and simplifying to a quadratic equation of the form \(2x^2 - 9x - 11 = 0\) (allow one sign error). M1 for solving their quadratic equation by factorising, formula, or completing the square. A1 for both solutions \(x = 5.5\) and \(x = -1\).

First, find the length of the base diagonal \(AC\) using Pythagoras' theorem in the horizontal right-angled triangle \(\triangle ABC\): \(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\). Next, consider the vertical right-angled triangle \(\triangle ACG\), where \(CG = AE = 12\text{ cm}\) is the vertical height. The angle between the diagonal \(AG\) and the base \(ABCD\) is \(\angle CAG\). Using trigonometry: \(\tan(\angle CAG) = \frac{CG}{AC} = \frac{12}{10} = 1.2\). Therefore, \(\angle CAG = \tan^{-1}(1.2) \approx 50.194\dots^\circ\). Rounding to 1 decimal place, the angle is \(50.2^\circ\).

M1 for using Pythagoras' theorem to find \(AC = \sqrt{8^2 + 6^2}\). A1 for \(AC = 10\). M1 for using \(\tan(\theta) = \frac{12}{\text{their } AC}\) (or other correct trigonometric ratio). A1 for \(50.2\) (accept answers in range \(50.19\) to \(50.20\)).

Rewrite the third term \(\frac{5}{3}\) as \(\frac{15}{9}\) so that the terms of the sequence are written with a consistent pattern in the numerators and denominators: \(\frac{3}{5}\), \(\frac{8}{7}\), \(\frac{15}{9}\), \(\frac{24}{11}\), ... Step 1: Find the \(n\)-th term of the numerators (3, 8, 15, 24, ...). First differences: 5, 7, 9, ... Second differences: 2, 2, ... Since the second difference is constant, the sequence of numerators is quadratic of the form \(a n^2 + b n + c\) where \(2a = 2\), so \(a = 1\). Subtracting \(n^2\) from the terms of the sequence gives the linear sequence: 2, 4, 6, 8, ... which has the \(n\)-th term \(2n\). Thus, the numerators have the \(n\)-th term \(n^2 + 2n\). Step 2: Find the \(n\)-th term of the denominators (5, 7, 9, 11, ...). This is an arithmetic progression with first term 5 and common difference 2. Its \(n\)-th term is \(5 + 2(n - 1) = 2n + 3\). Step 3: Combine both parts. The \(n\)-th term of the sequence is \(\frac{n^2 + 2n}{2n + 3}\).

M1 for rewriting the third term as \(\frac{15}{9}\) or showing the pattern of numerators. M1 for finding the quadratic expression \(n^2 + 2n\) for the numerators. M1 for finding the linear expression \(2n + 3\) for the denominators. A1 for the correct final answer.

The time for the outward journey is \(\frac{36}{x}\) hours. The time for the return journey is \(\frac{36}{x - 3}\) hours. Since the return journey takes 24 minutes longer, and 24 minutes is \(\frac{24}{60} = 0.4\) hours, we can write: \(\frac{36}{x - 3} - \frac{36}{x} = 0.4\). Multiplying both sides by the common denominator \(x(x - 3)\) gives: \(36x - 36(x - 3) = 0.4x(x - 3)\). Simplifying this yields: \(108 = 0.4(x^2 - 3x)\). Dividing by 0.4 gives: \(x^2 - 3x = 270\), which rearranges to: \(x^2 - 3x - 270 = 0\). Factoring the quadratic equation: \((x - 18)(x + 15) = 0\). Since speed must be positive, we reject \(x = -15\). Hence, \(x = 18\).

M1 for setting up the equation \(\frac{36}{x - 3} - \frac{36}{x} = \frac{24}{60}\) or equivalent. M1 for simplifying to a 3-term quadratic equation e.g., \(x^2 - 3x - 270 = 0\). M1 for solving their quadratic equation to find two roots. A1 for \(x = 18\) (rejecting the negative root).

The total volume is the sum of the volume of the cone and the volume of the hemisphere. The volume of the cone is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). The volume of the hemisphere is \(V_{\text{hemi}} = \frac{2}{3}\pi r^3\). The total volume is: \(V_{\text{total}} = \pi r^3 + \frac{2}{3}\pi r^3 = \frac{5}{3}\pi r^3\). We are given \(V_{\text{total}} = 360\pi\), so: \(\frac{5}{3}\pi r^3 = 360\pi\). Dividing both sides by \(\pi\) gives: \(\frac{5}{3}r^3 = 360\). Solving for \(r^3\): \(r^3 = 360 \times \frac{3}{5} = 216\). Taking the cube root of both sides gives: \(r = \sqrt[3]{216} = 6\).

M1 for volume of cone expressed as \(\pi r^3\). M1 for volume of hemisphere expressed as \(\frac{2}{3}\pi r^3\). M1 for setting up the equation \(\frac{5}{3}\pi r^3 = 360\pi\) and solving for \(r^3\). A1 for \(r = 6\).

We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 5\), \(b = -50\), and \(c = 48\). This gives \(x = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(5)(48)}}{2(5)}\). Simplifying under the square root, we get \(x = \frac{50 \pm \sqrt{2500 - 960}}{10} = \frac{50 \pm \sqrt{1540}}{10}\). Since \(\sqrt{1540} \approx 39.2428\), we have \(x = \frac{50 + 39.2428}{10} \approx 8.92\) or \(x = \frac{50 - 39.2428}{10} \approx 1.08\).

M1 for correct substitution into the quadratic formula. A1 for identifying the discriminant \(1540\). A1 for \(1.08\) (correct to 2 d.p.). A1 for \(8.92\) (correct to 2 d.p.).

The total number of counters is \(T = (3n + 2) + (n^2 + n + 1) = n^2 + 4n + 3\). Setting \(T = 323\) gives \(n^2 + 4n + 3 = 323\). Rearranging to form a quadratic equation equal to zero: \(n^2 + 4n - 320 = 0\). Factoring this quadratic, we seek two numbers that multiply to \(-320\) and add to \(4\). These are \(20\) and \(-16\). Thus, \((n + 20)(n - 16) = 0\), giving \(n = -20\) or \(n = 16\). Since the diagram number \(n\) must be positive, \(n = 16\).

M1 for writing the simplified total expression \(n^2 + 4n + 3\). M1 for setting up the equation \(n^2 + 4n - 320 = 0\). M1 for solving by factorisation to find \((n+20)(n-16)=0\) (or equivalent method). A1 for the final answer \(16\) (rejecting \(-20\)).

The arc length of the sector is given by \(\frac{\theta}{360} \times 2\pi r = 6\pi\), which simplifies to \(\frac{\theta}{360} \times r = 3\). The area of the sector is given by \(\frac{\theta}{360} \times \pi r^2 = 45\pi\), which simplifies to \(\frac{\theta}{360} \times r^2 = 45\). We can substitute \(\frac{\theta}{360} = \frac{3}{r}\) from the first equation into the second equation: \(\frac{3}{r} \times r^2 = 45 \implies 3r = 45 \implies r = 15\). Substituting \(r = 15\) back into \(\frac{\theta}{360} \times r = 3\) gives \(\frac{\theta}{360} \times 15 = 3 \implies \frac{\theta}{360} = \frac{1}{5} \implies \theta = 72\).

M1 for the arc length equation \(\frac{\theta}{360} \times 2\pi r = 6\pi\). M1 for the area equation \(\frac{\theta}{360} \times \pi r^2 = 45\pi\). A1 for finding \(r = 15\). A1 for finding \(\theta = 72\).

(a) Area of the rectangle is given by:
\(\text{Area} = \text{length} \times \text{width}\)
\(33 = (2x + 5)(x - 1)\)
\(33 = 2x^2 - 2x + 5x - 5\)
\(33 = 2x^2 + 3x - 5\)
Subtracting 33 from both sides:
\(2x^2 + 3x - 38 = 0\) [Shown]

(b) Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where \(a = 2\), \(b = 3\), and \(c = -38\):
\(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-38)}}{2(2)}\)
\(x = \frac{-3 \pm \sqrt{9 + 304}}{4}\)
\(x = \frac{-3 \pm \sqrt{313}}{4}\)

Using the positive root since length cannot be negative:
\(x = \frac{-3 + 17.6918}{4} \approx 3.673\)

The width of the garden is \(x - 1\):
\(\text{width} = 3.673 - 1 = 2.67\text{ m}\) (correct to 2 decimal places).

M1 for setting up the equation: \((2x+5)(x-1) = 33\)
M1 for expanding and simplifying to show \(2x^2 + 3x - 38 = 0\)
M1 for correct substitution into the quadratic formula: \(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-38)}}{2(2)}\)
A1 for finding the width \(2.67\)

(a) Write down equations using the given terms:
For \(n = 3\):
\(u_3 = a(3)^2 + b(3) = 21 \implies 9a + 3b = 21\)
Dividing by 3 gives:
\(3a + b = 7\) (Equation 1)

For \(n = 5\):
\(u_5 = a(5)^2 + b(5) = 55 \implies 25a + 5b = 55\)
Dividing by 5 gives:
\(5a + b = 11\) (Equation 2)

Subtract Equation 1 from Equation 2:
\((5a + b) - (3a + b) = 11 - 7\)
\(2a = 4 \implies a = 2\)

Substitute \(a = 2\) into Equation 1:
\(3(2) + b = 7 \implies 6 + b = 7 \implies b = 1\)

(b) The formula for the sequence is \(u_n = 2n^2 + n\).
For the 10th term (\(n = 10\)):
\(u_{10} = 2(10)^2 + 10 = 200 + 10 = 210\).

M1 for setting up two simultaneous equations: \(9a + 3b = 21\) and \(25a + 5b = 55\)
M1 for a valid method to solve the simultaneous equations to find \(a\) and \(b\)
A1 for both \(a = 2\) and \(b = 1\)
A1 for finding the 10th term \(210\)

Find the volume of the cylinder:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 4.5^2 \times 12\)
\(V_{\text{cylinder}} = 243\pi\text{ cm}^3\) (or approximately \(763.41\text{ cm}^3\))

The volume of the sphere is given by:
\(V_{\text{sphere}} = \frac{4}{3}\pi R^3\)

Since the volume remains constant:
\(\frac{4}{3}\pi R^3 = 243\pi\)

Divide both sides by \(\pi\):
\(\frac{4}{3}R^3 = 243\)

Solve for \(R^3\):
\(R^3 = 243 \times \frac{3}{4} = 182.25\)

Find the cube root:
\(R = \sqrt[3]{182.25} \approx 5.6696\text{ cm}\)

Rounding to 3 significant figures gives \(5.67\).

M1 for correct formula and substitution for volume of cylinder: \(\pi \times 4.5^2 \times 12\)
M1 for equating their cylinder volume to sphere volume formula: \(\frac{4}{3}\pi R^3 = 243\pi\) (or \(763.41\))
M1 for isolating \(R^3\) or \(R\): \(R = \sqrt[3]{182.25}\)
A1 for \(5.67\) (accept \(5.669\) to \(5.671\))

First, factorise the numerator: \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\). Next, factorise the denominator: \(2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5)\). Cancel out the common factor of \((x - 5)\): \(\frac{(3x + 1)(x - 5)}{2(x - 5)(x + 5)} = \frac{3x + 1}{2(x + 5)}\) (which can also be written as \(\frac{3x + 1}{2x + 10}\)).

M1 for factorising the numerator to \((3x + 1)(x - 5)\). M1 for factorising the denominator to \(2(x - 5)(x + 5)\) or \(2(x^2 - 25)\). A1 for final simplified fraction \(\frac{3x + 1}{2(x + 5)}\) or \(\frac{3x + 1}{2x + 10}\).

Find the first differences: \(11 - 4 = 7\), \(22 - 11 = 11\), \(37 - 22 = 15\). Find the second differences: \(11 - 7 = 4\), \(15 - 11 = 4\). Since the second differences are constant and equal to 4, the sequence is quadratic with leading term \(\frac{4}{2}n^2 = 2n^2\). Subtract \(2n^2\) from each term to find the linear component: For \(n=1\): \(4 - 2(1)^2 = 2\); For \(n=2\): \(11 - 2(2)^2 = 3\); For \(n=3\): \(22 - 2(3)^2 = 4\); For \(n=4\): \(37 - 2(4)^2 = 5\). The remaining linear sequence is \(2, 3, 4, 5, \dots\) which has \(n\)-th term \(n + 1\). Thus, the overall \(n\)-th term of the sequence is \(2n^2 + n + 1\).

M1 for finding the second differences are constant (equal to 4). M1 for identifying the leading term is \(2n^2\). M1 for subtracting \(2n^2\) to get the linear sequence \(2, 3, 4, 5\) (or for setting up correct simultaneous equations). A1 for the final expression \(2n^2 + n + 1\).

First, calculate the angle \(QPR\). The angle between bearing \(075^\circ\) and bearing \(130^\circ\) is \(130^\circ - 75^\circ = 55^\circ\). Use the cosine rule on triangle \(PQR\) to find the distance \(QR\): \(QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(55^\circ)\). Substitute the given values: \(QR^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos(55^\circ)\). Simplify the expression: \(QR^2 = 144 + 225 - 360 \cos(55^\circ) = 369 - 360(0.573576) = 369 - 206.4875 = 162.5125\). Thus, \(QR = \sqrt{162.5125} \approx 12.748\) km. Correct to 3 significant figures, the distance is \(12.7\) km.

M1 for finding angle \(QPR = 55^\circ\). M1 for correct substitution into the Cosine Rule: \(12^2 + 15^2 - 2 \times 12 \times 15 \times \cos(55^\circ)\). A1 for \(162.5\) or \(\sqrt{162.5\dots}\). A1 for final answer \(12.7\) (accept answers in range \(12.7\) to \(12.75\)).

Find the first and second differences:
Sequence: 5, 15, 31, 53, 81
First differences: 10, 16, 22, 28
Second differences: 6, 6, 6

Since the second difference is constant and equal to 6, the sequence contains a term in \(3n^2\) (since \(a = \frac{6}{2} = 3\)).

Subtract \(3n^2\) from the terms of the original sequence:
For \(n = 1\): \(5 - 3(1)^2 = 2\)
For \(n = 2\): \(15 - 3(2)^2 = 3\)
For \(n = 3\): \(31 - 3(3)^2 = 4\)
For \(n = 4\): \(53 - 3(4)^2 = 5\)
For \(n = 5\): \(81 - 3(5)^2 = 6\)

The remaining linear sequence is 2, 3, 4, 5, 6, ..., which has the \(n\)-th term of \(n + 1\).

Combine the quadratic and linear parts to get the final expression:
\(3n^2 + n + 1\).

M1 for finding first differences (10, 16, 22, 28) and second differences (6)
M1 for setting up the quadratic coefficient as \(3n^2\)
M1 for establishing the linear component \(n + 1\) (e.g. by subtracting \(3n^2\) from sequence values or setting up simultaneous equations)
A0.8 for the fully correct final expression \(3n^2 + n + 1\) (or equivalent)

We use the Cosine Rule to find the missing side \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)

Substitute the given values:
\(AC^2 = 7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(62^\circ)\)
\(AC^2 = 54.76 + 84.64 - 136.16\cos(62^\circ)\)
\(AC^2 = 139.4 - 136.16 \times 0.46947...\)
\(AC^2 = 139.4 - 63.923...\)
\(AC^2 = 75.477...\)

Taking the square root:
\(AC = \sqrt{75.477...} \approx 8.6877...\text{ cm}\)

To 3 significant figures, this is \(8.69\text{ cm}\).

M2 for correct substitution into the Cosine Rule: \(7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(62^\circ)\) (M1 for partial/incorrectly placed terms in a correct Cosine Rule formula)
A1 for obtaining \(AC^2\) in the range \([75.4, 75.5]\)
A0.8 for final answer of 8.69 (accept 8.68 to 8.69)

Multiply all terms by the common denominator \((x-3)(x+1)\):
\(2(x+1) + 3(x-3) = 1(x-3)(x+1)\)

Expand both sides:
\(2x + 2 + 3x - 9 = x^2 + x - 3x - 3\)
\(5x - 7 = x^2 - 2x - 3\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\(x^2 - 7x + 4 = 0\)

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(4)}}{2(1)}\)
\(x = \frac{7 \pm \sqrt{49 - 16}}{2}\)
\(x = \frac{7 \pm \sqrt{33}}{2}\)

Calculate both values:
\(x_1 = \frac{7 + 5.74456}{2} \approx 6.372 \implies 6.37\)
\(x_2 = \frac{7 - 5.74456}{2} \approx 0.627 \implies 0.63\)

M1 for multiplying by the common denominator to reach \(2(x+1) + 3(x-3) = (x-3)(x+1)\)
M1 for expanding and correctly simplifying to standard quadratic form \(x^2 - 7x + 4 = 0\)
M1 for correctly substituting their quadratic coefficients into the quadratic formula
A0.8 for both answers correct to 2 decimal places: 0.63 and 6.37 (A0.4 for one correct answer)

First, calculate the length of \(AC\) using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(AC^2 = 12^2 + 15^2 - 2(12)(15)\cos(74^\circ)\)
\(AC^2 = 144 + 225 - 360(0.275637)\)
\(AC^2 \approx 369 - 99.229 = 269.771\)
\(AC \approx 16.425\text{ m}\)

Next, find the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(74^\circ) \approx 86.514\text{ m}^2\)

Let \(h\) be the shortest distance from \(B\) to \(AC\). The area can also be expressed as:
\(\text{Area} = \frac{1}{2} \cdot AC \cdot h\)
\(86.514 = \frac{1}{2} \cdot 16.425 \cdot h\)
\(h = \frac{2 \cdot 86.514}{16.425} \approx 10.53\text{ m}\)

Rounding to 3 significant figures gives \(10.5\text{ m}\).

[M1] for a correct substitution into the Cosine Rule to find \(AC\): \(AC^2 = 12^2 + 15^2 - 2 \cdot 12 \cdot 15 \cdot \cos(74^\circ)\) (or reaching \(AC \approx 16.4\))
[M1] for finding the area of the triangle: \(\text{Area} = 0.5 \cdot 12 \cdot 15 \cdot \sin(74^\circ)\) (or reaching \(\text{Area} \approx 86.5\))
[M1] for equating their area to \(0.5 \cdot \text{their } AC \cdot h\)
[A1] for \(10.5\) (or \(10.53\dots\))

First, factorize the quadratic expression in the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Next, factorize the difference of two squares in the denominator:
\(4x^2 - 1 = (2x + 1)(2x - 1)\)

Now, rewrite the fraction and cancel the common factor of \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} =
\frac{x - 3}{2x - 1}\)

[B2] for a fully factorized numerator: \((2x + 1)(x - 3)\) (or [B1] for factorizing with one error like \((2x - 1)(x + 3)\))
[B1] for factorizing the denominator: \((2x + 1)(2x - 1)\)
[B1] for final simplified fraction: \(\frac{x - 3}{2x - 1}\)

First, determine the differences between successive terms:
Terms: \(4, \quad 7, \quad 14, \quad 25\)
First differences: \(3, \quad 7, \quad 11\)
Second differences: \(4, \quad 4\)

Since the second differences are constant, the sequence is quadratic of the form \(an^2 + bn + c\).
The coefficient \(a\) is half of the second difference:
\(a = \frac{4}{2} = 2\)

Now, subtract \(2n^2\) from the original terms to find the linear component:
For \(n=1\): \(4 - 2(1)^2 = 2\)
For \(n=2\): \(7 - 2(2)^2 = -1\)
For \(n=3\): \(14 - 2(3)^2 = -4\)
For \(n=4\): \(25 - 2(4)^2 = -7\)

The linear sequence is \(2,\ -1,\ -4,\ -7,\ \dots\)
This linear sequence has a first term of \(2\) and a common difference of \(-3\).
Its \(n\)-th term is:
\(2 + (n - 1)(-3) = -3n + 5\)

Combining both components, the overall \(n\)-th term is:
\(2n^2 - 3n + 5\)

[M1] for identifying that the second differences are constant and equal to \(4\)
[A1] for establishing the term \(2n^2\)
[M1] for attempting to find the linear sequence by subtracting \(2n^2\) from the original terms
[A1] for the correct final formula \(2n^2 - 3n + 5\)

First, form the equation for the area of the rectangle:
\((2x + 3)(x - 1) = 35\)

Expand the brackets:
\(2x^2 - 2x + 3x - 3 = 35\)
\(2x^2 + x - 3 = 35\)

Rearrange to form a quadratic equation equal to zero:
\(2x^2 + x - 38 = 0\)

Apply the quadratic formula, where \(a = 2\), \(b = 1\), and \(c = -38\):
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-38)}}{2(2)}\)
\(x = \frac{-1 \pm \sqrt{1 + 304}}{4}\)
\(x = \frac{-1 \pm \sqrt{305}}{4}\)

Calculate the two possible values of \(x\):
\(x \approx \frac{-1 + 17.4642}{4} = 4.116\)
\(x \approx \frac{-1 - 17.4642}{4} = -4.616\)

Since \(x\) represents a physical measurement (and the width is \(x - 1\)), \(x\) must be greater than 1. Thus, we discard the negative root.

Therefore, \(x \approx 4.12\) (correct to 2 decimal places).

M1 for setting up the initial equation: \((2x + 3)(x - 1) = 35\)
M1 for expanding and simplifying to a standard quadratic equation: \(2x^2 + x - 38 = 0\)
M1 for correctly substituting into the quadratic formula: \(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-38)}}{2(2)}\)
A1 for \(4.12\)

1. Find the volume of the cylinder:
\(V_{\text{cylinder}} = \pi \times \text{radius}^2 \times \text{height} = \pi \times (3x)^2 \times (2x) = \pi \times 9x^2 \times 2x = 18\pi x^3\)

2. Find the volume of the sphere:
\(V_{\text{sphere}} = \frac{4}{3}\pi r^3\)

3. Set the two volumes equal to each other:
\(18\pi x^3 = \frac{4}{3}\pi r^3\)

4. Divide both sides by \(\pi\):
\(18x^3 = \frac{4}{3}r^3\)

5. Rearrange to find the ratio of \(\frac{r^3}{x^3}\):
\(\frac{r^3}{x^3} = 18 \times \frac{3}{4} = \frac{54}{4} = \frac{27}{2}\)

6. Take the cube root of both sides to find \(\frac{r}{x}\):
\(\frac{r}{x} = \sqrt[3]{\frac{27}{2}} = \frac{3}{\sqrt[3]{2}}\)

7. Since the ratio is written as \(x : r = 1 : k\), we have:
\(k = \frac{r}{x}\)
\(k = \frac{3}{\sqrt[3]{2}} \approx 2.3811\)

Correct to 3 significant figures, \(k = 2.38\).

M1 for writing correct volume of cylinder: \(18\pi x^3\)
M1 for equating the two volumes: \(18\pi x^3 = \frac{4}{3}\pi r^3\)
M1 for isolating the ratio of the cubes: \(\frac{r^3}{x^3} = 13.5\) (or equivalent)
A1 for \(2.38\)

Using the Sine Rule on triangle \(ABC\):
\(\frac{\sin(ACB)}{AB} = \frac{\sin(BAC)}{BC}\)

Substitute the known values:
\(\frac{\sin(ACB)}{7.2} = \frac{\sin(38^\circ)}{5.4}\)

Rearrange to solve for \(\sin(ACB)\):
\(\sin(ACB) = \frac{7.2 \times \sin(38^\circ)}{5.4}\)
\(\sin(ACB) \approx 1.3333 \times 0.61566 = 0.82088\)

Find the acute angle first:
\(\text{Acute angle } ACB = \arcsin(0.82088) \approx 55.17^\circ\)

Since angle \(ACB\) is obtuse, find the second possible solution:
\(ACB = 180^\circ - 55.17^\circ = 124.83^\circ\)

Checking if the angle is valid:
\(38^\circ + 124.83^\circ = 162.83^\circ < 180^\circ\), which is possible.

Thus, the size of angle \(ACB\) is \(124.8^\circ\) correct to 1 decimal place.

M1 for correct substitution into Sine Rule: \(\frac{\sin(ACB)}{7.2} = \frac{\sin(38^\circ)}{5.4}\)
M1 for calculating \(\sin(ACB) \approx 0.821\)
M1 for finding the acute angle \(\approx 55.2^\circ\) or subtracting their acute angle from \(180^\circ\)
A1 for \(124.8\)

First, express the area of the rectangle in terms of \(x\):
\[\text{Area} = \text{length} \times \text{width}\]
\[(2x + 5)(x - 1) = 42\]

Expand the brackets:
\[2x^2 - 2x + 5x - 5 = 42\]
\[2x^2 + 3x - 5 = 42\]

Subtract 42 from both sides to form the quadratic equation:
\[2x^2 + 3x - 47 = 0\]

Next, solve the equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), and \(c = -47\):
\[x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-47)}}{2(2)}\]
\[x = \frac{-3 \pm \sqrt{9 + 376}}{4}\]
\[x = \frac{-3 \pm \sqrt{385}}{4}\]

This gives two possible values for \(x\):
\[x \approx \frac{-3 + 19.6214}{4} \approx 4.155\]
\[x \approx \frac{-3 - 19.6214}{4} \approx -5.655\]

Since \(x\) represents a physical length, the width \(x - 1\) must be greater than 0, meaning \(x > 1\). Therefore, we reject the negative solution.

\[x \approx 4.16\text{ (to 2 decimal places)}\]

M1: for setting up the equation \((2x + 5)(x - 1) = 42\)
A1: for correct expansion and rearranging to \(2x^2 + 3x - 47 = 0\)
M1: for correct substitution into the quadratic formula, i.e. \(\frac{-3 \pm \sqrt{3^2 - 4(2)(-47)}}{2(2)}\) (allow one sign error)
A1: for \(4.16\) (accept \(4.16\) and \(-5.66\) if both are presented, but only \(4.16\) is valid for the context of the garden)

To find the length of the side \(AC\) in a non-right-angled triangle where two sides and the included angle are known, we use the Cosine Rule:
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\]

Substitute the given values into the formula:
\[AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(62^\circ)\]
\[AC^2 = 49 + 100 - 140\cos(62^\circ)\]
\[AC^2 = 149 - 140(0.46947...)\]
\[AC^2 \approx 149 - 65.726\]
\[AC^2 \approx 83.274\]

Take the square root to find \(AC\):
\[AC \approx \sqrt{83.274} \approx 9.125\text{ cm}\]

Rounding to 3 significant figures gives:
\[AC \approx 9.13\text{ cm}\]

M1: for correct statement of the Cosine Rule: \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(62^\circ)\)
A1: for showing evaluation of \(AC^2 \approx 83.3\) or better
M1: for taking the square root: \(AC = \sqrt{83.274}\)
A1: for \(9.13\) (accept answers in the range \(9.12\) to \(9.13\))