2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

To factorise the expression \(12x^2y - 18xy^2\) completely, we first find the highest common factor (HCF) of the two terms.

- The HCF of the numerical coefficients \(12\) and \(18\) is \(6\).
- The HCF of the variable terms \(x^2y\) and \(xy^2\) is \(xy\).

Therefore, the highest common factor is \(6xy\).

Factoring this out of each term:
\(12x^2y = 6xy \times 2x\)
\(-18xy^2 = 6xy \times (-3y)\)

Combining these gives:
\(6xy(2x - 3y)\).

B1 for finding a partial common factor, e.g., \(3xy(4x - 6y)\) or \(6x(2xy - 3y^2)\).
M1 for \(6xy(ax + by)\) where \(a \neq 0\) and \(b \neq 0\).
A1 for the fully correct factorised expression: \(6xy(2x - 3y)\).

To find where the curve crosses the \(x\)-axis, set \(y = 0\):
\(x^2 - 4x - 5 = 0\)

Factorise the quadratic equation by finding two numbers that multiply to \(-5\) and add to \(-4\). These numbers are \(-5\) and \(1\).
\((x - 5)(x + 1) = 0\)

Solving for \(x\):
\(x - 5 = 0 \implies x = 5\)
\(x + 1 = 0 \implies x = -1\)

Therefore, the coordinates of the points are \((-1, 0)\) and \(5, 0\).

M1 for setting \(y = 0\) to get \(x^2 - 4x - 5 = 0\).
M1 for correct factorisation \((x - 5)(x + 1) = 0\) or for finding both roots \(x = 5\) and \(x = -1\).
A1 for both coordinates correct: \((-1, 0)\) and \(5, 0\) (accept in any order; also accept \(x = -1, y = 0\) and \(x = 5, y = 0\)).

The sum of the interior angles of an \(n\)-sided polygon is given by \((n - 2) \times 180^\circ\).
For a pentagon (\(n = 5\)), the sum is:
\((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)

Now, sum the given angles and set the equation equal to \(540^\circ\):
\(x + (x + 20) + (x + 40) + (x + 60) + (x + 80) = 540\)

Simplify the left side by collecting like terms:
\(5x + 200 = 540\)

Subtract \(200\) from both sides:
\(5x = 340\)

Divide by \(5\):
\(x = 68\).

M1 for correctly identifying the sum of the interior angles of a pentagon is \(540^\circ\).
M1 for setting up the equation \(5x + 200 = 540\) (or equivalent).
A1 for \(68\).

We can solve the simultaneous equations using the elimination method.

First, multiply the second equation by \(2\) so that the \(y\)-coefficients can cancel:
\(2 \times (4x - y) = 2 \times 19\)
\(8x - 2y = 38\)

Now add this new equation to the first equation:
\((3x + 2y) + (8x - 2y) = 17 + 38\)
\(11x = 55\)

Divide by \(11\):
\(x = 5\)

Substitute \(x = 5\) back into the second equation to find \(y\):
\(4(5) - y = 19\)
\(20 - y = 19\)
\(y = 1\)

Thus, the solutions are \(x = 5\) and \(y = 1\).

M1 for a correct method to eliminate one variable (e.g., multiplying the second equation by 2 to get \(8x - 2y = 38\) and adding it to the first equation).
A1 for finding one variable correctly (\(x = 5\) or \(y = 1\)).
A1 for finding both variables correctly (\(x = 5\) and \(y = 1\)).

Find the difference between consecutive terms:
\(10 - 3 = 7\)
\(17 - 10 = 7\)
\(24 - 17 = 7\)

Since the difference is constant and equal to \(7\), the term is of the form \(7n + c\).

To find \(c\), use the first term where \(n = 1\):
\(7(1) + c = 3\)
\(7 + c = 3\)
\(c = -4\)

Therefore, the expression for the \(n\)th term is \(7n - 4\).

M1 for identifying a common difference of \(7\) (or writing \(7n\) in the final expression).
M1 for using \(n = 1\) (or any other term) to find the constant term (e.g., \(7(1) + c = 3\)).
A1 for the correct expression \(7n - 4\) (or equivalent).

To make \(t\) the subject of the formula, isolate \(t\) step by step:

1. Multiply both sides of the equation by \(2\) to remove the fraction:
\(2s = 3t - 5\)

2. Add \(5\) to both sides to isolate the term with \(t\):
\(2s + 5 = 3t\)

3. Divide both sides by \(3\) to solve for \(t\):
\(t = \frac{2s + 5}{3}\)

M1 for multiplying by \(2\) to get \(2s = 3t - 5\).
M1 for adding \(5\) to get \(3t = 2s + 5\).
A1 for \(t = \frac{2s + 5}{3}\) (or equivalent, such as \(t = \frac{2s}{3} + \frac{5}{3}\)).

Since \(DE\) is parallel to \(BC\), the corresponding angles are equal. Therefore, the angle \(ABC\) is equal to angle \(ADE\):
\(\text{angle } ABC = 55^\circ\)

The sum of angles in a triangle is \(180^\circ\). In triangle \(ABC\):
\(\text{angle } BAC + \text{angle } ABC + \text{angle } ACB = 180^\circ\)

Substitute the known values:
\(\text{angle } BAC + 55^\circ + 75^\circ = 180^\circ\)
\(\text{angle } BAC + 130^\circ = 180^\circ\)

Subtract \(130^\circ\) from both sides:
\(\text{angle } BAC = 50^\circ\).

M1 for stating or using \(\text{angle } ABC = 55^\circ\) (corresponding angles).
M1 for using the angle sum in triangle \(ABC\) is \(180^\circ\) (e.g., \(180 - (55 + 75)\)).
A1 for \(50\) (or \(50^\circ\)).

First, find the gradient \(m\) of the line using the gradient formula:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Substitute the coordinates of the points \((2, 3)\) and \((6, 11)\):
\(m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\)

So the equation of the line is of the form:
\(y = 2x + c\)

To find \(c\), substitute the coordinates of one point, for example \((2, 3)\), into this equation:
\(3 = 2(2) + c\)
\(3 = 4 + c\)
\(c = -1\)

Thus, the equation of the line is \(y = 2x - 1\).

M1 for a correct method to find the gradient \(m\), resulting in \(m = 2\).
M1 for substituting their gradient and the coordinates of one point into \(y = mx + c\) to find \(c\).
A1 for the correct equation: \(y = 2x - 1\) (or equivalent).

First, find the highest common factor of \(15p^2q\) and \(10pq^2\). The highest common factor of the numbers 15 and 10 is 5. The highest common factor of \(p^2\) and \(p\) is \(p\). The highest common factor of \(q\) and \(q^2\) is \(q\). Thus, the overall highest common factor is \(5pq\). Divide each term by \(5pq\) to find the terms inside the brackets: \(15p^2q \div 5pq = 3p\) and \(-10pq^2 \div 5pq = -2q\). This gives \(5pq(3p - 2q)\).

M1 for finding a common factor of at least \(5p\), \(5q\), \(pq\) or \(5pq\). M1 for correctly identifying the terms inside the bracket as \(3p - 2q\). A1 for fully correct answer \(5pq(3p - 2q)\).

Expand the brackets: \(8x - 12 - 3x - 3 = 10\). Group like terms: \((8x - 3x) + (-12 - 3) = 10\), which simplifies to \(5x - 15 = 10\). Add 15 to both sides: \(5x = 25\). Divide by 5 to find \(x\): \(x = 5\).

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

To find the 6th term, substitute \(n = 6\) into the expression for the \(n\)-th term: \(3(6)^2 - 5 = 3(36) - 5\). Calculate the product: \(3 \times 36 = 108\). Subtract 5: \(108 - 5 = 103\).

M1 for substituting \(n=6\) into the expression. M1 for calculating \(3 \times 36\) correctly as 108. A1 for 103.

Since the interior and exterior angles of a polygon lie on a straight line, they sum to \(180^\circ\). The exterior angle is \(180^\circ - 144^\circ = 36^\circ\). The sum of all exterior angles in any polygon is \(360^\circ\). Thus, the number of sides is \(360^\circ \div 36^\circ = 10\).

M1 for finding the exterior angle: \(180^\circ - 144^\circ = 36^\circ\) or writing the equation \(180(n-2)/n = 144\). M1 for \(360 \div 36\) or solving the equation for \(n\). A1 for 10.

Substitute \(x = -3\) into the equation of the curve: \(y = (-3)^2 - 4(-3) + 1\). This simplifies to \(y = 9 + 12 + 1 = 22\). Therefore, the coordinates of the point are \((-3, 22)\).

M1 for substituting \(x = -3\) into the equation: \((-3)^2 - 4(-3) + 1\). M1 for evaluating \((-3)^2\) as 9 and \(-4(-3)\) as 12 (at least one arithmetic step shown). A1 for \((-3, 22)\) or \(x = -3, y = 22\).

Multiply both sides of the formula by 3 to eliminate the fraction: \(3V = \pi r^2 h\). Divide both sides by \(\pi r^2\) to isolate \(h\): \(h = \frac{3V}{\pi r^2}\).

M1 for multiplying both sides by 3: \(3V = \pi r^2 h\). M1 for dividing by \(\pi r^2\) (or \(\pi\) or \(r^2\)) as a correct step. A1 for \(h = \frac{3V}{\pi r^2}\) or any equivalent fully simplified form.

Multiply the second equation by 2 to get \(8x - 2y = 10\). Add this to the first equation: \((3x + 2y) + (8x - 2y) = 12 + 10\), which simplifies to \(11x = 22\). Dividing both sides by 11 gives \(x = 2\). Substitute \(x = 2\) back into the second equation: \(4(2) - y = 5 \Rightarrow 8 - y = 5 \Rightarrow y = 3\). Thus, the solution is \(x = 2\) and \(y = 3\).

M1 for multiplying the second equation by 2 to get \(8x - 2y = 10\) (or equivalent method to equate coefficients). M1 for adding the equations to eliminate \(y\) (or subtracting to eliminate \(x\)). A1 for both \(x = 2\) and \(y = 3\).

Simplify the coefficients and the variable parts separately: \(\frac{18}{6} = 3\). For the \(x\) term: \(\frac{x^5}{x^2} = x^{5-2} = x^3\). For the \(y\) term: \(\frac{y^3}{y^1} = y^{3-1} = y^2\). Combining these terms gives the fully simplified expression \(3x^3 y^2\).

M1 for simplifying the numerical coefficients to 3, or simplifying either the \(x\) variable part to \(x^3\) or the \(y\) variable part to \(y^2\). M1 for a second correct component (e.g. \(3x^3\) or \(3y^2\) or \(x^3 y^2\)). A1 for \(3x^3 y^2\).

First, expand each bracket carefully: \( 3(2x - 5) = 6x - 15 \) and \( -2(x - 4) = -2x + 8 \). Now, group the like terms together: \( 6x - 2x - 15 + 8 \). Simplify the terms to get \( 4x - 7 \).

M1 for expansion of the first bracket to \( 6x - 15 \). M1 for expansion of the second bracket to \( -2x + 8 \) (or seeing \( 6x - 15 - 2x + 8 \)). A1 for the correct final simplified expression \( 4x - 7 \).

First, multiply both sides of the equation by 2 to clear the fraction: \( 3y - 5 = 2(y + 4) \). Expand the bracket on the right side: \( 3y - 5 = 2y + 8 \). Subtract \( 2y \) from both sides: \( y - 5 = 8 \). Finally, add 5 to both sides: \( y = 13 \).

M1 for multiplying both sides by 2 to get \( 3y - 5 = 2(y + 4) \) or \( 3y - 5 = 2y + 8 \). M1 for collecting like terms on opposite sides, such as \( 3y - 2y = 8 + 5 \). A1 for 13.

The interior angle and the exterior angle at any vertex of a polygon sum to \( 180^{\circ} \). So, the exterior angle is \( 180^{\circ} - 144^{\circ} = 36^{\circ} \). The sum of all exterior angles in any polygon is \( 360^{\circ} \). The number of sides is \( 360^{\circ} \div 36^{\circ} = 10 \).

M1 for finding the exterior angle: \( 180 - 144 = 36 \). M1 for \( 360 \div \text{their exterior angle} \). A1 for 10.

First, find the common difference between consecutive terms: \( 11 - 5 = 6 \). This means the sequence is related to the \( 6n \) times table. The first term of \( 6n \) is 6, but our sequence starts at 5. We subtract 1 from \( 6n \) to get the correct term. Thus, the \( n \)-th term is \( 6n - 1 \).

M1 for identifying the common difference of 6 or writing an expression of the form \( 6n + c \). M1 for solving for \( c \) using any term: \( 6(1) + c = 5 \) hence \( c = -1 \). A1 for \( 6n - 1 \).

Identify the highest common factor of the coefficients 12 and 18, which is 6. Identify the highest common factors of the variable parts: the HCF of \( a^2 \) and \( a \) is \( a \), and the HCF of \( b \) and \( b^2 \) is \( b \). Therefore, the overall highest common factor is \( 6ab \). Divide each term by \( 6ab \) to find the terms inside the bracket: \( 12a^2b \div 6ab = 2a \) and \( -18ab^2 \div 6ab = -3b \). This gives \( 6ab(2a - 3b) \).

M1 for identifying any common factor such as \( 6 \), \( a \), \( b \), \( 6a \), or \( ab \). M1 for extracting \( 6ab \) with one term incorrect inside the bracket. A1 for the fully factorised expression \( 6ab(2a - 3b) \).

The equation of a straight line is \( y = mx + c \). Since the gradient is 4, we substitute \( m = 4 \) to get \( y = 4x + c \). Now, substitute the coordinates of the point \( (3, -2) \) into this equation: \( -2 = 4(3) + c \). This simplifies to \( -2 = 12 + c \). Subtracting 12 from both sides gives \( c = -14 \). Therefore, the equation of the line is \( y = 4x - 14 \).

M1 for substituting \( m = 4 \) to get \( y = 4x + c \). M1 for substituting \( x = 3 \) and \( y = -2 \) to find \( c \). A1 for \( y = 4x - 14 \).

The sum of angles in a triangle is always \( 180^{\circ} \). First, find the total number of parts in the ratio: \( 2 + 3 + 5 = 10 \). Next, find the value of one part: \( 180^{\circ} \div 10 = 18^{\circ} \). The largest angle corresponds to the largest share in the ratio, which is 5 parts: \( 5 \times 18^{\circ} = 90^{\circ} \).

M1 for summing the parts of the ratio: \( 2 + 3 + 5 = 10 \). M1 for calculating the size of one share \( 180 \div 10 \) or the fraction for the largest angle \( \frac{5}{10} \times 180 \). A1 for 90.

Substitute the coordinates of the point \( (3, 8) \) into the equation of the curve, where \( x = 3 \) and \( y = 8 \): \( 8 = 2(3)^2 - k(3) + 5 \). Simplify the equation: \( 8 = 2(9) - 3k + 5 \), which becomes \( 8 = 18 - 3k + 5 \), or \( 8 = 23 - 3k \). Rearrange to solve for \( k \): \( 3k = 23 - 8 \), so \( 3k = 15 \). Dividing by 3 gives \( k = 5 \).

M1 for substituting \( x = 3 \) and \( y = 8 \) into the equation. M1 for simplifying to a linear equation in \( k \), such as \( 3k = 15 \) or \( 8 = 23 - 3k \). A1 for 5.

First, find the common difference by subtracting consecutive terms: \(10 - 3 = 7\). This means the sequence increases by 7 each time, so the formula for the \(n\)-th term has the form \(7n + c\). Since the first term is 3 when \(n = 1\), we have \(7(1) + c = 3\), which gives \(c = -4\). The formula for the \(n\)-th term is \(7n - 4\). To find the 50th term, substitute \(n = 50\) into the formula: \(7(50) - 4 = 350 - 4 = 346\).

M1 for finding the common difference of 7 (or identifying the expression as \(7n + c\)) M1 for finding the correct \(n\)-th term formula \(7n - 4\) (or equivalent) A1 for 346

First, expand the double brackets: \((x + 4)(x - 3) = x^2 - 3x + 4x - 12 = x^2 + x - 12\). Next, expand the second term, paying attention to the negative sign: \(-x(x - 2) = -x^2 + 2x\). Now, combine the two parts: \(x^2 + x - 12 - x^2 + 2x\). Grouping like terms, the \(x^2\) terms cancel out: \((x^2 - x^2) + (x + 2x) - 12 = 3x - 12\).

M1 for expanding \((x + 4)(x - 3)\) to get at least 3 terms correct (e.g., \(x^2 + x - 12\)) M1 for expanding \(-x(x - 2)\) correctly to get \(-x^2 + 2x\) A1 for the correct simplified expression \(3x - 12\) or \(3(x - 4)\)

The interior angle and exterior angle of a regular polygon lie on a straight line, so they add up to \(180^\circ\). First, find the exterior angle: \(180^\circ - 144^\circ = 36^\circ\). The sum of all exterior angles in any polygon is \(360^\circ\). Therefore, the number of sides, \(n\), is given by dividing the sum of the exterior angles by the size of one exterior angle: \(n = 360^\circ \div 36^\circ = 10\).

M1 for calculating the exterior angle: \(180 - 144 = 36\) (or setting up the equation \(((n - 2) \times 180) \div n = 144\)) M1 for dividing 360 by their exterior angle (or attempting to solve their equation) A1 for 10

Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Substitute these back into the fraction to get \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancel the common factor of \(2x + 1\) to get \(\frac{x - 3}{2x - 1}\).

M1 for correctly factorising the numerator. M1 for correctly factorising the denominator. A1.5 for the final simplified fraction.

Find the first differences of the sequence: 7, 11, 15, 19, ... Then find the second differences: 4, 4, 4, ... Since the second differences are constant, the sequence is quadratic and the coefficient of \(n^2\) is half of the second difference, which is \(2\). Subtracting \(2n^2\) from the terms of the sequence yields the sequence: 3, 4, 5, 6, ... This is an arithmetic sequence with \(nth\) term \(n + 2\). Therefore, the overall \(nth\) term is \(2n^2 + n + 2\).

M1 for finding second differences are constant and equal to 4. M1 for writing down \(2n^2\) as part of the formula. A1.5 for the correct final answer.

The sum of the exterior angles of any polygon is \(360^\circ\). The exterior angles corresponding to the three given interior angles are \(180^\circ - 150^\circ = 30^\circ\), \(180^\circ - 160^\circ = 20^\circ\), and \(180^\circ - 170^\circ = 10^\circ\). The remaining \(n-3\) exterior angles are each \(180^\circ - 150^\circ = 30^\circ\). The sum of all exterior angles is: \(30^\circ + 20^\circ + 10^\circ + 30^\circ(n-3) = 360^\circ\). This simplifies to \(60 + 30(n-3) = 360\), which gives \(30(n-3) = 300\). Thus, \(n-3 = 10\), which means \(n = 13\).

M1 for finding individual exterior angles or writing the interior angle sum equation. M1 for setting up the equation \(60 + 30(n-3) = 360\) or equivalent. A1.5 for the correct value of 13.

Equating the two expressions for \(y\) gives: \(3x - 5 = x^2 - 2x - 1\). Rearranging this quadratic equation yields \(x^2 - 5x + 4 = 0\). Factorising gives \((x-1)(x-4) = 0\), which has solutions \(x = 1\) and \(x = 4\). Substituting \(x = 1\) into the first equation gives \(y = 3(1) - 5 = -2\). Substituting \(x = 4\) into the first equation gives \(y = 3(4) - 5 = 7\). Thus, the solutions are \(x = 1, y = -2\) and \(x = 4, y = 7\).

M1 for substituting to obtain a single quadratic equation in terms of \(x\). M1 for correctly factorising or solving to find both \(x\) values. A1.5 for both correct pairs of \(x\) and \(y\).

Since the curve crosses the x-axis at \(x = -2\) and \(x = 5\), its equation can be expressed in factored form as \(y = a(x + 2)(x - 5)\). Substituting the y-intercept \((0, -20)\) into this equation gives: \(-20 = a(0 + 2)(0 - 5)\), which simplifies to \(-20 = -10a\), so \(a = 2\). Substituting \(a = 2\) back gives: \(y = 2(x + 2)(x - 5) = 2(x^2 - 3x - 10) = 2x^2 - 6x - 20\). Therefore, \(a = 2, b = -6, c = -20\).

M1 for setting up the equation using the roots as factors or finding \(c = -20\). M1 for substituting the coordinates to find \(a\). A1.5 for all three correct values.

Multiply both sides by \(5 - 2x\) to get: \(y(5 - 2x) = 3x + 2\). Expand the left side: \(5y - 2xy = 3x + 2\). Rearrange to group all terms with \(x\) on one side: \(5y - 2 = 3x + 2xy\). Factorise \(x\) out of the terms on the right: \(5y - 2 = x(3 + 2y)\). Finally, divide by \(3 + 2y\) to isolate \(x\): \(x = \frac{5y - 2}{3 + 2y}\).

M1 for clearing the fraction. M1 for isolating terms containing \(x\). A1.5 for the correct final formula.

To write as a single fraction, find a common denominator, which is \((x - 2)(x + 1)\). Combine the fractions: \(\frac{4(x + 1) - 3(x - 2)}{(x - 2)(x + 1)}\). Expand the numerator: \(4x + 4 - 3x + 6\). Simplify the numerator to get \(x + 10\). Thus, the single fraction is \(\frac{x + 10}{(x - 2)(x + 1)}\).

M1 for writing over a common denominator with correct structure. M1 for correct expansion of the numerator. A1.5 for the final simplified fraction (accept expanded denominator \(x^2 - x - 2\)).

First, factorise the coefficient of \(x^2\) from the first two terms: \(2(x^2 - 6x) + 11\). Complete the square inside the brackets: \(x^2 - 6x = (x - 3)^2 - 9\). Substitute this back and expand: \(2[(x - 3)^2 - 9] + 11 = 2(x - 3)^2 - 18 + 11 = 2(x - 3)^2 - 7\). From the vertex form \(y = a(x - h)^2 + k\), the turning point is \((h, k)\). Therefore, the turning point is \((3, -7)\).

M1 for attempting to complete the square by factorising out 2 or creating a squared term. M1 for obtaining the correct vertex form \(2(x - 3)^2 - 7\). A1.5 for the correct turning point coordinates.

First, factorise the numerator and the denominator completely:
Numerator: \(3a^2 - 12b^2 = 3(a^2 - 4b^2) = 3(a - 2b)(a + 2b)\) (using the difference of two squares).
Denominator: \(2a^2 - 4ab = 2a(a - 2b)\).

Now, rewrite the fraction and cancel the common factor of \((a - 2b)\):
\(\frac{3(a - 2b)(a + 2b)}{2a(a - 2b)} =
\frac{3(a + 2b)}{2a}\) (or \(\frac{3a + 6b}{2a}\)).

M1 for factorising the numerator: \(3(a - 2b)(a + 2b)\)
M1 for factorising the denominator: \(2a(a - 2b)\)
A1.5 for the correct final simplified expression \(\frac{3(a+2b)}{2a}\) or \(\frac{3a+6b}{2a}\)

Let's find the first and second differences of the sequence:
Terms: 6, 13, 26, 45, 70
First differences: 7, 13, 19, 25
Second differences: 6, 6, 6

Since the second differences are constant, the sequence is quadratic and has the general form \(an^2 + bn + c\).
The coefficient of \(n^2\) is \(a = \frac{\text{second difference}}{2} = \frac{6}{2} = 3\).

Now, subtract \(3n^2\) from each term of the original sequence to find the linear part:
For \(n=1\): \(6 - 3(1)^2 = 3\)
For \(n=2\): \(13 - 3(2)^2 = 1\)
For \(n=3\): \(26 - 3(3)^2 = -1\)
For \(n=4\): \(45 - 3(4)^2 = -3\)
For \(n=5\): \(70 - 3(5)^2 = -5\)

The resulting sequence 3, 1, -1, -3, -5, ... is an arithmetic sequence with a first term of 3 and a common difference of \(-2\).
Its \(n\)th term is: \(3 + (n - 1)(-2) = -2n + 5\).

Combining the quadratic and linear parts gives the final \(n\)th term expression:
\(3n^2 - 2n + 5\).

M1 for identifying the second difference is 6 and establishing the \(3n^2\) term
M1 for attempting to find the linear part by subtracting \(3n^2\) or by setting up simultaneous equations
A1.5 for the correct expression \(3n^2 - 2n + 5\)

From the first equation, express \(y\) in terms of \(x\):
\(y = 2x + 1\)

Substitute this into the second equation:
\(x^2 + (2x + 1)^2 = 13\)
\(x^2 + 4x^2 + 4x + 1 = 13\)
\(5x^2 + 4x - 12 = 0\)

Solve this quadratic equation by factorisation:
We need two numbers that multiply to \(5 \times (-12) = -60\) and add to 4. These numbers are 10 and \(-6\).
\(5x^2 + 10x - 6x - 12 = 0\)
\(5x(x + 2) - 6(x + 2) = 0\)
\((5x - 6)(x + 2) = 0\)

This gives:
\(x = 1.2\) (or \(\frac{6}{5}\)) or \(x = -2\).

Now, find the corresponding \(y\) values using \(y = 2x + 1\):
If \(x = -2\), then \(y = 2(-2) + 1 = -3\).
If \(x = 1.2\), then \(y = 2(1.2) + 1 = 3.4\) (or \(\frac{17}{5}\)).

The solutions are \(x = -2, y = -3\) and \(x = 1.2, y = 3.4\).

M1 for substituting \(y = 2x + 1\) into the second equation and expanding correctly to get a quadratic equation equivalent to \(5x^2 + 4x - 12 = 0\)
M1 for factorising or using the quadratic formula to find at least one correct value for \(x\) (or \(y\))
A1.5 for both correct coordinate pairs: \(x = -2, y = -3\) and \(x = 1.2, y = 3.4\) (or equivalent fractions)

Let the size of each exterior angle be \(x^\circ\).
Then each interior angle is \((x + 120)^\circ\).

Since the interior and exterior angles at any vertex lie on a straight line, they sum to \(180^\circ\):
\(x + (x + 120) = 180\)
\(2x + 120 = 180\)
\(2x = 60\)
\(x = 30\)

So each exterior angle is \(30^\circ\).

The sum of the exterior angles of any regular polygon is \(360^\circ\). Therefore, the number of sides \(n\) is:
\(n = \frac{360}{30} = 12\).

M1 for setting up an equation linking the interior and exterior angle: \(x + x + 120 = 180\) (or equivalent)
M1 for calculating the exterior angle as \(30^\circ\) or the interior angle as \(150^\circ\)
A1.5 for the correct value of \(n = 12\)

Multiply both sides of the formula by \((5 - t)\) to clear the fraction:
\(w(5 - t) = 3t + 2\)

Expand the left-hand side:
\(5w - wt = 3t + 2\)

Rearrange to group all terms containing \(t\) on one side and the remaining terms on the other:
\(5w - 2 = 3t + wt\)

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

Divide both sides by \((w + 3)\) to isolate \(t\):
\(t = \frac{5w - 2}{w + 3}\) (or equivalent like \(t = \frac{2 - 5w}{-w - 3}\)).

M1 for multiplying by \((5 - t)\) to clear the fraction: \(w(5 - t) = 3t + 2\)
M1 for gathering all \(t\) terms on one side and factorising: \(t(3 + w) = 5w - 2\)
A1.5 for \(t = \frac{5w - 2}{w + 3}\) or equivalent

We can find the turning point by completing the square or by finding the axis of symmetry \(x = -\frac{b}{2a}\).

Method 1: Completing the square
\(y = 2(x^2 - 6x) + 7\)
\(y = 2[(x - 3)^2 - 9] + 7\)
\(y = 2(x - 3)^2 - 18 + 7\)
\(y = 2(x - 3)^2 - 11\)

This is in the form \(y = a(x - h)^2 + k\), where the vertex (turning point) is at \((h, k)\).
Therefore, the turning point is \((3, -11)\).

Method 2: Using formulas
The \(x\)-coordinate of the turning point is:
\(x = -\frac{b}{2a} = -\frac{-12}{2(2)} = 3\)

Substitute \(x = 3\) back into the curve's equation to find the \(y\)-coordinate:
\(y = 2(3)^2 - 12(3) + 7\)
\(y = 18 - 36 + 7 = -11\)

So the turning point is \((3, -11)\).

M1 for correctly identifying the \(x\)-coordinate of the vertex as 3, or attempting to complete the square by write expression in the form \(2(x-3)^2 + c\)
M1 for substituting their \(x\)-value back into the function to find \(y\), or completing the square to find \(c = -11\)
A1.5 for the final coordinates \((3, -11)\)

Express both bases, 27 and 9, as powers of 3:
\(27 = 3^3\) and \(9 = 3^2\)

Substitute these into the equation:
\((3^3)^{x - 1} = (3^2)^{2x + 3}\)

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

Since the bases are identical, equate the exponents:
\(3x - 3 = 4x + 6\)

Solve the linear equation:
\(-3 - 6 = 4x - 3x\)
\(x = -9\).

M1 for expressing both sides of the equation in terms of base 3: \(3^{3(x-1)}\) and \(3^{2(2x+3)}\)
M1 for equating indices and writing the linear equation: \(3x - 3 = 4x + 6\)
A1.5 for the correct value \(x = -9\)

Since \(DE\) is parallel to \(BC\):
1) Corresponding angles are equal, so \(\text{Angle } ADE = \text{Angle } DBC\):
\(2x + 10 = 3x - 20\)
\(3x - 2x = 10 + 20\)
\(x = 30\)

2) Corresponding angles are also equal for \(\text{Angle } AED = \text{Angle } ACB\):
\(y + 15 = 65\)
\(y = 65 - 15\)
\(y = 50\)

Thus, \(x = 30\) and \(y = 50\).

M1 for setting up the equation \(2x + 10 = 3x - 20\) and solving to find \(x = 30\)
M1 for setting up the equation \(y + 15 = 65\) and solving to find \(y = 50\)
A1.5 for both correct values of \(x = 30\) and \(y = 50\) (A0.5 if only one is correct)

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Divide the numerator by the denominator, cancelling the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

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

Find the first differences: 7, 9, 11, 13. Find the second differences: 2, 2, 2. Since the second difference is constant and equal to 2, the \(n^2\) coefficient is \(2 / 2 = 1\). Subtract \(n^2\) from the terms of the sequence: \(4 - 1 = 3\), \(11 - 4 = 7\), \(20 - 9 = 11\), \(31 - 16 = 15\). This gives the linear sequence: 3, 7, 11, 15, ... which has a first term of 3 and common difference of 4, so its \(n\)-th term is \(4n - 1\). Adding the two parts gives the \(n\)-th term of the original sequence: \(n^2 + 4n - 1\).

M1 for finding second difference of 2 (leading to \(n^2\) term). M1 for setting up equations or finding the linear sequence \(3, 7, 11, ...\) with formula \(4n - 1\). A1.5 for the correct final term \(n^2 + 4n - 1\).

The interior angle of a regular polygon is \(156^\circ\). The exterior angle 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 given by \(n = \frac{360^\circ}{24^\circ} = 15\).

M1 for finding the exterior angle: \(180 - 156 = 24\) (or for setting up \(\frac{(n-2) \times 180}{n} = 156\)). M1 for setting up \(\frac{360}{24}\) (or simplifying the equation to \(24n = 360\)). A1.5 for 15.

Substitute \(y = x - 2\) into the second equation: \(x^2 + (x - 2)^2 = 10\). Expand the bracket: \(x^2 + x^2 - 4x + 4 = 10\). Simplify to get: \(2x^2 - 4x - 6 = 0\). Divide by 2: \(x^2 - 2x - 3 = 0\). Factorise: \((x - 3)(x + 1) = 0\). This gives \(x = 3\) or \(x = -1\). If \(x = 3\), \(y = 3 - 2 = 1\). If \(x = -1\), \(y = -1 - 2 = -3\).

M1 for correct substitution of \(y\). M1 for simplified quadratic equation \(x^2 - 2x - 3 = 0\) (or equivalent). M1 for solving to find both \(x\) values: \(x = 3\) and \(x = -1\). A0.5 for both correct corresponding \(y\) values.

Express both sides as powers of 5: \(25 = 5^2\), so \(25^{x - 1} = (5^2)^{x - 1} = 5^{2x - 2}\). \(\frac{1}{5} = 5^{-1}\), so \(\left(\frac{1}{5}\right)^{3x - 8} = (5^{-1})^{3x - 8} = 5^{-3x + 8}\). Equating the indices: \(2x - 2 = -3x + 8\). Solving for \(x\): \(5x = 10\), which gives \(x = 2\).

M1 for expressing LHS as \(5^{2x - 2}\) or RHS as \(5^{-3x + 8}\). M1 for equating the exponents: \(2x - 2 = -3x + 8\). A1.5 for the correct answer \(x = 2\).

Complete the square for the expression \(2x^2 - 12x + 11\): Factorise out 2 from the \(x\) terms: \(2(x^2 - 6x) + 11\). Complete the square inside the bracket: \(2[(x - 3)^2 - 9] + 11\). Expand and simplify: \(2(x - 3)^2 - 18 + 11 = 2(x - 3)^2 - 7\). The turning point occurs when the squared term is zero, which is at \(x = 3\), giving \(y = -7\). Therefore, the turning point is \((3, -7)\). Alternatively, use the axis of symmetry formula \(x = -\frac{b}{2a} = -\frac{-12}{2 \times 2} = 3\). Substitute \(x = 3\) back into the equation to find \(y = 2(3)^2 - 12(3) + 11 = 18 - 36 + 11 = -7\).

M1 for finding the x-coordinate of the turning point, e.g., using completed square form \(2(x-3)^2\) or axis of symmetry \(x = -b/(2a) = 3\). M1 for substituting their x-coordinate to find \(y\). A1.5 for the correct coordinates \((3, -7)\).

First, find the critical values by solving the quadratic equation \(3x^2 - 10x - 8 = 0\). Factorise the quadratic expression: \((3x + 2)(x - 4) = 0\). The critical values are \(x = -\frac{2}{3}\) and \(x = 4\). Since the coefficient of \(x^2\) is positive, the graph of \(y = 3x^2 - 10x - 8\) is a U-shaped curve. The inequality is \(< 0\), which corresponds to the region below the x-axis, between the two critical values. Thus, \(-\frac{2}{3} < x < 4\).

M1 for factorising the quadratic expression to \((3x + 2)(x - 4)\). M1 for identifying correct critical values \(x = -2/3\) and \(x = 4\). A1.5 for the correct inequality range \(-2/3 < x < 4\) (or equivalent).

Tangents \(TA\) and \(TC\) meet the radii \(OA\) and \(OC\) at right angles (\(90^\circ\)). In the quadrilateral \(OATC\), the sum of angles is \(360^\circ\), so the angle at the centre is \(angle AOC = 180^\circ - 48^\circ = 132^\circ\). The angle subtended by an arc at the centre is twice the angle subtended at any point on the circumference. Since \(B\) lies on the major arc \(AC\), angle \(ABC\) is subtended by the minor arc \(AC\). Therefore, \(angle ABC = \frac{1}{2} \times angle AOC = \frac{1}{2} \times 132^\circ = 66^\circ\).

M1 for recognizing that \(angle OAT = angle OCT = 90^\circ\) or that \(angle AOC + angle ATC = 180^\circ\). M1 for finding \(angle AOC = 132^\circ\). A1.5 for \(angle ABC = 66\).

Express each term with a base of 3: \( 3^{2x-1} \times (3^2)^{x+2} = 3^{-3} \) which simplifies to \( 3^{2x-1} \times 3^{2x+4} = 3^{-3} \). Using the laws of indices to multiply, we add the exponents: \( 3^{(2x-1) + (2x+4)} = 3^{-3} \) which becomes \( 3^{4x+3} = 3^{-3} \). Equating the powers gives the linear equation: \( 4x + 3 = -3 \). Subtract 3 from both sides: \( 4x = -6 \). Divide by 4: \( x = -1.5 \) (or \( -\frac{3}{2} \)).

M1 for expressing 9 as \( 3^2 \) or 27 as \( 3^3 \). M1 for correctly applying index laws to form the linear equation \( 4x + 3 = -3 \) or equivalent. A1.5 for \( -1.5 \) or \( -\frac{3}{2} \).

First, factorise the numerator: \( 2x^2 - 5x - 3 = (2x + 1)(x - 3) \). Next, factorise the denominator as a difference of two squares: \( 4x^2 - 1 = (2x - 1)(2x + 1) \). Write the expression with these factors: \( \frac{(2x+1)(x-3)}{(2x-1)(2x+1)} \). Cancelling the common factor \( 2x + 1 \) from the numerator and denominator gives \( \frac{x-3}{2x-1} \).

M1 for factorising the numerator to \( (2x+1)(x-3) \). M1 for factorising the denominator to \( (2x-1)(2x+1) \). A1.5 for fully simplified expression \( \frac{x-3}{2x-1} \).

Find the differences between consecutive terms: First differences are \( 7, 11, 15, 19 \). Second differences are \( 4, 4, 4 \). Since the second difference is constant, the sequence is quadratic and has the form \( an^2 + bn + c \), where \( a = \frac{4}{2} = 2 \). Subtract \( 2n^2 \) from the original terms of the sequence to find the linear part: For \( n=1 \): \( 5 - 2(1)^2 = 3 \). For \( n=2 \): \( 12 - 2(2)^2 = 4 \). For \( n=3 \): \( 23 - 2(3)^2 = 5 \). The resulting sequence is \( 3, 4, 5, \dots \), which has the formula \( n + 2 \). Therefore, the overall \( n \)-th term is \( 2n^2 + n + 2 \).

M1 for finding second differences are constant and writing down \( 2n^2 \). M1 for finding the remaining linear sequence \( 3, 4, 5, \dots \) has the term \( n + 2 \) or setting up simultaneous equations. A1.5 for the correct final term \( 2n^2 + n + 2 \).

Let the exterior angle be \( x^\circ \). The interior angle is then \( (x + 140)^\circ \). Since the interior and exterior angles of a regular polygon add up to \( 180^\circ \) on a straight line: \( x + (x + 140) = 180 \) which simplifies to \( 2x + 140 = 180 \). Solving this gives \( 2x = 40 \) so \( x = 20 \). Each exterior angle is \( 20^\circ \). The number of sides of the polygon is \( \frac{360}{20} = 18 \).

M1 for setting up the equation \( x + (x + 140) = 180 \) or equivalent. M1 for finding the exterior angle is \( 20^\circ \) or the interior angle is \( 160^\circ \). A1.5 for \( 18 \).

Find the midpoint of the two points: \( M = \left(\frac{2+6}{2}, \frac{-3+7}{2}\right) = (4, 2) \). Rearrange the line equation \( 3x - 2y = 8 \) to find its gradient: \( 2y = 3x - 8 \) which becomes \( y = \frac{3}{2}x - 4 \), so the gradient of the given line is \( \frac{3}{2} \). The perpendicular line has a gradient of \( -\frac{2}{3} \). Using the midpoint \( (4, 2) \) and the perpendicular gradient: \( y - 2 = -\frac{2}{3}(x - 4) \). Expanding and simplifying gives: \( y = -\frac{2}{3}x + \frac{8}{3} + 2 \), which simplifies to \( y = -\frac{2}{3}x + \frac{14}{3} \).

M1 for finding the midpoint \( (4, 2) \). M1 for finding the gradient of the perpendicular line \( -\frac{2}{3} \). A1.5 for the correct final equation \( y = -\frac{2}{3}x + \frac{14}{3} \) or equivalent form.

First, expand each of the brackets carefully:
\(5(2x - 3) = 10x - 15\)
\(-3(x - 4) = -3x + 12\)

Now, group and combine the like terms together:
\(10x - 3x - 15 + 12\)
\(= 7x - 3\)

M1 for expansion showing either \(10x - 15\) or \(-3x + 12\) (or both)
A1.75 for correct fully simplified expression \(7x - 3\)

Alternate angles are equal. Therefore, we can set up the equation:
\(3x - 10 = 2x + 15\)

Subtract \(2x\) from both sides of the equation:
\(x - 10 = 15\)

Add \(10\) to both sides:
\(x = 25\)

M1 for setting up the equation \(3x - 10 = 2x + 15\)
A1.75 for correct answer \(25\)

Multiply both sides of the equation by \(3\) to eliminate the fraction:
\(2y + 5 = 7 \times 3\)
\(2y + 5 = 21\)

Subtract \(5\) from both sides:
\(2y = 21 - 5\)
\(2y = 16\)

Divide both sides by \(2\):
\(y = 8\)

M1 for \(2y + 5 = 21\)
A1.75 for correct answer \(8\)

Find the differences between consecutive terms:
\(11 - 4 = 7\)
\(18 - 11 = 7\)
\(25 - 18 = 7\)

Since the difference is constant, the sequence is linear with a common difference of \(7\). Thus, the formula is of the form \(7n + c\).

To find \(c\), substitute \(n = 1\) (for the first term):
\(7(1) + c = 4\)
\(7 + c = 4\)
\(c = -3\)

Therefore, the \(n\)th term is \(7n - 3\).

M1 for identifying a common difference of \(7\) (e.g., writing \(7n + c\) or similar)
A1.75 for the correct expression \(7n - 3\) (or equivalent)

The curve crosses the y-axis where \(x = 0\). Substitute \(x = 0\) into the equation of the curve:
\(y = (0)^2 - 4(0) - 5\)
\(y = -5\)

Therefore, the coordinate point is \((0, -5)\), which means \(c = -5\).

M1 for substituting \(x = 0\) into the quadratic equation
A1.75 for \(-5\)

To factorise the expression fully, find the highest common factor (HCF) of both terms:
- The HCF of the coefficients \(12\) and \(8\) is \(4\).
- The HCF of \(a\) and \(a^2\) is \(a\).
- The HCF of \(b^2\) and \(b\) is \(b\).

Combining these gives the highest common factor of \(4ab\).

Divide each term of the expression by \(4ab\):
\(12ab^2 \div 4ab = 3b\)
\(-8a^2b \div 4ab = -2a\)

So, the fully factorised expression is:
\(4ab(3b - 2a)\)

M1 for any partial factorisation, such as finding a common factor of \(4\), \(a\), or \(b\) (e.g., \(4(3ab^2 - 2a^2b)\) or \(ab(12b - 8a)\))
A1.75 for the fully factorised correct answer \(4ab(3b - 2a)\)

The sum of the interior angle and exterior angle of a polygon at any vertex is \(180^\circ\).
Exterior angle \(= 180^\circ - 140^\circ = 40^\circ\).

The sum of the exterior angles of any regular polygon is \(360^\circ\).
To find the number of sides, \(n\), divide \(360^\circ\) by the size of one exterior angle:
\(n = \frac{360^\circ}{40^\circ} = 9\).

Alternatively, use the interior angle formula:
\(\frac{(n - 2) \times 180}{n} = 140\)
\(180n - 360 = 140n\)
\(40n = 360\)
\(n = 9\).

M1 for finding the exterior angle \(180 - 140 = 40\) or for setting up the interior angle equation \((n-2)\times180 = 140n\)
A1.75 for the correct answer of \(9\)

Add the two equations together to eliminate \(y\):
\((3x + 2y) + (x - 2y) = 11 + 9\)
\(4x = 20\)

Divide by \(4\):
\(x = 5\)

Now, substitute \(x = 5\) into the second equation to find \(y\):
\(5 - 2y = 9\)
\(-2y = 9 - 5\)
\(-2y = 4\)

Divide by \(-2\):
\(y = -2\)

Thus, the solution is \(x = 5\) and \(y = -2\).

M1 for an algebraic method to eliminate one variable (e.g. adding the equations together to obtain \(4x = 20\) or isolating \(x = 9 + 2y\) and substituting)
A1.75 for both correct answers: \(x = 5\) and \(y = -2\) (Award A0.75 if only one variable value is calculated correctly)

To factorise fully, find the highest common factor (HCF) of the terms.
The HCF of \(12\) and \(18\) is \(6\).
The HCF of \(x^2\) and \(x\) is \(x\).
The HCF of \(y\) and \(y^2\) is \(y\).
So, the overall HCF is \(6xy\).
Divide each term by the HCF:
\(\frac{12x^2y}{6xy} = 2x\) and \(\frac{18xy^2}{6xy} = 3y\).
Thus, the fully factorised expression is \(6xy(2x - 3y)\).

M1 for finding any common factor such as \(2\), \(3\), \(x\), \(y\), or any combination (e.g., \(3xy(4x - 6y)\))
A1.75 for the fully correct factorised expression \(6xy(2x - 3y)\)

Substitute \(x = -2.5\) into the given equation:
\(y = 2(-2.5)^2 - 3(-2.5) - 5\)
\(y = 2(6.25) + 7.5 - 5\)
\(y = 12.5 + 7.5 - 5\)
\(y = 15\)

M1 for showing substitution of \(-2.5\) into the equation: \(2(-2.5)^2 - 3(-2.5) - 5\) (condone missing brackets around \(-2.5\) for method mark if final computation shows awareness that \((-2.5)^2 = 6.25\))
A1.75 for the correct final answer of 15

First, find the interior angle \(ACB\). Since angles on a straight line add up to \(180^\circ\):
\(\angle ACB = 180^\circ - 124^\circ = 56^\circ\).
Next, find \(\text{angle } ABC\) using the fact that the sum of angles in a triangle is \(180^\circ\):
\(\angle ABC = 180^\circ - (52^\circ + 56^\circ) = 180^\circ - 108^\circ = 72^\circ\).
Alternatively, using the exterior angle theorem, the exterior angle of a triangle equals the sum of the two opposite interior angles:
\(\angle ABC = 124^\circ - 52^\circ = 72^\circ\).

M1 for finding \(\angle ACB = 56^\circ\) or for writing the calculation \(124 - 52\)
A1.75 for the correct answer 72

To solve the equation:
Multiply both sides by \(4\):
\(3(x - 2) = 36\)
Expand the brackets:
\(3x - 6 = 36\)
Add \(6\) to both sides:
\(3x = 42\)
Divide by \(3\):
\(x = 14\)

M1 for correctly removing the fraction by multiplying by 4 to get \(3(x - 2) = 36\) or \(3x - 6 = 36\) (or for dividing by 3 first to get \(\frac{x - 2}{4} = 3\))
A1.75 for the correct final answer of 14

First, find the differences between consecutive terms:
\(11 - 5 = 6\)
\(17 - 11 = 6\)
\(23 - 17 = 6\)
Since the difference is a constant \(6\), the sequence is arithmetic, and the \(n\)-th term is of the form \(6n + c\).
To find \(c\), use the first term (where \(n=1\)):
\(6(1) + c = 5 \implies 6 + c = 5 \implies c = -1\).
Therefore, the \(n\)-th term is \(6n - 1\).

M1 for identifying the common difference of 6 (or for writing an expression of the form \(6n + c\))
A1.75 for the correct expression \(6n - 1\)

Let \(a\) be the unknown side, \(b = 8.4\text{ cm}\) be the known shorter side, and \(c = 15.6\text{ cm}\) be the hypotenuse.
By Pythagoras' theorem:
\(a^2 + b^2 = c^2\)
\(a^2 + 8.4^2 = 15.6^2\)
\(a^2 = 15.6^2 - 8.4^2\)
\(a^2 = 243.36 - 70.56\)
\(a^2 = 172.8\)
\(a = \sqrt{172.8} \approx 13.1453...\text{ cm}\)
Rounding to 1 decimal place gives \(13.1\text{ cm}\).

M1 for setting up the correct Pythagorean statement \(15.6^2 - 8.4^2\) or \(\sqrt{15.6^2 - 8.4^2}\)
A1.75 for the correct answer of 13.1 (accept 13.15 or 13.14 if due to alternative rounding, but 13.1 is the standard 1 d.p. answer)

The formula for the area of a trapezium is:
\(Area = \frac{1}{2}(a + b)h\)
where \(a\) and \(b\) are the lengths of the parallel sides, and \(h\) is the perpendicular height.
Substitute the given values into the formula:
\(Area = \frac{1}{2}(7.5 + 11.5) \times 6.4\)
\(Area = \frac{1}{2}(19) \times 6.4\)
\(Area = 9.5 \times 6.4\)
\(Area = 60.8\text{ cm}^2\)

M1 for correct substitution of the values into the trapezium area formula: \(\frac{1}{2}(7.5 + 11.5) \times 6.4\)
A1.75 for the correct final answer of 60.8

Let \(x\) be the original price of the laptop.
Since the price is reduced by \(15\%\), the sale price represents \(100\% - 15\% = 85\%\) of the original price.
Therefore, we can write the equation:
\(0.85x = 476\)
Solve for \(x\):
\(x = \frac{476}{0.85} = 560\)
So, the original price of the laptop was \(\$560\).

M1 for associating \(\$476\) with \(85\%\) (e.g., \(476 \div 0.85\) or setting up the equation \(85\% = 476\))
A1.75 for the correct final answer of 560

Expand the brackets first:
\(3(2x - 5) = 6x - 15\)
\(-4(x - 3) = -4x + 12\)

Now combine like terms by grouping the \(x\) terms and the constant terms together:
\(6x - 4x - 15 + 12 = 2x - 3\).

M1 for expanding at least one bracket correctly, i.e., showing \(6x - 15\) or \(-4x + 12\) (or \(-4x - (-12)\)).
A1.75 for the fully correct simplified expression \(2x - 3\).

The exterior angle of a triangle is equal to the sum of the two opposite interior angles:
\(\text{Angle } ACD = \text{angle } BAC + \text{angle } ABC\)
\(\text{Angle } ACD = 47^\circ + 68^\circ = 115^\circ\).

Alternatively, calculate the third interior angle first:
\(\text{Angle } ACB = 180^\circ - (47^\circ + 68^\circ) = 65^\circ\)
Then find the exterior angle on a straight line:
\(\text{Angle } ACD = 180^\circ - 65^\circ = 115^\circ\).

M1 for \(180 - (47 + 68)\) or for \(47 + 68\).
A1.75 for correct final answer \(115\).

Substitute \(x = -4\) into the equation:
\(y = (-4)^2 - 5(-4) + 3\)

Evaluate each term:
\((-4)^2 = 16\)
\(-5(-4) = 20\)

Add the terms together:
\(y = 16 + 20 + 3 = 39\).

M1 for substituting \(-4\) into the quadratic expression, showing \((-4)^2 - 5(-4) + 3\) (condone missing brackets around \(-4\) for the method mark).
A1.75 for the final answer \(39\).

Multiply both sides of the equation by 5 to clear the fraction:
\(2x - 3 = 35\)

Add 3 to both sides:
\(2x = 38\)

Divide by 2 to solve for \(x\):
\(x = 19\).

M1 for multiplying by 5 to get \(2x - 3 = 35\) (or for adding \(3/5\) to both sides to get \(2x/5 = 38/5\)).
A1.75 for the correct solution \(19\).

Find the common difference between consecutive terms:
\(11 - 7 = 4\)
\(15 - 11 = 4\)

Since the common difference is constant, this is an arithmetic sequence with common difference \(d = 4\). Therefore, the formula contains \(4n\).

When \(n = 1\), \(4(1) = 4\). To get the first term of \(7\), we must add \(3\).
Thus, the \(n\)-th term is \(4n + 3\).

M1 for finding the common difference is 4, or writing an expression of the form \(4n + k\) (where \(k\) is any constant).
A1.75 for the fully correct expression \(4n + 3\) (or equivalent).

Use the formula for the area of a trapezium:
\(\text{Area} = \frac{1}{2}(a + b)h\)

Substitute the given values, where the parallel sides are \(a = 8\) and \(b = 12\), and the height is \(h = 5.5\):
\(\text{Area} = \frac{1}{2}(8 + 12) \times 5.5\)
\(\text{Area} = \frac{1}{2}(20) \times 5.5 = 10 \times 5.5 = 55\text{ cm}^2\).

M1 for substituting the correct values into the trapezium area formula, i.e., \(\frac{1}{2}(8 + 12) \times 5.5\).
A1.75 for the correct final answer \(55\).

First, find the total number of parts in the ratio:
\(2 + 3 + 4 = 9\text{ parts}\)

Next, find the value of one part by dividing the total money by the total number of parts:
\(\$135 \div 9 = \$15\)

The largest share corresponds to the largest number in the ratio, which is \(4\) parts:
\(\text{Largest share} = 4 \times \$15 = \$60\).

M1 for dividing 135 by the sum of the ratio parts, i.e., \(135 \div (2+3+4)\).
A1.75 for the correct final answer \(60\).

In the right-angled triangle \(PQR\), with the right angle at \(Q\), the side \(PQ\) is opposite to angle \(PRQ\), and the side \(QR\) is adjacent to angle \(PRQ\).

Use the tangent ratio:
\(\tan(PRQ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{PQ}{QR}\)
\(\tan(PRQ) = \frac{6}{11}\)

Calculate angle \(PRQ\):
\(\text{Angle } PRQ = \tan^{-1}\left(\frac{6}{11}\right) \approx 28.610^\circ\)

Rounding to 1 decimal place gives \(28.6^\circ\).

M1 for writing a correct trigonometric ratio, e.g., \(\tan(PRQ) = \frac{6}{11}\).
A1.75 for the correct answer \(28.6\) (accept \(28.6^\circ\)).

To factorise the expression completely, find the highest common factor (HCF) of the terms.

1. The HCF of the coefficients \( 15 \) and \( 10 \) is \( 5 \).
2. The highest common factor of \( a^2 \) and \( a \) is \( a \).
3. The highest common factor of \( b \) and \( b^2 \) is \( b \).

Combine these to get the overall HCF: \( 5ab \).

Divide each term by \( 5ab \) to find the remaining terms inside the brackets:
\( \frac{15a^2b}{5ab} = 3a \)
\( \frac{-10ab^2}{5ab} = -2b \)

Thus, the fully factorised expression is:
\( 5ab(3a - 2b) \)

M1 for identifying a common factor and writing a partially factorised expression, such as \( 5(3a^2b - 2ab^2) \) or \( ab(15a - 10b) \).
A1.75 for the fully correct factorised form: \( 5ab(3a - 2b) \).

Substitute \( x = -2.5 \) into the given equation:

\( y = 2(-2.5)^2 - 3(-2.5) - 5 \)

First, calculate the square term:
\( (-2.5)^2 = 6.25 \)
\( 2(6.25) = 12.5 \)

Next, calculate the linear term:
\( -3(-2.5) = +7.5 \)

Now, substitute these back into the equation:
\( y = 12.5 + 7.5 - 5 \)
\( y = 20 - 5 = 15 \)

M1 for a correct substitution step showing brackets used correctly: \( 2(-2.5)^2 - 3(-2.5) - 5 \).
A1.75 for the correct final value of \( 15 \).

The sum of the interior angles of a triangle is \( 180^\circ \).

1. Find the total number of parts in the ratio:
\( 3 + 4 + 5 = 12 \text{ parts} \)

2. Find the value of one part:
\( 180^\circ \div 12 = 15^\circ \)

3. The largest angle corresponds to the largest part of the ratio (which is \( 5 \)):
\( 5 \times 15^\circ = 75^\circ \)

M1 for dividing \( 180 \) by the sum of ratio parts, i.e., \( \frac{180}{3+4+5} \) or \( 180 \div 12 \).
A1.75 for the correct answer \( 75 \) (or \( 75^\circ \)).

To find the value of \( x \), isolate \( x \) by performing inverse operations:

Multiply both sides of the equation by \( 4 \):
\( 2x - 3 = 5 \times 4 \)
\( 2x - 3 = 20 \)

Add \( 3 \ to both sides of the equation:
\) 2x = 20 + 3 \)
\( 2x = 23 \)

Divide by \( 2 \):
\( x = \frac{23}{2} \)
\( x = 11.5 \)

M1 for multiplying both sides by \( 4 \) to get \( 2x - 3 = 20 \) (or equivalent first algebraic step).
A1.75 for the correct answer \( 11.5 \) (accept \( \frac{23}{2} \) or \( 11\frac{1}{2} \)).

1. Find the difference between consecutive terms in the sequence:
\( 11 - 7 = 4 \)
\( 15 - 11 = 4 \)
\( 19 - 15 = 4 \)

Since the difference is constant, the sequence is linear and has the term \( 4n \).

2. Find the zero-th term (the term before the first term):
\( 7 - 4 = 3 \)

Alternatively, substitute \( n = 1 \) into the general form \( 4n + c \):
\( 4(1) + c = 7 \implies c = 3 \)

3. Combine these parts to write the expression for the \( n \)-th term:
\( 4n + 3 \)

M1 for finding the constant first difference of \( 4 \) and identifying \( 4n \) as part of the formula.
A1.75 for the fully correct algebraic expression \( 4n + 3 \) (or equivalent).

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

[M1] for factorising numerator into \((2x+1)(x-3)\). [M1] for factorising denominator into \((2x-1)(2x+1)\). [A1.8] for correct final simplified fraction.

Equating the equations: \(x^2 - 4x - 5 = 2x + 2\). Rearranging to get \(x^2 - 6x - 7 = 0\). Factorising gives \((x - 7)(x + 1) = 0\), which leads to \(x = 7\) and \(x = -1\). Substituting back to find \(y\): when \(x = -1\), \(y = 2(-1) + 2 = 0\); when \(x = 7\), \(y = 2(7) + 2 = 16\). The intersection points are \((-1, 0)\) and \((7, 16)\).

[M1] for setting up the equation \(x^2 - 6x - 7 = 0\). [M1] for solving to find \(x = 7\) and \(x = -1\). [A1.8] for both correct coordinate pairs.

The exterior angle is \(180^\circ - 156^\circ = 24^\circ\). The sum of all exterior angles in any polygon is \(360^\circ\). Thus, the number of sides is \(n = \frac{360^\circ}{24^\circ} = 15\).

[M1] for finding the exterior angle \(24^\circ\). [M1] for calculating \(360 / 24\). [A1.8] for correct answer 15.

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = -8\), and \(c = -2\). This yields \(x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(-2)}}{2(3)} = \frac{8 \pm \sqrt{64 + 24}}{6} = \frac{8 \pm \sqrt{88}}{6}\). Evaluating this gives \(x \approx 2.90\) and \(x \approx -0.23\).

[M1] for substitution of correct values into the quadratic formula. [M1] for simplifying to \(\frac{8 \pm \sqrt{88}}{6}\) or equivalent. [A1.8] for correct answers of 2.90 and -0.23.

The differences between consecutive terms are \(5, 7, 9\). The second differences are constant at \(2\), which indicates a quadratic term of \(n^2\). Subtracting \(n^2\) from each term gives \(2, 4, 6, 8\), which corresponds to the linear sequence \(2n\). Combining these, the \(n\)th term is \(n^2 + 2n\).

[M1] for identifying a second difference of 2 or attempting to use a quadratic form. [M1] for finding the linear part \(2n\). [A1.8] for the correct expression \(n^2 + 2n\).

Multiply both sides by \(2 - x\): \(y(2 - x) = 3x + 5\). Expand: \(2y - xy = 3x + 5\). Rearrange to isolate \(x\): \(2y - 5 = 3x + xy\). Factorise \(x\): \(2y - 5 = x(3 + y)\). Divide by \(3 + y\): \(x = \frac{2y - 5}{3 + y}\).

[M1] for multiplying both sides by \((2-x)\) and expanding correctly. [M1] for collecting all \(x\) terms on one side and factorising. [A1.8] for the correct final formula.

Find the diagonal of the base \(AC\) using Pythagoras' theorem: \(AC = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\). The angle \(\theta\) between \(EC\) and the base is the angle \(\angle ECA\) in the right-angled triangle \(EAC\). Using trigonometry, \(\tan(\theta) = \frac{AE}{AC} = \frac{5}{10} = 0.5\). Thus, \(\theta = \tan^{-1}(0.5) \approx 26.6^\circ\).

[M1] for calculating the base diagonal \(AC = 10\). [M1] for setting up the trig equation \(\tan(\theta) = 0.5\) or equivalent. [A1.8] for the correct angle of \(26.6\) (accept \(26.6^\circ\)).

The midpoint of \(AB\) is \(M = \left(\frac{2+6}{2}, \frac{3+11}{2}\right) = (4, 7)\). The gradient of \(AB\) is \(m_{AB} = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\). The gradient of the perpendicular line is \(m_{\perp} = -\frac{1}{2} = -0.5\). Using the point-slope form with \(M(4, 7)\): \(y - 7 = -0.5(x - 4)\), which simplifies to \(y = -0.5x + 9\).

[M1] for finding the correct midpoint \((4, 7)\). [M1] for calculating the perpendicular gradient of \(-0.5\). [A1.8] for finding the equation of the perpendicular bisector in the correct form.

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Substitute these back into the fraction to get \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancel the common factor \((2x + 1)\) to get the final simplified expression: \(\frac{x - 3}{2x - 1}\).

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

The sum of the interior angles of a pentagon with \(n = 5\) sides is \((5 - 2) \times 180^\circ = 540^\circ\). The sum of the parts of the ratio is \(3 + 4 + 5 + 5 + 7 = 24\). The value of one part is \(\frac{540^\circ}{24} = 22.5^\circ\). The largest angle corresponds to the largest part in the ratio, which is 7 parts: \(7 \times 22.5^\circ = 157.5^\circ\).

M1 for calculating sum of interior angles: \((5-2) \times 180 = 540^\circ\). M1 for dividing sum by the total parts: \(\frac{540}{24}\). A1 for the correct answer: \(157.5\).

First, find the differences between consecutive terms. First differences: \(11 - 3 = 8\), \(23 - 11 = 12\), \(39 - 23 = 16\). Second differences: \(12 - 8 = 4\), \(16 - 12 = 4\). Since the second differences are constant, the sequence is quadratic with a leading term of \(a n^2\), where \(2a = 4 \implies a = 2\). Subtract \(2n^2\) from each term of the original sequence to find the linear part: for \(n = 1\), \(3 - 2(1) = 1\); for \(n = 2\), \(11 - 2(4) = 3\); for \(n = 3\), \(23 - 2(9) = 5\); for \(n = 4\), \(39 - 2(16) = 7\). The resulting linear sequence is \(1, 3, 5, 7\), which has the formula \(2n - 1\). Combining these, the \(n\)-th term is \(2n^2 + 2n - 1\).

M1 for finding second difference of 4, leading to \(2n^2\). M1 for subtracting \(2n^2\) to get linear sequence \(1, 3, 5, 7\). A1 for \(2n^2 + 2n - 1\).

To find the stationary points, we find the first derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - 12\). Set \(\frac{dy}{dx} = 0\) to get \(3x^2 - 12 = 0\), which gives \(x^2 = 4\), so \(x = 2\) or \(x = -2\). For \(x = 2\), \(y = 2^3 - 12(2) + 5 = -11\). For \(x = -2\), \(y = (-2)^3 - 12(-2) + 5 = 21\). The second derivative is \(\frac{d^2y}{dx^2} = 6x\). For \(x = -2\), \(\frac{d^2y}{dx^2} = -12 < 0\), which confirms a local maximum. Thus, the local maximum point is \((-2, 21)\).

M1 for differentiating: \(\frac{dy}{dx} = 3x^2 - 12\). M1 for setting to 0 and solving to find \(x = \pm 2\). A1 for finding the correct coordinates of the local maximum: \((-2, 21)\).

Rearrange the linear equation to get \(y = 3 - x\). Substitute this into the quadratic equation: \(x^2 + (3 - x)^2 = 29\). Expand and simplify: \(x^2 + 9 - 6x + x^2 = 29\), which simplifies to \(2x^2 - 6x - 20 = 0\). Divide by 2 to get \(x^2 - 3x - 10 = 0\). Factorise the quadratic equation: \((x - 5)(x + 2) = 0\), which gives \(x = 5\) or \(x = -2\). Since we require \(x > 0\), we select \(x = 5\). Substitute \(x = 5\) into \(y = 3 - x\) to get \(y = -2\). The coordinates are \((5, -2)\).

M1 for substituting \(y = 3 - x\) into the quadratic equation. M1 for forming the simplified quadratic equation \(2x^2 - 6x - 20 = 0\) (or equivalent). M1 for factorising and solving to get \(x = 5\). A1 for the coordinates \((5, -2)\).

Multiply both sides by \((2 - t)\) to get \(w(2 - t) = 3t + 5\). Expand the left side: \(2w - wt = 3t + 5\). Collect all terms containing \(t\) on one side: \(2w - 5 = 3t + wt\). Factorise \(t\) on the right side: \(2w - 5 = t(3 + w)\). Finally, divide by \(w + 3\) to get \(t = \frac{2w - 5}{w + 3}\).

M1 for multiplying by \(2 - t\) to get \(w(2 - t) = 3t + 5\). M1 for isolating terms with \(t\): \(2w - 5 = 3t + wt\). M1 for factorising \(t\): \(t(3 + w)\). A1 for final answer: \(t = \frac{2w - 5}{w + 3}\).

Let us model the journey using triangle ABC. Port A is the start and point B is 12 km North of A, meaning BA points South. The bearing from B to C is \(120^\circ\). The interior angle \(\angle ABC\) is the angle between South (\(180^\circ\)) and the bearing of \(120^\circ\), which is \(\angle ABC = 180^\circ - 120^\circ = 60^\circ\). Now use the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substituting the values, we get: \(AC^2 = 12^2 + 15^2 - 2 \cdot 12 \cdot 15 \cdot \cos(60^\circ) = 144 + 225 - 360 \cdot 0.5 = 189\). Thus, \(AC = \sqrt{189} \approx 13.7477\) km, which is \(13.7\) km to 3 significant figures.

M1 for finding the angle \(\angle ABC = 180^\circ - 120^\circ = 60^\circ\). M1 for correctly substituting into the Cosine Rule: \(12^2 + 15^2 - 2(12)(15)\cos(60^\circ)\). A1 for getting \(13.7\) (accept 13.7 to 13.75).

Let \(P\) be the original price of the laptop. A 15% reduction means the sale price is \(100\% - 15\% = 85\%\) of the original price. This gives the equation \(0.85 \times P = 612\). Solving for \(P\), we get \(P = \frac{612}{0.85} = 720\). Thus, the original price of the laptop was $720.

M1 for identifying that 85% corresponds to $612. M1 for calculating \(\frac{612}{0.85}\). A1 for 720.

First, factorise the numerator and the denominator:
Numerator: \(2x^2 - 7x - 15 = (2x + 3)(x - 5)\)
Denominator: \(x^2 - 25 = (x + 5)(x - 5)\)

Now, cancel the common factor of \(x - 5\) from both the numerator and the denominator:
\(\frac{(2x + 3)(x - 5)}{(x + 5)(x - 5)} = \frac{2x + 3}{x + 5}\)

M1 for factorising the numerator: \((2x + 3)(x - 5)\)
M1 for factorising the denominator: \((x + 5)(x - 5)\)
A1.8 for the final simplified expression: \(\frac{2x + 3}{x + 5}\)

To find the gradient of the curve, we differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 3x^2 - 6x - 9\)

Set this gradient equal to \(15\):
\(3x^2 - 6x - 9 = 15\)
\(3x^2 - 6x - 24 = 0\)

Divide the entire equation by 3:
\(x^2 - 2x - 8 = 0\)

Factorise the quadratic equation:
\((x - 4)(x + 2) = 0\)

Thus, the \(x\)-coordinates are \(x = 4\) and \(x = -2\).

M1 for differentiating to obtain \(3x^2 - 6x - 9\)
M1 for setting their derivative equal to 15 to form a quadratic equation
M1 for factorising or solving their quadratic equation
A0.8 for both correct values: \(x = -2\) and \(x = 4\)

The interior angle and the exterior angle of a regular polygon sum to \(180^\circ\).
Exterior angle = \(180^\circ - 162^\circ = 18^\circ\)

The sum of the exterior angles of any polygon is always \(360^\circ\).
Number of sides = \(\frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{18^\circ} = 20\).

M1 for finding the exterior angle: \(180 - 162 = 18\)
M1 for using \(360 / \text{exterior angle}\)
A1.8 for \(20\)

Substitute \(y = 2x - 5\) into the quadratic equation:
\(x^2 + (2x - 5)^2 = 25\)
\(x^2 + (4x^2 - 20x + 25) = 25\)
\(5x^2 - 20x = 0\)

Factorise the resulting quadratic equation:
\(5x(x - 4) = 0\)
This gives the \(x\)-values:
\(x = 0\) or \(x = 4\)

Now find the corresponding \(y\)-values using \(y = 2x - 5\):
For \(x = 0\): \(y = 2(0) - 5 = -5\)
For \(x = 4\): \(y = 2(4) - 5 = 3\)

M1 for substituting \(y = 2x - 5\) into \(x^2 + y^2 = 25\)
M1 for expanding and simplifying to a quadratic equation: \(5x^2 - 20x = 0\) (or equivalent)
M1 for finding both \(x\)-values (\(0\) and \(4\))
A0.8 for both correct pairs of coordinates: \(x = 0, y = -5\) and \(x = 4, y = 3\)

Find the first and second differences:
First terms: \(5, 14, 29, 50\)
First differences: \(14 - 5 = 9\), \(29 - 14 = 15\), \(50 - 29 = 21\)
Second differences: \(15 - 9 = 6\), \(21 - 15 = 6\)

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

Subtract \(3n^2\) from the original terms to find the linear or constant part:
For \(n = 1\): \(5 - 3(1)^2 = 2\)
For \(n = 2\): \(14 - 3(2)^2 = 2\)
For \(n = 3\): \(29 - 3(3)^2 = 2\)

Since the result is a constant \(2\), the formula for the \(n\)th term is \(3n^2 + 2\).

M1 for finding second differences are constant and equal to 6
M1 for starting with \(3n^2\) as the leading term
A1.8 for the correct expression \(3n^2 + 2\) (or equivalent)

Using the cosine rule on triangle \(ABC\) to find angle \(A\) (which is angle \(BAC\)):
\(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)\)

Substitute the given side lengths:
\(10^2 = 7^2 + 12^2 - 2(7)(12)\cos(\angle BAC)\)
\(100 = 49 + 144 - 168\cos(\angle BAC)\)
\(100 = 193 - 168\cos(\angle BAC)\)

Rearrange to solve for \(\cos(\angle BAC)\):
\(168\cos(\angle BAC) = 193 - 100\)
\(168\cos(\angle BAC) = 93\)
\(\cos(\angle BAC) = \frac{93}{168}\)

Find the inverse cosine:
\(\angle BAC = \cos^{-1}\left(\frac{93}{168}\right) \approx 56.384^\circ\)

Rounding to 1 decimal place gives \(56.4^\circ\).

M1 for correct substitution into the Cosine Rule: \(10^2 = 7^2 + 12^2 - 2(7)(12)\cos(A)\)
M1 for rearranging to make \(\cos(A)\) the subject: \(\cos(A) = \frac{93}{168}\) (or equivalent)
A1.8 for \(56.4\) (or \(56.4^\circ\))

The total number of sweets initially is \(5 + 7 = 12\).

There are two possible ways to select two sweets of different colours:
1) A red sweet first, then a blue sweet:
\(P(\text{Red then Blue}) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)

2) A blue sweet first, then a red sweet:
\(P(\text{Blue then Red}) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\)

Add these two probabilities to find the total probability:
\(P(\text{different colours}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\) (or approximately \(0.530\)).

M1 for finding either product: \(\frac{5}{12} \times \frac{7}{11}\) or \(\frac{7}{12} \times \frac{5}{11}\)
M1 for adding both possible products together: \(\frac{35}{132} + \frac{35}{132}\)
A1.8 for \(\frac{35}{66}\) (or any equivalent fraction or decimal rounded to at least 3 significant figures: \(0.530\))

The compound interest formula is:
\(A = P\left(1 + \frac{r}{100}\right)^n\)

Substitute the given values:
\(A = 3500 \times (1 + 0.024)^5\)
\(A = 3500 \times (1.024)^5\)
\(A \approx 3500 \times 1.1258999\)
\(A \approx 3940.64966\)

To find the interest earned, subtract the initial principal from the total amount:
\(\text{Interest} = 3940.64966 - 3500 = 440.64966\)

Rounding to the nearest cent gives \(\$440.65\).

M1 for correct substitute into the compound interest formula: \(3500 \times (1.024)^5\)
M1 for finding the total value of the investment: \(3940.65\)
A1.8 for subtracting the principal to find the interest: \(440.65\)

To write \(\frac{3}{2x - 1} - \frac{4}{x + 3}\) as a single fraction, first find a common denominator, which is \((2x - 1)(x + 3)\). Express both fractions with this denominator: \(\frac{3(x + 3) - 4(2x - 1)}{(2x - 1)(x + 3)}\). Expand the numerator: \(3(x + 3) = 3x + 9\) and \(-4(2x - 1) = -8x + 4\). Combine the terms in the numerator: \(3x + 9 - 8x + 4 = 13 - 5x\). Thus, the simplified single fraction is \(\frac{13 - 5x}{(2x - 1)(x + 3)}\) or \(\frac{13 - 5x}{2x^2 + 5x - 3}\).

M1 for expressing fractions with a common denominator \((2x - 1)(x + 3)\). M1 for expanding the numerator correctly to obtain \(3x + 9 - 8x + 4\) or equivalent. A1 for the correct numerator \(13 - 5x\) (or \(-5x + 13\)). A1 for the correct denominator \((2x - 1)(x + 3)\) or \(2x^2 + 5x - 3\).

Find the first-order differences: \(11 - 4 = 7\), \(22 - 11 = 11\), \(37 - 22 = 15\), \(56 - 37 = 19\). Find the second-order 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 \(2a = 4\), so \(a = 2\). Subtracting \(2n^2\) from the terms of the sequence: For \(n=1\): \(4 - 2(1)^2 = 2\); For \(n=2\): \(11 - 2(2)^2 = 3\); For \(n=3\): \(22 - 2(3)^2 = 4\). The remaining sequence is 2, 3, 4, ... which is an arithmetic sequence with \(n\)-th term \(n + 1\). Therefore, the overall \(n\)-th term is \(2n^2 + n + 1\).

M1 for finding second-order differences of 4. M1 for setting up quadratic term as \(2n^2\). M1 for attempting to find the linear component \(n + 1\) (e.g. by subtracting \(2n^2\) or using simultaneous equations). A1 for the correct expression \(2n^2 + n + 1\).