2025 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

Apply the power to both the coefficient and the variable part:
\((27)^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9\)
\((w^9)^{\frac{2}{3}} = w^{9 \times \frac{2}{3}} = w^6\)
Combining the terms gives \(9w^6\).

B2 for \(9w^6\)

OR
B1 for \(9w^k\) \((k \neq 6)\) or \(k w^6\) \((k \neq 9)\) or \((3w^3)^2\) seen

Expand the brackets on both sides:
\[8x - 12 = 3x + 18\]
Rearrange to collect the \(x\) terms on one side and constants on the other:
\[8x - 3x = 18 + 12\]
\[5x = 30\]
Divide by 5:
\[x = 6\]

M1 for correct expansion of both brackets to \(8x - 12 = 3x + 18\), or for correct rearrangement to \(ax = b\) following one expansion error
A1 for \(6\) (or \(x = 6\))

To factorise the expression completely, find the highest common factor (HCF) of the numerical coefficients and each variable term:

1. The HCF of $15$ and $25$ is $5$.
2. The highest common power of $p$ is $p^1 = p$.
3. The highest common power of $q$ is $q^1 = q$.

Hence, the overall common factor is $5pq$.

Factor out $5pq$ from each term:
\[15p^2q \div 5pq = 3p\]
\[-25pq^3 \div 5pq = -5q^2\]

Combine the factors:
\[5pq(3p - 5q^2)\]

B2 for \(5pq(3p - 5q^2)\) final answer

B1 for a correct partial factorisation such as \(5p(3pq - 5q^3)\), \(5q(3p^2 - 5pq^2)\), or \(pq(15p - 25q^2)\), or for \(5pq(3p - 5q^2)\) seen then incorrectly expanded.

1. Multiply both sides by 12 (or cross-multiply):
\[3(3x - 2) = 4(x + 5)\]

2. Expand the brackets:
\[9x - 6 = 4x + 20\]

3. Collect terms in $x$ on one side and numerical constants on the other:
\[9x - 4x = 20 + 6\]
\[5x = 26\]

4. Solve for $x$:
\[x = \frac{26}{5} = 5.2\text{ (or } 5\frac{1}{5}\text{)}\]

M1 for correctly clearing fractions to obtain \(3(3x - 2) = 4(x + 5)\) or \(9x - 6 = 4x + 20\)

A1 for \(5.2\) or \(\frac{26}{5}\) or \(5\frac{1}{5}\)

Apply the power \(-\frac{3}{4}\) to each term inside the bracket:

\(16^{-\frac{3}{4}} = \left(16^{\frac{1}{4}}\right)^{-3} = 2^{-3} = \frac{1}{8}\)

\((x^8)^{-\frac{3}{4}} = x^{8 \times \left(-\frac{3}{4}\right)} = x^{-6}\)

\((y^{-4})^{-\frac{3}{4}} = y^{-4 \times \left(-\frac{3}{4}\right)} = y^3\)

Combining the simplified parts:
\(\frac{1}{8} x^{-6} y^3 = \frac{y^3}{8x^6}\)

B2 for \(\frac{y^3}{8x^6}\) or \(\frac{1}{8}x^{-6}y^3\) or \(0.125x^{-6}y^3\)

OR
B1 for any two of \(\frac{1}{8}\), \(x^{-6}\), \(y^3\) correct in a single product

Multiply all terms by the common denominator, 12, to eliminate fractions:

\(3(3x - 1) - 4(x + 2) = 12 \times 2\)

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

Collect like terms:
\(5x - 11 = 24\)

\(5x = 35\)

\(x = 7\)

M1 for correctly writing algebraic fractions over a common denominator or multiplying throughout by 12, e.g. \(3(3x - 1) - 4(x + 2) = 24\) or \(\frac{9x - 3 - 4x - 8}{12} = 2\)
A1 for \(x = 7\) (or \(7\) on the answer line)

Apply the power to both the numerical coefficient and the variable term:

\[ (64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16 \]

\[ (w^{12})^{\frac{2}{3}} = w^{12 \times \frac{2}{3}} = w^8 \]

Multiplying the components gives \(16w^8\).

B1 for \(16\) or \(w^8\) seen in a product of the form \(k w^8\) or \(16 w^n\) (where \(k, n \neq 0\))
B1 for \(16w^8\) cao

Multiply both sides by 4:
\[ 4m = 3p - 7 \]

Add 7 to both sides:
\[ 4m + 7 = 3p \]

Divide both sides by 3:
\[ p = \frac{4m + 7}{3} \]

M1 for correctly multiplying by 4 to get \(4m = 3p - 7\) or correctly isolating \(3p\) as a first step
A1 for \(p = \frac{4m + 7}{3}\) or \(p = \frac{4}{3}m + \frac{7}{3}\) oe

Divide the numerical coefficients and subtract the indices for like bases: \(\dfrac{15}{3} = 5\), \(x^{7-2} = x^5\), and \(y^{-2-4} = y^{-6} = \dfrac{1}{y^6}\). Combining these gives \(5x^5 y^{-6}\) or \(\dfrac{5x^5}{y^6}\).

B2 for \(\dfrac{5x^5}{y^6}\) or \(5x^5 y^{-6}\)
(M1 for any two of \(5\), \(x^5\), or \(y^{-6}\) (or \(\dfrac{1}{y^6}\)) correct in a single term)

Group terms with common factors: \(12px - 8qx + 15py - 10qy = 4x(3p - 2q) + 5y(3p - 2q)\). Factor out the common binomial factor \((3p - 2q)\): \((3p - 2q)(4x + 5y)\).

B2 for \((3p - 2q)(4x + 5y)\) oe
(M1 for partial factorisation by grouping, e.g. \(4x(3p - 2q) + 5y(3p - 2q)\) or \(3p(4x + 5y) - 2q(4x + 5y)\))

Find the highest common factor of the coefficients and the algebraic terms:
- The highest common factor of $18$ and $24$ is $6$.
- The common variable terms are $x$ and $y$.

Factoring out $6xy$:
$$18x^2y - 24xy^3 = 6xy(3x - 4y^2)$$

B2 for $6xy(3x - 4y^2)$

(B1 for a correct partial factorisation with at least two common factors extracted, e.g. $3xy(6x - 8y^2)$, $6x(3xy - 4y^3)$, $6y(3x^2 - 4xy^2)$, or $xy(18x - 24y^2)$)

Factorise the quadratic numerator:
$$2x^2 + 5x - 12 = (2x - 3)(x + 4)$$

Factorise the denominator as a difference of two squares:
$$4x^2 - 9 = (2x - 3)(2x + 3)$$

Cancel the common factor of \((2x - 3)\):
$$\frac{(2x - 3)(x + 4)}{(2x - 3)(2x + 3)} = \frac{x + 4}{2x + 3}$$

M1 for factorising the numerator into \((2x - 3)(x + 4)\) (or two brackets giving two correct terms when expanded)
M1 for factorising the denominator into \((2x - 3)(2x + 3)\)
A1 for \(\frac{x + 4}{2x + 3}\) cao

Find \(\vec{OM}\) and \(\vec{MA}\):
Since \(OM : MA = 2 : 1\), \(\vec{OM} = \frac{2}{3}\vec{OA} = \frac{2}{3}(6\mathbf{a}) = 4\mathbf{a}\) and \(\vec{MA} = \frac{1}{3}(6\mathbf{a}) = 2\mathbf{a}\).

Find \(\vec{AB}\):
$$\vec{AB} = \vec{AO} + \vec{OB} = -6\mathbf{a} + 10\mathbf{b}$$

Find \(\vec{AN}\):
Since \(AN : NB = 3 : 2\), \(\vec{AN} = \frac{3}{5}\vec{AB} = \frac{3}{5}(-6\mathbf{a} + 10\mathbf{b}) = -\frac{18}{5}\mathbf{a} + 6\mathbf{b}\).

Find \(\vec{MN}\):
$$\vec{MN} = \vec{MA} + \vec{AN} = 2\mathbf{a} + \left(-\frac{18}{5}\mathbf{a} + 6\mathbf{b}\right) = \left(2 - \frac{18}{5}\right)\mathbf{a} + 6\mathbf{b} = -\frac{8}{5}\mathbf{a} + 6\mathbf{b}$$

B1 for \(\vec{OM} = 4\mathbf{a}\) or \(\vec{MA} = 2\mathbf{a}\) soi
M1 for \(\vec{AB} = -6\mathbf{a} + 10\mathbf{b}\) oe
M1 for \(\vec{MN} = \vec{MA} + \frac{3}{5}\vec{AB}\) or \(\vec{MN} = \vec{ON} - \vec{OM}\) oe with correct vector path
A1 for \(-\frac{8}{5}\mathbf{a} + 6\mathbf{b}\) or \(6\mathbf{b} - 1.6\mathbf{a}\) or \(\frac{1}{5}(30\mathbf{b} - 8\mathbf{a})\) oe in simplest form

Factorise the numerator:
\[2x^2 + 3x - 20 = 2x^2 + 8x - 5x - 20 = 2x(x + 4) - 5(x + 4) = (2x - 5)(x + 4)\]

Factorise the denominator as a difference of two squares:
\[4x^2 - 25 = (2x - 5)(2x + 5)\]

Cancel the common factor \((2x - 5)\):
\[\frac{(2x - 5)(x + 4)}{(2x - 5)(2x + 5)} = \frac{x + 4}{2x + 5}\]

M1 for \((2x - 5)(x + 4)\)
M1 for \((2x - 5)(2x + 5)\)
A1 for \(\frac{x + 4}{2x + 5}\) or \(\frac{x+4}{2x+5}\) cao

First, find \(\vec{AB}\):
\[\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\]

Since \(AP : PB = 2 : 3\), \(\vec{AP} = \frac{2}{5}\vec{AB} = \frac{2}{5}(\mathbf{b} - \mathbf{a})\).

Since \(M\) is the midpoint of \(OA\), \(\vec{MA} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}\).

Now, express \(\vec{MP}\) along the path \(M \to A \to P\):
\[\vec{MP} = \vec{MA} + \vec{AP}\]
\[\vec{MP} = \frac{1}{2}\mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\]
\[\vec{MP} = \frac{1}{2}\mathbf{a} - \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\]
\[\vec{MP} = \left(\frac{5}{10} - \frac{4}{10}\right)\mathbf{a} + \frac{2}{5}\mathbf{b} = \frac{1}{10}\mathbf{a} + \frac{2}{5}\mathbf{b}\]

B1 for \(\vec{AB} = \mathbf{b} - \mathbf{a}\) oe soi
B1 for \(\vec{MA} = \frac{1}{2}\mathbf{a}\) or \(\vec{OM} = \frac{1}{2}\mathbf{a}\) soi
M1 for a correct vector route for \(\vec{MP}\), e.g. \(\vec{MA} + \frac{2}{5}\vec{AB}\) or \(-\vec{OM} + \vec{OA} + \frac{2}{5}\vec{AB}\)
A1 for \(\frac{1}{10}\mathbf{a} + \frac{2}{5}\mathbf{b}\) or \(\frac{\mathbf{a} + 4\mathbf{b}}{10}\) oe in simplest form

1. Factorise the quadratic numerator:
$$3x^2 - 10x - 8 = (3x + 2)(x - 4)$$

2. Factorise the denominator as a difference of two squares:
$$9x^2 - 4 = (3x - 2)(3x + 2)$$

3. Cancel the common factor \((3x + 2)\):
$$\frac{(3x + 2)(x - 4)}{(3x - 2)(3x + 2)} = \frac{x - 4}{3x - 2}$$

M1 for factorising the numerator to $(3x + 2)(x - 4)$ or $(3x + a)(x + b)$ where $ab = -8$ or $3b + a = -10$
M1 for factorising the denominator to $(3x - 2)(3x + 2)$
A1 for $\dfrac{x - 4}{3x - 2}$ final answer cao

1. Find $\vec{AB}$:
$$\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$$

2. Express $\vec{AP}$ using the ratio $AP : PB = 2 : 3$:
$$\vec{AP} = \frac{2}{5}\vec{AB} = \frac{2}{5}(\mathbf{b} - \mathbf{a})$$

3. Since $M$ is the midpoint of $OA$, $\vec{MA} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}$.

4. Express $\vec{MP}$ via path $M \to A \to P$:
$$\vec{MP} = \vec{MA} + \vec{AP} = \frac{1}{2}\mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})$$
$$\vec{MP} = \left(\frac{1}{2} - \frac{2}{5}\right)\mathbf{a} + \frac{2}{5}\mathbf{b} = \frac{1}{10}\mathbf{a} + \frac{2}{5}\mathbf{b}$$

B1 for $\vec{AB} = \mathbf{b} - \mathbf{a}$ or $\vec{BA} = \mathbf{a} - \mathbf{b}$ seen or implied
M1 for a correct vector path for $\vec{MP}$, e.g. $\vec{MA} + \vec{AP}$ or $\vec{MO} + \vec{OP}$
M1 for correct substitution of vector expressions, e.g. $\frac{1}{2}\mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})$ or $-\frac{1}{2}\mathbf{a} + \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})$
A1 for $\frac{1}{10}\mathbf{a} + \frac{2}{5}\mathbf{b}$ or $\frac{\mathbf{a} + 4\mathbf{b}}{10}$ or equivalent in simplest form

1. Factorise each numerator and denominator:
- \(3x^2 - 11x - 4 = (3x + 1)(x - 4)\)
- \(9x^2 - 1 = (3x - 1)(3x + 1)\)
- \(6x^2 - 2x = 2x(3x - 1)\)
- \(x^2 - 16 = (x - 4)(x + 4)\)

2. Substitute the factorised expressions into the product:
\[\frac{(3x + 1)(x - 4)}{(3x - 1)(3x + 1)} \times \frac{2x(3x - 1)}{(x - 4)(x + 4)}\]

3. Cancel common terms \((3x + 1)\), \((3x - 1)\), and \((x - 4)\):
\[= \frac{2x}{x + 4}\]

M1 for factorising \(3x^2 - 11x - 4 = (3x + 1)(x - 4)\)
M1 for factorising \(9x^2 - 1 = (3x - 1)(3x + 1)\), \(6x^2 - 2x = 2x(3x - 1)\), or \(x^2 - 16 = (x - 4)(x + 4)\)
A1 for \(\frac{2x}{x + 4}\) or \(\frac{2x}{4 + x}\) cao

(a) In parallelogram \(OACB\), \(\vec{AC} = \mathbf{b}\) and \(\vec{BC} = \mathbf{a}\).
\(\vec{AM} = \frac{1}{2}\vec{AC} = \frac{1}{2}\mathbf{b}\), so \(\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}\mathbf{b}\).
\(\vec{BN} = \frac{1}{3}\vec{BC} = \frac{1}{3}\mathbf{a}\), so \(\vec{ON} = \vec{OB} + \vec{BN} = \mathbf{b} + \frac{1}{3}\mathbf{a}\).
Therefore:
\(\vec{MN} = \vec{ON} - \vec{OM} = \left(\frac{1}{3}\mathbf{a} + \mathbf{b}\right) - \left(\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b) Since \(X\) lies on the line \(MN\), \(\vec{OX} = \vec{OM} + k\vec{MN}\) for some constant \(k\):
\(\vec{OX} = \left(\mathbf{a} + \frac{1}{2}\mathbf{b}\right) + k\left(-\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \left(1 - \frac{2}{3}k\right)\mathbf{a} + \left(\frac{1}{2} + \frac{1}{2}k\right)\mathbf{b}\).
Since \(X\) lies on the line \(OB\), the \(\mathbf{a}\)-component must equal 0:
\(1 - \frac{2}{3}k = 0 \implies k = \frac{3}{2}\).
Substitute \(k = \frac{3}{2}\) into the \(\mathbf{b}\)-component:
\(\frac{1}{2} + \frac{1}{2}\left(\frac{3}{2}\right) = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}\).
Hence, \(\vec{OX} = \frac{5}{4}\mathbf{b}\).

(a)
B1 for \(\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{b}\) or \(\vec{ON} = \frac{1}{3}\mathbf{a} + \mathbf{b}\) or \(\vec{MC} = \frac{1}{2}\mathbf{b}\) or \(\vec{CN} = -\frac{2}{3}\mathbf{a}\) soi
B1 for \(-\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\) oe in simplest form

(b)
M1 for setting up vector expression \(\vec{OX} = \mathbf{a} + \frac{1}{2}\mathbf{b} + k\left(-\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\right)\) and equating the \(\mathbf{a}\)-coefficient to 0 to find \(k = \frac{3}{2}\) oe
A1 for \(\frac{5}{4}\mathbf{b}\) or \(1.25\mathbf{b}\) oe

Multiply both sides by the denominator \(2 - 5x\):
\[ y(2 - 5x) = 4x - 3 \]

Expand the brackets:
\[ 2y - 5xy = 4x - 3 \]

Collect all terms containing \(x\) on one side and terms without \(x\) on the other side:
\[ 2y + 3 = 4x + 5xy \]

Factorise out \(x\):
\[ 2y + 3 = x(4 + 5y) \]

Divide both sides by \(4 + 5y\):
\[ x = \frac{2y + 3}{5y + 4} \]

M1 for multiplying by \((2 - 5x)\) and expanding correctly: \(2y - 5xy = 4x - 3\) oe
M1 for isolating all terms in \(x\) on one side and factorising: \(x(4 + 5y) = 2y + 3\) oe
A1 for \(x = \frac{2y + 3}{5y + 4}\) oe (e.g. \(x = \frac{-2y - 3}{-5y - 4}\))

(a)
\[ \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a} \]

(b)
Since \(M\) is the midpoint of \(AB\),
\[ \vec{MB} = \frac{1}{2}\vec{AB} = \frac{1}{2}(\mathbf{b} - \mathbf{a}) \]
Since \(ON : NB = 2 : 1\),
\[ \vec{BN} = -\frac{1}{3}\vec{OB} = -\frac{1}{3}\mathbf{b} \]
Using the vector path \(\vec{MN} = \vec{MB} + \vec{BN}\):
\[ \vec{MN} = \frac{1}{2}(\mathbf{b} - \mathbf{a}) - \frac{1}{3}\mathbf{b} \]
\[ \vec{MN} = -\frac{1}{2}\mathbf{a} + \left(\frac{1}{2} - \frac{1}{3}\right)\mathbf{b} = -\frac{1}{2}\mathbf{a} + \frac{1}{6}\mathbf{b} \]

(a) B1 for \(\mathbf{b} - \mathbf{a}\) oe
(b) M1 for a valid vector path for \(\vec{MN}\), e.g. \(\vec{MB} + \vec{BN}\) or \(\vec{MO} + \vec{ON}\)
M1 for substituting correct vector expressions in terms of \(\mathbf{a}\) and \(\mathbf{b}\), e.g. \(\frac{1}{2}(\mathbf{b} - \mathbf{a}) - \frac{1}{3}\mathbf{b}\) or \(-\frac{1}{2}(\mathbf{a} + \mathbf{b}) + \frac{2}{3}\mathbf{b}\)
A1 for \(-\frac{1}{2}\mathbf{a} + \frac{1}{6}\mathbf{b}\) or \(\frac{\mathbf{b} - 3\mathbf{a}}{6}\) oe in its simplest form

Step 1: Use the cosine rule in triangle \(ABC\) to find the length of \(AC\).
\[ AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \]
Substitute the known values:
\[ AC^2 = 3^2 + 8^2 - 2(3)(8)\cos 60^\circ \]
Since \(\cos 60^\circ = \dfrac{1}{2}\):
\[ AC^2 = 9 + 64 - 48\left(\dfrac{1}{2}\right) = 73 - 24 = 49 \]
\[ AC = \sqrt{49} = 7\text{ cm} \]

Step 2: Use the sine rule in triangle \(ACD\) to find the length of \(CD\).
\[ \dfrac{CD}{\sin(\angle CAD)} = \dfrac{AC}{\sin(\angle ADC)} \]
\[ \dfrac{CD}{\sin 45^\circ} = \dfrac{7}{\sin 30^\circ} \]
Substitute exact values \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\):
\[ CD = \dfrac{7 \times \dfrac{\sqrt{2}}{2}}{\dfrac{1}{2}} = 7\sqrt{2}\text{ cm} \]

M1 for correct substitution into cosine rule: \(3^2 + 8^2 - 2(3)(8)\cos 60^\circ\)
M1 for using \(\cos 60^\circ = \frac{1}{2}\) to reach \(AC^2 = 49\)
A1 for \(AC = 7\)
M1 for correct substitution into sine rule: \(\frac{CD}{\sin 45^\circ} = \frac{7}{\sin 30^\circ}\) oe
A1 for \(7\sqrt{2}\) cao

Apply the negative power by reciprocating the fraction:
\[ \left( \frac{81 x^{8}}{16 y^{12}} \right)^{-\frac{3}{4}} = \left( \frac{16 y^{12}}{81 x^{8}} \right)^{\frac{3}{4}} \]

Take the fourth root of each term:
\[ 16^{\frac{1}{4}} = 2, \quad 81^{\frac{1}{4}} = 3, \quad (y^{12})^{\frac{1}{4}} = y^3, \quad (x^8)^{\frac{1}{4}} = x^2 \]

Raise each to the power of 3:
\[ \frac{2^3 (y^3)^3}{3^3 (x^2)^3} = \frac{8 y^9}{27 x^6} \]

M1 for applying root or reciprocating power correctly to at least two terms (e.g. \(y^9\) or \(x^{-6}\) or \(\frac{8}{27}\) seen).
A0.5 for complete correct simplification \(\dfrac{8y^9}{27x^6}\) or \(\dfrac{8}{27} x^{-6} y^9\).

Let the original price be \(P\).
After a \(15\%\) reduction, \(85\%\) of the original price remains:
\[ 0.85 P = 663 \]
\[ P = \frac{663}{0.85} = 780 \]

M1 for \(663 \div 0.85\) or \(663 \times \frac{100}{85}\) oe.
A0.5 for 780.

First apply the negative exponent by inverting the fraction:
\[\left(\frac{16x^8}{81y^{-12}}\right)^{-\frac{3}{4}} = \left(\frac{81y^{-12}}{16x^8}\right)^{\frac{3}{4}} = \left(\frac{81}{16x^8y^{12}}\right)^{\frac{3}{4}}\]
Take the fourth root of each component:
\[\left(\frac{81}{16x^8y^{12}}\right)^{\frac{1}{4}} = \frac{3}{2x^2y^3}\]
Now cube the result:
\[\left(\frac{3}{2x^2y^3}\right)^3 = \frac{27}{8x^6y^9} \quad \text{or rewrite with } y^{12} \text{ inverted initially to give } \frac{27y^9}{8x^6}\]
Checking step-by-step:
\[16^{-\frac{3}{4}} = (16^{\frac{1}{4}})^{-3} = 2^{-3} = \frac{1}{8}\]
\[81^{-\frac{3}{4}} = (81^{\frac{1}{4}})^{-3} = 3^{-3} = \frac{1}{27}\]
\[(x^8)^{-\frac{3}{4}} = x^{-6} = \frac{1}{x^6}\]
\[(y^{-12})^{-\frac{3}{4}} = y^9\]
Multiplying together:
\[\frac{\frac{1}{8}x^{-6}}{\frac{1}{27}y^{-9}} = \frac{27y^9}{8x^6}\]

M1 for applying index laws to at least two terms correctly (e.g. \(16^{-\frac{3}{4}} = \frac{1}{8}\), \(81^{-\frac{3}{4}} = \frac{1}{27}\), \((x^8)^{-\frac{3}{4}} = x^{-6}\), or \((y^{-12})^{-\frac{3}{4}} = y^9\)) or reaching an unsimplified correct single fraction.
A1 for \(\frac{27y^9}{8x^6}\) oe in simplest fraction form with positive indices (e.g. \(\frac{27}{8}x^{-6}y^9\) scores M1 only).

Find the upper bound for each measurement:
Upper bound of length = \(3.8 + 0.05 = 3.85\text{ m}\)
Upper bound of width = \(2.45 + 0.005 = 2.455\text{ m}\)

Calculate the upper bound of the perimeter:
\[\text{Perimeter}_{\text{UB}} = 2 \times (\text{Length}_{\text{UB}} + \text{Width}_{\text{UB}})\]
\[\text{Perimeter}_{\text{UB}} = 2 \times (3.85 + 2.455) = 2 \times 6.305 = 12.61\text{ m}\]

B1 for either \(3.85\) or \(2.455\) seen as an upper bound.
B1 for \(12.61\) cao.

Apply the power of $-\frac{3}{4}$ to each component:

1. Coefficient:
$$16^{-\frac{3}{4}} = (16^{\frac{1}{4}})^{-3} = 2^{-3} = \frac{1}{8}$$

2. Numerator variable:
$$(a^8)^{-\frac{3}{4}} = a^{8 \times \left(-\frac{3}{4}\right)} = a^{-6} = \frac{1}{a^6}$$

3. Denominator variable:
$$(b^4)^{-\frac{3}{4}} = b^{4 \times \left(-\frac{3}{4}\right)} = b^{-3}$$

Dividing by $b^{-3}$ gives $b^3$ in the numerator.

Combining these:
$$\frac{b^3}{8a^6}$$

M1 for applying fractional power correctly to at least two terms, e.g. finding $\frac{1}{8}$ or $a^{-6}$ or $b^3$
A1 for $\frac{b^3}{8a^6}$ or $\frac{1}{8}a^{-6}b^3$ oe

1. Find the actual profit:
$$\text{Profit} = 576 - 450 = 126$$

2. Calculate the percentage profit based on the original cost price:
$$\text{Percentage profit} = \frac{126}{450} \times 100 = 28\%$$

M1 for $\frac{576 - 450}{450} \times 100$ or $\frac{576}{450} \times 100 - 100$ oe
A1 for 28

First, invert the fraction to deal with the negative power:
$$\left( \frac{64x^6}{y^9} \right)^{-\frac{2}{3}} = \left( \frac{y^9}{64x^6} \right)^{\frac{2}{3}}$$

Apply the power of $\frac{2}{3}$ to the numerator and each factor in the denominator:
$$(y^9)^{\frac{2}{3}} = y^{9 \times \frac{2}{3}} = y^6$$
$$(64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16$$
$$(x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4$$

Combine the terms:
$$\frac{y^6}{16x^4}$$

M1 for applying negative index or fractional power correctly to at least two terms (e.g. $64^{2/3} = 16$, $x^4$, or $y^6$ seen)
A0.5 for $\frac{y^6}{16x^4}$ or $\frac{1}{16}x^{-4}y^6$ in simplest form

Find the upper bounds for the dimensions:
$$\text{Upper bound of length} = 24.6 + 0.05 = 24.65\text{ m}$$
$$\text{Upper bound of width} = 15.4 + 0.05 = 15.45\text{ m}$$

Calculate the upper bound of the perimeter:
$$\text{Perimeter} = 2 \times (\text{length} + \text{width})$$
$$\text{Upper bound of perimeter} = 2 \times (24.65 + 15.45) = 2 \times 40.10 = 80.2\text{ m}$$

M1 for $24.65$ or $15.45$ seen, or for $2 \times (\text{UB of length} + \text{UB of width})$
A0.5 for $80.2$ cao

First, apply the negative exponent by inverting the fraction: \(\left(\dfrac{64x^6}{y^{12}}\right)^{-\frac{2}{3}} = \left(\dfrac{y^{12}}{64x^6}\right)^{\frac{2}{3}}\). Next, apply the power of \(\frac{2}{3}\) to the numerator and denominator: \((y^{12})^{\frac{2}{3}} = y^{12 \times \frac{2}{3}} = y^8\), \(64^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\), and \((x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\). Combining these gives \(\dfrac{y^8}{16x^4}\).

M1 for applying negative power or fractional power correctly to at least two terms (e.g. \(64^{-\frac{2}{3}} = \frac{1}{16}\), \(x^{-4}\), or \(y^8\)). A0.5 for fully simplified final answer \(\dfrac{y^8}{16x^4}\) or \(\frac{1}{16}x^{-4}y^8\).

Gradient of line \(L = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-2 - 14}{5 - (-3)} = \dfrac{-16}{8} = -2\). The gradient of a perpendicular line is \(-\dfrac{1}{m} = -\dfrac{1}{-2} = \dfrac{1}{2} = 0.5\).

M1 for finding the gradient of \(L\) as \(\dfrac{-2 - 14}{5 - (-3)} = -2\) or for using \(m_1 \times m_2 = -1\). A0.5 for \(0.5\) or \(\frac{1}{2}\) oe.

1. Find the midpoint of the line segment \(AB\):
\[\text{Midpoint } M = \left(\frac{-2 + 4}{2}, \frac{7 + (-5)}{2}\right) = (1, 1)\]

2. Find the gradient of the line passing through \(A\) and \(B\):
\[m_{AB} = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2\]

3. Find the gradient of the perpendicular bisector:
\[m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-2} = \frac{1}{2} = 0.5\]

4. Use the point-slope form with midpoint \((1, 1)\):
\[y - 1 = \frac{1}{2}(x - 1)\]
\[y = \frac{1}{2}x - \frac{1}{2} + 1\]
\[y = \frac{1}{2}x + \frac{1}{2} \quad \text{or} \quad y = 0.5x + 0.5\]

M1 for gradient of \(AB = \frac{-5-7}{4-(-2)}\) oe (\(= -2\))
M1 for midpoint of \(AB = \left(\frac{-2+4}{2}, \frac{7-5}{2}\right) = (1, 1)\)
M1 for gradient of perpendicular \(= -\frac{1}{\text{their } m_{AB}} = \frac{1}{2}\) oe
A1 for \(y = \frac{1}{2}x + \frac{1}{2}\) or \(y = 0.5x + 0.5\) oe

1. Express the total surface area in terms of \(r\) and \(l\):
\[\text{Total Surface Area} = \pi r^2 + \pi r l\]
Given that \(l = r + 5\) and the total surface area is \(102\pi\):
\[\pi r^2 + \pi r (r + 5) = 102\pi\]

2. Divide the entire equation by \(\pi\):
\[r^2 + r(r + 5) = 102\]
\[r^2 + r^2 + 5r = 102\]
\[2r^2 + 5r - 102 = 0\]

3. Solve the quadratic equation by factorisation or formula:
\[(2r + 17)(r - 6) = 0\]
\[r = -\frac{17}{2} \quad \text{or} \quad r = 6\]

Since a length must be positive, \(r = 6\).

M1 for \(\pi r^2 + \pi r(r + 5) = 102\pi\) oe
M1 for reducing to 3-term quadratic \(2r^2 + 5r - 102 = 0\) oe
M1 for factorising \((2r + 17)(r - 6) = 0\) or correct substitution into quadratic formula \(r = \frac{-5 \pm \sqrt{5^2 - 4(2)(-102)}}{2(2)}\)
A1 for \(r = 6\) cao

1. Multiply through by the common denominator \((2x - 1)(x + 3)\):
\[5(x + 3) - 2(2x - 1) = (2x - 1)(x + 3)\]

2. Expand both sides:
\[5x + 15 - 4x + 2 = 2x^2 + 6x - x - 3\]
\[x + 17 = 2x^2 + 5x - 3\]

3. Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\[2x^2 + 4x - 20 = 0\]
Dividing by 2:
\[x^2 + 2x - 10 = 0\]

4. Apply the quadratic formula:
\[x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-10)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 40}}{2} = \frac{-2 \pm \sqrt{44}}{2} = -1 \pm \sqrt{11}\]

5. Calculate decimal values to 2 decimal places:
\[x = -1 + 3.3166... \approx 2.32\]
\[x = -1 - 3.3166... \approx -4.32\]

M1 for multiplying by \((2x - 1)(x + 3)\) to clear fractions: \(5(x + 3) - 2(2x - 1) = (2x - 1)(x + 3)\) oe
M1 for expanding brackets and reducing to standard 3-term quadratic form: \(2x^2 + 4x - 20 = 0\) or \(x^2 + 2x - 10 = 0\) oe
M1 for correct substitution into quadratic formula \(\frac{-2 \pm \sqrt{2^2 - 4(1)(-10)}}{2}\) or \(\frac{-4 \pm \sqrt{4^2 - 4(2)(-20)}}{4}\) or correct completion of the square \((x + 1)^2 = 11\)
A1 for \(x = 2.32\) and \(x = -4.32\) (both correct to 2 dp, cao)

1. Find the coordinates of the midpoint of \(AB\):
\[M = \left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2)\]

2. Find the gradient of the line segment \(AB\):
\[m_{AB} = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}\]

3. Find the gradient of the perpendicular bisector:
\[m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}\]

4. Find the equation of the line passing through \((1, 2)\) with gradient \(\frac{2}{3}\):
\[y - 2 = \frac{2}{3}(x - 1)\]
\[y - 2 = \frac{2}{3}x - \frac{2}{3}\]
\[y = \frac{2}{3}x + \frac{4}{3}\]

B1 for midpoint \((1, 2)\) seen or used
M1 for gradient of \(AB = \frac{-4 - 8}{5 - (-3)} = -\frac{3}{2}\) oe
M1 for perpendicular gradient \(m = \frac{2}{3}\) and substituting their midpoint into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\)
A1 for \(y = \frac{2}{3}x + \frac{4}{3}\) or \(y = \frac{2}{3}x + 1\frac{1}{3}\) oe

1. Find the midpoint of \(AB\):
$$\text{Midpoint} = \left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right) = (1, 2)$$

2. Find the gradient of line \(L_1\):
$$m_1 = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}$$

3. Find the gradient of the perpendicular line \(L_2\):
$$m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$$

4. Form the equation of \(L_2\) using point \((1, 2)\):
$$y - 2 = \frac{2}{3}(x - 1)$$
$$y = \frac{2}{3}x - \frac{2}{3} + 2$$
$$y = \frac{2}{3}x + \frac{4}{3}$$

M1 for finding the gradient of \(L_1\): \(\frac{-4 - 8}{5 - (-3)}\) oe, soi by \(-\frac{3}{2}\)
M1 for finding the midpoint of \(AB\): \(\left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right)\), soi by \((1, 2)\)
M1 for using \(m_2 = -\frac{1}{\text{their } m_1}\) and substituting their midpoint into \(y = mx + c\) or \(y - y_1 = m(x - x_1)\)
A1 for \(y = \frac{2}{3}x + \frac{4}{3}\) oe (e.g. \(y = \frac{2}{3}x + 1\frac{1}{3}\) or \(y = 0.667x + 1.33\))

1. Total time for the journey:
$$T = 10 + 35 + 15 = 60\text{ s}$$

2. The speed-time graph forms a trapezium with parallel sides of length \(35\text{ s}\) and \(60\text{ s}\).
The area under the graph represents the total distance travelled:
$$\text{Distance} = \frac{1}{2}(35 + 60) \times v = 1140$$
$$47.5 v = 1140$$
$$v = \frac{1140}{47.5} = 24\text{ m/s}$$

3. Calculate deceleration in the final 15 seconds:
$$\text{Deceleration} = \frac{\text{Change in speed}}{\text{Time}} = \frac{24 - 0}{15} = 1.6\text{ m/s}^2$$

M1 for attempting to find the area under the speed-time graph in terms of \(v\), e.g. \(\frac{1}{2}(35 + 60)v\) or \(\frac{1}{2}(10)v + 35v + \frac{1}{2}(15)v\)
M1 for equating their total area expression to 1140, e.g. \(47.5v = 1140\)
A1 for \(v = 24\)
A1 for \(1.6\) oe (e.g. \(\frac{8}{5}\))

1. Find the gradient of line \(L_1\):
\[ \text{Gradient of } L_1 = \frac{9 - (-3)}{8 - 2} = \frac{12}{6} = 2 \]

2. Find the gradient of line \(L_2\), which is perpendicular to \(L_1\):
\[ m = -\frac{1}{2} = -0.5 \]

3. Use the point \((4, 1)\) to determine the \(y\)-intercept \(c\):
\[ y = mx + c \implies 1 = -0.5(4) + c \]
\[ 1 = -2 + c \implies c = 3 \]

4. State the equation of line \(L_2\):
\[ y = -0.5x + 3 \quad \text{or} \quad y = -\frac{1}{2}x + 3 \]

M1: for \(\frac{9 - (-3)}{8 - 2}\) oe
M1: for \(-\frac{1}{\text{their gradient of } L_1}\)
M1: for substituting \((4, 1)\) into \(y = (\text{their } m)x + c\) oe
A1: for \(y = -0.5x + 3\) oe (e.g. \(y = -\frac{1}{2}x + 3\))

1. Set the two equations equal to each other:
\[ 2x^2 - 5x - 5 = x + 3 \]

2. Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\[ 2x^2 - 6x - 8 = 0 \implies x^2 - 3x - 4 = 0 \]

3. Factorise the quadratic equation:
\[ (x - 4)(x + 1) = 0 \]
\[ x = 4 \quad \text{or} \quad x = -1 \]

4. Substitute the \(x\)-values back into \(y = x + 3\) to find the corresponding \(y\)-values:
- When \(x = 4\): \(y = 4 + 3 = 7\)
- When \(x = -1\): \(y = -1 + 3 = 2\)

Therefore, the solutions are \(x = 4, y = 7\) and \(x = -1, y = 2\).

M1: for equating equations: \(2x^2 - 5x - 5 = x + 3\) oe
A1: for correct 3-term quadratic in standard form: \(2x^2 - 6x - 8 = 0\) or \(x^2 - 3x - 4 = 0\)
M1: for factorising or correctly using quadratic formula to find \(x = 4\) and \(x = -1\)
A1: for correct pairs \(x = 4, y = 7\) and \(x = -1, y = 2\)

1. Write the left-hand side over a common denominator:
\[\frac{5(x + 4) - 3(2x - 1)}{(2x - 1)(x + 4)} = 1\]

2. Expand the numerator:
\[5x + 20 - 6x + 3 = 23 - x\]

3. Multiply both sides by the denominator \((2x - 1)(x + 4) = 2x^2 + 7x - 4\):
\[23 - x = 2x^2 + 7x - 4\]

4. Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\[2x^2 + 8x - 27 = 0\]

5. Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[x = \frac{-8 \pm \sqrt{8^2 - 4(2)(-27)}}{2(2)}\]
\[x = \frac{-8 \pm \sqrt{64 + 216}}{4} = \frac{-8 \pm \sqrt{280}}{4}\]
\[x = \frac{-8 \pm 16.7332\dots}{4}\]

6. Calculate the two solutions:
\[x = \frac{-8 + 16.7332}{4} \approx 2.18\]
\[x = \frac{-8 - 16.7332}{4} \approx -6.18\]

M1 for correctly writing the algebraic fractions over a common denominator, e.g., \(5(x + 4) - 3(2x - 1)\) soi
M1 for expanding and rearranging to reach a 3-term quadratic equation \(2x^2 + 8x - 27 = 0\) oe
M1 for correct substitution into the quadratic formula for their quadratic equation \(ax^2 + bx + c = 0\), e.g., \(\frac{-8 \pm \sqrt{8^2 - 4(2)(-27)}}{2(2)}\)
A1 for \(x = 2.18\) and \(x = -6.18\) cao (both correct to 2 decimal places)

1. Find the midpoint of the line segment \(PQ\):
\[M = \left(\frac{-5 + 3}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{-2}{2}, \frac{4}{2}\right) = (-1, 2)\]

2. Find the gradient of the line segment \(PQ\):
\[m_{PQ} = \frac{-4 - 8}{3 - (-5)} = \frac{-12}{8} = -\frac{3}{2}\]

3. Find the gradient of the perpendicular bisector:
\[m_{\perp} = -\frac{1}{m_{PQ}} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}\]

4. Use the point-slope form with midpoint \((-1, 2)\) and gradient \(m = \frac{2}{3}\):
\[y - 2 = \frac{2}{3}(x - (-1))\]
\[y - 2 = \frac{2}{3}x + \frac{2}{3}\]
\[y = \frac{2}{3}x + \frac{8}{3}\] (or \(y = \frac{2}{3}x + 2\frac{2}{3}\))

M1 for finding the coordinates of the midpoint of \(PQ\), \((-1, 2)\) soi
M1 for finding the gradient of \(PQ\), \(-\frac{3}{2}\) oe, and using \(m_{\perp} = -\frac{1}{m}\) to find the perpendicular gradient \(\frac{2}{3}\) oe
A1 for \(y = \frac{2}{3}x + \frac{8}{3}\) oe (e.g. \(y = \frac{2}{3}x + 2\frac{2}{3}\) or \(y = 0.667x + 2.67\))

1. Find the midpoint of the line segment \(AB\):
\[\text{Midpoint} = \left( \frac{-3 + 5}{2}, \frac{11 + (-5)}{2} \right) = (1, 3)\]

2. Find the gradient of line segment \(AB\):
\[m_{AB} = \frac{-5 - 11}{5 - (-3)} = \frac{-16}{8} = -2\]

3. Find the gradient of the perpendicular line:
\[m_{\perp} = -\frac{1}{-2} = \frac{1}{2} = 0.5\]

4. Find the equation of the perpendicular bisector passing through \((1, 3)\):
\[y - 3 = 0.5(x - 1)\]
\[y - 3 = 0.5x - 0.5\]
\[y = 0.5x + 2.5\]

B1 for midpoint \((1, 3)\) seen or implied.
M1 for gradient of \(AB = \frac{-5 - 11}{5 - (-3)} = -2\) oe.
M1 for perpendicular gradient \(= -\frac{1}{\text{their } m_{AB}}\) (e.g. \(\frac{1}{2}\)).
A1 for \(y = 0.5x + 2.5\) or \(y = \frac{1}{2}x + \frac{5}{2}\) oe.

1. Calculate the volume of the cylinder:
\[V = \pi r^2 h = \pi \times 3.6^2 \times 14.5 = 187.92\pi \approx 590.37\text{ cm}^3\]

2. Equate the volume of the sphere to the volume of the cylinder:
\[\frac{4}{3}\pi r^3 = 187.92\pi\]

3. Solve for \(r\):
\[r^3 = \frac{3}{4} \times 187.92 = 140.94\]
\[r = \sqrt[3]{140.94} \approx 5.20409...\]

Correct to 3 significant figures, \(r = 5.20\).

M1 for volume of cylinder \(= \pi \times 3.6^2 \times 14.5\) (\(= 187.92\pi\) or \(590.3\ldots\) or \(590.4\)).
M1 for equating their volume to \(\frac{4}{3}\pi r^3\) and rearranging to \(r = \sqrt[3]{\frac{3 \times \text{volume}}{4\pi}}\).
A1 for \(5.20\) or \(5.204\ldots\) (must round to \(5.20\) to 3 s.f.).

1. Find the coordinates of the midpoint of \(AB\):
\[ \text{Midpoint } M = \left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right) = (1, 2) \]

2. Calculate the gradient of the line segment \(AB\):
\[ m_{AB} = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2} \]

3. Determine the gradient of the perpendicular bisector:
\[ m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3} \]

4. Use the point-slope formula with \(M(1, 2)\) and \(m = \frac{2}{3}\):
\[ y - 2 = \frac{2}{3}(x - 1) \]
\[ y = \frac{2}{3}x - \frac{2}{3} + 2 = \frac{2}{3}x + \frac{4}{3} \]

M1 for finding the midpoint of \(AB\) as \((1, 2)\)
M1 for calculating the gradient of \(AB\) as \(-\frac{3}{2}\) oe
M1 for using \(m_1 m_2 = -1\) to find perpendicular gradient \(\frac{2}{3}\) oe (FT their gradient of \(AB\))
A1 for \(y = \frac{2}{3}x + \frac{4}{3}\) or \(y = \frac{2}{3}x + 1\frac{1}{3}\) oe

1. The motion forms a speed-time graph with a trapezoidal shape.
Total time = \(15 + 25 + 10 = 50\text{ s}\).
Time spent at constant speed = \(25\text{ s}\).

2. Set up the equation for the total distance using the area of the trapezium:
\[ \text{Distance} = \frac{1}{2} \times (50 + 25) \times v = 675 \]
\[ \frac{1}{2} \times 75 \times v = 675 \]
\[ 37.5 v = 675 \implies v = \frac{675}{37.5} = 18\text{ m/s} \]

3. Calculate the deceleration during the final 10 seconds:
\[ \text{Deceleration} = \frac{v - 0}{t} = \frac{18}{10} = 1.8\text{ m/s}^2 \]

M1 for setting up a correct area expression for total distance, e.g. \(\frac{1}{2}(50 + 25)v = 675\) or \(\frac{1}{2}(15)v + 25v + \frac{1}{2}(10)v = 675\) oe
A1 for \(v = 18\)
A1 for \(1.8\) (or \(-1.8\) if acceleration is stated)

(a) The sequence of grey tiles is 5, 9, 13, 17, ...
This is a linear sequence with a common difference of \(d = 4\).
First term \(a = 5\).
\(n\)th term \(= 4n + c\)
When \(n = 1\), \(4(1) + c = 5 \implies c = 1\).
So, the number of grey tiles in Pattern \(n\) is \(4n + 1\).

(b) Total tiles: 8, 17, 28, 41, ...
First differences: \(17 - 8 = 9\), \(28 - 17 = 11\), \(41 - 28 = 13\)
Second differences: \(11 - 9 = 2\), \(13 - 11 = 2\)
Since the second difference is 2, the coefficient of \(n^2\) is \(a = \frac{2}{2} = 1\).
Subtract \(n^2\) from the sequence terms:
\(n = 1: 8 - 1^2 = 7\)
\(n = 2: 17 - 2^2 = 13\)
\(n = 3: 28 - 3^2 = 19\)
\(n = 4: 41 - 4^2 = 25\)
The linear sequence is 7, 13, 19, 25, ... which has a common difference of 6.
Linear term \(= 6n + 1\).
Therefore, the total number of tiles is \(n^2 + 6n + 1\).

(c) Set the total tiles expression equal to 668:
\(k^2 + 6k + 1 = 668\)
\(k^2 + 6k - 667 = 0\)
Factorising gives:
\((k + 29)(k - 23) = 0\)
Since \(k\) must be a positive integer, \(k = 23\).

(a) B2 for \(4n + 1\) oe
(B1 for \(4n + c\) where \(c \ne 1\) or \(k n + 1\) where \(k \ne 4\))

(b) B2 for \(n^2 + 6n + 1\) oe
(M1 for second differences = 2 or \(a = 1\) seen, or for \((4n + 1) + n(n + 2)\))

(c) M1 for equating their quadratic expression to 668 and attempting to solve, e.g. \(k^2 + 6k - 667 = 0\)
A1 for 23 cao

(a) Assembly time condition: \(3x + 5y \le 150\).

(b) At least 10 standard tables: \(x \ge 10\).
Deluxe tables not exceeding standard tables: \(y \le x\).

(c) Profit function: \(P = 40x + 70y\).
We need to find non-negative integer coordinates \((x, y)\) in the feasible region defined by:
1) \(3x + 5y \le 150\)
2) \(x \ge 10\)
3) \(y \le x\)
4) \(y \ge 0\)

To maximise profit \(P = 40x + 70y\), we check integer boundary points near the maximum values of \(y\):
- At intersection of \(y = x\) and \(3x + 5y = 150\):
\(8x = 150 \implies x = 18.75\).
Integer points near this boundary with \(y \le x\) and \(3x + 5y \le 150\):
- \((18, 18)\): \(3(18) + 5(18) = 144 \le 150\). Profit \(= 40(18) + 70(18) = 720 + 1260 = \$1980\).
- \((19, 18)\): \(3(19) + 5(18) = 57 + 90 = 147 \le 150\). Profit \(= 40(19) + 70(18) = 760 + 1260 = \$2020\).
- \((20, 18)\): \(3(20) + 5(18) = 60 + 90 = 150 \le 150\). Profit \(= 40(20) + 70(18) = 800 + 1260 = \$2060\).
- \((21, 17)\): \(3(21) + 5(17) = 63 + 85 = 148 \le 150\). Profit \(= 40(21) + 70(17) = 840 + 1190 = \$2030\).
- \((22, 16)\): \(3(22) + 5(16) = 66 + 80 = 146 \le 150\). Profit \(= 40(22) + 70(16) = 880 + 1120 = \$2000\).

The maximum profit occurs at \(x = 20\), \(y = 18\), giving a profit of $2060.

(a) B1 for \(3x + 5y \le 150\) oe

(b) B1 for \(x \ge 10\)
B1 for \(y \le x\) oe

(c) M1 for establishing profit expression \(40x + 70y\) soi
M1 for identifying critical region / intersection point \((18.75, 18.75)\) or testing integer points near the boundary
M1 for testing candidate point \((20, 18)\) yielding \(40(20) + 70(18)\)
A1 for [$]2060 cao