2024 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

Using the cosine rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substitute the given values: \(AC^2 = 7.2^2 + 5.5^2 - 2(7.2)(5.5)\cos(118^\circ)\). Calculate the terms: \(AC^2 = 51.84 + 30.25 - 79.2 \cdot (-0.46947...)\), which gives \(AC^2 = 82.09 + 37.182... = 119.272...\). Thus, \(AC = \sqrt{119.272...} \approx 10.921\text{ cm}\). To 3 significant figures, the length is \(10.9\text{ cm}\).

M1 for correct substitution into the cosine rule: \(7.2^2 + 5.5^2 - 2(7.2)(5.5)\cos(118)\). A1 for \(AC^2 = 119\) to \(119.3\) or \(\sqrt{119.3}\). A1 for \(10.9\) (accept \(10.92\) to \(10.922\)).

Let the original price of the television be \(P\). After a \(15\%\) increase, the price is \(1.15P\). After a \(10\%\) reduction on this new price, the price is: \(1.15P \times (1 - 0.10) = 1.15P \times 0.90 = 1.035P\). We are given that this sale price is \(\$496.80\), so: \(1.035P = 496.80\). Solving for \(P\): \(P = \frac{496.80}{1.035} = 480\). The original price of the television was \(\$480\).

M1 for writing an equation or expression involving multipliers, e.g., \(P \times 1.15 \times 0.9 = 496.80\). M1 for \(P = \frac{496.80}{1.035}\). A1 for \(480\).

First, factorise the numerator \(2x^2 - 5x - 3\): Identify two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\). Rewrite and group: \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator \(4x^2 - 1\) as a difference of two squares: \(4x^2 - 1 = (2x + 1)(2x - 1)\). Simplify the fraction by 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: \((2x + 1)(x - 3)\). M1 for factorising the denominator: \((2x + 1)(2x - 1)\). A1 for \(\frac{x - 3}{2x - 1}\) or \((x - 3)/(2x - 1)\).

Multiply the first equation by 3 and the second equation by 4 to eliminate \(y\): \(3(3x - 4y) = 3(18) \implies 9x - 12y = 54\) and \(4(2x + 3y) = 4(-5) \implies 8x + 12y = -20\). Add the two new equations: \((9x + 8x) + (-12y + 12y) = 54 + (-20) \implies 17x = 34 \implies x = 2\). Substitute \(x = 2\) into the second equation: \(2(2) + 3y = -5 \implies 4 + 3y = -5 \implies 3y = -9 \implies y = -3\).

M1 for a correct method to eliminate one variable (e.g., multiplying equations to equate coefficients of x or y). A1 for \(x = 2\) or \(y = -3\). A1 for both \(x = 2\) and \(y = -3\).

Find the volume of the cone: \(V_{\text{cone}} = \frac{1}{3} \pi \times (4.5)^2 \times 12 = 4\pi \times 20.25 = 81\pi\text{ cm}^3\). Since the metal is recast, the sphere has the same volume: \(\frac{4}{3}\pi r^3 = 81\pi\). Divide both sides by \(\pi\): \(\frac{4}{3} r^3 = 81\). Solve for \(r^3\): \(r^3 = 81 \times \frac{3}{4} = 60.75\). Find \(r\): \(r = \sqrt[3]{60.75} \approx 3.930979...\text{ cm}\). To 3 significant figures, the radius of the sphere is \(3.93\text{ cm}\).

M1 for calculating the volume of the cone as \(81\pi\) or \(254.47...\). M1 for equating their cone volume to the sphere volume formula: \(\frac{4}{3}\pi r^3 = 81\pi\) (or equivalent). A1 for \(3.93\) (accept \(3.93\) to \(3.931\)).

Express both sides as powers of 2: \(\frac{1}{8} = 2^{-3}\) and \(4 = 2^2\). Substitute these into the equation: \((2^{-3})^{p-1} = (2^2)^{2p+5}\). Apply the laws of indices: \(2^{-3(p-1)} = 2^{2(2p+5)}\). Since the bases are now identical, equate the exponents: \(-3(p-1) = 2(2p+5)\). Expand both sides: \(-3p + 3 = 4p + 10\). Solve for \(p\): \(-7p = 7 \implies p = -1\).

M1 for writing both sides with a common base of 2 (or equivalent): \((2^{-3})^{p-1} = (2^2)^{2p+5}\). M1 for setting up a correct linear equation: \(-3p + 3 = 4p + 10\). A1 for \(p = -1\).

First, calculate the numerator by converting to the same power of 10: \(4.2 \times 10^7 - 0.85 \times 10^7 = (4.2 - 0.85) \times 10^7 = 3.35 \times 10^7\). Next, divide by the denominator: \(\frac{3.35 \times 10^7}{5.0 \times 10^{-3}} = \left(\frac{3.35}{5.0}\right) \times 10^{7 - (-3)} = 0.67 \times 10^{10}\). Convert this into standard form: \(0.67 \times 10^{10} = 6.7 \times 10^9\).

M1 for calculating the numerator as \(3.35 \times 10^7\) or \(33,500,000\). M1 for dividing by \(5.0 \times 10^{-3}\) to get \(6,700,000,000\) or \(0.67 \times 10^{10}\). A1 for \(6.7 \times 10^9\).

Group the terms in pairs: \((12ax - 8ay) - (3bx - 2by)\). Factorise each pair by finding their highest common factors: \(4a(3x - 2y) - b(3x - 2y)\). Take out the common binomial factor \((3x - 2y)\): \((4a - b)(3x - 2y)\).

M1 for factorising a common factor from one pair, e.g., \(4a(3x - 2y)\), \(-b(3x - 2y)\), \(3x(4a - b)\), or \(-2y(4a - b)\). A1 for the complete factorisation: \((4a - b)(3x - 2y)\) or \((3x - 2y)(4a - b)\).

First, factorise the numerator by grouping: \(2x^2 - 7x - 4 = 2x^2 - 8x + x - 4 = 2x(x - 4) + 1(x - 4) = (2x + 1)(x - 4)\). Next, factorise the denominator using the difference of two squares: \(x^2 - 16 = (x - 4)(x + 4)\). Now, substitute these factorised forms back into the fraction: \(\frac{(2x + 1)(x - 4)}{(x - 4)(x + 4)}\). Cancel the common factor of \((x - 4)\) from the numerator and denominator to get the simplified fraction: \(\frac{2x + 1}{x + 4}\).

M1 for factorising the numerator correctly: \((2x + 1)(x - 4)\)
M1 for factorising the denominator correctly: \((x - 4)(x + 4)\)
A1 for the final simplified answer: \(\frac{2x + 1}{x + 4}\)

Let \(C\) be the original cost price. A 20% profit gives a price of \(1.20C\). A 15% reduction on this price gives a sale price of \(1.20C \times (1 - 0.15) = 1.20C \times 0.85 = 1.02C\). We are given that the sale price is \(\$122.40\). Set up the equation: \(1.02C = 122.40\). Solve for \(C\): \(C = \frac{122.40}{1.02} = 120\). Therefore, the original cost price was \(\$120\).

M1 for expressing the final price in terms of \(C\), e.g., \(1.20 \times 0.85 \times C = 122.40\)
M1 for calculating the price before the sale reduction: \(122.40 \div 0.85 = 144\)
A1 for the correct cost price: 120

The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\). For a sphere of radius 6 cm, \(V = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi\text{ cm}^3\). The volume of a cone is given by \(V = \frac{1}{3}\pi R^2 h\). For a cone with base radius 4 cm and height \(h\), \(V = \frac{1}{3}\pi (4)^2 h = \frac{16}{3}\pi h\). Since the metal volume remains the same, equate the two volumes: \(\frac{16}{3}\pi h = 288\pi\). Cancel \(\pi\) from both sides: \(\frac{16}{3}h = 288\). Multiply both sides by 3: \(16h = 864\). Divide by 16: \(h = 54\text{ cm}\).

M1 for calculating the volume of the sphere: \(288\pi\) (or approx 905)
M1 for setting up the equation: \(\frac{1}{3}\pi (4)^2 h = 288\pi\) (or equivalent with decimals)
A1 for the correct height: 54

Multiply all terms by the common denominator \(2x(x-1)\) to clear the fractions: \(2x(x-1)\left(\frac{6}{x-1}\right) - 2x(x-1)\left(\frac{6}{x}\right) = 2x(x-1)\left(\frac{1}{2}\right)\). This simplifies to: \(12x - 12(x-1) = x(x-1)\). Expand both sides: \(12x - 12x + 12 = x^2 - x\), which simplifies to \(12 = x^2 - x\). Rearrange into a standard quadratic equation: \(x^2 - x - 12 = 0\). Factorise the quadratic equation: \((x - 4)(x + 3) = 0\). This gives the solutions: \(x = 4\) or \(x = -3\).

M1 for multiplying by a common denominator to clear fractions, e.g., \(12x - 12(x-1) = x(x-1)\)
M1 for obtaining the simplified quadratic equation: \(x^2 - x - 12 = 0\)
A1 for both correct solutions: \(x = 4\) and \(x = -3\)

Using the Cosine Rule to find side \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). Substitute the given values: \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(64^\circ)\). Calculate the terms: \(AC^2 = 49 + 100 - 140\cos(64^\circ)\). Compute the numerical values: \(AC^2 = 149 - 140 \times 0.43837 = 149 - 61.372 = 87.628\). Take the square root: \(AC = \sqrt{87.628} \approx 9.361\text{ cm}\). To 3 significant figures, the length is \(9.36\text{ cm}\).

M1 for substituting correctly into the Cosine Rule: \(7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(64^\circ)\)
M1 for finding \(AC^2 \approx 87.6\) or \(AC = \sqrt{87.628}\)
A1 for the correct length to 3 s.f.: 9.36 (accept 9.361 to 9.362)

First, multiply both sides by the denominator \((2 - t)\): \(w(2 - t) = 3t + 5\). Expand the left side: \(2w - wt = 3t + 5\). Next, move all terms containing \(t\) to one side of the equation and all other terms to the opposite side: \(2w - 5 = 3t + wt\). Factorise \(t\) from the right side: \(2w - 5 = t(3 + w)\). Finally, divide both sides by \((3 + w)\) to isolate \(t\): \(t = \frac{2w - 5}{3 + w}\).

M1 for clearing the fraction: \(w(2 - t) = 3t + 5\)
M1 for grouping terms with \(t\) and factorising: \(t(3 + w) = 2w - 5\) (or equivalent)
A1 for the correct rearranged formula: \(t = \frac{2w - 5}{3 + w}\) (or equivalent, e.g., \(t = \frac{5 - 2w}{-3 - w}\))

First, simplify the expression inside the parentheses: \(\frac{81x^8}{y^{-4}} = 81x^8y^4\). Now apply the power of \(-\frac{3}{4}\) to each component of the product: \(\left( 81x^8y^4 \right)^{-\frac{3}{4}} = 81^{-\frac{3}{4}} \times (x^8)^{-\frac{3}{4}} \times (y^4)^{-\frac{3}{4}}\). Evaluate each part individually: \(81^{-\frac{3}{4}} = \frac{1}{(\sqrt[4]{81})^3} = \frac{1}{3^3} = \frac{1}{27}\); \((x^8)^{-\frac{3}{4}} = x^{8 \times -\frac{3}{4}} = x^{-6} = \frac{1}{x^6}\); \((y^4)^{-\frac{3}{4}} = y^{4 \times -\frac{3}{4}} = y^{-3} = \frac{1}{y^3}\). Combine these terms together as a single fraction: \(\frac{1}{27x^6y^3}\).

M1 for dealing with the negative index inside the brackets or outside, e.g., \(\left(81x^8y^4\right)^{-\frac{3}{4}}\) or \(\left(\frac{y^{-4}}{81x^8}\right)^{\frac{3}{4}}\)
M1 for correctly applying the power to at least two elements, e.g., finding \(81^{-\frac{3}{4}} = \frac{1}{27}\) or \((x^8)^{-\frac{3}{4}} = x^{-6}\)
A1 for the final simplified fraction with positive indices: \(\frac{1}{27x^6y^3}\)

First, determine the gradient of the given line, \(3x - 2y = 8\). Rearrange this equation into slope-intercept form \(y = mx + c\): \(2y = 3x - 8 \implies y = \frac{3}{2}x - 4\). So, the gradient of this line is \(m_1 = \frac{3}{2}\). The gradient of a perpendicular line, \(m_2\), satisfies the condition \(m_1 \times m_2 = -1\). Therefore, \(m_2 = -\frac{2}{3}\). Use the point-slope formula with the point \((6, -1)\) and gradient \(-\frac{2}{3}\): \(y - (-1) = -\frac{2}{3}(x - 6)\). Simplify: \(y + 1 = -\frac{2}{3}x + 4\). Subtract 1 from both sides to obtain the final equation: \(y = -\frac{2}{3}x + 3\).

M1 for finding the gradient of the original line: \(\frac{3}{2}\) (or 1.5)
M1 for calculating the perpendicular gradient: \(-\frac{2}{3}\) and substituting point \((6, -1)\) to solve for \(c\)
A1 for the correct equation: \(y = -\frac{2}{3}x + 3\) (accept equivalent fractions, but must be in \(y = mx + c\) form)

First, factorise the numerator: \(3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)\). Next, factorise the denominator by splitting the middle term: \(2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6 = 2x(x+2) + 3(x+2) = (2x+3)(x+2)\). Now, divide the numerator and the denominator by their common factor \((x+2)\). This gives \(\frac{3(x-2)}{2x+3}\), which can also be written as \(\frac{3x-6}{2x+3}\).

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

Using the Sine Rule, we have \(\frac{\sin(ACB)}{AB} = \frac{\sin(BAC)}{BC}\). Substituting the given values gives \(\frac{\sin(ACB)}{7} = \frac{\sin(115^\circ)}{12}\). Solving for \(\sin(ACB)\) gives \(\sin(ACB) = \frac{7 \sin(115^\circ)}{12}\). Calculating this value gives \(\sin(ACB) \approx 0.5287\). Taking the inverse sine, we find \(ACB \approx 31.91^\circ\). Correct to 1 decimal place, the angle is \(31.9^\circ\).

M1 for correct substitution into the Sine Rule: \(\frac{\sin(ACB)}{7} = \frac{\sin(115^\circ)}{12}\). M1 for rearranging to find \(\sin(ACB) = \frac{7 \sin(115^\circ)}{12}\). A1 for final answer 31.9 (accept answers in range 31.9 to 31.92).

The volume of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). Substituting \(h = 3r\), we get \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). The volume of a sphere of radius \(R\) is \(V_{\text{sphere}} = \frac{4}{3}\pi R^3\). Equating the two volumes gives \(\pi r^3 = \frac{4}{3}\pi R^3\), which simplifies to \(r^3 = \frac{4}{3}R^3\). Dividing by \(R^3\) gives \(\left(\frac{r}{R}\right)^3 = \frac{4}{3}\), so \(\frac{r}{R} = \sqrt[3]{\frac{4}{3}}\). We want to write the ratio \(r : R\) in the form \(1 : k\), which means \(1 : \frac{R}{r}\). Thus, \(k = \frac{R}{r} = \sqrt[3]{\frac{3}{4}} \approx 0.90856\). Correct to 2 decimal places, \(k = 0.91\).

M1 for expressing the volume of the cone as \(\pi r^3\). M1 for equating volumes to find \(\frac{r}{R} = \sqrt[3]{\frac{4}{3}}\) or \(\frac{R}{r} = \sqrt[3]{\frac{3}{4}}\). A1 for final answer 0.91.

Let \(P\) be the original price of the jacket. An increase of \(15\%\) corresponds to a multiplier of \(1.15\). A decrease of \(20\%\) corresponds to a multiplier of \(0.80\). The price after both changes is \(P \times 1.15 \times 0.80 = 0.92P\). We are given that the final sale price is \(\$73.60\), so \(0.92P = 73.60\). Dividing both sides by \(0.92\) gives \(P = 80\). Therefore, the original price of the jacket was \(\$80\).

M1 for finding the combined percentage multiplier: \(1.15 \times 0.80 = 0.92\). M1 for setting up the equation \(0.92P = 73.60\) or calculating \(\frac{73.60}{0.92}\). A1 for 80.

To eliminate \(y\), multiply the first equation by 5 and the second equation by 2. This gives \(15x - 10y = 70\) and \(4x + 10y = -32\). Adding these two equations together eliminates \(y\), giving \(19x = 38\), which simplifies to \(x = 2\). Substitute \(x = 2\) back into the first equation: \(3(2) - 2y = 14\), which simplifies to \(6 - 2y = 14\). Subtracting 6 from both sides gives \(-2y = 8\), so \(y = -4\). The solution is \(x = 2\) and \(y = -4\).

M1 for a correct method to eliminate one variable, e.g., showing coefficients of \(y\) as \(-10\) and \(10\) or coefficients of \(x\) as \(6\) and \(-6\). A1 for \(x = 2\) or \(y = -4\). A1 for both \(x = 2\) and \(y = -4\).

First, multiply both sides by the denominator to clear the fraction: \(y(3 - x) = 2x + 5\). Expanding the left side gives \(3y - xy = 2x + 5\). Next, collect all terms containing \(x\) on one side of the equation and the other terms on the opposite side: \(3y - 5 = 2x + xy\). Factorise out \(x\) on the right side: \(3y - 5 = x(2 + y)\). Finally, divide both sides by \(y + 2\) to isolate \(x\): \(x = \frac{3y - 5}{y + 2}\).

M1 for eliminating the fraction: \(y(3 - x) = 2x + 5\). M1 for isolating terms in \(x\) and factorising: \(x(2 + y) = 3y - 5\). A1 for final answer \(x = \frac{3y - 5}{y + 2}\) or any algebraically equivalent expression.

We apply the power of \(-\frac{2}{3}\) to each factor inside the bracket: \((8x^6)^{-\frac{2}{3}} = 8^{-\frac{2}{3}} \times (x^6)^{-\frac{2}{3}}\). Evaluating the numerical part: \(8^{-\frac{2}{3}} = \frac{1}{8^{2/3}} = \frac{1}{(\sqrt[3]{8})^2} = \frac{1}{2^2} = \frac{1}{4}\). Evaluating the algebraic part using the power of a power rule: \((x^6)^{-\frac{2}{3}} = x^{6 \times -2/3} = x^{-4} = \frac{1}{x^4}\). Multiplying the two parts together gives the final simplified expression: \(\frac{1}{4x^4}\).

M1 for simplifying \(8^{-\frac{2}{3}} = \frac{1}{4}\) or 0.25. M1 for simplifying \((x^6)^{-\frac{2}{3}} = x^{-4}\) or \(\frac{1}{x^4}\). A1 for final answer \(\frac{1}{4x^4}\) or \(0.25x^{-4}\).

First, calculate the vector \(2\vec{a}\): \(2\vec{a} = 2 \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ 6 \end{pmatrix}\). Next, calculate the vector \(2\vec{a} - \vec{b}\): \(2\vec{a} - \vec{b} = \begin{pmatrix} 8 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -6 \end{pmatrix} = \begin{pmatrix} 8 - 3 \\ 6 - (-6)
\end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\). Finally, find the magnitude of this resulting vector using Pythagoras' theorem: \(|2\vec{a} - \vec{b}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).

M1 for calculating the vector \(2\vec{a} - \vec{b} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\). M1 for using Pythagoras' theorem: \(\sqrt{5^2 + 12^2}\). A1 for 13.

First, factorise the numerator: \(4x^2 - 9\) is a difference of two squares, which gives \((2x - 3)(2x + 3)\). Next, factorise the denominator: \(2x^2 + 5x - 12 = 2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\). Now, substitute these back into the fraction to get \(\frac{(2x - 3)(2x + 3)}{(2x - 3)(x + 4)}\). Cancelling the common factor \(2x - 3\) from the numerator and denominator gives the simplified fraction: \(\frac{2x + 3}{x + 4}\).

M1 for factorising the numerator: \((2x - 3)(2x + 3)\). M1 for factorising the denominator: \((2x - 3)(x + 4)\). A1 for the final simplified fraction: \(\frac{2x + 3}{x + 4}\).

Let the population in 2018 be \(P_{2018}\). Since the population in 2021 is a 15% increase on 2018, we have: \(P_{2018} \times 1.15 = 21160\), which gives \(P_{2018} = \frac{21160}{1.15} = 18400\). Let the population in 2015 be \(P_{2015}\). Since the population in 2018 is an 8% decrease on 2015, we have: \(P_{2015} \times (1 - 0.08) = 18400\), which simplifies to \(P_{2015} \times 0.92 = 18400\). This gives \(P_{2015} = \frac{18400}{0.92} = 20000\). Alternatively, we can write a single equation: \(P_{2015} \times 0.92 \times 1.15 = 21160\), which simplifies to \(P_{2015} \times 1.058 = 21160\), so \(P_{2015} = \frac{21160}{1.058} = 20000\).

M1 for finding the population in 2018: \(\frac{21160}{1.15}\) or 18400. M1 for setting up the second stage: \(\frac{\text{their } 18400}{0.92}\) or for setting up the combined equation: \(P \times 0.92 \times 1.15 = 21160\). A1 for 20000.

(a) Using the compound interest formula: \(V = P(1 + \frac{r}{100})^n\)
\(V = 8500 \times (1.032)^6\)
\(V = 8500 \times 1.20803... = 10268.273...\)
Value = $10 268.27.

(b) Let the original value be \(x\).
\(x \times (1 - 0.12)^3 = 13628.16\)
\(x \times (0.88)^3 = 13628.16\)
\(x \times 0.681472 = 13628.16\)
\(x = 20000\).
Original value = $20 000.

(c) Profit in 2021 = $450 000.
Profit in 2022 = \(450 000 \times 1.15 = 517 500\).
Profit in 2023 = \(517 500 \times 0.92 = 476 100\).
Overall percentage increase = \(\frac{476 100 - 450 000}{450 000} \times 100 = 5.8\%\).

(d) Percentage = \(\frac{476 100}{517 500} \times 100 = 92\%\) (or simply \(100\% - 8\% = 92\%\)).

M1 for \(8500 \times 1.032^6\)
A1 for 10268.27

M1 for \(x \times 0.88^3 = 13628.16\)
A1 for 20000

M1 for \(450000 \times 1.15 \times 0.92\) or 476100
M1 for \(\frac{476100 - 450000}{450000} \times 100\)
A1 for 5.8%

M1 for \(\frac{476100}{517500} \times 100\) or \(100 - 8\)
A1 for 92%

(a) Multiply both sides by \(2t + 4\):
\(s(2t + 4) = 3t - 5\)
\(2st + 4s = 3t - 5\)
Rearrange to group \(t\) terms on one side:
\(4s + 5 = 3t - 2st\)
Factorise \(t\):
\(4s + 5 = t(3 - 2s)\)
Divide by \(3 - 2s\):
\(t = \frac{4s + 5}{3 - 2s}\).

(b) Factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
Factorise the denominator using difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\).
Simplify:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

(c) Write with a common denominator of \((y - 2)(y + 1)\):
\(\frac{3(y + 1) - 2(y - 2)}{(y - 2)(y + 1)}\)
\(= \frac{3y + 3 - 2y + 4}{(y - 2)(y + 1)}\)
\(= \frac{y + 7}{(y - 2)(y + 1)}\).

Part (a):
M1 for multiplying by \(2t + 4\)
M1 for collecting terms in \(t\) on one side
M1 for factorising \(t\)
A1 for \(t = \frac{4s + 5}{3 - 2s}\)

Part (b):
M1 for factorising numerator into \((2x + 1)(x - 3)\)
M1 for factorising denominator into \((2x - 1)(2x + 1)\)
A1 for \(\frac{x - 3}{2x - 1}\)

Part (c):
M1 for common denominator \((y-2)(y+1)\)
M1 for \(3(y + 1) - 2(y - 2)\) as numerator
A1 for \(\frac{y + 7}{(y - 2)(y + 1)}\)

(a) Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{36}{x}\).

(b) The return journey speed is \(x - 3\) km/h. Time taken is \(\frac{36}{x - 3}\).
Since the difference is 40 minutes (which is \(\frac{40}{60} = \frac{2}{3}\) hours):
\(\frac{36}{x - 3} - \frac{36}{x} = \frac{2}{3}\)
Multiply all terms by the common denominator \(3x(x - 3)\):
\(3 \times 36 \times x - 3 \times 36 \times (x - 3) = 2x(x - 3)\)
\(108x - 108(x - 3) = 2x^2 - 6x\)
\(108x - 108x + 324 = 2x^2 - 6x\)
\(324 = 2x^2 - 6x\)
\(2x^2 - 6x - 324 = 0\).

(c) Divide the entire equation by 2:
\(x^2 - 3x - 162 = 0\)
Factorise by finding two numbers that multiply to \(-162\) and add to \(-3\):
\((x - 18)(x + 15) = 0\)
So, \(x = 18\) or \(x = -15\).
Since speed must be positive, \(x = 18\) km/h.

(d) Return journey speed = \(18 - 3 = 15\) km/h.
Time taken = \(\frac{36}{15} = 2.4\) hours.
Since \(0.4 \times 60 = 24\) minutes, the time is 2 hours 24 minutes.

Part (a):
B1 for \(\frac{36}{x}\)

Part (b):
M1 for \(\frac{36}{x-3}\)
M1 for setting up \(\frac{36}{x - 3} - \frac{36}{x} = \frac{2}{3}\) (or equivalent)
M1 for multiplying by \(3x(x-3)\) correctly
A1 for complete correct algebraic derivation leading to \(2x^2 - 6x - 324 = 0\)

Part (c):
M1 for factorisation to \((x-18)(x+15) = 0\) or correct substitution in quadratic formula
A1 for \(x = 18\) (rejecting \(x = -15\))

Part (d):
M1 for dividing 36 by their \((x-3)\)
A1 for 2 hours 24 minutes (or 2.4 hours)

(a) Volume of cylinder = \(\pi r^2 h_1 = \pi \times 6^2 \times 10 = 360\pi\) cm\(^3\).
Volume of cone = \(\frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \pi \times 6^2 \times 8 = 96\pi\) cm\(^3\).
Total Volume = \(360\pi + 96\pi = 456\pi\) cm\(^3\).
In decimals, \(456 \times \pi \approx 1432.57\) cm\(^3\), which is 1430 cm\(^3\) to 3 s.f.

(b) Using Pythagoras' theorem for the cone's slant height:
\(l^2 = r^2 + h_2^2\)
\(l^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(l = \sqrt{100} = 10\) cm.

(c) Total surface area consists of:
1. The circular base of the cylinder: \(\pi r^2 = \pi \times 6^2 = 36\pi\) cm\(^2\).
2. The curved surface area of the cylinder: \(2\pi r h_1 = 2\pi \times 6 \times 10 = 120\pi\) cm\(^2\).
3. The curved surface area of the cone: \(\pi r l = \pi \times 6 \times 10 = 60\pi\) cm\(^2\).
Total Surface Area = \(36\pi + 120\pi + 60\pi = 216\pi\) cm\(^2\).
In decimals, \(216 \times \pi \approx 678.58\) cm\(^2\), which is 679 cm\(^2\) to 3 s.f.

(d) Mass = Volume \(\times\) Density
\(M = 456\pi \times 0.75 = 342\pi \approx 1074.4\) g, which is 1070 g to 3 s.f. (using exact volume). If using 1430: \(1430 \times 0.75 = 1072.5\) g, which also rounds to 1070 g.

Part (a):
M1 for \(\pi \times 6^2 \times 10\) or \(360\pi\)
M1 for \(\frac{1}{3} \pi \times 6^2 \times 8\) or \(96\pi\)
A1 for \(456\pi\) or 1430

Part (b):
M1 for \(\sqrt{6^2 + 8^2}\)
A1 for showing \(10\) clearly

Part (c):
M1 for \(\pi \times 6^2\) or \(36\pi\)
M1 for \(2 \pi \times 6 \times 10\) or \(120\pi\)
M1 for \(\pi \times 6 \times 10\) or \(60\pi\)
A1 for 679 or 678.5 to 678.6

Part (d):
M1 for multiplying their volume by 0.75
A1 for 1070 or 1074

(a) The angle \(BAC\) is the difference in bearings:
\(BAC = 140^\circ - 65^\circ = 75^\circ\).
Use the Cosine Rule to find \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 12^2 + 15^2 - 2(12)(15)\cos(75^\circ)\)
\(BC^2 = 144 + 225 - 360 \times 0.258819\)
\(BC^2 = 369 - 93.1748 = 275.825\)
\(BC = \sqrt{275.825} \approx 16.608\) km.
Distance = 16.6 km (to 3 s.f.).

(b) Use the Sine Rule to find angle \(ABC\) (let this be \(B\)):
\(\frac{\sin(B)}{AC} = \frac{\sin(75^\circ)}{BC}\)
\(\frac{\sin(B)}{15} = \frac{\sin(75^\circ)}{16.608}\)
\(\sin(B) = \frac{15 \times \sin(75^\circ)}{16.608} = \frac{15 \times 0.965926}{16.608} \approx 0.8724\)
\(B = \arcsin(0.8724) \approx 60.74^\circ\).
Angle \(ABC = 60.7^\circ\).

(c) Let the North line at \(B\) be drawn. The back-bearing of \(A\) from \(B\) is \(65^\circ + 180^\circ = 245^\circ\).
Since \(C\) lies to the south-east of \(A\), looking from \(B\), the direction of \(C\) is to the south-west.
Specifically, the bearing of \(C\) from \(B\) is:
\(\text{Bearing} = 245^\circ - \text{angle } ABC = 245^\circ - 60.74^\circ = 184.26^\circ \approx 184.3^\circ\) (or \(184^\circ\) to 3 s.f.).

(d) Area of triangle \(ABC = \frac{1}{2} \times AB \times AC \times \sin(BAC)\)
\(\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin(75^\circ) = 90 \times 0.965926 = 86.933\) km\(^2\).
Area = 86.9 km\(^2\) (to 3 s.f.).

Part (a):
M1 for angle \(BAC = 75^\circ\)
M1 for Cosine Rule formula: \(BC^2 = 12^2 + 15^2 - 2(12)(15)\cos(75)\)
A1 for 275.8
A1 for 16.6 (or 16.61)

Part (b):
M1 for Sine Rule formula: \(\frac{\sin B}{15} = \frac{\sin 75}{\text{their } BC}\)
M1 for \(\sin B = 0.8724\)
A1 for 60.7 or 60.74

Part (c):
M1 for back-bearing calculation: \(65 + 180 = 245\)
A1 for 184.3 or 184

Part (d):
M1 for Area = \(0.5 \times 12 \times 15 \times \sin(75)\)
A1 for 86.9

(a) Factorise out the common factor of \(2x\):
\(18x^3 - 50xy^2 = 2x(9x^2 - 25y^2)\)
Recognise the difference of two squares inside the brackets:
\(9x^2 - 25y^2 = (3x - 5y)(3x + 5y)\)
So, completely factorised: \(2x(3x - 5y)(3x + 5y)\).

(b) First, expand the first two brackets:
\((2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12\)
Now multiply this result by the third bracket:
\((2x^2 + 5x - 12)(3x - 1)\)
\t= \(2x^2(3x - 1) + 5x(3x - 1) - 12(3x - 1)\)
\t= \(6x^3 - 2x^2 + 15x^2 - 5x - 36x + 12\)
\t= \(6x^3 + 13x^2 - 41x + 12\).

(c) Factorise the numerator (difference of two squares):
\(x^2 - 9 = (x - 3)(x + 3)\)
Factorise the denominator:
\(x^2 - 2x - 15 = (x - 5)(x + 3)\)
Simplify the fraction by cancelling the common factor of \((x + 3)\):
\(\frac{(x - 3)(x + 3)}{(x - 5)(x + 3)} = \frac{x - 3}{x - 5}\).

Part (a):
M1 for \(2x(9x^2 - 25y^2)\)
M1 for factoring difference of squares \((3x - 5y)(3x + 5y)\)
A1 for \(2x(3x-5y)(3x+5y)\)

Part (b):
M1 for expanding first two brackets: \(2x^2 + 5x - 12\)
M1 for multiplying their quadratic by \(3x - 1\)
A1 for obtaining at least 3 correct terms of the final expansion
A1 for final simplified answer: \(6x^3 + 13x^2 - 41x + 12\)

Part (c):
M1 for factoring numerator: \((x - 3)(x + 3)\)
M1 for factoring denominator: \((x - 5)(x + 3)\)
A1 for \(\frac{x - 3}{x - 5}\)

(a) Substitute \(y = 2x + 5\) into \(x^2 + y^2 = 25\):
\(x^2 + (2x + 5)^2 = 25\)
\(x^2 + (4x^2 + 20x + 25) = 25\)
\(5x^2 + 20x + 25 = 25\)
Subtract 25 from both sides:
\(5x^2 + 20x = 0\).

(b) Factorise the quadratic equation:
\(5x(x + 4) = 0\)
So, \(x = 0\) or \(x = -4\).

(c) Substitute the \(x\) values back into the linear equation \(y = 2x + 5\):
When \(x = 0\): \(y = 2(0) + 5 = 5\) \(\implies (0, 5)\).
When \(x = -4\): \(y = 2(-4) + 5 = -3\) \(\implies (-4, -3)\).
The intersection points are \((0, 5)\) and \((-4, -3)\).

(d) Use the distance formula between the points \((0, 5)\) and \((-4, -3)\):
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(d = \sqrt{(-4 - 0)^2 + (-3 - 5)^2}\)
\(d = \sqrt{(-4)^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} \approx 8.9442...\)
Distance = 8.94 to 3 s.f.

Part (a):
M1 for substitution \(x^2 + (2x+5)^2 = 25\)
M1 for expanding \((2x+5)^2 = 4x^2 + 20x + 25\)
A1 for correct completion to \(5x^2 + 20x = 0\)

Part (b):
M1 for factoring \(5x(x+4) = 0\)
A1 for \(x = 0\) and \(x = -4\)

Part (c):
M1 for substituting one \(x\) value to find \(y\)
A1 for \((0, 5)\)
A1 for \((-4, -3)\)

Part (d):
M1 for \((-4 - 0)^2 + (-3 - 5)^2\) (or equivalent)
M1 for \(\sqrt{80}\)
A1 for 8.94 (or 8.944...)

(a) First find \(f(3)\):
\(f(3) = \frac{3}{3 - 2} = 3\).
Now find \(f(f(3)) = f(3)\):
\(ff(3) = 3\).

(b) Set \(y = f(x)\) and swap variables to find the inverse:
\(y = \frac{3}{x - 2}\)
\(y(x - 2) = 3\)
\(xy - 2y = 3\)
\(xy = 3 + 2y\)
\(x = \frac{3 + 2y}{y}\).
So, \(f^{-1}(x) = \frac{3 + 2x}{x}\).

(c) Substitute \(h(x)\) into \(g(x)\):
\(g(h(x)) = 2(x^2 - 1) + 5\)
\(g(h(x)) = 2x^2 - 2 + 5 = 2x^2 + 3\).

(d) Solve \(\frac{3}{x - 2} + (2x + 5) = 0\):
Multiply all terms by \((x - 2)\):
\(3 + (2x + 5)(x - 2) = 0\)
\(3 + 2x^2 - 4x + 5x - 10 = 0\)
\(2x^2 + x - 7 = 0\).
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-7)}}{2(2)}\)
\(x = \frac{-1 \pm \sqrt{1 + 56}}{4} = \frac{-1 \pm \sqrt{57}}{4}\)
\(x_1 = \frac{-1 + 7.54983}{4} \approx 1.637\)
\(x_2 = \frac{-1 - 7.54983}{4} \approx -2.137\).
Solutions are \(1.64\) and \(-2.14\) (to 3 s.f.).

Part (a):
M1 for finding \(f(3) = 3\)
A1 for 3

Part (b):
M1 for rearranging \(y(x - 2) = 3\)
M1 for making the term in \(x\) the subject (e.g., \(xy = 3 + 2y\))
A1 for \(\frac{3+2x}{x}\) (or equivalent, e.g. \(\frac{3}{x} + 2\))

Part (c):
M1 for \(2(x^2 - 1) + 5\)
A1 for \(2x^2 + 3\)

Part (d):
M1 for setting up equation \(\frac{3}{x - 2} + 2x + 5 = 0\)
M1 for clearing fractions to get \(2x^2 + x - 7 = 0\)
M1 for using quadratic formula with their coefficients
A1 for 1.64
A1 for -2.14

\((a)\) To subtract the fractions, find a common denominator, which is \((x - 2)(2x + 3)\): \(\frac{3(2x + 3) - 5(x - 2)}{(x - 2)(2x + 3)}\) = \(\frac{6x + 9 - (5x - 10)}{(x - 2)(2x + 3)}\) = \(\frac{6x + 9 - 5x + 10}{(x - 2)(2x + 3)}\) = \(\frac{x + 19}{(x - 2)(2x + 3)}\). \((b)\) Factorise the numerator and the denominator: Numerator: \(3x^2 - 14x - 5 = 3x^2 - 15x + x - 5 = (3x + 1)(x - 5)\). Denominator: \(x^2 - 25 = (x - 5)(x + 5)\). Simplify by cancelling the common factor \((x - 5)\): \(\frac{(3x + 1)(x - 5)}{(x - 5)(x + 5)} = \frac{3x + 1}{x + 5}\).

\((a)\) [Total: 5 marks] - M1: Attempting to find a common denominator of \((x-2)(2x+3)\) - M1: Multiplying the numerators correctly: \(3(2x+3)\) and \(5(x-2)\) - M1: Correct expansion of numerators: \(6x+9\) and \(5x-10\) - A1: Correctly subtracting to obtain the numerator \(x+19\) - A1: Correct final fraction. \((b)\) [Total: 5.83 marks] - M2: Factorising the numerator to \((3x+1)(x-5)\) (M1 for one correct term) - M1: Factorising the denominator to \((x-5)(x+5)\) - A2: Correct simplified fraction \(\frac{3x+1}{x+5}\) (A1 for partial cancellation)

\((a)\) Outward time \(= \frac{45}{x}\) hours. \((b)\) Return time \(= \frac{45}{x - 3}\) hours. \((c)\) Since the return journey takes \(45\text{ minutes} = \frac{3}{4}\) hours longer: \(\frac{45}{x - 3} - \frac{45}{x} = \frac{3}{4}\). Divide both sides by 3: \(\frac{15}{x - 3} - \frac{15}{x} = \frac{1}{4}\). Multiply by \(4x(x - 3)\): \(60x - 60(x - 3) = x(x - 3)\) which simplifies to \(60x - 60x + 180 = x^2 - 3x\), so \(180 = x^2 - 3x\), which gives \(x^2 - 3x - 180 = 0\). \((d)\) Factorising \(x^2 - 3x - 180 = 0\) gives \((x - 15)(x + 12) = 0\). Thus \(x = 15\) or \(x = -12\). Since speed is positive, \(x = 15\text{ km/h}\).

\((a)\) [1 mark] - B1: Correct expression \(\frac{45}{x}\). \((b)\) [1 mark] - B1: Correct expression \(\frac{45}{x - 3}\). \((c)\) [4.83 marks] - M1: Convert 45 minutes to hours, i.e., \(\frac{3}{4}\) - M1: Setting up \(\frac{45}{x - 3} - \frac{45}{x} = \frac{3}{4}\) - M1: Attempting to clear denominators - A1: Clear algebraic steps to arrive at \(x^2 - 3x - 180 = 0\). \((d)\) [4 marks] - M1: Factorising to \((x-15)(x+12)\) - A1: Roots \(15\) and \(-12\) - A1: Choosing positive root - B1: Correct units.

\((a)\) \(V_{\text{cylinder}} = \pi r^2 (3r) = 3\pi r^3\). \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\). Total volume \(= 3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\). \((b)\) \(\frac{11}{3}\pi r^3 = 240\pi\) implies \(r^3 = \frac{720}{11}\), so \(r = \sqrt[3]{\frac{720}{11}} \approx 4.0298\text{ cm}\) which rounds to \(4.03\text{ cm}\). \((c)\) Surface area is the sum of: base of cylinder \(\pi r^2\), curved cylinder \(6\pi r^2\), and curved hemisphere \(2\pi r^2\). Total surface area \(= 9\pi r^2\). Substituting \(r = 4.0298\) gives \(9\pi (4.0298)^2 \approx 459.16\text{ cm}^2\) which rounds to \(459\text{ cm}^2\).

\((a)\) [3 marks] - M1: Cylinder volume \(3\pi r^3\) - M1: Hemisphere volume \(\frac{2}{3}\pi r^3\) - A1: Sum \(\frac{11}{3}\pi r^3\). \((b)\) [3.83 marks] - M1: Equating to \(240\pi\) - M1: Solving for \(r^3\) - A1: Unrounded value \(4.0298\) - A0.83: Rounded to 3 sf \(4.03\). \((c)\) [4 marks] - M1: Surface area parts identified - M1: Sum \(9\pi r^2\) - M1: Substitution of \(r\) - A1: \(459\) (allow 458 to 460).

\((a)\) Angle \(\angle BAC = 135^\circ - 75^\circ = 60^\circ\). \((b)\) By Cosine Rule: \(BC^2 = 14^2 + 20^2 - 2(14)(20)\cos(60^\circ) = 196 + 400 - 280 = 316\). \(BC = \sqrt{316} \approx 17.776\text{ km}\), which is \(17.8\text{ km}\) to 3 sf. \((c)\) By Sine Rule: \(\frac{\sin(\angle ABC)}{20} = \frac{\sin(60^\circ)}{\sqrt{316}}\), giving \(\angle ABC \approx 77.0^\circ\). Back bearing of \(A\) from \(B\) is \(255^\circ\). Bearing of \(C\) from \(B\) is \(255^\circ - 77.0^\circ = 178^\circ\). \((d)\) Area \(= \frac{1}{2} \times 14 \times 20 \times \sin(60^\circ) \approx 121.24\text{ km}^2\), which is \(121\text{ km}^2\) to 3 sf.

\((a)\) [1 mark] - B1: \(60^\circ\). \((b)\) [3.83 marks] - M1: Cosine rule formula substitution - A1: \(BC^2 = 316\) - A1.83: \(17.8\). \((c)\) [4 marks] - M1: Sine rule formula substitution - A1: \(\angle ABC \approx 77.0^\circ\) - M1: Find back bearing \(255^\circ\) or equivalent - A1: Bearing \(178^\circ\). \((d)\) [2 marks] - M1: Area formula - A1: \(121\).