2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

First, expand both sets of brackets: \(4(2x - 3) = 8x - 12\) and \(-3(x - 5) = -3x + 15\). Combining like terms gives \(8x - 12 - 3x + 15 = (8x - 3x) + (-12 + 15) = 5x + 3\).

M1 for correct expansion of one bracket (e.g. \(8x - 12\) or \(-3x + 15\)). M1 for combining like terms with at least one correct term in the final answer (e.g. \(5x + c\) or \(kx + 3\)). A1 for \(5x + 3\).

Multiply both sides of the equation by 5: \(4x - 3 = 5(x - 2)\). Expand the bracket on the right side: \(4x - 3 = 5x - 10\). Rearrange to solve for \(x\): subtract \(4x\) from both sides to get \(-3 = x - 10\), then add 10 to both sides to get \(x = 7\).

M1 for multiplying by 5 to clear the fraction: \(4x - 3 = 5(x - 2)\). M1 for collecting terms in \(x\) on one side and numerical terms on the other side (e.g. \(5x - 4x = -3 + 10\)). A1 for \(7\) (or \(x = 7\)).

A reduction of 20% means the sale price is 80% of the original price. Let \(P\) be the original price: \(0.80 \times P = 144\). Solving for \(P\): \(P = \frac{144}{0.80} = \frac{1440}{8} = 180\).

M1 for establishing that 80% corresponds to 144 (e.g., \(80\% = 144\)). M1 for a complete method to find the original price: \(\frac{144}{80} \times 100\) or \(\frac{144}{0.8}\). A1 for \(180\).

The formula for the total surface area of a cuboid is \(2(lw + lh + wh)\). Substituting the given values: \(2(6 \times 4 + 6 \times 3 + 4 \times 3) = 2(24 + 18 + 12) = 2 \times 54 = 108\text{ cm}^2\).

M1 for calculating the area of at least one face (e.g., 24, 18, or 12). M1 for a complete expression for the total surface area of all six faces: \(2(24 + 18 + 12)\). A1 for \(108\).

Multiply both sides of the equation by 2: \(2H = 3w - 5y\). Add \(5y\) to both sides to isolate the term with \(w\): \(2H + 5y = 3w\). Divide both sides by 3 to get \(w\) on its own: \(w = \frac{2H + 5y}{3}\).

M1 for multiplying both sides by 2: \(2H = 3w - 5y\). M1 for correctly isolating the term containing \(w\): \(3w = 2H + 5y\). A1 for \(\frac{2H + 5y}{3}\) or equivalent correct expression.

Using the simple interest formula \(I = \frac{P \times R \times T}{100}\), where \(P = 800\), \(R = 4.5\), and \(T = 3\): \(I = \frac{800 \times 4.5 \times 3}{100} = 8 \times 4.5 \times 3 = 36 \times 3 = 108\).

M1 for a correct substitution into the simple interest formula: \(\frac{800 \times 4.5 \times 3}{100}\). M1 for a correct partial calculation (e.g., \(8 \times 4.5\) or \(13.5\%\) of 800). A1 for \(108\).

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substituting \(r = 5\) and \(h = 8\): \(V = \pi \times 5^2 \times 8 = \pi \times 25 \times 8 = 200\pi\text{ cm}^3\).

M1 for recalling the volume formula \(\pi r^2 h\). M1 for substituting the given values correctly: \(\pi \times 5^2 \times 8\). A1 for \(200\pi\).

First, simplify the numerical coefficients: \(\frac{15}{10} = \frac{3}{2}\). Next, apply the laws of indices for the variable terms: \(\frac{a^5}{a^2} = a^{5-2} = a^3\) and \(\frac{b^2}{b^6} = b^{2-6} = b^{-4} = \frac{1}{b^4}\). Combining these gives \(\frac{3a^3}{2b^4}\).

M1 for simplifying the coefficient to \(\frac{3}{2}\) (or 1.5). M1 for applying index laws correctly to either the \(a\) or \(b\) terms (e.g., obtaining \(a^3\) or \(b^{-4}\)). A1 for \(\frac{3a^3}{2b^4}\) or \(\frac{3}{2}a^3b^{-4}\).

Expand the brackets: \(3(2x - 5) = 6x - 15\) and \(-4(x - 3) = -4x + 12\). Combine like terms: \((6x - 4x) + (-15 + 12) = 2x - 3\).

M1 for correct expansion of the first term to \(6x - 15\). M1 for correct expansion of the second term to \(-4x + 12\). A1 for final correct answer of \(2x - 3\).

The discounted price represents 85% of the original price (100% - 15% = 85%). Let the original price be \(x\). So, \(0.85x = 102\). Dividing both sides gives \(x = \frac{102}{0.85} = \frac{10200}{85} = 120\).

M1 for associating 102 with 85% (e.g., \(0.85x = 102\)). M1 for a complete correct division method to find 100% (e.g., \(102 \div 0.85\)). A1 for 120.

First, calculate the cross-sectional area of the triangular end: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 8 = 20\text{ cm}^2\). Then, multiply by the length of the prism to find the volume: \(\text{Volume} = \text{Area} \times \text{length} = 20 \times 12 = 240\text{ cm}^3\).

M1 for calculating the area of the triangular cross-section: \(\frac{1}{2} \times 5 \times 8\) (or 20). M1 for multiplying their cross-sectional area by the length 12. A1 for 240.

Rearrange the second equation to express \(y\) in terms of \(x\): \(y = 2x - 8\). Substitute this expression into the first equation: \(3x + 2(2x - 8) = 19\). Expand and simplify: \(3x + 4x - 16 = 19\), which gives \(7x = 35\), so \(x = 5\). Substitute \(x = 5\) back into the expression for \(y\): \(y = 2(5) - 8 = 2\). Thus, \(x = 5\) and \(y = 2\).

M1 for a correct attempt to eliminate one variable (e.g., multiplying the second equation by 2 to get \(4x - 2y = 16\) and adding). A1 for finding either \(x = 5\) or \(y = 2\). A1 for both \(x = 5\) and \(y = 2\).

Find the highest common factor of both terms. The highest common factor of 6 and 9 is 3. The highest common factor of \(x^2 y\) and \(xy^2\) is \(xy\). Therefore, the highest common factor of both terms is \(3xy\). Divide both terms by \(3xy\) to get the term inside the brackets: \(3xy(2x - 3y)\).

B1 for finding a partial common factor (e.g., \(3x(2xy - 3y^2)\) or \(xy(6x - 9y)\)). M1 for identifying the full highest common factor \(3xy\). A1 for final correct fully factorised expression: \(3xy(2x - 3y)\).

First, calculate the interest earned in one year: \(450 \times 0.04 = 18\). Multiply by 3 to find the total interest earned over three years: \(18 \times 3 = 54\). Add the total interest to the original principal to find the total value: \(450 + 54 = 504\).

M1 for calculating the interest for one year: \(450 \times \frac{4}{100}\) (or 18). M1 for multiplying their annual interest by 3 and adding it to 450. A1 for 504.

A cuboid has three pairs of identical faces. Calculate the area of each unique face: Face 1: \(3 \times 4 = 12\text{ cm}^2\), Face 2: \(4 \times 6 = 24\text{ cm}^2\), Face 3: \(3 \times 6 = 18\text{ cm}^2\). Sum the areas and multiply by 2 to find the total surface area: \(\text{Total Surface Area} = 2 \times (12 + 24 + 18) = 2 \times 54 = 108\text{ cm}^2\).

M1 for finding the area of at least two unique faces (e.g., 12, 24, or 18). M1 for a complete method to find the sum of all six faces: \(2 \times (12 + 24 + 18)\). A1 for 108.

Start with \(t = \frac{3v - 5}{2}\). Multiply both sides by 2 to clear the fraction: \(2t = 3v - 5\). Add 5 to both sides to isolate the term containing \(v\): \(2t + 5 = 3v\). Divide both sides by 3 to make \(v\) the subject: \(v = \frac{2t + 5}{3}\).

M1 for multiplying by 2: \(2t = 3v - 5\). M1 for adding 5: \(2t + 5 = 3v\) (or a correct consecutive rearrangement step). A1 for \(v = \frac{2t + 5}{3}\).

First, factor out the common factor of \(3\): \(3(x^2 - 4)\). Next, recognise that \(x^2 - 4\) is a difference of two squares, which can be factorised as \((x - 2)(x + 2)\). Thus, the completely factorised expression is \(3(x - 2)(x + 2)\).

M1 for extracting the common factor \(3(x^2 - 4)\). M1 for factorising the difference of two squares to \((x - 2)(x + 2)\). A1 for the correct final answer: \(3(x - 2)(x + 2)\) (or equivalent with different bracket order).

Let the original price be \(P\). The price after a \(15\%\) increase is \(1.15 \times P\). We have \(1.15P = 276\), so \(P = \frac{276}{1.15} = \frac{27600}{115}\). Simplifying the fraction by dividing numerator and denominator by \(5\) gives \(\frac{5520}{23}\). Performing the division: \(5520 \div 23 = 240\). Thus, the original price was \(\$240\).

M1 for expressing the relationship \(1.15P = 276\) or writing \(276 \div 1.15\). M1 for a correct division method, such as simplifying to \(\frac{5520}{23}\) or using long division. A1 for the correct final answer of \(240\).

First, find the area of the triangular cross-section: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\). Next, find the volume of the prism by multiplying the cross-sectional area by the length: \(\text{Volume} = 30 \times 15 = 450\text{ cm}^3\).

M1 for calculating the area of the triangular cross-section: \(\frac{1}{2} \times 5 \times 12\) or \(30\). M1 for multiplying their cross-sectional area by the length: \(\text{their } 30 \times 15\). A1 for the correct final answer of \(450\).

To clear the fractions, multiply every term in the equation by the lowest common multiple of the denominators, which is \(4\): \(2(3x - 5) - (x - 1) = 2(3)\). Expand the brackets: \(6x - 10 - x + 1 = 6\). Simplify the left-hand side: \(5x - 9 = 6\). Add \(9\) to both sides: \(5x = 15\). Divide by \(5\) to get \(x = 3\).

M1 for multiplying both sides by a common denominator (e.g. 4) to eliminate fractions, with at most one sign error: \(2(3x - 5) - (x - 1) = 6\). M1 for expanding brackets and collecting like terms to get \(5x - 9 = 6\) or \(5x = 15\) or equivalent. A1 for \(x = 3\).

First, multiply both sides of the formula by \(3\) to clear the fraction: \(3v = u + at\). Next, subtract \(u\) from both sides to isolate the term with \(t\): \(3v - u = at\). Finally, divide both sides by \(a\) to make \(t\) the subject: \(t = \frac{3v - u}{a}\).

M1 for multiplying by 3 to get \(3v = u + at\). M1 for subtracting \(u\) to get \(3v - u = at\) or making \(at\) the subject. A1 for \(t = \frac{3v - u}{a}\) or an equivalent correct algebraic fraction.

The formula for simple interest is \(I = \frac{P \times R \times T}{100}\), where \(P = 600\), \(R = 4\), and \(T = 3\). Substituting these values: \(I = \frac{600 \times 4 \times 3}{100} = 6 \times 4 \times 3 = 72\). Thus, the total interest earned is \(\$72\).

M1 for finding the interest for 1 year: \(600 \times 0.04\) or \(24\). M1 for multiplying their yearly interest by 3: \(\text{their } 24 \times 3\). A1 for \(72\).

Multiply the second equation by \(2\) to align the \(y\)-coefficients: \(8x - 2y = 16\). Now add this new equation to the first equation to eliminate \(y\): \((3x + 2y) + (8x - 2y) = 17 + 16\), which simplifies to \(11x = 33\). Dividing by \(11\) gives \(x = 3\). Substitute \(x = 3\) back into the second original equation: \(4(3) - y = 8\), so \(12 - y = 8\), which gives \(y = 4\). The solution is \(x = 3\) and \(y = 4\).

M1 for a correct method to eliminate one variable (e.g., multiplying the second equation by 2 to get \(8x - 2y = 16\)). M1 for substituting their found variable back to find the second variable, or for finding one variable correctly (e.g., \(x=3\) or \(y=4\)). A1 for both \(x = 3\) and \(y = 4\) correct.

First, expand the double brackets: \((2x - 3)(3x + 4) = 6x^2 + 8x - 9x - 12 = 6x^2 - x - 12\). Next, expand the second part: \(-5x(x - 2) = -5x^2 + 10x\). Now, combine and simplify the like terms: \(6x^2 - x - 12 - 5x^2 + 10x = (6x^2 - 5x^2) + (-x + 10x) - 12 = x^2 + 9x - 12\).

M1 for expanding \((2x-3)(3x+4)\) to at least 3 correct terms (e.g. \(6x^2 + 8x - 9x - 12\)). M1 for expanding \(-5x(x-2)\) correctly to \(-5x^2 + 10x\). A1 for the fully simplified expression: \(x^2 + 9x - 12\).

We group the terms in pairs: \(6ax - 9ay + 8bx - 12by = (6ax - 9ay) + (8bx - 12by)\). Factorise out the common factor \(3a\) from the first group and \(4b\) from the second group: \(3a(2x - 3y) + 4b(2x - 3y)\). Now, factorise out the common bracket \((2x - 3y)\) to get the final answer: \((3a + 4b)(2x - 3y)\).

M1 for finding a common factor of two terms, e.g., \(3a(2x - 3y)\) or \(4b(2x - 3y)\) or \(2x(3a + 4b)\). M1 for \(3a(2x - 3y) + 4b(2x - 3y)\) or equivalent grouping. A1 for \((3a + 4b)(2x - 3y)\) or equivalent.

The increased price of $69 represents 115% of the original price. Let \(x\) be the original price: \(1.15x = 69\). Solving for \(x\): \(x = \frac{69}{1.15} = \frac{6900}{115}\). We can simplify this fraction by dividing the numerator and denominator by 23: \(69 \div 23 = 3\) and \(115 \div 23 = 5\). This gives \(x = \frac{300}{5} = 60\). Thus, the original price was $60.

M1 for equating 115% to 69 (e.g., \(1.15x = 69\) or 115% = 69). M1 for a complete method to find the original price: \(\frac{69}{1.15}\) or \(\frac{69}{115} \times 100\). A1 for 60.

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

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

Let the original price be \(P\). The sale price is a 20% reduction, so \(0.8P = 360\), which gives \(P = \frac{360}{0.8} = 450\). The new price is a 15% increase on $360, so New Price = \(360 \times 1.15 = 360 + 54 = 414\). The overall reduction is from $450 to $414. Reduction amount = \(450 - 414 = 36\). Percentage reduction = \(\frac{36}{450} \times 100 = 8\%\).

M1 for finding the original price: \(360 \div 0.8 = 450\). M1 for finding the new price: \(360 \times 1.15 = 414\). M1 for setting up the percentage reduction calculation: \(\frac{450 - 414}{450} \times 100\). A1 for 8%.

The volume of a sphere is \(V = \frac{4}{3}\pi R^3\). For this sphere with radius \(3r\): \(V_{\text{sphere}} = \frac{4}{3}\pi (3r)^3 = \frac{4}{3}\pi (27r^3) = 36\pi r^3\). The volume of a cylinder is \(V = \pi R_{\text{cyl}}^2 h\). For this cylinder with radius \(2r\): \(V_{\text{cylinder}} = \pi (2r)^2 h = 4\pi r^2 h\). Equating the two volumes: \(36\pi r^3 = 4\pi r^2 h\). Dividing both sides by \(4\pi r^2\) gives \(h = 9r\).

M1 for expressing the volume of the sphere: \(\frac{4}{3}\pi (3r)^3\). M1 for expressing the volume of the cylinder: \(\pi (2r)^2 h\). M1 for equating and simplifying the volumes: \(36\pi r^3 = 4\pi r^2 h\). A1 for \(h = 9r\).

Equate the two expressions for \(y\): \(2x - 3 = x^2 - 3x - 9\). Rearrange the equation into a standard quadratic form: \(x^2 - 5x - 6 = 0\). Factorise the quadratic equation: \((x - 6)(x + 1) = 0\). This gives two values of \(x\): \(x = 6\) or \(x = -1\). Substitute these values back into the linear equation \(y = 2x - 3\) to find the corresponding values of \(y\): For \(x = 6\), \(y = 2(6) - 3 = 9\). For \(x = -1\), \(y = 2(-1) - 3 = -5\).

M1 for equating the equations: \(2x - 3 = x^2 - 3x - 9\). M1 for solving the quadratic: \((x-6)(x+1)=0\) leading to \(x=6\) or \(x=-1\). M1 for substituting \(x\) values to find \(y\). A1 for both correct pairs: \((6, 9)\) and \((-1, -5)\).

Multiply both sides by \((2 - x)\): \(y(2 - x) = 3x + 5\). Expand the brackets: \(2y - xy = 3x + 5\). Collect all terms containing \(x\) on one side: \(2y - 5 = 3x + xy\). Factorise out \(x\) on the right-hand side: \(2y - 5 = x(3 + y)\). Divide both sides by \((3 + y)\) to make \(x\) the subject: \(x = \frac{2y - 5}{y + 3}\).

M1 for multiplying by the denominator: \(y(2-x) = 3x + 5\). M1 for expanding and rearranging to group \(x\) terms: \(2y - 5 = 3x + xy\). M1 for factorising \(x\): \(x(3 + y)\). A1 for \(x = \frac{2y - 5}{y + 3}\) or equivalent.

Using the compound interest formula: \(A = P\left(1 + \frac{r}{100}\right)^n\). Substitute the given values into the formula: \(4410 = 4000\left(1 + \frac{r}{100}\right)^2\). Divide both sides by 4000: \(\frac{4410}{4000} = \left(1 + \frac{r}{100}\right)^2 \Rightarrow \frac{441}{400} = \left(1 + \frac{r}{100}\right)^2\). Take the square root of both sides (since \(r > 0\)): \(\sqrt{\frac{441}{400}} = 1 + \frac{r}{100} \Rightarrow \frac{21}{20} = 1 + \frac{r}{100}\). Convert to decimals: \(1.05 = 1 + \frac{r}{100}\). Therefore, \(\frac{r}{100} = 0.05\), which gives \(r = 5\).

M1 for setting up the equation: \(4000(1 + r/100)^2 = 4410\). M1 for dividing by 4000 and simplifying to \(\frac{441}{400}\). M1 for taking the square root: \(1 + r/100 = 21/20\). A1 for \(r = 5\).

The total surface area of a cone consists of the base area and the curved surface area: \(A = \pi r^2 + \pi r l\). Given the radius \(r = 5\text{ cm}\) and the slant height \(l = 13\text{ cm}\): Base Area = \(\pi \times 5^2 = 25\pi\). Curved Surface Area = \(\pi \times 5 \times 13 = 65\pi\). Total Surface Area = \(25\pi + 65\pi = 90\pi\text{ cm}^2\).

M1 for calculation of base area: \(\pi \times 5^2\) or \(25\pi\). M1 for calculation of curved surface area: \(\pi \times 5 \times 13\) or \(65\pi\). M1 for adding base area and curved surface area. A1 for \(90\pi\).

Find a common denominator for the fractions on the left side: \(\frac{3(x-2) - 2(x-3)}{(x-3)(x-2)} = \frac{1}{2}\). Simplify the numerator: \(3x - 6 - 2x + 6 = x\). So the equation becomes: \(\frac{x}{(x-3)(x-2)} = \frac{1}{2}\). Cross-multiply to eliminate the fractions: \(2x = (x-3)(x-2)\). Expand the right-hand side: \(2x = x^2 - 5x + 6\). Rearrange into standard quadratic form: \(x^2 - 7x + 6 = 0\). Factorise the quadratic equation: \((x-6)(x-1) = 0\). Therefore, the solutions are \(x = 6\) or \(x = 1\).

M1 for combining the fractions on the LHS: \(\frac{3(x-2) - 2(x-3)}{(x-3)(x-2)}\). M1 for cross-multiplying and expanding: \(2x = x^2 - 5x + 6\). M1 for factorising the quadratic equation: \((x-6)(x-1) = 0\). A1 for \(x = 1\) or \(x = 6\).

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 both the numerator and the denominator by the common factor \(2x + 1\) to get the simplified fraction: \(\frac{x - 3}{2x - 1}\).

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

Multiply both sides by \(2 - w\) to get \(P(2 - w) = 3w + 5\). Expand the left side: \(2P - Pw = 3w + 5\). Group all terms containing \(w\) on one side and the rest on the other: \(2P - 5 = 3w + Pw\). Factorise the right side: \(2P - 5 = w(3 + P)\). Finally, divide by \(3 + P\) to make \(w\) the subject: \(w = \frac{2P - 5}{P + 3}\).

M1 for multiplying by \(2 - w\) to clear the fraction. M1 for expanding the bracket and isolating terms with \(w\) on one side. M1 for factorising to obtain \(w(3 + P)\). A1 for the final answer: \(w = \frac{2P - 5}{P + 3}\) or equivalent.

To subtract the fractions, find a common denominator: \((2x - 1)(x + 3)\). Express both fractions over this denominator: \(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\). Expand the numerator: \(3x + 9 - 4x + 2\). Simplify the numerator: \(11 - x\). This gives the final simplified fraction: \(\frac{11 - x}{(2x - 1)(x + 3)}\).

M1 for a common denominator of \((2x - 1)(x + 3)\). M1 for expanding the numerator terms: \(3(x + 3) - 2(2x - 1)\). M1 for simplifying the numerator to \(11 - x\). A1 for the final answer.

Since there is a \(30\%\) discount, the sale price of $84 represents \(70\%\) of the original price. Let \(x\) be the original price. We can set up the equation: \(0.70 \times x = 84\). Solving for \(x\), we get \(x = \frac{84}{0.70} = \frac{840}{7} = 120\). Therefore, the original price of the jacket was $120.

M1 for equating \(70\%\) to 84 (e.g., \(0.7x = 84\) or equivalent). M1 for division: \(\frac{84}{0.7}\). A1 for the correct original price of 120.

First, calculate the interest earned in the first year: \(4000 \times 0.025 = 100\). The value of the investment after 1 year is \(4000 + 100 = 4100\). Next, calculate the interest earned in the second year: \(4100 \times 0.025 = 102.50\). The total interest earned over the 2 years is \(100 + 102.50 = 202.50\).

M1 for calculating the first-year interest: \(4000 \times 0.025 = 100\) or finding the first-year total value of 4100. M1 for calculating the second-year interest: \(4100 \times 0.025 = 102.5\). M1 for adding both interests to find the total interest: \(100 + 102.5\). A1 for 202.50 or 202.5.

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\). Substituting \(r = 3\), the volume is \(V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi\text{ cm}^3\). The volume of a cylinder is given by \(V = \pi R^2 h\). Substituting \(R = 2\), the volume is \(V = \pi (2)^2 h = 4\pi h\text{ cm}^3\). Since the volume remains the same, set the two volumes equal: \(4\pi h = 36\pi\). Divide both sides by \(4\pi\) to find the height: \(h = 9\text{ cm}\).

M1 for the volume of the sphere: \(\frac{4}{3}\pi \times 3^3 = 36\pi\). M1 for setting up the cylinder volume: \(\pi \times 2^2 \times h = 4\pi h\). M1 for equating the volumes: \(4\pi h = 36\pi\). A1 for the correct height of 9.

Equate the two expressions for \(y\): \(2x - 3 = x^2 - 3x + 1\). Rearrange the equation into a standard quadratic form: \(x^2 - 5x + 4 = 0\). Factorise the quadratic equation: \((x - 1)(x - 4) = 0\). This gives two values for \(x\): \(x = 1\) or \(x = 4\). Substitute these values back into \(y = 2x - 3\) to find the corresponding \(y\) values. For \(x = 1\), \(y = 2(1) - 3 = -1\). For \(x = 4\), \(y = 2(4) - 3 = 5\). Thus, the solutions are \(x = 1, y = -1\) and \(x = 4, y = 5\).

M1 for equating both expressions: \(2x - 3 = x^2 - 3x + 1\). M1 for forming a three-term quadratic equation: \(x^2 - 5x + 4 = 0\). M1 for factorising or solving to find \(x = 1\) and \(x = 4\). A1 for both pairs of correct answers: \(x = 1, y = -1\) and \(x = 4, y = 5\).

The area of a rectangle is length multiplied by width: \(x(2x + 3) = 35\). Expand and rearrange the equation to standard quadratic form: \(2x^2 + 3x - 35 = 0\). To solve by factorisation, find two numbers that multiply to \(-70\) and add to \(3\). These are \(10\) and \(-7\). Rewrite and group terms: \(2x^2 + 10x - 7x - 35 = 0 \implies 2x(x + 5) - 7(x + 5) = 0 \implies (2x - 7)(x + 5) = 0\). This gives \(x = 3.5\) or \(x = -5\). Since the width must be positive, we discard \(x = -5\). Thus, the width is \(3.5\text{ cm}\).

M1 for setting up the equation: \(x(2x + 3) = 35\). M1 for rearranging to a standard quadratic equation: \(2x^2 + 3x - 35 = 0\). M1 for factorising or using the quadratic formula to find the roots \(x = 3.5\) and \(x = -5\). A1 for the final answer 3.5 (accept 7/2) and explicitly discarding the negative solution.

First, factorise the numerator, \(3x^2 - 10x + 8\):
Find two numbers that multiply to \(3 \times 8 = 24\) and add to \(-10\). These numbers are \(-6\) and \(-4\).
\(3x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2)\)

Next, factorise the denominator, \(9x^2 - 16\), using the difference of two squares:
\(9x^2 - 16 = (3x - 4)(3x + 4)\)

Now rewrite the fraction with the factorised forms and simplify by cancelling the common factor:
\(\frac{(3x - 4)(x - 2)}{(3x - 4)(3x + 4)} = \frac{x - 2}{3x + 4}\)

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

Let \(P\) be the original price of the bicycle.
A discount of \(15\%\) means the sale price is \(85\%\) of the original price.
\(0.85 \times P = 272\)
\(P = \frac{272}{0.85}\)
Convert the decimal to a fraction to make non-calculator division easier:
\(0.85 = \frac{17}{20}\)
\(P = \frac{272}{\frac{17}{20}} = \frac{272 \times 20}{17}\)
Calculate \(272 \div 17\):
\(17 \times 10 = 170\)
\(272 - 170 = 102\)
\(17 \times 6 = 102\)
So, \(272 \div 17 = 16\).
Now, multiply by 20:
\(P = 16 \times 20 = 320\).
The original price of the bicycle was \(\$320\).

M1 for setting up the equation \(0.85P = 272\) or equivalent
M1 for a correct division method to find \(P\), e.g., \(272 \div \frac{17}{20}\) or reaching \(1\% = 3.2\)
A1 for 320

Calculate the volume of the cylinder:
\(V_{\text{cylinder}} = \pi \times \text{radius}^2 \times \text{height}\)
\(V_{\text{cylinder}} = \pi \times (3y)^2 \times (4y)\)
\(V_{\text{cylinder}} = \pi \times 9y^2 \times 4y = 36\pi y^3\)

Set the volume of the sphere equal to the volume of the cylinder:
\(\frac{4}{3}\pi r^3 = 36\pi y^3\)

Divide both sides by \(\pi\):
\(\frac{4}{3} r^3 = 36 y^3\)

Multiply both sides by \(\frac{3}{4}\) to isolate \(r^3\):
\(r^3 = 36 y^3 \times \frac{3}{4}\)
\(r^3 = 27 y^3\)

Take the cube root of both sides to find \(r\):
\(r = \sqrt[3]{27 y^3} = 3y\)

M1 for finding the volume of the cylinder in terms of \(\pi\) and \(y\): \(36\pi y^3\)
M1 for equating the cylinder volume to the sphere volume: \(\frac{4}{3}\pi r^3 = 36\pi y^3\)
M1 for simplifying to \(r^3 = 27y^3\)
A1 for \(r = 3y\)

Substitute the first equation into the second:
\(x^2 + (2x - 3)^2 = 26\)

Expand the brackets:
\(x^2 + (4x^2 - 12x + 9) = 26\)
\(5x^2 - 12x + 9 = 26\)

Rearrange to form a quadratic equation equal to zero:
\(5x^2 - 12x - 17 = 0\)

Factorise the quadratic equation:
\((5x - 17)(x + 1) = 0\)

This gives two solutions for \(x\):
\(x = \frac{17}{5} = 3.4\) or \(x = -1\)

Now, substitute these back into the linear equation to find the corresponding \(y\) values:
When \(x = -1\):
\(y = 2(-1) - 3 = -5\)

When \(x = 3.4\):
\(y = 2(3.4) - 3 = 6.8 - 3 = 3.8\)

So the solutions are:
\(x = -1, y = -5\) and \(x = 3.4, y = 3.8\) (or \(x = \frac{17}{5}, y = \frac{19}{5}\)).

M1 for substituting \(y = 2x - 3\) into the quadratic equation
M1 for expanding and simplifying to a 3-term quadratic: \(5x^2 - 12x - 17 = 0\)
M1 for solving their quadratic to find both values of \(x\) (e.g., \(x = -1\), \(x = 3.4\))
A1 for both correct coordinate pairs: \(x = -1, y = -5\) and \(x = 3.4, y = 3.8\) (or fraction equivalents)

Multiply both sides by the denominator \(2w + 7\):
\(A(2w + 7) = 3w - 5\)

Expand the brackets on the left-hand side:
\(2Aw + 7A = 3w - 5\)

Group all terms containing \(w\) on one side, and the other terms on the opposite side:
\(7A + 5 = 3w - 2Aw\)

Factorise \(w\) out of the right-hand side:
\(7A + 5 = w(3 - 2A)\)

Divide by \(3 - 2A\) to make \(w\) the subject:
\(w = \frac{7A + 5}{3 - 2A}\)
(Alternatively, \(w = \frac{-7A - 5}{2A - 3}\) is also correct).

M1 for multiplying by \(2w + 7\) to obtain \(A(2w + 7) = 3w - 5\)
M1 for expanding brackets and collecting terms with \(w\) on one side: e.g., \(3w - 2Aw = 7A + 5\)
M1 for factorising \(w\) to obtain \(w(3 - 2A) = 7A + 5\) or \(w(2A - 3) = -7A - 5\)
A1 for \(w = \frac{7A + 5}{3 - 2A}\) or \(w = \frac{-7A - 5}{2A - 3}\)

Let the initial population of birds be represented by \(P\).
After a decrease of \(10\%\) in the first year, the population is:
\(P \times (1 - 0.10) = 0.9P\)

In the second year, this population increases by \(20\%\):
\(0.9P \times (1 + 0.20) = 0.9P \times 1.2 = 1.08P\)

The overall change from the initial population \(P\) to \(1.08P\) is a factor of \(1.08\).
This corresponds to an overall percentage increase of:
\((1.08 - 1) \times 100 = 8\%\)

M1 for applying the \(10\%\) decrease to get a multiplier of \(0.9\)
M1 for applying the \(20\%\) increase to get a combined multiplier of \(0.9 \times 1.2\) or \(1.08\)
A1 for \(8\%\) (or \(8\))

Multiply the entire equation by the common denominator \((x+2)(x-1)\):
\(3(x-1) + 2(x+2) = 1 \times (x+2)(x-1)\)

Expand the brackets on both sides:
\(3x - 3 + 2x + 4 = x^2 + x - 2\)

Simplify both sides:
\(5x + 1 = x^2 + x - 2\)

Rearrange to form a standard quadratic equation equal to zero:
\(x^2 - 4x - 3 = 0\)

Since this does not factorise easily, solve by completing the square:
\((x - 2)^2 - 2^2 - 3 = 0\)
\((x - 2)^2 - 4 - 3 = 0\)
\((x - 2)^2 - 7 = 0\)
\((x - 2)^2 = 7\)

Take the square root of both sides:
\(x - 2 = \pm\sqrt{7}\)

Solve for \(x\):
\(x = 2 \pm \sqrt{7}\)

Here, \(a = 2\) and \(b = 7\).

M1 for multiplying by the common denominator to obtain \(3(x-1) + 2(x+2) = (x+2)(x-1)\)
M1 for simplifying to the standard quadratic equation \(x^2 - 4x - 3 = 0\)
M1 for correctly completing the square to get \((x-2)^2 = 7\) or applying the quadratic formula
A1 for the final answer \(x = 2 \pm \sqrt{7}\)

First, expand the brackets: \(4(2x - 3) = 8x - 12\) and \(-3(x - 5) = -3x + 15\). Next, collect and combine like terms: \(8x - 3x - 12 + 15 = 5x + 3\).

M1 for correct expansion of at least one bracket, i.e., \(8x - 12\) or \(-3x + 15\). A1 for \(5x + 3\).

Let \(x\) be the original price of the laptop. A \(10\%\) discount means the selling price is \(90\%\) of the original price. Therefore, \(0.90 \times x = 414\). Solving for \(x\) gives \(x = \frac{414}{0.90} = 460\). The original price of the laptop is \(\$460\).

M1 for \(414 \div 0.90\) or setting up \(90\% = 414\). A1 for \(460\).

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substituting the given values: \(V = \pi \times 3^2 \times 8 = 72\pi \approx 226.19467\text{ cm}^3\). Rounding to 1 decimal place gives \(226.2\text{ cm}^3\).

M1 for \(\pi \times 3^2 \times 8\) or equivalent. A1 for \(226.2\) (accept answers in range \(226.1\) to \(226.3\)).

Add the two equations together to eliminate \(y\): \((3x + 2y) + (x - 2y) = 17 + 3\) which simplifies to \(4x = 20\), so \(x = 5\). Substitute \(x = 5\) into the second equation: \(5 - 2y = 3\), which gives \(-2y = -2\), so \(y = 1\). Thus, the solution is \(x = 5, y = 1\).

M1 for a valid method to eliminate one variable (e.g., adding the equations or substituting \(x = 2y + 3\)). A1 for \(x = 5\) and \(y = 1\).

The highest common factor of \(6a^2b\) and \(9ab^2\) is \(3ab\). Dividing each term by \(3ab\) gives: \(6a^2b \div 3ab = 2a\) and \(-9ab^2 \div 3ab = -3b\). Thus, the completely factorised form is \(3ab(2a - 3b)\).

M1 for finding a common factor of at least \(3\), \(a\), \(b\), \(3a\), \(3b\) or \(ab\) and showing it outside parentheses, e.g., \(3(2a^2b - 3ab^2)\). A1 for \(3ab(2a - 3b)\).

Using the compound interest formula \(A = P(1 + \frac{r}{100})^t\) with \(P = 1500\), \(r = 2.5\), and \(t = 3\), we get \(A = 1500 \times (1.025)^3 \approx 1500 \times 1.076890625 = 1615.3359375\). Rounding to the nearest cent gives \(\$1615.34\).

M1 for \(1500 \times 1.025^3\) or equivalent. A1 for \(1615.34\) (accept \(1615.33\)).

The total surface area \(A\) of a cuboid is given by \(A = 2(lw + lh + wh)\). Substituting the given dimensions: \(A = 2(8 \times 5 + 8 \times 4 + 5 \times 4) = 2(40 + 32 + 20) = 2(92) = 184\text{ cm}^2\).

M1 for \(2(8 \times 5 + 8 \times 4 + 5 \times 4)\) or for finding at least two correct areas of the faces: \(40\), \(32\), \(20\). A1 for \(184\).

Multiply both sides of the equation by \(5\) to get \(2x - 3 = 15\). Then add \(3\) to both sides to get \(2x = 18\). Finally, divide both sides by \(2\) to get \(x = 9\).

M1 for multiplying by \(5\) to get \(2x - 3 = 15\). A1 for \(9\).

First, expand both brackets: \( 4(3x - 5) = 12x - 20 \) and \( -2(x - 7) = -2x + 14 \). Next, combine like terms: \( 12x - 2x - 20 + 14 = 10x - 6 \).

M1 for expanding at least one bracket correctly: \(12x - 20\) or \(-2x + 14\). A1 for final simplified answer \(10x - 6\).

Find 15% of $85: \( 0.15 \times 85 = 12.75 \). Subtract this from the original price: \( 85 - 12.75 = 72.25 \). Alternatively, calculate 85% of the original price directly: \( 85 \times 0.85 = 72.25 \).

M1 for showing a complete method to calculate the reduced price, e.g., \(85 \times 0.85\) or \(85 - 0.15 \times 85\). A1 for 72.25.

The formula for the total surface area of a rectangular prism is \( 2(lw + lh + wh) \). Substituting the given values: \( 2(8 \times 5 + 8 \times 3 + 5 \times 3) = 2(40 + 24 + 15) = 2(79) = 158 \text{ cm}^2 \).

M1 for showing a correct expression for the surface area of at least three different faces, e.g., \(8 \times 5 + 8 \times 3 + 5 \times 3\). A1 for 158.

First, expand the left side of the equation: \( 5x - 15 = 2x + 9 \). Rearrange to collect the \( x \) terms on one side and the numerical terms on the other: \( 5x - 2x = 9 + 15 \), which simplifies to \( 3x = 24 \). Dividing both sides by 3 gives \( x = 8 \).

M1 for correct expansion of the bracket: \(5x - 15\). M1 for collecting like terms to get \(3x = 24\) or equivalent. A1 for 8.

Identify the highest common factor of the coefficients 12 and 18, which is 6. The highest common factor of \( a^2b \) and \( ab^2 \) is \( ab \). Thus, the highest common factor of the entire expression is \( 6ab \). Divide each term by \( 6ab \) to find the remaining terms inside the bracket: \( 6ab(2a - 3b) \).

M1 for finding a partial common factor, e.g., \(2ab(6a - 9b)\) or \(6a(2ab - 3b^2)\), or for a correct factorisation with one minor error. A1 for \(6ab(2a - 3b)\).

Calculate the increase in population: \( 13375 - 12500 = 875 \). Express this increase as a percentage of the original population: \( \frac{875}{12500} \times 100 = 7 \% \).

M1 for calculating the change divided by original, i.e., \(\frac{13375 - 12500}{12500}\) or \(\frac{13375}{12500} \times 100\). A1 for 7.

The formula for the volume of a cylinder is \( V = \pi r^2 h \). Substituting the values: \( V = \pi \times 4^2 \times 10 = 160\pi \approx 502.6548 \text{ cm}^3 \). Rounding to 1 decimal place gives \( 502.7 \text{ cm}^3 \).

M1 for substituting correctly into the volume formula: \(\pi \times 4^2 \times 10\). A1 for 502.7 (accept 503 if correct method is shown).

Multiply the second equation by 2: \( 8x - 2y = 38 \). Add this to the first equation: \( (3x + 2y) + (8x - 2y) = 17 + 38 \), which simplifies to \( 11x = 55 \), hence \( x = 5 \). Substitute \( x = 5 \) into the second equation: \( 4(5) - y = 19 \implies 20 - y = 19 \implies y = 1 \).

M1 for a valid algebraic method to eliminate one variable. A1 for either \(x = 5\) or \(y = 1\). A1 for both \(x = 5\) and \(y = 1\).

First expand the brackets: \(5(2x - 3) = 10x - 15\) and \(-3(x - 4) = -3x + 12\). Combine like terms: \(10x - 3x - 15 + 12 = 7x - 3\).

M1 for correct expansion of at least one bracket (e.g. \(10x - 15\) or \(-3x + 12\)) A1 for \(7x - 3\)

Let \(x\) be the original price. Since it is reduced by 15%, the sale price is 85% of the original price: \(0.85x = 408\). Thus, \(x = 408 \div 0.85 = 480\).

M1 for \(408 \div 0.85\) or \(408 \div 85 \times 100\) A1 for 480

Use the formula for the volume of a cylinder: \(V = \pi r^2 h\). Substituting the values: \(V = \pi \times 4^2 \times 15 = 240\pi \approx 753.98\text{ cm}^3\). Rounded to 1 decimal place, this is 754.0.

M1 for \(V = \pi \times 4^2 \times 15\) A1 for 754.0 (accept 754)

Multiply both sides by 4: \(3x - 2 = 28\). Add 2 to both sides: \(3x = 30\). Divide both sides by 3: \(x = 10\).

M1 for multiplying by 4: \(3x - 2 = 28\) A1 for 10

Find the highest common factor of \(12a^2b\) and \(18ab^2\), which is \(6ab\). Factor out \(6ab\) to get: \(6ab(2a - 3b)\).

M1 for any partial factorisation, e.g., \(6(2a^2b - 3ab^2)\) or \(ab(12a - 18b)\) A1 for \(6ab(2a - 3b)\)

The increase in population is \(3584 - 3200 = 384\). The percentage increase is \(\frac{384}{3200} \times 100 = 12\\%\).

M1 for \(\frac{3584 - 3200}{3200} \times 100\) or \(\frac{3584}{3200} \times 100\) A1 for 12

The total surface area is the sum of the areas of all 6 faces: \(SA = 2(lw + lh + wh) = 2(8 \times 5 + 8 \times 3 + 5 \times 3) = 2(40 + 24 + 15) = 2 \times 79 = 158\text{ cm}^2\).

M1 for \(2(8 \times 5 + 8 \times 3 + 5 \times 3)\) or finding at least two correct areas of different faces (40, 24, 15) A1 for 158

Multiply the second equation by 2 to get \(8x - 2y = 38\). Add this to the first equation: \((3x + 2y) + (8x - 2y) = 17 + 38 \implies 11x = 55 \implies x = 5\). Substitute \(x = 5\) into the second equation: \(4(5) - y = 19 \implies 20 - y = 19 \implies y = 1\).

M1 for a correct method to eliminate one variable (e.g., multiplying second equation by 2) M1 for finding one correct coordinate (\(x = 5\) or \(y = 1\)) A1 for both \(x = 5\) and \(y = 1\)

Expand the brackets first: \(4(3x - 2) = 12x - 8\) and \(-3(x - 5) = -3x + 15\). Combining the like terms gives \(12x - 3x - 8 + 15 = 9x + 7\).

M1 for correct expansion of at least one bracket, such as \(12x - 8\) or \(-3x + 15\). A1 for final answer \(9x + 7\).

First find the profit: \(113.40 - 84 = 29.40\) dollars. Then calculate the percentage profit: \(\frac{29.40}{84} \times 100 = 35\%\).

M1 for finding the profit of \(29.40\) or setting up the fraction \(\frac{113.40 - 84}{84}\). A1 for final answer of 35.

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substituting the given values: \(V = \pi \times 4.5^2 \times 12 = 243\pi \approx 763.407\text{ cm}^3\). Correct to 1 decimal place, this is \(763.4\).

M1 for \(\pi \times 4.5^2 \times 12\). A1 for \(763.4\) or answer in range \([763.0, 764.0]\).

Expand the bracket: \(5x + 15 = 2x - 9\). Subtract \(2x\) from both sides: \(3x + 15 = -9\). Subtract 15 from both sides: \(3x = -24\). Divide by 3: \(x = -8\).

M1 for expanding the bracket correctly to get \(5x + 15\). M1 for collecting like terms correctly to obtain \(3x = k\) or \(cx = -24\). A1 for \(-8\).

Find the highest common factor of \(18a^2b\) and \(12ab^2\), which is \(6ab\). Dividing each term by this factor gives \(18a^2b / (6ab) = 3a\) and \(-12ab^2 / (6ab) = -2b\). Thus, the factorised expression is \(6ab(3a - 2b)\).

M1 for identifying a common factor (e.g., \(2ab\), \(3ab\), \(6a\), or \(6b\)) and attempting to factorise, or partial factorisation like \(6(3a^2b - 2ab^2)\). A1 for final answer of \(6ab(3a - 2b)\).

Let the original price be \(x\). A reduction of \(15\%\) means the sale price is \(85\%\) of the original price: \(0.85x = 306\). Dividing both sides by 0.85 gives \(x = 306 / 0.85 = 360\).

M1 for recognizing that \(85\%\) corresponds to \(\$306\), e.g., writing \(0.85x = 306\) or \(306 / 0.85\). A1 for final answer of 360.

Part (a): \(3y^2 - 75 = 3(y^2 - 25)\). Using the difference of two squares, \(y^2 - 25 = (y - 5)(y + 5)\). Therefore, \(3y^2 - 75 = 3(y - 5)(y + 5)\).
Part (b): Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator: \(x^2 - 9 = (x - 3)(x + 3)\). Divide out the common factor \(x - 3\): \(\frac{(2x+1)(x-3)}{(x-3)(x+3)} = \frac{2x+1}{x+3}\).

Part (a): M1 for factorising out 3: \(3(y^2 - 25)\). A1 for final answer \(3(y-5)(y+5)\).
Part (b): M1 for factoring numerator: \((2x+1)(x-3)\). M1 for factoring denominator: \((x-3)(x+3)\). A1 for fully simplified fraction: \(\frac{2x+1}{x+3}\).

Part (a): Let the original price be \(x\). \(0.85x = 646 \implies x = \frac{646}{0.85} = 760\).
Part (b): Using the compound interest formula: \(1200 \left(1 + \frac{r}{100}\right)^3 = 1311.12 \implies \left(1 + \frac{r}{100}\right)^3 = 1.0926 \implies 1 + \frac{r}{100} = \sqrt[3]{1.0926} = 1.03 \implies r = 3\).

Part (a): M1 for setting up \(0.85x = 646\) or \(646 \div 0.85\). A1 for $760.
Part (b): M1 for setting up equation \(1200 \times (1 + r/100)^3 = 1311.12\). M1 for taking the cube root to find \(1.03\). A1 for \(r = 3\).

Part (a): Volume of a cylinder = \(\pi r^2 \times \text{height} = \pi r^2 (2r) = 2\pi r^3\). Volume of a cone = \(\frac{1}{3}\pi r^2 h\). Total Volume \(V = 2\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{\pi r^2}{3}(6r + h)\).
Part (b): Substitute \(r = 4\) and \(V = 144\pi\) into the formula: \(144\pi = 2\pi(4)^3 + \frac{1}{3}\pi(4)^2 h \implies 144\pi = 128\pi + \frac{16\pi}{3} h \implies 16\pi = \frac{16\pi}{3} h \implies 1 = \frac{h}{3} \implies h = 3\).

Part (a): M1 for cylinder volume formula \(2\pi r^3\). M1 for cone volume formula \(\frac{1}{3}\pi r^2 h\). A1 for correct combined expression.
Part (b): M1 for substituting \(r=4\) and equating to \(144\pi\). A1 for correct algebraic simplification leading to \(h = 3\).

Part (a): Area is length multiplied by width: \(A = (2x + 5)(x - 2) = 2x(x) - 4x + 5x - 10 = 2x^2 + x - 10\).
Part (b): Set \(A = 50\) to get \(2x^2 + x - 10 = 50 \implies 2x^2 + x - 60 = 0\). Solve using the quadratic formula: \(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-60)}}{4} = \frac{-1 \pm \sqrt{481}}{4}\). Since width must be positive, \(x > 2\), so we take the positive root: \(x = \frac{-1 + \sqrt{481}}{4} \approx 5.23\).

Part (a): M1 for expanding \((2x+5)(x-2)\). A1 for establishing \(2x^2 + x - 10\).
Part (b): M1 for setting up \(2x^2 + x - 10 = 50\) and rearranging. M1 for correct substitution into quadratic formula. A1 for \(x = 5.23\) (rejecting negative root).

Part (a): Gradient of \(AB\) is \(m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). Using \(y - y_1 = m(x - x_1)\): \(y - 5 = 2(x - 6) \implies y = 2x - 7\).
Part (b): The midpoint of \(AB\) is \(\left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\). The gradient of the perpendicular line is \(m' = -\frac{1}{2} = -0.5\). Using the midpoint in \(y - y_1 = m'(x - x_1)\): \(y - 1 = -0.5(x - 4) \implies y = -0.5x + 3\).

Part (a): M1 for finding the gradient of \(AB\) as 2. A1 for \(y = 2x - 7\).
Part (b): M1 for finding the midpoint \((4, 1)\). M1 for perpendicular gradient \(-0.5\). A1 for \(y = -0.5x + 3\) or equivalent.

Part (a): In the right-angled triangle \(APT\), \(\tan(36^\circ) = \frac{TP}{AP} = \frac{TP}{15}\). So, \(TP = 15 \tan(36^\circ) \approx 10.898\text{ m}\), which is \(10.9\text{ m}\) (to 3 s.f.).
Part (b): In the right-angled triangle \(BPT\), \(\tan(\theta) = \frac{TP}{BP} = \frac{10.898}{22} \approx 0.49537\). Thus, \(\theta = \arctan(0.49537) \approx 26.35^\circ\), which is \(26.4^\circ\) (to 1 d.p.).

Part (a): M1 for \(\tan(36^\circ) = \frac{TP}{15}\). A1 for \(10.9\text{ m}\) (accept 10.89 to 10.9).
Part (b): M1 for \(\tan(\theta) = \frac{\text{their } TP}{22}\). A1 for \(26.4^\circ\) (accept 26.3 to 26.4).

Part (a): Total counters = 8. Probability of Red = \(\frac{5}{8}\), Blue = \(\frac{3}{8}\). Both same colour is \(P(R, R) + P(B, B) = \left(\frac{5}{8}\right)^2 + \left(\frac{3}{8}\right)^2 = \frac{25}{64} + \frac{9}{64} = \frac{34}{64} = \frac{17}{32} = 0.53125\).
Part (b): Probability of at least one Red is \(1 - P(B, B)\) without replacement. \(P(B, B) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}\). So, \(P(\text{at least one R}) = 1 - \frac{3}{28} = \frac{25}{28} \approx 0.893\).

Part (a): M1 for adding the correct probability products: \((5/8)^2 + (3/8)^2\). A1 for \(17/32\) or 0.531.
Part (b): M1 for product of two blue probabilities without replacement: \(3/8 \times 2/7\). M1 for subtracting from 1: \(1 - 6/56\). A1 for \(25/28\) or 0.893.

Part (a): Sequence A increases by 5 each time. So it is of the form \(5n + c\). Since the first term is 7, \(5(1) + c = 7 \implies c = 2\). The \(n\)-th term is \(5n + 2\).
Part (b): Sequence B terms are \(2^2, 3^2, 4^2, 5^2, \dots\) which are \((n+1)^2\). So the \(n\)-th term is \((n+1)^2\).
Part (c): Substitute \(n = 50\) into the \(n\)-th term of Sequence A: \(5(50) + 2 = 252\).

Part (a): M1 for finding the common difference of 5 (e.g., \(5n\)). A1 for \(5n + 2\).
Part (b): A1 for \((n+1)^2\) or \(n^2 + 2n + 1\).
Part (c): B1 for 252 (or follow-through of their part (a) for \(n = 50\)).

(a) Let \(P\) be the original price of the television.
The sale price represents an 88% value of the original price because of the 12% reduction:
\(0.88P = 616\)
\(P = \frac{616}{0.88} = 700\)
So, the original price is $700.

(b) The customer pays a deposit of 20% of $616:
\(\text{Deposit} = 0.20 \times 616 = 123.20\)
The remaining amount to be paid is:
\(\text{Remainder} = 616 - 123.20 = 492.80\)
This remaining amount is divided into 8 equal installments:
\(\text{Installment} = \frac{492.80}{8} = 61.60\)
Each monthly installment is $61.60.

Part (a):
- M1 for \(616 \div 0.88\)
- A1 for 700

Part (b):
- M1 for calculating the remaining amount: \(616 \times 0.80 = 492.8\)
- A1.54 for dividing by 8 to obtain 61.6

First, factorize the quadratic expression in the numerator:
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

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

Rewrite the rational expression with these factored terms:
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)}\)

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

- M1 for factoring the numerator: \((2x + 1)(x - 3)\)
- M1 for factoring the denominator: \((2x + 1)(2x - 1)\)
- A2.54 for the fully simplified fraction: \(\frac{x - 3}{2x - 1}\)

(a) The volume of a sphere of radius \(R\) is:
\(V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3} \pi (4.5)^3 = \frac{4}{3} \pi (91.125) = 121.5\pi\text{ cm}^3\)

The volume of a cone of radius \(r\) and height \(h\) is:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \pi (3)^2 h = 3\pi h\)

Since the volume of the metal remains constant:
\(3\pi h = 121.5\pi\)
\(h = 40.5\text{ cm}\) (Showed)

(b) The total surface area of a cone is \(\text{TSA} = \pi r l + \pi r^2\), where \(l\) is the slant height.
Using Pythagoras' theorem to find the slant height \(l\):
\(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 40.5^2} = \sqrt{9 + 1640.25} = \sqrt{1649.25} \approx 40.611\text{ cm}\)

Now, calculate the total surface area:
\(\text{TSA} = \pi (3)(40.611) + \pi (3)^2 = 121.833\pi + 9\pi = 130.833\pi \approx 411.03\text{ cm}^2\)
To 3 significant figures, the total surface area is 411 \(\text{cm}^2\).

Part (a):
- M1 for volume of sphere formula with radius substituted: \(\frac{4}{3}\pi(4.5)^3\)
- A1 for establishing the volume equivalence and showing \(h = 40.5\)

Part (b):
- M1 for finding slant height: \(l = \sqrt{3^2 + 40.5^2}\)
- M1 for total surface area formula: \(\pi r l + \pi r^2\)
- A0.54 for 411 (accept 411.0 to 411.1)

(a) Time is distance divided by speed.
Time taken for the first journey: \(\frac{36}{x}\)
Time taken for the second journey: \(\frac{45}{x - 3}\)
Total time is 4.5 hours:
\(\frac{36}{x} + \frac{45}{x - 3} = 4.5\)

Divide the entire equation by 4.5:
\(\frac{8}{x} + \frac{10}{x - 3} = 1\)

Multiply by the common denominator \(x(x - 3)\):
\(8(x - 3) + 10x = x(x - 3)\)
\(8x - 24 + 10x = x^2 - 3x\)
\(18x - 24 = x^2 - 3x\)

Rearranging to make a quadratic equation equal to zero:
\(x^2 - 21x + 24 = 0\) (Showed)

(b) Solve using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(24)}}{2}\)
\(x = \frac{21 \pm \sqrt{441 - 96}}{2} = \frac{21 \pm \sqrt{345}}{2}\)

Evaluating the roots:
\(x = \frac{21 + 18.574}{2} \approx 19.79\)
\(x = \frac{21 - 18.574}{2} \approx 1.21\)

Since the speed in the second part is \(x - 3\) km/h, \(x\) must be greater than 3. Thus, the cyclist's speed is \(x = 19.79\) km/h.

Part (a):
- M1 for setting up the time sum: \(\frac{36}{x} + \frac{45}{x - 3} = 4.5\)
- A1 for correct algebraic manipulation leading to the given quadratic equation

Part (b):
- M1 for using the quadratic formula with correct coefficients: \(\frac{21 \pm \sqrt{(-21)^2 - 4(1)(24)}}{2}\)
- A1 for obtaining roots 19.79 and 1.21
- A0.54 for identifying 19.79 as the only realistic speed value (as \(x > 3\))

Start by multiplying both sides by the denominator \((5 - mw)\):
\(T(5 - mw) = 3w + 2\)

Expand the brackets on the left side:
\(5T - mwT = 3w + 2\)

Rearrange to collect all terms containing \(w\) on one side:
\(5T - 2 = 3w + mwT\)

Factorize out \(w\) from the right-hand side:
\(5T - 2 = w(3 + mT)\)

Divide both sides by \((3 + mT)\) to isolate \(w\):
\(w = \frac{5T - 2}{3 + mT}\)

- M1 for eliminating the fraction: \(T(5 - mw) = 3w + 2\)
- M1 for expanding brackets and regrouping: \(5T - 2 = 3w + mwT\)
- M1 for factoring out \(w\): \(w(3 + mT) = 5T - 2\)
- A1.54 for the final correct expression: \(w = \frac{5T - 2}{3 + mT}\)

(a) The depreciation equation over 3 years is:
\(24000 \times \left(1 - \frac{r}{100}\right)^3 = 17496\)

Divide both sides by 24,000:
\(\left(1 - \frac{r}{100}\right)^3 = \frac{17496}{24000} = 0.729\)

Take the cube root of both sides:
\(1 - \frac{r}{100} = \sqrt[3]{0.729} = 0.9\)

Solve for \(r\):
\(\frac{r}{100} = 0.1\)
\(r = 10\)

(b) To find the value at the start of 2025 (which is 2 years after 2023, or 5 years after 2020):
Using the value from 2023 with a depreciation rate of 10% per year for 2 years:
\(\text{Value} = 17496 \times (1 - 0.10)^2\)
\(\text{Value} = 17496 \times (0.9)^2\)
\(\text{Value} = 17496 \times 0.81 = 14171.76\)

Rounding to the nearest dollar gives $14,172.

Part (a):
- M1 for setting up the equation: \(24000(1 - r/100)^3 = 17496\)
- A1.27 for \(r = 10\)

Part (b):
- M1 for calculating \(17496 \times (0.9)^2\) or \(24000 \times (0.9)^5\)
- A1.27 for 14172

The total surface area includes:
1. The circular base of the cylinder: \(\text{Area}_{\text{base}} = \pi r^2\)
2. The curved surface area of the cylinder: \(\text{Area}_{\text{cylinder}} = 2\pi r h\)
3. The curved surface area of the hemisphere: \(\text{Area}_{\text{hemisphere}} = 2\pi r^2\)

Total Surface Area (TSA):
\(\text{TSA} = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\)

Substitute the given dimensions \(r = 2.4\) and \(h = 8.5\):
\(\text{TSA} = 3\pi(2.4)^2 + 2\pi(2.4)(8.5)\)
\(\text{TSA} = 3\pi(5.76) + \pi(40.8)\)
\(\text{TSA} = 17.28\pi + 40.8\pi = 58.08\pi\)

Calculate the numerical value:
\(\text{TSA} \approx 58.08 \times 3.14159 = 182.46\text{ m}^2\)

Rounding to 3 significant figures, we get 182 \(\text{m}^2\).

- M1 for finding the base or hemisphere curved area: \(\pi (2.4)^2\) or \(2\pi (2.4)^2\)
- M1 for finding the cylinder curved area: \(2\pi(2.4)(8.5)\)
- M1 for summing the correct three components: \(3\pi (2.4)^2 + 2\pi (2.4)(8.5)\)
- A1.54 for 182 (accept answers from 182.4 to 182.5)

Substitute the expression for \(y\) from the first equation into the second equation:
\(x^2 + (2x - 3)^2 = 41\)

Expand the brackets:
\(x^2 + (4x^2 - 12x + 9) = 41\)
\(5x^2 - 12x + 9 = 41\)

Rearrange to equal 0:
\(5x^2 - 12x - 32 = 0\)

Solve the quadratic equation using factorization:
Multiply to \(5 \times (-32) = -160\) and add to \(-12\). The numbers are \(-20\) and \(8\).
\(5x^2 - 20x + 8x - 32 = 0\)
\(5x(x - 4) + 8(x - 4) = 0\)
\((5x + 8)(x - 4) = 0\)

This gives two values for \(x\):
\(x = 4\)
\(x = -1.6\)

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

For \(x = -1.6\):
\(y = 2(-1.6) - 3 = -6.2\)

So the solutions are:
\(x = 4, y = 5\) and \(x = -1.6, y = -6.2\).

- M1 for substituting \(y = 2x - 3\) into the second equation: \(x^2 + (2x - 3)^2 = 41\)
- M1 for expanding and simplifying to a 3-term quadratic: \(5x^2 - 12x - 32 = 0\)
- M1 for solving the quadratic to find \(x = 4\) and \(x = -1.6\)
- A1.54 for both correct coordinate pairs: \((4, 5)\) and \((-1.6, -6.2)\)

(a) Volume of the cylinder, \(V_{\text{cyl}} = \pi r^2 h = \pi r^2 (3r) = 3\pi r^3\). Volume of 12 spheres, \(V_{\text{spheres}} = 12 \times \frac{4}{3}\pi R^3 = 16\pi R^3\). Since the cylinder is melted down to make the spheres, their volumes are equal: \(3\pi r^3 = 16\pi R^3\), which simplifies to \(R^3 = \frac{3}{16}r^3\) and thus \(R = r \sqrt[3]{\frac{3}{16}}\). (b) The total surface area of a cylinder is \(A = 2\pi r^2 + 2\pi rh\). Since \(h = 3r\), we get \(A = 2\pi r^2 + 2\pi r(3r) = 8\pi r^2\). Given \(A = 100\pi\), we have \(8\pi r^2 = 100\pi \implies r^2 = 12.5 \implies r = \sqrt{12.5} \approx 3.5355\text{ cm}\). Using the formula from part (a): \(R = 3.5355 \times \sqrt[3]{\frac{3}{16}} \approx 3.5355 \times 0.572357 \approx 2.0235\text{ cm}\). To 3 significant figures, \(R = 2.02\text{ cm}\).

M1: for equating volume of cylinder to volume of 12 spheres: \(3\pi r^3 = 16\pi R^3\) M1: for expression of total surface area: \(8\pi r^2 = 100\pi\) A1: for finding \(r = \sqrt{12.5}\) or 3.54 A1: for the final radius \(R = 2.02\) (accept 2.02 to 2.03)

(a) We find two numbers that multiply to \(6 \times (-10) = -60\) and add to \(-11\). These numbers are \(-15\) and \(4\). Rewriting the expression: \(6x^2 - 15x + 4x - 10 = 3x(2x - 5) + 2(2x - 5) = (3x + 2)(2x - 5)\). (b) To subtract the fractions, we find a common denominator, which is \((2x - 5)(3x + 2) = 6x^2 - 11x - 10\). The numerator becomes: \(4(3x + 2) - 3(2x - 5) = 12x + 8 - 6x + 15 = 6x + 23\). Thus, the single fraction is \(\frac{6x + 23}{6x^2 - 11x - 10}\).

M1: for factorising the quadratic to \((3x + 2)(2x - 5)\) M1: for writing the fraction with a common denominator of \((2x - 5)(3x + 2)\) M1: for correct expansion of the numerator: \(12x + 8 - 6x + 15\) A1: for final correct fraction: \(\frac{6x + 23}{6x^2 - 11x - 10}\) or \(\frac{6x + 23}{(2x - 5)(3x + 2)}\)

(a) Using the compound interest formula: \(4500 \left(1 + \frac{r}{100}\right)^5 = 5310 \implies \left(1 + \frac{r}{100}\right)^5 = 1.18 \implies 1 + \frac{r}{100} = (1.18)^{0.2} \approx 1.033623\). This gives \(r = 3.36\%\) to 2 decimal places. (b) Using the simple interest formula, the total value after \(t\) years is: \(4500 + 4500 \times 0.032 \times t = 4500 + 144t\). To exceed $6000: \(4500 + 144t > 6000 \implies 144t > 1500 \implies t > 10.42\) years. Since we need the number of complete years, we round up to 11 years.

M1: for compound interest equation: \(4500(1 + r/100)^5 = 5310\) A1: for \(r = 3.36\) M1: for simple interest equation or inequality: \(4500 + 4500 \times 0.032 \times t > 6000\) A1: for 11 complete years

(a) The lawn has dimensions \(2x + 5\) by \(x + 3\). With a path of width 1 m added to all sides, the new length is \((2x + 5) + 2(1) = 2x + 7\) m, and the new width is \((x + 3) + 2(1) = x + 5\) m. The combined area is \((2x + 7)(x + 5) = 2x^2 + 10x + 7x + 35 = 2x^2 + 17x + 35\) \(\text{m}^2\). (b) Setting the combined area to 120: \(2x^2 + 17x + 35 = 120 \implies 2x^2 + 17x - 85 = 0\). Using the quadratic formula: \(x = \frac{-17 \pm \sqrt{17^2 - 4(2)(-85)}}{2(2)} = \frac{-17 \pm \sqrt{289 + 680}}{4} = \frac{-17 \pm \sqrt{969}}{4}\). Since \(x\) must be positive, we have \(x = \frac{-17 + 31.1288}{4} \approx 3.53\) m.

M1: for showing new dimensions are \(2x + 7\) and \(x + 5\) M1: for correct expansion to obtain \(2x^2 + 17x + 35\) M1: for setting up the quadratic equation \(2x^2 + 17x - 85 = 0\) M1: for correct substitution into the quadratic formula A1: for \(x = 3.53\)

(a) Multiply both sides by \(2 - p\): \(q(2 - p) = 3p + 5 \implies 2q - pq = 3p + 5\). Rearranging to group all terms with \(p\) on one side: \(2q - 5 = 3p + pq\). Factorising out \(p\): \(2q - 5 = p(3 + q)\). Dividing by \(3 + q\) gives: \(p = \frac{2q - 5}{3 + q}\). (b) Substituting \(q = -4\) into the rearranged formula: \(p = \frac{2(-4) - 5}{3 + (-4)} = \frac{-8 - 5}{-1} = 13\).

M1: for multiplying by \(2 - p\): \(q(2 - p) = 3p + 5\) M1: for expanding and collecting terms in \(p\): \(2q - 5 = 3p + pq\) M1: for factorising out \(p\): \(p(3 + q) = 2q - 5\) A1: for \(p = \frac{2q - 5}{3 + q}\) (or equivalent) A1: for substituting \(q = -4\) to find \(p = 13\)

(a) Let the original price be \(P\). After a \(15\%\) increase, the price is \(1.15P\). After an \(8\%\) tax, the price is \(1.15P \times 1.08 = 1.242P\). Given \(1.242P = 223.56\), we solve for \(P = \frac{223.56}{1.242} = 180\). The original price was $180. (b) The value of the car after the first year is: \(18000 \times (1 - 0.12) = 15840\). During the second year, the value depreciates by \(x\%\): \(15840 \times \left(1 - \frac{x}{100}\right) = 13939.20\). This gives \(1 - \frac{x}{100} = \frac{13939.20}{15840} = 0.88\), so \(\frac{x}{100} = 0.12\) and thus \(x = 12\).

M1: for representing the two increases as \(P \times 1.15 \times 1.08 = 223.56\) A1: for $180 M1: for calculating the value of the car after 1 year as \(18000 \times 0.88 = 15840\) M1: for setting up the equation \(15840 \times (1 - x/100) = 13939.20\) A1: for \(x = 12\)