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

Multiply the first equation by 5 and the second equation by 2:
\(15x - 10y = 65\)

\(4x + 10y = -8\)

Add the two equations:
\(19x = 57\)

\(x = 3\)

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

\(6 + 5y = -4\)

\(5y = -10\)

\(y = -2\)

M1 for a correct method to eliminate one variable (e.g. multiplying equations to get coefficients equal and adding/subtracting).
A1 for both \(x = 3\) and \(y = -2\).

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 \((x - 3)\):
\(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\)

M1 for factorising either numerator into \((2x+1)(x-3)\) or denominator into \((x-3)(x+3)\).
A1 for \(\frac{2x+1}{x+3}\).

Use the formula for the area of a triangle:
\(\text{Area} = \frac{1}{2}ab\sin(C)\)

Here, \(a = 8\), \(b = 15\), and \(C = 150^\circ\).
\(\text{Area} = \frac{1}{2} \times 8 \times 15 \times \sin(150^\circ)\)

Since \(\sin(150^\circ) = \sin(30^\circ) = \frac{1}{2}\):
\(\text{Area} = 4 \times 15 \times \frac{1}{2} = 30\text{ cm}^2\).

M1 for substituting correctly into the area formula, e.g., \(\frac{1}{2} \times 8 \times 15 \times \sin(150^\circ)\).
A1 for 30.

First find the vertical height \(h\) of the cone using Pythagoras' theorem:
\(h^2 + r^2 = l^2\) where \(r = 5\) and \(l = 13\).
\(h^2 + 5^2 = 13^2 \implies h^2 = 169 - 25 = 144 \implies h = 12\text{ cm}\).

Now substitute \(r = 5\) and \(h = 12\) into the volume formula:
\(V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 5^2 \times 12 = \frac{1}{3} \pi \times 25 \times 12 = 100\pi\text{ cm}^3\).

M1 for finding height \(h = 12\) (either shown directly or through Pythagoras' theorem: \(\sqrt{13^2 - 5^2}\)).
A1 for \(100\pi\).

Multiply both sides by \((x - 5)\):
\(y(x - 5) = 3x + 2\)

Expand the bracket:
\(xy - 5y = 3x + 2\)

Rearrange to get all terms with \(x\) on one side:
\(xy - 3x = 5y + 2\)

Factorise out \(x\):
\(x(y - 3) = 5y + 2\)

Divide by \((y - 3)\):
\(x = \frac{5y + 2}{y - 3}\)

M1 for a correct first step of multiplying by denominator and expanding, e.g. \(xy - 5y = 3x + 2\) or rearranging to isolate \(x\) terms on one side, e.g. \(x(y-3) = 5y+2\).
A1 for \(x = \frac{5y+2}{y-3}\) (or equivalent).

Apply the negative fractional exponent:
\((64w^{12})^{-\frac{2}{3}} = \frac{1}{(64w^{12})^{\frac{2}{3}}}\)

Apply the exponent \(\frac{2}{3}\) to each factor:
\((64)^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\)
\((w^{12})^{\frac{2}{3}} = w^{12 \times \frac{2}{3}} = w^8\)

Combine the parts:
\(\frac{1}{16w^8}\)

M1 for correctly evaluating either the numerical coefficient to \(\frac{1}{16}\) or the algebraic part to \(w^{-8}\).
A1 for \(\frac{1}{16w^8}\).

Multiply every term by \(x(x-1)\) to clear the denominators:
\(6(x - 1) - 2x = x(x - 1)\)

Expand the brackets:
\(6x - 6 - 2x = x^2 - x\)

Simplify both sides:
\(4x - 6 = x^2 - x\)

Rearrange into a quadratic equation equal to 0:
\(x^2 - 5x + 6 = 0\)

Factorise the quadratic:
\((x - 2)(x - 3) = 0\)

Solve for \(x\):
\(x = 2\) or \(x = 3\)

M1 for multiplying by the common denominator to get \(6(x - 1) - 2x = x(x - 1)\) or better, or for expanding to \(x^2 - 5x + 6 = 0\).
A1 for \(x = 2\) and \(x = 3\).

Group the terms into two parts:
\((9a^2 - 4b^2) - (3a - 2b)\)

Factorise the first part using the difference of two squares:
\(9a^2 - 4b^2 = (3a - 2b)(3a + 2b)\)

Rewrite the entire expression:
\((3a - 2b)(3a + 2b) - 1(3a - 2b)\)

Factorise out the common term \((3a - 2b)\):
\((3a - 2b)(3a + 2b - 1)\)

M1 for factorising the first two terms as a difference of two squares: \((3a-2b)(3a+2b)\).
A1 for \((3a - 2b)(3a + 2b - 1)\).

Multiply both sides by \((x + 1)(2x - 5)\) to get \(2(2x - 5) = 3(x + 1)\). Expanding the brackets gives \(4x - 10 = 3x + 3\). Subtracting \(3x\) from both sides gives \(x - 10 = 3\). Adding 10 to both sides gives \(x = 13\).

M1 for \(2(2x - 5) = 3(x + 1)\) or better. A1 for \(13\).

Find the highest common factor of \(12a^2b\) and \(18ab^2\). The HCF of 12 and 18 is 6. The HCF of \(a^2\) and \(a\) is \(a\). The HCF of \(b\) and \(b^2\) is \(b\). Thus, the highest common factor is \(6ab\). Dividing both terms by \(6ab\) gives \(12a^2b \div 6ab = 2a\) and \(-18ab^2 \div 6ab = -3b\). Thus, the fully factorised expression is \(6ab(2a - 3b)\).

B1 for any correct partial factorisation (e.g., \(2ab(6a - 9b)\) or \(3ab(4a - 6b)\) or \(6(2a^2b - 3ab^2)\)). B2 for \(6ab(2a - 3b)\).

The formula for the total surface area of a solid cylinder is \(A = 2\pi r^2 + 2\pi rh\). Substituting the given values \(r = 3\) and \(h = 8\) into the formula gives \(A = 2\pi(3)^2 + 2\pi(3)(8)\). This simplifies to \(A = 18\pi + 48\pi = 66\pi\).

M1 for \(2\pi(3)^2 + 2\pi(3)(8)\) or for either area of two bases \(18\pi\) or curved surface area \(48\pi\) seen. A1 for \(66\pi\).

Using the cosine rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\). Substituting the given values: \(AC^2 = 6^2 + 5^2 - 2(6)(5)\left(\frac{3}{5}\right)\). Simplifying this gives \(AC^2 = 36 + 25 - 36 = 25\). Since length must be positive, \(AC = \sqrt{25} = 5\text{ cm}\).

M1 for correct substitution into cosine rule: \(6^2 + 5^2 - 2(6)(5)\left(\frac{3}{5}\right)\). A1 for \(5\).

Apply the power of \(-\frac{2}{3}\) to each term inside the bracket: \(\left(64x^6\right)^{-\frac{2}{3}} = (64)^{-\frac{2}{3}} \cdot (x^6)^{-\frac{2}{3}}\). First, simplify the numerical part: \((64)^{-\frac{2}{3}} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}\). Next, simplify the variable part: \((x^6)^{-\frac{2}{3}} = x^{-4} = \frac{1}{x^4}\). Multiplying these together gives \(\frac{1}{16x^4}\).

M1 for \(\frac{1}{16}\) or \(x^{-4}\) (or \(\frac{1}{x^4}\)) seen in a product. A1 for \(\frac{1}{16x^4}\) or \(\frac{1}{16}x^{-4}\).

Since \(y\) is inversely proportional to \(x^2\), we can write the equation as \(y = \frac{k}{x^2}\). Using the values \(x = 3\) and \(y = 4\), we get \(4 = \frac{k}{3^2}\), which gives \(k = 36\). Thus, the equation is \(y = \frac{36}{x^2}\). Substituting \(x = 6\) into the equation gives \(y = \frac{36}{6^2} = 1\).

M1 for \(y = \frac{k}{x^2}\) and substituting \(x=3, y=4\) to find \(k = 36\). A1 for \(1\).

Let \(x\) be the number of students who study both Spanish and French, so \(n(S \cap F) = x\). The number of students studying at least one of the two languages is \(n(S \cup F) = 30 - 5 = 25\). Using the set addition formula: \(n(S \cup F) = n(S) + n(F) - n(S \cap F)\), we get \(25 = 18 + 15 - x\), which simplifies to \(25 = 33 - x\). This gives \(x = 8\).

M1 for \(18 + 15 - x = 25\) or \(18 + 15 - 25\) seen. A1 for \(8\).

From the first equation, express \(y\) in terms of \(x\):
\(y = 7 - 2x\)

Substitute this into the second equation:
\(x^2 + (7 - 2x)^2 = 10\)

Expand the brackets:
\(x^2 + 49 - 28x + 4x^2 = 10\)

Simplify and rearrange into a quadratic equation equal to zero:
\(5x^2 - 28x + 39 = 0\)

Factorise the quadratic equation:
\((5x - 13)(x - 3) = 0\)

This gives the solutions for \(x\):
\(x = 3\) or \(x = 2.6\) (or \(\frac{13}{5}\))

Substitute these back into \(y = 7 - 2x\) to find \(y\):
When \(x = 3\), \(y = 7 - 2(3) = 1\).
When \(x = 2.6\), \(y = 7 - 2(2.6) = 1.8\).

So the solutions are \(x = 3, y = 1\) and \(x = 2.6, y = 1.8\).

M1: Substituting \(y = 7 - 2x\) (or equivalent) into the quadratic equation.
M1: Correctly simplifying to a 3-term quadratic equation, e.g., \(5x^2 - 28x + 39 = 0\).
M1: Factorising or using the quadratic formula to find at least one value of \(x\) or \(y\).
A1: Fully correct solutions: \(x = 3, y = 1\) and \(x = 2.6, y = 1.8\).

Factorise each quadratic expression completely:
1. \(2x^2 - 7x + 6 = (2x - 3)(x - 2)\)
2. \(2x^2 - x - 3 = (2x - 3)(x + 1)\)
3. \(x^2 - 1 = (x - 1)(x + 1)\)
4. \(x^2 - 4 = (x - 2)(x + 2)\)

Substitute these factored forms back into the expression:
\(\frac{(2x - 3)(x - 2)}{(2x - 3)(x + 1)} \times \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}\)

Cancel out common terms in the numerator and denominator:
- Cancel \((2x - 3)\)
- Cancel \((x - 2)\)
- Cancel \((x + 1)\)

This simplifies to:
\(\frac{x - 1}{x + 2}\)

M1: Factorising the first numerator to \((2x - 3)(x - 2)\).
M1: Factorising the first denominator to \((2x - 3)(x + 1)\).
M1: Factorising \(x^2 - 1\) to \((x - 1)(x + 1)\) and \(x^2 - 4\) to \((x - 2)(x + 2)\).
A1: Correct final simplified fraction \(\frac{x - 1}{x + 2}\).

First, find the length of side \(BC\) using the Cosine Rule:
\(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(BAC)\)
\(BC^2 = 8^2 + 5^2 - 2(8)(5)\cos(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\):
\(BC^2 = 64 + 25 - 80(0.5)\)
\(BC^2 = 89 - 40 = 49\)
\(BC = 7\text{ cm}\)

Next, use the Sine Rule to find \(\sin(\angle ABC)\):
\(\frac{\sin(\angle ABC)}{AC} = \frac{\sin(BAC)}{BC}\)
\(\frac{\sin(\angle ABC)}{5} = \frac{\sin(60^\circ)}{7}\)

Since \(\text{sin}(60^\circ) = \frac{\sqrt{3}}{2}\):
\(\sin(\angle ABC) = \frac{5 \cdot \frac{\sqrt{3}}{2}}{7} = \frac{5\sqrt{3}}{14}\)

M1: Correct substitution into Cosine Rule: \(8^2 + 5^2 - 2(8)(5)\cos(60^\circ)\).
A1: Correctly finding \(BC = 7\).
M1: Correct substitution into Sine Rule: \(\frac{\sin(\angle ABC)}{5} = \frac{\sin(60^\circ)}{7}\).
A1: Correctly simplifying to \(\frac{5\sqrt{3}}{14}\).

The total surface area of a sphere of radius \(R = 3x\) is:
\(A_{\text{sphere}} = 4 \pi R^2 = 4 \pi (3x)^2 = 36 \pi x^2\)

The curved surface area of a cone with base radius \(r = 2x\) and slant height \(l\) is:
\(A_{\text{cone}} = \pi r l = \pi (2x) l = 2 \pi x l\)

Since the two areas are equal:
\(36 \pi x^2 = 2 \pi x l\)

Divide both sides by \(2 \pi x\) to find \(l\):
\(l = 18x\)

Using Pythagoras\' theorem in the cone to find the height \(h\):
\(r^2 + h^2 = l^2\)

Substitute \(r = 2x\) and \(l = 18x\):
\((2x)^2 + h^2 = (18x)^2\)

\(4x^2 + h^2 = 324x^2\)

\(h^2 = 320x^2\)

\(h = \sqrt{320}x = \sqrt{64 \times 5}x = 8\sqrt{5}x\)

M1: Correctly finding total surface area of the sphere as \(36 \pi x^2\).
M1: Equating sphere area to cone curved area to find slant height \(l = 18x\).
M1: Applying Pythagoras\' theorem \((2x)^2 + h^2 = l^2\) with their \(l\).
A1: Correctly simplifying to find \(h = 8\sqrt{5}x\).

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

Expand the expressions on both sides:
\(3x + 3 - 2x + 2 = x^2 - 1\)

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

Rearrange into a quadratic equation equal to zero:
\(x^2 - x - 6 = 0\)

Factorise the quadratic equation:
\((x - 3)(x + 2) = 0\)

So, \(x = 3\) or \(x = -2\).

M1: Clearing the denominators correctly to get \(3(x + 1) - 2(x - 1) = (x - 1)(x + 1)\).
M1: Expanding and simplifying to a 3-term quadratic equation, e.g., \(x^2 - x - 6 = 0\).
M1: Factorising their 3-term quadratic equation to find the roots.
A1: Correct answers: \(x = 3\) or \(x = -2\) (both required).

We can rewrite the equation in terms of \(2^x\):
\(4^{x+1} = 4 \cdot 4^x = 4(2^x)^2\)

Substitute \(u = 2^x\) into the equation:
\(4u^2 - 9u + 2 = 0\)

Factorise the quadratic equation:
\((4u - 1)(u - 2) = 0\)

This gives:
\(u = \frac{1}{4}\) or \(u = 2\)

Now, substitute back \(u = 2^x\):
- If \(2^x = \frac{1}{4} = 2^{-2}\), then \(x = -2\).
- If \(2^x = 2 = 2^1\), then \(x = 1\).

So the solutions are \(x = -2\) or \(x = 1\).

M1: Rewriting \(4^{x+1}\) as \(4(2^x)^2\) or substituting \(u = 2^x\) to get a quadratic.
M1: Solving the quadratic equation to find \(u = \frac{1}{4}\) and \(u = 2\) (or equivalent values for \(2^x\)).
M1: Setting up \(2^x = \frac{1}{4}\) and \(2^x = 2\) to solve for \(x\).
A1: Correct final answers: \(x = -2\) or \(x = 1\) (both required).

Square both sides to eliminate the square root:
\(w^2 = \frac{2p + 3}{5 - p}\)

Multiply both sides by \(5 - p\):
\(w^2(5 - p) = 2p + 3\)

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

Rearrange the equation to group all terms with \(p\) on one side:
\(5w^2 - 3 = 2p + p w^2\)

Factorise the right-hand side to isolate \(p\):
\(5w^2 - 3 = p(2 + w^2)\)

Divide by \(2 + w^2\) to make \(p\) the subject:
\(p = \frac{5w^2 - 3}{2 + w^2}\)

M1: Squaring both sides correctly to get \(w^2 = \frac{2p + 3}{5 - p}\).
M1: Multiplying by the denominator and expanding correctly to get \(5w^2 - p w^2 = 2p + 3\).
M1: Grouping terms and factorising to get \(p(2 + w^2) = 5w^2 - 3\).
A1: Correct final formula \(p = \frac{5w^2 - 3}{2 + w^2}\) (or equivalent).

First, rearrange the inequality so that one side is zero:
\(2x^2 + 5x - 12 \le 0\)

Solve the corresponding quadratic equation \(2x^2 + 5x - 12 = 0\) to find the critical values.
Factorise the quadratic expression:
We look for numbers that multiply to \(-24\) and add to \(5\). These are \(8\) and \(-3\).
\(2x^2 + 8x - 3x - 12 = 0\)
\(2x(x + 4) - 3(x + 4) = 0\)
\((2x - 3)(x + 4) = 0\)

This gives the critical values:
\(x = 1.5\) or \(x = -4\)

Since the inequality is \(\le 0\), the solution is the region between the critical values:
\(-4 \le x \le 1.5\)

M1: Rearranging to \(2x^2 + 5x - 12 \le 0\).
M1: Attempting to factorise the quadratic expression.
A1: Finding correct critical values of \(x = 1.5\) and \(x = -4\).
A1: Writing the correct inequality range \(-4 \le x \le 1.5\) (or equivalent).

Multiply both sides of the equation by the common denominator \((x+3)(2x-1)\): \(2(2x-1) + 5(x+3) = (x+3)(2x-1)\). Expand the brackets on both sides: \(4x - 2 + 5x + 15 = 2x^2 - x + 6x - 3\). Simplify and collect like terms: \(9x + 13 = 2x^2 + 5x - 3\). Rearrange the equation to equal zero: \(2x^2 - 4x - 16 = 0\). Divide the entire equation by 2: \(x^2 - 2x - 8 = 0\). Factorise the quadratic expression: \((x-4)(x+2) = 0\). This gives the solutions: \(x = 4\) or \(x = -2\).

M1 for multiplying by the common denominator: \(2(2x-1) + 5(x+3) = (x+3)(2x-1)\). M1 for expanding and simplifying to a three-term quadratic equation, e.g., \(2x^2 - 4x - 16 = 0\) or \(x^2 - 2x - 8 = 0\). M1 for factorising their quadratic equation, e.g., \((x-4)(x+2) = 0\), or correct use of the quadratic formula. A1 for both correct solutions: \(x = 4\) or \(x = -2\).

First, apply the Cosine Rule to find the length of the side \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substitute the known values into the formula: \(AC^2 = 5^2 + 8^2 - 2 \cdot 5 \cdot 8 \cdot \cos(60^\circ)\). Recall that \(\cos(60^\circ) = \frac{1}{2}\): \(AC^2 = 25 + 64 - 80 \cdot \frac{1}{2} = 89 - 40 = 49\). Thus, \(AC = \sqrt{49} = 7\text{ cm}\). Now, calculate the perimeter of the triangle: \(\text{Perimeter} = AB + BC + AC = 5 + 8 + 7 = 20\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\). B1 for \(\cos(60^\circ) = 0.5\) (used or implied). M1 for finding side \(AC = 7\). A1 for final perimeter \(= 20\).

The area of a rectangle is given by \(\text{length} \times \text{width}\). Therefore, we can set up the equation: \((x + 5)(2x - 1) = 63\). Expanding the left-hand side: \(2x^2 - x + 10x - 5 = 63\), which simplifies to \(2x^2 + 9x - 5 = 63\). Rearranging into standard quadratic form: \(2x^2 + 9x - 68 = 0\). To solve this quadratic equation, we factorise it. We find two numbers that multiply to \(2 \times (-68) = -136\) and add to \(9\). These numbers are \(17\) and \(-8\). Splitting the middle term: \(2x^2 - 8x + 17x - 68 = 0\), which factorises to \(2x(x - 4) + 17(x - 4) = 0\) and hence \((2x + 17)(x - 4) = 0\). This gives the solutions: \(x = -8.5\) or \(x = 4\). Since the dimensions of the flower bed must be positive, we select \(x = 4\). The dimensions are: \(\text{Length} = 4 + 5 = 9\text{ m}\) and \(\text{Width} = 2(4) - 1 = 7\text{ m}\). The perimeter is: \(\text{Perimeter} = 2 \times (9 + 7) = 32\text{ m}\).

M1 for setting up the equation \((x + 5)(2x - 1) = 63\) M1 for expanding and rearranging to \(2x^2 + 9x - 68 = 0\) M1 for correctly factorising to \((2x + 17)(x - 4) = 0\) (or equivalent use of the quadratic formula) A1 for \(x = 4\) (and discarding \(x = -8.5\)) A1 for finding the perimeter is 32

We factorise each quadratic term in the expression: 1) \(6x^2 - 7x - 3 = (3x + 1)(2x - 3)\). 2) \(4x^2 - 9 = (2x - 3)(2x + 3)\). 3) \(2x^2 + 5x - 12 = (2x - 3)(x + 4)\). 4) \(x^2 + 4x = x(x + 4)\). Rewriting the division as multiplication by the reciprocal of the second fraction: \(\frac{(3x + 1)(2x - 3)}{(2x - 3)(2x + 3)} \times \frac{x(x + 4)}{(2x - 3)(x + 4)}\). Cancelling the common factors \((2x - 3)\) and \((x + 4)\) leaves: \(\frac{3x + 1}{2x + 3} \times \frac{x}{2x - 3}\). Multiplying the remaining terms gives: \(\frac{x(3x + 1)}{(2x + 3)(2x - 3)} = \frac{3x^2 + x}{4x^2 - 9}\).

M1 for factorising \(6x^2 - 7x - 3\) as \((3x + 1)(2x - 3)\) M1 for factorising \(4x^2 - 9\) as \((2x - 3)(2x + 3)\) M1 for factorising \(2x^2 + 5x - 12\) as \((2x - 3)(x + 4)\) M1 for factorising \(x^2 + 4x\) as \(x(x + 4)\) and multiplying by the reciprocal A1 for the final simplified fraction \(\frac{3x^2 + x}{4x^2 - 9}\) or \(\frac{x(3x + 1)}{(2x + 3)(2x - 3)}\)

First, we use the Cosine Rule to find the length of the side \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\). Substituting the given values: \(AC^2 = 8^2 + 5^2 - 2(8)(5)\cos(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), we have \(AC^2 = 64 + 25 - 40 = 49\), which gives \(AC = 7\text{ cm}\). Next, we find the exact area of triangle \(ABC\): \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(60^\circ)\). Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have \(\text{Area} = \frac{1}{2} \times 8 \times 5 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\). We can also express the area using \(AC\) as the base and \(BD\) as the height: \(\text{Area} = \frac{1}{2} \times AC \times BD\). This gives \(10\sqrt{3} = \frac{1}{2} \times 7 \times BD\). Solving for \(BD\), we get \(BD = \frac{20\sqrt{3}}{7}\text{ cm}\).

M1 for substituting into the Cosine Rule: \(AC^2 = 8^2 + 5^2 - 2(8)(5)\cos(60^\circ)\) A1 for finding \(AC = 7\) M1 for substituting into the Area formula: \(\text{Area} = \frac{1}{2} \times 8 \times 5 \times \sin(60^\circ)\) M1 for equating the calculated area to \(\frac{1}{2} \times 7 \times BD\) A1 for the final answer \(\frac{20\sqrt{3}}{7}\)

The total volume of the toy is the sum of the volumes of the cylinder and the cone: \(V = V_{\text{cylinder}} + V_{\text{cone}} = \pi r^2(r) + \frac{1}{3}\pi r^2(r) = \frac{4}{3}\pi r^3\). Setting this equal to the given volume: \(\frac{4}{3}\pi r^3 = 288\pi \implies r^3 = 216 \implies r = 6\text{ cm}\). The slant height \(l\) of the cone is \(l = \sqrt{r^2 + r^2} = r\sqrt{2} = 6\sqrt{2}\text{ cm}\). The external surface area of the toy consists of: 1) the bottom base of the cylinder: \(\pi r^2\), 2) the curved surface of the cylinder: \(2\pi r h = 2\pi r^2\), 3) the curved surface of the cone: \(\pi r l = \pi r^2\sqrt{2}\). Total surface area \(A = \pi r^2 + 2\pi r^2 + \pi r^2\sqrt{2} = \pi r^2(3 + \sqrt{2})\). Substituting \(r = 6\): \(A = 36\pi(3 + \sqrt{2}) = \pi(108 + 36\sqrt{2})\text{ cm}^2\).

M1 for setting up the volume equation: \(\pi r^3 + \frac{1}{3}\pi r^3 = 288\pi\) A1 for finding \(r = 6\) M1 for finding the slant height \(l = 6\sqrt{2}\) M1 for summing the external surface areas: \(\pi r^2 + 2\pi r^2 + \pi r l\) A1 for the final answer \(\pi(108 + 36\sqrt{2})\)

We substitute the linear equation \(y = 2x - 3\) into the quadratic equation: \(x^2 + (2x - 3)^2 = 26\). Expanding the bracket: \(x^2 + 4x^2 - 12x + 9 = 26 \implies 5x^2 - 12x + 9 = 26\). Subtracting 26 from both sides gives the standard form: \(5x^2 - 12x - 17 = 0\). We factorise this quadratic: \(5x^2 + 5x - 17x - 17 = 0 \implies 5x(x + 1) - 17(x + 1) = 0 \implies (5x - 17)(x + 1) = 0\). This gives \(x = -1\) or \(x = 3.4\). Substituting these values back into the linear equation: For \(x = -1\), \(y = 2(-1) - 3 = -5\). For \(x = 3.4\), \(y = 2(3.4) - 3 = 3.8\). The solutions are \(x = -1, y = -5\) and \(x = 3.4, y = 3.8\).

M1 for substituting \(y = 2x - 3\) into the quadratic equation A1 for expanding and simplifying to \(5x^2 - 12x - 17 = 0\) M1 for factorising to \((5x - 17)(x + 1) = 0\) A1 for finding both \(x = -1\) and \(x = 3.4\) A1 for pairing with \(y = -5\) and \(y = 3.8\) respectively

First, factorise the denominator of the third fraction: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). The common denominator is \((2x - 1)(x + 3)\). Expressing all three terms with the common denominator: \(\frac{3(x + 3) - 2(2x - 1) + (5x - 7)}{(2x - 1)(x + 3)}\). Expanding and simplifying the numerator: \(3x + 9 - 4x + 2 + 5x - 7 = (3x - 4x + 5x) + (9 + 2 - 7) = 4x + 4\). Thus, the simplified single fraction is \(\frac{4x + 4}{(2x - 1)(x + 3)}\) or \(\frac{4(x + 1)}{(2x - 1)(x + 3)}\).

M1 for factorising \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\) M1 for putting all three fractions over the common denominator M1 for expanding the numerator terms correctly to \(3x + 9\) and \(-4x + 2\) A1 for simplifying the numerator to \(4x + 4\) A1 for the final simplified fraction \(\frac{4x + 4}{(2x - 1)(x + 3)}\) or \(\frac{4(x + 1)}{(2x - 1)(x + 3)}\)

Multiply both sides by 15 (cross-multiply) to get \(3(4x - 3) = 5(2x + 7)\). Expanding the brackets gives \(12x - 9 = 10x + 35\). Subtracting \(10x\) from both sides gives \(2x - 9 = 35\). Adding 9 to both sides gives \(2x = 44\). Dividing by 2 gives \(x = 22\).

M1 for a correct cross-multiplication or expansion step, e.g., \(3(4x - 3) = 5(2x + 7)\). A0.5 for the correct final answer of 22.

Find the highest common factor of both terms. For the coefficients 12 and 18, the highest common factor is 6. For the variables \(p^2q\) and \(pq^2\), the highest common factor is \(pq\). Factoring out the combined highest common factor \(6pq\) gives \(6pq(2p - 3q)\).

M1 for finding a partial common factor, such as \(3pq(4p - 6q)\) or \(6(2p^2q - 3pq^2)\). A0.5 for the fully factorised correct answer \(6pq(2p - 3q)\).

Apply the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substitute the given values: \(AC^2 = 8.4^2 + 6.5^2 - 2(8.4)(6.5)\cos(54^\circ)\). Calculating each part: \(AC^2 = 70.56 + 42.25 - 109.2 \cdot 0.587785\), which simplifies to \(AC^2 = 112.81 - 64.186 = 48.624\). Taking the square root gives \(AC \approx 6.973\text{ cm}\). To 3 significant figures, this is 6.97.

M1 for correct substitution into the Cosine Rule: \(8.4^2 + 6.5^2 - 2(8.4)(6.5)\cos(54^\circ)\). A0.5 for 6.97.

The total surface area of a cylinder is given by the formula \(A = 2\pi r^2 + 2\pi rh\). Substituting the given values \(r = 4\) and \(h = 15\) yields: \(A = 2\pi(4)^2 + 2\pi(4)(15) = 32\pi + 120\pi = 152\pi\). Evaluating this gives \(A \approx 477.522\text{ cm}^2\). Rounding to 1 decimal place, we obtain 477.5.

M1 for a correct substitution into the total surface area formula, e.g., \(2\pi(4)^2 + 2\pi(4)(15)\). A0.5 for the correct final answer of 477.5.

Apply the power of \(\frac{1}{3}\) to each term inside the bracket: \(27^{\frac{1}{3}} \cdot (x^9)^{\frac{1}{3}} \cdot (y^{-6})^{\frac{1}{3}}\). Simplifying each part: \(27^{\frac{1}{3}} = 3\), \((x^9)^{\frac{1}{3}} = x^3\), and \((y^{-6})^{\frac{1}{3}} = y^{-2}\). Combining these results gives the simplified expression \(3x^3y^{-2}\) (or \(\frac{3x^3}{y^2}\)).

M1 for simplifying at least two out of the three terms correctly (e.g., getting 3 and \(x^3\), or \(x^3\) and \(y^{-2}\)). A0.5 for the fully simplified correct answer \(3x^3y^{-2}\) or \(\frac{3x^3}{y^2}\).

The sale price of \(\$369\) represents \(100\% - 18\% = 82\%\) of the original price. Let the original price be \(x\). Then, \(0.82x = 369\). Solving for \(x\), we find \(x = \frac{369}{0.82} = 450\). Thus, the original price of the bicycle was \(\$450\).

M1 for dividing the sale price by the correct percentage factor, e.g., \(369 \div 0.82\). A0.5 for the correct original price of 450.

First calculate the vector \(\mathbf{a} + 2\mathbf{b}\): \(\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + 2\begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 - 2 \\ -2 + 10 \end{pmatrix} = \begin{pmatrix} 1 \\ 8 \end{pmatrix}\). Next, find the magnitude of this vector using Pythagoras' theorem: \(|\mathbf{a} + 2\mathbf{b}| = \sqrt{1^2 + 8^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8.062\). Correct to 2 decimal places, the magnitude is 8.06.

M1 for finding the vector \(\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 1 \\ 8 \end{pmatrix}\). A0.5 for the correct magnitude of 8.06 (or \(\sqrt{65}\)).

Since the mean of 5 numbers is 14.8, the total sum of the five numbers is \(14.8 \times 5 = 74\). The sum of the four given numbers is \(9 + 17 + 12 + 15 = 53\). Let the fifth number be \(x\). Then, \(53 + x = 74\), which gives \(x = 74 - 53 = 21\).

M1 for calculating the required total sum of 74 (e.g., \(14.8 \times 5\)) or the sum of the four numbers of 53. A0.5 for the correct fifth number of 21.

To solve \(\frac{3x - 1}{4} - \frac{x + 2}{3} = 2\), multiply all terms by the lowest common multiple of 4 and 3, which is 12. This gives \(3(3x - 1) - 4(x + 2) = 24\). Expanding the brackets yields \(9x - 3 - 4x - 8 = 24\). Simplifying the left side gives \(5x - 11 = 24\). Adding 11 to both sides gives \(5x = 35\). Dividing by 5 yields \(x = 7\).

M1 for a correct method to clear the denominators, e.g. \(3(3x - 1) - 4(x + 2) = 24\) or expressing the left hand side as a single fraction with a correct numerator. A0.5 for \(7\).

Using the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), we have \(\frac{4}{3}\pi r^3 = 288\pi\). Dividing both sides by \(\pi\) gives \(\frac{4}{3}r^3 = 288\). Multiplying both sides by \(\frac{3}{4}\) gives \(r^3 = 216\). Taking the cube root of both sides gives \(r = \sqrt[3]{216} = 6\text{ cm}\).

M1 for setting up the equation \(\frac{4}{3}\pi r^3 = 288\pi\) and finding \(r^3 = 216\). A0.5 for \(6\).

Multiply the entire equation by the common denominator \(x(x+2)\) to eliminate the fractions:

\(2(x+2) + 3x = 1 \cdot x(x+2)\)

Expand both sides:

\(2x + 4 + 3x = x^2 + 2x\)

\(5x + 4 = x^2 + 2x\)

Rearrange into a standard quadratic equation form \(ax^2 + bx + c = 0\):

\(x^2 - 3x - 4 = 0\)

Factorise the quadratic expression:

\((x - 4)(x + 1) = 0\)

This gives the solutions:

\(x = 4\) or \(x = -1\)

M1 for multiplying by \(x(x+2)\) correctly to clear fractions: \(2(x+2) + 3x = x(x+2)\) (or equivalent)
M1 for simplifying to a correct quadratic equation, e.g., \(x^2 - 3x - 4 = 0\)
A1 for both correct solutions: \(x = 4\) and \(x = -1\)

Using the Cosine Rule to find the side opposite to the given angle:

\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)

Substitute the given values into the formula:

\(AC^2 = 7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(112^\circ)\)

Calculate the terms:

\(7.4^2 = 54.76\)

\(9.2^2 = 84.64\)

\(2 \times 7.4 \times 9.2 \times \cos(112^\circ) \approx 136.16 \times (-0.3746) \approx -51.01\)

Now add these together:

\(AC^2 = 54.76 + 84.64 - (-51.01) \approx 190.41\)

Find the square root to get \(AC\):

\(AC = \sqrt{190.41} \approx 13.80\text{ cm}\)

To 3 significant figures, the length is \(13.8\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(7.4^2 + 9.2^2 - 2(7.4)(9.2)\cos(112^\circ)\)
A1 for \(AC^2\) value in the range \([190.0, 190.5]\)
A1 for \(13.8\) (accept 13.79 to 13.81)

The total surface area \(A\) of a solid cylinder is the sum of the curved surface area and the areas of the two circular bases:

\(A = 2\pi r^2 + 2\pi r h\)

Substitute \(r = 3.5\) and \(h = 12\):

\(A = 2\pi(3.5)^2 + 2\pi(3.5)(12)\)

Calculate each part:

Area of the two bases: \(2\pi(12.25) = 24.5\pi \approx 76.97\text{ cm}^2\)

Curved surface area: \(2\pi(42) = 84\pi \approx 263.89\text{ cm}^2\)

Total surface area:

\(A = 24.5\pi + 84\pi = 108.5\pi \approx 340.86\text{ cm}^2\)

Rounding to 1 decimal place yields \(340.9\text{ cm}^2\).

M1 for a correct expression for either the curved surface area (\(2\pi \times 3.5 \times 12\)) or the area of both circular ends (\(2\pi \times 3.5^2\))
M1 for summing both components: \(2\pi(3.5)^2 + 2\pi(3.5)(12)\)
A1 for \(340.9\) (accept 340.8 to 341.0)

Multiply both sides by the denominator \((3 - 2x)\) to clear the fraction:

\(y(3 - 2x) = 5x - 2\)

Expand the left-hand side:

\(3y - 2xy = 5x - 2\)

Rearrange the equation to group all terms containing \(x\) on one side and the rest on the other:

\(3y + 2 = 5x + 2xy\)

Factorise \(x\) from the right-hand side:

\(3y + 2 = x(5 + 2y)\)

Divide both sides by \((5 + 2y)\) to make \(x\) the subject:

\(x = \frac{3y + 2}{2y + 5}\)

M1 for multiplying by \((3 - 2x)\) and expanding correctly: \(3y - 2xy = 5x - 2\) (or equivalent)
M1 for collecting all terms with \(x\) on one side and factorising: \(x(5 + 2y) = 3y + 2\) (or equivalent)
A1 for \(x = \frac{3y+2}{2y+5}\) (accept equivalent fractions)

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

An increase of 15% means the price is multiplied by \(1.15\).

A subsequent decrease of 20% means this new price is multiplied by \(1 - 0.20 = 0.80\).

Form an equation representing both percentage changes:

\(P \times 1.15 \times 0.80 = 184\)

Simplify the multipliers:

\(1.15 \times 0.80 = 0.92\)

So,

\(0.92P = 184\)

Solve for \(P\):

\(P = \frac{184}{0.92} = 200\)

The original price was $200.

M1 for dividing the final price by the second multiplier: \(184 \div 0.80 = 230\) (or equivalent)
M1 for dividing that result by the first multiplier: \(230 \div 1.15\) (or writing the combined equation \(P \times 0.92 = 184\))
A1 for \(200\)

Express both bases as powers of 3:

\(27 = 3^3\) and \(9 = 3^2\)

Substitute these into the equation:

\((3^3)^{2x - 1} = (3^2)^{x + 4}\)

Apply the index law \((a^m)^n = a^{mn}\):

\(3^{3(2x - 1)} = 3^{2(x + 4)}\)

Equate the exponents since the bases are now the same:

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

Expand the brackets:

\(6x - 3 = 2x + 8\)

Rearrange to solve for \(x\):

\(4x = 11\)

\(x = \frac{11}{4} = 2.75\)

M1 for expressing both sides as powers of the same base: \((3^3)^{2x-1} = (3^2)^{x+4}\) (or equivalent)
M1 for equating exponents and forming a linear equation: \(3(2x-1) = 2(x+4)\) (or equivalent)
A1 for \(2.75\) (or \(\frac{11}{4}\))

The total number of marbles in the bag is \(5 + 3 = 8\).

There are two possible outcomes where the marbles are different colours:
1. First marble is red and the second is blue (RB).
2. First marble is blue and the second is red (BR).

Since the marbles are drawn without replacement, calculate the probability of each scenario:

\(P(RB) = P(\text{1st Red}) \times P(\text{2nd Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

\(P(BR) = P(\text{1st Blue}) \times P(\text{2nd Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Add the probabilities of these two mutually exclusive outcomes:

\(P(\text{different}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56}\)

Simplify the fraction:

\(\frac{30}{56} = \frac{15}{28}\)

M1 for calculating the probability of a single combined outcome, e.g., \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
M1 for summing the two correct scenarios: \(\frac{15}{56} + \frac{15}{56}\) (or multiplying by 2)
A1 for \(\frac{15}{28}\) (must be fully simplified)

The formula for the gradient \(m\) of a straight line passing through points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Substitute the given values \(P(3, -1)\), \(Q(a, 7)\), and \(m = \frac{4}{3}\) into the formula:

\(\frac{7 - (-1)}{a - 3} = \frac{4}{3}\)

Simplify the numerator:

\(\frac{8}{a - 3} = \frac{4}{3}\)

Cross-multiply to eliminate the fractions:

\(8 \times 3 = 4(a - 3)\)

\(24 = 4a - 12\)

Add 12 to both sides:

\(36 = 4a\)

Divide by 4:

\(a = 9\)

M1 for substituting coordinates into the gradient formula: \(\frac{7 - (-1)}{a - 3}\) (or equivalent)
M1 for setting up the equation \(\frac{8}{a - 3} = \frac{4}{3}\) and showing a correct step of cross-multiplication, e.g., \(24 = 4(a - 3)\)
A1 for \(9\)

Multiply both sides by \((x - 3)(x + 2)\) to clear the denominators: \(4(x + 2) + 3(x - 3) = 2(x - 3)(x + 2)\). Expand both sides: \(4x + 8 + 3x - 9 = 2(x^2 - x - 6)\). This simplifies to \(7x - 1 = 2x^2 - 2x - 12\). Rearranging to standard quadratic form: \(2x^2 - 9x - 11 = 0\). Factorising the quadratic gives \((2x - 11)(x + 1) = 0\). Solving this gives \(x = 5.5\) or \(x = -1\).

M1 for clearing the fractions to get \(4(x + 2) + 3(x - 3) = 2(x - 3)(x + 2)\) or equivalent. M1 for reducing to standard quadratic form \(2x^2 - 9x - 11 = 0\) or equivalent. A1 for both solutions \(x = 5.5\) (or \(\frac{11}{2}\)) and \(x = -1\).

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Dividing both the numerator and the denominator by the common factor \((2x + 1)\) gives 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 final answer \(\frac{x - 3}{2x - 1}\) or equivalent.

Using the cosine rule: \(a^2 = b^2 + c^2 - 2bc \cos(A)\), where \(a = 9.2\), \(b = 12.5\), and \(c = 7.4\). Substituting these values gives: \(9.2^2 = 12.5^2 + 7.4^2 - 2(12.5)(7.4) \cos(BAC)\). Simplifying this yields: \(84.64 = 156.25 + 54.76 - 185 \cos(BAC)\), which becomes \(84.64 = 211.01 - 185 \cos(BAC)\). Rearranging to find the cosine of the angle: \(185 \cos(BAC) = 126.37\), so \( \cos(BAC) = \frac{126.37}{185}\). Calculating the inverse cosine: \(BAC = \cos^{-1}(0.68308...) \approx 46.9^\circ\).

M1 for correct substitution into the cosine rule: \(9.2^2 = 12.5^2 + 7.4^2 - 2(12.5)(7.4)\cos(BAC)\). M1 for rearranging to get \(\cos(BAC) = \frac{126.37}{185}\). A1 for \(46.9\) or \(46.91...\) (accept range \(46.9\) to \(46.92\)).

The total volume of the toy is the sum of the volume of the cylinder and the volume of the cone. Volume of the cylinder = \(\pi r^2 h = \pi r^2(3r) = 3\pi r^3\). Volume of the cone = \(\frac{1}{3}\pi r^2 H = \frac{1}{3}\pi r^2(2r) = \frac{2}{3}\pi r^3\). Total volume = \(3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\). Setting this equal to the given total volume: \(\frac{11}{3}\pi r^3 = 440\). Solving for \(r^3\) gives: \(r^3 = \frac{440 \times 3}{11\pi} = \frac{120}{\pi} \approx 38.197\). Taking the cube root yields: \(r = \sqrt[3]{38.197} \approx 3.37 \text{ cm}\).

M1 for expressing total volume as \(3\pi r^3 + \frac{2}{3}\pi r^3\) or equivalent. M1 for setting up the equation \(\frac{11}{3}\pi r^3 = 440\) and solving for \(r^3\) to get \(r^3 = \frac{120}{\pi}\) or \(r^3 \approx 38.2\). A1 for \(3.37\) or \(3.367...\) (accept range \(3.36\) to \(3.37\)).

Substitute the expression for \(y\) from the first equation into the second equation: \(x^2 + (2x - 3)^2 = 5\). Expand and simplify: \(x^2 + 4x^2 - 12x + 9 = 5\), which reduces to \(5x^2 - 12x + 4 = 0\). Solve this quadratic equation using the quadratic formula or factorisation: \((5x - 2)(x - 2) = 0\). This gives \(x = 2\) or \(x = 0.4\). Substitute these values of \(x\) back into the linear equation to find the corresponding values of \(y\): If \(x = 2\), \(y = 2(2) - 3 = 1\). If \(x = 0.4\), \(y = 2(0.4) - 3 = -2.2\). Thus, the solutions are \(x = 2, y = 1\) and \(x = 0.4, y = -2.2\).

M1 for substituting \(y = 2x - 3\) into the quadratic equation to obtain \(x^2 + (2x - 3)^2 = 5\). M1 for simplifying to a standard 3-term quadratic equation \(5x^2 - 12x + 4 = 0\) and attempting to solve. A1 for both pairs of solutions: \((2, 1)\) and \((0.4, -2.2)\).

First, multiply both sides by the denominator to eliminate the fraction: \(y(5 - 2x) = 3x + 2\). Expand the left side: \(5y - 2xy = 3x + 2\). Collect all terms involving \(x\) on one side and the rest of the terms on the other side: \(5y - 2 = 3x + 2xy\). Factorise out \(x\) on the right side: \(5y - 2 = x(3 + 2y)\). Finally, divide both sides by \((3 + 2y)\) to isolate \(x\): \(x = \frac{5y - 2}{3 + 2y}\).

M1 for clearing the fraction to obtain \(y(5 - 2x) = 3x + 2\) or equivalent. M1 for grouping \(x\) terms together and factorising to get \(x(3 + 2y) = 5y - 2\) or equivalent. A1 for final answer \(x = \frac{5y - 2}{3 + 2y}\) or equivalent.

Using the sine rule: \(\frac{PR}{\sin(PQR)} = \frac{QR}{\sin(QPR)}\). Substituting the given values gives: \(\frac{PR}{\sin 58^\circ} = \frac{8.5}{\sin 47^\circ}\). Rearranging to make \(PR\) the subject: \(PR = \frac{8.5 \sin 58^\circ}{\sin 47^\circ}\). Calculating this gives: \(PR \approx \frac{8.5 \times 0.84805}{0.73135} \approx 9.856 \text{ cm}\). To 3 significant figures, \(PR = 9.86 \text{ cm}\).

M1 for correct substitution into the sine rule: \(\frac{PR}{\sin 58^\circ} = \frac{8.5}{\sin 47^\circ}\). M1 for rearranging to make \(PR\) the subject: \(PR = \frac{8.5 \sin 58^\circ}{\sin 47^\circ}\). A1 for \(9.86\) or \(9.856...\) (accept range \(9.85\) to \(9.86\)).

1. Set up the equation for the area of the garden:
\((x + 5)(2x - 3) = 150\)
2. Expand and simplify the equation:
\(2x^2 - 3x + 10x - 15 = 150\)
\(2x^2 + 7x - 15 = 150\)
\(2x^2 + 7x - 165 = 0\)
3. Use the quadratic formula to solve for \(x\):
\(x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-165)}}{2(2)}
= \frac{-7 \pm \sqrt{49 + 1320}}{4}
= \frac{-7 \pm \sqrt{1369}}{4}
= \frac{-7 \pm 37}{4}\)
This gives \(x = \frac{30}{4} = 7.5\) or \(x = \frac{-44}{4} = -11\).
Since the dimensions must be positive, \(x = 7.5\).
4. Find the length and width of the garden:
Length = \(7.5 + 5 = 12.5\text{ m}\)
Width = \(2(7.5) - 3 = 12\text{ m}\)
5. Calculate the perimeter:
Perimeter = \(2 \times (\text{length} + \text{width}) = 2 \times (12.5 + 12) = 2 \times 24.5 = 49\text{ m}\).

- **M1**: for writing \((x + 5)(2x - 3) = 150\)
- **M1**: for expanding and simplifying to a three-term quadratic equation, e.g., \(2x^2 + 7x - 165 = 0\)
- **M1**: for solving the quadratic equation to find \(x = 7.5\) (or \(x = -11\))
- **M1**: for calculating the dimensions as \(12.5\) and \(12\)
- **A1**: for the final answer \(49\)

1. Substitute \(y = 2x - 5\) into the second equation:
\(x^2 + (2x - 5)^2 = 10\)
2. Expand the squared term:
\(x^2 + 4x^2 - 20x + 25 = 10\)
3. Simplify and form a quadratic equation:
\(5x^2 - 20x + 15 = 0\)
4. Divide the equation by 5:
\(x^2 - 4x + 3 = 0\)
5. Factorise:
\((x - 1)(x - 3) = 0\)
So, \(x = 1\) or \(x = 3\).
6. Substitute the \(x\) values back into \(y = 2x - 5\):
When \(x = 1\), \(y = 2(1) - 5 = -3\).
When \(x = 3\), \(y = 2(3) - 5 = 1\).
Thus, the two solutions are \((1, -3)\) and \((3, 1)\).

- **M1**: for substituting \(y = 2x - 5\) into \(x^2 + y^2 = 10\)
- **M1**: for expanding and simplifying to a 3-term quadratic, e.g., \(5x^2 - 20x + 15 = 0\) (or \(x^2 - 4x + 3 = 0\))
- **M1**: for factorising or using the quadratic formula to find \(x = 1\) and \(x = 3\)
- **M1**: for substituting both \(x\) values back to find \(y = -3\) and \(y = 1\)
- **A1**: for the final answers written correctly as \((1, -3)\) and \((3, 1)\)

1. Find a common denominator for the two fractions, which is \((x - 3)(2x + 1)\).
2. Rewrite each fraction with the common denominator:
\(\frac{4(2x + 1)}{(x - 3)(2x + 1)} - \frac{3(x - 3)}{(x - 3)(2x + 1)}
= \frac{4(2x + 1) - 3(x - 3)}{(x - 3)(2x + 1)}\)
3. Expand the terms in the numerator:
\(4(2x + 1) = 8x + 4\)
\(-3(x - 3) = -3x + 9\)
4. Simplify the numerator:
\(8x + 4 - 3x + 9 = 5x + 13\)
5. The single simplified fraction is:
\(\frac{5x + 13}{(x - 3)(2x + 1)}\) (or \(\frac{5x + 13}{2x^2 - 5x - 3}\)).

- **M1**: for identifying the common denominator as \((x - 3)(2x + 1)\)
- **M1**: for multiplying the numerators correctly: \(4(2x + 1)\) and \(3(x - 3)\)
- **M1**: for expanding the numerator terms to \(8x + 4\) and \(-3x + 9\)
- **A1**: for the correct simplified numerator \(5x + 13\)
- **A1**: for the correct denominator \((x - 3)(2x + 1)\) or \(2x^2 - 5x - 3\)

1. Square both sides of the equation to eliminate the square root:
\(w^2 = \frac{3t + 5}{t - 2}\)
2. Multiply both sides by \((t - 2)\) to clear the fraction:
\(w^2(t - 2) = 3t + 5\)
3. Expand the left-hand side:
\(w^2 t - 2w^2 = 3t + 5\)
4. Collect all terms containing \(t\) on one side, and all other terms on the opposite side:
\(w^2 t - 3t = 2w^2 + 5\)
5. Factorise \(t\) on the left-hand side:
\(t(w^2 - 3) = 2w^2 + 5\)
6. Divide both sides by \((w^2 - 3)\) to isolate \(t\):
\(t = \frac{2w^2 + 5}{w^2 - 3}\)

- **M1**: for squaring both sides: \(w^2 = \frac{3t + 5}{t - 2}\)
- **M1**: for multiplying by \((t - 2)\): \(w^2(t - 2) = 3t + 5\)
- **M1**: for expanding and rearranging to collect \(t\) terms on one side: \(w^2 t - 3t = 2w^2 + 5\)
- **M1**: for factorising \(t\): \(t(w^2 - 3) = 2w^2 + 5\)
- **A1**: for the final correct formula \(t = \frac{2w^2 + 5}{w^2 - 3}\) (or equivalent such as \(t = \frac{-2w^2 - 5}{3 - w^2}\))

1. Use the Cosine rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(72^circ)\)
\(AC^2 = 14400 + 22500 - 36000 \times \cos(72^circ)\)
Using a calculator: \(\cos(72^circ) \approx 0.309017\)
\(AC^2 \approx 36900 - 11124.61 = 25775.39\)
\(AC \approx \sqrt{25775.39} \approx 160.547\text{ m}\)
2. Find the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(72^circ) = 9000 \times \sin(72^circ) \approx 8559.51\text{ m}^2\)
3. Let \(h\) be the shortest distance (perpendicular height) from \(B\) to the side \(AC\):
\(\text{Area} = \frac{1}{2} \times AC \times h\)
\(8559.51 \approx \frac{1}{2} \times 160.547 \times h\)
\(h = \frac{2 \times 8559.51}{160.547} \approx 106.629\text{ m}\)
Rounding to 1 decimal place gives \(106.6\text{ m}\).

- **M1**: for using the Cosine rule to find \(AC\): \(120^2 + 150^2 - 2(120)(150)\cos(72^circ)\)
- **A1**: for \(AC \approx 160.5\) (or \(160.55\))
- **M1**: for calculating the area of the triangle: \(\frac{1}{2} \times 120 \times 150 \times \sin(72^circ)\) [or finding \(\text{Area} \approx 8560\)]
- **M1**: for setting up the height equation: \(\frac{1}{2} \times 160.5 \times h = 8560\) (or alternative method using sine of angle \(A\) or \(C\))
- **A1**: for the final answer \(106.6\) (accept \(106.5\) to \(106.7\))

1. Use the area formula to find angle \(PQR\):
\(\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(PQR)\)
\(42 = \frac{1}{2} \times 8 \times 13 \times \sin(PQR)\)
\(42 = 52 \sin(PQR)\)
\(\sin(PQR) = \frac{42}{52} = \frac{21}{26} \approx 0.807692\)
2. Since angle \(PQR\) is obtuse:
\(PQR = 180^\circ - \arcsin\left(\frac{21}{26}\right) \approx 180^\circ - 53.868^\circ = 126.132^\circ\)
3. Use the Cosine rule to calculate the length of \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\)
\(PR^2 = 8^2 + 13^2 - 2 \times 8 \times 13 \times \cos(126.132^\circ)\)
\(PR^2 = 64 + 169 - 208 \times \cos(126.132^\circ)\)
Using a calculator: \(\cos(126.132^\circ) \approx -0.589626\)
\(PR^2 \approx 233 - 208 \times (-0.589626) = 233 + 122.64 = 355.64\)
\(PR \approx \sqrt{355.64} \approx 18.858\text{ cm}\)
Rounding to 3 significant figures gives \(18.9\text{ cm}\).

- **M1**: for setting up the area equation: \(\frac{1}{2} \times 8 \times 13 \times \sin(PQR) = 42\)
- **A1**: for finding angle \(PQR \approx 126.1^\circ\) (or \(126.13^\circ\))
- **M1**: for using the Cosine rule with their angle: \(PR^2 = 8^2 + 13^2 - 2(8)(13)\cos(126.1^\circ)\)
- **M1**: for calculating \(PR^2 \approx 355.6\)
- **A1**: for the final answer \(18.9\) (accept \(18.8\) to \(19.0\))

1. Write down the expression for the total volume of the toy:
\(\text{Volume} = \text{Volume of cylinder} + \text{Volume of hemisphere}\)
\(\text{Volume} = \pi r^2 h + \frac{2}{3}\pi r^3\)
Since \(h = 3r\):
\(\text{Volume} = \pi r^2(3r) + \frac{2}{3}\pi r^3 = 3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\)
2. Set the volume equal to \(1000\) and solve for \(r\):
\(\frac{11}{3}\pi r^3 = 1000\)
\(r^3 = \frac{3000}{11\pi} \approx 86.812\)
\(r \approx \sqrt[3]{86.812} \approx 4.428\text{ cm}\)
3. Formulate the total surface area of the toy:
\(\text{Total Surface Area} = \text{Area of base} + \text{Curved area of cylinder} + \text{Curved area of hemisphere}\)
\(\text{Total Surface Area} = \pi r^2 + 2\pi r h + 2\pi r^2 = \pi r^2 + 2\pi r(3r) + 2\pi r^2 = 9\pi r^2\)
4. Substitute \(r \approx 4.428\) into the surface area formula:
\(\text{Total Surface Area} = 9 \times \pi \times 4.428^2 \approx 554.39\text{ cm}^2\)
Rounding to 3 significant figures gives \(554\text{ cm}^2\).

- **M1**: for writing the volume equation: \(3\pi r^3 + \frac{2}{3}\pi r^3 = 1000\) or \(\frac{11}{3}\pi r^3 = 1000\)
- **A1**: for finding \(r \approx 4.43\) (or \(4.428\))
- **M1**: for the total surface area formula: \(\pi r^2 + 6\pi r^2 + 2\pi r^2 = 9\pi r^2\)
- **M1**: for substituting their value of \(r\) into their surface area expression
- **A1**: for the final answer \(554\) (accept \(553\) to \(555\))

1. Find the slant height, \(L\), of the large cone:
\(L = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = 26\text{ cm}\)
2. Determine the dimensions of the small cone that was removed:
The height of the frustum is \(12\text{ cm}\), so the height of the small cone is:
\(h = 24 - 12 = 12\text{ cm}\).
By similarity, the scale factor between the small cone and the large cone is:
\(k = \frac{12}{24} = \frac{1}{2}\).
So, the radius of the small cone is:
\(r = 10 \times \frac{1}{2} = 5\text{ cm}\).
The slant height of the small cone is:
\(l = 26 \times \frac{1}{2} = 13\text{ cm}\).
3. Calculate the curved surface area (CSA) of the frustum:
\(\text{CSA of frustum} = \text{CSA of large cone} - \text{CSA of small cone}\)
\(\text{CSA of frustum} = \pi R L - \pi r l = \pi(10)(26) - \pi(5)(13) = 260\pi - 65\pi = 195\pi\)
4. Calculate the areas of the two circular bases of the frustum:
\(\text{Area of large base} = \pi R^2 = \pi (10)^2 = 100\pi\)
\(\text{Area of small base} = \pi r^2 = \pi (5)^2 = 25\pi\)
5. Calculate the total surface area of the frustum:
\(\text{Total Surface Area} = 195\pi + 100\pi + 25\pi = 320\pi\text{ cm}^2\).

- **M1**: for finding the slant height of the large cone: \(L = \sqrt{24^2 + 10^2} = 26\)
- **M1**: for finding the radius \(r = 5\) and slant height \(l = 13\) of the small cone using similarity
- **M1**: for calculating the curved surface area of the frustum: \(\pi(10)(26) - \pi(5)(13) = 195\pi\)
- **M1**: for adding the areas of the two circular bases: \(\pi(10)^2 + \pi(5)^2 = 125\pi\)
- **A1**: for the final answer \(320\pi\)