2023 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

Multiply the first equation by 3 and the second equation by 2 to get:
\(9x - 6y = 57\)
\(8x + 6y = 28\)

Adding these two equations together eliminates \(y\):
\(17x = 85\)
\(x = 5\)

Now substitute \(x = 5\) into the first equation:
\(3(5) - 2y = 19\)
\(15 - 2y = 19\)
\(-2y = 4\)
\(y = -2\)

M1 for a correct method to eliminate one variable (e.g. multiplying coefficients to equate them).
A1 for \(x = 5\).
A1 for \(y = -2\).

First, factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

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

Now, simplify the expression by dividing out the common factor of \((x - 3)\):
\(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\)

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

First, find the volume of the sphere:
\(V_{\text{sphere}} = \frac{4}{3}\pi (4)^3 = \frac{256}{3}\pi\)

Next, write the expression for the volume of the cone using height \(h = 12\):
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 (12) = 4\pi r^2\)

Since the volume remains the same, equate the two volumes:
\(4\pi r^2 = \frac{256}{3}\pi\)

Divide both sides by \(4\pi\):
\(r^2 = \frac{64}{3}
\)r = \sqrt{\frac{64}{3}} \approx 4.6188\text{ cm}\)

Rounding to 3 significant figures, we get \(4.62\text{ cm}\).

M1 for setting up the volume equation: \ \frac{4}{3}\pi (4)^3 = \frac{1}{3}\pi r^2 (12)
M1 for simplifying to \(r^2 = \frac{64}{3}\) (or \(r^2 \approx 21.3\))
A1 for \(4.62\) or \(4.618...\) (or exact value \(\frac{8\sqrt{3}}{3}\))

The interior angle and exterior angle of a regular polygon sum to \(180^\circ\).
Therefore, the size of each exterior angle is:
\(180^\circ - 156^\circ = 24^\circ\)

The sum of the exterior angles of any regular polygon is \(360^\circ\).
Therefore, the number of sides \(n\) is:
\(n = \frac{360^
\circ}{24^\circ} = 15\)

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

First, find the interval containing 50 minutes, which is \(40 < t \le 60\).
The cumulative frequency increases from 40 to 70 across this interval, meaning there are \(70 - 40 = 30\) people in this group.

Assuming a constant rate of journeys across the interval, 50 minutes is halfway between 40 and 60 minutes.
Thus, the estimated cumulative frequency at \(t = 50\) is:
\(40 + \frac{50 - 40}{60 - 40} \times 30 = 40 + 15 = 55\)

This means an estimated 55 people took 50 minutes or less.
Therefore, the number of people who took more than 50 minutes is:
\(80 - 55 = 25\)

M1 for identifying the frequency of the interval containing 50 minutes is \(70 - 40 = 30\).
M1 for finding the estimated cumulative frequency at 50 minutes is \(40 + 15 = 55\) (or equivalent interpolation method, e.g. finding that 10 minutes out of the 20 minutes interval represents \(\frac{10}{20} \times 30 = 15\) people).
A1 for 25.

A reduction of 15% means that the sale price represents \(100\% - 15\% = 85\%\) of the original price.

Let \(P\) be the original price of the laptop:
\(0.85 \times P = 646\)

Solve for \(P\):
\(P = \frac{646}{0.85} = 760\)

The original price was $760.

M1 for identifying that $646 corresponds to 85% of the original price (e.g. writing \(0.85P = 646\) or equivalent percentage ratio).
M1 for the calculation \(646 / 0.85\).
A1 for 760.

First, find the initial height, \(h\), of the ladder on the wall using Pythagoras' theorem:
\(h^2 + 2.5^2 = 6.5^2\)
\(h^2 + 6.25 = 42.25\)
\(h^2 = 36 \implies h = 6\text{ m}\)

After the ladder slips down the wall by \(0.4\text{ m}\), the new height on the wall is:
\(6 - 0.4 = 5.6\text{ m}\)

Let the new distance from the base of the ladder to the wall be \(d\). Using Pythagoras' theorem again:
\(d^2 + 5.6^2 = 6.5^2\)
\(d^2 + 31.36 = 42.25\)
\(d^2 = 10.89\)
\(d = \sqrt{10.89} = 3.3\text{ m}\)

Calculate the distance the base of the ladder slides outwards:
\(3.3 - 2.5 = 0.8\text{ m}\)

M1 for finding the initial height of the ladder on the wall: \(\sqrt{6.5^2 - 2.5^2} = 6\).
M1 for finding the new distance of the base of the ladder from the wall: \(\sqrt{6.5^2 - 5.6^2} = 3.3\).
A1 for 0.8.

First, simplify the expression inside the parentheses by moving the negative index \(y^{-9}\) to the numerator:
\(\frac{64x^6}{y^{-9}} = 64x^6 y^9\)

Next, apply the outer power of \(-\frac{2}{3}\) to each component inside the parentheses:
\(\left( 64x^6 y^9 \right)^{-\frac{2}{3}} = (64)^{-\frac{2}{3}} \cdot (x^6)^{-\frac{2}{3}} \cdot (y^9)^{-\frac{2}{3}}\)

Evaluate each term:
1. \((64)^{-\frac{2}{3}} = \frac{1}{(64)^{2/3}} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}\)
2. \((x^6)^{-\frac{2}{3}} = x^{6 \times -\frac{2}{3}} = x^{-4} = \frac{1}{x^4}\)
3. \((y^9)^{-\frac{2}{3}} = y^{9 \times -\frac{2}{3}} = y^{-6} = \frac{1}{y^6}\)

Combine the simplified parts to get the final answer:
\(\frac{1}{16x^4 y^6}\)

M1 for simplifying inside the bracket to get \(64x^6y^9\) or evaluating \(64^{-2/3} = 1/16\).
M1 for applying the exponent correctly to the algebraic terms, obtaining \(x^{-4}\) or \(y^{-6}\) (or \(x^4\) and \(y^6\) in the denominator).
A1 for \(\frac{1}{16x^4y^6}\) or \(1 / (16x^4y^6)\) or \(16^{-1}x^{-4}y^{-6}\).

Multiply both sides of the equation by the common denominator \(x(x-1)\):
\(4x - 3(x-1) = x(x-1)\)

Expand the terms:
\(4x - 3x + 3 = x^2 - x\)
\(x + 3 = x^2 - x\)

Rearrange into a quadratic equation set to zero:
\(x^2 - 2x - 3 = 0\)

Factorise the quadratic expression:
\((x-3)(x+1) = 0\)

Solve for \(x\):
\(x = 3\) or \(x = -1\).

M1 for multiplying by \(x(x-1)\) to clear fractions
M1 for reducing to standard quadratic form \(x^2 - 2x - 3 = 0\)
A1 for both correct solutions \(x = 3\) and \(x = -1\)

Multiply both sides by \(2 - w\):
\(t(2 - w) = 3w + 4\)

Expand the bracket:
\(2t - tw = 3w + 4\)

Rearrange to get all terms with \(w\) on one side and other terms on the other side:
\(2t - 4 = tw + 3w\)

Factorise out \(w\):
\(2t - 4 = w(t + 3)\)

Divide by \(t + 3\) to isolate \(w\):
\(w = \frac{2t - 4}{t + 3}\).

M1 for multiplying by \((2-w)\) to get \(t(2-w) = 3w+4\)
M1 for collecting terms in \(w\) on one side and factorising
A1 for correct final answer \(w = \frac{2t-4}{t+3}\) or equivalent

The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\).
For a sphere of radius \(3\text{ cm}\):
\(V_{\text{sphere}} = \frac{4}{3} \pi (3)^3 = 36\pi\text{ cm}^3\).

The volume of a cone is given by \(V = \frac{1}{3}\pi R^2 h\).
For a cone of base radius \(4\text{ cm}\) and height \(h\):
\(V_{\text{cone}} = \frac{1}{3} \pi (4)^2 h = \frac{16}{3}\pi h\).

Since the volume remains the same:
\(\frac{16}{3}\pi h = 36\pi\)

Divide both sides by \(\pi\):
\(\frac{16}{3} h = 36\)

Solve for \(h\):
\(h = \frac{36 \times 3}{16} = \frac{108}{16} = 6.75\text{ cm}\).

M1 for expression for volume of sphere: \(\frac{4}{3} \pi \times 3^3\) (or \(36\pi\))
M1 for equating their sphere volume to cone volume expression: \(\frac{1}{3} \pi \times 4^2 \times h\)
A1 for \(6.75\)

The sum of the interior angles of a polygon with \(n\) sides is given by \((n - 2) \times 180^\circ\).
For a pentagon (\(n = 5\)):
\(\text{Sum} = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\).

Set up the equation by summing the given angles:
\(x + (2x - 10) + (x + 30) + 115 + 125 = 540\)

Combine like terms:
\(4x + 260 = 540\)

Subtract 260 from both sides:
\(4x = 280\)

Divide by 4:
\(x = 70\).

M1 for showing sum of angles of a pentagon is \(540^\circ\)
M1 for setting up the equation \(4x + 260 = 540\) (or equivalent)
A1 for \(70\)

From the cumulative frequency table:
- The number of parcels with weight \(w \le 8\) is 72.
- The number of parcels with weight \(w \le 4\) is 30.

To find the number of parcels with weight \(w\) in the range \(4 < w \le 8\), subtract the cumulative frequency at \(w = 4\) from the cumulative frequency at \(w = 8\):
\(\text{Number of parcels} = 72 - 30 = 42\).

M1 for identifying the cumulative frequency values 72 and 30
M1 for subtracting the two cumulative frequencies: \(72 - 30\)
A1 for 42

Since \(ABCD\) is a rectangle:
- The opposite side \(BC = AD = 8\text{ cm}\).
- The opposite side \(CD = AB = 15\text{ cm}\).

The point \(P\) lies on \(CD\) with \(DP = 9\text{ cm}\). Therefore, the remaining segment \(PC\) is:
\(PC = CD - DP = 15 - 9 = 6\text{ cm}\).

In the right-angled triangle \(BCP\) (with the right angle at \(C\)), we apply Pythagoras' theorem:
\(BP^2 = BC^2 + PC^2\)
\(BP^2 = 8^2 + 6^2\)
\(BP^2 = 64 + 36 = 100\)
\(BP = \sqrt{100} = 10\text{ cm}\).

M1 for finding \(PC = 15 - 9 = 6\text{ cm}\)
M1 for applying Pythagoras' theorem: \(BP^2 = 8^2 + 6^2\)
A1 for 10

Since \(ACT\) is a straight line, the angles \(ACD\) and \(DCT\) lie on a straight line and sum to \(180^\circ\):
\(\text{Angle } ACD = 180^\circ - 112^\circ = 68^\circ\).

Since \(AC\) is a diameter of the circle, the angle subtended by the diameter at the circumference, \(ADC\), is a right angle:
\(\text{Angle } ADC = 90^\circ\).

In triangle \(ACD\), the sum of the angles is \(180^\circ\):
\(\text{Angle } CAD = 180^\circ - (90^\circ + 68^\circ) = 22^\circ\).

M1 for finding \(\text{Angle } ACD = 180^\circ - 112^\circ = 68^\circ\)
M1 for stating \(\text{Angle } ADC = 90^\circ\) (angle in a semicircle)
A1 for 22

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

After a \(15\%\) increase, the price becomes:
\(1.15 \times P\)

After a \(20\%\) reduction, this new price is multiplied by \(1 - 0.20 = 0.80\):
\(0.80 \times 1.15 \times P = 0.92 \times P\)

We are given that the final sale price is \(\$73.60\):
\(0.92 \times P = 73.60\)

Solve for \(P\):
\(P = \frac{73.60}{0.92} = 80\).

The original price of the jacket was \(\$80\).

M1 for expressing the overall multiplier as \(1.15 \times 0.80\) (or \(0.92\))
M1 for setting up the equation \(0.92P = 73.60\) (or equivalent)
A1 for 80

Multiply the first equation by 5 and the second equation by 2:
\(15x - 10y = 55\)
\(4x + 10y = 2\)

Add the two equations to eliminate \(y\):
\(19x = 57\)
\(x = 3\)

Substitute \(x = 3\) into the second equation:
\(2(3) + 5y = 1\)
\(6 + 5y = 1\)
\(5y = -5\)
\(y = -1\)

So, \(x = 3\) and \(y = -1\).

M1 for a valid method to eliminate one variable (e.g., equating coefficients)
A1 for \(x = 3\)
A1 for \(y = -1\)

The 20th percentile corresponds to:
\(20\% \text{ of } 120 = 0.20 \times 120 = 24\) students.

The 80th percentile corresponds to:
\(80\% \text{ of } 120 = 0.80 \times 120 = 96\) students.

The number of students taking more than 14 minutes but at most 32 minutes is the difference between these two ranks:
\(96 - 24 = 72\) students.

M1 for finding either the 20th percentile rank (24) or the 80th percentile rank (96)
M1 for subtracting their two ranks
A1 for 72

Let the exterior angle be \(x^{\circ}\). Then the interior angle is \(11x^{\circ}\).
Since the interior and exterior angles on any vertex lie on a straight line, they sum to \(180^{\circ}\):
\(x + 11x = 180\)
\(12x = 180\)
\(x = 15\)

The sum of the exterior angles of any polygon is \(360^{\circ}\). Therefore, the number of sides \(n\) is:
\(n = \frac{360}{15} = 24\).

M1 for setting up the equation \(x + 11x = 180\) or equivalent
A1 for finding the exterior angle is \(15^{\circ}\)
A1 for 24

The volume of the cylinder is given by:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi \approx 226.19\text{ cm}^3\)

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

Equating the volumes:
\(\frac{4}{3}\pi R^3 = 72\pi\)
\(\frac{4}{3} R^3 = 72\)
\(R^3 = 72 \times \frac{3}{4} = 54\)
\(R = \sqrt[3]{54} \approx 3.77976\text{ cm}\)

Correct to 3 significant figures, the radius is \(3.78\text{ cm}\).

M1 for correct expression for the volume of the cylinder: \(\pi \times 3^2 \times 8\) (or \(72\pi\))
M1 for equating their cylinder volume to \(\frac{4}{3}\pi R^3\) and solving for \(R\)
A1 for 3.78 (accept 3.779 to 3.78)

First, factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Next, factorise the denominator using the difference of two squares:
\(4x^2 - 1 = (2x + 1)(2x - 1)\)

Now, simplify the fraction by dividing out the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}\)

M1 for factorising the numerator to \((2x + 1)(x - 3)\)
M1 for factorising the denominator to \((2x + 1)(2x - 1)\)
A1 for \(\frac{x - 3}{2x - 1}\) or equivalent

First, find an expression for \(\vec{a} + 2\vec{b}\):
\(\vec{a} + 2\vec{b} = \begin{pmatrix} k \\ 5 \end{pmatrix} + 2\begin{pmatrix} -2 \\ k+1 \end{pmatrix} = \begin{pmatrix} k \\ 5 \end{pmatrix} + \begin{pmatrix} -4 \\ 2k+2 \end{pmatrix} = \begin{pmatrix} k - 4 \\ 2k + 7 \end{pmatrix}\)

Since this vector is parallel to \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\), the ratio of the components must be equal:
\(\frac{2k + 7}{k - 4} = 3\)
\(2k + 7 = 3(k - 4)\)
\(2k + 7 = 3k - 12\)
\(k = 19\)

M1 for finding \(\vec{a} + 2\vec{b} = \begin{pmatrix} k - 4 \\ 2k + 7 \end{pmatrix}\) or equivalent
M1 for setting up a correct equation in \(k\) from the parallel condition (e.g., \(2k + 7 = 3(k - 4)\))
A1 for 19

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

Substitute the given values into the formula:
\(5310 = 4500\left(1 + \frac{r}{100}\right)^5\)

Divide both sides by 4500:
\(\left(1 + \frac{r}{100}\right)^5 = \frac{5310}{4500} = 1.18\)

Take the fifth root of both sides:
\(1 + \frac{r}{100} = \sqrt[5]{1.18}\)
\(1 + \frac{r}{100} \approx 1.033618\)
\(\frac{r}{100} \approx 0.033618\)
\(r \approx 3.3618\)

Correct to 2 decimal places, \(r = 3.36\).

M1 for setting up the equation \(4500\left(1 + \frac{r}{100}\right)^5 = 5310\) or equivalent
M1 for rearranging to find \(1 + \frac{r}{100} = \sqrt[5]{1.18}\) (or 1.0336...)
A1 for 3.36

(a) Total height is \(14\text{ cm}\). The hemisphere has radius \(6\text{ cm}\), so the height of the cone is \(h = 14 - 6 = 8\text{ cm}\).
Volume of hemisphere = \(\frac{2}{3}\pi r^3 = \frac{2}{3}\pi(6^3) = 144\pi\text{ cm}^3\).
Volume of cone = \(\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6^2)(8) = 96\pi\text{ cm}^3\).
Total volume = \(144\pi + 96\pi = 240\pi\text{ cm}^3\).

(b) Slant height of cone \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}\).
Curved surface area of cone = \(\pi r l = \pi(6)(10) = 60\pi\text{ cm}^2\).
Curved surface area of hemisphere = \(2\pi r^2 = 2\pi(6^2) = 72\pi\text{ cm}^2\).
Total surface area = \(60\pi + 72\pi = 132\pi\text{ cm}^2\) (or \(414.69... \approx 415\text{ cm}^2\)).

(c)(i) Mass = Volume \(\times\) Density
Mass = \(240\pi \times 8.4 = 2016\pi \approx 6333.45\text{ g}\).
Convert to kilograms: \(6333.45 / 1000 = 6.33345\text{ kg} \approx 6.33\text{ kg}\).

(ii) Volume and mass ratio for same density is \(k^3 = \frac{50.668}{6.33345} \approx 8.00\).
Therefore, the linear scale factor is \(k = \sqrt[3]{8} = 2\).
Total height of the giant ornament = \(14 \times 2 = 28\text{ cm}\).

(a)
M1 for \(\frac{2}{3}\pi(6)^3\) or \(\frac{1}{3}\pi(6)^2(8)\)
M1 for summing their hemisphere and cone volumes (with correct cone height \(8\))
A1 for \(240\pi\) (or \(754\))

(b)
M1 for \(l = \sqrt{6^2 + 8^2} = 10\)
M1 for \(\pi(6)(10)\) (curved area of cone)
M1 for \(2\pi(6)^2\) (curved area of hemisphere)
M1 for summing their two curved areas
A1 for \(132\pi\) or \(415\)

(c)(i)
M1 for their \(240\pi \times 8.4\)
M1 for dividing their mass in grams by 1000
A1 for \(6.33\) (or \(6.333\) to \(6.334\))

(c)(ii)
M1 for \(k^3 = 50.668 / \text{their mass in kg}\) (leading to \(k=2\))
A1 for \(28\)

(a) Factorise numerator: \(3(x^2 - 4) = 3(x - 2)(x + 2)\).
Factorise denominator: \((2x + 5)(x - 2)\).
Cancel common factor \(x - 2\):
\(\frac{3(x + 2)}{2x + 5} = \frac{3x + 6}{2x + 5}\).

(b) Multiply by \(t^3 + 4\):
\(w(t^3 + 4) = 5 - 2t^3\)
\(wt^3 + 4w = 5 - 2t^3\)
Rearrange to group \(t^3\) terms:
\(wt^3 + 2t^3 = 5 - 4w\)
Factorise \(t^3\):
\(t^3(w + 2) = 5 - 4w\)
Divide and take cube root:
\(t^3 = \frac{5 - 4w}{w + 2} \implies t = \sqrt[3]{\frac{5 - 4w}{w + 2}}\).

(c) Multiply by common denominator \((y-1)(y+2)\):
\(6(y+2) - 3(y-1) = 2(y-1)(y+2)\)
\(6y + 12 - 3y + 3 = 2(y^2 + y - 2)\)
\(3y + 15 = 2y^2 + 2y - 4\)
Rearrange to quadratic form:
\(2y^2 - y - 19 = 0\)
Use the quadratic formula:
\(y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-19)}}{2(2)}\)
\(y = \frac{1 \pm \sqrt{1 + 152}}{4} = \frac{1 \pm \sqrt{153}}{4}\)
\(y = \frac{1 \pm 12.3693}{4}\)
\(y \approx 3.34\) or \(y \approx -2.84\).

(a)
M1 for factorising numerator: \(3(x - 2)(x + 2)\)
M2 for factorising denominator: \((2x + 5)(x - 2)\) (or M1 for partial factorisation or identifying factor \((x-2)\))
A1 for \(\frac{3x+6}{2x+5}\) or \(\frac{3(x+2)}{2x+5}\)

(b)
M1 for \(w(t^3 + 4) = 5 - 2t^3\)
M1 for gathering \(t^3\) terms: \(t^3(w + 2) = 5 - 4w\)
M1 for isolating \(t^3 = \frac{5 - 4w}{w + 2}\)
A1 for \(t = \sqrt[3]{\frac{5-4w}{w+2}}\)

(c)
M1 for common denominator and expanding numerator: \(6(y+2) - 3(y-1)\)
M1 for expanding RHS: \(2(y^2+y-2)\)
M1 for forming quadratic equation: \(2y^2 - y - 19 = 0\)
M1 for applying quadratic formula correctly: \(y = \frac{1 \pm \sqrt{153}}{4}\)
A1 for \(3.34\) and \(-2.84\) (both correct to 3 s.f.)

(a) Time taken = \(\frac{\text{Distance}}{\text{Speed}} = \frac{180}{x}\).

(b) Speed is increased by \(15\text{ km/h}\), so new speed is \(x+15\).
Time taken = \(\frac{180}{x+15}\).

(c)(i) Write the difference in hours: \(40\text{ minutes} = \frac{40}{60} = \frac{2}{3}\text{ hours}\).
\(\frac{180}{x} - \frac{180}{x+15} = \frac{2}{3}\)
Multiply all terms by the common denominator \(3x(x+15)\):
\(3 \times 180(x+15) - 3 \times 180x = 2x(x+15)\)
\(540x + 8100 - 540x = 2x^2 + 30x\)
\(8100 = 2x^2 + 30x\)
\(2x^2 + 30x - 8100 = 0\)
Divide by 2:
\(x^2 + 15x - 4050 = 0\).

(ii) Factorise the quadratic equation:
We need two numbers that multiply to \(-4050\) and add to \(15\).
These are \(75\) and \(-60\) (or \(90\) and \(-45\)? Let's check: \(90 \times -45 = -4050\), and \(90 - 45 = 45\). No, we want \(x^2 + 15x - 4050 = 0\) so \(a + b = 15\). Let's search: \(45 \times 90\)? No, \(45 \times 90 = 4050\), but difference is \(45\).
What about \(x = \frac{-15 \pm \sqrt{15^2 - 4(1)(-4050)}}{2} = \frac{-15 \pm \sqrt{225 + 16200}}{2} = \frac{-15 \pm \sqrt{16425}}{2}\)? No, let's re-multiply: \(45 \times 90 = 4050\)? No, \(45 \times 90 = 4050\) is correct. Wait, we want \(x^2 + 15x - 4050 = 0\). Two factors: \(x^2 + 15x - 4050 = 0\). Let's factorise: \(45 \times 90 = 4050\)? No, \(45 \times 90 = 4050\) but difference is 45. What about \(x = 45\)? \(45^2 + 15(45) - 4050 = 2025 + 675 - 4050 = -1350\)? No, wait!
Let's check: \(8100 = 2x^2 + 30x \implies 2x^2 + 30x - 8100 = 0 \implies x^2 + 15x - 4050 = 0\).
Let's find the roots of \(x^2 + 15x - 4050 = 0\).
Using quadratic formula:
\(x = \frac{-15 \pm \sqrt{225 + 16200}}{2} = \frac{-15 \pm \sqrt{16425}}{2}\).
Wait, \(\sqrt{16425} = 128.16\).
Ah, did we calculate correctly? Let's check: \(180 \times 3 = 540\). \(540 \times 15 = 8100\). Correct.
Let's check roots: \((x - 45)(x + 90)\)? No, \(-45 \times 90 = -4050\), and \(90 - 45 = 45\). We want a coefficient of 15. Wait, \(60 \times 75 = 4500\) and \(75 - 60 = 15\)? No, wait, if \(a \times b = -4050\) and \(a+b = 15\).
Let's try \(x = -45\)? No, what about \(x = 60\)? \(60^2 + 15(60) = 3600 + 900 = 4500\). What about \(x = -75\)? No. Let's find factors: \(4050 = 2 \times 3^4 \times 5^2\).
Factors with difference 15:
Try \(45\)? No, let's test \(x = 45\): \(45^2 + 15(45) = 2025 + 675 = 2700\). No.
Try \(x = 54\)? \(54^2 + 15(54) = 2916 + 810 = 3726\).
Try \(x = 60\)? \(60^2 + 15(60) = 4500\). Wait, the factors of \(4050\) are \(x = 45\)? No, if the equation is \(x^2 + 15x - 4050 = 0\), using formula:
\(x = \frac{-15 \pm 128.16}{2}\).
So \(x = 56.58\) or \(x = -71.58\).
Wait, let's check: \(56.58 \times (-71.58) \approx -4050\).
Let's re-verify the setup: \(\frac{180}{x} - \frac{180}{x+15} = \frac{2}{3}\).
\(180(x+15) - 180x = \frac{2}{3}x(x+15)\)
\(2700 = \frac{2}{3}(x^2 + 15x)\)
Multiply by 3:
\(8100 = 2(x^2 + 15x)\)
\(4050 = x^2 + 15x\)
So \(x^2 + 15x - 4050 = 0\). This is correct.
Let's solve: \(x = \frac{-15 \pm \sqrt{16425}}{2}\).
\(\sqrt{16425} \approx 128.16\).
So \(x = \frac{-15 + 128.16}{2} = 56.58 \approx 56.6\text{ km/h}\).
(iii) Outward journey time = \(\frac{180}{56.58} \approx 3.18\text{ hours}\).
Return journey time = \(\frac{180}{56.58+15} = \frac{180}{71.58} \approx 2.51\text{ hours}\).
Total time = \(3.18 + 2.51 = 5.69\text{ hours}\) (or exactly \(3.181... + 2.514... = 5.695\text{ hours}\)).
Total distance = \(360\text{ km}\).
Average speed = \(\frac{360}{5.695} \approx 63.2\text{ km/h}\).

(a)
B1 for \(\frac{180}{x}\)

(b)
B1 for \(\frac{180}{x+15}\)

(c)(i)
M1 for \(\frac{180}{x} - \frac{180}{x+15} = \frac{40}{60}\) (or \(\frac{2}{3}\))
M1 for clearing fractions: \(180(x+15) - 180x = \frac{2}{3}x(x+15)\)
M1 for expanding and grouping: \(2700 = \frac{2}{3}x^2 + 10x\)
A1 for showing given equation \(x^2 + 15x - 4050 = 0\) with no steps omitted

(c)(ii)
M1 for identifying correct quadratic formula components or completing the square
M1 for \(x = \frac{-15 \pm \sqrt{15^2 - 4(1)(-4050)}}{2}\)
A1 for \(\sqrt{16425}\) or \(128.2\)
A1 for \(56.6\) (and rejecting negative root)

(c)(iii)
M1 for finding outward time and return time using their \(x\)
M1 for total distance (360) divided by total time
A1 for \(63.2\) (or consistent with their \(x\))

(a) The exterior angle is \(180^\circ - 156^\circ = 24^\circ\).
The number of sides is \(n = \frac{360^\circ}{24^\circ} = 15\).

(b) The sum of interior angles in a quadrilateral is \(360^\circ\).
\((3x+5) + (2x-15) + (4x+10) + (3x+12) = 360\)
\(12x + 12 = 360\)
\(12x = 348 \implies x = 29\).
Now find all angles:
- Angle \(P = 3(29) + 5 = 92^\circ\)
- Angle \(Q = 2(29) - 15 = 43^\circ\)
- Angle \(R = 4(29) + 10 = 126^\circ\)
- Angle \(S = 3(29) + 12 = 99^\circ\)
The largest angle is \(126^\circ\).

(c)(i) Bearing of \(C\) from \(A\) is \(150^\circ\).
Bearing of \(A\) from \(C\) is \(150^\circ + 180^\circ = 330^\circ\).

(ii) Since \(B\) is due East of \(A\), the bearing of \(B\) from \(A\) is \(090^\circ\).
The angle \(\angle BAC\) is the angle between East and \(150^\circ\), which is \(150^\circ - 90^\circ = 60^\circ\).
Using the Cosine Rule in \(\triangle ABC\) to find \(BC\):
\(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)\)
\(BC^2 = 12^2 + 8^2 - 2 \times 12 \times 8 \times \cos(60^\circ)\)
\(BC^2 = 144 + 64 - 192 \times 0.5\)
\(BC^2 = 208 - 96 = 112\)
\(BC = \sqrt{112} \approx 10.583\text{ km} \approx 10.6\text{ km}\).

(a)
M1 for \(180 - 156 = 24\)
M1 for \(360 / 24\)
A1 for \(15\)

(b)
M1 for summing and setting to 360: \(12x + 12 = 360\)
A1 for \(x = 29\)
M1 for calculating individual angles (at least one)
A1 for \(126^\circ\)

(c)(i)
M1 for \(150 + 180\) or a diagram showing north lines
A1 for \(330^\circ\)

(c)(ii)
M1 for \(\angle BAC = 150 - 90 = 60^\circ\)
M1 for applying Cosine Rule: \(12^2 + 8^2 - 2(12)(8)\cos(60)\)
A1 for \(BC^2 = 112\)
A1 for \(10.6\)

(a)(i) From the table, for \(t \le 30\), cumulative frequency is 56.

(ii) Total students = 120. Students taking \(t \le 40\) is 90.
So, students taking more than 40 minutes = \(120 - 90 = 30\).
Percentage = \(\frac{30}{120} \times 100\% = 25\%\).

(b)(i) Deduct successive values:
- \(10 < t \le 20\): \(24 - 8 = 16\)
- \(20 < t \le 30\): \(56 - 24 = 32\)
- \(30 < t \le 40\): \(90 - 56 = 34\)
- \(40 < t \le 50\): \(112 - 90 = 22\)
- \(50 < t \le 60\): \(120 - 112 = 8\)

(ii) Midpoints are: 5, 15, 25, 35, 45, 55.
Estimated Mean = \(\frac{(8 \times 5) + (16 \times 15) + (32 \times 25) + (34 \times 35) + (22 \times 45) + (8 \times 55)}{120}\)
\(= \frac{40 + 240 + 800 + 1190 + 990 + 440}{120} = \frac{3700}{120} \approx 30.83\text{ minutes}\).

(c) Number of students taking \(30 < t \le 50\) is \(112 - 56 = 56\).
Probability = \(\frac{56}{120} = \frac{7}{15} \approx 0.467\).

(d) Total number of students with \(t > 40\) is \(120 - 90 = 30\).
Those among them with \(t > 50\) is \(120 - 112 = 8\).
Probability = \(\frac{8}{30} \times \frac{7}{29} = \frac{56}{870} = \frac{28}{435} \approx 0.0644\).

(a)(i) B1 for 56
(a)(ii) M1 for \(120-90\) or 30 seen; A1 for 25%

(b)(i) B2 for all 5 correct frequencies (16, 32, 34, 22, 8). If not all correct, B1 for 3 or 4 correct.
(b)(ii)
M1 for at least 4 correct midpoints seen (5, 15, 25, 35, 45, 55)
M1 for \(\sum f x\) calculation using their frequencies and midpoints
M1 for dividing their sum by 120
A1 for 30.8 (or 30.83...)

(c)
M1 for identifying 56 students; A1 for \(\frac{56}{120}\) or \(\frac{7}{15}\) (or 0.467)

(d)
M1 for product of fractions with decreasing denominators: \(\frac{8}{30} \times \frac{7}{29}\)
A1 for \(\frac{28}{435}\) or 0.0644

(a)(i) The common difference is 3. The next term is \(14 + 3 = 17\).
(ii) Since the common difference is 3, the \(n\)-th term is of the form \(3n + c\).
For \(n = 1\), \(3(1) + c = 5 \implies c = 2\).
So the \(n\)-th term is \(3n + 2\).

(b) Sequence B terms are \(3, 8, 15, 24, \dots\).
The first differences are 5, 7, 9. The second difference is constant at 2.
So, the term is \(an^2 + bn + c\) with \(a = 2/2 = 1\).
Let \(u_n = n^2 + bn + c\).
For \(n = 1\), \(1 + b + c = 3 \implies b + c = 2\).
For \(n = 2\), \(4 + 2b + c = 8 \implies 2b + c = 4\).
Subtracting these equations gives \(b = 2\), which means \(c = 0\).
So the \(n\)-th term is \(n^2 + 2n\).

(c)(i) Sequence C is geometric with common ratio 3.
Terms: \(u_1 = 2\), \(u_2 = 6\), \(u_3 = 18\), \(u_4 = 54\), \(u_5 = 162\), \(u_6 = 486\).

(ii) The general form of a geometric sequence is \(a r^{n-1}\), where \(a = 2\) and \(r = 3\).
So the \(n\)-th term is \(2 \times 3^{n-1}\).

(d)(i) Sequence D has numerators from Sequence A and denominators from Sequence B.
So the \(n\)-th term is \(\frac{3n+2}{n^2+2n}\).

(ii) For \(n = 10\), \(u_{10} = \frac{3(10) + 2}{10^2 + 2(10)} = \frac{32}{120} = \frac{4}{15}\).

(a)(i) B1 for 17
(a)(ii) B2 for \(3n+2\) (B1 for \(3n+k\))

(b)
M1 for recognizing quadratic relation (second differences constant at 2)
M1 for substituting points into \(n^2 + bn + c\)
A1 for \(n^2 + 2n\) or \(n(n+2)\)

(c)(i)
M1 for continuing sequence to find 5th term (162); A1 for 486
(c)(ii)
M1 for \(k \cdot 3^{n-1}\); A1 for \(2 \cdot 3^{n-1}\)

(d)(i)
B1 for \(\frac{3n+2}{n^2+2n}\) (or using their parts (a)(ii) and (b))

(d)(ii)
M1 for substituting \(n=10\) into their expression; A1 for \(4/15\) (must be simplest form)

(a)
- \(p = f(-2) = (-2)^2 - \frac{4}{-2} = 4 + 2 = 6\).
- \(q = f(0.5) = (0.5)^2 - \frac{4}{0.5} = 0.25 - 8 = -7.75\).
- \(r = f(4) = (4)^2 - \frac{4}{4} = 16 - 1 = 15\).

(b)(i) The derivative of the function is \(f'(x) = 2x + \frac{4}{x^2}\).
At \(x = 2\), the exact gradient is \(2(2) + \frac{4}{4} = 5\).
Using a tangent, drawing it carefully at \((2, 2)\) and measuring \(\frac{\text{rise}}{\text{run}}\) yields a value close to 5.

(ii) Using gradient \(m = 5\) and point \((2, 2)\):
\(y - 2 = 5(x - 2) \implies y = 5x - 8\).
We accept other equations consistent with their gradient from (b)(i).

(c)(i) Set the two equations equal:
\(x^2 - \frac{4}{x} = 2x + 1\)
Multiply by \(x\) (since \(x \ne 0\)):
\(x(x^2 - \frac{4}{x}) = x(2x + 1)\)
\(x^3 - 4 = 2x^2 + x\)
Rearrange to make the right side 0:
\(x^3 - 2x^2 - x - 4 = 0\).

(ii) Reading the intersection point of the line \(y = 2x + 1\) and the curve, the \(x\)-coordinate is approximately \(2.85\) (accept any value in the range \([2.8, 2.9]\)).

(a)
B1 for \(p = 6\)
B1 for \(q = -7.75\)
B1 for \(r = 15\)

(b)(i)
M1 for attempting to draw a tangent at \(x = 2\) on a mock grid
M1 for rise/run attempt using their tangent line
A1 for value in range \([4.0, 6.0]\)

(b)(ii)
M1 for using \(y - 2 = m(x - 2)\) with their gradient \(m\)
A1 for \(y = 5x - 8\) (or consistent with their gradient)

(c)(i)
M1 for setting \(x^2 - \frac{4}{x} = 2x + 1\)
M1 for multiplying correctly by \(x\)
A1 for algebraic steps leading to the given cubic equation \(x^3 - 2x^2 - x - 4 = 0\)

(c)(ii)
B2 for answer in range \([2.8, 2.9]\) (B1 for identifying intersection point of curve and line)

(a) Tree diagram structure:
- First counter branches:
- Red: \(\frac{7}{12}\)
- Blue: \(\frac{5}{12}\)
- Second counter branches (after Red first):
- Red: \(\frac{6}{11}\)
- Blue: \(\frac{5}{11}\)
- Second counter branches (after Blue first):
- Red: \(\frac{7}{11}\)
- Blue: \(\frac{4}{11}\)

(b)(i) \(P(\text{Red, Red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22} \approx 0.318\).

(ii) \(P(\text{Different}) = P(\text{Red, Blue}) + P(\text{Blue, Red})\)
\(= \left(\frac{7}{12} \times \frac{5}{11}\right) + \left(\frac{5}{12} \times \frac{7}{11}\right) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \approx 0.530\).

(c) To have at least two reds out of three, we have four successful outcomes:
1) Red, Red, Red: \(\frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{210}{1320}\)
2) Red, Red, Blue: \(\frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{210}{1320}\)
3) Red, Blue, Red: \(\frac{7}{12} \times \frac{5}{11} \times \frac{6}{10} = \frac{210}{1320}\)
4) Blue, Red, Red: \(\frac{5}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{210}{1320}\)

Sum of these probabilities = \(\frac{210 + 210 + 210 + 210}{1320} = \frac{840}{1320} = \frac{7}{11} \approx 0.636\).

(a)
B1 for first set of branches labeled correctly with probabilities \(\frac{7}{12}\) and \(\frac{5}{12}\)
B2 for second set of branches with correct conditional probabilities (B1 if only 1 branch set is correct)

(b)(i)
M1 for \(\frac{7}{12} \times \frac{6}{11}\)
A1 for \(\frac{7}{22}\) or 0.318

(b)(ii)
M1 for \(\frac{7}{12} \times \frac{5}{11}\) or \(\frac{5}{12} \times \frac{7}{11}\)
M1 for summing both possible orders
A1 for \(\frac{35}{66}\) or 0.530

(c)
M1 for listing successful cases: RRR, RRB, RBR, BRR
M2 for multiplying probabilities of three combined events (e.g., any one product correct)
M1 for summing the four outcomes
A1 for \(\frac{7}{11}\) or 0.636

(a) The volume of the cylinder is:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi \approx 502.65\text{ cm}^3\)

The volume of the cone is:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 H = \frac{1}{3}\pi \times 4^2 \times 3 = 16\pi \approx 50.27\text{ cm}^3\)

Total volume:
\(V_{\text{total}} = 160\pi + 16\pi = 176\pi \approx 552.92\text{ cm}^3\)

Correct to 3 significant figures, the total volume is \(553\text{ cm}^3\).

(b) First, calculate the slant height, \(l\), of the cone:
\(l = \sqrt{r^2 + H^2} = \sqrt{4^2 + 3^2} = 5\text{ cm}\)

The total surface area of the solid consists of:
1. The curved surface area of the cone: \(\pi r l = \pi \times 4 \times 5 = 20\pi \approx 62.83\text{ cm}^2\)
2. The curved surface area of the cylinder: \(2\pi r h = 2\pi \times 4 \times 10 = 80\pi \approx 251.33\text{ cm}^2\)
3. The flat circular base of the cylinder: \(\pi r^2 = \pi \times 4^2 = 16\pi \approx 50.27\text{ cm}^2\)

Total surface area:
\(A_{\text{total}} = 20\pi + 80\pi + 16\pi = 116\pi \approx 364.42\text{ cm}^2\)

Correct to 3 significant figures, the total surface area is \(364\text{ cm}^2\).

(c) The volume of one sphere is:
\(V_{\text{sphere}} = \frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi \times (1.5)^3 = 4.5\pi \approx 14.137\text{ cm}^3\)

To find the number of spheres:
\(N = \frac{176\pi}{4.5\pi} = \frac{176}{4.5} = 39.111...\)

Therefore, the maximum number of complete spheres that can be made is 39.

(a)
M1 for \(\pi \times 4^2 \times 10\) (or 503)
M1 for \(\frac{1}{3} \times \pi \times 4^2 \times 3\) (or 50.3)
M1 for adding their cylinder and cone volumes
A1 for 553 (allow 552.9 to 553.1)

(b)
M1 for slant height of cone \(l = \sqrt{4^2 + 3^2} = 5\)
M1 for curved area of cone \(\pi \times 4 \times 5\) (or 62.8)
M1 for curved area of cylinder \(2 \times \pi \times 4 \times 10\) (or 251)
M1 for base area \(\pi \times 4^2\) (or 50.3)
A1 for 364 (allow 364.2 to 364.5)

(c)
M1 for volume of sphere formula with \(r = 1.5\): \(\frac{4}{3}\pi \times 1.5^3\)
A1 for \(4.5\pi\) or 14.1 to 14.14
M1 for \(\text{their volume from (a)} \div \text{their sphere volume}\)
A1 for 39 (must be an integer, dependent on previous M1)

(a) Time taken by the coach is \(\frac{150}{x}\) hours.
Time taken by the train is \(\frac{150}{x + 15}\) hours.
The difference in time is 35 minutes, which is \(\frac{35}{60} = \frac{7}{12}\) hours.
Therefore:
\(\frac{150}{x} - \frac{150}{x + 15} = \frac{7}{12}\)
Multiply the entire equation by \(12x(x + 15)\) to clear the denominators:
\(12 \times 150(x + 15) - 12 \times 150x = 7x(x + 15)\)
\(1800(x + 15) - 1800x = 7x^2 + 105x\)
\(1800x + 27000 - 1800x = 7x^2 + 105x\)
\(27000 = 7x^2 + 105x\)
Rearranging this gives:
\(7x^2 + 105x - 27000 = 0\) (as required).

(b) Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where \(a = 7\), \(b = 105\), and \(c = -27000\).
\(x = \frac{-105 \pm \sqrt{105^2 - 4(7)(-27000)}}{2(7)}
\)x = \frac{-105 \pm \sqrt{11025 + 756000}}{14}\)
\(x = \frac{-105 \pm \sqrt{767025}}{14}\)
Since \(\sqrt{767025} \approx 875.799635\):
\(x_1 = \frac{-105 + 875.799635}{14} \approx 55.057\)
\(x_2 = \frac{-105 - 875.799635}{14} \approx -70.057\)

Giving answers correct to 2 decimal places:
\(x = 55.06\) or \(x = -70.06\)

(c) Since speed must be positive, we use \(x \approx 55.057\text{ km/h}\).
Time taken by the coach = \(\frac{150}{55.057}\text{ hours} \approx 2.7244\text{ hours}\).
Converting the decimal part to minutes:
\(0.7244 \times 60 = 43.46\text{ minutes}\).
To the nearest minute, this is 2 hours and 43 minutes.

(a)
M1 for expressions \(\frac{150}{x}\) or \(\frac{150}{x + 15}\)
M1 for \(\frac{150}{x} - \frac{150}{x + 15} = \frac{35}{60}\) (or equivalent)
M1 for removing fractions: \(150 \times 12(x + 15) - 150 \times 12x = 7x(x + 15)\) (or with 60 and 35)
M1 for expanding brackets: \(1800x + 27000 - 1800x = 7x^2 + 105x\) (or equivalent)
A1 for fully correct simplification leading to the given equation without errors.

(b)
M1 for substitution into formula: \(\frac{-105 \pm \sqrt{105^2 - 4(7)(-27000)}}{2(7)}\) (allow one error)
B1 for \(\sqrt{767025}\) or 875.8...
A1 for 55.06 (allow 55.05 to 55.06)
A1 for -70.06 (allow -70.05 to -70.06)

(c)
M1 for \(\frac{150}{\text{their positive } x}\)
A1 for 2.72 hours (or equivalent decimal representation)
M1 for converting the fractional part of their hours into minutes (multiplying by 60)
A1 for 2 hours 43 minutes