2024 IB DP Mathematics - Applications and Interpretation Practice Paper with Answers

(a) Acceleration is the derivative of velocity:
\(a(t) = v'(t) = 6t - 12\)
At \(t = 3\):
\(a(3) = 6(3) - 12 = 6 \text{ m s}^{-2}\)

(b) To find the total distance, we find when the particle changes direction:
\(3t^2 - 12t + 9 = 0 \implies 3(t-1)(t-3) = 0\)
So the particle changes direction at \(t = 1\) and \(t = 3\).

The total distance is given by:
\(\int_{0}^{4} |v(t)| \, dt = \int_{0}^{1} v(t)\,dt - \int_{1}^{3} v(t)\,dt + \int_{3}^{4} v(t)\,dt\)

Using the antiderivative \(s(t) = t^3 - 6t^2 + 9t\):
\(s(0) = 0\)
\(s(1) = 1 - 6 + 9 = 4\)
\(s(3) = 27 - 54 + 27 = 0\)
\(s(4) = 64 - 96 + 36 = 4\)

Distance in interval \([0, 1]\): \(|4 - 0| = 4\)
Distance in interval \([1, 3]\): \(|0 - 4| = 4\)
Distance in interval \([3, 4]\): \(|4 - 0| = 4\)

Total distance = \(4 + 4 + 4 = 12 \text{ m}\).

(a)
- **M1**: Attempt to differentiate \(v(t)\).
- **A1**: Correct acceleration expression \(6t - 12\) and final value \(6 \text{ m s}^{-2}\).

(b)
- **M1**: Finding critical points where \(v(t) = 0\) (i.e., \(t = 1, 3\)).
- **M1**: Set up integrals or sum of absolute displacements.
- **A1**: Correct individual displacements (4, -4, 4).
- **A1.33**: Correct final total distance of \(12 \text{ m}\).

(a) To find the intersection points, set \(f(x) = g(x)\):
\(-x^2 + 4x = x \implies x^2 - 3x = 0 \implies x(x-3) = 0\)
So, the \(x\)-coordinates of the points of intersection are \(x = 0\) and \(x = 3\).

(b) The area of the region is given by:
\(\text{Area} = \int_{0}^{3} ((-x^2 + 4x) - x) \, dx = \int_{0}^{3} (-x^2 + 3x) \, dx\)
\(\text{Area} = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} \right]_{0}^{3}\)
\(\text{Area} = \left( -\frac{27}{3} + \frac{27}{2} \right) - 0 = -9 + 13.5 = 4.5 \text{ cm}^2\).

(a)
- **M1**: Setting the equations equal to each other.
- **A2**: Both correct \(x\)-values \(x = 0\) and \(x = 3\) (1 mark each).

(b)
- **M1**: Setting up the correct definite integral with appropriate limits.
- **A1**: Correct integration of the terms.
- **A2.33**: Correct evaluation of the area as \(4.5 \text{ cm}^2\) (or equivalent fraction \(\frac{9}{2}\)).

(a) Using the trapezoidal rule with interval width \(h = 10\):
\(\text{Area} \approx \frac{10}{2} [W(0) + 2W(10) + 2W(20) + 2W(30) + W(40)]\)
\(\text{Area} \approx 5 [0 + 2(15) + 2(22) + 2(18) + 0]\)
\(\text{Area} \approx 5 [0 + 30 + 44 + 36 + 0] = 5 \times 110 = 550 \text{ m}^2\).

(b) The area predicted by the surveyor's model is:
\(\text{Area}_{\text{model}} = \int_{0}^{40} -0.055x(x-40) \, dx\)
\(= \int_{0}^{40} (-0.055x^2 + 2.2x) \, dx\)
\(= \left[ -\frac{0.055x^3}{3} + 1.1x^2 \right]_{0}^{40}\)
\(= \left( -\frac{0.055(64000)}{3} + 1.1(1600) \right) - 0\)
\(= -\frac{3520}{3} + 1760 = \frac{1760}{3} \approx 586.67 \text{ m}^2\).

Percentage error:
\(\text{Percentage Error} = \frac{|550 - 586.67|}{586.67} \times 100\% \approx 6.25\%\).

(a)
- **M1**: Attempt to apply the trapezoidal rule formula with \(h=10\).
- **A2**: Correct substitution and calculation giving \(550 \text{ m}^2\).

(b)
- **M1**: Set up the definite integral for the model.
- **A1**: Obtain the model area \(586.67 \text{ m}^2\) (or \(\frac{1760}{3}\)).
- **M1**: Apply the percentage error formula.
- **A1.33**: Correct final percentage error of \(6.25\%\).

(a) Deposit paid: \(0.10 \times 25\,000 = \$2\,500\).
Loan amount (Present Value, \(PV\)): \(25\,000 - 2\,500 = \$22\,500\).
Number of periods (\(N\)): \(5 \times 12 = 60\).
Interest rate (\(I\%\)): \(4.8\%\).
Future Value (\(FV\)): \(0\).
Periods per year (\(P/Y\)) & Compounding per year (\(C/Y\)): \(12\).

Using a GDC TVM Solver:
\(PMT \approx -422.462...\)
So, the monthly payment is \(\$422.46\).

(b) Total amount paid over the 5 years:
\(\text{Total Paid} = 60 \times 422.4622 = \$25\,347.73\).

Total interest paid:
\(\text{Interest} = \text{Total Paid} - PV = 25\,347.73 - 22\,500 = \$2\,847.73\).

(a)
- **A1**: Correct loan amount of \(\$22\,500\).
- **M1**: Attempt to use TVM solver on GDC with appropriate parameters (\(N=60\), \(I\%=4.8\), \(PV=22500\), \(FV=0\)).
- **A2**: Correct monthly payment of \(\$422.46\) (accept \(\$422.46\) or \(\$422.50\) depending on rounding conventions, but standard finance rounding yields \(422.46\)).

(b)
- **M1**: Attempt to find total payments by multiplying monthly payment by 60.
- **A2.33**: Correct total interest paid of \(\$2\,847.73\) (accept \(\$2\,847.60\) if using rounded monthly payment of \(422.46\)).

(a) The \(n\)-th term of an arithmetic sequence is given by:
\(u_n = u_1 + (n-1)d\)
For \(n = 10\):
\(u_{10} = 12 + (10-1)(3.5) = 12 + 31.5 = 43.5\).

(b) We require \(v_n > u_n\).
\(v_n = 3 \times (1.2)^{n-1}\)
\(u_n = 12 + (n-1)(3.5) = 3.5n + 8.5\)

We set up the inequality:
\(3 \times (1.2)^{n-1} > 3.5n + 8.5\)
Using a GDC to find the intersection or testing integer values:
For \(n = 18\):
\(v_{18} = 3 \times (1.2)^{17} \approx 66.56\)
\(u_{18} = 12 + 17(3.5) = 71.5\) (Here \(v_{18} < u_{18}\))

For \(n = 19\):
\(v_{19} = 3 \times (1.2)^{18} \approx 79.88\)
\(u_{19} = 12 + 18(3.5) = 75\) (Here \(v_{19} > u_{19}\))

Thus, the smallest value of \(n\) is \(19\).

(a)
- **M1**: Substitution into the correct arithmetic sequence formula.
- **A1**: Correct value of \(43.5\).

(b)
- **M1**: Formulating the inequality or equation \(3 \times (1.2)^{n-1} > 3.5n + 8.5\).
- **M2**: Graphical or tabular approach shown via GDC values around the critical region.
- **A1.33**: Correct integer answer \(n = 19\).

Let \(X\) be the mass of an apple, where \(X \sim N(150, 12^2)\).

(a) Using a GDC to find \(P(140 < X < 165)\):
\(P(140 < X < 165) = \text{normalcdf}(140, 165, 150, 12) \approx 0.69123...\)
To 3 significant figures, the probability is \(0.691\).

(b) We want to find \(k\) such that \(P(X < k) = 0.05\).
Using the inverse normal cumulative distribution function on a GDC:
\(k = \text{invNorm}(0.05, 150, 12) \approx 130.26...\)
To 3 significant figures, the maximum mass is \(130 \text{ grams}\).

(a)
- **M1**: Clearly identifying normal distribution boundaries \(140\) and \(165\).
- **A2**: Correct probability of \(0.691\) (accept \(0.69123...\)).

(b)
- **M2**: Identifying the equation \(P(X < k) = 0.05\) and attempting to use inverse normal function.
- **A2.33**: Correct value of \(130 \text{ grams}\) (accept \(130.26...\) or \(130\) to 3 s.f.).

(a) The null hypothesis, \(H_0\): Preferred music genre and age group are independent.

(b) The expected frequency is calculated as:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)
For "Under 30" and "Classical":
\(E = \frac{90 \times 55}{200} = \frac{4950}{200} = 24.75\). (Shown)

(c) Using a GDC to perform a \(\chi^2\) two-way test on the matrix:
\(\text{Observed} = \begin{pmatrix} 45 & 35 & 10 \\ 25 & 40 & 45 \end{pmatrix}\)
We find the test statistic \(\chi^2 \approx 26.59\) with \(2\) degrees of freedom.
This yields a \(p\)-value of \(1.69 \times 10^{-6}\) (or \(0.00000169\)).

(a)
- **A1**: Correct statement of null hypothesis (must mention independence of the two variables).

(b)
- **M1**: Correct substitution into the expected value formula \(\frac{90 \times 55}{200}\).
- **A1**: Clear arithmetic calculation showing it equals \(24.75\).

(c)
- **M2**: Correct entry of matrix into GDC and choosing \(\chi^2\) test.
- **A2.33**: Correct \(p\)-value of \(1.69 \times 10^{-6}\) (or \(1.685... \times 10^{-6}\)).

(a) The base \(ABCD\) is a rectangle. Using Pythagoras' theorem on right-angled triangle \(ABC\):
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}\).

(b) The center of the base is \(M\), which is the midpoint of diagonal \(AC\).
\(AM = \frac{1}{2} AC = \frac{10}{2} = 5 \text{ cm}\).
The vertical height \(VM = 12 \text{ cm}\).
Using Pythagoras' theorem on right-angled triangle \(VAM\):
\(VA = \sqrt{AM^2 + VM^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}\).

(c) The angle \(\theta\) that the edge \(VA\) makes with the base is the angle \(\angle VAM\).
Using trigonometry in triangle \(VAM\):
\(\tan \theta = \frac{VM}{AM} = \frac{12}{5} = 2.4\)
\(\theta = \arctan(2.4) \approx 67.380...^{\circ}\)
To 3 significant figures, the angle is \(67.4^{\circ}\) (or \(1.18 \text{ radians}\)).

(a)
- **M1**: Applying Pythagoras' theorem to find \(AC\).
- **A1**: Correct diagonal length of \(10 \text{ cm}\).

(b)
- **M1**: Finding \(AM = 5 \text{ cm}\).
- **M1**: Applying Pythagoras' theorem to find \(VA\).
- **A1**: Correct edge length of \(13 \text{ cm}\).

(c)
- **M1**: Identifying the angle \(\angle VAM\) and using an appropriate trigonometric ratio.
- **A1.33**: Correct angle of \(67.4^{\circ}\) (accept \(1.18 \text{ radians}\)).

(a) Let \(h\) be the height of the cylindrical can. The volume is given by:
\(V = \pi r^2 h = 350 \implies h = \frac{350}{\pi r^2}\)

The cost \(C\) of materials is:
\(C = 2 \times 0.02 \times \text{Area}_{\text{base}} + 0.015 \times \text{Area}_{\text{side}}\)
\(C = 0.04 \pi r^2 + 0.015(2\pi r h)\)

Substitute \(h = \frac{350}{\pi r^2}\):
\(C(r) = 0.04 \pi r^2 + 0.03\pi r \left(\frac{350}{\pi r^2}\right)\)
\(C(r) = 0.04 \pi r^2 + \frac{10.5}{r}\) [AG]

(b) To minimize the cost, differentiate \(C(r)\) with respect to \(r\) and set it equal to zero:
\(C'(r) = 0.08 \pi r - \frac{10.5}{r^2} = 0\)
\(0.08 \pi r^3 = 10.5 \implies r^3 = \frac{10.5}{0.08\pi} \approx 41.778\)
\(r = \sqrt[3]{41.778} \approx 3.47\text{ cm}\) (to 3 s.f.)

(c) The minimum cost is:
\(C(3.47) = 0.04\pi (3.47)^2 + \frac{10.5}{3.47} \approx 1.513 + 3.026 = 4.54\)
Thus, the minimum cost is \(\$4.54\) (to 3 s.f.)

(a) [3 marks]
M1 for expressing \(h\) in terms of \(r\): \(h = \frac{350}{\pi r^2}\).
A1 for setting up the cost equation: \(C = 0.04\pi r^2 + 0.03\pi r h\).
A1 for substituting \(h\) and simplifying to obtain the given expression \(C(r) = 0.04\pi r^2 + \frac{10.5}{r}\).

(b) [2.33 marks]
M1 for differentiating \(C(r)\): \(C'(r) = 0.08\pi r - \frac{10.5}{r^2}\).
M1 for setting \(C'(r) = 0\) and solving for \(r\).
A0.33 for \(r \approx 3.47\text{ cm}\).

(c) [2 marks]
M1 for substituting their value of \(r\) back into the cost function.
A1 for \(\$4.54\) (accept \(4.54\)).

Let \(X\) be the mass of a bag of sugar, where \(X \sim N(502, 3.5^2)\).

(a) We want to find \(P(X < 500)\).
Using a graphic display calculator with lower bound \(-\infty\), upper bound \(500\), \(\mu = 502\), and \(\sigma = 3.5\):
\(P(X < 500) \approx 0.28385 \approx 0.284\) (to 3 s.f.)

(b) Let \(Y\) be the number of bags in a box of 8 that weigh less than \(500\text{ g}\).
Then \(Y \sim B(8, 0.28385)\).
We want to find \(P(Y \le 2)\).
Using binomial cumulative probability function on GDC with \(n = 8\), \(p = 0.28385\), and \(x = 2\):
\(P(Y \le 2) \approx 0.58784 \approx 0.588\) (to 3 s.f.)

(a) [3.33 marks]
M1 for identifying normal distribution parameters: \(\mu = 502\), \(\sigma = 3.5\).
M1 for setting up the probability expression: \(P(X < 500)\).
A1.33 for \(0.284\) (accept \(0.28385\)).

(b) [4 marks]
M1 for identifying binomial distribution: \(Y \sim B(8, p)\).
A1 for using their \(p\) value from part (a).
M1 for setting up \(P(Y \le 2)\).
A1 for \(0.588\) (accept \(0.58784\)).

(a) Using the compound interest formula:
\(FV = PV \left(1 + \frac{r}{100k}\right)^{kn}\)
Here, \(PV = 12000\), \(r = 3.6\), \(k = 12\) (compounded monthly), and \(n = 4\).
\(FV = 12000 \left(1 + \frac{3.6}{1200}\right)^{12 \times 4} = 12000 (1.003)^{48}\)
\(FV \approx 13852.66\) USD (accept \(\$13900\) rounded to 3 s.f.)

(b) Olivia withdraws \(\$4000\), so the new principal is:
\(PV_{\text{new}} = 13852.66 - 4000 = 9852.66\) USD.

This is reinvested for 3 years compounded quarterly (\(k=4\), \(n=3\)):
\(11500 = 9852.66 \left(1 + \frac{r}{400}\right)^{12}\)

Divide by 9852.66:
\(\left(1 + \frac{r}{400}\right)^{12} \approx 1.16720\)
Take the 12th root:
\(1 + \frac{r}{400} \approx (1.16720)^{1/12} \approx 1.01294\)
\(\frac{r}{400} \approx 0.01294 \implies r \approx 5.176 \approx 5.18\%\)

(a) [3.33 marks]
M1 for correct formula with correct substitutions: \(12000 (1.003)^{48}\).
A1 for \(13852.66\) (accept \(13900\)).
A1.33 for showing correct monthly interest rate or power \(48\).

(b) [4 marks]
M1 for finding the new principal: \(13852.66 - 4000 = 9852.66\).
M1 for setting up the equation: \(11500 = 9852.66 (1 + r/400)^{12}\).
M1 for solving the equation (using GDC TVM solver or algebra).
A1 for \(r = 5.18\) (accept \(5.176\)).

(a) Using the Cosine Rule on \(\triangle PQR\):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(P\hat{Q}R)\)
\(PR^2 = 55^2 + 75^2 - 2(55)(75)\cos(115^\circ)\)
\(PR^2 = 3025 + 5625 - 8250(-0.422618)\)
\(PR^2 = 8650 + 3486.60 = 12136.60\)
\(PR = \sqrt{12136.60} \approx 110.166 \approx 110\text{ m}\) (to 3 s.f.)

(b) First, find the area of \(\triangle PQR\):
\(\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(P\hat{Q}R)\)
\(\text{Area} = \frac{1}{2} \times 55 \times 75 \times \sin(115^\circ) \approx 2062.5 \times 0.906308 = 1869.26\text{ m}^2\)

Alternatively, the area can be written using the base \(PR\) and height \(QS\):
\(\text{Area} = \frac{1}{2} \times PR \times QS\)
\(1869.26 = \frac{1}{2} \times 110.166 \times QS\)

Solve for \(QS\):
\(QS = \frac{2 \times 1869.26}{110.166} \approx 33.935 \approx 33.9\text{ m}\) (to 3 s.f.)

(a) [3.33 marks]
M1 for choosing the Cosine Rule.
A1 for correct substitution: \(55^2 + 75^2 - 2(55)(75)\cos(115^\circ)\).
A1.33 for \(110\text{ m}\) (accept \(110.166\)).

(b) [4 marks]
M1 for using the area of a triangle formula with \(\sin(115^\circ)\).
A1 for calculating the correct area: \(1869.26\text{ m}^2\).
M1 for equating this to \(\frac{1}{2} \times PR \times QS\).
A1 for \(33.9\text{ m}\) (accept \(33.935\)).

(a) Acceleration \(a(t)\) is the derivative of velocity \(v(t)\):
\(a(t) = \frac{dv}{dt} = 12t^2 - 36t + 20\)

At \(t = 2.5\):
\(a(2.5) = 12(2.5)^2 - 36(2.5) + 20 = 75 - 90 + 20 = 5\text{ m s}^{-2}\)

(b) The car is at rest when \(v(t) = 0\):
\(4t^3 - 18t^2 + 20t = 0 \implies 2t(2t^2 - 9t + 10) = 0\)
\(2t(2t - 5)(t - 2) = 0\)
Since \(0 \le t \le 3\), the car is at rest at \(t = 0\), \(t = 2\), and \(t = 2.5\) seconds.

(c) The total distance travelled is given by the integral of the speed:
\(\text{Distance} = \int_0^3 |v(t)| dt = \int_0^3 |4t^3 - 18t^2 + 20t| dt\)

Using GDC to evaluate the integral directly:
\(\int_0^3 |4t^3 - 18t^2 + 20t| dt = 9.375\text{ m}\)
(to 3 s.f. \(9.38\text{ m}\)).

Alternatively, partitioning the integrals:
\(\int_0^2 (4t^3 - 18t^2 + 20t) dt - \int_2^{2.5} (4t^3 - 18t^2 + 20t) dt + \int_{2.5}^3 (4t^3 - 18t^2 + 20t) dt\)
\(= 8 - (-0.1875) + 1.1875 = 9.375\text{ m}\).

(a) [2 marks]
M1 for differentiating \(v(t)\) to find \(a(t)\).
A1 for \(5\text{ m s}^{-2}\).

(b) [2 marks]
M1 for setting \(v(t) = 0\).
A1 for non-trivial solutions \(t = 2\) and \(t = 2.5\) (accept \(t=0\) as well).

(c) [3.33 marks]
M1 for setting up the integral: \(\int_0^3 |v(t)| dt\).
M1 for split-interval integration or correct GDC method.
A1.33 for \(9.38\text{ m}\) (accept \(9.375\)).

Using a graphic display calculator with the given data:

(a) Pearson's correlation coefficient is \(r \approx -0.994\) (to 3 s.f.)

(b) Since \(r\) is very close to \(-1\), there is a **strong, negative correlation** between exercising hours and resting heart rate.

(c) The linear regression line \(y\) on \(x\) is of the form \(y = mx + c\):
\(m \approx -2.17\) (to 3 s.f.)
\(c = 87.0\) (to 3 s.f.)
So the equation is \(y = -2.17x + 87.0\).

(d) For \(x = 9\):
\(y = -2.1667(9) + 87 = -19.5 + 87 = 67.5\text{ bpm}\).

(a) [2 marks]
M1 for inputting data into GDC lists.
A1 for \(r \approx -0.994\).

(b) [1 mark]
A1 for describing as 'strong' and 'negative' correlation.

(c) [3 marks]
M1 for identifying linear regression function on GDC.
A1 for \(m = -2.17\).
A1 for \(c = 87.0\).

(d) [1.33 marks]
M1 for substituting \(x = 9\) into their regression equation.
A0.33 for \(67.5\text{ bpm}\) (accept \(67.5\)).

This situation can be modeled by an arithmetic sequence with \(u_1 = 18\) and common difference \(d = 2\).

(a) The number of seats in row \(n\) is \(u_n = u_1 + (n-1)d\).
\(u_{15} = 18 + (15-1) \times 2 = 18 + 28 = 46\text{ seats}\).

(b) The total capacity is the sum of the first 25 rows:
\(S_n = \frac{n}{2}(2u_1 + (n-1)d)\)
\(S_{25} = \frac{25}{2}(2 \times 18 + 24 \times 2) = 12.5 (36 + 48) = 12.5 \times 84 = 1050\text{ seats}\).

(c) The number of seats in the first 10 rows is:
\(S_{10} = \frac{10}{2}(2 \times 18 + 9 \times 2) = 5 (36 + 18) = 5 \times 54 = 270\text{ seats}\).

The remaining seats are in rows 11 to 25:
\(\text{Remaining seats} = S_{25} - S_{10} = 1050 - 270 = 780\text{ seats}\).

Total revenue is:
\(\text{Revenue} = (270 \times 25) + (780 \times 15)\)
\(\text{Revenue} = 6750 + 11700 = \$18\,450\).

(a) [2 marks]
M1 for using \(u_n = u_1 + (n-1)d\).
A1 for \(46\).

(b) [2 marks]
M1 for using the arithmetic series sum formula.
A1 for \(1050\).

(c) [3.33 marks]
M1 for calculating the capacity of the first 10 rows: \(270\).
M1 for calculating the remaining seats: \(1050 - 270 = 780\).
M1 for setting up the revenue calculation: \((270 \times 25) + (780 \times 15)\).
A0.33 for \(\$18\,450\) (accept \(18450\)).

(a) \(H_0\): Preferred exercise type and age group are independent. \(H_1\): Preferred exercise type and age group are not independent.

(b) Expected Frequency \(= \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 90}{340} \approx 26.4706 \approx 26.5\).

(c) Degrees of Freedom \(= (r - 1)(c - 1) = (3 - 1)(3 - 1) = 4\).

(d) Using a GDC: \(\chi^2 \approx 18.9359 \approx 18.9\) and \(p\text{-value} \approx 0.0008101 \approx 0.000810\).

(e) Since the \(p\)-value \((0.000810)\) is less than the significance level \((0.05)\), we reject the null hypothesis \(H_0\). There is sufficient evidence to suggest that preferred exercise type and age group are not independent.

(f) \(P(\text{Under 30} \mid \text{Strength}) = \frac{55}{120} = \frac{11}{24} \approx 0.458\).

(g) Total Cardio \(= 130\), of which 35 are Over 50 and 95 are not.
\(P(\text{at least one Over 50}) = 1 - P(\text{neither is Over 50}) = 1 - \frac{95}{130} \times \frac{94}{129} = 1 - \frac{8930}{16770} = 1 - 0.5325 = 0.4675 \approx 0.468\).

(a) M1 for stating H0 and H1 clearly. [2 marks]
(b) M1 for setting up the expected formula, A1 for 26.5 (or 26.47). [2 marks]
(c) A1 for 4. [1 mark]
(d) A1 for chi-squared value, A2 for p-value (award A1 for 0.0008). [3 marks]
(e) R1 for comparing p-value to 0.05, A1 for correct conclusion. [2 marks]
(f) M1 for dividing by 120, A1 for 11/24 or 0.458. [2 marks]
(g) M1 for using complement or tree diagram, M1 for product without replacement, A1 for correct fraction/decimal, A1 for final answer 0.468. [4 marks]

(a) The area of the square base is \(x^2\), so its cost is \(0.08x^2\).
Each of the four sides has area \(xh\), so their combined cost is \(4 \times 0.05xh = 0.2xh\).
The total cost is \(120\), so:
\(0.08x^2 + 0.2xh = 120\)
\(0.2xh = 120 - 0.08x^2\)
\(h = \frac{120 - 0.08x^2}{0.2x}\) (as required).

(b) The volume of the box is:
\(V = x^2 h = x^2 \left( \frac{120 - 0.08x^2}{0.2x} \right) = \frac{x(120 - 0.08x^2)}{0.2} = \frac{120x - 0.08x^3}{0.2} = 600x - 0.4x^3\) (as required).

(c) \(\frac{\text{d}V}{\text{d}x} = 600 - 3(0.4)x^2 = 600 - 1.2x^2\).

(d) To maximize volume, set \(\frac{\text{d}V}{\text{d}x} = 0\):
\(600 - 1.2x^2 = 0 \implies 1.2x^2 = 600 \implies x^2 = 500 \implies x = 10\sqrt{5} \approx 22.4\) cm.

(e) Substituting \(x = 10\sqrt{5}\) into the volume expression:
\(V = 600(10\sqrt{5}) - 0.4(10\sqrt{5})^3 = 4000\sqrt{5} \approx 8944.27 \approx 8940\) \(\text{cm}^3\).

(f) Substituting \(x = 10\sqrt{5}\) into the height expression:
\(h = \frac{120 - 0.08(500)}{0.2(10\sqrt{5})} = \frac{80}{2\sqrt{5}} = 8\sqrt{5} \approx 17.9\) cm.
The ratio is:
\(\frac{h}{x} = \frac{8\sqrt{5}}{10\sqrt{5}} = 0.8\).

(a) M1 for finding cost of base, M1 for finding cost of 4 sides, A1 for equating and rearrangement to obtain the given expression. [3 marks]
(b) M1 for volume formula, M1 for substituting h, A1 for simplifying to show given expression. [3 marks]
(c) M1 for power rule, A1 for correct derivative. [2 marks]
(d) M1 for setting derivative to 0, A1 for finding x^2, A1 for correct value of x. [3 marks]
(e) M1 for substituting x into V(x), A1 for correct final answer. [2 marks]
(f) M1 for finding h, A1 for correct h value, A1 for correct ratio 0.8. [3 marks]

(a) Using GDC TVM Solver:
\(N = 60\)
\(I\% = 4.8\)
\(PV = 40000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\) yields \(PMT = -751.189... \approx \$751.19\).

(b) \(\text{Total paid} = 751.19 \times 60 = \$45\,071.40\).

(c) \(\text{Total interest} = 45\,071.40 - 40\,000 = \$5\,071.40\).

(d) After 3 years, 24 payments remain. Using GDC TVM Solver with:
\(N = 24\)
\(I\% = 4.8\)
\(PMT = -751.19\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PV\) yields \(PV \approx 17154.51\).
Thus, the remaining balance is \(\$17\,154.51\).

(e) Without paying early, the total remaining payments would be:
\(24 \times 751.19 = \$18\,028.56\).
By paying the lump sum, she pays only \(\$17\,154.51\).
Interest saved \(= 18\,028.56 - 17\,154.51 = \$874.05\).

(f) Using compound interest formula:
\(FV = 5000 \times \left(1 + \frac{0.052}{4}\right)^{20} \approx 5000 \times 1.294759 = 6473.80\).
To the nearest dollar, the value is \(\$6\,474\).

(a) M2 for correct TVM inputs, A1 for 751.19. [3 marks]
(b) M1 for multiplying by 60, A1 for 45,071.40. [2 marks]
(c) A1 for 5,071.40. [1 mark]
(d) M2 for correct TVM setup (e.g. N=24), A1 for 17,154.51 (accept answers around 17,153 to 17,156 depending on rounding). [3 marks]
(e) M1 for finding remaining payments total, M1 for subtracting the remaining balance, A1 for 874.05. [3 marks]
(f) M1 for using correct quarterly rate, M1 for formula setup, A1 for 6474. [3 marks]

(a) \(OA = \sqrt{(-30)^2 + 40^2} = \sqrt{900 + 1600} = 50\) m.
\(OB = 50\) m.
\(OC = \sqrt{20^2 + (-60)^2} = \sqrt{400 + 3600} = \sqrt{4000} \approx 63.2\) m.

(b) In right-angled triangle \(OAP\), \(\tan(35^\circ) = \frac{h}{OA} = \frac{h}{50}\).
\(h = 50 \tan(35^\circ) \approx 35.01 \approx 35.0\) m.

(c) (i) In right-angled triangle \(OBP\):
\(BP = \sqrt{OB^2 + h^2} = \sqrt{50^2 + 35.01^2} = \sqrt{2500 + 1225.7} = \sqrt{3725.7} \approx 61.0\) m.
(ii) In right-angled triangle \(OCP\):
\(\tan(\theta_C) = \frac{h}{OC} = \frac{35.01}{\sqrt{4000}} \approx 0.55356\).
\(\theta_C = \arctan(0.55356) \approx 29.0^\circ\).

(d) \(AC = \sqrt{(20 - (-30))^2 + (-60 - 40)^2} = \sqrt{50^2 + (-100)^2} = \sqrt{12500} \approx 111.8 \approx 112\) m.

(e) Using coordinate geometry area formula (shoelace formula) for \(A(-30, 40)\), \(B(50, 0)\), and \(C(20, -60)\):
\(\text{Area} = \frac{1}{2} | x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) |\)
\(\text{Area} = \frac{1}{2} | -30(0 - (-60)) + 50(-60 - 40) + 20(40 - 0) | = \frac{1}{2} | -1800 - 5000 + 800 | = \frac{1}{2} | -6000 | = 3000\) \(\text{m}^2\).

(a) A1 for OA=50, A1 for OB=50, A1 for OC=63.2. [3 marks]
(b) M1 for correct trig ratio setup, A1 for 35.0 m. [2 marks]
(c) (i) M1 for Pythagoras, A1 for 61.0 m. (ii) M1 for arctan setup, A1 for 29.0 degrees. [4 marks]
(d) M1 for distance formula, A1 for 112 m. [2 marks]
(e) M2 for setting up shoelace formula or other valid area calculation, M1 for calculation step, A1 for 3000. [4 marks]

(a) \(P(W > 20) = 0.3196 \approx 0.320\) (using GDC normal CDF with lower=20, upper=\(\infty\), mean=18.5, SD=3.2).

(b) \(P(W > k) = 0.15 \implies P(W \le k) = 0.85\).
Using GDC inverse normal with area=0.85, mean=18.5, SD=3.2:
\(k \approx 21.816 \approx 21.8\) kg.

(c) \(P(15 < W < 21) = 0.6425 \approx 0.643\) (using GDC normal CDF with lower=15, upper=21).

(d) Let \(X\) be the number of highly productive beehives. \(X \sim B(45, 0.15)\).
\(E(X) = n p = 45 \times 0.15 = 6.75\).

(e) (i) \(P(X = 8) = \binom{45}{8} (0.15)^8 (0.85)^{37} \approx 0.129\).
(ii) \(P(X \ge 10) = 1 - P(X \le 9) \approx 1 - 0.87186 = 0.128\).

(f) The sample mean weight \(\bar{W}\) is normally distributed: \(\bar{W} \sim N\left(18.5, \frac{3.2^2}{45}\right)\).
The standard error is \(\frac{3.2}{\sqrt{45}} \approx 0.47703\).
We want \(P(\bar{W} > 19.0)\). Using GDC normal CDF with lower=19.0, upper=\(\infty\), mean=18.5, SD=0.47703:
\(P(\bar{W} > 19.0) \approx 0.1473 \approx 0.147\).

(a) M1 for normal CDF setup, A1 for 0.320. [2 marks]
(b) M1 for setting up inverse normal equation, M1 for correct input (area 0.85), A1 for 21.8. [3 marks]
(c) M1 for correct boundary inputs, A1 for 0.643. [2 marks]
(d) M1 for formula np, A1 for 6.75. [2 marks]
(e) (i) M1 for binomial setup, A1 for 0.129. (ii) M1 for using 1 - P(X <= 9), A2 for 0.128 (award A1 for 0.872). [5 marks]
(f) M1 for using standard error 3.2/sqrt(45), A1 for 0.147. [2 marks]

(a) \(f(0) = 0.5\) m.
\(f(4) = 0.05(64) - 0.3(16) + 0.8(4) + 0.5 = 3.2 - 4.8 + 3.2 + 0.5 = 2.1\) m.

(b) \(f'(x) = 3(0.05)x^2 - 2(0.3)x + 0.8 = 0.15x^2 - 0.6x + 0.8\).

(c) For \(f'(x) = 0.15x^2 - 0.6x + 0.8\), the discriminant \(\Delta\) is:
\(\Delta = (-0.6)^2 - 4(0.15)(0.8) = 0.36 - 0.48 = -0.12\).
Since \(\Delta < 0\) and the coefficient of \(x^2\) is positive \((0.15 > 0)\), the quadratic is positive for all real \(x\).
Thus, \(f'(x) > 0\) for all \(0 \le x \le 4\), proving the ramp is strictly increasing.

(d) \(\text{Area} = \int_{0}^{4} (0.05x^3 - 0.3x^2 + 0.8x + 0.5) \,\text{d}x\)
\(= \left[ 0.0125x^4 - 0.1x^3 + 0.4x^2 + 0.5x \right]_{0}^{4}\)
\(= (0.0125(256) - 0.1(64) + 0.4(16) + 0.5(4)) - 0 = 3.2 - 6.4 + 6.4 + 2 = 5.2\) \(\text{m}^2\).

(e) (i) \(h = \frac{4 - 0}{4} = 1\) m.
(ii) Evaluating \(f(x)\) at interval boundaries:
\(f(0) = 0.5\), \(f(1) = 1.05\), \(f(2) = 1.3\), \(f(3) = 1.55\), \(f(4) = 2.1\).
Using the trapezoidal rule:
\(\text{Area Estimate} = \frac{1}{2} [ 0.5 + 2.1 + 2(1.05 + 1.3 + 1.55) ] = \frac{1}{2} [ 2.6 + 2(3.9) ] = 5.2\) \(\text{m}^2\).

(f) Since both the exact area and the estimate are \(5.2\) \(\text{m}^2\):
\(\text{Percentage Error} = \frac{|5.2 - 5.2|}{5.2} \times 100\% = 0\%\).

(a) A1 for f(0)=0.5, A1 for f(4)=2.1. [2 marks]
(b) M1 for differentiating terms, A1 for correct f'(x). [2 marks]
(c) M1 for finding discriminant, R1 for showing discriminant is negative, R1 for stating why this implies strictly increasing. [3 marks]
(d) M1 for integrating, A1 for correct expression, A1 for 5.2. [3 marks]
(e) (i) A1 for 1. (ii) M1 for finding values at x=1,2,3, M1 for applying trapezoidal formula, A1 for 5.2. [4 marks]
(f) M1 for percentage error formula setup, A1 for 0%. [2 marks]

(a) It represents the probability that a member who attended a Spin class in one week switches to a Yoga class the following week.

(b) (i) \(\mathbf{s}_0 = \begin{pmatrix} 120 \\ 80 \\ 100 \end{pmatrix}\).
(ii) \(\mathbf{s}_1 = \mathbf{T}\mathbf{s}_0 = \begin{pmatrix} 0.7 & 0.1 & 0.1 \\ 0.2 & 0.8 & 0.1 \\ 0.1 & 0.1 & 0.8 \end{pmatrix}\begin{pmatrix} 120 \\ 80 \\ 100 \end{pmatrix} = \begin{pmatrix} 102 \\ 98 \\ 100 \end{pmatrix}\).
So, 102 in Spin, 98 in Yoga, 100 in Pilates.
(iii) Using GDC: \(\mathbf{s}_3 = \mathbf{T}^3\mathbf{s}_0 = \begin{pmatrix} 0.7 & 0.1 & 0.1 \\ 0.2 & 0.8 & 0.1 \\ 0.1 & 0.1 & 0.8 \end{pmatrix}^3 \begin{pmatrix} 120 \\ 80 \\ 100 \end{pmatrix} \approx \begin{pmatrix} 84.72 \\ 115.28 \\ 100 \end{pmatrix}\).
So, 84.7 in Spin, 115.3 in Yoga, 100 in Pilates.

(c) Let the steady-state vector be \(\mathbf{s} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) where \(x+y+z=1\) and \(\mathbf{T}\mathbf{s}=\mathbf{s}\).
This yields the system:
\(-0.3x + 0.1y + 0.1z = 0 \implies 3x - y - z = 0 \implies y + z = 3x\).
Since \(x+y+z=1 \implies x + 3x = 1 \implies x = 0.25\).
Thus, \(y + z = 0.75\).
From \(0.1x + 0.1y - 0.2z = 0 \implies 0.25 + y - 2z = 0 \implies y - 2z = -0.25\).
Subtracting these two equations: \(3z = 1 \implies z = \frac{1}{3} \approx 0.333\).
Then \(y = 0.75 - \frac{1}{3} = \frac{5}{12} \approx 0.417\).
Steady-state vector is \(\begin{pmatrix} 0.25 \\ 0.417 \\ 0.333 \end{pmatrix}\).

(d) In the long run, the proportion of members in Pilates is \(z = \frac{1}{3}\).
With 300 members, the number attending Pilates is \(300 \times \frac{1}{3} = 100\).
Since \(100 < 150\), the club's goal will not be met.

(a) R1 for correct explanation. [1 mark]
(b) (i) A1 for correct vector. (ii) M1 for matrix multiplication setup, A1 for correct values. (iii) M1 for setting up T^3 * s0, A2 for correct values. [6 marks]
(c) M1 for setting up Ts = s, M1 for equation x+y+z=1, M1 for algebraic substitution/elimination, A1 for finding one variable, A1 for complete correct vector. [5 marks]
(d) M1 for finding long-run number of members (300 * 1/3), A1 for 100, R1 for comparing 100 < 150, A1 for correct conclusion. [4 marks]

**(a)**
\[\frac{dx}{dt} = -0.8x \implies \int \frac{1}{x} \, dx = \int -0.8 \, dt\]
\[\ln|x| = -0.8t + C \implies x(t) = Ae^{-0.8t}\]
Since \(x(0) = 100\):
\[100 = A e^0 \implies A = 100\]
Therefore, \(x(t) = 100e^{-0.8t}\).

**(b)**
**(i)**
Using Euler's method formulas:
\[x_{n+1} = x_n + h(-0.8x_n) = 0.6x_n\]
\[y_{n+1} = y_n + h(0.8x_n - 0.2y_n) = y_n + 0.5(0.8x_n - 0.2y_n) = y_n + 0.4x_n - 0.1y_n\]
At \(t_0 = 0\): \(x_0 = 100\), \(y_0 = 0\).
\[y_1 = 0 + 0.4(100) - 0.1(0) = 40\text{ mg}\]

**(ii)**
At \(t_1 = 0.5\): \(x_1 = 0.6(100) = 60\), \(y_1 = 40\).
\[x_2 = 0.6(60) = 36\text{ mg}\]
\[y_2 = 40 + 0.4(60) - 0.1(40) = 40 + 24 - 4 = 60\text{ mg}\]
So \(x(1.0) \approx 36\text{ mg}\) and \(y(1.0) \approx 60\text{ mg}\).

**(c)**
Substituting \(x(t)\) into the second equation:
\[\frac{dy}{dt} + 0.2y = 80e^{-0.8t}\]
This is a first-order linear differential equation. The integrating factor is:
\[I(t) = e^{\int 0.2 \, dt} = e^{0.2t}\]
Multiplying both sides by \(e^{0.2t}\):
\[\frac{d}{dt}\left( y e^{0.2t} \right) = 80 e^{-0.6t}\]
Integrating both sides with respect to \(t\):
\[y e^{0.2t} = \int 80 e^{-0.6t} \, dt = -\frac{80}{0.6}e^{-0.6t} + C = -\frac{400}{3}e^{-0.6t} + C\]
\[y(t) = -\frac{400}{3}e^{-0.8t} + C e^{-0.2t}\]
Using the initial condition \(y(0) = 0\):
\[0 = -\frac{400}{3} + C \implies C = \frac{400}{3}\]
Thus,
\[y(t) = \frac{400}{3} \left( e^{-0.2t} - e^{-0.8t} \right)\]

**(d)**
**(i)**
To find the maximum, set \(\frac{dy}{dt} = 0\):
\[\frac{400}{3} \left( -0.2e^{-0.2t} + 0.8e^{-0.8t} \right) = 0\]
\[0.2e^{-0.2t} = 0.8e^{-0.8t} \implies e^{0.6t} = 4\]
\[0.6t = \ln 4 \implies t_{\text{max}} = \frac{\ln 4}{0.6} = \frac{5}{3}\ln 2 \approx 2.31\text{ hours}\]

**(ii)**
Substituting \(t = \frac{5}{3}\ln 2\) back into the equation for \(y(t)\):
\[y\left(\frac{5}{3}\ln 2\right) = \frac{400}{3} \left( e^{-1/3 \ln 2} - e^{-4/3 \ln 2} \right) = \frac{400}{3} \left( 2^{-1/3} - 2^{-4/3} \right)\]
\[y_{\text{max}} = \frac{200}{3 \cdot 2^{1/3}} \approx 52.91\text{ mg}\]

**(e)**
**(i)**
For the eigenvalues, solve \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[\det \begin{pmatrix} -1.6 - \lambda & 0.4 \\ 0.6 & -0.6 - \lambda \end{pmatrix} = 0\]
\[(-1.6 - \lambda)(-0.6 - \lambda) - 0.24 = 0\]
\[\lambda^2 + 2.2\lambda + 0.72 = 0 \implies (\lambda + 1.8)(\lambda + 0.4) = 0\]
So, the eigenvalues are \(\lambda_1 = -1.8\) and \(\lambda_2 = -0.4\).

**(ii)**
For \(\lambda_1 = -1.8\):
\[\begin{pmatrix} 0.2 & 0.4 \\ 0.6 & 1.2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies v_1 + 2v_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\]

For \(\lambda_2 = -0.4\):
\[\begin{pmatrix} -1.2 & 0.4 \\ 0.6 & -0.2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -3v_1 + v_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\]

**(iii)**
The general solution is:
\[\mathbf{u}(t) = c_1 e^{-1.8t} \begin{pmatrix} -2 \\ 1 \end{pmatrix} + c_2 e^{-0.4t} \begin{pmatrix} 1 \\ 3 \end{pmatrix}\]
Using \(y(0) = 100\), \(z(0) = 0\):
\[-2c_1 + c_2 = 100\]
\[c_1 + 3c_2 = 0 \implies c_1 = -3c_2\]
Substituting gives \(7c_2 = 100 \implies c_2 = \frac{100}{7}\), \(c_1 = -\frac{300}{7}\).
Thus, the general solution is:
\[y(t) = \frac{600}{7} e^{-1.8t} + \frac{100}{7} e^{-0.4t}\]
\[z(t) = -\frac{300}{7} e^{-1.8t} + \frac{300}{7} e^{-0.4t}\]
As \(t \to \infty\), since both eigenvalues are negative, both \(y(t) \to 0\) and \(z(t) \to 0\).

**(a)**
**M1** for separation of variables and attempting integration.
**A1** for obtaining \(\ln x = -0.8t + C\) (or equivalent).
**A1** for successfully applying initial conditions to show the given formula.

**(b)**
**(i)**
**M1** for setting up the Euler recurrence relations for \(x\) and \(y\).
**A1** for obtaining \(y(0.5) \approx 40\).
**(ii)**
**A1** for \(x(1.0) \approx 36\).
**M1** for the formula for \(y(1.0)\).
**A1** for \(y(1.0) \approx 60\).

**(c)**
**M1** for substituting \(x(t)\) and identifying the Integrating Factor method.
**A1** for the correct integrating factor \(e^{0.2t}\).
**M1** for integrating \(80e^{-0.6t}\).
**A1** for obtaining the general solution \(y(t) = -\frac{400}{3} e^{-0.8t} + C e^{-0.2t}\).
**M1** for using the initial conditions to find \(C\).
**A1** for arriving at the correct shown expression.

**(d)**
**(i)**
**M1** for setting \(dy/dt = 0\).
**A1** for simplifying to \(e^{0.6t} = 4\).
**A1** for \(t_{\text{max}} = \frac{5}{3}\ln 2\) (or equivalent accurate to 3sf: 2.31).
**(ii)**
**M1** for substituting \(t_{\text{max}}\) into \(y(t)\).
**A1** for correct algebraic simplification.
**A1** for \(52.91\text{ mg}\) (must be rounded to 2 decimal places).

**(e)**
**(i)**
**M1** for writing down the determinant equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\).
**A1** for the quadratic equation \(\lambda^2 + 2.2\lambda + 0.72 = 0\).
**A1** for eigenvalues \(-1.8\) and \(-0.4\).
**(ii)**
**M1** for attempting to solve \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}\) for both eigenvalues.
**A1** for the first eigenvector proportional to \([-2, 1]^T\).
**A1** for the second eigenvector proportional to \([1, 3]^T\).
**(iii)**
**A1** for the correct analytical system of equations for \(y(t)\) and \(z(t)\).
**A1** for describing that both tend to 0 because both eigenvalues are negative.

**(a)**
**(i)**
The transition matrix (where columns represent 'from' and rows represent 'to') is:
\[\mathbf{P} = \begin{pmatrix} 0.5 & 0.2 & 0.1 \\ 0.3 & 0.6 & 0.3 \\ 0.2 & 0.2 & 0.6 \end{pmatrix}\]

**(ii)**
The initial state is \(\mathbf{s}_0 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).
After 3 steps:
\[\mathbf{s}_3 = \mathbf{P}^3 \mathbf{s}_0\]
Using GDC to compute \(\mathbf{P}^3\):
\[\mathbf{P}^3 = \begin{pmatrix} 0.279 & 0.232 & 0.209 \\ 0.417 & 0.432 & 0.417 \\ 0.304 & 0.336 & 0.374 \end{pmatrix}\]
The probability of being in Zone B after 3 trips is \(0.417\).

**(b)**
Solve \(\mathbf{P}\mathbf{s} = \mathbf{s}\):
\[0.5\pi_A + 0.2\pi_B + 0.1\pi_C = \pi_A \implies -0.5\pi_A + 0.2\pi_B + 0.1\pi_C = 0\]
\[0.3\pi_A + 0.6\pi_B + 0.3\pi_C = \pi_B \implies 0.3\pi_A - 0.4\pi_B + 0.3\pi_C = 0\]
\[0.2\pi_A + 0.2\pi_B + 0.6\pi_C = \pi_C \implies 0.2\pi_A + 0.2\pi_B - 0.4\pi_C = 0\]
From the second equation:
\[0.3\pi_A + 0.3\pi_C = 0.4\pi_B \implies \pi_A + \pi_C = \frac{4}{3}\pi_B\]
Substitute into \(\pi_A + \pi_B + \pi_C = 1\):
\[\frac{4}{3}\pi_B + \pi_B = 1 \implies \frac{7}{3}\pi_B = 1 \implies \pi_B = \frac{3}{7}\]
Now substitute \(\pi_B = \frac{3}{7}\) into the first equation multiplied by 10:
\[-5\pi_A + 2\pi_B + \pi_C = 0 \implies -5\pi_A + \pi_C = -\frac{6}{7}\]
We also have:
\[\pi_A + \pi_C = 1 - \pi_B = \frac{4}{7}\]
Subtracting the first equation from the second:
\[6\pi_A = \frac{10}{7} \implies \pi_A = \frac{5}{21}\]
Thus,
\[\pi_C = \frac{4}{7} - \frac{5}{21} = \frac{7}{21} = \frac{1}{3}\]
So the steady-state probability vector is:
\[\mathbf{s} = \begin{pmatrix} 5/21 \\ 9/21 \\ 7/21 \end{pmatrix} \approx \begin{pmatrix} 0.238 \\ 0.429 \\ 0.333 \end{pmatrix}\]

**(c)**
**(i)**
\(H_0\): The distribution of drivers across the zones matches the steady-state probabilities.
\(H_1\): The distribution does not match the steady-state probabilities.

**(ii)**
Expected frequencies:
\[E_A = 500 \times \frac{5}{21} \approx 119.05\]
\[E_B = 500 \times \frac{9}{21} \approx 214.29\]
\[E_C = 500 \times \frac{7}{21} \approx 166.67\]

**(iii)**
Calculate the \(\chi^2\) test statistic:
\[\chi^2 = \frac{(125-119.05)^2}{119.05} + \frac{(215-214.29)^2}{214.29} + \frac{(160-166.67)^2}{166.67} \approx 0.2976 + 0.0024 + 0.2667 = 0.567\]
Degrees of freedom \(df = 3 - 1 = 2\).
From GDC, \(p\)-value \(\approx 0.753\).
Since \(0.753 > 0.05\), we fail to reject \(H_0\). There is insufficient evidence to suggest the distribution of drivers differs from the steady-state distribution.

**(d)**
**(i)**
\[\mathbf{P}_x = \begin{pmatrix} 0.5 & 0.2 + 2x & 0.1 + x \\ 0.3 & 0.6 - x & 0.3 \\ 0.2 & 0.2 - x & 0.6 - x \end{pmatrix}\]

**(ii)**
Solve \(\mathbf{P}_x \mathbf{s} = \mathbf{s}\):
From Row 2:
\[0.3\pi_A + (0.6-x)\pi_B + 0.3\pi_C = \pi_B \implies 0.3(\pi_A + \pi_C) = (0.4+x)\pi_B \implies \pi_A + \pi_C = \frac{4+10x}{3}\pi_B\]
Substitute into \(\pi_A + \pi_B + \pi_C = 1\):
\[\frac{4+10x}{3}\pi_B + \pi_B = 1 \implies \frac{7+10x}{3}\pi_B = 1 \implies \pi_B = \frac{3}{7+10x}\]
From Row 1:
\[-0.5\pi_A + (0.2+2x)\pi_B + (0.1+x)\pi_C = 0\]
Substitute \(\pi_C = \frac{4+10x}{3}\pi_B - \pi_A\):
\[-0.5\pi_A + (0.2+2x)\pi_B + (0.1+x)\left(\frac{4+10x}{3}\pi_B - \pi_A\right) = 0\]
\[-(0.6+x)\pi_A + \left( (0.2+2x) + \frac{0.4+5x+10x^2}{3} \right) \pi_B = 0\]
\[(0.6+x)\pi_A = \left( \frac{10x^2+11x+1}{3} \right) \pi_B\]
Substitute \(\pi_B = \frac{3}{7+10x}\):
\[(0.6+x)\pi_A = \frac{10x^2+11x+1}{7+10x} \implies \pi_A = \frac{10x^2+11x+1}{(0.6+x)(7+10x)}\]
Multiply numerator and denominator by 5:
\[\pi_A(x) = \frac{50x^2 + 55x + 5}{5(0.6+x)(7+10x)} = \frac{50x^2 + 55x + 5}{(3+5x)(7+10x)} = \frac{50x^2 + 55x + 5}{50x^2 + 65x + 21}\]

**(iii)**
Set \(\pi_A(x) \ge 0.40\):
\[\frac{50x^2 + 55x + 5}{50x^2 + 65x + 21} \ge 0.40 \implies 50x^2 + 55x + 5 \ge 20x^2 + 26x + 8.4\]
\[30x^2 + 29x - 3.4 \ge 0\]
Using GDC or the quadratic formula to solve \(30x^2 + 29x - 3.4 = 0\):
\[x \approx 0.1057 \text{ or } x \approx -1.072\]
Since \(0 \le x \le 0.2\), the minimum value of \(x\) is \(0.106\) (to 3 decimal places).

**(a)**
**(i)**
**A2** for correct transition matrix (award **A1** if rows/columns are transposed).\\
**(ii)**
**M1** for attempting to find \(\mathbf{P}^3\mathbf{s}_0\).\\
**A1** for finding \(\mathbf{P}^3\).\\
**A1** for correct probability of \(0.417\).

**(b)**
**M1** for setting up system of equations from \(\mathbf{P}\mathbf{s} = \mathbf{s}\).\\
**M1** for incorporating \(\pi_A + \pi_B + \pi_C = 1\).\\
**A1** for \(\pi_B = \frac{3}{7}\).\\
**A1** for \(\pi_A = \frac{5}{21}\).\\
**A1** for \(\pi_C = \frac{7}{21}\).\\
**A1** for writing as a vector: \(\begin{pmatrix} 0.238 \\ 0.429 \\ 0.333 \end{pmatrix}\).

**(c)**
**(i)**
**A1** for both hypotheses stated correctly in context.\\
**(ii)**
**A1** for \(E_A \approx 119.05, E_B \approx 214.29\).\\
**A1** for \(E_C \approx 166.67\).\\
**(iii)**
**M1** for attempting to calculate \(\chi^2\).\\
**A1** for \(p\)-value \(\approx 0.753\).\\
**R1** for correct contextual conclusion based on their \(p\)-value.

**(d)**
**(i)**
**A2** for correct matrix \(\mathbf{P}_x\) in terms of \(x\).\\
**(ii)**
**M1** for setting up system \(\mathbf{P}_x \mathbf{s} = \mathbf{s}\).\\
**A1** for expressing \(\pi_B\) in terms of \(x\).\\
**M1** for substituting \(\pi_C\) to eliminate it.\\
**A1** for showing the given fraction successfully.\\
**(iii)**
**M1** for setting up inequality \(\pi_A(x) \ge 0.40\).\\
**A1** for obtaining the simplified quadratic inequality \(30x^2 + 29x - 3.4 \ge 0\).\\
**M1** for finding roots of the equation.\\
**A1** for \(x \approx 0.106\) (must be 3 decimal places).