2025 IB DP Mathematics - Applications and Interpretation Practice Paper with Answers

(a) The number of employees in year \(n\) is given by the geometric sequence term \(u_n = u_1 \cdot r^{n-1}\), with \(u_1 = 120\) and \(r = 1.15\). For the fifth year: \(u_5 = 120 \cdot (1.15)^4 \approx 209.88\). Since the number of employees is a discrete value, we round to 210 (or accept 209.88).

(b) The sum of a geometric series is \(S_n = \frac{u_1(r^n - 1)}{r - 1}\). For the first 8 years: \(S_8 = \frac{120(1.15^8 - 1)}{1.15 - 1} \approx \frac{120(3.059023 - 1)}{0.15} = 800 \cdot 2.059023 \approx 1647.22\). Rounding to the nearest whole number gives 1647.

(a)
M1 for writing a correct term expression: \(120 \times 1.15^4\).
A1 for \(210\) (accept \(209.88\) or \(209\)).

(b)
M1 for substituting correct values into the geometric sum formula: \(\frac{120(1.15^8 - 1)}{1.15 - 1}\).
A1 for \(1647.22...\).
A1 for rounding to \(1647\) (accept \(1650\) if using 3 significant figures rounding).

(a) The midpoint of \(AB\) is \(\left(\frac{1+5}{2}, \frac{2+2}{2}\right) = (3, 2)\). Since both points \(A\) and \(B\) have the same y-coordinate, the line segment \(AB\) is horizontal. The perpendicular bisector is a vertical line passing through the midpoint's x-coordinate, which gives \(x = 3\).

(b) The kiosk must be located at the circumcenter of the triangle \(ABC\), which is the intersection of the perpendicular bisectors. Substituting \(x = 3\) into \(y = -0.5x + 5\): \(y = -0.5(3) + 5 = 3.5\). The coordinates of the kiosk are \((3, 3.5)\).

(a)
M1 for finding the midpoint \((3, 2)\) or recognizing that \(AB\) is horizontal.
A1 for the correct equation \(x = 3\) (must be an equation).

(b)
M1 for realizing the kiosk is at the intersection of the perpendicular bisectors.
M1 for substituting \(x = 3\) into the given equation.
A1 for \((3, 3.5)\) (accept \(x = 3, y = 3.5\)).

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

(a) We seek \(P(X > 165)\). Standardizing or using a GDC: \(P(X > 165) = P\left(Z > \frac{165-150}{12}\right) = P(Z > 1.25) \approx 0.105649\). To 3 significant figures, this is \(0.106\).

(b) We seek the value \(w\) such that \(P(X < w) = 0.15\). Using the inverse normal function on a GDC: \(w \approx 137.56\). To 3 significant figures, this is \(138\) grams.

(a)
M1 for writing a correct probability expression, e.g. \(P(X > 165)\) or \(P(Z > 1.25)\).
A1 for \(0.106\) (accept \(10.6\%\) or \(0.1056\)).

(b)
M1 for stating the inverse relation, e.g. \(P(X < w) = 0.15\) or \(Z = -1.036\).
M1 for solving for \(w\) using GDC or standardization formula.
A1 for \(138\) (accept \(137.6\) or \(137.56\)).

(a) Differentiating each term of \(f(x)\) using the power rule: \(f'(x) = 6x^2 - 10x + 3\).

(b) At \(x = 2\):
The y-coordinate is \(f(2) = 2(2)^3 - 5(2)^2 + 3(2) + 1 = 16 - 20 + 6 + 1 = 3\).
The gradient of the tangent is \(f'(2) = 6(2)^2 - 10(2) + 3 = 24 - 20 + 3 = 7\).
Using the equation of a straight line: \(y - y_1 = m(x - x_1) \implies y - 3 = 7(x - 2) \implies y - 3 = 7x - 14 \implies y = 7x - 11\).

(a)
M1 for applying the power rule to at least two terms.
A1 for \(6x^2 - 10x + 3\).

(b)
M1 for finding the correct y-coordinate: \(f(2) = 3\) and gradient: \(f'(2) = 7\).
M1 for substituting their coordinates and gradient into a straight-line formula.
A1 for \(y = 7x - 11\).

(a) The initial temperature is found when \(t = 0\): \(T(0) = 20 + 65\text{e}^{0} = 20 + 65 = 85^\circ\text{C}\).

(b) Given \(T(5) = 55\): \(20 + 65\text{e}^{-5k} = 55 \implies 65\text{e}^{-5k} = 35 \implies \text{e}^{-5k} = \frac{7}{13}\). Taking the natural logarithm: \(-5k = \ln\left(\frac{7}{13}\right) \implies k \approx 0.123808\). To 3 significant figures, \(k = 0.124\).

(c) Substituting \(t = 12\) and the value of \(k\) back into the formula: \(T(12) = 20 + 65\text{e}^{-12(0.123808)} \approx 20 + 65(0.226344) \approx 20 + 14.712 = 34.7^\circ\text{C}\).

(a)
A1 for \(85\) (or \(85^\circ\text{C}\)).

(b)
M1 for substituting \(t = 5\) and \(T = 55\) into the function.
A1 for \(k \approx 0.124\) (accept exact \(-\frac{1}{5}\ln(7/13)\) or \(0.1238...\)).

(c)
M1 for substituting \(t = 12\) and their \(k\) value into the formula.
A1 for \(34.7\) (accept answers rounding to \(34.7\) or \(35\)).

(a) Using Pythagoras' theorem on the rectangular base:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\).

(b) The point \(O\) is the midpoint of diagonal \(AC\), so:
\(AO = \frac{1}{2} \times AC = 5\text{ cm}\).
Using Pythagoras' theorem on the right-angled triangle \(VOA\):
\(VA = \sqrt{AO^2 + VO^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\).

(c) The angle that \(VA\) makes with the base is \(\angle VAO\).
Using right-angled trigonometry in triangle \(VOA\):
\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{12}{5} = 2.4\)
\(\angle VAO = \arctan(2.4) \approx 67.3801...^\circ\)
To three significant figures, the angle is \(67.4^\circ\) (or \(1.18\text{ radians}\)).

(a)
(M1) for an attempt to use Pythagoras' theorem to find \(AC\).
(A1) for \(10\text{ cm}\).

(b)
(M1) for finding \(AO = 5\) and attempting to use Pythagoras' theorem on triangle \(VOA\).
(A1) for \(13\text{ cm}\).

(c)
(M1) for a correct trigonometric ratio to find angle \(VAO\) (e.g., \(\tan(\theta) = \frac{12}{5}\)).
(A1) for \(67.4^\circ\) (accept \(67.3^\circ\) to \(67.4^\circ\)) or \(1.18\text{ rad}\) (accept \(1.17\) to \(1.18\)).

Let \(W\) be the weight of a bag of apples, where \(W \sim N(1.5, 0.1^2)\).

(a) We wish to find \(P(W < 1.4)\).
Using a graphic display calculator or standard normal table:
\(P(W < 1.4) = P\left(Z < \frac{1.4 - 1.5}{0.1}\right) = P(Z < -1) \approx 0.158655...\)
To three significant figures, this is \(0.159\).

(b) Let \(X\) be the number of bags weighing less than \(1.4\text{ kg}\) out of 10.
Then \(X\) follows a binomial distribution, \(X \sim B(10, p)\), where \(p \approx 0.158655...\).
Using a binomial probability calculation:
\(P(X = 2) = \binom{10}{2} (0.158655...)^2 (1 - 0.158655...)^8 \approx 0.282542...\)
To three significant figures, this is \(0.283\).

(c) We want to find the conditional probability \(P(W < 1.4 \mid W < 1.6)\).
\(P(W < 1.4 \mid W < 1.6) = \frac{P(W < 1.4 \cap W < 1.6)}{P(W < 1.6)} = \frac{P(W < 1.4)}{P(W < 1.6)}\)
We have \(P(W < 1.4) \approx 0.158655...\).
Using a graphic display calculator:
\(P(W < 1.6) = P(Z < 1) \approx 0.841344...\)
Therefore:
\(P(W < 1.4 \mid W < 1.6) = \frac{0.158655...}{0.841344...} \approx 0.188573...\)
To three significant figures, this is \(0.189\).

(a)
(M1) for setting up the normal distribution calculation (e.g., standardizing or writing a GDC expression like \(\text{normalcdf}(-\infty, 1.4, 1.5, 0.1)\)).
(A1) for \(0.159\) (accept \(0.158\) to \(0.159\)).

(b)
(M1) for recognizing the binomial distribution model with \(n = 10\) and their \(p\) from part (a).
(A1) for \(0.283\) (accept \(0.282\) to \(0.283\)).

(c)
(M1) for expressing the conditional probability as a fraction of two probabilities (e.g., \(\frac{P(W < 1.4)}{P(W < 1.6)}\)).
(A1) for \(0.189\) (accept \(0.188\) to \(0.190\)).

(a) Midpoint of \(AB\) is \(M = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7)\).
Gradient of \(AB\) is \(m_{AB} = \frac{6-8}{8-2} = -\frac{1}{3}\).
Thus, the gradient of the perpendicular bisector is \(m_{\perp} = 3\).
Using the point-slope form: \(y - 7 = 3(x - 5)\) which simplifies to \(y = 3x - 8\).

(b) To find the circumcentre, we find the intersection of the two perpendicular bisectors:
\(y = 3x - 8\) and \(x + y = 10\).
Substituting the first equation into the second:
\(x + (3x - 8) = 10 \Rightarrow 4x = 18 \Rightarrow x = 4.5\).
Substitute \(x = 4.5\) back to find \(y\):
\(y = 10 - 4.5 = 5.5\).
So the circumcentre is \((4.5, 5.5)\).

(c) The distance from the circumcentre \((4.5, 5.5)\) to kiosk \(A(2, 8)\) is:
\(d = \sqrt{(4.5 - 2)^2 + (5.5 - 8)^2} = \sqrt{2.5^2 + (-2.5)^2} = \sqrt{12.5} \approx 3.54\text{ km}\) (to 3 significant figures).

(a)
[M1] for finding the midpoint of \(AB\) as \((5, 7)\).
[M1] for finding the gradient of the perpendicular bisector as \(3\).
[A1] for the correct final equation \(y = 3x - 8\).

(b)
[M1] for attempting to solve the system of equations of the two perpendicular bisectors.
[A1] for the correct coordinates \((4.5, 5.5)\).

(c)
[M1] for correctly substituting their circumcentre and point \(A\) into the distance formula.
[A1] for \(3.54\text{ km}\) (accept \(\sqrt{12.5}\) or \(3.5355...\)).

(a) Let \(X\) be the weight of a coffee package, where \(X \sim N(\mu, 12^2)\).
We are given \(P(X < 480) = 0.08\).
Using the standard normal variable \(Z = \frac{X - \mu}{\sigma}\):
\(P\left(Z < \frac{480 - \mu}{12}\right) = 0.08\).
From the standard normal distribution table or a GDC, the critical value for a lower-tail probability of \(0.08\) is \(z \approx -1.40507\).
Therefore,
\(\frac{480 - \mu}{12} = -1.40507\)
\(480 - \mu = -16.8608 \Rightarrow \mu \approx 496.86\text{ g}\) (or \(497\text{ g}\) to 3 significant figures).

(b) Using \(\mu = 496.86\text{ g}\):
We want to find \(P(X > 510)\).
\(Z = \frac{510 - 496.86}{12} \approx 1.095\).
\(P(X > 510) = P(Z > 1.095) \approx 0.1368\) (or \(0.137\) to 3 significant figures).
(If using the rounded mean of \(497\text{ g}\), then \(P(X > 510) \approx 0.139\)).

(c) The expected number of packages is:
\(E(Y) = n \times p = 250 \times 0.1368 \approx 34.2\) packages (or \(34.8\) if using \(0.139\)).
We accept \(34\), \(35\), \(34.2\), or \(34.8\).

(a)
[M1] for attempting to standardize or write down a correct probability equation.
[M1] for finding the correct z-value of \(-1.405\) (or \(-1.41\)).
[A1] for \(\mu = 497\text{ g}\) (or \(496.86\text{ g}\)).

(b)
[M1] for attempting to find the probability \(P(X > 510)\) using their mean.
[A1] for \(0.137\) (accept \(0.139\) from using 3sf rounded mean).

(c)
[M1] for multiplying their probability by 250.
[A1] for \(34.2\) (accept \(34\), \(35\), or \(34.8\)).

(a) Differentiating \(P(x)\) with respect to \(x\):
\(\frac{\mathrm{d}P}{\mathrm{d}x} = -3x^2 + 18x - 15\).

(b) To maximize the profit, we set the derivative to zero:
\(-3x^2 + 18x - 15 = 0\)
Dividing by \(-3\):
\(x^2 - 6x + 5 = 0\)
\((x - 1)(x - 5) = 0\)
So the stationary points are at \(x = 1\) and \(x = 5\).
To determine which maximizes profit, we can check the second derivative:
\(\frac{\mathrm{d}^2P}{\mathrm{d}x^2} = -6x + 18\).
At \(x = 5\): \(-6(5) + 18 = -12 < 0\), which confirms a local maximum.
(At \(x = 1\), \(-6(1) + 18 = 12 > 0\), which is a local minimum).
Thus, \(x = 5\) hours maximizes the daily profit.

(c) Substituting \(x = 5\) into the profit function:
\(P(5) = -(5)^3 + 9(5)^2 - 15(5) - 5\)
\(P(5) = -125 + 225 - 75 - 5 = 20\).
Since the profit is in thousands of dollars, the maximum daily profit is \(20\) thousand dollars (or \(\$20,000\)).

(a)
[M1] for attempting to differentiate at least one term correctly.
[A1] for the correct derivative \(-3x^2 + 18x - 15\).

(b)
[M1] for setting their derivative equal to 0.
[M1] for solving the quadratic equation to obtain \(x = 1\) and \(x = 5\).
[A1] for identifying \(x = 5\) as the maximum (must show some justification, e.g., second derivative test, or evaluating both values).

(c)
[M1] for substituting their \(x = 5\) back into the original profit function.
[A1] for \(\$20,000\) or \(20\) (thousand dollars).

The annual profit forms a geometric sequence where the first term is \(u_1 = 45000\) and the common ratio is \(r = 1.055\). To find the total profit over the first 10 years, we use the sum of the first \(n\) terms of a geometric sequence formula: \(S_n = \frac{u_1(r^n - 1)}{r - 1}\). Substituting \(n = 10\), \(u_1 = 45000\), and \(r = 1.055\): \(S_{10} = \frac{45000(1.055^{10} - 1)}{1.055 - 1}\). Calculating the value: \(S_{10} = \frac{45000(1.708144... - 1)}{0.055} = \frac{45000(0.708144...)}{0.055} \approx 579390.92\). Rounding to the nearest dollar gives \(\$579\,391\).

(M1) for identifying that the situation represents a geometric series with \(u_1 = 45000\) and \(r = 1.055\). (M1) for substituting the correct values into the geometric series sum formula. (A1) for a correct unrounded value of \(579390.92...\) seen. (A1) for the correct final answer rounded to the nearest dollar (\(579391\)).

(a) First, find the midpoint of \(AB\):
\(M_{AB} = \left(\frac{7+0}{2}, \frac{7+6}{2}\right) = (3.5, 6.5)\)

Next, find the gradient of \(AB\):
\(m_{AB} = \frac{6-7}{0-7} = \frac{-1}{-7} = \frac{1}{7}\)

The gradient of the perpendicular bisector is the negative reciprocal:
\(m_{\perp} = -7\)

Use the point-slope form with midpoint \((3.5, 6.5)\):
\(y - 6.5 = -7(x - 3.5)\)
\(y - 6.5 = -7x + 24.5\)
\(y = -7x + 31\)

(b) The circumcentre is the intersection of the perpendicular bisectors of the sides of the triangle.
Set the two perpendicular bisector equations equal to each other:
\(-7x + 31 = 3x - 9\)
\(10x = 40\)
\(x = 4\)

Substitute \(x = 4\) back to find \(y\):
\(y = 3(4) - 9 = 3\)

So the coordinates of the circumcentre are \((4, 3)\).

(c) The distance from the circumcentre \(P(4, 3)\) to any of the vertices (for example, \(A(7, 7)\)) is:
\(d = \sqrt{(7 - 4)^2 + (7 - 3)^2}\)
\(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ km}\)

(a)
(M1) for attempting to find the midpoint of \(AB\).
(M1) for attempting to find the perpendicular gradient.
(A1) for the correct final equation: \(y = -7x + 31\).

(b)
(M1) for setting up a system of equations using the two perpendicular bisectors.
(A1) for finding \(x = 4\).
(A1) for the correct coordinates \((4, 3)\).

(c)
(M1) for substituting the circumcentre and one of the points into the distance formula.
(A1) for a correct intermediate numerical expression (e.g., \(\sqrt{9+16}\)).
(A1) for \(5\) (accept \(5\text{ km}\)).

(a) The monthly interest rate is \(r = \frac{4.8\%}{12} = 0.4\% = 0.004\).
The total number of monthly compounding periods over 10 years is \(n = 10 \times 12 = 120\).
Using the future value of an ordinary annuity formula:
\(FV = 500 \times \frac{(1 + 0.004)^{120} - 1}{0.004}\)
\(FV \approx 500 \times 153.53549\)
\(FV \approx \$76767.75\) (or \(\$76800\) to 3sf).

(b) The initial amount for the second investment is \(PV = 76767.75\).
The interest rate is \(6\%\) per annum compounded semi-annually, so the interest rate per period is \(r = \frac{0.06}{2} = 0.03\).
Let \(k\) be the number of semi-annual periods. We want:
\(76767.75 \times (1.03)^k \ge 100000\)
\((1.03)^k \ge \frac{100000}{76767.75}\)
\((1.03)^k \ge 1.30263\)
\(k \ln(1.03) \ge \t\ln(1.30263)\)
\(k \ge \frac{\ln(1.30263)}{\ln(1.03)}\)
\(k \ge 8.94\)

Since interest is paid at the end of each compounding period, we round up to the next integer: \(k = 9\) periods.
Since compounding is semi-annual, \(9\) periods correspond to \(4.5\) years.

(c) At the end of \(9\) periods, the value of the investment is:
\(76767.75 \times (1.03)^9 = \$100163.01\)
Leah's total cash contributions over the entire period were the 120 monthly deposits:
\(\text{Total Deposits} = 120 \times 500 = \$60000\)

The total interest earned is:
\(\text{Interest} = \text{Final Value} - \text{Total Deposits} = 100163.01 - 60000 = \$40163.01\) (or \(\$40200\) to 3sf).

(a)
(M1) for substituting the parameters into the TVM solver or annuity formula.
(A1) for correct substitution: \(PMT = 500\), \(N = 120\), \(I\% = 4.8\%\) (or equivalent values in geometric series formula).
(A1) for \(\$76767.75\) (accept \(\$76800\)).

(b)
(M1) for setting up the equation or inequality using their answer from part (a): \(76767.75(1.03)^k \ge 100000\).
(A1) for using the correct period interest rate of \(3\%\) or \(0.03\).
(M1) for solving for the number of periods (e.g., finding \(k \approx 8.94\) or demonstrating a trial and error step showing \(k=8\) is insufficient and \(k=9\) is sufficient).
(A1) for \(4.5\text{ years}\) (or 9 semi-annual periods).

(c)
(M1) for calculating the final investment value at the end of the required periods (\(\$100163.01\)) and subtracting the total contributions (\(\$60000\)).
(A1) for \(\$40163.01\) (accept answers rounding to \(\$40200\)).

(a) (i) Let \(X\) be the weight of a member. \(X \sim N(75, 8^2)\).
Using GDC: \(P(X > 80) \approx 0.265985\)
\(P(X > 80) \approx 0.266\) (3 s.f.)

(ii) Let \(Y\) be the number of members weighing more than 80 kg. \(Y \sim B(5, 0.265985)\).
Using GDC: \(P(Y = 2) = \binom{5}{2} (0.265985)^2 (1 - 0.265985)^3 \approx 0.27978\)
\(P(Y = 2) \approx 0.280\) (3 s.f.)

(b) (i) \(H_0\): Preferred workout time and membership type are independent (or: There is no association between preferred workout time and membership type).

(ii) \(\text{Expected Frequency} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)
Row Total (Premium) = \(35 + 15 + 20 = 70\)
Column Total (Evening) = \(55 + 20 = 75\)
Grand Total = 200
\(\text{Expected Frequency} = \frac{70 \times 75}{200} = 26.25\)

(iii) Using GDC for the \(\chi^2\) test of independence on the \(2 \times 3\) contingency table:
\(\chi^2 \approx 5.0366\), with \(df = 2\)
\(p\text{-value} \approx 0.080598 \approx 0.0806\) (3 s.f.)

(iv) Since \(p\text{-value} \approx 0.0806 > 0.05\), we fail to reject the null hypothesis (or we accept \(H_0\)).
There is insufficient evidence at the 5% significance level to suggest that preferred workout time and membership type are associated.

(c) (i) Entering the data into the GDC:
\(r \approx 0.98524 \approx 0.985\) (3 s.f.)

(ii) Regression line equation:
\(y = 3.8015x + 21.7884\)
\(y = 3.80x + 21.8\) (3 s.f.)

(iii) Substitute \(x = 7\) into the regression equation:
\(y = 3.8015(7) + 21.7884 = 48.4\) USD (or 48.4 USD using 3 s.f.)
Reliability: The estimate is highly reliable because \(x = 7\) lies within the range of the given data \([2, 15]\) (interpolation) and the correlation coefficient \(r \approx 0.985\) shows a very strong linear correlation.

(a) (i)
M1 for setting up the normal distribution calculation.
A1 for 0.266.
(ii)
M1 for identifying a binomial model with parameters \(n=5\) and \(p\approx0.266\).
A1 for 0.280.

(b) (i)
A1 for stating the null hypothesis correctly (must contain "independent" or "no association" and refer to the context).
(ii)
M1 for finding the row total, column total, and substituting into the expected frequency formula.
A1 for 26.25.
(iii)
A1 for 0.0806.
(iv)
R1 for comparing the \(p\)-value to the significance level (0.05).
A1 for a consistent conclusion in context.

(c) (i)
M1 for attempting to find \(r\) using a GDC.
A1 for 0.985.
(ii)
M1 for attempting a linear regression on GDC.
A1 for \(y = 3.80x + 21.8\) (accept equivalent 3 s.f. forms).
(iii)
A1 for 48.4 USD (or 48.4).
R1 for a valid reason about reliability (must mention both interpolation/within range and strong correlation).

(a) The volume of a rectangular box is given by:
\(V = \text{length} \times \text{width} \times \text{height}\)
\(36 = (2x) \cdot (x) \cdot h\)
\(36 = 2x^2 h\)
\(h = \frac{36}{2x^2} = \frac{18}{x^2}\)

(b) The box is open, so it has one base and four sides.
\(\text{Base Area} = (2x) \cdot (x) = 2x^2\)
The sides consist of two faces with area \(x \cdot h\) and two faces with area \(2x \cdot h\).
\(\text{Side Area} = 2(xh) + 2(2xh) = 6xh\)
\(A(x) = 2x^2 + 6xh\)
Substituting \(h = \frac{18}{x^2}\):
\(A(x) = 2x^2 + 6x \left(\frac{18}{x^2}\right)\)
\(A(x) = 2x^2 + \frac{108}{x}\)

(c) \(A(x) = 2x^2 + 108x^{-1}\)
\(A'(x) = 4x - 108x^{-2} = 4x - \frac{108}{x^2}\)

(d) To find the minimum surface area, set \(A'(x) = 0\):
\(4x - \frac{108}{x^2} = 0\)
\(4x^3 = 108\)
\(x^3 = 27\)
\(x = 3\) meters

(e) Substitute \(x = 3\) into \(A(x)\):
\(A(3) = 2(3)^2 + \frac{108}{3} = 18 + 36 = 54\text{ m}^2\)

(f) The cost function \(C(x)\) is:
\(C(x) = 5 \cdot (\text{Base Area}) + 3 \cdot (\text{Side Area})\)
\(C(x) = 5(2x^2) + 3\left(\frac{108}{x}\right)\)
\(C(x) = 10x^2 + \frac{324}{x}\)
To find the minimum cost, set \(C'(x) = 0\):
\(C'(x) = 20x - \frac{324}{x^2} = 0\)
\(20x^3 = 324\)
\(x^3 = 16.2\)
\(x = \sqrt[3]{16.2} \approx 2.53\) meters (or 2.5305...)

(a)
M1 for writing a correct volume formula using \(2x\), \(x\), and \(h\).
A1 for fully showing the steps to arrive at \(h = \frac{18}{x^2}\).

(b)
M1 for finding the area of the base as \(2x^2\).
M1 for finding the total side area as \(6xh\).
A1 for substituting \(h\) and obtaining the final expression for \(A(x)\).

(c)
M1 for attempting to differentiate with power rule.
A1 for \(4x - \frac{108}{x^2}\).

(d)
M1 for setting \(A'(x) = 0\).
M1 for solving \(4x^3 = 108\).
A1 for \(x = 3\).

(e)
M1 for substituting their \(x\) value from part (d) into \(A(x)\).
A1 for 54 (or \(54\text{ m}^2\)).

(f)
M1 for setting up the cost function \(C(x) = 10x^2 + \frac{324}{x}\).
M1 for equating \(C'(x) = 0\) (or using GDC to find the minimum point of \(C(x)\)).
A1 for \(x \approx 2.53\) (or 2.53 meters).

(a) Substitute \(t = 6\) into the function \(T(t)\): \(T(6) = -0.05(6)^3 + 1.8(6)^2 - 19.2(6) + 82 = -10.8 + 64.8 - 115.2 + 82 = 20.8\). Thus, the temperature is \(20.8^\circ\text{C}\). (b) Differentiating \(T(t)\) with respect to \(t\) using the power rule: \(T'(t) = -0.15t^2 + 3.6t - 19.2\). (c) Setting \(T'(t) = 0\): \(-0.15t^2 + 3.6t - 19.2 = 0\). Dividing by \(-0.15\) gives \(t^2 - 24t + 128 = 0\). Factoring gives \((t-8)(t-16) = 0\), so \(t = 8\) and \(t = 16\). Both times are within the range \(4 \le t \le 18\). (d) Differentiating \(T'(t)\) to find the second derivative: \(T''(t) = -0.3t + 3.6\). Evaluating at \(t = 8\): \(T''(8) = -0.3(8) + 3.6 = 1.2\). Since \(T'(8) = 0\) and \(T''(8) > 0\), the graph of \(T(t)\) is concave up, confirming a local minimum at \(t = 8\). (Alternatively, checking the sign of \(T'(t)\) around \(t = 8\): for \(t = 7\), \(T'(7) = -1.35 < 0\) and for \(t = 9\), \(T'(9) = 1.05 > 0\). Since \(T'(t)\) changes from negative to positive, a local minimum is verified.) (e) At the local minimum \(t = 8\), the temperature is \(T(8) = -0.05(8)^3 + 1.8(8)^2 - 19.2(8) + 82 = 18^\circ\text{C}\). Checking boundaries: \(T(4) = 30.8^\circ\text{C}\) and \(T(18) = 28^\circ\text{C}\). Thus, the absolute minimum temperature is \(18^\circ\text{C}\). (f) The ventilation system is on when \(T'(t) > 1.5\). This gives: \(-0.15t^2 + 3.6t - 19.2 > 1.5 \Rightarrow -0.15t^2 + 3.6t - 20.7 > 0\). Solving the equation \(-0.15t^2 + 3.6t - 20.7 = 0\) using a GDC or the quadratic formula gives: \(t = \frac{-3.6 \pm \sqrt{3.6^2 - 4(-0.15)(-20.7)}}{-0.3} = 12 \pm \sqrt{6}\). Therefore, \(t \approx 9.55\) and \(t \approx 14.4\) (to 3 significant figures). Since the coefficient of \(t^2\) is negative, the inequality holds between these roots, giving the range \(9.55 < t < 14.4\) (or \(12 - \sqrt{6} < t < 12 + \sqrt{6}\)).

(a) M1 for substituting \(t = 6\) into the function \(T(t)\), A1 for \(20.8^\circ\text{C}\) (or 20.8). (b) M1 for attempting to use the power rule on at least one term, A1 for \(-0.15t^2 + 3.6t\), A1 for \(-19.2\). (c) M1 for setting their \(T'(t) = 0\), A1 for \(t = 8\), A1 for \(t = 16\). (d) M1 for finding the second derivative \(T''(t) = -0.3t + 3.6\) (or attempting a sign diagram), A1 for evaluating \(T''(8) = 1.2\) (or evaluating \(T'(t)\) on both sides of 8), R1 for a valid concluding statement linking \(T''(8) > 0\) (or change in sign of \(T'\)) to a local minimum. (e) A1 for \(18^\circ\text{C}\) (or 18). (f) M1 for setting up the inequality \(T'(t) > 1.5\), M1 for attempting to find the roots of the quadratic equation \(-0.15t^2 + 3.6t - 20.7 = 0\), A1 for \(t \approx 9.55\) (or \(12-\sqrt{6}\)), A1 for \(t \approx 14.4\) (or \(12+\sqrt{6}\)), A1 for the correct interval \(9.55 < t < 14.4\) (accept \(9.55 \le t \le 14.4\) or equivalent).

(a) Using Pythagoras' theorem in the right-angled triangle formed by the river width and the distance \(x\), we have \(AP^2 = x^2 + 400^2\), which gives \(AP = \sqrt{x^2 + 160000}\). (b) The total cost \(C(x)\) is the sum of the underwater cost and the land cost. The distance along the bank is \(1500 - x\). Thus, \(C(x) = 50 \times \sqrt{x^2 + 160000} + 30(1500 - x) = 50\sqrt{x^2 + 160000} + 45000 - 30x\). (c) Using the chain rule, \(C'(x) = 50 \cdot \frac{1}{2}(x^2 + 160000)^{-1/2} \cdot 2x - 30 = \frac{50x}{\sqrt{x^2 + 160000}} - 30\). (d) To find the minimum cost, set \(C'(x) = 0\): \(\frac{50x}{\sqrt{x^2 + 160000}} - 30 = 0 \implies 50x = 30\sqrt{x^2 + 160000} \implies 5x = 3\sqrt{x^2 + 160000}\). Squaring both sides gives \(25x^2 = 9(x^2 + 160000) \implies 25x^2 = 9x^2 + 1440000 \implies 16x^2 = 1440000 \implies x^2 = 90000 \implies x = 300\) meters (since \(x \ge 0\)). (e) The minimum cost is \(C(300) = 50\sqrt{300^2 + 160000} + 45000 - 30(300) = 50\sqrt{250000} + 45000 - 9000 = 50(500) + 36000 = 25000 + 36000 = \$61000\). (f) With underwater cost at \(\$60\), the cost function derivative is \(C'(x) = \frac{60x}{\sqrt{x^2 + 160000}} - 30\). Setting \(C'(x) = 0\) gives \(60x = 30\sqrt{x^2 + 160000} \implies 2x = \sqrt{x^2 + 160000} \implies 4x^2 = x^2 + 160000 \implies 3x^2 = 160000 \implies x = \sqrt{\frac{160000}{3}} \approx 230.94\text{ m}\). Since the new optimal distance is \(231\text{ m}\) (to 3 sig figs), which is less than \(300\text{ m}\), the point \(P\) should be moved closer to the power station's downstream level (meaning \(x\) decreases).

(a) Award A1 for \(\sqrt{x^2 + 160000}\). (b) Award M1 for writing the cost equation with the two components: \(50 \times \text{underwater distance} + 30 \times \text{land distance}\). Award A1 for substituting \(1500 - x\) for the land distance and successfully simplifying to show the given expression. (c) Award M1 for an attempt to differentiate using the chain rule. Award A1 for the correct derivative of the square root term, \(\frac{50x}{\sqrt{x^2 + 160000}}\). Award A1 for the correct complete derivative, \(\frac{50x}{\sqrt{x^2 + 160000}} - 30\). (d) Award M1 for setting their \(C'(x) = 0\). Award A1 for isolating the radical term, e.g., \(5x = 3\sqrt{x^2 + 160000}\). Award M1 for squaring both sides and solving the resulting quadratic equation. Award A1 for \(x = 300\). (e) Award M1 for substituting \(x = 300\) into their cost function. Award A1 for \(\$61000\) (or \(61000\)). (f) Award A1 for stating 'closer' (or '\(x\) decreases'). Award A1 for the correct new value of \(x \approx 231\) (accept \(230.94...\) or \(\frac{400}{\sqrt{3}}\)).

**Part (a)(i)**
Using the interior angle properties of parallel lines and bearings, we find the angle \(\angle ABC\):
- The bearing of \(B\) from \(A\) is \(080^\circ\), so the bearing of \(A\) from \(B\) is \(260^\circ\).
- The bearing of \(C\) from \(B\) is \(150^\circ\).
- Thus, \(\angle ABC = 260^\circ - 150^\circ = 110^\circ\).

Using the Cosine Rule in \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(AC^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos(110^\circ)\)
\(AC^2 = 144 + 324 - 432 \cos(110^\circ) \approx 615.75\)
\(AC \approx 24.814\text{ km} = 24.8\text{ km}\) (to 3 s.f.)

**Part (a)(ii)**
Using the Sine Rule in \(\triangle ABC\) to find \(\angle BAC\):
\(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}\)
\(\sin(\angle BAC) = \frac{18 \cdot \sin(110^\circ)}{24.814} \approx 0.6816\)
\(\angle BAC \approx 42.97^\circ\)

The bearing of \(C\) from \(A\) is:
\(\text{Bearing} = 080^\circ + 42.97^\circ = 122.97^\circ \approx 123^\circ\) (to 3 s.f.)

**Part (b)**
The patrol boat \(P\) is equidistant from \(A\), \(B\), and \(C\), meaning \(P\) is the circumcentre of \(\triangle ABC\). The distance from \(P\) to any of the stations is the circumradius \(R\).
Using the formula for the circumradius:
\(2R = \frac{AC}{\sin(\angle ABC)}\)
\(2R = \frac{24.814}{\sin(110^\circ)} \approx 26.407\)
\(R \approx 13.204\text{ km} = 13.2\text{ km}\) (to 3 s.f.)

**Part (c)**
Let \(M\) be the midpoint of \(AC\). Since \(\triangle APC\) is isosceles with \(AP = CP = R\), the shortest distance from \(P\) to the flight path \(AC\) is the perpendicular height \(PM\).
\(AM = \frac{AC}{2} \approx 12.407\text{ km}\)
Using Pythagoras' theorem in \(\triangle AMP\):
\(PM^2 + AM^2 = AP^2\)
\(PM^2 = 13.204^2 - 12.407^2 \approx 20.395\)
\(PM \approx 4.516\text{ km} = 4.52\text{ km}\) (to 3 s.f.)

**Part (d)**
*Method 1: Bearings and Trigonometry*
1. Find \(\angle BCA = 180^\circ - 110^\circ - 42.97^\circ = 27.03^\circ\).
2. The bearing of \(B\) from \(C\) is \(150^\circ + 180^\circ = 330^\circ\).
3. The line \(CA\) is at an angle of \(27.03^\circ\) counter-clockwise from \(CB\), so the bearing of \(A\) from \(C\) is \(330^\circ - 27.03^\circ = 302.97^\circ\).
4. In \(\triangle APC\), \(\cos(\angle PCA) = \frac{AM}{R} = \frac{12.407}{13.204} \implies \angle PCA \approx 20.00^\circ\).
5. Since \(P\) lies on the opposite side of \(AC\) as \(B\) (because \(\angle ABC > 90^\circ\)), and \(B\) is clockwise from \(CA\), \(P\) lies counter-clockwise from \(CA\). Thus, the bearing of \(P\) from \(C\) is:
\(\text{Bearing of } P = 302.97^\circ - 20.00^\circ = 282.97^\circ\).
6. Now, find the angle \(\angle PCD\):
\(\angle PCD = 282.97^\circ - 200^\circ = 82.97^\circ\).
7. Using the Cosine Rule in \(\triangle PCD\) to find \(PD\):
\(PD^2 = PC^2 + CD^2 - 2 \cdot PC \cdot CD \cdot \cos(\angle PCD)\)
\(PD^2 = 13.204^2 + 8^2 - 2 \cdot 13.204 \cdot 8 \cdot \cos(82.97^\circ) \approx 212.46\)
\(PD \approx 14.576\text{ km}\).

*Method 2: Coordinate Geometry*
Let \(A\) be the origin \((0,0)\).
\(B = (12 \sin 80^\circ, 12 \cos 80^\circ) \approx (11.818, 2.084)\)
\(C = (11.818 + 18 \sin 150^\circ, 2.084 + 18 \cos 150^\circ) \approx (20.818, -13.505)\)
By solving the perpendicular bisectors of \(AB\) and \(BC\), we find the circumcentre \(P\):
\(P \approx (7.952, -10.541)\)
Coordinates of \(D\):
\(D = (20.818 + 8 \sin 200^\circ, -13.505 + 8 \cos 200^\circ) \approx (18.082, -21.023)\)
\(PD = \sqrt{(18.082 - 7.952)^2 + (-21.023 - (-10.541))^2} \approx \sqrt{10.130^2 + (-10.482)^2} \approx 14.576\text{ km}\)

To find the minimum travel time:
\(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{14.576}{24} \approx 0.6073\text{ hours}\)
\(\text{Time in minutes} = 0.6073 \times 60 \approx 36.4\text{ minutes}\) (accept \(36.4 - 36.5\)).

**Part (a)(i)**
- **M1**: For calculating \(\angle ABC = 110^\circ\) (can be implied by substitution in the Cosine Rule).
- **M1**: For substitution into the Cosine Rule:
\(AC^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos(110^\circ)\)
- **A1**: \(24.8\text{ km}\) (accept \(24.814\dots\))

**Part (a)(ii)**
- **M1**: For substitution into the Sine Rule (or Cosine Rule) to find \(\angle BAC\):
\(\frac{\sin(\angle BAC)}{18} = \frac{\sin(110^\circ)}{24.814}\)
- **A1**: \(\angle BAC = 43.0^\circ\) (accept \(42.97^\circ\))
- **A1**: \(123^\circ\) (accept \(122.97^\circ\))

**Part (b)**
- **M1**: For recognizing that the distance is the circumradius \(R\) of \(\triangle ABC\).
- **M1**: For correct substitution into the circumradius formula or setting up perpendicular bisector equations:
\(2R = \frac{24.814}{\sin(110^\circ)}\)
- **A1**: \(2R \approx 26.4\text{ km}\) (or equivalent intermediate step)
- **A1**: \(13.2\text{ km}\) (accept \(13.204\dots\))

**Part (c)**
- **M1**: For recognizing that the shortest distance is the perpendicular height \(PM\) from \(P\) to \(AC\).
- **A1**: Identifying \(AM = 12.4\text{ km}\) (accept \(12.407\text{ km}\)).
- **M1**: For applying Pythagoras' theorem:
\(PM^2 = 13.204^2 - 12.407^2\)
- **A1**: \(4.52\text{ km}\) (accept \(4.516\dots\))

**Part (d)**
- **M1**: For finding the bearing of \(P\) from \(C\) (\(283^\circ\)) OR finding the coordinates of \(P\) and \(D\).
- **M1**: For calculating the angle \(\angle PCD = 83.0^\circ\) (accept \(82.97^\circ\)) OR setting up distance formula with coordinates.
- **A1**: For finding the distance \(PD = 14.6\text{ km}\) (accept \(14.576\dots\)).
- **M1**: For dividing their distance by \(24\) and multiplying by \(60\).
- **A1**: \(36.4\text{ minutes}\) (accept \(36.4 - 36.5\)).