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

(a) Let \(X\) be the mass of an apple. \(X \sim N(150, 12^2)\).
We want to find \(P(X > 165)\).
Using a GDC normal CDF with lower bound \(165\), upper bound \(\infty\), \(\mu = 150\), and \(\sigma = 12\):
\(P(X > 165) \approx 0.10565\).
To 3 significant figures, this is \(0.106\).

(b) The expected number of Premium apples is given by:
\(E = n \times p = 500 \times 0.10565 = 52.825\).
To 3 significant figures, this is \(52.8\).

(c) Let \(Y\) be the number of Premium apples in a sample of 10. \(Y \sim B(10, 0.10565)\).
We want to find \(P(Y \ge 2) = 1 - P(Y \le 1)\).
Using binomial CDF on a GDC:
\(P(Y \le 1) \approx 0.71439\).
Therefore, \(P(Y \ge 2) = 1 - 0.71439 = 0.28561\).
To 3 significant figures, this is \(0.286\).

(a)
[M1] for setting up the normal distribution probability calculation.
[A1] for correct parameters in the GDC.
[A1] for the final answer 0.106.

(b)
[A1] for 52.8 (accept 53 from using 0.106).

(c)
[M1] for identifying the binomial distribution with parameters \(n = 10\) and \(p \approx 0.106\).
[M1] for setting up the calculation \(1 - P(Y \le 1)\).
[A1] for the final answer 0.286.

(a) The volume of a cylinder is given by \(V = \pi r^2 h = 500\).
Rearranging for height \(h\):
\(h = \frac{500}{\pi r^2}\).

The cost of manufacturing the cylinder is:
\(C = 0.05 \times (2\pi r^2) + 0.03 \times (2\pi r h)\)
\(C = 0.1\pi r^2 + 0.06\pi r h\).

Substituting \(h\) into the equation:
\(C(r) = 0.1\pi r^2 + 0.06\pi r \left(\frac{500}{\pi r^2}\right)\)
\(C(r) = 0.1\pi r^2 + \frac{30}{r}\).

(b) To minimize the cost, we find the derivative \(C'(r)\) and set it to 0:
\(C'(r) = 0.2\pi r - \frac{30}{r^2} = 0\)
\(0.2\pi r^3 = 30\)
\(r^3 = \frac{150}{\pi}\)
\(r = \sqrt[3]{\frac{150}{\pi}} \approx 3.6278\text{ cm}\).
To 3 significant figures, the radius is \(3.63\text{ cm}\).

(a)
[M1] for expressing \(h\) in terms of \(r\).
[M1] for setting up the cost equation with both lid and curved surface costs.
[A1] for algebraic simplification leading to the given expression.

(b)
[M1] for differentiating \(C(r)\).
[M1] for setting their derivative equal to zero.
[A1] for the final answer 3.63 cm.

(a) The null hypothesis \(H_0\) is: Favorite music genre is independent of age group (or there is no association between favorite music genre and age group).

(b) The expected frequency is calculated as:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{160 \times 50}{300} = \frac{80}{3} \approx 26.67\).
To 3 significant figures, this is \(26.7\).

(c) Using a GDC to perform a \(\chi^2\) test of independence on the observed contingency matrix:
\(\chi^2 \approx 14.833\), with \(df = 2\).
\(p\)-value \(\approx 0.000601\) (or \(6.01 \times 10^{-4}\)).

(d) Since the \(p\)-value (\(0.000601\)) is less than the significance level (\(0.05\)), we reject the null hypothesis. There is significant evidence to suggest that favorite music genre is associated with age group.

(a)
[A1] for stating a correct null hypothesis stating independence.

(b)
[M1] for showing the correct expected frequency formula.
[A1] for 26.7.

(c)
[A2] for the correct \(p\)-value (accept 0.000601 or 0.00060).

(d)
[R1] for comparing the \(p\)-value to 0.05 and concluding to reject \(H_0\).

(a) This is a geometric sequence with \(u_1 = 150000\) and common ratio \(r = 1.06\).
The profit in the 5th year is:
\(u_5 = 150000 \times 1.06^4 = 189371.54\).
To 3 significant figures, this is \(\$189,000\).

(b) The total profit over 10 years is given by the sum formula:
\(S_{10} = \frac{150000(1.06^{10} - 1)}{1.06 - 1} \approx 1977119.24\).
To 3 significant figures, this is \(\$1,980,000\).

(c) We want to find the smallest integer \(n\) such that:
\(150000 \times 1.06^{n-1} > 250000\)
\(1.06^{n-1} > \frac{5}{3}\)
Taking logarithms:
\((n-1) \ln(1.06) > \ln\left(\frac{5}{3}\right)\)
\(n-1 > \frac{0.510825}{0.0582689} \approx 8.766\)
\(n > 9.766\)
Thus, \(n = 10\).
The 10th year corresponds to the year 2029.

(a)
[M1] for attempting to use the geometric sequence term formula.
[A1] for \(\$189,000\) (or \(\$189,372\)).

(b)
[M1] for attempting to use the geometric series sum formula.
[A1] for \(\$1,980,000\) (or \(\$1,977,119\)).

(c)
[M1] for setting up an inequality or equation to find \(n\).
[A1] for finding \(n = 10\).
[A1] for identifying the year as 2029.

(a) Using Pythagoras' theorem in the base triangle \(ABC\):
\(AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \approx 14.142\text{ m}\).
To 3 significant figures, \(AC \approx 14.1\text{ m}\).

(b) The point \(O\) is the midpoint of diagonal \(AC\), so:
\(AO = \frac{1}{2} AC = 5\sqrt{2} \approx 7.071\text{ m}\).
Using Pythagoras' theorem in the vertical right triangle \(VOA\):
\(VA = \sqrt{VO^2 + AO^2} = \sqrt{12^2 + (5\sqrt{2})^2} = \sqrt{144 + 50} = \sqrt{194} \approx 13.928\text{ m}\).
To 3 significant figures, \(VA \approx 13.9\text{ m}\).

(c) The angle \(\theta\) between the edge \(VA\) and the base \(ABCD\) is \(\angle VAO\).
\(\tan \theta = \frac{VO}{AO} = \frac{12}{5\sqrt{2}} \approx 1.697\)
\(\theta = \arctan(1.697) \approx 59.489^\circ\).
To 3 significant figures, the angle is \(59.5^\circ\) (or \(1.04\) radians).

(a)
[M1] for using Pythagoras' theorem in 2D.
[A1] for 14.1 m (or \(10\sqrt{2}\)).

(b)
[M1] for finding \(AO\) and applying Pythagoras' theorem to triangle \(VOA\).
[A1] for 13.9 m (or \(\sqrt{194}\)).

(c)
[M1] for identifying the correct angle and writing down a trigonometric ratio.
[A1] for 59.5° (or 1.04 radians).

(a) Using the GDC, we enter the lists into the statistics editor.
The product-moment correlation coefficient is \(r \approx 0.9833\).
To 3 significant figures, this is \(0.983\).

(b) Using GDC linear regression (\(y = ax + b\)):
\(a \approx 4.01515\)
\(b \approx 39.2538\)
So the regression line equation is \(y = 4.02x + 39.3\).

(c) Substituting \(x = 8\) into the regression equation:
\(y = 4.01515(8) + 39.2538 = 32.1212 + 39.2538 = 71.375\).
To 3 significant figures, the estimated score is \(71.4\).

(a)
[A2] for the correct correlation coefficient of 0.983.

(b)
[A1] for gradient 4.02, [A1] for y-intercept 39.3. (Award [A1] total if the variable names \(x\) and \(y\) are missing or reversed).

(c)
[M1] for substituting \(x = 8\) into their equation.
[A1] for 71.4 (accept 71.5 if rounded values from (b) were used).

(a) The exact area is represented by the definite integral:
\(\int_{0}^{8} (4e^{-0.2x} + 2) \, dx\).

(b) Integrating with respect to \(x\):
\(\int (4e^{-0.2x} + 2) \, dx = \left[ \frac{4}{-0.2}e^{-0.2x} + 2x \right] = \left[ -20e^{-0.2x} + 2x \right]\).

Evaluating between the limits \(0\) and \(8\):
\(\left( -20e^{-1.6} + 16 \right) - \left( -20e^0 + 0 \right)\)
\(= -20(0.201897) + 16 + 20\)
\(= -4.03793 + 36 = 31.962\).
To 3 significant figures, the exact area is \(32.0\text{ m}^2\).

(c) The interval width is \(h = 2\).
The grid points are \(x = 0, 2, 4, 6, 8\).
Let's calculate the function values:
* \(y_0 = f(0) = 4e^0 + 2 = 6\)
* \(y_1 = f(2) = 4e^{-0.4} + 2 \approx 4.68128\)
* \(y_2 = f(4) = 4e^{-0.8} + 2 \approx 3.79732\)
* \(y_3 = f(6) = 4e^{-1.2} + 2 \approx 3.20478\)
* \(y_4 = f(8) = 4e^{-1.6} + 2 \approx 2.80759\)

Applying the trapezoidal rule:
\(\text{Area} \approx \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]\)
\(\text{Area} \approx \frac{2}{2} [6 + 2.80759 + 2(4.68128 + 3.79732 + 3.20478)]\)
\(\text{Area} \approx 8.80759 + 2(11.68338) = 32.174\).
To 3 significant figures, the approximate area is \(32.2\text{ m}^2\).

(a)
[A1] for the correct expression including limits.

(b)
[M1] for a correct attempt at integration of both terms.
[M1] for substituting limits into their integrated expression.
[A1] for 32.0.

(c)
[M1] for evaluating function values at correct grid points.
[A1] for the correct substitution into the trapezoidal formula, and the final answer 32.2.

(a) First, we assign ranks to the scores for both judges (with rank 1 being the highest score):

| Participant | Judge 1 Rank | Judge 2 Rank | \(d\) | \(d^2\) |
|---|---|---|---|---|
| A | 5 | 4 | 1 | 1 |
| B | 2 | 2 | 0 | 0 |
| C | 6 | 5 | 1 | 1 |
| D | 1 | 1 | 0 | 0 |
| E | 3 | 3 | 0 | 0 |
| F | 4 | 6 | -2 | 4 |

\(\sum d^2 = 1 + 0 + 1 + 0 + 0 + 4 = 6\).
Using the formula for Spearman's rank correlation coefficient:
\(r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} = 1 - \frac{6(6)}{6(35)} = 1 - \frac{36}{210} \approx 0.82857\).
To 3 significant figures, \(r_s \approx 0.829\).

(b) Entering the raw scores into a GDC, we find the Pearson product-moment correlation coefficient:
\(r \approx 0.925\).

(c) There is a strong, positive agreement (or correlation) between the rankings of the participants by the two judges.

(a)
[M1] for a correct attempt to rank both sets of data.
[M1] for finding the differences \(d\) and their squares \(d^2\).
[A1] for \(\sum d^2 = 6\).
[A1] for \(r_s \approx 0.829\).

(b)
[A2] for the correct Pearson coefficient of 0.925 (accept 0.92).

(c)
[R1] for interpreting both the strength (strong) and direction (positive) of the correlation.

Let \(W\) be the weight of an apple, where \(W \sim N(150, 12^2)\). (a) Using a GDC, we find \(P(W < 130) \approx 0.047790\) which rounds to 0.0478. (b) Using a GDC, we find \(P(W > 165) \approx 0.105649\) which rounds to 0.106. (c) The expected number of premium apples is given by \(E(X) = n \times p = 500 \times 0.105649 \approx 52.8\) (or 53).

(a) M1 for attempting to use normal cumulative distribution with correct parameters, A1 for 0.0478. (b) M1 for normal cumulative distribution with lower bound 165, A1 for 0.106. (c) M1 for multiplying 500 by their probability from part (b), A1 for 52.8 (accept 53 from integer rounding).

(a) Differentiating the profit function gives \(P'(x) = -1.5x^2 + 12x - 18\). (b) To find the maximum, we set \(P'(x) = 0 \Rightarrow -1.5x^2 + 12x - 18 = 0\). This simplifies to \(x^2 - 8x + 12 = 0\), which factors as \((x-2)(x-6)=0\). Thus \(x=2\) or \(x=6\). Testing the nature of the critical point using the second derivative \(P''(x) = -3x + 12\), we find \(P''(6) = -6 < 0\), showing a maximum at \(x=6\). (c) Substituting \(x=6\) into the profit function gives \(P(6) = -0.5(6)^3 + 6(6)^2 - 18(6) + 50 = 50\) hundred dollars, which is \(\$5000\).

(a) M1 for applying power rule to at least two terms, A1 for correct derivative. (b) M1 for setting their derivative to 0, A1 for finding x = 6, R1 for verifying it is a maximum (e.g., using second derivative or sketch). (c) M1 for substituting x = 6 into the profit function, A1 for $5000 (accept 50 hundred dollars if units clearly stated).

(a) Using the compound interest formula: \(FV = PV \left(1 + \frac{r}{100k}\right)^{kn} = 8000 \left(1 + \frac{4.5}{1200}\right)^{12 \times 5} = 8000(1.00375)^{60} \approx 10014.38\). (b) For the investment to double, \(FV = 16000\). So we solve \(16000 = 8000(1.00375)^{12t}\) where \(t\) is in years. This simplifies to \(2 = (1.00375)^{12t}\). Solving using logarithms: \(12t = \frac{\ln 2}{\ln 1.00375} \approx 185.15\) months, which is \(t \approx 15.43\) years. Since we require the number of complete years, we round up to \(16\) years.

(a) M1 for substituting correct values into compound interest formula, A1 for correct intermediate step or setup, A1 for $10014.38. (b) M1 for setting up the doubling equation, M1 for solving for t (obtaining t approx 15.4), A1 for 16.

(a) The bearing from B to A is \(060^\circ + 180^\circ = 240^\circ\). The bearing from B to C is \(150^\circ\). Therefore, the interior angle \(\angle ABC = 240^\circ - 150^\circ = 90^\circ\). (b) Since \(\triangle ABC\) is right-angled, we use Pythagoras' theorem: \(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 12^2} = \sqrt{208} \approx 14.422\text{ km}\), which is \(14.4\text{ km}\). (c) Let \(\theta = \angle BAC\). Then \(\tan \theta = \frac{BC}{AB} = \frac{12}{8} = 1.5 \Rightarrow \theta \approx 56.31^\circ\). The bearing of C from A is the bearing of B from A plus \(\angle BAC\): \(\text{Bearing} = 60^\circ + 56.31^\circ = 116.31^\circ \approx 116^\circ\).

(a) M1 for finding the back bearing of 240 degrees (or drawing a diagram with correct angle components), A1 for showing angle ABC = 90 degrees. (b) M1 for Pythagoras' theorem, A1 for 14.4 km (accept 14.42). (c) M1 for using tangent to find angle BAC, A1 for 56.3 degrees, A1 for bearing of 116 degrees.

(a) \(H_0\): Gym member exercise preference is independent of their age group.
\(H_1\): Gym member exercise preference is associated with (not independent of) their age group.

(b) (i) Row total for Over 50 is 70. Column total for Strength is 70. Total sample size is 250.
\(E = \frac{70 \times 70}{250} = 19.6\).
(ii) Degrees of freedom \(df = (r-1)(c-1) = (3-1)(3-1) = 4\).

(c) Using GDC for the Chi-squared test on the contingency table:
\(\chi^2 \approx 5.51\) (5.5075...)
\(p\text{-value} \approx 0.239\) (0.23891...)

(d) Since the \(p\)-value \((0.239) > 0.05\) (the significance level), we fail to reject (accept) the null hypothesis \(H_0\). There is no significant evidence of an association between exercise preference and age group.

(e) Let \(T \sim N(62, 12^2)\).
Using GDC normalCDF(75, infinity, 62, 12):
\(P(T > 75) \approx 0.139\) (0.13926...)

(f) We need the conditional probability \(P(T < 75 \mid T > 50) = \frac{P(50 < T < 75)}{P(T > 50)}\).
\(P(50 < T < 75) = 0.70201...\) (using normalCDF(50, 75, 62, 12))
\(P(T > 50) = 0.84134...\) (using normalCDF(50, infinity, 62, 12))

\(P(T < 75 \mid T > 50) = \frac{0.70201}{0.84134} \approx 0.834\) (0.83439...)

(a) M1 for stating H0 correctly, M1 for stating H1 correctly.
(b) (i) M1 for formula setup, A1 for 19.6. (ii) A1 for 4.
(c) M1 A1 for \(\chi^2 \approx 5.51\), A1 for \(p \approx 0.239\).
(d) R1 for comparing p-value with 0.05, A1 for correct conclusion consistent with their comparison.
(e) M1 for standardizing or setting up normalCDF, A1 for 0.139.
(f) M1 for setting up conditional probability formula, A1 for \(P(50 < T < 75)\), A1 for \(P(T > 50)\), A1 for final answer 0.834.

(a) Using GDC with the 2-variable statistics/regression function:
(i) \(r \approx -0.993\) (-0.99285...)
(ii) \(y = -2.86x + 98.6\) (specifically \(y = -2.8648x + 98.591\))

(b) Substitute \(x = 14\) into the regression line equation:
Using precise values: \(y = -2.8648(14) + 98.591 \approx 58.5\) (using 3s.f. coefficients: \(-2.86(14) + 98.6 = 58.6\)).
Reliability comment: The estimate is highly reliable because \(x = 14\) lies within the domain of the sample data (\([3, 18]\)), meaning it is interpolation, and the correlation coefficient \(r\) is extremely strong (very close to \(-1\)).

(c) (i) Ranks from smallest (1) to largest (8):

| Student | A | B | C | D | E | F | G | H |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Rank of \(x\) | 2 | 6 | 4 | 7 | 1 | 8 | 5 | 3 |
| Rank of \(y\) | 7 | 3 | 5 | 2 | 8 | 1 | 4 | 6 |
| Difference \(d\) | -5 | 3 | -1 | 5 | -7 | 7 | 1 | -3 |
| \(d^2\) | 25 | 9 | 1 | 25 | 49 | 49 | 1 | 9 |

\(\sum d^2 = 25 + 9 + 1 + 25 + 49 + 49 + 1 + 9 = 168\).
\(r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 168}{8(63)} = 1 - 2 = -1\).
(ii) The value \(r_s = -1\) suggests a perfect negative monotonic relationship between the hours spent on social media and English test scores (as social media hours increase, English test scores always decrease).

(d) (i) Students with score \(\ge 80\) are A (85), E (92), H (80). There are exactly 3 out of 8 students. \(P(\text{reward}) = \frac{3}{8} = 0.375\).
(ii) Let \(X\) be the number of rewarded students, \(X \sim B(5, 0.375)\).
\(P(X = 2) = \binom{5}{2} (0.375)^2 (0.625)^3 \approx 0.343\) (0.34332...)

(a) (i) A1 for \(r \approx -0.993\). (ii) A1 for slope \(-2.86\), A1 for intercept \(98.6\).
(b) M1 for substituting 14, A1 for 58.5 (or 58.6), R1 for correct justification of reliability (interpolation and strong correlation).
(c) (i) M1 for correct ranking of \(x\), M1 for correct ranking of \(y\), M1 for calculating differences and formula substitution, A1 for \(r_s = -1\). (ii) R1 for stating 'perfect negative' and 'monotonic'.
(d) (i) M1 for identifying the 3 students with score \(\ge 80\), A1 for showing \(3/8 = 0.375\). (ii) M1 for recognizing binomial distribution, M1 for formula setup or GDC syntax, A1 for 0.343.

(a) Width \(= x\), Length \(= 2x\). Area of base \(= \text{Length} \times \text{Width} = 2x \times x = 2x^2\).

(b) The box has 4 vertical sides: two of size \(x \times h\) and two of size \(2x \times h\).
Total surface area \(S = \text{Base Area} + \text{Side Areas} = 2x^2 + 2(xh) + 2(2xh) = 2x^2 + 6xh\).
Given \(S = 1200\):
\(2x^2 + 6xh = 1200 \implies 6xh = 1200 - 2x^2 \implies h = \frac{1200 - 2x^2}{6x}\).

(c) Volume \(V = \text{Length} \times \text{Width} \times \text{Height} = 2x^2 h\).
Substituting \(h\):
\(V = 2x^2 \left( \frac{1200 - 2x^2}{6x} \right) = \frac{x(1200 - 2x^2)}{3} = 400x - \frac{2}{3}x^3\).

(d) \(\frac{\text{d}V}{\text{d}x} = 400 - 2x^2\).

(e) To maximize volume, set \(\frac{\text{d}V}{\text{d}x} = 0\):
\(400 - 2x^2 = 0 \implies x^2 = 200 \implies x = \sqrt{200} = 10\sqrt{2} \approx 14.1\text{ cm}\) (since \(x > 0\)).
Using the second derivative: \(\frac{\text{d}^2V}{\text{d}x^2} = -4x\), which is negative for \(x = 10\sqrt{2}\), confirming a maximum.

(f) Maximum Volume \(V = 400(10\sqrt{2}) - \frac{2}{3}(10\sqrt{2})^3 = 4000\sqrt{2} - \frac{4000}{3}\sqrt{2} = \frac{8000\sqrt{2}}{3} \approx 3771.24\text{ cm}^3\).
To the nearest cubic centimetre, the maximum volume is \(3771\text{ cm}^3\).

(g) For \(x^2 = 200\):
Base Area \(= 2x^2 = 400\text{ cm}^2\).
Cost of base \(= 400 \times 0.05 = 20\) dollars.
Side Area \(= 6xh = 1200 - 2x^2 = 1200 - 400 = 800\text{ cm}^2\).
Cost of sides \(= 800 \times 0.02 = 16\) dollars.
Total cost \(= 20 + 16 = 36\) dollars.

(a) A1 for multiplying length by width and showing \(2x^2\).
(b) M1 for finding total side area \(6xh\), M1 for setting up \(2x^2 + 6xh = 1200\), A1 for correct rearrangement.
(c) M1 for substituting \(h\) into \(V = 2x^2h\), A1 for simplifying to show given result.
(d) M1 for differentiating \(400x\), M1 for differentiating \(-\frac{2}{3}x^3\) (A1 for overall correct expression \(400-2x^2\)).
(e) M1 for setting derivative to 0, A1 for finding \(x = \sqrt{200}\) (or \(14.1\)), R1 for showing it is a maximum.
(f) M1 for substituting their \(x\) into \(V(x)\), A1 for \(3771\).
(g) M1 for calculating base cost (20), M1 for calculating side cost (16), A1 for 36 dollars.

(a) For Plan A, let \(u_n\) be the number of panels in year \(n\), where \(u_1 = 250\).
\(u_5 = u_1 + 4d = 410 \implies 250 + 4d = 410\)
\(4d = 160 \implies d = 40\).

(b) \(u_{10} = u_1 + 9d = 250 + 9(40) = 250 + 360 = 610\) panels.

(c) \(S_{10} = \frac{10}{2}(u_1 + u_{10}) = 5(250 + 610) = 5(860) = 4300\) panels.

(d) For Plan B, let \(v_n\) be the number of panels in year \(n\), where \(v_1 = 250\).
\(v_5 = v_1 \times g^4 = 410 \implies 250 \times g^4 = 410\)
\(g^4 = 1.64 \implies g = (1.64)^{0.25} \approx 1.1317\) (to 4 s.f.).
Therefore, the percentage increase \(r\% = (1.1317 - 1) \times 100\% = 13.17\% \approx 13.2\%\).

(e) \(S_{10} = \frac{v_1(g^{10} - 1)}{g - 1}\)
Using the precise value of \(g = 1.131707...\):
\(S_{10} = \frac{250(1.131707^{10} - 1)}{1.131707 - 1} \approx 4642\) panels (or \(4640\) to 3 s.f.).

(f) We seek the smallest integer \(n\) such that \(v_n > u_n\):
\(250 \times (1.131707)^{n-1} > 250 + 40(n-1)\)
Using GDC table or solving:
For \(n = 5\): \(u_5 = 410\), \(v_5 = 410\)
For \(n = 6\): \(u_6 = 450\), \(v_6 = 250 \times 1.131707^5 \approx 464.0\)
Thus, at \(n = 6\), \(v_n > u_n\).
Year 1 is 2018, so Year 6 corresponds to the calendar year 2023.

(a) M1 for setting up the equation \(250 + 4d = 410\), A1 for showing \(d = 40\).
(b) M1 for substituting into the arithmetic sequence formula, A1 for 610.
(c) M1 for using the arithmetic sum formula, M1 for substituting correct values, A1 for 4300.
(d) M1 for setting up \(250 \times g^4 = 410\), A1 for \(g \approx 1.132\) (or 1.1317), A1 for \(r = 13.2\) (or 13.17).
(e) M1 for geometric sum formula, A1 for substitution of their \(g\), A1 for 4642 (or 4640).
(f) M1 for setting up inequality or comparing values for \(n = 5\) and \(n = 6\), A1 for finding the 6th year, A1 for calendar year 2023.

(a) The diagram should show Tower A at the origin, Tower B on a bearing of \(075^\circ\) (dist. 12 km), and Tower C on a bearing of \(135^\circ\) (dist. 15 km), with the angle \(\angle BAC = 135^\circ - 75^\circ = 60^\circ\).

(b) Using the cosine rule in triangle ABC:
\(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)\)
\(BC^2 = 12^2 + 15^2 - 2(12)(15)\cos(60^\circ)\)
\(BC^2 = 144 + 225 - 360(0.5) = 189\)
\(BC = \sqrt{189} \approx 13.7\text{ km}\) (13.7477...)

(c) To find the bearing of C from B, first calculate the interior angle \(\angle ABC\) using the sine rule:
\(\frac{\sin(\angle ABC)}{15} = \frac{\sin(60^\circ)}{\sqrt{189}}\)
\(\sin(\angle ABC) = \frac{15 \sin(60^\circ)}{\sqrt{189}} \approx 0.9449\)
Since \(AC > BC\), \(\angle ABC\) is acute (as verified by cosine rule) and \(\angle ABC \approx 70.89^\circ\).
Now, the bearing of A from B is \(75^\circ + 180^\circ = 255^\circ\).
Since C is south-west from B, the bearing of C from B is:
\(\text{Bearing} = 255^\circ - \angle ABC \approx 255^\circ - 70.89^\circ = 184.11^\circ \approx 184^\circ\).

(d) \(\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC) = \frac{1}{2} \times 12 \times 15 \times \sin(60^\circ) = 45\sqrt{3} \approx 77.9\text{ km}^2\) (77.942...)

(e) (i) Let \(d\) be the shortest distance from A to road BC. Using the area of the triangle:
\(\text{Area} = \frac{1}{2} \times BC \times d \implies 77.942 = \frac{1}{2} \times 13.748 \times d\)
\(d = \frac{2 \times 77.942}{13.748} \approx 11.3\text{ km}\) (11.338...)
(ii) Geometrically, this represents the perpendicular height (altitude) of triangle ABC from A to the base BC.

(f) Since the drone is directly above A, the triangle formed by the drone, A, and B is right-angled at A.
\(\tan(8^\circ) = \frac{\text{Height}}{AB} \implies \text{Height} = 12 \times \tan(8^\circ) \approx 1.6865\text{ km}\).
In metres: \(1.6865 \times 1000 = 1686.5\text{ m} \approx 1686\text{ m}\) (to the nearest metre).

(a) A1 for showing relative locations with bearings, A1 for labelling the angle \(60^\circ\) and distances.
(b) M1 for identifying the angle \(60^\circ\), M1 for cosine rule setup, A1 for \(BC \approx 13.7\text{ km}\).
(c) M1 for using the sine (or cosine) rule, A1 for finding \(\angle ABC \approx 70.9^\circ\), M1 for using alternate angles / bearings (e.g. \(255^\circ - 70.9^\circ\)), A1 for the bearing \(184^\circ\).
(d) M1 for the area formula, A1 for \(77.9\text{ km}^2\).
(e) (i) M1 for equating area to \(\frac{1}{2} \times \text{base} \times \text{height}\), A1 for \(11.3\text{ km}\). (ii) R1 for stating it is the perpendicular height / altitude.
(f) M1 for using \(\tan(8^\circ)\) with 12 km, A1 for \(1686\text{ m}\).