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

(a) The null hypothesis is that the die is fair (or all outcomes have an equal probability of \(\frac{1}{6}\)).

(b) Since there are 120 rolls and 6 possible outcomes, the expected frequency for each outcome is:
\(E = \frac{120}{6} = 20\).

(c) Using a GDC for a \(\chi^2\) Goodness of Fit test with \(6 - 1 = 5\) degrees of freedom with the observed frequencies \([15, 24, 18, 16, 22, 25]\) and expected frequencies \([20, 20, 20, 20, 20, 20]\):
\(\chi^2 = 4.5\)
\(p\text{-value} \approx 0.480\) (to 3 s.f.).

(d) Since the \(p\)-value (\(0.480\)) is greater than the significance level (\(0.05\)), we fail to reject the null hypothesis. There is insufficient evidence to suggest the die is unfair.

(a) [1 mark]
\(M1\) for stating the null hypothesis in context (e.g., 'the die is fair' or 'the probability of each outcome is equal').

(b) [1 mark]
\(A1\) for 20.

(c) [2 marks]
\(M1\) for setting up the GDC test with 5 degrees of freedom.
\(A1\) for \(p = 0.480\) (accept 0.48).

(d) [2.15 marks]
\(R1\) for comparing their \(p\)-value with 0.05.
\(A1\) for a consistent conclusion (fail to reject \(H_0\), the die is fair).

(a) Let \(X\) be the mass of an apple. \(X \sim N(150, 12^2)\).
We want to find \(P(140 < X < 165)\).
Using a GDC with lower bound 140, upper bound 165, \(\mu = 150\), and \(\sigma = 12\):
\(P(140 < X < 165) \approx 0.692\) (to 3 s.f.).

(b) We want to find \(P(X < 130)\).
Using a GDC with lower bound \(-\infty\), upper bound 130, \(\mu = 150\), and \(\sigma = 12\):
\(P(X < 130) \approx 0.04779\).
As a percentage, this is \(4.78\%\) (to 3 s.f.).

(c) Let \(M\) be the minimum mass for premium apples.
We require \(P(X > M) = 0.10\), which is equivalent to \(P(X < M) = 0.90\).
Using the inverse normal function on a GDC with area 0.90, \(\mu = 150\), and \(\sigma = 12\):
\(M \approx 165.378...\)
So, the minimum mass of a premium apple is \(165\) grams (to 3 s.f.).

(a) [2 marks]
\(M1\) for setting up the normal cumulative probability with correct boundaries.
\(A1\) for \(0.692\).

(b) [2 marks]
\(M1\) for finding \(P(X < 130)\).
\(A1\) for \(4.78\%\) (accept \(4.78\) or \(0.0478\)).

(c) [2.15 marks]
\(M1\) for setting up the equation \(P(X > M) = 0.10\) or using inverse normal with area 0.90.
\(A1\) for \(165\) (accept \(165.38\) or \(165.4\)).

(a) First, find the midpoint of \(AB\):
\(M_{AB} = \left(\frac{1+5}{2}, \frac{4+8}{2}\right) = (3, 6)\).
Next, find the gradient of \(AB\):
\(m_{AB} = \frac{8-4}{5-1} = 1\).
The gradient of the perpendicular bisector is \(m_{\perp} = -1\).
Using the point-slope formula with \((3, 6)\):
\(y - 6 = -1(x - 3) \Rightarrow y = -x + 9\).

(b) Find the midpoint of \(BC\):
\(M_{BC} = \left(\frac{5+7}{2}, \frac{8+2}{2}\right) = (6, 5)\).
Next, find the gradient of \(BC\):
\(m_{BC} = \frac{2-8}{7-5} = -3\).
The gradient of the perpendicular bisector is \(m_{\perp} = \frac{1}{3}\).
Using the point-slope formula with \((6, 5)\):
\(y - 5 = \frac{1}{3}(x - 6) \Rightarrow y = \frac{1}{3}x + 3\).

(c) The circumcentre is the point where the perpendicular bisectors intersect.
Equating the two expressions for \(y\):
\(-x + 9 = \frac{1}{3}x + 3\)
Multiply by 3:
\(-3x + 27 = x + 9\)
\(4x = 18 \Rightarrow x = 4.5\).
Substitute \(x = 4.5\) back into \(y = -x + 9\):
\(y = -4.5 + 9 = 4.5\).
Thus, the coordinates of the hospital site are \((4.5, 4.5)\).

(a) [2 marks]
\(M1\) for finding the midpoint \((3, 6)\) and gradient \(1\) of \(AB\).
\(A1\) for \(y = -x + 9\) (or equivalent form).

(b) [2 marks]
\(M1\) for finding the midpoint \((6, 5)\) and gradient \(-3\) of \(BC\).
\(A1\) for \(y = \frac{1}{3}x + 3\) (or equivalent form).

(c) [2.15 marks]
\(M1\) for setting up and solving a system of linear equations using their answers to (a) and (b).
\(A1\) for \(x = 4.5\) and \(y = 4.5\) (accept \((4.5, 4.5)\)).

Let us represent the path of the ship using a triangle \(PQR\).

(a) Let's find the angle \(\angle PQR\).
At point \(Q\), the direction back to \(P\) is on a bearing of \(060^\circ + 180^\circ = 240^\circ\). This means the line \(QP\) makes an angle of \(60^\circ\) with the South-bound line at \(Q\).
The bearing to \(R\) is \(150^\circ\). This line makes an angle of \(180^\circ - 150^\circ = 30^\circ\) with the South-bound line at \(Q\).
Therefore, the interior angle at \(Q\) is:
\(\angle PQR = 60^\circ + 30^\circ = 90^\circ\).

Since \(\angle PQR = 90^\circ\), the triangle is right-angled at \(Q\).
Using Pythagoras' theorem:
\(PR = \sqrt{PQ^2 + QR^2} = \sqrt{15^2 + 22^2} = \sqrt{225 + 484} = \sqrt{709} \approx 26.627...\text{ km}\).
So the distance is \(26.6\text{ km}\) (to 3 s.f.).

(b) To find the bearing of \(R\) from \(P\), we first find the interior angle \(\angle QPR\).
Using right-angled trigonometry:
\(\tan(\angle QPR) = \frac{QR}{PQ} = \frac{22}{15}\)
\(\angle QPR = \arctan\left(\frac{22}{15}\right) \approx 55.713^\circ\).

The bearing of \(Q\) from \(P\) is \(060^\circ\).
Therefore, the bearing of \(R\) from \(P\) is:
\(\text{Bearing} = 60^\circ + 55.713^\circ = 115.713^\circ\).
This is \(116^\circ\) (to 3 s.f.).

(a) [3 marks]
\(M1\) for determining that \(\angle PQR = 90^\circ\) (or using the cosine rule with \(90^\circ\)).
\(M1\) for using Pythagoras' theorem \(PR^2 = 15^2 + 22^2\).
\(A1\) for \(26.6\text{ km}\) (accept \(26.627...\)).

(b) [3.15 marks]
\(M1\) for using trigonometry to find the angle \(\angle QPR\) (e.g., \(\tan(\angle QPR) = \frac{22}{15}\)).
\(A1\) for finding the angle \(\approx 55.7^\circ\).
\(M1\) for adding the bearing of \(060^\circ\) to \(\angle QPR\).
\(A1\) for \(116^\circ\) (accept \(115.7^\circ\) or \(115.713^\circ\)).

(a) The total length of the fencing is 80 metres, which is used for two sides of length \(x\) and one side of length \(y\):
\(2x + y = 80 \Rightarrow y = 80 - 2x\).

(b) The area \(A\) of the rectangular garden is:
\(A = x \times y\)
Substitute \(y = 80 - 2x\):
\(A = x(80 - 2x) = 80x - 2x^2\).

(c) Differentiating \(A\) with respect to \(x\):
\(\frac{dA}{dx} = 80 - 4x\).

(d) To find the maximum area, set \(\frac{dA}{dx} = 0\):
\(80 - 4x = 0 \Rightarrow 4x = 80 \Rightarrow x = 20\text{ m}\).

To find the maximum area, substitute \(x = 20\) back into the area formula:
\(A = 80(20) - 2(20)^2 = 1600 - 800 = 800\text{ m}^2\).

(a) [1 mark]
\(A1\) for \(y = 80 - 2x\).

(b) [1.15 marks]
\(M1\) for substituting \(y\) into \(A = xy\) to obtain \(A = 80x - 2x^2\).

(c) [1 mark]
\(A1\) for \(80 - 4x\).

(d) [3 marks]
\(M1\) for setting their derivative to 0.
\(A1\) for \(x = 20\text{ m}\).
\(A1\) for \(800\text{ m}^2\).

(a) Substitute \(t = 3\) into the rate of flow equation:
\(R(3) = 3(3)^2 - 12(3) + 15 = 27 - 36 + 15 = 6\text{ litres per minute}\).

(b) To find the minimum rate of flow, find the derivative \(R'(t)\):
\(R'(t) = 6t - 12\).
Set \(R'(t) = 0\):
\(6t - 12 = 0 \Rightarrow t = 2\text{ minutes}\).
Since the second derivative is \(R''(t) = 6 > 0\), this point is a minimum.

(c) The total volume of water is the integral of the rate of flow from \(t = 0\) to \(t = 5\):
\(V = \int_{0}^{5} (3t^2 - 12t + 15) \, dt\)
\(V = \left[ t^3 - 6t^2 + 15t \right]_{0}^{5}\)
\(V = (5^3 - 6(5^2) + 15(5)) - (0)\)
\(V = 125 - 150 + 75 = 50\text{ litres}\).

(a) [1 mark]
\(A1\) for 6 (litres per minute).

(b) [2.15 marks]
\(M1\) for finding the derivative \(6t - 12\).
\(A1\) for \(t = 2\) (minutes).

(c) [3 marks]
\(M1\) for setting up the definite integral \(\int_{0}^{5} R(t) \, dt\).
\(M1\) for correct integration to find \(t^3 - 6t^2 + 15t\).
\(A1\) for 50 (litres).

(a) We can use a GDC with a TVM Solver with the following parameters:
\(N = 5 \times 12 = 60\)
\(I\% = 4.5\)
\(PV = 24000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\), we find:
\(PMT \approx -447.32\).
Thus, the monthly payment is \(\$447.32\) (or \(\$447\) to 3 s.f.).

(b) The total amount paid is the monthly payment multiplied by the total number of payments:
\(\text{Total paid} = 447.32 \times 60 = \$26,839.20\).
(If using the 3 s.f. value of \(\$447\), the total is \(447 \times 60 = \$26,820\)).

(c) The total interest paid is the total amount paid minus the original loan amount:
\(\text{Total interest} = 26,839.20 - 24,000 = \$2,839.20\).
(If using the 3 s.f. value, \(26,820 - 24,000 = \$2,820\)).

(a) [2.15 marks]
\(M1\) for attempting to use a TVM solver with appropriate parameters (e.g., \(N=60, I\%=4.5, PV=24000\)).
\(A1\) for \(\$447.32\) (accept \(\$447\)).

(b) [2 marks]
\(M1\) for multiplying their monthly payment by 60.
\(A1\) for \(\$26,839.20\) (accept \(\$26,820\) or \(\$26,800\)).

(c) [2 marks]
\(M1\) for subtracting 24000 from their answer to (b).
\(A1\) for \(\$2,839.20\) (accept \(\$2,820\) or \(\$2,840\)).

(a) The initial temperature is when \(t = 0\):
\(T(0) = 22 + 68e^0 = 22 + 68 = 90^\circ\text{C}\).

(b) Substitute \(t = 15\) into the temperature equation:
\(T(15) = 22 + 68e^{-0.05 \times 15} = 22 + 68e^{-0.75} \approx 22 + 32.12 = 54.1^\circ\text{C}\) (to 3 s.f.).

(c) Set the temperature equation equal to 50:
\(22 + 68e^{-0.05t} = 50\)
\(68e^{-0.05t} = 28\)
\(e^{-0.05t} = \frac{28}{68}\)
\(-0.05t = \ln\left(\frac{7}{17}\right)\)
\(t = \frac{\ln(7/17)}{-0.05} \approx 17.746...\text{ minutes}\).
So, it takes approximately \(17.7\) minutes (to 3 s.f.) to reach this temperature.

(a) [1 mark]
\(A1\) for \(90^\circ\text{C}\) (or 90).

(b) [2 marks]
\(M1\) for substituting \(15\) into the formula.
\(A1\) for \(54.1^\circ\text{C}\) (accept \(54.1\)).

(c) [3.15 marks]
\(M1\) for setting up the equation \(22 + 68e^{-0.05t} = 50\).
\(M1\) for attempting to solve the equation (using logarithms or a GDC solver).
\(A1\) for \(17.7\) minutes (accept \(17.746...\)).

(a) Using a graphic display calculator (GDC), enter the \(x\) values into List 1 and the \(y\) values into List 2. Running a linear regression calculation gives:
\(r \approx 0.99398...\)
To 3 significant figures, \(r \approx 0.994\).

(b) From the GDC linear regression output:
\(a \approx 3.36170...\)
\(b \approx 36.5851...\)
Therefore, the equation of the regression line is:
\(y = 3.36x + 36.6\) (where coefficients are given to 3 significant figures).

(c) Substitute \(x = 10\) into the regression equation:
\(y = 3.36170 \times 10 + 36.5851 = 70.202...\)
To 3 significant figures, the estimated exam score is \(70.2\).

(a)
(M1) for attempting to use a GDC to find \(r\).
(A1) for \(0.994\) (accept \(0.99398...\)).

(b)
(A1) for \(a \approx 3.36\) (accept \(3.36170...\)).
(A1) for \(b \approx 36.6\) (accept \(36.5851...\)).

(c)
(M1) for substituting \(x = 10\) into their regression equation.
(A1) for \(70.2\) (accept \(70.202...\) or a correct value following through from their equation in part b).

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

(b) The point \(O\) is the center of the rectangular base, so it is the midpoint of the diagonal \(AC\). Therefore, the distance from \(A\) to \(O\) is:
\(AO = \frac{AC}{2} = \frac{10}{2} = 5\text{ cm}\).
Triangle \(VOA\) is right-angled at \(O\). Using Pythagoras' theorem in triangle \(VOA\):
\(VA = \sqrt{VO^2 + AO^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\).

(c) The angle between the edge \(VA\) and the base \(ABCD\) is the angle \(\angle VAO\). In the right-angled triangle \(VOA\):
\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{12}{5} = 2.4\)
\(\angle VAO = \arctan(2.4) \approx 67.3801...^\circ\)
To 3 significant figures, the angle is \(67.4^\circ\) (or \(1.18\text{ radians}\)).

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

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

(c)
(M1) for a correct trigonometric ratio for angle \(\angle VAO\).
(A1) for \(67.4^\circ\) (accept \(67.3801...^\circ\)) or \(1.18\text{ radians}\) (accept \(1.176...\)).

(a) The velocity is the derivative of the displacement function with respect to time:
\(v(t) = s'(t) = \frac{d}{dt}(2t^3 - 9t^2 + 12t + 4)\)
\(v(t) = 6t^2 - 18t + 12\).

(b) The particle is at rest when its velocity is zero:
\(6t^2 - 18t + 12 = 0\)
Dividing by 6:
\(t^2 - 3t + 2 = 0\)
\((t - 1)(t - 2) = 0\)
Thus, \(t = 1\) and \(t = 2\) seconds.

(c) Acceleration is the derivative of the velocity function with respect to time:
\(a(t) = v'(t) = 12t - 18\).
The velocity is first equal to zero at \(t = 1\) second.
Substituting \(t = 1\) into the acceleration function:
\(a(1) = 12(1) - 18 = -6\text{ m/s}^2\).

(a)
(M1) for attempting to differentiate \(s(t)\) (at least one non-zero term differentiated correctly).
(A1) for \(6t^2 - 18t + 12\).

(b)
(M1) for setting their \(v(t) = 0\).
(A1) for \(t = 1\) and \(t = 2\).

(c)
(M1) for differentiating their \(v(t)\) to find \(a(t) = 12t - 18\) and substituting \(t = 1\).
(A1) for \(-6\text{ m/s}^2\) (accept \(-6\)).

(a) We use the compound interest formula:
\(A = P \left(1 + \frac{r}{100k}\right)^{kt}\)
where \(P = 5000\), \(r = 4.5\), \(k = 4\) (compounded quarterly), and \(t = 5\).
\(A = 5000 \left(1 + \frac{4.5}{400}\right)^{4 \times 5} = 5000 \left(1.01125\right)^{20} \approx 6253.75\)
To the nearest dollar, the amount is $6254.

(b) We want to find the smallest integer number of years, \(t\), such that the amount is at least twice the initial deposit:
\(5000 \left(1.01125\right)^{4t} \ge 10000\)
\(\left(1.01125\right)^{4t} \ge 2\)
Taking the natural logarithm of both sides:
\(4t \ln(1.01125) \ge \ln(2)\)
\(4t \ge \frac{\ln(2)}{\ln(1.01125)} \approx 61.96...\)
\(t \ge 15.49...\) years.
Since we require the number of complete years, we round up to the next integer:
\(t = 16\) years.
(Alternatively, using a GDC TVM solver, \(N = 61.96\) quarters, which is \(15.49\) years, meaning it will double during the 16th year, so 16 complete years are required).

(a)
(M1) for substituting the correct values into the compound interest formula.
(A1) for $6254.

(b)
(M1) for setting up the inequality or equation \(5000 \left(1 + \frac{4.5}{400}\right)^{4t} = 10000\) (or equivalent TVM formulation).
(M1) for attempting to solve for \(t\) (using logs, GDC solver, or systematic trial and error).
(A1) for \(15.5\) (or \(15.49...\)).
(A1) for 16.

(a) When \(t = 0\), \(T(0) = 85\):
\(85 = A e^{0} + 20\)
\(85 = A + 20 \implies A = 65\).

(b) When \(t = 10\), \(T(10) = 50\):
\(50 = 65 e^{-10k} + 20\)
\(30 = 65 e^{-10k}\)
\(e^{-10k} = \frac{30}{65} = \frac{6}{13}\)
\(-10k = \ln\left(\frac{6}{13}\right) \approx -0.77319...\)
\(k = \frac{-0.77319...}{-10} \approx 0.077319...\)
To four decimal places, \(k = 0.0773\).

(c) Substitute \(A = 65\), \(k = 0.077319...\) (or \(0.0773\)), and \(t = 25\) into the temperature function:
\(T(25) = 65 e^{-0.077319 \times 25} + 20\)
\(T(25) = 65 e^{-1.932975} + 20 \approx 65 \times 0.14472... + 20 \approx 9.407... + 20 = 29.407...^\circ\text{C}\)
To 3 significant figures, the temperature of the coffee is \(29.4^\circ\text{C}\).

(a)
(A1) for \(A = 65\).

(b)
(M1) for substituting \(A = 65\), \(t = 10\), and \(T = 50\) into the equation.
(M1) for a valid algebraic or graphical method to solve for \(k\).
(A1) for \(0.0773\) (must be to 4 decimal places).

(c)
(M1) for substituting \(t = 25\) and their values of \(A\) and \(k\) into the function.
(A1) for \(29.4\) (accept \(29.4^\circ\text{C}\), accept answers which use \(0.0773\) to give \(29.4\)).

**(a)**
Using GDC normalcdf with lower bound 1350, upper bound \(\infty\), \(\mu = 1200\), \(\sigma = 150\):
\(P(X > 1350) \approx 0.158655 \approx 0.159\)

**(b)**
Using GDC normalcdf with lower bound 1000, upper bound 1150, \(\mu = 1200\), \(\sigma = 150\):
\(P(1000 < X < 1150) \approx 0.280604 \approx 0.281\)

**(c)**
We want to find \(k\) such that \(P(X < k) = 0.05\).
Using inverse normal on GDC:
\(k \approx 953.27 \approx 953\) hours.

**(d)**
This follows a Binomial distribution \(Y \sim B(5, p)\) where \(p = P(X > 1350) \approx 0.158655\).
We want to find \(P(Y = 2)\).
Using GDC binomialpdf(5, 0.158655, 2):
\(P(Y = 2) = \binom{5}{2} (0.158655)^2 (0.841345)^3 \approx 0.14992 \approx 0.150\)

**(e)**
Let \(\mu\) be the population mean lifetime of bulbs from the new production line.
Hypotheses:
\(H_0: \mu = 1200\)
\(H_1: \mu > 1200\)

Using GDC for a one-sample t-test on the given sample data:
Sample mean \(\bar{x} = 1245\)
Sample standard deviation \(s_{n-1} \approx 47.1939\)
Sample size \(n = 12\)

The t-test gives:
\(t \approx 3.30\n\)p\)-value \approx 0.00348

Since the \(p\)-value (0.00348) is less than the significance level (0.05), we reject the null hypothesis \(H_0\).
There is significant evidence at the 5% level to suggest that the mean lifetime of the bulbs has increased.

**(a)**
M1: for setting up the normal distribution parameters or writing \(P(X > 1350)\).
A1: for correct answer 0.159 (accept 0.158). [2 marks]

**(b)**
M1: for setting up the normal probability interval \(P(1000 < X < 1150)\).
A1: for correct answer 0.281 (accept 0.280). [2 marks]

**(c)**
M1: for setting up the equation \(P(X < k) = 0.05\) (or equivalent diagram).
A1: for attempting to use inverse normal.
A1: for correct answer 953 (accept 953.3). [3 marks]

**(d)**
M1: for recognizing binomial distribution \(B(5, p)\).
A1: for substituting correct values into binomial formula or GDC.
A1: for correct answer 0.150 (accept 0.15). [3 marks]

**(e)**
A1A1: for stating both hypotheses correctly (\(H_0: \mu = 1200\) and \(H_1: \mu > 1200\)).
A1: for finding the correct sample mean (1245) or standard deviation (47.2).
A2: for finding the correct \(p\)-value \approx 0.00348 (accept 0.0035).
R1: for a consistent conclusion in context (rejecting \(H_0\) with reasoning comparing \(p\)-value to 0.05). [6 marks]

**(a)**
\(\text{Angle } ACB = 180^\circ - 48^\circ - 72^\circ = 60^\circ\).

**(b)**
Using the Sine Rule in triangle \(ABC\):
\(\frac{AC}{\sin 72^\circ} = \frac{AB}{\sin 60^\circ}\)
\(AC = \frac{120 \sin 72^\circ}{\sin 60^\circ}\)
\(AC \approx 131.78 \approx 132\) m.

**(c)**
\(\text{Area} = \frac{1}{2} \times AB \times AC \times \sin 48^\circ\)
\(\text{Area} = \frac{1}{2} \times 120 \times 131.78 \times \sin 48^\circ \approx 5876.04 \approx 5880\) m\(^2\).

**(d) (i)**
In the right-angled triangle \(ACD\):
\(\tan 15^\circ = \frac{CD}{AC}\)
\(CD = AC \times \tan 15^\circ \approx 131.78 \times \tan 15^\circ \approx 35.31 \approx 35.3\) m.

**(d) (ii)**
First, find the length \(BC\) using the Sine Rule:
\(\frac{BC}{\sin 48^\circ} = \frac{120}{\sin 60^\circ}\)
\(BC = \frac{120 \sin 48^\circ}{\sin 60^\circ} \approx 102.97\) m.
Since flagpole \(CD\) is vertical, triangle \(BCD\) is right-angled at \(C\).
By Pythagoras' theorem:
\(BD = \sqrt{BC^2 + CD^2} = \sqrt{102.97^2 + 35.31^2} = \sqrt{10602.8 + 1246.8} = \sqrt{11849.6} \approx 108.86 \approx 109\) m.

**(e)**
Let \(h\) be the perpendicular distance from \(C\) to \(AB\).
Using right-angled triangle \(ACH\) (where \(H\) is on \(AB\) and \(CH \perp AB\)):
\(\sin 48^\circ = \frac{h}{AC}\)
\(h = 131.78 \times \sin 48^\circ \approx 97.93 \approx 97.9\) m.
(Alternatively, using \(\text{Area} = \frac{1}{2} \times AB \times h \implies 5876.04 = \frac{1}{2} \times 120 \times h \implies h \approx 97.9\) m).

**(a)**
A1: for correct angle \(60^\circ\). [1 mark]

**(b)**
M1: for substituting into the Sine Rule formula.
A1: for correct expression \(AC = \frac{120 \sin 72^\circ}{\sin 60^\circ}\).
A1: for correct length \(132\) m (accept 131.8). [3 marks]

**(c)**
M1: for substituting into the area formula \(\frac{1}{2}ab\sin C\).
A1: for correct values in the formula.
A1: for correct area \(5880\) m\(^2\) (accept 5876). [3 marks]

**(d) (i)**
M1: for identifying right-angled triangle \(ACD\) and using tangent.
A1: for correct height \(35.3\) m (accept 35.31). [2 marks]

**(d) (ii)**
M1: for finding \(BC \approx 102.97\) m (using Sine Rule or other correct method).
M1: for recognizing that triangle \(BCD\) has a right angle at \(C\).
A1: for substituting into Pythagoras' theorem.
A1: for correct final distance \(109\) m (accept 108.9). [4 marks]

**(e)**
M1: for a correct setup using area or basic trigonometry.
A1: for correct substitution.
A1: for correct path length \(97.9\) m (accept 97.93). [3 marks]

**(a)**
The volume of a box with a square base of side \(x\) and height \(h\) is:
\(V = x^2 h\)
Since \(V = 4000\):
\(x^2 h = 4000 \implies h = \frac{4000}{x^2}\).

**(b)**
The box has a square base (area \(x^2\)) and 4 vertical sides (each of area \(xh\)).
Total external surface area \(A = x^2 + 4xh\).
Substitute \(h = \frac{4000}{x^2}\):
\(A = x^2 + 4x \left(\frac{4000}{x^2}\right)\)
\(A = x^2 + \frac{16000}{x}\).

**(c)**
\(\frac{dA}{dx} = 2x - 16000x^{-2} = 2x - \frac{16000}{x^2}\).

**(d)**
To find the minimum, set \(\frac{dA}{dx} = 0\):
\(2x - \frac{16000}{x^2} = 0 \implies 2x = \frac{16000}{x^2} \implies 2x^3 = 16000 \implies x^3 = 8000 \implies x = 20\).

To show it is a minimum, find the second derivative:
\(\frac{d^2A}{dx^2} = 2 + \frac{32000}{x^3}\)
At \(x = 20\):
\(\frac{d^2A}{dx^2} = 2 + \frac{32000}{8000} = 2 + 4 = 6\)
Since \(\frac{d^2A}{dx^2} > 0\), \(x = 20\) indeed yields a minimum surface area.

**(e)**
Substitute \(x = 20\) into the expression for \(A\):
\(A = 20^2 + \frac{16000}{20} = 400 + 800 = 1200\) cm\(^2\).

**(f)**
The total cost \(C\) is:
\(C = 0.05 \times x^2 + 0.03 \times (4xh) = 0.05 x^2 + 0.12 x \left(\frac{4000}{x^2}\right) = 0.05 x^2 + \frac{480}{x}\).
Differentiating with respect to \(x\):
\(\frac{dC}{dx} = 0.1x - \frac{480}{x^2}\)
Set \(\frac{dC}{dx} = 0 \implies 0.1x = \frac{480}{x^2} \implies 0.1x^3 = 480 \implies x^3 = 4800\)
\(x = \sqrt[3]{4800} \approx 16.8687 \approx 16.9\) cm.
The corresponding height \(h\) is:
\(h = \frac{4000}{16.8687^2} \approx 14.057 \approx 14.1\) cm.

**(a)**
A1: for correct algebraic derivation from volume formula. [1 mark]

**(b)**
M1: for stating the correct general formula for surface area of an open box \(A = x^2 + 4xh\).
M1: for substituting the expression for \(h\).
A1: for completing the proof to reach the given result. [3 marks]

**(c)**
M1: for differentiating at least one term correctly.
A1: for completely correct derivative \(2x - \frac{16000}{x^2}\). [2 marks]

**(d)**
M1: for setting their derivative equal to 0.
A1: for finding \(x = 20\).
M1: for obtaining the second derivative expression \(2 + \frac{32000}{x^3}\) (or setting up a first derivative sign diagram).
A1: for substituting \(x = 20\) into the second derivative.
R1: for a correct conclusion based on the sign of the second derivative (or sign change in first derivative). [5 marks]

**(e)**
M1: for substituting \(x = 20\) into the formula for \(A\).
A1: for correct answer 1200 cm\(^2\). [2 marks]

**(f)**
M1: for constructing the cost function \(C = 0.05x^2 + \frac{480}{x}\).
A1: for finding the critical value \(x \approx 16.9\) cm.
A1: for finding the corresponding height \(h \approx 14.1\) cm. [3 marks]

**(a)**
This is an arithmetic sequence with \(u_1 = 20\) and \(d = 4\).
The number of seats in the 15th row is:
\(u_{15} = u_1 + 14d = 20 + 14(4) = 20 + 56 = 76\).

**(b)**
The total number of seats in \(n\) rows is given by the sum of an arithmetic series:
\(S_n = \frac{n}{2} [2u_1 + (n-1)d]\)
\(S_n = \frac{n}{2} [2(20) + (n-1)4] = \frac{n}{2} [40 + 4n - 4] = \frac{n}{2} [36 + 4n] = n(18 + 2n) = 2n^2 + 18n\)
Since the total number of seats is 840:
\(2n^2 + 18n = 840 \implies 2n^2 + 18n - 840 = 0\).

Solving this quadratic equation using GDC or factoring:
\(n^2 + 9n - 420 = 0 \implies (n - 15)(n + 28) = 0\)
Since the number of rows must be positive, \(n = 15\).
There are 15 rows in the auditorium.

**(c)**
Using the GDC TVM Solver for Option 1:
\(N = 36\) (3 years \(\times 12\) months)
\(I\% = 4.8\)
\(PV = 0\)
\(PMT = ?\)
\(FV = 25000\)
\(P/Y = 12\)
\(C/Y = 12\)
This gives \(PMT \approx -647.07\).
So the monthly deposit is $647.07.

**(d)**
Using the GDC TVM Solver for Option 2:
\(N = 36\)
\(I\% = 6.2\)
\(PV = 25000\)
\(PMT = ?\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
This gives \(PMT \approx -762.77\).
So the monthly repayment is $762.77.

**(e)**
Total amount paid over 3 years = \(762.77 \times 36 = 27459.72\)
Total interest paid = \(27459.72 - 25000 = 2459.72\) dollars.

**(f)**
For Option 1: Total out-of-pocket deposits = \(647.07 \times 36 = 23294.52\) dollars.
For Option 2: Total out-of-pocket payments = \(762.77 \times 36 = 27459.72\) dollars.
Option 1 has the lower net cost.
Difference = \(27459.72 - 23294.52 = 4165.20\) dollars.

**(a)**
M1: for substituting correct values into arithmetic sequence term formula.
A1: for correct answer 76. [2 marks]

**(b)**
M1: for substituting into arithmetic sum formula.
A1: for showing intermediate algebraic steps to obtain \(2n^2 + 18n - 840 = 0\).
M1: for solving the quadratic equation (using GDC or algebraic factoring).
A1: for selecting positive solution \(n = 15\) and rejecting negative solution. [4 marks]

**(c)**
M1: for choosing standard TVM solver settings with \(N = 36\), \(I\% = 4.8\), \(FV = 25000\).
A1: for setting \(PV = 0\).
A1: for correct monthly deposit $647.07. [3 marks]

**(d)**
M1: for choosing standard TVM solver settings with \(N = 36\), \(I\% = 6.2\), \(PV = 25000\).
A1: for setting \(FV = 0\).
A1: for correct monthly repayment $762.77. [3 marks]

**(e)**
M1: for calculating total payments \(762.77 \times 36\).
A1: for subtracting principal to find correct total interest of $2459.72 (accept values between $2455 and $2461 depending on rounding of monthly payments). [2 marks]

**(f)**
M1: for finding total out-of-pocket cost of Option 1 ($23,294.52).
A1: for identifying Option 1 as cheaper and calculating the difference as $4165.20 (accept consistent variations due to rounding). [2 marks]

**(a)**
The value 22 represents the ambient room temperature (the horizontal asymptote as \(t \to \infty\)).

**(b)**
The initial temperature is \(T(0) = 85\):
\(A e^0 + 22 = 85 \implies A + 22 = 85 \implies A = 63\).

**(c)**
We are given \(T(10) = 50\):
\(63 e^{-10k} + 22 = 50\)
\(63 e^{-10k} = 28 \implies e^{-10k} = \frac{28}{63} = \frac{4}{9}\)
\(-10k = \ln\left(\frac{4}{9}\right) \approx -0.81093\)
\(k = \frac{-0.81093}{-10} \approx 0.081093\)
Thus, \(k \approx 0.0811\) to three significant figures.

**(d)**
Substitute \(t = 25\) into the temperature model:
\(T(25) = 63 e^{-0.081093 \times 25} + 22 \approx 63 e^{-2.0273} + 22 \approx 8.296 + 22 = 30.296 \approx 30.3^\circ\)C.

**(e)**
Set \(T(t) = 30\):
\(63 e^{-0.081093 t} + 22 = 30 \implies 63 e^{-0.081093 t} = 8 \implies e^{-0.081093 t} = \frac{8}{63}\)
\(-0.081093 t = \ln\left(\frac{8}{63}\right) \approx -2.0637\)
\(t = \frac{-2.0637}{-0.081093} \approx 25.45 \approx 25.5\) minutes.

**(f)**
The sketch should include:
- A starting point at \((0, 85)\).
- A smooth decreasing exponential curve.
- A labeled horizontal asymptote at \(T = 22\) (drawn with a dashed line).

**(g)**
We want to solve \(T(t) = G(t)\) for \(t > 0\):
\(63 e^{-0.081093 t} + 22 = 85(0.98)^t\)
Using the intersection feature on a GDC to find the non-zero intersection of \(y = 63 e^{-0.081093 x} + 22\) and \(y = 85(0.98)^x\):
\(t \approx 66.3\) minutes.

**(a)**
R1: for identifying 22 as room/ambient temperature. [1 mark]

**(b)**
M1: for substituting \(t = 0\) and setting the equation equal to 85.
A1: for finding \(A = 63\). [2 marks]

**(c)**
M1: for substituting \(T(10) = 50\) and \(A = 63\).
A1: for isolating the exponential term \(e^{-10k} = \frac{4}{9}\) (or equivalent).
A1: for taking natural logarithms and showing \(k \approx 0.0811\). [3 marks]

**(d)**
M1: for substituting \(t = 25\) into the formula.
A1: for correct temperature \(30.3^\circ\)C (accept 30.3). [2 marks]

**(e)**
M1: for setting \(T(t) = 30\).
A1: for taking natural logs to solve for \(t\).
A1: for correct time \(25.5\) minutes (accept 25.45). [3 marks]

**(f)**
A1: for correct shape starting at \((0, 85)\).
A1: for drawing and labeling horizontal asymptote at \(T = 22\).
A1: for overall correct scale and labeling axes. [3 marks]

**(g)**
M1: for setting up the equation \(T(t) = G(t)\).
A1: for correct GDC intersection value \(t \approx 66.3\) minutes (accept 66.26). [2 marks]