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

(a) Using the compound interest formula with \( PV = 5000 \), \( r = 4.5 \), \( k = 12 \), and \( n = 5 \): \( FV = 5000 \left(1 + \frac{4.5}{1200}\right)^{60} \approx 6258.98 \). Thus, the value after 5 years is \( \$6258.98 \) (or \( \$6260 \) to 3 s.f.). (b) We set up the inequality \( 5000 \left(1 + \frac{4.5}{1200}\right)^m > 7500 \), which simplifies to \( (1.00375)^m > 1.5 \). Taking the natural logarithm of both sides gives \( m > \frac{\ln(1.5)}{\ln(1.00375)} \approx 108.31 \). Since \( m \) must be an integer, it requires 109 complete months.

(a) M1 for substituting correctly into the compound interest formula, A1 for \( \$6258.98 \) (or \( \$6260 \)). (b) M1 for setting up the inequality or using a GDC table, A1 for finding \( m \approx 108.31 \), A1 for rounding up to the next complete month (109).

(a) Substitute \( x = 15 \) into the profit function: \( P(15) = -2(15)^2 + 120(15) - 800 = -450 + 1800 - 800 = 550 \). (b) The maximum profit occurs at the vertex of the parabola, where \( x = -\frac{b}{2a} = -\frac{120}{2(-2)} = 30 \). (c) The bakery makes a profit when \( P(x) > 0 \). Solving \( -2x^2 + 120x - 800 = 0 \) yields roots at \( x = 30 \pm 10\sqrt{5} \), which are approximately \( 7.64 \) and \( 52.4 \). Since \( x \) must be an integer, the range of cakes is \( 8 \le x \le 52 \).

(a) M1 for substituting 15 into the formula, A1 for 550. (b) M1 for using the vertex formula or differentiating and setting to 0, A1 for 30. (c) M1 for finding the roots of the quadratic equation, A1 for the interval \( 7.64 < x < 52.4 \) (or integer range \( 8 \le x \le 52 \)).

(a) Using Pythagoras' theorem on the base: \( AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ cm} \). (b) The distance from the center of the base to corner \( A \) is \( AO = \frac{1}{2} AC = 5 \text{ cm} \). Using Pythagoras' theorem on triangle \( VAO \): \( VA = \sqrt{AO^2 + VO^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \text{ cm} \). (c) The angle \( \theta \) between the edge \( VA \) and the base is \( \angle VAO \). Using trigonometry: \( \tan(\theta) = \frac{VO}{AO} = \frac{12}{5} = 2.4 \). Hence, \( \theta = \tan^{-1}(2.4) \approx 67.4^\circ \) (or 1.18 radians).

(a) M1 for Pythagoras on the base, A1 for 10. (b) M1 for dividing \( AC \) by 2, M1 for applying Pythagoras to find \( VA \), A1 for 13. (c) M1 for an appropriate trigonometric ratio, A1 for \( 67.4^\circ \) (accept \( 67.38^\circ \) or 1.18 radians).

(a) Let \( X \sim N(150, 12^2) \). Using a GDC to find \( P(X > 165) \) yields \( 0.105649... \), which rounds to 0.106 to 3 s.f. (b) We need to find the weight \( w \) such that \( P(X > w) = 0.15 \), which is equivalent to \( P(X \le w) = 0.85 \). Using the inverse normal cumulative function on a GDC, we get \( w \approx 162.437... \) grams. To the nearest gram, the minimum weight is 162 grams.

(a) M1 for setting up the normal distribution parameters or standardized Z-value, A1 for 0.106. (b) M1 for setting up the equation \( P(X > w) = 0.15 \) or \( P(X \le w) = 0.85 \), M1 for applying the inverse normal method, A1 for 162.

(a) Differentiating term by term: \( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - 6x + 8 \). (b) Set the derivative equal to zero to find stationary points: \( x^2 - 6x + 8 = 0 \implies (x-2)(x-4) = 0 \), so \( x = 2 \) or \( x = 4 \). Checking the second derivative: \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2x - 6 \). At \( x=4 \), \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0 \), confirming \( x=4 \) is a local minimum. The y-coordinate is \( y(4) = \frac{1}{3}(4)^3 - 3(4)^2 + 8(4) + 5 = \frac{64}{3} - 48 + 32 + 5 = \frac{31}{3} \approx 10.3 \). The coordinates are \( (4, 10.3) \). (c) At \( x=0 \), \( y = 5 \) and the gradient is \( m = \frac{\mathrm{d}y}{\mathrm{d}x}\big|_{x=0} = 8 \). The equation of the tangent is \( y = 8x + 5 \).

(a) M1 for power rule differentiation, A1 for \( x^2 - 6x + 8 \). (b) M1 for setting their derivative to 0, M1 for identifying \( x=4 \) as the local minimum, A1 for \( (4, 10.3) \) (or \( (4, \frac{31}{3}) \)). (c) M1 for finding the gradient at \( x=0 \), A1 for \( y = 8x + 5 \).

(a) The null hypothesis \( H_0 \) is: Gender and preferred sport are independent. (b) The row total for Male is 100, the column total for Football is 70, and the grand total is 200. Expected frequency \( = \frac{\text{Row Total} \times \text{Col Total}}{\text{Grand Total}} = \frac{100 \times 70}{200} = 35 \). (c) Entering the contingency table into a GDC and running a standard two-way chi-squared test gives a test statistic of \( \chi^2 \approx 9.5604 \) with 2 degrees of freedom. This yields a \( p \)-value of \( 0.00839 \). Since the \( p \)-value (0.00839) is less than the significance level (0.05), we reject the null hypothesis.

(a) A1 for stating that gender and preferred sport are independent. (b) M1 for showing the product of the totals divided by the grand total, A1 for confirming it equals 35. (c) M1 for finding the correct \( p \)-value of 0.00839, M1 for comparing with 0.05, A1 for rejecting the null hypothesis.

(a) The initial temperature is when \( t = 0 \): \( T(0) = 22 + 68 \mathrm{e}^{0} = 22 + 68 = 90^\circ\text{C} \). (b) For \( t = 15 \): \( T(15) = 22 + 68 \mathrm{e}^{-0.05 \times 15} = 22 + 68 \mathrm{e}^{-0.75} \approx 22 + 32.12 = 54.1^\circ\text{C} \). (c) Set \( T(t) = 40 \): \( 22 + 68 \mathrm{e}^{-0.05t} = 40 \implies 68 \mathrm{e}^{-0.05t} = 18 \implies \mathrm{e}^{-0.05t} = \frac{18}{68} \). Solving for \( t \): \( -0.05t = \ln\left(\frac{9}{34}\right) \implies t = -20 \ln\left(\frac{9}{34}\right) \approx 26.6 \) minutes.

(a) A1 for 90. (b) M1 for substituting \( t = 15 \), A1 for 54.1. (c) M1 for setting the equation to 40, M1 for algebraic work isolating the exponential term (or GDC equivalent), A1 for 26.6.

(a) The minimum height occurs when \( \cos(bt) = 1 \), giving \( h_{\min} = -15(1) + 17 = 2 \text{ m} \). The maximum height occurs when \( \cos(bt) = -1 \), giving \( h_{\max} = -15(-1) + 17 = 32 \text{ m} \). (b) Since the period \( T = 8 \) minutes, we have \( \frac{2\pi}{b} = 8 \implies b = \frac{2\pi}{8} = \frac{\pi}{4} \approx 0.785 \). (c) Substitute \( t = 3 \) and \( b = \frac{\pi}{4} \) into the equation: \( h(3) = -15 \cos\left(\frac{3\pi}{4}\right) + 17 \approx -15(-0.7071) + 17 = 10.61 + 17 = 27.6 \text{ m} \).

(a) A1 for maximum height of 32, A1 for minimum height of 2. (b) M1 for setting period equation \( \frac{2\pi}{b} = 8 \), A1 for \( 0.785 \) (or \( \frac{\pi}{4} \)). (c) M1 for substituting 3 into their model, A1 for 27.6.

(a) Calculate the difference in ranks \(d = A - B\):
- P: \(1 - 2 = -1 \Rightarrow d^2 = 1\)
- Q: \(3 - 1 = 2 \Rightarrow d^2 = 4\)
- R: \(2 - 4 = -2 \Rightarrow d^2 = 4\)
- S: \(5 - 3 = 2 \Rightarrow d^2 = 4\)
- T: \(4 - 6 = -2 \Rightarrow d^2 = 4\)
- U: \(6 - 5 = 1 \Rightarrow d^2 = 1\)

Sum of \(d^2 = 1 + 4 + 4 + 4 + 4 + 1 = 18\).

Using the formula with \(n = 6\):
\(r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6(18)}{6(35)} = 1 - \frac{18}{35} = \frac{17}{35} \approx 0.486\).

(b) This represents a moderate positive correlation, showing moderate agreement between the rankings of Arthur and Beatrice.

(c) Yes, because a correlation of \(0.82\) is closer to \(+1\) than \(0.486\), indicating a stronger positive association (closer to perfect agreement).

(a)
M1: Attempt to find differences \(d\) and calculate \(\sum d^2\).
A1: Correct value of \(\sum d^2 = 18\).
A1.15: Correct value of \(r_s = \frac{17}{35} \approx 0.486\).

(b)
R1: Moderate positive correlation / agreement.

(c)
R1: States 'Yes'.
R1: Valid reasoning comparing proximity to 1.

(a) Midpoint of \(AB = \left(\frac{1+5}{2}, \frac{2+2}{2}\right) = (3, 2)\).
Since \(A\) and \(B\) share the same \(y\)-coordinate, the line segment \(AB\) is horizontal.
Thus, the perpendicular bisector is a vertical line passing through \((3, 2)\), which has the equation \(x = 3\).

(b) Midpoint of \(AC = \left(\frac{1+3}{2}, \frac{2+6}{2}\right) = (2, 4)\).
Gradient of \(AC = \frac{6 - 2}{3 - 1} = 2\).
Gradient of the perpendicular bisector of \(AC = -\frac{1}{2}\).
Equation of the perpendicular bisector of \(AC\):
\(y - 4 = -\frac{1}{2}(x - 2) \Rightarrow y = -\frac{1}{2}x + 5\).

We now find the intersection of the two perpendicular bisectors by substituting \(x = 3\) into the equation:
\(y = -\frac{1}{2}(3) + 5 = 3.5\).

So, the coordinates of the circumcenter are \((3, 3.5)\).

(a)
M1: Identify the midpoint of \(AB\) as \((3, 2)\).
A1.15: State the correct vertical line equation \(x = 3\).

(b)
M1: Correctly find the midpoint \((2, 4)\) and gradient \(2\) of \(AC\).
M1: Identify perpendicular gradient as \(-\frac{1}{2}\).
A1: Formulate equation \(y = -\frac{1}{2}x + 5\).
A1: Find the intersection point \((3, 3.5)\).

(a) When \(t = 0\), \(T(0) = 85\).
\(20 + a e^0 = 85 \Rightarrow 20 + a = 85 \Rightarrow a = 65\).

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

(c) We want to find \(t\) when \(T(t) = 30\).
\(20 + 65 e^{-0.077319 t} = 30\)
\(65 e^{-0.077319 t} = 10\)
\(e^{-0.077319 t} = \frac{10}{65} = \frac{2}{13}\)
\(-0.077319 t = \ln\left(\frac{2}{13}\right) \approx -1.8718\)
\(t = \frac{-1.8718}{-0.077319} \approx 24.2\) minutes.

(a)
A1.15: Correct value \(a = 65\).

(b)
M1: Correct substitution of values into formula.
A1: Clear working showing logarithmic step leading to \(k \approx 0.0773\).

(c)
M1: Formulate the equation \(20 + 65 e^{-0.0773 t} = 30\).
M1: Attempt to solve for \(t\) using logarithms or GDC.
A1: Correct answer of \(24.2\) (accept \(24.2\) to \(24.3\) depending on rounding).

(a) The width of each interval is \(h = \frac{10 - 0}{5} = 2\) meters.
Using the trapezoidal rule:
\(\text{Area} \approx \frac{h}{2} [y_0 + y_5 + 2(y_1 + y_2 + y_3 + y_4)]\)
\(\text{Area} \approx \frac{2}{2} [0 + 0 + 2(1.2 + 1.8 + 2.1 + 1.5)]\)
\(\text{Area} \approx 1 \times [0 + 2(6.6)] = 13.2\text{ m}^2\).

(b) Volume flow rate per second: \(Q = \text{Area} \times \text{speed}\).
\(Q \approx 13.2 \times 0.4 = 5.28\text{ m}^3\text{ s}^{-1}\).
In 1 minute (60 seconds):
\(V = 5.28 \times 60 = 316.8 \approx 317\text{ m}^3\) (to 3 significant figures).

(a)
M1: Identify interval width \(h = 2\).
M1: Correct substitution into trapezoidal rule formula.
A1.15: Correct answer of \(13.2\) (\(\text{m}^2\)).

(b)
M1: Identify volume rate formula \(\text{Area} \times \text{speed}\).
M1: Multiply rate by 60 seconds.
A1: Correct answer of \(317\) (accept \(316.8\)).

(a) Compound interest formula:
\(FV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}\)
With \(PV = 5000\), \(r = 4.2\), \(k = 12\), \(n = 5\):
\(FV = 5000 \times \left(1 + \frac{4.2}{1200}\right)^{60} = 5000 \times (1.0035)^{60} \approx 6166.38\).
To the nearest dollar, the amount is \(\$6166\).

(b) Continuous compounding formula:
\(FV = PV \times e^{rt}\)
For doubling, \(2PV = PV \times e^{0.045t} \Rightarrow e^{0.045t} = 2\).
Taking natural logarithms:
\(0.045t = \ln(2)\)
\(t = \frac{\ln(2)}{0.045} \approx 15.403...\) years.
To 3 significant figures, this is \(15.4\) years.

(a)
M1: Correct substitution into the compound interest formula.
A1.15: \(6166\) (accept \(6170\) if rounded to 3 s.f.).

(b)
M1: Formulate equation \(e^{0.045t} = 2\).
M1: Attempt to solve using natural logs or GDC.
A1: \(15.403...\)
A1: \(15.4\) (years).

(a) At \(t = 0\), \(P(0) = a \cdot e^0 \cos(0) + b = a + b = 1200\). At \(t = 6\), \(P(6) = a \cdot e^{-0.3} \cos(\pi) + b = -a e^{-0.3} + b = 590.71\). Subtracting the second equation from the first: \(a(1 + e^{-0.3}) = 1200 - 590.71 = 609.29\). Since \(1 + e^{-0.3} \approx 1.740818\), we find \(a = \frac{609.29}{1.740818} \approx 350.00\). Thus \(a = 350\) and \(b = 1200 - 350 = 850\). (b) At \(t = 3\), \(P(3) = 350 e^{-0.15} \cos(\pi/2) + 850\). Since \(\cos(\pi/2) = 0\), \(P(3) = 850\). (c) We find \(P'(10)\) using the GDC derivative solver. \(P'(t) = 350 e^{-0.05t} [-0.05\cos(\frac{\pi}{6}t) - \frac{\pi}{6}\sin(\frac{\pi}{6}t)]\). At \(t = 10\), \(P'(10) \approx 91.0\) birds per month. (d) Graphing \(P(t)\) on the interval \([0, 24]\), we see that the function decreases initially since \(P'(0) = -17.5 < 0\). The subsequent peak is at \(t \approx 12\), where \(P(12) \approx 1042\). Therefore, the global maximum occurs at the boundary \(t = 0\) with a population of 1200. (e) We solve \(350 e^{-0.05t} \cos\left(\frac{\pi}{6} t\right) + 850 < 800\) using GDC. The intersection points with \(P(t) = 800\) are at \(t_1 \approx 3.32\), \(t_2 \approx 8.64\), \(t_3 \approx 15.60\), and \(t_4 \approx 20.23\). The population is below 800 in the intervals \([3.32, 8.64]\) and \([15.60, 20.23]\). Total duration = \((8.64 - 3.32) + (20.23 - 15.60) = 5.32 + 4.63 = 9.95\) months.

(a) M1 for setting up \(a + b = 1200\), M1 for setting up \(-a e^{-0.3} + b = 591\), A1 for solving to get \(a \approx 350\), A1 for concluding \(b = 850\). (b) M1 for substituting \(t = 3\) into the formula, A1 for \(850\). (c) M1 for attempting to find the derivative, A2 for \(91.0\) (accept 91). (d) M1 for evaluating at boundary \(t=0\), M1 for checking other local extrema, A1 for identifying maximum population of 1200 at \(t=0\). (e) M1 for setting up inequality or equation \(P(t) = 800\), A1 for finding the first set of roots \(3.32\) and \(8.64\), A1 for finding the second set of roots \(15.60\) and \(20.23\), A1 for final answer \(9.95\) (accept answers between \(9.9\) and \(10.0\)).

(a) Let \(W \sim N(100, 4^2)\). (i) \(P(W < 95) = \text{normCDF}(-\infty, 95, 100, 4) \approx 0.1056\). (ii) \(P(98 < W < 104) = \text{normCDF}(98, 104, 100, 4) \approx 0.5328\). (b) We solve \(P(W < k) = 0.025\). Using inverse normal: \(k = \text{invNorm}(0.025, 100, 4) \approx 92.16 \approx 92.2\) grams. (c) (i) Let \(p\) be the probability that a bar weighs more than 98 grams: \(p = P(W > 98) = \text{normCDF}(98, \infty, 100, 4) \approx 0.69146 \approx 0.691\). (ii) Let \(X\) be the number of bars weighing more than 98 grams in a box of 12. \(X \sim B(12, 0.69146)\). A box is premium if \(X \ge 10\). \(P(X \ge 10) = 1 - P(X \le 9) = 1 - \text{binomCDF}(12, 0.69146, 9) \approx 1 - 0.76865 = 0.23135 \approx 0.231\). (d) Let \(Y\) be the number of premium boxes out of 5. \(Y \sim B(5, 0.23135)\). We want \(P(Y = 2) = \text{binomPDF}(5, 0.23135, 2) \approx 0.2428 \approx 0.243\). (e) This is a conditional probability: \(P(X = 9 \mid X < 10) = \frac{P(X = 9)}{P(X < 10)}\). We have \(P(X = 9) = \text{binomPDF}(12, 0.69146, 9) \approx 0.22227\), and \(P(X < 10) = P(X \le 9) \approx 0.76865\). Thus, \(P(X = 9 \mid X < 10) = \frac{0.22227}{0.76865} \approx 0.2891 \approx 0.289\).

(a) (i) M1 for setup using normal model, A1 for \(0.106\). (ii) A1 for \(0.533\). (b) M1 for inverse normal setup, A1 for \(92.2\). (c) (i) A1 for \(0.691\). (ii) M1 for binomial model setup with \(n=12\), M1 for evaluating \(P(X \ge 10)\), A1 for \(0.231\). (d) M1 for binomial setup with \(n=5\) and their premium probability, A1 for formula or GDC entry, A1 for \(0.243\). (e) M1 for conditional probability formula, A1 for \(0.289\).

(a) Midpoint of \(AB\) is \(M = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). Gradient of \(AB\) is \(m = \frac{10-8}{8-2} = \frac{1}{3}\). The gradient of the perpendicular bisector is \(-3\). The equation is \(y - 9 = -3(x - 5) \implies 3x + y = 24\). (b) We solve the system of equations: \(3x + y = 24\) and \(x - 4y = -15\). Substituting \(x = 4y - 15\) into the first equation: \(3(4y - 15) + y = 24 \implies 13y = 69 \implies y = 5.31\) and \(x = 6.23\). So \(V_1 = (6.23, 5.31)\). (c) (i) Sketch should show points \(A, B, C, D\), vertices \(V_1, V_2, V_3\), and lines connecting the vertices to form the boundaries of the Voronoi cells. (ii) By calculating distances from \(P(6, 7)\) to each station: \(d(P, A) = \sqrt{4^2 + (-1)^2} = \sqrt{17} \approx 4.12\), \(d(P, B) = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13} \approx 3.61\), \(d(P, C) = \sqrt{(-4)^2 + 5^2} = \sqrt{41} \approx 6.40\), \(d(P, D) = \sqrt{2^2 + 6^2} = \sqrt{40} \approx 6.32\). Since \(d(P, B)\) is the smallest, Station B is closest to the hiker. (d) (i) Distance from \(V_1(6.23, 5.31)\) to \(A(2, 8)\) is \(d = \sqrt{(6.23-2)^2 + (5.31-8)^2} = \sqrt{4.23^2 + (-2.69)^2} \approx 5.01\) km. (ii) A Voronoi vertex is equidistant from the three nearest sites and represents the point furthest from any station in that region (the largest empty circle center). Placed at \(V_1\), the watchtower is optimally located to monitor the region most remote from the existing ranger stations.

(a) M1 for finding midpoint \((5, 9)\), M1 for perpendicular gradient \(-3\), A1 for showing \(3x + y = 24\). (b) M1 for attempting to solve the simultaneous equations, A1 for \((6.23, 5.31)\) (accept fractional coordinates \((81/13, 69/13)\)). (c) (i) A1 for plotting stations and vertices correctly, A1 for drawing boundaries, A1 for correct labelling. (ii) M1 for attempting distance calculations or checking the position of \(P\) relative to boundary equations, A1 for Station B. (d) (i) M1 for using the distance formula, A1 for substituting coordinates, A1 for \(5.01\) km (accept \(5.01 - 5.02\)). (ii) R1 for stating \(V_1\) is equidistant to \(A, B, C\), R1 for explaining it is the point furthest away from any station, R1 for connecting to the purpose of monitoring remote area.

(a) Volume \(V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2 h\). Since \(V = 36\), \(2x^2 h = 36 \implies h = \frac{18}{x^2}\). (b) The open box has 1 base and 4 sides. \(A = \text{Area of base} + 2(\text{width} \times h) + 2(\text{length} \times h) = 2x^2 + 2xh + 2(2x)h = 2x^2 + 6xh\). Substituting \(h = \frac{18}{x^2}\): \(A(x) = 2x^2 + 6x\left(\frac{18}{x^2}\right) = 2x^2 + \frac{108}{x}\). (c) \(A'(x) = 4x - 108x^{-2} = 4x - \frac{108}{x^2}\). (d) Set \(A'(x) = 0 \implies 4x - \frac{108}{x^2} = 0 \implies 4x^3 = 108 \implies x^3 = 27 \implies x = 3\) meters. (e) Minimum surface area \(A(3) = 2(3)^2 + \frac{108}{3} = 18 + 36 = 54 \text{ m}^2\). (f) Let \(C(x)\) be the cost: \(C(x) = 15 \times (2x^2) + 10 \times \left(\frac{108}{x}\right) = 30x^2 + \frac{1080}{x}\). Finding the derivative: \(C'(x) = 60x - \frac{1080}{x^2}\). Set \(C'(x) = 0 \implies 60x^3 = 1080 \implies x^3 = 18 \implies x = \sqrt[3]{18} \approx 2.62\) m. The dimensions are: width \(x \approx 2.62\) m, length \(2x \approx 5.24\) m, and height \(h = \frac{18}{x^2} \approx \frac{18}{18^{2/3}} = 18^{1/3} \approx 2.62\) m.

(a) M1 for formula \(V = 2x^2 h\), A1 for showing \(h = 18/x^2\). (b) M1 for identifying correct 5 faces, A1 for \(2x^2 + 6xh\), M1 for substituting \(h\), A1 for showing the final expression. (c) A2 for \(4x - \frac{108}{x^2}\) (M1 for one correct term). (d) M1 for setting derivative to 0, A1 for \(x^3 = 27\), A1 for \(x = 3\) m. (e) M1 for substituting \(x = 3\) into \(A(x)\), A1 for \(54\). (f) M1 for writing cost function \(C(x) = 30x^2 + 1080/x\), M1 for finding \(C'(x)\) and setting to 0, A1 for dimensions: width = \(2.62\) m, length = \(5.24\) m, height = \(2.62\) m.

(a) (i) Using GDC TVM Solver: \(N = 60\), \(I\% = 4.2\), \(PV = 0\), \(PMT = -800\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(FV\) yields \(\$53,263.53\). (ii) Total deposited = \(800 \times 60 = \$48,000\). Interest = \(53,263.53 - 48,000 = \$5,263.53\). (b) (i) Loan amount \(PV = 350,000 - 110,000 = \$240,000\). Using TVM Solver: \(N = 240\), \(I\% = 5.4\), \(PV = 240000\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(PMT\) yields \(\$1,637.59\). (ii) Total payments = \(1,637.59 \times 240 = \$393,021.60\). (c) (i) Remaining term is \(20 - 7 = 13\) years, which is \(156\) payments. Using TVM Solver: \(N = 156\), \(I\% = 5.4\), \(PMT = -1637.59\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(PV\) (outstanding balance) yields \(\$183,329.80\) (or \(\$183,329.13\) using unrounded payment). (ii) Total paid over 7 years = \(84 \times 1,637.59 = \$137,557.56\). Plus the lump sum: \(137,557.56 + 183,329.80 = \$320,887.36\). Total paid under original plan = \(\$393,021.60\). Interest saved = \(393,021.60 - 320,887.36 = \$72,134.24\).

(a) (i) M1 for GDC TVM setup with correct inputs, A1 for \(\$53,263.53\). (ii) M1 for subtracting total deposits, A1 for \(\$5,263.53\). (b) (i) M1 for identifying loan amount \(\$240,000\), M1 for GDC mortgage solver entry, A1 for \(\$1,637.59\). (ii) M1 for multiplying payment by 240, A1 for \(\$393,021.60\). (c) (i) M1 for finding remaining balance using TVM Solver with \(N=156\) (or finding FV after \(N=84\)), A2 for \(\$183,329.80\) (accept \(\$183,329\) to \(\$183,330\)). (ii) M1 for calculating total amount paid under early payoff, M1 for comparing to total payments of full term, A1 for \(\$72,134.24\) (accept surrounding values due to rounding, e.g. \(\$72,135\)).