An original Thinka practice paper modelled on the structure and difficulty of the May 2023 SL (TZ1) IB Diploma Programme Mathematics - Applications and Interpretation paper. Not affiliated with or reproduced from IB.
Paper 1
Answer all questions in the boxes provided. A GDC is required. Numerical answers should be given exactly or to three significant figures unless specified otherwise.
13 PastPaper.question · 79.95 PastPaper.marks
PastPaper.question 1 · Short Answer
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The time \(T\) in minutes to assemble a toy is normally distributed with mean \(\mu = 14.2\) and standard deviation \(\sigma = 2.5\). (a) Find the probability that a randomly chosen toy takes less than 12 minutes to assemble. (b) Given that a toy takes more than 12 minutes to assemble, find the probability that it takes less than 15 minutes. (c) A batch of 500 toys is assembled. Find the expected number of toys that take between 12 and 15 minutes to assemble.
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PastPaper.workedSolution
(a) Using the normal cumulative distribution function on a GDC with mean \(\mu = 14.2\) and standard deviation \(\sigma = 2.5\), we find \(P(T < 12) \approx 0.189429\). To three significant figures, this is 0.189. (b) We want to find the conditional probability \(P(T < 15 \mid T > 12) = \frac{P(12 < T < 15)}{P(T > 12)}\). First, we find \(P(12 < T < 15) = \text{normalCDF}(12, 15, 14.2, 2.5) \approx 0.436087\). Then, we find \(P(T > 12) = 1 - P(T < 12) \approx 1 - 0.189429 = 0.810571\). Thus, the conditional probability is \(\frac{0.436087}{0.810571} \approx 0.537999\), which is 0.538 to three significant figures. (c) The expected number of toys is \(500 \times P(12 < T < 15) = 500 \times 0.436087 \approx 218.043\), which is 218 to the nearest integer.
PastPaper.markingScheme
(a) [2 marks] M1 for setting up normal cumulative distribution on GDC, A1 for 0.189. (b) [3 marks] M1 for identifying the conditional probability formula, M1 for calculating both individual probabilities, A1 for 0.538. (c) [1.15 marks] M1 for multiplying 500 by the probability from part (b), A0.15 for 218.
PastPaper.question 2 · Short Answer
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A rectangular garden is built against a long stone wall, so only three sides need fencing. The homeowner has 80 meters of fencing material and wants to partition the garden into two equal parts with an additional fence parallel to the shorter side. Let the length of the shorter side (which is partitioned) be \(x\) meters. (a) Write down an expression for the area \(A\) of the garden in terms of \(x\). (b) Find \(\frac{\mathrm{d}A}{\mathrm{d}x}\). (c) Find the value of \(x\) that maximizes the area of the garden. (d) Calculate the maximum possible area of the garden.
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PastPaper.workedSolution
(a) Let the length of the shorter side be \(x\). The three parallel fences of length \(x\) and the one long side of length \(y\) use a total of 80 meters of fencing, so \(3x + y = 80 \implies y = 80 - 3x\). The area is \(A(x) = x \cdot y = x(80 - 3x) = 80x - 3x^2\). (b) Differentiating \(A(x)\) with respect to \(x\) gives \(\frac{\mathrm{d}A}{\mathrm{d}x} = 80 - 6x\). (c) For a maximum area, set \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0 \implies 80 - 6x = 0 \implies x = \frac{80}{6} = \frac{40}{3} \approx 13.3\) meters. (d) The maximum area is \(A\left(\frac{40}{3}\right) = \frac{40}{3}\left(80 - 3\left(\frac{40}{3}\right)\right) = \frac{40}{3}(40) = \frac{1600}{3} \approx 533\) square meters.
PastPaper.markingScheme
(a) [1.15 marks] M1 for establishing \(y = 80 - 3x\), A0.15 for \(A(x) = 80x - 3x^2\). (b) [1 mark] A1 for \(80 - 6x\). (c) [2 marks] M1 for setting their derivative to 0, A1 for \(13.3\) (or \(\frac{40}{3}\)). (d) [2 marks] M1 for substituting their value of \(x\) into their area function, A1 for \(533\) (or \(\frac{1600}{3}\)).
PastPaper.question 3 · Short Answer
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The population of wild rabbits on an island is modeled by the function \(P(t) = \frac{1200}{1 + 5\mathrm{e}^{-0.15t}}\), where \(t \geq 0\) is the time in years since the population was first recorded. (a) Find the initial population of the rabbits. (b) Find the population of rabbits after 10 years. (c) Find the time \(t\) when the population reaches 1000 rabbits.
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PastPaper.workedSolution
(a) At \(t = 0\), the initial population is \(P(0) = \frac{1200}{1 + 5\mathrm{e}^0} = \frac{1200}{6} = 200\). (b) At \(t = 10\), the population is \(P(10) = \frac{1200}{1 + 5\mathrm{e}^{-1.5}} \approx \frac{1200}{2.11565} \approx 567.199\). To the nearest integer, there are 567 rabbits. (c) Set \(P(t) = 1000 \implies 1000 = \frac{1200}{1 + 5\mathrm{e}^{-0.15t}} \implies 1 + 5\mathrm{e}^{-0.15t} = 1.2 \implies 5\mathrm{e}^{-0.15t} = 0.2 \implies \mathrm{e}^{-0.15t} = 0.04\). Taking the natural logarithm of both sides gives \(-0.15t = \ln(0.04) \implies t = \frac{\ln(0.04)}{-0.15} \approx 21.459\) years. To three significant figures, \(t = 21.5\) years.
PastPaper.markingScheme
(a) [2 marks] M1 for substituting \(t=0\) into the model, A1 for 200. (b) [2 marks] M1 for substituting \(t=10\) into the model, A1 for 567 (accept 567.199). (c) [2.15 marks] M1 for setting up the equation \(P(t) = 1000\), M1 for solving using logarithms or GDC solver, A0.15 for 21.5.
PastPaper.question 4 · Short Answer
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A study was conducted to investigate if there is an association between a person's preferred exercise type (Cardio, Strength, Yoga) and their age group (Under 30, 30 and over). A random sample of 250 participants was surveyed, and the observed frequencies are: Under 30: 50 Cardio, 45 Strength, 15 Yoga; 30 and over: 40 Cardio, 35 Strength, 65 Yoga. (a) State the null hypothesis \(H_0\) for this test. (b) Show that the expected number of participants under 30 who prefer Yoga is 35.2. (c) Find the \(p\)-value for this test. (d) State the conclusion of the test at the 5% significance level. Give a reason for your answer.
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PastPaper.workedSolution
(a) The null hypothesis \(H_0\) is that preferred exercise type and age group are independent. (b) The total number of participants under 30 is \(50 + 45 + 15 = 110\). The total number of participants who prefer Yoga is \(15 + 65 = 80\). The grand total is 250. The expected frequency is \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{110 \times 80}{250} = 35.2\). (c) Using a chi-squared test on a GDC with the observed 2x3 table, we find the \(p\)-value is \(2.20 \times 10^{-7}\) (or \(2.195 \times 10^{-7}\)). (d) Since the \(p\)-value \((2.20 \times 10^{-7})\) is less than the significance level of 0.05, we reject the null hypothesis. There is a significant association between exercise preference and age group.
PastPaper.markingScheme
(a) [1 mark] A1 for stating that preferred exercise type and age group are independent. (b) [1.15 marks] M1 for identifying row total (110) and column total (80), A0.15 for showing the calculation resulting in 35.2. (c) [2 marks] M1 for setting up the matrix in GDC, A1 for \(2.20 \times 10^{-7}\). (d) [2 marks] R1 for comparing \(p\)-value to 0.05, A1 for concluding to reject \(H_0\).
PastPaper.question 5 · Short Answer
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A delivery company has three hubs located at coordinates \(A(2, 8)\), \(B(8, 10)\), and \(C(6, 4)\) on a grid where 1 unit represents 1 km. (a) Find the equation of the perpendicular bisector of the line segment \(AB\). Write your answer in the form \(y = mx + c\). (b) The equation of the perpendicular bisector of the line segment \(AC\) is \(x - y = -2\). Find the coordinates of the circumcenter of triangle \(ABC\), which represents the vertex where the cells of \(A\), \(B\), and \(C\) meet in the Voronoi diagram. (c) A new parcel needs to be delivered to a customer located at \(P(4, 7)\). Determine which hub is closest to this customer.
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PastPaper.workedSolution
(a) The midpoint of \(AB\) is \(\left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). The gradient of \(AB\) is \(\frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\). The gradient of the perpendicular bisector is the negative reciprocal, \(-3\). The equation is \(y - 9 = -3(x - 5) \implies y = -3x + 24\). (b) The circumcenter is the intersection of \(y = -3x + 24\) and \(x - y = -2\). Substituting \(y = x + 2\) into the first equation: \(x + 2 = -3x + 24 \implies 4x = 22 \implies x = 5.5\). Then \(y = 5.5 + 2 = 7.5\). The coordinates of the circumcenter are \((5.5, 7.5)\). (c) The distance from \(P(4, 7)\) to \(A(2, 8)\) is \(\sqrt{(4-2)^2 + (7-8)^2} = \sqrt{5} \approx 2.24\). The distance to \(B(8, 10)\) is \(\sqrt{(4-8)^2 + (7-10)^2} = 5\). The distance to \(C(6, 4)\) is \(\sqrt{(4-6)^2 + (7-4)^2} = \sqrt{13} \approx 3.61\). Since Hub A is the closest, the customer lies within the cell of Hub A.
PastPaper.markingScheme
(a) [3 marks] M1 for finding midpoint (5, 9), M1 for finding perpendicular gradient \(-3\), A1 for \(y = -3x + 24\). (b) [2.15 marks] M1 for setting up a system of equations, M1 for solving the system, A0.15 for \((5.5, 7.5)\). (c) [1 mark] A1 for concluding Hub A based on distances or cell location.
PastPaper.question 6 · Short Answer
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Water is leaking from a tank. The rate of leakage, \(R(t)\), in liters per hour, is modeled by the function \(R(t) = 15 - 15\mathrm{e}^{-0.2t}\), where \(t \geq 0\) is the time in hours since the leak started. (a) Find the rate of leakage after 5 hours. (b) Find the total volume of water that leaked from the tank during the first 5 hours. (c) Write down the value that the rate of leakage approaches as \(t\) becomes very large.
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PastPaper.workedSolution
(a) Substituting \(t = 5\) into the rate function: \(R(5) = 15 - 15\mathrm{e}^{-0.2 \times 5} = 15 - 15\mathrm{e}^{-1} \approx 9.48\) liters per hour. (b) The total volume is given by \(\int_0^5 R(t) \,\mathrm{d}t = \int_0^5 (15 - 15\mathrm{e}^{-0.2t}) \,\mathrm{d}t = \left[ 15t + 75\mathrm{e}^{-0.2t} \right]_0^5 = (75 + 75\mathrm{e}^{-1}) - (0 + 75) = 75\mathrm{e}^{-1} \approx 27.6\) liters. (c) As \(t \to \infty\), \(\mathrm{e}^{-0.2t} \to 0\), so \(R(t) \to 15\) liters per hour.
PastPaper.markingScheme
(a) [2 marks] M1 for substituting \(t=5\) into \(R(t)\), A1 for 9.48. (b) [3.15 marks] M1 for expressing total volume as a definite integral from 0 to 5, M1 for integration resulting in \(15t + 75\mathrm{e}^{-0.2t}\), M1 for applying limits, A0.15 for 27.6. (c) [1 mark] A1 for 15.
PastPaper.question 7 · Short Answer
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A ball is thrown vertically upwards from the top of a cliff. Its height, \(h\), in meters above sea level, \(t\) seconds after being thrown is modeled by the quadratic function: \(h(t) = -4.9t^2 + 19.6t + 58.8\). (a) State the height of the cliff. (b) Find the maximum height reached by the ball. (c) Find the time it takes for the ball to hit the sea.
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PastPaper.workedSolution
(a) At \(t = 0\), the height is \(h(0) = 58.8\) meters, which is the height of the cliff. (b) The maximum height occurs at the vertex of the parabola, where \(t = -\frac{19.6}{2(-4.9)} = 2\) seconds. Substituting \(t = 2\) into the equation gives \(h(2) = -4.9(2)^2 + 19.6(2) + 58.8 = -19.6 + 39.2 + 58.8 = 78.4\) meters. (c) The ball hits the sea when \(h(t) = 0\), so \(-4.9t^2 + 19.6t + 58.8 = 0\). Dividing by \(-4.9\) gives \(t^2 - 4t - 12 = 0 \implies (t-6)(t+2) = 0\). Since \(t \geq 0\), the solution is \(t = 6\) seconds.
PastPaper.markingScheme
(a) [1 mark] A1 for 58.8. (b) [2.15 marks] M1 for finding the time of maximum height (\(t = 2\)), M1 for calculating the maximum height, A0.15 for 78.4. (c) [3 marks] M1 for setting \(h(t) = 0\), M1 for factorizing or using quadratic formula, A1 for 6.
PastPaper.question 8 · Short Answer
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A researcher investigates the relationship between the daily average temperature, \(x\) (°C), and the number of cups of hot coffee sold, \(y\), at a local cafe. The data from 8 randomly selected days is: \(x\): 12, 15, 8, 20, 5, 18, 10, 14; \(y\): 140, 110, 180, 75, 210, 95, 150, 125. (a) Find the Pearson's product-moment correlation coefficient, \(r\), for this data. (b) Write down the equation of the regression line \(y\) on \(x\) in the form \(y = ax + b\). (c) Use your regression equation to estimate the number of cups of hot coffee sold on a day when the average temperature is 13 °C.
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PastPaper.workedSolution
(a) Entering the data into a GDC, the Pearson's product-moment correlation coefficient is \(r \approx -0.99197\), which rounds to -0.992. (b) Using linear regression on the GDC, the slope is \(a \approx -8.7535\) and the y-intercept is \(b \approx 247.23\). The equation of the regression line is \(y = -8.75x + 247\) (to three significant figures). (c) Substituting \(x = 13\) into the regression line gives \(y = -8.7535(13) + 247.23 = 133.43\). Rounding to the nearest whole cup gives 133 cups.
PastPaper.markingScheme
(a) [2 marks] M1 for entering data into GDC, A1 for \(-0.992\). (b) [2.15 marks] M1 for regression calculation on GDC, A1 for \(a = -8.75\), A0.15 for \(b = 247\). (c) [2 marks] M1 for substituting 13 into their equation, A1 for 133.
PastPaper.question 9 · Short Answer
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A survey was conducted to investigate if there is an association between a person's preferred choice of hot beverage (Coffee, Tea, Hot Chocolate) and their age group (Under 30, 30 and over).
The results of the survey of 220 people are shown in the table below:
| Age Group | Coffee | Tea | Hot Chocolate | | :--- | :---: | :---: | :---: | | Under 30 | 45 | 30 | 25 | | 30 and over | 55 | 50 | 15 |
A \(\chi^2\) independence test is conducted at the 5% significance level.
(a) State the null hypothesis, \(H_0\).
(b) Calculate the expected number of people under 30 who prefer hot chocolate.
(c) Find the \(p\)-value for this test.
(d) State the conclusion of the test. Give a reason for your answer.
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PastPaper.workedSolution
(a) \(H_0\): Preferred choice of hot beverage and age group are independent (or there is no association between preferred choice of hot beverage and age group).
(c) From GDC, the \(p\)-value is \(0.0344\) (3 s.f.) (or \(0.034418...\))
(d) Since \(0.0344 < 0.05\) (or \(p\text{-value} < 0.05\)), we reject the null hypothesis \(H_0\). There is sufficient evidence to suggest that preferred choice of hot beverage and age group are not independent.
PastPaper.markingScheme
(a) M1 for stating that the preferred beverage and age group are independent (do not accept 'not related' or 'no correlation'). [1 mark]
(b) M1 for setting up the calculation: \(\frac{100 \times 40}{220}\). A1 for \(18.2\) (or \(18.18...\)). [2 marks]
(c) A1.15 for \(0.0344\). [1.15 marks]
(d) R1 for comparing the \(p\)-value to \(0.05\) (e.g., \(0.0344 < 0.05\)). A1 for rejecting the null hypothesis (or concluding they are not independent). [2 marks]
PastPaper.question 10 · Short Answer
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The concentration of a drug in a patient's bloodstream, \(C(t)\) in mg/L, \(t\) hours after injection, is modeled by:
\[C(t) = 15t e^{-0.4t}, \quad t \ge 0\]
(a) Find the concentration of the drug in the patient's bloodstream after 2 hours.
(b) Find the rate of change of the concentration of the drug with respect to time after 2 hours.
(c) Find the time \(t\) at which the concentration of the drug is at its maximum.
(b) Using the product rule to differentiate \(C(t)\): \(C'(t) = 15 e^{-0.4t} + 15t (-0.4) e^{-0.4t} = (15 - 6t) e^{-0.4t}\) Substitute \(t = 2\): \(C'(2) = (15 - 12) e^{-0.8} = 3 e^{-0.8} \approx 1.35\) mg/L/hour (1.34798...) (Note: GDC numerical derivative can also be used directly)
(c) Set \(C'(t) = 0\): \(15 - 6t = 0 \implies t = 2.5\) hours.
PastPaper.markingScheme
(a) M1 for substituting \(t = 2\) into the function. A1.15 for \(13.5\) (accept \(30e^{-0.8}\)). [1.15 marks]
(b) M1 for attempt to use product rule or chain rule (or finding derivative numerically). A1 for \(C'(t) = (15 - 6t) e^{-0.4t}\) or equivalent. A1 for substituting \(t = 2\) to get \(1.35\) (accept \(3e^{-0.8}\)). [3 marks]
(c) M1 for setting their derivative equal to 0 (or finding the maximum of the curve on GDC). A1 for \(t = 2.5\). [2 marks]
PastPaper.question 11 · Short Answer
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A population of a certain species of birds in a nature reserve is modeled by the function:
\[P(t) = \frac{800}{1 + A e^{-kt}}\]
where \(t\) is the number of years since the start of 2010.
At the start of 2010 (\(t = 0\)), the population was 200.
(a) Show that \(A = 3\).
At the start of 2020 (\(t = 10\)), the population was 500.
(b) Find the value of \(k\), giving your answer to four decimal places.
(c) According to this model, find the limiting population size of the birds as \(t\) becomes very large.
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PastPaper.workedSolution
(a) At \(t = 0\), \(P(0) = 200\): \(\frac{800}{1 + A e^0} = 200 \implies \frac{800}{1 + A} = 200\) \(1 + A = 4 \implies A = 3\).
(c) As \(t \to \infty\), \(e^{-kt} \to 0\) since \(k > 0\). Therefore, \(P(t) \to \frac{800}{1 + 0} = 800\). The limiting population size is 800.
PastPaper.markingScheme
(a) M1 for substituting \(t = 0\) and \(P = 200\) into the equation. A1.15 for correctly showing that \(A = 3\). [1.15 marks]
(b) M1 for substituting \(t = 10, P = 500\) and \(A = 3\) into the equation. M1 for an attempt to solve for \(k\) (algebraically or using GDC). A1 for \(k = 0.1609\). [3 marks]
(c) M1 for analyzing the limit as \(t \to \infty\) (or evaluating for a very large value of \(t\)). A1 for \(800\). [2 marks]
PastPaper.question 12 · Short Answer
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Consider three sites on a coordinate plane representing the locations of communication towers: \(A(2, 8)\), \(B(8, 6)\), and \(C(4, 2)\). A Voronoi diagram is constructed to divide the plane into regions closest to each tower.
(a) Find the equation of the boundary line between the cells of site \(A\) and site \(B\). Give your answer in the form \(y = mx + c\).
(b) Find the coordinates of the vertex \(V(x, y)\) where the boundaries of all three cells meet.
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PastPaper.workedSolution
(a) The boundary line between \(A\) and \(B\) is the perpendicular bisector of the line segment \(AB\). Midpoint of \(AB\) is: \(M_{AB} = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7)\) Gradient of \(AB\) is: \(m_{AB} = \frac{6-8}{8-2} = -\frac{2}{6} = -\frac{1}{3}\) Gradient of the perpendicular bisector is: \(m_{\perp} = -\frac{1}{m_{AB}} = 3\) Equation of the perpendicular bisector: \(y - 7 = 3(x - 5) \implies y = 3x - 8\)
(b) To find the vertex \(V\), we find the equation of another boundary line, e.g., the perpendicular bisector of \(BC\). Midpoint of \(BC\) is: \(M_{BC} = \left(\frac{8+4}{2}, \frac{6+2}{2}\right) = (6, 4)\) Gradient of \(BC\) is: \(m_{BC} = \frac{2-6}{4-8} = 1\) Gradient of the perpendicular bisector of \(BC\) is: \(m_{\perp} = -1\) Equation of this perpendicular bisector: \(y - 4 = -1(x - 6) \implies y = -x + 10\)
Solving the system of equations for the two boundary lines: \(3x - 8 = -x + 10\) \(4x = 18 \implies x = 4.5\) Substitute \(x = 4.5\) into \(y = -x + 10\): \(y = -4.5 + 10 = 5.5\) Therefore, the coordinates of \(V\) are \((4.5, 5.5)\).
PastPaper.markingScheme
(a) M1 for finding the midpoint of \(AB\): \((5, 7)\). M1 for finding the gradient of \(AB\) as \(-\frac{1}{3}\) and identifying the perpendicular gradient as \(3\). A1 for the correct equation \(y = 3x - 8\). [3 marks]
(b) M1 for finding the midpoint of \(BC\) (or \(AC\)) and its gradient. M1 for finding the equation of the perpendicular bisector of \(BC\) (or \(AC\)) (e.g., \(y = -x + 10\) or \(y = \frac{1}{3}x + 4\)). A1.15 for solving the two equations simultaneously to find the coordinates \((4.5, 5.5)\). [3.15 marks]
PastPaper.question 13 · Short Answer
6.15 PastPaper.marks
Elena invests $5000 in a savings account that offers a nominal annual interest rate of 4.5%, compounded monthly.
(a) Find the value of her investment after 5 years, assuming no further deposits or withdrawals are made.
At the end of the 5 years, Elena decides to withdraw the total amount and invest it in another account that pays an annual interest rate of \(r\\%\) compounded half-yearly. She wants the investment to grow to $8000 in a further 6 years.
(b) Find the value of \(r\).
(c) Find the equivalent effective annual interest rate for this second account.
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PastPaper.workedSolution
(a) Using the compound interest formula: \(FV = 5000 \times \left(1 + \frac{4.5}{100 \times 12}\right)^{5 \times 12} = 5000 \times (1.00375)^{60} \approx 6258.98\) USD (or $6260 to 3 s.f.)
(b) Elena's new principal is \(PV = 6258.98\). The investment grows to $8000 in 6 years, compounded half-yearly. Using the compound interest formula: \(8000 = 6258.98 \times \left(1 + \frac{r}{200}\right)^{12}\) \(\left(1 + \frac{r}{200}\right)^{12} = \frac{8000}{6258.98} \approx 1.278164\) \(1 + \frac{r}{200} = (1.278164)^{1/12} \approx 1.020664\) \(\frac{r}{200} = 0.020664 \implies r \approx 4.13\\%\) (or \(4.1328...\\%\)) (Values from GDC TVM Solver: \(N=12\), \(I\\%=4.1328...\), \(PV=-6258.98\), \(FV=8000\), \(P/Y=2\), \(C/Y=2\))
(a) M1 for substituting values into the compound interest formula or TVM solver. A1 for \(6258.98\) (or \(6260\)). [2 marks]
(b) M1 for substituting their answer from (a) as the principal and setting up the equation for \(r\) (or entering values in TVM solver). M1 for an attempt to solve for \(r\). A1.15 for \(4.13\) (or \(4.1328...\)). [3.15 marks]
(c) A1 for \(4.18\\%\) (accept \(4.17\\%\) from premature rounding). [1 mark]
Paper 2
Answer all questions in the answer booklet provided. Start each question on a new page. A GDC is required.
A company packages organic coffee beans. The weight of a bag, Equ_1, is normally distributed with mean Equ_2 and standard deviation Equ_3. (a) Find the probability that a randomly selected bag of coffee weighs: (i) more than Equ_4, (ii) between Equ_5 and Equ_4. (b) A bag is rejected if it weighs less than Equ_6. Find the probability that a randomly chosen bag is rejected. (c) The company sells bags in boxes of Equ_7. Find the probability that in a randomly selected box, there is at least one rejected bag. (d) The quality control manager suspects the machine's mean filling weight has decreased. She takes a random sample of Equ_8 bags and finds their weights (in grams): Equ_9. Assuming the standard deviation remains Equ_3: (i) State the null and alternative hypotheses to test the manager's suspicion, (ii) Find the p-value for a one-tailed z-test at the Equ_10 significance level, (iii) State the conclusion of the test in context, justifying your answer.
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PastPaper.workedSolution
(a) Let Equ_11. (i) Using GDC normalcdf(255, infinity, 252, 3.5), we get Equ_12. (ii) Using GDC normalcdf(250, 255, 252, 3.5), we get Equ_13. (b) P(W < 246) = normalcdf(-infinity, 246, 252, 3.5) = 0.043238... = 0.0432. (c) Let X be the number of rejected bags. Equ_14. We need Equ_15. (d) (i) Equ_16 and Equ_17. (ii) Sample mean Equ_18. With Equ_19, Equ_20, and sample size Equ_21, we compute the z-statistic: Equ_22. The corresponding one-tailed p-value is Equ_23. (iii) Since the p-value Equ_24, we reject the null hypothesis. There is significant evidence at the 5% level to suggest that the mean filling weight has decreased.
PastPaper.markingScheme
(a)(i) M1 for setting up normal distribution, A1 for 0.196. (ii) M1 for interval probability, A1 for 0.520. (b) M1 for setting up P(W < 246), A1 for 0.0432. (c) M1 for using Binomial distribution, M1 for 1 - P(X=0), A1 for 0.358. (d)(i) A1 for H_0, A1 for H_1. (ii) M1 for calculating sample mean 249.867, M1 for calculating z-statistic or setting up z-test, A1 for p-value = 0.00914. (iii) R1 for comparing p-value with 0.05, A1 for correct conclusion in context.
A coastal region has three rescue stations located at Equ_25, Equ_26, and Equ_27, where coordinates are in kilometers. (a) Find the equation of the perpendicular bisector of the line segment joining Equ_28 and Equ_29. (b) Show that the equation of the perpendicular bisector of the line segment joining Equ_26 and Equ_29 is Equ_30. (c) The site of a new radar tower, Equ_31, is at the circumcentre of the triangle Equ_32, which is the intersection of these two perpendicular bisectors. (i) Find the coordinates of Equ_31. (ii) Hence, calculate the distance from the radar tower Equ_31 to rescue station Equ_26. (d) A ship is in distress at point Equ_33. (i) Determine which of the three rescue stations is closest to the ship. Justify your answer using distances. (ii) A helicopter leaves station Equ_28 to fly directly to the ship at Equ_33. Find the bearing of the helicopter's flight path from Equ_28.
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(a) Midpoint of AC is Equ_34. Gradient of AC is Equ_35. The perpendicular gradient is Equ_36. Equation: Equ_37, which simplifies to Equ_38. (b) Midpoint of BC is Equ_39. Gradient of BC is Equ_40. The perpendicular gradient is Equ_41. Equation: Equ_42, which simplifies to Equ_30. (c) (i) Equating the bisectors: Equ_43. Thus Equ_44. Substituting gives Equ_45. So Equ_46. (ii) Distance from T(7, 12) to B(14, 11) is Equ_47. (d) (i) Distance Equ_48. Distance Equ_49. Distance Equ_50. Since DA = DC = 4.47 km, both stations A and C are the closest. (ii) Displacement vector from A to D is Equ_51. The direction is East and South. The angle with the South axis is Equ_52. Thus the bearing is Equ_53.
PastPaper.markingScheme
(a) M1 for finding midpoint (5, 6), M1 for perpendicular gradient 3, A1 for equation y = 3x - 9. (b) M1 for midpoint (11, 8), M1 for perpendicular gradient -1, A1 for showing clearly that y - 8 = -1(x - 11) reduces to y = -x + 19. (c)(i) M1 for setting up system of equations, A1 for T(7, 12). (ii) M1 for distance formula, A1 for 7.07. (d)(i) M1 for calculating at least two distances, A1 for calculating all three correctly, A1 for stating both A and C are closest with justification. (ii) M1 for angle calculation, M1 for bearing angle conversion, A1 for 153 degrees.
A manufacturing company designed a new storage tank. The volume of the tank, Equ_54, is given by the function Equ_55, where Equ_56 is the width in meters, for Equ_57. (a) Find the rate of change of the volume with respect to the width, Equ_58. (b) Find the value of Equ_56 that gives: (i) a local maximum volume, (ii) a local minimum volume. (c) Sketch the graph of Equ_59 for Equ_57, showing clearly the coordinates of the local maximum, local minimum, and the start and end points of the domain. (d) The rate of change of volume is Equ_60. (i) Set up an integral to represent the average rate of change of volume over the interval Equ_61. (ii) Calculate this average rate of change.
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(a) Equ_62. (b) Set Equ_63 to get Equ_64, which gives Equ_65. (i) Second derivative Equ_66. At Equ_67, Equ_68, which is a local maximum. Thus, local maximum is at Equ_67. (ii) At Equ_69, Equ_70, which is a local minimum. Thus, local minimum is at Equ_69. (c) Key points: Start Equ_71, Max Equ_72, Min Equ_73, End Equ_74. The graph starts at 58.5, rises to 61 at x=1, drops to 34 at x=4, and rises to 45 at x=5. (d) (i) The average value is given by Equ_75. (ii) Evaluating the integral: Equ_76. Substituting limits: Equ_77.
PastPaper.markingScheme
(a) M1 for power rule application, A1 for 6x^2 - 30x + 24. (b)(i) M1 for setting dV/dx = 0, A1 for factorizing/solving to get critical values, A1 for verifying x = 1 is the maximum. (ii) A1 for x = 4. (c) A1 for correct shape of cubic, A1 for labeling start/end points, A1 for labeling local maximum, A1 for labeling local minimum. (d)(i) M1 for average value formula structure, A1 for correct integral. (ii) M1 for integrating term-by-term, M1 for substituting bounds, A1 for -9 (accepting direct formula V(4)-V(1) / 3 if fully explained).
The temperature of a hot beverage, Equ_78 in Equ_79, cooling in a room of constant temperature is modeled by the function Equ_80, where Equ_81 is the time in minutes since the beverage was poured, and Equ_82 are positive constants with Equ_83. (a) State the physical meaning of the constant Equ_84 in the context of this model. (b) The initial temperature of the beverage when poured (Equ_85) is Equ_86. (i) Find the value of Equ_87. (ii) After Equ_88 minutes, the temperature of the beverage is Equ_89. Show that Equ_90, correct to Equ_91 significant figures. (c) Find the temperature of the beverage after Equ_92 minutes. (d) Find the time taken, in minutes, for the beverage to reach a temperature of Equ_93. (e) An alternative model proposed is a linear model Equ_94, valid for Equ_95. (i) Find the value of Equ_81 for which the two models give the same temperature. (ii) State, with a reason, which model is more appropriate for very large values of Equ_81.
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(a) The constant 20 represents the room temperature, which is the horizontal asymptote that the beverage's temperature approaches over time. (b) (i) At Equ_85, Equ_96. (ii) At Equ_88, Equ_97. This gives Equ_98, so Equ_99, which yields Equ_100 (3 s.f.). (c) Using Equ_101: Equ_102. (Using exact Equ_103, Equ_104). (d) Set Equ_105. Thus Equ_106, which gives Equ_107. Solving for t: Equ_108. (Using exact Equ_109, Equ_110). (e) (i) Equating the models: Equ_111. Using a GDC to find the intersection other than t=0, we find Equ_112 (or Equ_113 if using b=1.13). (ii) The exponential model is more appropriate. As Equ_114, the exponential model approaches 20 degrees C, whereas the linear model decreases indefinitely, predicting impossible negative temperatures.
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(a) A1 for stating room temperature. (b)(i) M1 for substituting t=0, A1 for a = 65. (ii) M1 for substituting t=5 and T=55, M1 for rearranging to make b^5 or b^-5 the subject, A1 for showing b = 1.13. (c) M1 for substituting t=12, A1 for 35.0 (or 34.7). (d) M1 for setting up the equation T(t) = 35, M1 for logarithmic step or GDC solve, A1 for 12.0 (or 11.8). (e)(i) M1 for setting up equation T(t) = L(t), M1 for using GDC solver, A1 for 10.3 (or 10.4). (ii) R1 for mentioning behaviour as t approaches infinity, R1 for concluding the exponential model is better.
A study was conducted to investigate the relationship between the weekly hours spent studying mathematics (Equ_56) and the exam score obtained (Equ_59, out of 100) by a sample of 8 students. The data is shown in the table: Student A: (2, 45); Student B: (4, 58); Student C: (5, 65); Student D: (6, 62); Student E: (7, 75); Student F: (9, 80); Student G: (10, 85); Student H: (12, 95). (a) Write down: (i) the Pearson's product-moment correlation coefficient, Equ_115, (ii) the equation of the regression line of Equ_59 on Equ_56. (b) Use your regression line to estimate the exam score of a student who studies for Equ_116 hours. (c) State, with a reason, whether this estimate is reliable. (d) The teacher decides to classify the students' weekly study hours into two categories: 'Low study hours' (Equ_117 hours) and 'High study hours' (Equ_118 hours). The exam scores are also classified into 'Pass' (Equ_119) and 'Fail' (Equ_120). (i) Construct a Equ_121 contingency table for this classification. (ii) Perform a chi-squared test of independence at the Equ_10 significance level to determine if the category of study hours is independent of the exam outcome. State your null and alternative hypotheses, the p-value, and your conclusion in context.
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(a) (i) Entering the data into a GDC, we obtain Equ_122. (ii) The regression line equation is Equ_123. (b) For Equ_124, Equ_125 (or Equ_126 using rounded coefficients). (c) The estimate is reliable because 8 hours is within the domain of the data (interpolation between 2 and 12 hours) and the correlation coefficient r is very close to 1, indicating a very strong positive linear relationship. (d) (i) Contingency table: Pass (score >= 70) consists of E, F, G, H (4 students, all High). Fail (score < 70) consists of A, B, C, D (4 students, all Low). The table is: Pass: Low=0, High=4 (Total=4); Fail: Low=4, High=0 (Total=4); Column Totals: Low=4, High=4. (ii) Hypotheses: H_0: Category of study hours and exam outcome are independent. H_1: Category of study hours and exam outcome are dependent. Using a GDC on the 2x2 matrix, we obtain a chi-squared test statistic of Equ_127 and p-value = 0.00468. Since p-value = 0.00468 < 0.05, we reject the null hypothesis. There is strong evidence that study hours and exam outcome are dependent.
PastPaper.markingScheme
(a)(i) A2 for r = 0.986. (ii) A1 for slope 4.81, A1 for intercept 37.6. (b) M1 for substituting x = 8 into regression equation, A1 for 76.0 (or 76.1). (c) R1 for stating it is interpolation / within the data range, R1 for noting the very strong correlation, concluding it is reliable. (d)(i) A3 for completely correct 2x2 contingency table (deduct 1 mark for each incorrect cell). (ii) A1 for H_0 and H_1, M2 for calculating the p-value (or chi-squared test statistic), A1 for p-value = 0.00468, R1 for comparing with 0.05 and giving a valid contextual conclusion.