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

(a) Using the compound interest formula: \(A = P \left(1 + \frac{r}{100k}\right)^{kt}\) where \(P = 15000\), \(r = 4.2\), \(k = 12\), and \(t = 5\).

\(A = 15000 \left(1 + \frac{4.2}{1200}\right)^{12 \times 5}\)

\(A = 15000 (1.0035)^{60} \approx 18497.66\)

To the nearest dollar, the value of the investment is $18,498.

(b) We set up the inequality: \(15000 (1.0035)^{12t} \ge 22000\)

\(1.0035^{12t} \ge \frac{22}{15}\)

Taking the natural logarithm of both sides: \(12t \ln(1.0035) \ge \ln\left(\frac{22}{15}\right)\)

\(12t \ge 109.475...\)

\(t \ge 9.12\) years.

Since we require the minimum number of complete years, we round up to 10 years.

(a) M1 for substituting correct values into the compound interest formula. A1 for 18497.66. A1 for 18498.

(b) M1 for setting up a correct inequality or equation. A1 for 9.12. A1 for 10.

(a) The gradient of \(AB\) is \(m_{AB} = \frac{8 - 2}{5 - 1} = \frac{6}{4} = 1.5\).

(b) The gradient of the perpendicular bisector of \(AB\) is \(m_{\perp} = -\frac{1}{1.5} = -\frac{2}{3}\).

The midpoint of \(AB\) is \(M\left(\frac{1+5}{2}, \frac{2+8}{2}\right) = (3, 5)\).

The equation of the perpendicular bisector is: \(y - 5 = -\frac{2}{3}(x - 3) \implies y = -\frac{2}{3}x + 7\).

(c) The vertex is the intersection of the two perpendicular bisectors:

\(-\frac{2}{3}x + 7 = \frac{2}{3}x + \frac{1}{3}\n\implies \frac{4}{3}x = \frac{20}{3} \implies 4x = 20 \implies x = 5\).

Substituting \(x = 5\) into the equation of the perpendicular bisector of \(AB\):

\(y = -\frac{2}{3}(5) + 7 = \frac{11}{3} \approx 3.67\).

The coordinates of the vertex are \((5, \frac{11}{3})\).

(a) A1 for gradient 1.5 (or \\frac{3}{2}\)).

(b) M1 for finding the midpoint (3,5) and perpendicular gradient. A1 for \(y = -\frac{2}{3}x + 7\).

(c) M1 for equating the two perpendicular bisectors. A1 for finding \(x = 5\). A1 for \(y = \frac{11}{3}\) (or 3.67).

Let \(W\) be the weight of an orange, with \(W \sim N(185, 12^2)\).

(a) We need to find \(P(175 \le W \le 195)\).

Using a GDC normal cumulative distribution function (normalCDF) with lower limit = 175, upper limit = 195, \(\mu = 185\), and \(\sigma = 12\):

\(P(175 \le W \le 195) \approx 0.59534...\)

To 3 significant figures, the probability is 0.595.

(b) We require the value \(w\) such that \(P(W \ge w) = 0.15\), which is equivalent to \(P(W < w) = 0.85\).

Using a GDC inverse normal function (inverseNormal) with area = 0.85, \(\mu = 185\), and \(\sigma = 12\):

\(w \approx 197.43...\) grams.

To 3 significant figures, the minimum weight is 197 grams.

(a) M1 for setting up the normal CDF expression with correct limits and parameters. A1 for 0.595.

(b) M1 for setting up \(P(W < w) = 0.85\) or equivalent. A1 for inverse normal setup. A2 for 197 (accept 197.4).

(a) \(H_0\): Preferred music genre and age group are independent (or there is no association between preferred music genre and age group).

(b) The number of degrees of freedom is \(df = (r - 1)(c - 1) = (2 - 1)(3 - 1) = 2\).

(c) Using the GDC to conduct a \(\chi^2\) two-way test on the matrix of observed frequencies:

\(\chi^2 \approx 26.2\)

\(p\)-value \\approx 2.05 \\times 10^{-6}\) (or 0.00000205).

(d) Since the \(p\)-value \((2.05 \times 10^{-6}) < 0.05\), we reject the null hypothesis \(H_0\).

There is strong evidence at the 5% significance level to suggest that preferred music genre and age group are not independent (there is an association).

(a) A1 for stating preferred music genre and age group are independent (or equivalent).

(b) A1 for 2.

(c) A2 for \(2.05 \times 10^{-6}\) (accept standard decimal notation, e.g., 0.00000205).

(d) R1 for comparing the \(p\)-value with 0.05. R1 for a consistent conclusion in context.

(a) As \(t \to \infty\), \(b^{-t} \to 0\) (since \(b > 1\)), so \(T(t) \to 22\). Therefore, the room temperature is \(22^{\circ}\text{C}\).

(b) At \(t = 0\), \(T(0) = 85\).

\(22 + a b^0 = 85 \implies 22 + a = 85 \implies a = 63\).

(c) At \(t = 10\), \(T(10) = 48\).

\(22 + 63 b^{-10} = 48 \implies 63 b^{-10} = 26\)

\(b^{-10} = \frac{26}{63} \implies b^{10} = \frac{63}{26}\)

\(b = \left(\frac{63}{26}\right)^{0.1} \approx 1.0925...\)

To 3 significant figures, \(b = 1.09\).

(a) A1 for 22.

(b) M1 for substituting \(t=0\) and equating to 85. A1 for \(a = 63\).

(c) M1 for substituting \(t=10, a=63\) and equating to 48. M1 for solving for \(b\) (logarithms or GDC). A1 for \(b = 1.09\).

(a) The total surface area of a closed rectangular box with a square base is given by:

\(A = 2x^2 + 4xh\)

Setting \(A = 300\):

\(2x^2 + 4xh = 300\)

\(4xh = 300 - 2x^2 \implies h = \frac{300 - 2x^2}{4x}\).

(b) The volume of the box is given by:

\(V = x^2 h = x^2 \left(\frac{300 - 2x^2}{4x}\right) = \frac{x(300 - 2x^2)}{4} = 75x - 0.5x^3\)

Differentiating \(V\) with respect to \(x\):

\(\frac{\text{d}V}{\text{d}x} = 75 - 1.5x^2\)

To maximize volume, we set the derivative to 0:


75 - 1.5x^2 = 0 \\implies 1.5x^2 = 75 \\implies x^2 = 50 \\implies x = \\sqrt{50} = 5\\sqrt{2} \\approx 7.07\\text{ cm}\).

(c) The maximum volume is obtained by substituting \(x = \sqrt{50}\) back into the volume equation:

\(V = 75(\sqrt{50}) - 0.5(\sqrt{50})^3 = 250\sqrt{2} \approx 353.55... \approx 354\text{ cm}^3\).

(a) M1 for expressing correct surface area formula. A1 for algebraic rearrangement to show the given formula.

(b) M1 for expressing \(V\) in terms of \(x\) only. M1 for differentiating \(V\) and setting to 0. A1 for \(x = 7.07\) (or \(5\sqrt{2}\)).

(c) A1 for \(354\) (or \(250\sqrt{2}\)).

(a) Using Pythagoras' theorem on the square base \(ABCD\):

\(AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \approx 14.1\text{ cm}\).

(b) Since \(O\) is the center of the base, the length of \(AO\) is half of \(AC\):

\(AO = \frac{1}{2} AC = 5\sqrt{2} \approx 7.071\text{ cm}\).

The triangle \(AOP\) is a right-angled triangle at \(O\). Using Pythagoras' theorem:

\(AP = \sqrt{AO^2 + OP^2} = \sqrt{(5\sqrt{2})^2 + 12^2} = \sqrt{50 + 144} = \sqrt{194} \approx 13.9\text{ cm}\).

(c) The angle between the sloping edge \(AP\) and the base \(ABCD\) is \(\angle PAO\).

\(\tan(\angle PAO) = \frac{OP}{AO} = \frac{12}{5\sqrt{2}} \approx 1.697\).

\(\angle PAO = \arctan(1.697) \approx 59.5^{\circ}\) (or 1.04 radians).

(a) M1 for Pythagoras on base. A1 for \(14.1\) (or \(10\sqrt{2}\)).

(b) M1 for finding \(AO = 5\sqrt{2}\) or 7.07. M1 for Pythagoras on triangle \(AOP\). A1 for \(13.9\) (or \\sqrt{194}\)).

(c) M1 for correct trigonometric ratio. A1 for \(59.5^{\circ}\) (accept 1.04 radians).

(a) This is an arithmetic sequence with first term \(u_1 = 30\) and common difference \(d = 4\).

\(u_{25} = u_1 + 24d = 30 + 24(4) = 126\) seats.

(b) The total number of seats is the sum of the first 25 terms:

\(S_{25} = \frac{25}{2}(u_1 + u_{25}) = \frac{25}{2}(30 + 126) = 12.5 \times 156 = 1950\) seats.

(c) First, find the number of seats in the first 10 rows:

\(S_{10} = \frac{10}{2}(2(30) + 9(4)) = 5(60 + 36) = 480\) seats.

The number of seats in the remaining rows (11 to 25) is:


1950 - 480 = 1470\) seats.

The total revenue is:


Total Revenue = (480 \\times 45) + (1470 \\times 30) = 21600 + 44100 = 65700\) dollars.

(a) M1 for identifying arithmetic sequence and substituting values. A1 for 126.

(b) M1 for using arithmetic series sum formula. A1 for 1950.

(c) M1 for calculating \(S_{10} = 480\). M1 for calculating remaining seats (1470) and setting up revenue calculation. A1 for $65,700.

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

(b) The total cost is the sum of the cost of the top and bottom circles and the curved surface area. \(C = 2 \times (\pi r^2) \times 0.04 + (2\pi r h) \times 0.025 = 0.08\pi r^2 + 0.05\pi r h\). Substituting \(h\) from part (a): \(C(r) = 0.08\pi r^2 + 0.05\pi r \left(\frac{500}{\pi r^2}\right) = 0.08\pi r^2 + \frac{25}{r}\).

(c) Differentiating \(C(r)\) with respect to \(r\): \(\frac{dC}{dr} = 2(0.08\pi r) - 25r^{-2} = 0.16\pi r - \frac{25}{r^2}\).

(d) Setting \(\frac{dC}{dr} = 0\): \(0.16\pi r - \frac{25}{r^2} = 0 \implies 0.16\pi r^3 = 25 \implies r^3 = \frac{25}{0.16\pi} \approx 49.7359 \implies r = \sqrt[3]{49.7359} \approx 3.68\text{ cm}\).

(e) Find the second derivative: \(\frac{d^2C}{dr^2} = 0.16\pi + \frac{50}{r^3}\). When \(r \approx 3.68\), \(\frac{d^2C}{dr^2} = 0.16\pi + \frac{50}{(3.68)^3} \approx 1.51 > 0\). Since the second derivative is positive, this value of \(r\) minimizes the cost.

(f) Substitute \(r \approx 3.677\) back into the cost function: \(C(3.677) = 0.08\pi(3.677)^2 + \frac{25}{3.677} \approx 3.398 + 6.799 = 10.197 \approx \$10.20\).

(a) M1 for using the volume formula, A1 for rearranging correctly to get the show-that expression.
(b) M1 for the cost expression in terms of \(r\) and \(h\), M1 for substituting \(h\), A1 for obtaining the final correct equation.
(c) M1 for differentiating the first term, M1 for differentiating the second term, A1 for the correct derivative.
(d) M1 for setting their derivative to 0, M1 for algebraic manipulation to isolate \(r^3\), A1 for the final value \(r \approx 3.68\).
(e) M1 for finding the second derivative expression, A1 for demonstrating it is positive and stating it is a minimum.
(f) M1 for substituting their value of \(r\) into \(C(r)\), A1 for correct calculation, A1 for rounding to 2 decimal places (nearest cent).

(a) \(H_0\): Choice of primary exercise activity is independent of age group.

(b) Total flexibility = \(15 + 35 = 50\). Total "30 and over" = 100. Grand total = 200. Expected frequency = \(\frac{100 \times 50}{200} = 25\).

(c) Degrees of freedom = \((\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(3 - 1) = 2\).

(d) Entering the contingency table into a GDC:
(i) \(\chi^2 \approx 10.7\) (3 s.f.)
(ii) \(p \approx 0.00480\) (3 s.f.)

(e) Since the \(p\)-value (\(0.00480\)) is less than the significance level (\(0.05\)), we reject the null hypothesis. There is significant evidence that choice of primary exercise activity is not independent of age group.

(f) Let \(T \sim N(55, 12^2)\). Using GDC: \(P(T > 65) \approx 0.2023 \approx 0.202\).

(g) We need to find \(P(T < 75 | T > 60) = \frac{P(60 < T < 75)}{P(T > 60)}\). Using GDC:
\(P(60 < T < 75) \approx 0.29067\)
\(P(T > 60) \approx 0.33846\)
\(P(T < 75 | T > 60) = \frac{0.29067}{0.33846} \approx 0.8588 \approx 0.859\).

(a) A1 for stating independence correctly.
(b) M1 for formula setup, A1 for verifying the answer 25.
(c) A1 for 2.
(d) A1 for \(\chi^2 \approx 10.7\), A2 for \(p \approx 0.00480\).
(e) R1 for comparing \(p\)-value with 0.05, A1 for concluding rejection.
(f) M1 for normal cumulative distribution setup, A1 for \(0.202\).
(g) M1 for conditional probability formula, M1 for finding numerator \(0.291\), M1 for finding denominator \(0.338\), A1 for division step, A1 for \(0.859\).

(a) The diagram must show point \(A\) with a North line, point \(B\) at bearing \(065^\circ\) with distance \(12\text{ km}\), and point \(C\) at bearing \(135^\circ\) with distance \(18\text{ km}\). The angle \(BAC = 135^\circ - 65^\circ = 70^\circ\).

(b) Using the cosine rule in triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(70^\circ)\)
\(BC^2 = 12^2 + 18^2 - 2(12)(18)\cos(70^\circ) = 144 + 324 - 432\cos(70^\circ)\)
\(BC^2 \approx 468 - 147.753 = 320.247 \implies BC \approx 17.895 \approx 17.9\text{ km}\).

(c) First, find the interior angle \(ABC\) using the sine rule:
\(\frac{\sin(ABC)}{18} = \frac{\sin(70^\circ)}{17.895} \implies \sin(ABC) \approx 0.9452 \implies ABC \approx 70.93^\circ\).
Draw a North-South line at \(B\). The angle of the line \(BA\) with the South line at \(B\) is \(65^\circ\) (alternate interior angles). The line \(BC\) lies to the east of the South line at an angle of \(70.93^\circ - 65^\circ = 5.93^\circ\). Thus, the bearing from \(B\) to \(C\) is \(180^\circ - 5.93^\circ = 174.07^\circ \approx 174^\circ\).

(d) The triangle formed by \(B\), \(C\), and the helicopter \(H\) is right-angled at \(C\).
\(\tan(12.5^\circ) = \frac{h}{BC} \implies h = BC \tan(12.5^\circ) = 17.895 \times \tan(12.5^\circ) \approx 3.967 \approx 3.97\text{ km}\).

(e) The helicopter \(H\) is vertically above \(C\), so triangle \(ACH\) is right-angled at \(C\).
\(AH = \sqrt{AC^2 + h^2} = \sqrt{18^2 + 3.967^2} = \sqrt{324 + 15.737} \approx 18.43 \approx 18.4\text{ km}\).

(a) A1 for showing points \(A\), \(B\), \(C\) relative to North, A1 for indicating angle \(BAC = 70^\circ\) and distances 12 and 18.
(b) M1 for substituting into cosine rule, A1 for intermediate values, A1 for \(17.9\text{ km}\).
(c) M1 for using sine rule to find angle \(ABC\), A1 for \(ABC \approx 71^\circ\), M1 for using bearings geometry, A1 for \(174^\circ\).
(d) M1 for setting up tangent ratio, A1 for substituting values, A1 for \(3.97\text{ km}\).
(e) M1 for recognizing right-angled triangle \(ACH\), M1 for Pythagoras' theorem, A1 for correct substitutions, A1 for \(18.4\text{ km}\).

(a) Using a financial GDC or the amortization formula:
\(N = 300\), \(I\% = 4.8\), \(PV = 300000\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\).
Solving for monthly payment: \(PMT = \$1718.70\).

(b) Total amount paid = \(1718.695 \times 300 = \$515,608.50 \approx \$515,609\).

(c) Total interest paid = Total payments \(-\) Principal = \(515,608.50 - 300,000 = \$215,608.50 \approx \$215,609\).

(d) The remaining balance is the present value of the remaining 180 payments (15 years remaining):
\(N = 180\), \(I\% = 4.8\), \(PMT = -1718.695\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\).
Solving for \(PV\): \(PV \approx \$220,156.55 \approx \$220,157\).

(e) For the investment to double, \(FV = 2 \times 80000 = 160000\).
\(160000 = 80000 \left(1 + \frac{0.052}{4}\right)^{4t} \implies 2 = (1.013)^{4t}\).
Taking natural logarithms:
\(\ln(2) = 4t \ln(1.013) \implies 4t = \frac{\ln(2)}{\ln(1.013)} \approx 53.66\text{ quarters}\).
Since interest is compounded quarterly, it takes 54 quarters to exceed double. 54 quarters is \(13.5\) years. Since we need complete years, the number of complete years is 14.

(a) M1 for using a financial calculator or loan formula with correct inputs, A2 for \(\$1718.70\).
(b) M1 for multiplying payment by 300, A1 for \(\$515,609\).
(c) M1 for subtracting 300,000, A1 for \(\$215,609\).
(d) M1 for identifying 180 remaining payments, M2 for financial solver input setup, A1 for \(\$220,157\).
(e) M1 for setting up doubling equation, M1 for dividing rate by 4 and correct quarterly compounding form, M1 for solving using logs, A1 for \(13.5\) years (54 quarters), A1 for stating 14 complete years.

(a) (i) Amplitude \(a = \frac{\text{Max} - \text{Min}}{2} = \frac{32 - 16}{2} = 8\).
(ii) Vertical shift \(c = \frac{\text{Max} + \text{Min}}{2} = \frac{32 + 16}{2} = 24\).

(b) The period of the temperature cycle is 24 hours. The formula for the period is \(\frac{2\pi}{b}\). Thus, \(24 = \frac{2\pi}{b} \implies b = \frac{2\pi}{24} = \frac{\pi}{12}\).

(c) At \(06:00\), \(t = 6\).
\(T(6) = 8 \sin\left(\frac{\pi}{12}(6 - 8)\right) + 24 = 8 \sin\left(-\frac{\pi}{6}\right) + 24 = 8(-0.5) + 24 = -4 + 24 = 20{}^\circ\text{C\cdot}\)

(d) Set \(T(t) = 20\):
\(8 \sin\left(\frac{\pi}{12}(t - 8)\right) + 24 = 20 \implies \sin\left(\frac{\pi}{12}(t - 8)\right) = -0.5\).
Within \(0 \le t \le 24\), the solutions are:
\(\frac{\pi}{12}(t - 8) = -\frac{\pi}{6} \implies t - 8 = -2 \implies t = 6\) (which is \(06:00\)).
\(\frac{\pi}{12}(t - 8) = \frac{7\pi}{6} \implies t - 8 = 14 \implies t = 22\) (which is \(22:00\)).

(e) At \(t = 8\): \(T(8) = 8 \sin(0) + 24 = 24{}^\circ\text{C}\).
At \(t = 14\): \(T(14) = 32{}^\circ\text{C}\) (given).
Average rate of change = \frac{T(14) - T(8)}{14 - 8} = \frac{32 - 24}{6} = \frac{8}{6} \approx 1.33{}^\circ\text{C/hour}\).

(a) A1 for showing amplitude calculation, A2 for vertical shift calculation.
(b) M1 for using the period formula, A1 for simplifying to show \(\frac{\pi}{12}\).
(c) M1 for substituting \(t = 6\), A1 for obtaining \(20{}^\circ\text{C}\).
(d) M1 for setting the equation to 20, M1 for solving the sine equation, A1 for \(t = 6\), A1 for \(t = 22\), A1 for writing both as 24-hour times (06:00 and 22:00).
(e) M1 for finding \(T(8)\), M1 for finding \(T(14)\), M1 for the average rate of change quotient, A1 for \(1.33{}^\circ\text{C/hour}\).