SL AI · Practice Paper 2 · Interactive

IB Maths Applications & Interpretation — Interactive Practice Paper 2

Paper 2 is the calculator-required paper — real-world modelling, financial mathematics, regression and applied calculus. Grab your GDC (Casio fx-CG50 / TI-84) and go. Two modes: instant per-part feedback, or full timed exam.

32 marks · 5 questions · GDC required · Modelling & applications focus
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Question 1

7 marksGDC required
Financial Mathematics — annuities

Elena makes a deposit of $\$400$ at the end of each month into a savings annuity paying a nominal annual rate of $4.8\%$, compounded monthly. She does this for 10 years.

(a) Using your GDC, calculate the total future value of the account after 10 years. Give your answer to the nearest cent. [3]

Casio fx-CG50 · MENU F · CMPD: $N=120, I\%=4.8, PV=0, PMT=-400, P/Y=12, C/Y=12$.

(b) Calculate the total amount Elena deposited from her own pocket. [2]

(c) Calculate the total interest Elena earned over the 10 years. [2]

(a) GDC Compound Interest app: $N=120, I\%=4.8, PV=0, PMT=-400, P/Y=12, C/Y=12$. Solve for $FV \Rightarrow \mathbf{\$61\,461.76}$.
(b) Deposited $= 400 \times 120 = \mathbf{\$48\,000}$.
(c) Interest $= FV - \text{Deposited} = 61461.76 - 48000 = \mathbf{\$13\,461.76}$.

Question 2

7 marksGDC required
Coordinate Geometry — perpendicular bisector & Voronoi

Two schools are represented by points $A(2, 20)$ and $B(14, 24)$ on a map. A road, represented by line $R$ with equation $-x + y = 4$, passes near the schools. A town planner wants to place a bus stop on the road so it is exactly the same distance from both schools.

(a) Find the gradient of $[AB]$. [1]

(b) Find the midpoint of $[AB]$. Give as $(x, y)$. [1]

(c) Find the equation of the perpendicular bisector of $[AB]$ in the form $y = mx + c$. Give $c$ to the nearest integer. [3]

$m_\perp = -3$; passes through midpoint.

(d) Find the $x$-coordinate of the bus-stop location (intersection of $R$ with the perpendicular bisector). [2]

(a) $m_{AB} = \dfrac{24-20}{14-2} = \dfrac{4}{12} = \mathbf{\tfrac{1}{3}}$.
(b) Midpoint $= \left(\tfrac{2+14}{2}, \tfrac{20+24}{2}\right) = \mathbf{(8, 22)}$.
(c) $m_\perp = -3$. $y - 22 = -3(x - 8) \Rightarrow y = \mathbf{-3x + 46}$.
(d) Solve $-3x + 46 = x + 4 \Rightarrow 42 = 4x \Rightarrow x = \mathbf{10.5}$.

Question 3

6 marksGDC required
Statistics — combined mean & missing value

The mean of five distinct positive integers is $8.5$.

(a) Find the sum of these five integers. [1]

(b) A sixth integer, $x$, is added to the data set. The new mean of all six values is $9$. Find $x$. [2]

(c) Class A (20 students) achieved a mean score of $65$. Class B (30 students) achieved a mean score of $80$. Calculate the combined mean of all 50 students. [3]

(a) Sum $= 5 \times 8.5 = \mathbf{42.5}$.
(b) New sum $= 6 \times 9 = 54$; so $x = 54 - 42.5 = \mathbf{11.5}$.
(c) Combined $= \dfrac{20 \times 65 + 30 \times 80}{50} = \dfrac{1300 + 2400}{50} = \dfrac{3700}{50} = \mathbf{74}$.

Question 4

6 marksGDC required
Bivariate Data — regression line & percentage error

An environmentalist models the relationship between the distance from a highway ($d$, in metres) and the concentration of a pollutant ($C$, in parts per million). The regression line is $C = -0.04d + 8.5$.

(a) Estimate the concentration at $d = 50$ m. [2]

(b) The actual measured concentration at $50$ m was $7.1$ ppm. Calculate the percentage error of the estimate, to 3 significant figures. [3]

Percentage error $= \left|\dfrac{v_A - v_E}{v_E}\right| \times 100$.

(c) State whether $C$ or $d$ is the response variable. [1]

(a) $C = -0.04(50) + 8.5 = -2 + 8.5 = \mathbf{6.5}$ ppm.
(b) $\%\varepsilon = \left|\dfrac{6.5 - 7.1}{7.1}\right| \times 100 = \dfrac{0.6}{7.1} \times 100 = \mathbf{8.45\%}$.
(c) The concentration $C$ depends on the distance $d$, so $\mathbf{C}$ is the response.

Question 5

6 marksGDC required
Calculus — area under a curve

The curve $y = -x^2 + 6x - 5$ forms an enclosed region with the $x$-axis.

(a) Use your GDC to find the $x$-intercepts (roots) of the curve. Give both in the form $x =$. Smaller first. [2]

(b) Write down the definite integral for the area of the enclosed region. Give in the form int(f,a,b). [1]

(c) Evaluate the area. Give to 3 significant figures. [3]

Exact answer $= \dfrac{32}{3}$.
(a) GDC (MENU 5 → G-Solv → ROOT): $x = \mathbf{1}$ and $x = \mathbf{5}$.
(b) Area $= \displaystyle\int_{1}^{5} (-x^2 + 6x - 5) \, dx$.
(c) Evaluate: $\left[-\tfrac{x^3}{3} + 3x^2 - 5x\right]_1^5 = \left(-\tfrac{125}{3} + 75 - 25\right) - \left(-\tfrac{1}{3} + 3 - 5\right) = \tfrac{25}{3} + \tfrac{7}{3} = \tfrac{32}{3} \approx \mathbf{10.7}$.

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