HL AI · Practice Paper 3 · Interactive

IB Maths Applications & Interpretation — Interactive Practice Paper 3

Two modes, one paper: Practice gives instant feedback per part, or switch to Timed exam mode for a full simulated Paper 3 with a countdown and single "Submit paper" score. Five authentic HL AI style questions covering Geometry, Statistics, Hypothesis Testing, Optimisation and Inverse Functions.

36 marks · 5 questions · GDC required · 3-sf accuracy

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  • A graphic display calculator (GDC) is required. Unless stated, use exact values or 3 significant figures.
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Question 1

4 marks
Geometry & Trigonometry

A triangular field on a farm has $AB = 17\text{ m}$, $AC = 45\text{ m}$ and angle $BAC = 38^\circ$. A new fence is to be installed around its entire perimeter.

Calculate the total length of fencing required. [4]

Round to 3 significant figures.
Step 1 — find $BC$ using the cosine rule.
$BC^{2} = AB^{2} + AC^{2} - 2\cdot AB\cdot AC\cdot\cos(38^\circ)$
$BC^{2} = 17^{2} + 45^{2} - 2(17)(45)\cos 38^\circ = 289 + 2025 - 1530\cos 38^\circ \approx 1108.4$
$BC \approx \sqrt{1108.4} \approx 33.29\text{ m}$

Step 2 — add the three sides.
Perimeter $= AB + AC + BC = 17 + 45 + 33.29 \approx \mathbf{95.3\text{ m}}$ (3 s.f.).

Question 2

6 marks
Descriptive Statistics

Palvinder breeds Labrador puppies at his farm. Their weights (g) are shown in the five-number summary:

MinQ₁MedianQ₃Max
100190210240290

(a) Write down the median weight of the puppies. [1]

(b) Write down the upper quartile. [1]

(c) Find the interquartile range (IQR). [2]

(d) The weights are normally distributed. Find the weight of the heaviest possible puppy that is not considered an outlier. [2]

Upper outlier boundary $= Q_{3} + 1.5\times \text{IQR}$.
(a) Median $= 210$ g.
(b) $Q_{3} = 240$ g.
(c) $\text{IQR} = Q_{3} - Q_{1} = 240 - 190 = 50$ g.
(d) Upper outlier boundary $= Q_{3} + 1.5 \cdot \text{IQR} = 240 + 1.5(50) = 240 + 75 = \mathbf{315}\text{ g}$.

Question 3

6 marks
Hypothesis Testing

It is claimed that women from Japan are taller on average than women from India. The heights, in cm, of a random sample of 11 women from each country are shown below.

CountryHeights (cm)
Japan173.0158.2148.5150.6168.7149.8158.8155.3159.2158.9166.0
India155.2157.8156.0142.7149.6150.1152.6148.2151.3147.6168.0

A two-sample $t$-test is performed at the $5\%$ significance level.

(a) State the alternative hypothesis $H_{1}$ (use $\mu_J$ for Japan, $\mu_I$ for India). [2]

Type your answer using letters — e.g. muJ > muI.

(b) Calculate the $p$-value (to 3 s.f.). [2]

Use your GDC's 2-Sample T-Test, pooled = No, one-tailed.

(c) Is the claim justified at the $5\%$ level? Answer yes or no. [2]

(a) $H_{0}: \mu_{J} = \mu_{I}$    $H_{1}: \mu_{J} > \mu_{I}$ (one-tailed).
(b) Entering both lists into a GDC and running a 2-sample $t$-test (unpooled, one-tailed) gives $p \approx \mathbf{0.0794}$.
(c) Since $p \approx 0.0794 > 0.05$, we do not reject $H_{0}$. Therefore the claim that Japanese women are taller on average is not justified at the 5% level.

Question 4

14 marks
Calculus & Optimisation

A hollow chocolate box is manufactured as a right prism with a regular hexagonal base. Each hexagonal face has side length $x\text{ cm}$; the prism height is $h\text{ cm}$.

(a) Given that $\sin 60^\circ = \tfrac{\sqrt{3}}{2}$, show that the area of the base of the box is $\dfrac{3\sqrt{3}x^{2}}{2}$. [2] (Show-workings question — reveal solution.)

(b) Given the total external surface area is $1200\text{ cm}^{2}$, show that $V = 300\sqrt{3}\,x - \tfrac{9}{4}x^{3}$. [3] (Show-workings question — reveal solution.)

(c) Find $\dfrac{dV}{dx}$. Type your answer as a polynomial in $x$ (use sqrt(3) for $\sqrt{3}$). [2]

Whitespace is ignored. Accepted forms: 300sqrt(3)-27/4x^2 or 300√3-27x^2/4.

(d) Find the value of $x$ which maximises the volume (3 s.f.). [2]

(e) Find the maximum possible volume of the box (3 s.f., in $\text{cm}^{3}$). [2]

(f) Explain why simply dividing $V$ by the volume of a single spherical chocolate and rounding down is not a valid way to count the chocolates. [1] (Explanation question — reveal solution.)

(a) A regular hexagon is 6 equilateral triangles of side $x$. Area of one triangle $= \tfrac{1}{2}x^{2}\sin 60^\circ = \tfrac{\sqrt{3}}{4}x^{2}$. Six of these give $6\cdot \tfrac{\sqrt{3}}{4}x^{2} = \dfrac{3\sqrt{3}}{2}x^{2}$. ✓

(b) Surface area = 2 hexagon caps + 6 rectangular sides:
$2\cdot\tfrac{3\sqrt3}{2}x^{2} + 6xh = 1200 \;\;\Rightarrow\;\; 3\sqrt{3}x^{2} + 6xh = 1200 \;\;\Rightarrow\;\; h = \dfrac{1200-3\sqrt{3}x^{2}}{6x} = \dfrac{200}{x} - \dfrac{\sqrt{3}}{2}x$.
Then $V = \dfrac{3\sqrt{3}}{2}x^{2}\cdot h = \dfrac{3\sqrt{3}}{2}x^{2}\left(\dfrac{200}{x}-\dfrac{\sqrt{3}}{2}x\right) = 300\sqrt{3}\,x - \dfrac{9}{4}x^{3}$. ✓

(c) $\dfrac{dV}{dx} = 300\sqrt{3} - \dfrac{27}{4}x^{2}$.
(d) Set $\dfrac{dV}{dx}=0$: $\;x^{2} = \dfrac{1200\sqrt{3}}{27}$, so $x = \sqrt{\dfrac{1200\sqrt{3}}{27}} \approx \mathbf{8.77}$ cm.
(e) $V_{\max} = 300\sqrt{3}(8.77) - \tfrac{9}{4}(8.77)^{3} \approx \mathbf{3040}\text{ cm}^{3}$ (3 s.f.).
(f) Spheres do not tile space — the sphere-packing density in a container is at most $\approx 74\%$, so there is always empty space between chocolates. Dividing volumes ignores this packing inefficiency and would give an over-estimate of how many chocolates actually fit.

Question 5

6 marks
Functions & Modelling

The perimeter of a square is $P(A) = 4\sqrt{A}$, where $A \ge 0$ is the area of the square. The graph of $P$ is drawn for $0 \le A \le 25$.

📊 Figure: A concave-increasing curve of $P(A) = 4\sqrt{A}$ passing through $(0,0)$, $(1,4)$, $(4,8)$, $(9,12)$, $(16,16)$, $(25,20)$.

(a) Write down the value of $P(25)$. [1]

(b) The range of $P(A)$ on this domain is $0 \le P(A) \le n$. Write down the exact value of $n$. [1]

(c) Sketch the graph of $P^{-1}$ on the same axes. [4] (Sketch question — reveal solution.)

(d) Interpret the statement $P^{-1}(8) = 4$ in context. Type the missing area (in square units) that a square with perimeter 8 must have. [1]

(a) $P(25) = 4\sqrt{25} = 4\cdot 5 = 20$.
(b) The largest output on $[0,25]$ is at $A=25$, giving $n = 20$.
(c) $P^{-1}$ is the reflection of $P$ in the line $y = x$. It is a concave-increasing curve on the domain $0 \le P \le 20$, passing through $(0,0)$, $(4,1)$, $(8,4)$, $(12,9)$, $(16,16)$, $(20,25)$. Rule: $A = \left(\tfrac{P}{4}\right)^{2} = \tfrac{P^{2}}{16}$.
(d) $P^{-1}(8) = 4$ means: a square whose perimeter equals 8 units has area 4 square units. (Side length 2, so area $=2^{2} = 4$.)

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