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IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

28 marks

Air quality at a school: a Voronoi diagram, natural neighbour interpolation and a t-test

Geometry and trigonometry

(a)(i)

2 marks

$B (8, 9)$ and $C (8, 1)$ have the same $x$ -coordinate. The segment $[B C]$ is vertical and its midpoint is $(8, 5)$ , so the perpendicular bisector is a horizontal line.

y = 5

(a)(ii)

2 marks

The vertex $V$ is where $y = 5$ meets $y = 15 - 2 x$ .

5 = 15 - 2 x \Rightarrow x = 5 V = (5, 5)

(b)

1 mark

Compare the distances from $S (5, 6)$ to the stations: $S A = 25 + 1 = 5.10$ , $S B = 9 + 9 = 4.24$ and $S C = 9 + 25 = 5.83$ .

station B

(c)

1 mark

Using the reading from the nearest station is nearest neighbour interpolation.

(d)

1 mark

The school is also fairly close to stations $A$ and $C$ , which have much higher average values (132 and 125), so the air quality at the school could easily be worse than station $B$ 's average of 49. (Also, 49 is only an average, so readings above 50 may be common.)

(e)(i)

2 marks

The midpoint of $S (5, 6)$ and $B (8, 9)$ .

(\frac{5 + 8}{2}, \frac{6 + 9}{2}) = (6.5, 7.5)

(e)(ii)

3 marks

The gradient of $S B$ is $\frac{9 - 6}{8 - 5} = 1$ , so the perpendicular bisector has gradient $- 1$ and passes through $(6.5, 7.5)$ .

y - 7.5 = - (x - 6.5) \Rightarrow y = 14 - x

(e)(iii)

5 marks

Region $Q$ has corners where this bisector meets the other edges. It meets $y = 5$ at $x = 9$ (the point $(9, 5)$ ) and it meets $y = 15 - 2 x$ at the point where $14 - x = 15 - 2 x$ .
The region $Q$ is the triangle with corners $V (5, 5)$ , $(9, 5)$ and $(1, 13)$ . It has a base of length 4 along the line $y = 5$ and a height of $13 - 5 = 8$ .

x = 1 \Rightarrow (1, 13) Area = \frac{1}{2} ∣ 5 (5 - 13) + 9 (13 - 5) + 1 (5 - 5) ∣ = \frac{1}{2} ∣ - 40 + 72 + 0 ∣ = 16 a_{B} = 16 km^{2}

(f)

2 marks

The total area of the new cell is $T = 13.7 + 16 + 6.9 = 36.6$ .

W = \frac{132 ( 13.7 ) + 49 ( 16 ) + 125 ( 6.9 )}{36.6} = \frac{3453.5}{36.6} \approx 94.4

(g)

6 marks

$H_{0}$ : $μ = 94.4$ and $H_{1}$ : $μ > 94.4$ , where $μ$ is the population mean AQI value at the school.
Use a one-tailed one-sample $t$ -test with $\overset{x}{ˉ} = 97.8$ , $s_{n - 1} = 17.2$ and $n = 60$ .
Since $0.0655 < 0.10$ , reject $H_{0}$ . There is sufficient evidence, at the 10% level, that the mean AQI value at the school is greater than 94.4.

t = \frac{97.8 - 94.4}{17.2/ 60} \approx 1.53, p \approx 0.0655

(h)

3 marks

Model the AQI as $X \sim N (97.8, 17. 2^{2})$ . A day is good if $X < 50$ .
The expected number of good days a year is about 1.

P (X < 50) \approx 0.002726 365 \times 0.002726 \approx 0.995

Question 2

27 marks

Electric vehicles in the EU: a logistic model, its fit and charging points

Calculus

(a)

1 mark

As $t$ grows, $e^{- k t} \to 0$ and $N \to 310$ . So 310 million is the upper limit (carrying capacity) the model gives for the number of electric vehicles in the European Union.

(b)(i)

2 marks

At $t = 1$ , $N = 0.2$ .

0.2 = \frac{310}{1 + C e ^{- k}} \Rightarrow 1 + C e^{- k} = 1550 \Rightarrow C e^{- k} = 1549 C = 1549 e^{k}

(b)(ii)

3 marks

At $t = 7$ , $N = 3.1$ .

3.1 = \frac{310}{1 + C e ^{- 7 k}} \Rightarrow 1 + C e^{- 7 k} = 100 \Rightarrow C = 99 e^{7 k}

(c)(i)

2 marks

Equate the two expressions for $C$ .

1549 e^{k} = 99 e^{7 k} \Rightarrow e^{6 k} = \frac{1549}{99} \Rightarrow k = \frac{1}{6} ln \frac{1549}{99} \approx 0.458

(c)(ii)

1 mark

Substitute $k$ into $C = 99 e^{7 k}$ .

C \approx 2450

(d)

2 marks

Evaluate $N$ at $t = 4$ .

N (4) = \frac{310}{1 + 2449.7 e ^{- 0.4584 \times 4}} \approx 0.79 a = 0.8

(e)

2 marks

The data values for 2017 to 2021 are 0.3, 0.4, 0.6, 1.1, 1.9 and the predicted values are 0.3, 0.5, 0.8, 1.2, 2.0. The residuals are $0, 0.1, 0.2, 0.1, 0.1$ .

S S_{r es} = 0^{2} + 0. 1^{2} + 0. 2^{2} + 0. 1^{2} + 0. 1^{2} = 0.07

(f)

3 marks

There are $n = 5$ predictions.
Since $0.118 < 0.25$ , George will decide to use the model.

E = \frac{0.07}{5} \approx 0.118

(g)

2 marks

The end of 2035 is $t = 20$ .

N (20) = \frac{310}{1 + 2449.7 e ^{- 0.4584 \times 20}} \approx 247 million

(h)

2 marks

Of all EVs, 20% need public charging points, and there is one point for every 10 of these, so the number of points is $0.2 \times 0.1 = 0.02$ times the number of EVs.

0.02 N = \frac{6.2}{1 + 2449.7 e ^{- 0.4584 t}} million

(i)

1 mark

From the end of 2020 to the end of 2022 is 2 years.

\frac{0.54 - 0.22}{2} = 0.16 million per year

(j)

6 marks

The end of 2020 is $t = 5$ with 0.22 million points. With a linear increase of 0.16 million per year, the supply is
The points are not enough when the demand from part (h) is greater than the supply.
Solve with the calculator (find where the two graphs cross).
The time $t = 15$ is the end of 2030, so the shortage first occurs during the year 2031.

S (t) = 0.22 + 0.16 (t - 5) \frac{6.2}{1 + 2449.7 e ^{- 0.4584 t}} > 0.22 + 0.16 (t - 5) t \approx 15.15

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry28 marks · 51%
Q1
Calculus27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Voronoi vertex · Q1
The vertex is the same distance from all three stations, where the bisectors cross.
02
Natural neighbour · Q1
Weight each station's reading by the area of its old cell that the new school cell takes over.
03
Logistic limit · Q2
In $N = \frac{310}{1 + C e ^{- k t}}$ , the term $C e^{- k t}$ tends to 0, so $N$ tends to 310.
04
Fit measure · Q2
$E = \frac{S S _{r es}}{n}$ is small when the model predicts the data well.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

28 marks

Air quality at a school: a Voronoi diagram, natural neighbour interpolation and a t-test

Geometry and trigonometry

(a)(i)

2 marks

$B (8, 9)$ and $C (8, 1)$ have the same $x$ -coordinate. The segment $[B C]$ is vertical and its midpoint is $(8, 5)$ , so the perpendicular bisector is a horizontal line.

y = 5

(a)(ii)

2 marks

The vertex $V$ is where $y = 5$ meets $y = 15 - 2 x$ .

5 = 15 - 2 x \Rightarrow x = 5 V = (5, 5)

(b)

1 mark

Compare the distances from $S (5, 6)$ to the stations: $S A = 25 + 1 = 5.10$ , $S B = 9 + 9 = 4.24$ and $S C = 9 + 25 = 5.83$ .

station B

(c)

1 mark

Using the reading from the nearest station is nearest neighbour interpolation.

(d)

1 mark

(e)(i)

2 marks

The midpoint of $S (5, 6)$ and $B (8, 9)$ .

(\frac{5 + 8}{2}, \frac{6 + 9}{2}) = (6.5, 7.5)

(e)(ii)

3 marks

The gradient of $S B$ is $\frac{9 - 6}{8 - 5} = 1$ , so the perpendicular bisector has gradient $- 1$ and passes through $(6.5, 7.5)$ .

y - 7.5 = - (x - 6.5) \Rightarrow y = 14 - x

(e)(iii)

5 marks

Region $Q$ has corners where this bisector meets the other edges. It meets $y = 5$ at $x = 9$ (the point $(9, 5)$ ) and it meets $y = 15 - 2 x$ at the point where $14 - x = 15 - 2 x$ .
The region $Q$ is the triangle with corners $V (5, 5)$ , $(9, 5)$ and $(1, 13)$ . It has a base of length 4 along the line $y = 5$ and a height of $13 - 5 = 8$ .

x = 1 \Rightarrow (1, 13) Area = \frac{1}{2} ∣ 5 (5 - 13) + 9 (13 - 5) + 1 (5 - 5) ∣ = \frac{1}{2} ∣ - 40 + 72 + 0 ∣ = 16 a_{B} = 16 km^{2}

(f)

2 marks

The total area of the new cell is $T = 13.7 + 16 + 6.9 = 36.6$ .

W = \frac{132 ( 13.7 ) + 49 ( 16 ) + 125 ( 6.9 )}{36.6} = \frac{3453.5}{36.6} \approx 94.4

(g)

6 marks

$H_{0}$ : $μ = 94.4$ and $H_{1}$ : $μ > 94.4$ , where $μ$ is the population mean AQI value at the school.
Use a one-tailed one-sample $t$ -test with $\overset{x}{ˉ} = 97.8$ , $s_{n - 1} = 17.2$ and $n = 60$ .
Since $0.0655 < 0.10$ , reject $H_{0}$ . There is sufficient evidence, at the 10% level, that the mean AQI value at the school is greater than 94.4.

t = \frac{97.8 - 94.4}{17.2/ 60} \approx 1.53, p \approx 0.0655

(h)

3 marks

Model the AQI as $X \sim N (97.8, 17. 2^{2})$ . A day is good if $X < 50$ .
The expected number of good days a year is about 1.

P (X < 50) \approx 0.002726 365 \times 0.002726 \approx 0.995

Question 2

27 marks

Electric vehicles in the EU: a logistic model, its fit and charging points

Calculus

(a)

1 mark

As $t$ grows, $e^{- k t} \to 0$ and $N \to 310$ . So 310 million is the upper limit (carrying capacity) the model gives for the number of electric vehicles in the European Union.

(b)(i)

2 marks

At $t = 1$ , $N = 0.2$ .

0.2 = \frac{310}{1 + C e ^{- k}} \Rightarrow 1 + C e^{- k} = 1550 \Rightarrow C e^{- k} = 1549 C = 1549 e^{k}

(b)(ii)

3 marks

At $t = 7$ , $N = 3.1$ .

3.1 = \frac{310}{1 + C e ^{- 7 k}} \Rightarrow 1 + C e^{- 7 k} = 100 \Rightarrow C = 99 e^{7 k}

(c)(i)

2 marks

Equate the two expressions for $C$ .

1549 e^{k} = 99 e^{7 k} \Rightarrow e^{6 k} = \frac{1549}{99} \Rightarrow k = \frac{1}{6} ln \frac{1549}{99} \approx 0.458

(c)(ii)

1 mark

Substitute $k$ into $C = 99 e^{7 k}$ .

C \approx 2450

(d)

2 marks

Evaluate $N$ at $t = 4$ .

N (4) = \frac{310}{1 + 2449.7 e ^{- 0.4584 \times 4}} \approx 0.79 a = 0.8

(e)

2 marks

The data values for 2017 to 2021 are 0.3, 0.4, 0.6, 1.1, 1.9 and the predicted values are 0.3, 0.5, 0.8, 1.2, 2.0. The residuals are $0, 0.1, 0.2, 0.1, 0.1$ .

S S_{r es} = 0^{2} + 0. 1^{2} + 0. 2^{2} + 0. 1^{2} + 0. 1^{2} = 0.07

(f)

3 marks

There are $n = 5$ predictions.
Since $0.118 < 0.25$ , George will decide to use the model.

E = \frac{0.07}{5} \approx 0.118

(g)

2 marks

The end of 2035 is $t = 20$ .

N (20) = \frac{310}{1 + 2449.7 e ^{- 0.4584 \times 20}} \approx 247 million

(h)

2 marks

Of all EVs, 20% need public charging points, and there is one point for every 10 of these, so the number of points is $0.2 \times 0.1 = 0.02$ times the number of EVs.

0.02 N = \frac{6.2}{1 + 2449.7 e ^{- 0.4584 t}} million

(i)

1 mark

From the end of 2020 to the end of 2022 is 2 years.

\frac{0.54 - 0.22}{2} = 0.16 million per year

(j)

6 marks

The end of 2020 is $t = 5$ with 0.22 million points. With a linear increase of 0.16 million per year, the supply is
The points are not enough when the demand from part (h) is greater than the supply.
Solve with the calculator (find where the two graphs cross).
The time $t = 15$ is the end of 2030, so the shortage first occurs during the year 2031.

S (t) = 0.22 + 0.16 (t - 5) \frac{6.2}{1 + 2449.7 e ^{- 0.4584 t}} > 0.22 + 0.16 (t - 5) t \approx 15.15

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry28 marks · 51%
Q1
Calculus27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Voronoi vertex · Q1
The vertex is the same distance from all three stations, where the bisectors cross.
02
Natural neighbour · Q1
Weight each station's reading by the area of its old cell that the new school cell takes over.
03
Logistic limit · Q2
In $N = \frac{310}{1 + C e ^{- k t}}$ , the term $C e^{- k t}$ tends to 0, so $N$ tends to 310.
04
Fit measure · Q2
$E = \frac{S S _{r es}}{n}$ is small when the model predicts the data well.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers