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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

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 1–12 · 80 marks

Question 1

5 marks

A pyramid in coordinates: centre of the base, height and an edge

Geometry and trigonometry

(a)

2 marks

$G$ is the midpoint of $A (- 2, 0, 0)$ and $C (1, 3, 0)$ .

G = (\frac{- 2 + 1}{2}, \frac{0 + 3}{2}, 0) = (- 0.5, 1.5, 0)

(b)

1 mark

The vertex $H (- 0.5, 1.5, 5)$ is directly above $G$ .

H G = 5 cm

(c)

2 marks

Use the 3D distance formula.

C H = (1 + 0.5)^{2} + (3 - 1.5)^{2} + (0 - 5)^{2} = 2.25 + 2.25 + 25 \approx 5.43 cm

Question 2

5 marks

Household energy use: the mean and standard deviation from grouped data

Statistics and probability

(a)

4 marks

The midpoints are 600, 800, 1000, 1200, 1400 and 1600.
The standard deviation from the calculator.

\overset{x}{ˉ} = \frac{600 ( 38 ) + 800 ( 45 ) + 1000 ( 25 ) + 1200 ( 18 ) + 1400 ( 16 ) + 1600 ( 8 )}{150} \approx 937 kWh σ \approx 299 kWh

(b)

1 mark

The standard deviation in Eureka (95 kWh) is much smaller than in Helvetia (299 kWh), so the monthly energy use is much less spread out in Eureka, with the households closer together.

Question 3

6 marks

The level of water in a tank: a rational model and its inverse

Functions

(a)

2 marks

Substitute $t = 0.5$ .

h (0.5) = \frac{20}{2 ( 0.5 ) + 5} = \frac{20}{6} = \frac{10}{3} \approx 3.33 m

(b)

3 marks

(i) $h^{- 1} (2.5)$ is the $t$ with $h (t) = 2.5$ .

(ii) It takes 1.5 hours for the water level to reach 2.5 metres.

\frac{20}{2 t + 5} = 2.5 \Rightarrow 2 t + 5 = 8 \Rightarrow t = 1.5

(c)

1 mark

The range of the inverse is the domain of $h$ .

h^{- 1} (x) \geq 0

Question 4

5 marks

A cubic: turning points, decreasing interval and a parallel tangent

Calculus

(a)

2 marks

Differentiate $f^{'} (x) = 3 x^{2} + 4 x - 4 = (3 x - 2) (x + 2)$ and set it to zero.
(i) The local maximum is on the left.
(ii) The local minimum is on the right.

x = - 2 x = \frac{2}{3} \approx 0.667

(b)

1 mark

The function decreases between the local maximum and the local minimum.

- 2 < x < \frac{2}{3}

(c)

2 marks

(i) Parallel to $y = 3 x + 5$ means the same gradient.

(ii) The tangent passes through $(1, - 3)$ .

3 y + 3 = 3 (x - 1) \Rightarrow y = 3 x - 6

Question 5

9 marks

Cell towers and Voronoi diagrams

Geometry and trigonometry

(a)

4 marks

The midpoint of $A (1, 5)$ and $B (7, 8)$ is $(4, 6.5)$ . The gradient of $A B$ is $\frac{8 - 5}{7 - 1} = \frac{1}{2}$ , so the perpendicular bisector has gradient $- 2$ .

y - 6.5 = - 2 (x - 4) \Rightarrow y = - 2 x + 14.5

(b)

2 marks

The new cells for $D (8, 2)$ and $E (14, 10)$ come from the perpendicular bisectors with their nearest neighbours. The diagram that matches is

Option 3

(c)

3 marks

The distance from the parlour $(6, 4)$ to the new tower $D (8, 2)$ .
Since $2.83 < 3$ , the parlour is within the radius of excellent coverage of $D$ . Yes, excellent coverage is guaranteed.

(8 - 6)^{2} + (2 - 4)^{2} = 8 \approx 2.83 km

Question 6

9 marks

A triangle NPV: the cosine rule, the sine rule and a shortest distance

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $N P V$ .

N V^{2} = 2 0^{2} + 2 5^{2} - 2 (20) (25) cos 55. 5^{\circ} \Rightarrow N V \approx 21.4 m

(b)

3 marks

Use the sine rule.

\frac{sin P N V}{25} = \frac{sin 55. 5 ^{\circ}}{21.41} \Rightarrow P N V \approx 74. 2^{\circ}

(c)

3 marks

The shortest distance from $P$ to $[N V]$ is the perpendicular from $P$ to the line $N V$ . In the right-angled triangle with hypotenuse $N P = 20$ and angle $74. 2^{\circ}$ at $N$ ,

d = 20 sin 74.1 7^{\circ} \approx 19.2 m

Question 7

5 marks

Loudness in decibels from the intensity of a sound

Functions

(a)

2 marks

Substitute $I = 1 0^{- 5}$ and $I_{0} = 1 0^{- 12}$ .

L = 10 lo g_{10} \frac{1 0 ^{- 5}}{1 0 ^{- 12}} = 10 lo g_{10} 1 0^{7} = 70 dB

(b)

3 marks

Solve $185 = 10 lo g_{10} \frac{I}{1 0 ^{- 12}}$ .

\frac{I}{1 0 ^{- 12}} = 1 0^{18.5} \Rightarrow I = 1 0^{6.5} \approx 3.16 \times 1 0^{6} I \approx 3.16 \times 1 0^{6} W m^{- 2}

Question 8

7 marks

Customer satisfaction and type of service: a chi-squared test

Statistics and probability

(a)

1 mark

$H_{1}$ : the satisfaction level and the type of service interaction are not independent.

(b)

1 mark

The table is 3 by 3.

ν = (3 - 1) (3 - 1) = 4

(c)

2 marks

Enter the table in the calculator and run a $χ^{2}$ test.

χ_{c a l c}^{2} \approx 3.32

(d)

3 marks

(i) Since $3.32 < 9.488$ (the p-value is about 0.506, which is above 0.05), there is not enough evidence to reject $H_{0}$ . Prakash is not correct.

(ii) There is insufficient evidence at the 5% level to say that satisfaction level and type of service interaction are not independent.

Question 9

7 marks

Profit from the rate of change of profit

Calculus

(a)

5 marks

Integrate $\frac{d P}{d x} = - 10 x + 460$ .
Use $P = 3300$ when $x = 10$ .

P = - 5 x^{2} + 460 x + c 3300 = - 500 + 4600 + c \Rightarrow c = - 800 P = - 5 x^{2} + 460 x - 800

(b)

2 marks

The increase in profit is $P (50) - P (25)$ .

P (50) = - 12500 + 23000 - 800 = 9700, P (25) = - 3125 + 11500 - 800 = 7575 9700 - 7575 = 2125 MUR

Question 10

7 marks

Swimming times: quartiles, the interquartile range and an outlier

Statistics and probability

(a)

1 mark

The upper quartile is 24.70, with a quarter of the 16 swimmers above it.

\frac{1}{4} \times 16 = 4 swimmers

(b)

2 marks

From the box plot, $Q_{1} = 24.28$ and $Q_{3} = 24.70$ .

IQR = 24.70 - 24.28 = 0.42 seconds

(c)

4 marks

(i) The boundaries are

The fastest times 23.96 and 24.12 are above 23.65, and the slowest time 25.12 is below 25.33, so none of them are outliers. Only 25.40 is above 25.33.

(ii) Complete the plot: the lower whisker goes from $Q_{1}$ to 23.96, the upper whisker goes from $Q_{3}$ to 25.12, and the time 25.40 is marked separately as an outlier (with a cross), with no whisker joining it.

Q_{1} - 1.5 (0.42) = 23.65, Q_{3} + 1.5 (0.42) = 25.33

Question 11

8 marks

A 25 year loan paid off early

Number and algebra

(a)

2 marks

Use the finance solver with $N = 300$ , $I = 4.2%$ , $P V = 50000$ , $F V = 0$ , 12 payments a year.

P M T \approx 269.47 euros

(b)

6 marks

(i) After 8 years, $N = 96$ payments have been made. The balance still owed is the future value.

(ii) Without the early payment Sofia would pay $12 \times 25 \times 269.47 = 80 841.00$ . With it, she pays 96 payments plus the single payment.

N = 96, I = 4.2%, P V = 50000, P M T = - 269.47 \Rightarrow F V \approx 39 243.42 euros 12 \times 8 \times 269.47 + 39243.42 = 65 112.54 saved = 80841.00 - 65112.54 \approx 15728 euros

Question 12

7 marks

A happiness model: sampling, correlation and a regression prediction

Statistics and probability

(a)

1 mark

The countries are divided into three wealth groups and two are chosen at random from each group.

stratified sampling

(b)

3 marks

(i) Enter the six pairs in the calculator.

(ii) Since $0.854 > 0.8$ , the model can be regarded as a valid predictor.

r \approx 0.854

(c)

1 mark

The regression line of $y$ on $x$ .

y = 0.266 x + 2.40

(d)

2 marks

Substitute $x = 17.2$ .

y = 0.2659 (17.2) + 2.3957 \approx 6.97

Our analysis

Where the marks are

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

Statistics and probability26 marks · 33%
Q2, Q8, Q10, Q12
Geometry and trigonometry23 marks · 29%
Q1, Q5, Q6
Calculus12 marks · 15%
Q4, Q9
Functions11 marks · 14%
Q3, Q7
Number and algebra8 marks · 10%
Q11

Mentor insights

Traps to watch for

01
Perpendicular bisector · Q5
Midpoint and negative reciprocal gradient.
02
Shortest distance to a line · Q6
It is the perpendicular distance, which is a side of a right-angled triangle.
03
Test for independence · Q8
$ν = (r - 1) (c - 1)$ , and reject $H_{0}$ when $χ_{c a l c}^{2}$ exceeds the critical value.
04
Integrate a rate · Q9
$P = \int \frac{d P}{d x} d x$ , and use the known profit to find the constant.
05
Remaining balance · Q11
After 96 payments, the balance is the value found with the finance solver, with the payment entered.

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

Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

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 1–12 · 80 marks

Question 1

5 marks

A pyramid in coordinates: centre of the base, height and an edge

Geometry and trigonometry

(a)

2 marks

$G$ is the midpoint of $A (- 2, 0, 0)$ and $C (1, 3, 0)$ .

G = (\frac{- 2 + 1}{2}, \frac{0 + 3}{2}, 0) = (- 0.5, 1.5, 0)

(b)

1 mark

The vertex $H (- 0.5, 1.5, 5)$ is directly above $G$ .

H G = 5 cm

(c)

2 marks

Use the 3D distance formula.

C H = (1 + 0.5)^{2} + (3 - 1.5)^{2} + (0 - 5)^{2} = 2.25 + 2.25 + 25 \approx 5.43 cm

Question 2

5 marks

Household energy use: the mean and standard deviation from grouped data

Statistics and probability

(a)

4 marks

The midpoints are 600, 800, 1000, 1200, 1400 and 1600.
The standard deviation from the calculator.

\overset{x}{ˉ} = \frac{600 ( 38 ) + 800 ( 45 ) + 1000 ( 25 ) + 1200 ( 18 ) + 1400 ( 16 ) + 1600 ( 8 )}{150} \approx 937 kWh σ \approx 299 kWh

(b)

1 mark

The standard deviation in Eureka (95 kWh) is much smaller than in Helvetia (299 kWh), so the monthly energy use is much less spread out in Eureka, with the households closer together.

Question 3

6 marks

The level of water in a tank: a rational model and its inverse

Functions

(a)

2 marks

Substitute $t = 0.5$ .

h (0.5) = \frac{20}{2 ( 0.5 ) + 5} = \frac{20}{6} = \frac{10}{3} \approx 3.33 m

(b)

3 marks

(i) $h^{- 1} (2.5)$ is the $t$ with $h (t) = 2.5$ .

(ii) It takes 1.5 hours for the water level to reach 2.5 metres.

\frac{20}{2 t + 5} = 2.5 \Rightarrow 2 t + 5 = 8 \Rightarrow t = 1.5

(c)

1 mark

The range of the inverse is the domain of $h$ .

h^{- 1} (x) \geq 0

Question 4

5 marks

A cubic: turning points, decreasing interval and a parallel tangent

Calculus

(a)

2 marks

Differentiate $f^{'} (x) = 3 x^{2} + 4 x - 4 = (3 x - 2) (x + 2)$ and set it to zero.
(i) The local maximum is on the left.
(ii) The local minimum is on the right.

x = - 2 x = \frac{2}{3} \approx 0.667

(b)

1 mark

The function decreases between the local maximum and the local minimum.

- 2 < x < \frac{2}{3}

(c)

2 marks

(i) Parallel to $y = 3 x + 5$ means the same gradient.

(ii) The tangent passes through $(1, - 3)$ .

3 y + 3 = 3 (x - 1) \Rightarrow y = 3 x - 6

Question 5

9 marks

Cell towers and Voronoi diagrams

Geometry and trigonometry

(a)

4 marks

The midpoint of $A (1, 5)$ and $B (7, 8)$ is $(4, 6.5)$ . The gradient of $A B$ is $\frac{8 - 5}{7 - 1} = \frac{1}{2}$ , so the perpendicular bisector has gradient $- 2$ .

y - 6.5 = - 2 (x - 4) \Rightarrow y = - 2 x + 14.5

(b)

2 marks

The new cells for $D (8, 2)$ and $E (14, 10)$ come from the perpendicular bisectors with their nearest neighbours. The diagram that matches is

Option 3

(c)

3 marks

The distance from the parlour $(6, 4)$ to the new tower $D (8, 2)$ .
Since $2.83 < 3$ , the parlour is within the radius of excellent coverage of $D$ . Yes, excellent coverage is guaranteed.

(8 - 6)^{2} + (2 - 4)^{2} = 8 \approx 2.83 km

Question 6

9 marks

A triangle NPV: the cosine rule, the sine rule and a shortest distance

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $N P V$ .

N V^{2} = 2 0^{2} + 2 5^{2} - 2 (20) (25) cos 55. 5^{\circ} \Rightarrow N V \approx 21.4 m

(b)

3 marks

Use the sine rule.

\frac{sin P N V}{25} = \frac{sin 55. 5 ^{\circ}}{21.41} \Rightarrow P N V \approx 74. 2^{\circ}

(c)

3 marks

The shortest distance from $P$ to $[N V]$ is the perpendicular from $P$ to the line $N V$ . In the right-angled triangle with hypotenuse $N P = 20$ and angle $74. 2^{\circ}$ at $N$ ,

d = 20 sin 74.1 7^{\circ} \approx 19.2 m

Question 7

5 marks

Loudness in decibels from the intensity of a sound

Functions

(a)

2 marks

Substitute $I = 1 0^{- 5}$ and $I_{0} = 1 0^{- 12}$ .

L = 10 lo g_{10} \frac{1 0 ^{- 5}}{1 0 ^{- 12}} = 10 lo g_{10} 1 0^{7} = 70 dB

(b)

3 marks

Solve $185 = 10 lo g_{10} \frac{I}{1 0 ^{- 12}}$ .

\frac{I}{1 0 ^{- 12}} = 1 0^{18.5} \Rightarrow I = 1 0^{6.5} \approx 3.16 \times 1 0^{6} I \approx 3.16 \times 1 0^{6} W m^{- 2}

Question 8

7 marks

Customer satisfaction and type of service: a chi-squared test

Statistics and probability

(a)

1 mark

$H_{1}$ : the satisfaction level and the type of service interaction are not independent.

(b)

1 mark

The table is 3 by 3.

ν = (3 - 1) (3 - 1) = 4

(c)

2 marks

Enter the table in the calculator and run a $χ^{2}$ test.

χ_{c a l c}^{2} \approx 3.32

(d)

3 marks

(i) Since $3.32 < 9.488$ (the p-value is about 0.506, which is above 0.05), there is not enough evidence to reject $H_{0}$ . Prakash is not correct.

(ii) There is insufficient evidence at the 5% level to say that satisfaction level and type of service interaction are not independent.

Question 9

7 marks

Profit from the rate of change of profit

Calculus

(a)

5 marks

Integrate $\frac{d P}{d x} = - 10 x + 460$ .
Use $P = 3300$ when $x = 10$ .

P = - 5 x^{2} + 460 x + c 3300 = - 500 + 4600 + c \Rightarrow c = - 800 P = - 5 x^{2} + 460 x - 800

(b)

2 marks

The increase in profit is $P (50) - P (25)$ .

P (50) = - 12500 + 23000 - 800 = 9700, P (25) = - 3125 + 11500 - 800 = 7575 9700 - 7575 = 2125 MUR

Question 10

7 marks

Swimming times: quartiles, the interquartile range and an outlier

Statistics and probability

(a)

1 mark

The upper quartile is 24.70, with a quarter of the 16 swimmers above it.

\frac{1}{4} \times 16 = 4 swimmers

(b)

2 marks

From the box plot, $Q_{1} = 24.28$ and $Q_{3} = 24.70$ .

IQR = 24.70 - 24.28 = 0.42 seconds

(c)

4 marks

(i) The boundaries are

The fastest times 23.96 and 24.12 are above 23.65, and the slowest time 25.12 is below 25.33, so none of them are outliers. Only 25.40 is above 25.33.

Q_{1} - 1.5 (0.42) = 23.65, Q_{3} + 1.5 (0.42) = 25.33

Question 11

8 marks

A 25 year loan paid off early

Number and algebra

(a)

2 marks

Use the finance solver with $N = 300$ , $I = 4.2%$ , $P V = 50000$ , $F V = 0$ , 12 payments a year.

P M T \approx 269.47 euros

(b)

6 marks

(i) After 8 years, $N = 96$ payments have been made. The balance still owed is the future value.

(ii) Without the early payment Sofia would pay $12 \times 25 \times 269.47 = 80 841.00$ . With it, she pays 96 payments plus the single payment.

N = 96, I = 4.2%, P V = 50000, P M T = - 269.47 \Rightarrow F V \approx 39 243.42 euros 12 \times 8 \times 269.47 + 39243.42 = 65 112.54 saved = 80841.00 - 65112.54 \approx 15728 euros

Question 12

7 marks

A happiness model: sampling, correlation and a regression prediction

Statistics and probability

(a)

1 mark

The countries are divided into three wealth groups and two are chosen at random from each group.

stratified sampling

(b)

3 marks

(i) Enter the six pairs in the calculator.

(ii) Since $0.854 > 0.8$ , the model can be regarded as a valid predictor.

r \approx 0.854

(c)

1 mark

The regression line of $y$ on $x$ .

y = 0.266 x + 2.40

(d)

2 marks

Substitute $x = 17.2$ .

y = 0.2659 (17.2) + 2.3957 \approx 6.97

Our analysis

Where the marks are

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

Statistics and probability26 marks · 33%
Q2, Q8, Q10, Q12
Geometry and trigonometry23 marks · 29%
Q1, Q5, Q6
Calculus12 marks · 15%
Q4, Q9
Functions11 marks · 14%
Q3, Q7
Number and algebra8 marks · 10%
Q11

Mentor insights

Traps to watch for

01
Perpendicular bisector · Q5
Midpoint and negative reciprocal gradient.
02
Shortest distance to a line · Q6
It is the perpendicular distance, which is a side of a right-angled triangle.
03
Test for independence · Q8
$ν = (r - 1) (c - 1)$ , and reject $H_{0}$ when $χ_{c a l c}^{2}$ exceeds the critical value.
04
Integrate a rate · Q9
$P = \int \frac{d P}{d x} d x$ , and use the known profit to find the constant.
05
Remaining balance · Q11
After 96 payments, the balance is the value found with the finance solver, with the payment entered.

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