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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–11 · 80 marks

Question 1

7 marks

A savings plan and a second account

Number and algebra

(a)

3 marks

Use $N = 120$ , $I = 3.6%$ , $P V = 0$ , $P M T = - 1500$ , 12 payments a year.

F V \approx 216 278.58 EUR

(b)

4 marks

After the withdrawal the amount invested is
Compound quarterly for 15 years, which is 60 quarters.

216278.58 - 100000 = 116 278.58 116278.58 (1 + \frac{0.045}{4})^{60} \approx 227 515.92 EUR

Question 2

8 marks

The difference of two dice: a distribution, the mean and a conditional probability

Statistics and probability

(a)

2 marks

(i) The six doubles.

(ii) Only $(1, 6)$ and $(6, 1)$ give a difference of 5.

P (X = 0) = \frac{6}{36} = \frac{1}{6} P (X = 5) = \frac{2}{36} = \frac{1}{18}

(b)

2 marks

Count the outcomes for each difference: 10 for $X = 1$ , 8 for $X = 2$ and 4 for $X = 4$ .

P (X = 1) = \frac{10}{36}, P (X = 2) = \frac{8}{36}, P (X = 4) = \frac{4}{36}

(c)

2 marks

Use the distribution $\frac{6}{36}, \frac{10}{36}, \frac{8}{36}, \frac{6}{36}, \frac{4}{36}, \frac{2}{36}$ for $x = 0, \dots, 5$ .

E (X) = \frac{0 ( 6 ) + 1 ( 10 ) + 2 ( 8 ) + 3 ( 6 ) + 4 ( 4 ) + 5 ( 2 )}{36} = \frac{70}{36} \approx 1.94

(d)

2 marks

An odd difference means $X = 1, 3$ or 5, with probability $\frac{10 + 6 + 2}{36} = \frac{18}{36}$ .

P (X = 5 ∣ X odd) = \frac{2/36}{18/36} = \frac{1}{9} \approx 0.111

Question 3

6 marks

An ant colony growing by 15% each week

Number and algebra

(a)

3 marks

Ten weeks of growth from the start of week one to the start of week 11.

200 (1.15)^{10} \approx 809

(b)

3 marks

Find the first $n$ with $200 (1.15)^{n} > 3000$ .
Check: $200 (1.15)^{19} = 2846 < 3000$ and $200 (1.15)^{20} = 3273 > 3000$ .

1.1 5^{n} > 15 \Rightarrow n > \frac{ln 15}{ln 1.15} \approx 19.38 k = 20

Question 4

8 marks

Pet food flavours for cats and dogs: chi-squared at two significance levels

Statistics and probability

(a)

1 mark

$H_{0}$ : animal type and flavour preference are independent.
$H_{1}$ : animal type and flavour preference are not independent.

(b)

1 mark

A $2 \times 3$ table.

ν = (2 - 1) (3 - 1) = 2

(c)

2 marks

Run a $χ^{2}$ test on the table.

p \approx 0.0297

(d)

3 marks

(i) At the 1% level: $0.0297 > 0.01$ , so do not reject $H_{0}$ . There is not enough evidence that animal type and flavour preference are related.

(ii) At the 5% level: $0.0297 < 0.05$ , so reject $H_{0}$ . There is sufficient evidence at the 5% level that animal type and flavour preference are not independent.

(e)

1 mark

Jane's conclusion was that preference is not independent, which only happens at the 5% level.

5%

Question 5

8 marks

A ship's voyage: the cosine rule, the sine rule and a bearing

Geometry and trigonometry

(a)

1 mark

The ship sails south-east from $M$ to $S$ , then due east to $V$ . The angle at $S$ between $S M$ (heading back north-west) and $S V$ (east) is $13 5^{\circ}$ .

V S M = 13 5^{\circ}

(b)

3 marks

Use the cosine rule in triangle $M S V$ with $M S = 900$ , $S V = 300$ .

M V^{2} = 90 0^{2} + 30 0^{2} - 2 (900) (300) cos 13 5^{\circ} \Rightarrow M V \approx 1130 km

(c)

4 marks

Use the sine rule to find the angle at $V$ .
From $V$ , the direction to $S$ is due west (bearing $27 0^{\circ}$ ) and $M$ is $34. 2^{\circ}$ further round towards north.

\frac{sin M V S}{900} = \frac{sin 13 5 ^{\circ}}{1132.2} \Rightarrow M V S \approx 34. 2^{\circ} bearing = 270 + 34.2 \approx 30 4^{\circ}

Question 6

6 marks

A speed camera: is the car certainly speeding?

Number and algebra

6 marks

The speed is distance over time. The time is between 1.15 s and 1.25 s.
The largest possible time, 1.25 s, gives the smallest possible speed.
This is below the limit of 8.3 m s $^{- 1}$ , so the car could have been below the limit. (A time of 1.2 s gives $\frac{10}{1.2} = 8.33 > 8.3$ , so it could also have been above.)
It is not certain that the car was exceeding the speed limit.

speed = \frac{10}{1.25} = 8 m s^{- 1}

Question 7

9 marks

A Voronoi diagram: a boundary and two vertices

Geometry and trigonometry

(a)

2 marks

The sites are $D (5, 9)$ and $E (1, 5)$ . The midpoint is $(3, 7)$ and the gradient of $D E$ is $\frac{9 - 5}{5 - 1} = 1$ , so the boundary has gradient $- 1$ .

y - 7 = - (x - 3) \Rightarrow y = 10 - x

(b)

2 marks

(i) Read the vertex $X$ from the diagram.

(ii) The distance from $B (10, 5)$ .

X = (8, 4) B X = 2^{2} + 1^{2} = 5, n = 5

(c)

5 marks

(i) $Y$ is the same distance from $B$ and $F$ , which are at the same height, so it lies on their vertical perpendicular bisector.

(ii) The distances from $Y (8, b)$ to $B (10, 5)$ and to $D (5, 9)$ are equal.

a = 8 (8 - 10)^{2} + (b - 5)^{2} = (8 - 5)^{2} + (9 - b)^{2} \Rightarrow 4 + b^{2} - 10 b + 25 = 9 + 81 - 18 b + b^{2} 8 b = 61 \Rightarrow b = \frac{61}{8}

Question 8

8 marks

Speeds of 80 cars from a cumulative frequency curve

Statistics and probability

(a)

2 marks

From the curve, 72 of the 80 cars are at or below 80 km/h.

\frac{80 - 72}{80} \times 100 = 10%

(b)

3 marks

Each group holds 20 cars, so read the speeds at cumulative frequencies 20, 40 and 60.

p = 44, m = 60, q = 70

(c)

3 marks

The midpoints of the four classes are 22, 52, 65 and 85.

\overset{x}{ˉ} = \frac{20 ( 22 ) + 20 ( 52 ) + 20 ( 65 ) + 20 ( 85 )}{80} = \frac{22 + 52 + 65 + 85}{4} = 56 km h^{- 1}

Question 9

5 marks

A function on five integers and its inverse

Functions

(a)

1 mark

Find the $x$ with $f (x) = 4$ .

x = 3

(b)

1 mark

Find the $x$ with $f (x) = x$ . From the table, $f (1) = 1$ .

x = 1

(c)

3 marks

Read the table backwards: $f (2) = 0$ , $f (1) = 1$ , $f (4) = 2$ , $f (0) = 3$ , $f (3) = 4$ .

f^{- 1} (0) = 2, f^{- 1} (1) = 1, f^{- 1} (2) = 4, f^{- 1} (3) = 0, f^{- 1} (4) = 3

Question 10

6 marks

Two judges ranking eight sheepdogs: Spearman's coefficient

Statistics and probability

(a)

1 mark

Brett's ranking has $B$ in fourth place.

4

(b)

4 marks

Write the ranks of each dog.
Enter the two lists of ranks and find the correlation, or use $r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )}$ with $\sum d^{2} = 0 + 4 + 4 + 0 + 0 + 1 + 1 + 0 = 10$ .

A : (1, 1), B : (4, 2), C : (2, 4), D : (3, 3), E : (5, 5), F : (6, 7), G : (7, 6), H : (8, 8) r_{s} = 1 - \frac{6 ( 10 )}{8 ( 63 )} \approx 0.881

(c)

1 mark

Since $r_{s}$ is close to 1, the two judges ranked the dogs in a very similar order. Since it is not exactly 1, they did not agree on every rank.

Question 11

9 marks

A right-angled triangle in three dimensions

Number and algebra

(a)

2 marks

$A B = (2, 3, 1)$ .

A B = 2^{2} + 3^{2} + 1^{2} = 14 \approx 3.74

(b)

4 marks

The right angle is at $B$ , so $B A \cdot B C = 0$ . With $B A = (- 2, - 3, - 1)$ and $B C = (- 1, 3, k - 2)$ :

2 - 9 - (k - 2) = 0 \Rightarrow - 7 - k + 2 = 0 \Rightarrow k = - 5

(c)

3 marks

$A B = (2, 3, 1)$ and $A C = (1, 6, - 6)$ , with $∣ A C ∣ = 73$ .

cos B A C = \frac{2 + 18 - 6}{14 73} = \frac{14}{14 73} = \frac{14}{73} B A C \approx 64. 0^{\circ}

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 probability30 marks · 38%
Q2, Q4, Q8, Q10
Number and algebra28 marks · 35%
Q1, Q3, Q6, Q11
Geometry and trigonometry17 marks · 21%
Q5, Q7
Functions5 marks · 6%
Q9

Mentor insights

Traps to watch for

01
Differences · Q2
There are 36 equally likely outcomes. The difference $X = 0$ happens for the six doubles.
02
Degrees of freedom · Q4
$(2 - 1) (3 - 1) = 2$ for a $2 \times 3$ table.
03
Bearings · Q5
Bearings are measured clockwise from north.
04
Perpendicular bisector · Q7
The boundary between two sites is the perpendicular bisector of the segment joining them.
05
Right angle · Q11
The scalar product of $B A$ and $B C$ is zero.

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

TZ1TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–11 · 80 marks

Question 1

7 marks

A savings plan and a second account

Number and algebra

(a)

3 marks

Use $N = 120$ , $I = 3.6%$ , $P V = 0$ , $P M T = - 1500$ , 12 payments a year.

F V \approx 216 278.58 EUR

(b)

4 marks

After the withdrawal the amount invested is
Compound quarterly for 15 years, which is 60 quarters.

216278.58 - 100000 = 116 278.58 116278.58 (1 + \frac{0.045}{4})^{60} \approx 227 515.92 EUR

Question 2

8 marks

The difference of two dice: a distribution, the mean and a conditional probability

Statistics and probability

(a)

2 marks

(i) The six doubles.

(ii) Only $(1, 6)$ and $(6, 1)$ give a difference of 5.

P (X = 0) = \frac{6}{36} = \frac{1}{6} P (X = 5) = \frac{2}{36} = \frac{1}{18}

(b)

2 marks

Count the outcomes for each difference: 10 for $X = 1$ , 8 for $X = 2$ and 4 for $X = 4$ .

P (X = 1) = \frac{10}{36}, P (X = 2) = \frac{8}{36}, P (X = 4) = \frac{4}{36}

(c)

2 marks

Use the distribution $\frac{6}{36}, \frac{10}{36}, \frac{8}{36}, \frac{6}{36}, \frac{4}{36}, \frac{2}{36}$ for $x = 0, \dots, 5$ .

E (X) = \frac{0 ( 6 ) + 1 ( 10 ) + 2 ( 8 ) + 3 ( 6 ) + 4 ( 4 ) + 5 ( 2 )}{36} = \frac{70}{36} \approx 1.94

(d)

2 marks

An odd difference means $X = 1, 3$ or 5, with probability $\frac{10 + 6 + 2}{36} = \frac{18}{36}$ .

P (X = 5 ∣ X odd) = \frac{2/36}{18/36} = \frac{1}{9} \approx 0.111

Question 3

6 marks

An ant colony growing by 15% each week

Number and algebra

(a)

3 marks

Ten weeks of growth from the start of week one to the start of week 11.

200 (1.15)^{10} \approx 809

(b)

3 marks

Find the first $n$ with $200 (1.15)^{n} > 3000$ .
Check: $200 (1.15)^{19} = 2846 < 3000$ and $200 (1.15)^{20} = 3273 > 3000$ .

1.1 5^{n} > 15 \Rightarrow n > \frac{ln 15}{ln 1.15} \approx 19.38 k = 20

Question 4

8 marks

Pet food flavours for cats and dogs: chi-squared at two significance levels

Statistics and probability

(a)

1 mark

$H_{0}$ : animal type and flavour preference are independent.
$H_{1}$ : animal type and flavour preference are not independent.

(b)

1 mark

A $2 \times 3$ table.

ν = (2 - 1) (3 - 1) = 2

(c)

2 marks

Run a $χ^{2}$ test on the table.

p \approx 0.0297

(d)

3 marks

(i) At the 1% level: $0.0297 > 0.01$ , so do not reject $H_{0}$ . There is not enough evidence that animal type and flavour preference are related.

(ii) At the 5% level: $0.0297 < 0.05$ , so reject $H_{0}$ . There is sufficient evidence at the 5% level that animal type and flavour preference are not independent.

(e)

1 mark

Jane's conclusion was that preference is not independent, which only happens at the 5% level.

5%

Question 5

8 marks

A ship's voyage: the cosine rule, the sine rule and a bearing

Geometry and trigonometry

(a)

1 mark

The ship sails south-east from $M$ to $S$ , then due east to $V$ . The angle at $S$ between $S M$ (heading back north-west) and $S V$ (east) is $13 5^{\circ}$ .

V S M = 13 5^{\circ}

(b)

3 marks

Use the cosine rule in triangle $M S V$ with $M S = 900$ , $S V = 300$ .

M V^{2} = 90 0^{2} + 30 0^{2} - 2 (900) (300) cos 13 5^{\circ} \Rightarrow M V \approx 1130 km

(c)

4 marks

Use the sine rule to find the angle at $V$ .
From $V$ , the direction to $S$ is due west (bearing $27 0^{\circ}$ ) and $M$ is $34. 2^{\circ}$ further round towards north.

\frac{sin M V S}{900} = \frac{sin 13 5 ^{\circ}}{1132.2} \Rightarrow M V S \approx 34. 2^{\circ} bearing = 270 + 34.2 \approx 30 4^{\circ}

Question 6

6 marks

A speed camera: is the car certainly speeding?

Number and algebra

6 marks

The speed is distance over time. The time is between 1.15 s and 1.25 s.
The largest possible time, 1.25 s, gives the smallest possible speed.
This is below the limit of 8.3 m s $^{- 1}$ , so the car could have been below the limit. (A time of 1.2 s gives $\frac{10}{1.2} = 8.33 > 8.3$ , so it could also have been above.)
It is not certain that the car was exceeding the speed limit.

speed = \frac{10}{1.25} = 8 m s^{- 1}

Question 7

9 marks

A Voronoi diagram: a boundary and two vertices

Geometry and trigonometry

(a)

2 marks

The sites are $D (5, 9)$ and $E (1, 5)$ . The midpoint is $(3, 7)$ and the gradient of $D E$ is $\frac{9 - 5}{5 - 1} = 1$ , so the boundary has gradient $- 1$ .

y - 7 = - (x - 3) \Rightarrow y = 10 - x

(b)

2 marks

(i) Read the vertex $X$ from the diagram.

(ii) The distance from $B (10, 5)$ .

X = (8, 4) B X = 2^{2} + 1^{2} = 5, n = 5

(c)

5 marks

(i) $Y$ is the same distance from $B$ and $F$ , which are at the same height, so it lies on their vertical perpendicular bisector.

(ii) The distances from $Y (8, b)$ to $B (10, 5)$ and to $D (5, 9)$ are equal.

a = 8 (8 - 10)^{2} + (b - 5)^{2} = (8 - 5)^{2} + (9 - b)^{2} \Rightarrow 4 + b^{2} - 10 b + 25 = 9 + 81 - 18 b + b^{2} 8 b = 61 \Rightarrow b = \frac{61}{8}

Question 8

8 marks

Speeds of 80 cars from a cumulative frequency curve

Statistics and probability

(a)

2 marks

From the curve, 72 of the 80 cars are at or below 80 km/h.

\frac{80 - 72}{80} \times 100 = 10%

(b)

3 marks

Each group holds 20 cars, so read the speeds at cumulative frequencies 20, 40 and 60.

p = 44, m = 60, q = 70

(c)

3 marks

The midpoints of the four classes are 22, 52, 65 and 85.

\overset{x}{ˉ} = \frac{20 ( 22 ) + 20 ( 52 ) + 20 ( 65 ) + 20 ( 85 )}{80} = \frac{22 + 52 + 65 + 85}{4} = 56 km h^{- 1}

Question 9

5 marks

A function on five integers and its inverse

Functions

(a)

1 mark

Find the $x$ with $f (x) = 4$ .

x = 3

(b)

1 mark

Find the $x$ with $f (x) = x$ . From the table, $f (1) = 1$ .

x = 1

(c)

3 marks

Read the table backwards: $f (2) = 0$ , $f (1) = 1$ , $f (4) = 2$ , $f (0) = 3$ , $f (3) = 4$ .

f^{- 1} (0) = 2, f^{- 1} (1) = 1, f^{- 1} (2) = 4, f^{- 1} (3) = 0, f^{- 1} (4) = 3

Question 10

6 marks

Two judges ranking eight sheepdogs: Spearman's coefficient

Statistics and probability

(a)

1 mark

Brett's ranking has $B$ in fourth place.

4

(b)

4 marks

Write the ranks of each dog.
Enter the two lists of ranks and find the correlation, or use $r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )}$ with $\sum d^{2} = 0 + 4 + 4 + 0 + 0 + 1 + 1 + 0 = 10$ .

A : (1, 1), B : (4, 2), C : (2, 4), D : (3, 3), E : (5, 5), F : (6, 7), G : (7, 6), H : (8, 8) r_{s} = 1 - \frac{6 ( 10 )}{8 ( 63 )} \approx 0.881

(c)

1 mark

Since $r_{s}$ is close to 1, the two judges ranked the dogs in a very similar order. Since it is not exactly 1, they did not agree on every rank.

Question 11

9 marks

A right-angled triangle in three dimensions

Number and algebra

(a)

2 marks

$A B = (2, 3, 1)$ .

A B = 2^{2} + 3^{2} + 1^{2} = 14 \approx 3.74

(b)

4 marks

The right angle is at $B$ , so $B A \cdot B C = 0$ . With $B A = (- 2, - 3, - 1)$ and $B C = (- 1, 3, k - 2)$ :

2 - 9 - (k - 2) = 0 \Rightarrow - 7 - k + 2 = 0 \Rightarrow k = - 5

(c)

3 marks

$A B = (2, 3, 1)$ and $A C = (1, 6, - 6)$ , with $∣ A C ∣ = 73$ .

cos B A C = \frac{2 + 18 - 6}{14 73} = \frac{14}{14 73} = \frac{14}{73} B A C \approx 64. 0^{\circ}

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 probability30 marks · 38%
Q2, Q4, Q8, Q10
Number and algebra28 marks · 35%
Q1, Q3, Q6, Q11
Geometry and trigonometry17 marks · 21%
Q5, Q7
Functions5 marks · 6%
Q9

Mentor insights

Traps to watch for

01
Differences · Q2
There are 36 equally likely outcomes. The difference $X = 0$ happens for the six doubles.
02
Degrees of freedom · Q4
$(2 - 1) (3 - 1) = 2$ for a $2 \times 3$ table.
03
Bearings · Q5
Bearings are measured clockwise from north.
04
Perpendicular bisector · Q7
The boundary between two sites is the perpendicular bisector of the segment joining them.
05
Right angle · Q11
The scalar product of $B A$ and $B C$ is zero.

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