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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

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

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

5 marks

Mode, median, standard deviation and range of ten values

Statistics and probability

(a)

1 mark

The value 21 appears three times, more than any other.

mode = 21

(b)

1 mark

The middle two values of the ordered list are both 19.

median = 19

(c)

1 mark

Use the calculator with the whole set of ten values.

σ \approx 4.96

(d)

2 marks

The largest is 32 and the smallest is 12.

32 - 12 = 20

Question 2

6 marks

A logarithmic model for the number of words a child understands

Functions

(a)

2 marks

Substitute $x = 10$ .

y = - 167 + 6264 lo g_{10} 10 = - 167 + 6264 = 6097

(b)

2 marks

Solve $y = 5000$ .

- 167 + 6264 lo g_{10} x = 5000 \Rightarrow lo g_{10} x = \frac{5167}{6264} \approx 0.8249 x = 1 0^{0.8249} \approx 6.681 years

(c)

2 marks

At $x = 1$ , $y = - 167 + 6264 lo g_{10} 1 = - 167$ .
The model is not realistic, because it gives a negative number of words.

Question 3

6 marks

Main course and ice cream: a chi-squared test for independence

Statistics and probability

(a)

1 mark

Add every entry in the table.

53 + 50 + 47 = 150

(b)

1 mark

$H_{0}$ : the choice of main course is independent of the choice of ice cream flavour.

(c)

2 marks

Run a $χ^{2}$ independence test on the $3 \times 4$ table, with $ν = (3 - 1) (4 - 1) = 6$ .

p \approx 0.0926

(d)

2 marks

Since $0.0926 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that the choice of main course depends on the choice of ice cream flavour.

Question 4

5 marks

Normal heights of pine trees

Statistics and probability

(a)

1 mark

The mean is the middle of a normal distribution, so half the trees are shorter.

P (H < 35) = 0.5

(b)

2 marks

Use the calculator.

P (H > 25) \approx 0.965

(c)

2 marks

Use the inverse normal with an area of $0.08$ to the left.

P (H < k) = 0.08 \Rightarrow k \approx 27.3 m

Question 5

9 marks

Four bakeries: nearest locations and perpendicular bisectors

Geometry and trigonometry

(a)

1 mark

The distances from $O (0, 0)$ : $O P = 5$ , $O N = 116 \approx 10.8$ , $O Q = 116 \approx 10.8$ and $O M = 200 \approx 14.1$ .

nearest: P

(b)

3 marks

Without $P$ , the nearest are $N$ and $Q$ , which are both $116 \approx 10.8$ from $O$ , and $M$ is further away at $14.1$ . So she can go to either $N$ or $Q$ , as they are the same distance away.

(c)

1 mark

$M (10, 10)$ and $N (4, 10)$ are on the same horizontal line, so the line equidistant from them is vertical, through the midpoint $x = 7$ .

x = 7

(d)

4 marks

The gradient of $M P$ .
The perpendicular gradient is $- \frac{2}{3}$ . The midpoint of $M P$ .
The equation of the bisector.

\frac{10 - ( - 5 )}{10 - 0} = \frac{3}{2} (\frac{10 + 0}{2}, \frac{10 + ( - 5 )}{2}) = (5, \frac{5}{2}) y - \frac{5}{2} = - \frac{2}{3} (x - 5) \Rightarrow y = - \frac{2}{3} x + \frac{35}{6} 4 x + 6 y - 35 = 0

Question 6

6 marks

A reciprocal function: value, range and an inverse value

Functions

(a)

1 mark

Substitute.

f (10) = \frac{4}{10} - 3 = - 2.6

(b)

3 marks

For $- 6 \leq x < 0$ , $\frac{4}{x}$ takes values from $- \frac{2}{3}$ down towards $- \infty$ as $x \to 0^{-}$ , so $f (x) \leq - \frac{2}{3} - 3 = - \frac{11}{3} \approx - 3.67$ .
For $0 < x \leq 10$ , $\frac{4}{x} \geq 0.4$ , so $f (x) \geq 0.4 - 3 = - 2.6$ .

y \leq - 3.67 or y \geq - 2.6

(c)

2 marks

Solve $f (x) = - 2$ .

\frac{4}{x} - 3 = - 2 \Rightarrow \frac{4}{x} = 1 \Rightarrow x = 4 f^{- 1} (- 2) = 4

Question 7

7 marks

A tree diagram for morning and afternoon activities

Statistics and probability

(a)

1 mark

The three morning branches add to 1.

a = 1 - 0.5 - 0.3 = 0.2

(b)

2 marks

Multiply along the path ice-skating then climbing.

0.5 \times 0.3 = 0.15

(c)

1 mark

Multiply along the path go-karting then ice-skating.

0.3 \times b = 0.3 b

(d)

3 marks

(i) Ice-skating can be chosen in the morning, or in the afternoon after go-karting or climbing. Add the three paths.

(ii) The two afternoon branches after go-karting add to 1.

0.5 + 0.3 b + 0.2 \times 0.625 = 0.875 0.3 b + 0.625 = 0.875 \Rightarrow b = \frac{0.25}{0.3} \approx 0.833 c = 1 - b = \frac{1}{6} \approx 0.167

Question 8

6 marks

The path of water in a fountain modelled by a parabola

Functions

(a)

1 mark

The vertex $T$ is at height 13 on the $y$ -axis, so the curve passes through $(0, 13)$ .

c = 13

(b)

2 marks

(i) The $x$ -intercepts $R$ and $S$ are 3 apart and symmetric about the $y$ -axis, so $S$ is 1.5 to the right of the origin.

(ii) Substitute $S$ into $y = a x^{2} + b x + 13$ .

S (1.5, 0) 0 = 2.25 a + 1.5 b + 13

(c)

3 marks

The vertex is on the $y$ -axis, so $- \frac{b}{2 a} = 0$ and $b = 0$ .
Substitute in part (b)(ii).

2.25 a + 13 = 0 \Rightarrow a = - \frac{13}{2.25} = - \frac{52}{9} \approx - 5.78 a \approx - 5.78, b = 0

Question 9

6 marks

A cubic function and its tangent

Calculus

(a)

2 marks

Substitute $x = 1$ .

f (1) = 1 - 2 + 5 - 7 = - 3

(b)

2 marks

Differentiate each term.

f^{'} (x) = 3 x^{2} - 4 x + 5

(c)

2 marks

The gradient at $x = 1$ .
The tangent through $(1, - 3)$ .

f^{'} (1) = 3 - 4 + 5 = 4 y + 3 = 4 (x - 1) \Rightarrow y = 4 x - 7

Question 10

7 marks

Guessing on a job assessment: expected value and binomial probabilities

Statistics and probability

(a)

2 marks

Let $X \sim B (15, 0.2)$ .

E (X) = n p = 15 \times 0.2 = 3

(b)

3 marks

Use the binomial probability.

P (X = 3) = (3 15) (0.2)^{3} (0.8)^{12} \approx 0.250

(c)

2 marks

He is offered the job with 7 or more correct.

P (X \geq 7) = 1 - P (X \leq 6) \approx 0.0181

Question 11

6 marks

A cylindrical cake cut into sixteen equal pieces

Geometry and trigonometry

(a)

3 marks

The cake has radius 11 cm and height 10 cm.

V = \frac{1}{16} π (11)^{2} (10) = \frac{605 π}{8} \approx 238 cm^{3}

(b)

3 marks

The curved side is one sixteenth of the circumference, times the height.
The two flat sides are rectangles that are 11 cm by 10 cm.
Add.

\frac{2 π ( 11 )}{16} \times 10 \approx 43.2 cm^{2} 2 \times 11 \times 10 = 220 cm^{2} 43.2 + 220 \approx 263 cm^{2}

Question 12

6 marks

Radioactive decay of two samples

Functions

(a)

3 marks

(i) The initial amount in sample X.

(ii) After 138 days the amount is 50.

a = 100 50 = 100 b^{138} \Rightarrow b^{138} = 0.5 \Rightarrow b = 0. 5^{1/138} \approx 0.995

(b)

3 marks

For sample Y use $f (30) = 4$ with the same $b$ .

4 = a (0.99499)^{30} \Rightarrow a = \frac{4}{0.8602} \approx 4.65 micrograms

Question 13

5 marks

A gift box of fixed volume with minimum surface area

Functions

5 marks

The length is $l = 1.5 w$ , and the volume is $1.5 w \cdot w \cdot h = 10000$ , so
The surface area of a rectangular prism.
Substitute for $h$ .
Differentiate and set equal to zero.

h = \frac{10 000}{1.5 w ^{2}} = \frac{20 000}{3 w ^{2}} S = 2 l w + 2 l h + 2 w h = 3 w^{2} + 5 w h S = 3 w^{2} + 5 w \cdot \frac{20 000}{3 w ^{2}} = 3 w^{2} + \frac{100 000}{3 w} \frac{d S}{d w} = 6 w - \frac{100 000}{3 w ^{2}} = 0 \Rightarrow w^{3} = \frac{100 000}{18} w \approx 17.7 cm

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%
Q1, Q3, Q4, Q7, Q10
Functions29 marks · 36%
Q2, Q6, Q8, Q12, Q13
Geometry and trigonometry15 marks · 19%
Q5, Q11
Calculus6 marks · 8%
Q9

Mentor insights

Traps to watch for

01
Model · Q2
$y = - 167 + 6264 lo g_{10} x$ with $x$ in years.
02
Test · Q3
Compare the $p$ -value with the significance level.
03
Distance · Q5
$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
04
Branch probabilities · Q7
The branches from each point add to 1.
05
Binomial · Q10
15 independent questions, each correct with probability $0.2$ .

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

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

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

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

5 marks

Mode, median, standard deviation and range of ten values

Statistics and probability

(a)

1 mark

The value 21 appears three times, more than any other.

mode = 21

(b)

1 mark

The middle two values of the ordered list are both 19.

median = 19

(c)

1 mark

Use the calculator with the whole set of ten values.

σ \approx 4.96

(d)

2 marks

The largest is 32 and the smallest is 12.

32 - 12 = 20

Question 2

6 marks

A logarithmic model for the number of words a child understands

Functions

(a)

2 marks

Substitute $x = 10$ .

y = - 167 + 6264 lo g_{10} 10 = - 167 + 6264 = 6097

(b)

2 marks

Solve $y = 5000$ .

- 167 + 6264 lo g_{10} x = 5000 \Rightarrow lo g_{10} x = \frac{5167}{6264} \approx 0.8249 x = 1 0^{0.8249} \approx 6.681 years

(c)

2 marks

At $x = 1$ , $y = - 167 + 6264 lo g_{10} 1 = - 167$ .
The model is not realistic, because it gives a negative number of words.

Question 3

6 marks

Main course and ice cream: a chi-squared test for independence

Statistics and probability

(a)

1 mark

Add every entry in the table.

53 + 50 + 47 = 150

(b)

1 mark

$H_{0}$ : the choice of main course is independent of the choice of ice cream flavour.

(c)

2 marks

Run a $χ^{2}$ independence test on the $3 \times 4$ table, with $ν = (3 - 1) (4 - 1) = 6$ .

p \approx 0.0926

(d)

2 marks

Since $0.0926 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that the choice of main course depends on the choice of ice cream flavour.

Question 4

5 marks

Normal heights of pine trees

Statistics and probability

(a)

1 mark

The mean is the middle of a normal distribution, so half the trees are shorter.

P (H < 35) = 0.5

(b)

2 marks

Use the calculator.

P (H > 25) \approx 0.965

(c)

2 marks

Use the inverse normal with an area of $0.08$ to the left.

P (H < k) = 0.08 \Rightarrow k \approx 27.3 m

Question 5

9 marks

Four bakeries: nearest locations and perpendicular bisectors

Geometry and trigonometry

(a)

1 mark

The distances from $O (0, 0)$ : $O P = 5$ , $O N = 116 \approx 10.8$ , $O Q = 116 \approx 10.8$ and $O M = 200 \approx 14.1$ .

nearest: P

(b)

3 marks

Without $P$ , the nearest are $N$ and $Q$ , which are both $116 \approx 10.8$ from $O$ , and $M$ is further away at $14.1$ . So she can go to either $N$ or $Q$ , as they are the same distance away.

(c)

1 mark

$M (10, 10)$ and $N (4, 10)$ are on the same horizontal line, so the line equidistant from them is vertical, through the midpoint $x = 7$ .

x = 7

(d)

4 marks

The gradient of $M P$ .
The perpendicular gradient is $- \frac{2}{3}$ . The midpoint of $M P$ .
The equation of the bisector.

\frac{10 - ( - 5 )}{10 - 0} = \frac{3}{2} (\frac{10 + 0}{2}, \frac{10 + ( - 5 )}{2}) = (5, \frac{5}{2}) y - \frac{5}{2} = - \frac{2}{3} (x - 5) \Rightarrow y = - \frac{2}{3} x + \frac{35}{6} 4 x + 6 y - 35 = 0

Question 6

6 marks

A reciprocal function: value, range and an inverse value

Functions

(a)

1 mark

Substitute.

f (10) = \frac{4}{10} - 3 = - 2.6

(b)

3 marks

For $- 6 \leq x < 0$ , $\frac{4}{x}$ takes values from $- \frac{2}{3}$ down towards $- \infty$ as $x \to 0^{-}$ , so $f (x) \leq - \frac{2}{3} - 3 = - \frac{11}{3} \approx - 3.67$ .
For $0 < x \leq 10$ , $\frac{4}{x} \geq 0.4$ , so $f (x) \geq 0.4 - 3 = - 2.6$ .

y \leq - 3.67 or y \geq - 2.6

(c)

2 marks

Solve $f (x) = - 2$ .

\frac{4}{x} - 3 = - 2 \Rightarrow \frac{4}{x} = 1 \Rightarrow x = 4 f^{- 1} (- 2) = 4

Question 7

7 marks

A tree diagram for morning and afternoon activities

Statistics and probability

(a)

1 mark

The three morning branches add to 1.

a = 1 - 0.5 - 0.3 = 0.2

(b)

2 marks

Multiply along the path ice-skating then climbing.

0.5 \times 0.3 = 0.15

(c)

1 mark

Multiply along the path go-karting then ice-skating.

0.3 \times b = 0.3 b

(d)

3 marks

(i) Ice-skating can be chosen in the morning, or in the afternoon after go-karting or climbing. Add the three paths.

(ii) The two afternoon branches after go-karting add to 1.

0.5 + 0.3 b + 0.2 \times 0.625 = 0.875 0.3 b + 0.625 = 0.875 \Rightarrow b = \frac{0.25}{0.3} \approx 0.833 c = 1 - b = \frac{1}{6} \approx 0.167

Question 8

6 marks

The path of water in a fountain modelled by a parabola

Functions

(a)

1 mark

The vertex $T$ is at height 13 on the $y$ -axis, so the curve passes through $(0, 13)$ .

c = 13

(b)

2 marks

(i) The $x$ -intercepts $R$ and $S$ are 3 apart and symmetric about the $y$ -axis, so $S$ is 1.5 to the right of the origin.

(ii) Substitute $S$ into $y = a x^{2} + b x + 13$ .

S (1.5, 0) 0 = 2.25 a + 1.5 b + 13

(c)

3 marks

The vertex is on the $y$ -axis, so $- \frac{b}{2 a} = 0$ and $b = 0$ .
Substitute in part (b)(ii).

2.25 a + 13 = 0 \Rightarrow a = - \frac{13}{2.25} = - \frac{52}{9} \approx - 5.78 a \approx - 5.78, b = 0

Question 9

6 marks

A cubic function and its tangent

Calculus

(a)

2 marks

Substitute $x = 1$ .

f (1) = 1 - 2 + 5 - 7 = - 3

(b)

2 marks

Differentiate each term.

f^{'} (x) = 3 x^{2} - 4 x + 5

(c)

2 marks

The gradient at $x = 1$ .
The tangent through $(1, - 3)$ .

f^{'} (1) = 3 - 4 + 5 = 4 y + 3 = 4 (x - 1) \Rightarrow y = 4 x - 7

Question 10

7 marks

Guessing on a job assessment: expected value and binomial probabilities

Statistics and probability

(a)

2 marks

Let $X \sim B (15, 0.2)$ .

E (X) = n p = 15 \times 0.2 = 3

(b)

3 marks

Use the binomial probability.

P (X = 3) = (3 15) (0.2)^{3} (0.8)^{12} \approx 0.250

(c)

2 marks

He is offered the job with 7 or more correct.

P (X \geq 7) = 1 - P (X \leq 6) \approx 0.0181

Question 11

6 marks

A cylindrical cake cut into sixteen equal pieces

Geometry and trigonometry

(a)

3 marks

The cake has radius 11 cm and height 10 cm.

V = \frac{1}{16} π (11)^{2} (10) = \frac{605 π}{8} \approx 238 cm^{3}

(b)

3 marks

The curved side is one sixteenth of the circumference, times the height.
The two flat sides are rectangles that are 11 cm by 10 cm.
Add.

\frac{2 π ( 11 )}{16} \times 10 \approx 43.2 cm^{2} 2 \times 11 \times 10 = 220 cm^{2} 43.2 + 220 \approx 263 cm^{2}

Question 12

6 marks

Radioactive decay of two samples

Functions

(a)

3 marks

(i) The initial amount in sample X.

(ii) After 138 days the amount is 50.

a = 100 50 = 100 b^{138} \Rightarrow b^{138} = 0.5 \Rightarrow b = 0. 5^{1/138} \approx 0.995

(b)

3 marks

For sample Y use $f (30) = 4$ with the same $b$ .

4 = a (0.99499)^{30} \Rightarrow a = \frac{4}{0.8602} \approx 4.65 micrograms

Question 13

5 marks

A gift box of fixed volume with minimum surface area

Functions

5 marks

The length is $l = 1.5 w$ , and the volume is $1.5 w \cdot w \cdot h = 10000$ , so
The surface area of a rectangular prism.
Substitute for $h$ .
Differentiate and set equal to zero.

h = \frac{10 000}{1.5 w ^{2}} = \frac{20 000}{3 w ^{2}} S = 2 l w + 2 l h + 2 w h = 3 w^{2} + 5 w h S = 3 w^{2} + 5 w \cdot \frac{20 000}{3 w ^{2}} = 3 w^{2} + \frac{100 000}{3 w} \frac{d S}{d w} = 6 w - \frac{100 000}{3 w ^{2}} = 0 \Rightarrow w^{3} = \frac{100 000}{18} w \approx 17.7 cm

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%
Q1, Q3, Q4, Q7, Q10
Functions29 marks · 36%
Q2, Q6, Q8, Q12, Q13
Geometry and trigonometry15 marks · 19%
Q5, Q11
Calculus6 marks · 8%
Q9

Mentor insights

Traps to watch for

01
Model · Q2
$y = - 167 + 6264 lo g_{10} x$ with $x$ in years.
02
Test · Q3
Compare the $p$ -value with the significance level.
03
Distance · Q5
$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
04
Branch probabilities · Q7
The branches from each point add to 1.
05
Binomial · Q10
15 independent questions, each correct with probability $0.2$ .

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