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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 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 May 2025 TZ1 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

6 marks

Hot air balloons: correlation and a regression line

Statistics and probability

(a)

2 marks

Enter the six pairs on the calculator.

r \approx 0.999

(b)

2 marks

(i) The regression line.

(ii) The recommended minimum volume increases by about 340 m $^{3}$ for each additional passenger.

y = 340 x + 742

(c)

2 marks

Substitute $x = 10$ .

y = 340.38 (10) + 742.0 \approx 4150 m^{3}

Question 2

5 marks

Car colour preference by age group: a chi-squared test

Statistics and probability

(a)

1 mark

The row for $25 \leq Age < 45$ must add to 58.

b = 58 - 15 - 10 - 12 = 21

(b)

2 marks

Enter the $3 \times 4$ table into the calculator and run a $χ^{2}$ test for independence.

χ_{c a l c}^{2} \approx 18.3

(c)

2 marks

Since $18.3 > 16.81$ (the $p$ -value of 0.00546 is less than the 1% level, 0.01), reject $H_{0}$ . There is sufficient evidence at the 1% level that age and car colour preference are not independent.

Question 3

5 marks

Cupcake boxes: three equations in three unknowns

Number and algebra

(a)

3 marks

(i) Small boxes have 1 cupcake, medium boxes 6 and large boxes 12.

(ii) The prices are 4.50, 23.50 and 44.75 dollars.

x + 6 y + 12 z = 383 4.50 x + 23.50 y + 44.75 z = 1486

(b)

2 marks

Solve the three equations $x + y + z = 88$ , $x + 6 y + 12 z = 383$ and $4.5 x + 23.5 y + 44.75 z = 1486$ on the calculator.
The number of medium boxes sold was

x = 53, y = 15, z = 20 15

Question 4

5 marks

A swimming pool: gradient and inverse proportion

Geometry and trigonometry
Number and algebra

(a)

2 marks

The floor drops $2.0 - 1.2 = 0.8$ m over a horizontal distance of $7.5 - 3.0 - 1.8 = 2.7$ m.
Since $0.296 < \frac{1}{3} = 0.333$ , the pool satisfies the safety regulation.

gradient = \frac{0.8}{2.7} \approx 0.296

(b)

3 marks

The time is inversely proportional to the flow rate, so $w t$ is constant.

300 \times 4.5 = 170 t \Rightarrow t = \frac{1350}{170} \approx 7.94 hours

Question 5

7 marks

A dolphin's jump modelled by a quadratic

Functions

(a)

2 marks

The axis of symmetry is halfway between $d = 3$ and $d = 8.5$ .

d = \frac{3 + 8.5}{2} = 5.75

(b)

2 marks

The axis of symmetry of $- 0.2 d^{2} + b d$ is $d = - \frac{b}{2 ( - 0.2 )}$ .

\frac{b}{0.4} = 5.75 \Rightarrow b = 2.3

(c)

2 marks

The dolphin re-enters the water when $h = 0$ .

- 0.2 d^{2} + 2.3 d = 0 \Rightarrow d (- 0.2 d + 2.3) = 0 \Rightarrow d = 11.5 m

(d)

1 mark

A negative $h$ means the dolphin is below the surface of the water, which is outside the jump the model describes.

Question 6

6 marks

Snow paths: a sector area and a path length

Geometry and trigonometry

(a)

3 marks

The shaded region is one fifth of the area between the circles of radius 4 and 2.8.
Since $5.13 < 6$ , Maureen has enough dye.

A = \frac{1}{5} (π (4)^{2} - π (2.8)^{2}) = \frac{1}{5} π (16 - 7.84) \approx 5.13 m^{2}

(b)

3 marks

The route has three fifths of the outer circle, plus two radii of length 4.

\frac{3}{5} \times 2 π (4) + 2 \times 4 \approx 15.08 + 8 \approx 23.1 m

Question 7

5 marks

Eye colour of five children: conditions for a binomial and probabilities

Statistics and probability

(a)

1 mark

The number of children is fixed at 5. (Other criteria: each child has one of two outcomes, brown eyes or not, and the probability of brown eyes is the same for each child.)

(b)

4 marks

Let $X \sim B (5, 0.75)$ be the number of children with brown eyes.
(i) Exactly three.
(ii) At least four.

P (X = 3) = (3 5) (0.75)^{3} (0.25)^{2} \approx 0.264 P (X \geq 4) = P (X = 4) + P (X = 5) = 5 (0.75)^{4} (0.25) + (0.75)^{5} \approx 0.633

Question 8

7 marks

Finding a constant from a gradient and checking that a function increases

Calculus

(a)

5 marks

Differentiate $f (x) = 2 x^{- 2} + b x^{- 1} + c$ .
Substitute $x = 5$ and $f^{'} (5) = 0.208$ .

f^{'} (x) = - 4 x^{- 3} - b x^{- 2} 0.208 = - \frac{4}{125} - \frac{b}{25} \Rightarrow - \frac{b}{25} = 0.24 \Rightarrow b = - 6

(b)

2 marks

With $b = - 6$ , $f^{'} (x) = - \frac{4}{x ^{3}} + \frac{6}{x ^{2}}$ .
The gradient is positive, so $f$ is increasing at $x = 3.5$ .

f^{'} (3.5) = - \frac{4}{42.875} + \frac{6}{12.25} \approx 0.397 > 0

Question 9

4 marks

The roots of a cubic and an area under it

Calculus

(a)

2 marks

The roots are 1, 5 and $a$ .

1 + 5 + a = 9 \Rightarrow a = 3

(b)

2 marks

The curve is above the $x$ -axis between 1 and 3.

\int_{1}^{3} (x^{3} - 9 x^{2} + 23 x - 15) d x = 4

Question 10

7 marks

Hat prices with a discount after the first 15

Functions

(a)

1 mark

Fifteen hats at $p$ euros each.

15 p

(b)

3 marks

The remaining $n - 15$ hats cost $p - 5$ each.

C (n) = 15 p + (p - 5) (n - 15)

(c)

3 marks

(i) The total is the mean price times the number of hats.

(ii) Substitute $n = 100$ in the cost equation.

25.73 \times 100 = 2573 EUR 2573 = 15 p + (p - 5) (85) \Rightarrow 2573 = 100 p - 425 \Rightarrow p = 29.98

Question 11

5 marks

Wind chill with a percentage error

Functions

(a)

2 marks

Substitute $v = 13$ .

W = - 34.1 - 7.33 ln 13 \approx - 52.9

(b)

3 marks

The approximate value $- 52.9$ is within 6% of the exact value $E$ , so $- 52.9 = E (1 \pm 0.06)$ .
The wind chill index is between

E = \frac{- 52.90}{1.06} \approx - 49.9 or E = \frac{- 52.90}{0.94} \approx - 56.3 - 56.3 (minimum) and - 49.9 (maximum)

Question 12

9 marks

A bag whose length is three times its width: the maximum volume and the limit M

Calculus

(a)

2 marks

Set the derivative to zero.

690 w - 36 w^{2} = 0 \Rightarrow w = \frac{690}{36} \approx 19.17 cm

(b)

4 marks

Integrate $\frac{d V}{d w} = 690 w - 36 w^{2}$ .
The volume is zero when the width is zero, so $c = 0$ .
Substitute $w = 19.17$ .

V = 345 w^{2} - 12 w^{3} + c V = 345 (19.17)^{2} - 12 (19.17)^{3} \approx 42246 \approx 42200 cm^{3}

(c)

3 marks

The length is $3 w = 57.5$ cm and the volume is width times length times height.
The sum of the width, length and height is $M$ .

h = \frac{42 246}{19.17 \times 57.5} \approx 38.3 cm M = 19.17 + 57.5 + 38.3 \approx 115 cm

Question 13

9 marks

A boat off course: the cosine rule and bearings

Geometry and trigonometry

(a)

7 marks

(i) In triangle $D B C$ we have $D B = 2$ , $B C = 13$ and $D B C = 17 0^{\circ}$ . Use the cosine rule.

(ii) Use the sine rule to find the angle $B D C$ .

Moving from the due east direction at $D$ (bearing $09 0^{\circ}$ ) round towards the south by this angle gives

D C^{2} = 2^{2} + 1 3^{2} - 2 (2) (13) cos 17 0^{\circ} \Rightarrow D C \approx 14.97 \approx 15.0 km \frac{sin B D C}{13} = \frac{sin 17 0 ^{\circ}}{14.97} \Rightarrow B D C \approx 8.6 7^{\circ} bearing = 90 + 8.67 \approx 098. 7^{\circ}

(b)

2 marks

The points $D$ and $B$ are both 2 km west of $E$ and $C$ respectively (so $D B$ and $E C$ are parallel and equal in length). So $D B C E$ is a parallelogram and the opposite sides $D E$ and $B C$ are equal.

D E = 13 km

Our analysis

Where the marks are

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

Calculus20 marks · 25%
Q8, Q9, Q12
Functions19 marks · 24%
Q5, Q10, Q11
Geometry and trigonometry17 marks · 21%
Q4, Q6, Q13
Statistics and probability16 marks · 20%
Q1, Q2, Q7
Number and algebra8 marks · 10%
Q3, Q4

Mentor insights

Traps to watch for

01
Symmetry · Q5
The two points at the same height are symmetric about the axis of symmetry.
02
Power rule · Q8
$\frac{d}{d x} x^{- n} = - n x^{- n - 1}$ .
03
Piecewise cost · Q10
The first 15 hats cost $p$ each, the rest cost $p - 5$ each.
04
Maximum · Q12
Set $\frac{d V}{d w} = 0$ .
05
Angle at B · Q13
The bearing of $C$ from $B$ is $10 0^{\circ}$ , and $D$ is due west of $B$ (bearing $27 0^{\circ}$ ), so the angle $D B C$ is $27 0^{\circ} - 10 0^{\circ} = 17 0^{\circ}$ .

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

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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 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 May 2025 TZ1 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

6 marks

Hot air balloons: correlation and a regression line

Statistics and probability

(a)

2 marks

Enter the six pairs on the calculator.

r \approx 0.999

(b)

2 marks

(i) The regression line.

(ii) The recommended minimum volume increases by about 340 m $^{3}$ for each additional passenger.

y = 340 x + 742

(c)

2 marks

Substitute $x = 10$ .

y = 340.38 (10) + 742.0 \approx 4150 m^{3}

Question 2

5 marks

Car colour preference by age group: a chi-squared test

Statistics and probability

(a)

1 mark

The row for $25 \leq Age < 45$ must add to 58.

b = 58 - 15 - 10 - 12 = 21

(b)

2 marks

Enter the $3 \times 4$ table into the calculator and run a $χ^{2}$ test for independence.

χ_{c a l c}^{2} \approx 18.3

(c)

2 marks

Since $18.3 > 16.81$ (the $p$ -value of 0.00546 is less than the 1% level, 0.01), reject $H_{0}$ . There is sufficient evidence at the 1% level that age and car colour preference are not independent.

Question 3

5 marks

Cupcake boxes: three equations in three unknowns

Number and algebra

(a)

3 marks

(i) Small boxes have 1 cupcake, medium boxes 6 and large boxes 12.

(ii) The prices are 4.50, 23.50 and 44.75 dollars.

x + 6 y + 12 z = 383 4.50 x + 23.50 y + 44.75 z = 1486

(b)

2 marks

Solve the three equations $x + y + z = 88$ , $x + 6 y + 12 z = 383$ and $4.5 x + 23.5 y + 44.75 z = 1486$ on the calculator.
The number of medium boxes sold was

x = 53, y = 15, z = 20 15

Question 4

5 marks

A swimming pool: gradient and inverse proportion

Geometry and trigonometry
Number and algebra

(a)

2 marks

The floor drops $2.0 - 1.2 = 0.8$ m over a horizontal distance of $7.5 - 3.0 - 1.8 = 2.7$ m.
Since $0.296 < \frac{1}{3} = 0.333$ , the pool satisfies the safety regulation.

gradient = \frac{0.8}{2.7} \approx 0.296

(b)

3 marks

The time is inversely proportional to the flow rate, so $w t$ is constant.

300 \times 4.5 = 170 t \Rightarrow t = \frac{1350}{170} \approx 7.94 hours

Question 5

7 marks

A dolphin's jump modelled by a quadratic

Functions

(a)

2 marks

The axis of symmetry is halfway between $d = 3$ and $d = 8.5$ .

d = \frac{3 + 8.5}{2} = 5.75

(b)

2 marks

The axis of symmetry of $- 0.2 d^{2} + b d$ is $d = - \frac{b}{2 ( - 0.2 )}$ .

\frac{b}{0.4} = 5.75 \Rightarrow b = 2.3

(c)

2 marks

The dolphin re-enters the water when $h = 0$ .

- 0.2 d^{2} + 2.3 d = 0 \Rightarrow d (- 0.2 d + 2.3) = 0 \Rightarrow d = 11.5 m

(d)

1 mark

A negative $h$ means the dolphin is below the surface of the water, which is outside the jump the model describes.

Question 6

6 marks

Snow paths: a sector area and a path length

Geometry and trigonometry

(a)

3 marks

The shaded region is one fifth of the area between the circles of radius 4 and 2.8.
Since $5.13 < 6$ , Maureen has enough dye.

A = \frac{1}{5} (π (4)^{2} - π (2.8)^{2}) = \frac{1}{5} π (16 - 7.84) \approx 5.13 m^{2}

(b)

3 marks

The route has three fifths of the outer circle, plus two radii of length 4.

\frac{3}{5} \times 2 π (4) + 2 \times 4 \approx 15.08 + 8 \approx 23.1 m

Question 7

5 marks

Eye colour of five children: conditions for a binomial and probabilities

Statistics and probability

(a)

1 mark

The number of children is fixed at 5. (Other criteria: each child has one of two outcomes, brown eyes or not, and the probability of brown eyes is the same for each child.)

(b)

4 marks

Let $X \sim B (5, 0.75)$ be the number of children with brown eyes.
(i) Exactly three.
(ii) At least four.

P (X = 3) = (3 5) (0.75)^{3} (0.25)^{2} \approx 0.264 P (X \geq 4) = P (X = 4) + P (X = 5) = 5 (0.75)^{4} (0.25) + (0.75)^{5} \approx 0.633

Question 8

7 marks

Finding a constant from a gradient and checking that a function increases

Calculus

(a)

5 marks

Differentiate $f (x) = 2 x^{- 2} + b x^{- 1} + c$ .
Substitute $x = 5$ and $f^{'} (5) = 0.208$ .

f^{'} (x) = - 4 x^{- 3} - b x^{- 2} 0.208 = - \frac{4}{125} - \frac{b}{25} \Rightarrow - \frac{b}{25} = 0.24 \Rightarrow b = - 6

(b)

2 marks

With $b = - 6$ , $f^{'} (x) = - \frac{4}{x ^{3}} + \frac{6}{x ^{2}}$ .
The gradient is positive, so $f$ is increasing at $x = 3.5$ .

f^{'} (3.5) = - \frac{4}{42.875} + \frac{6}{12.25} \approx 0.397 > 0

Question 9

4 marks

The roots of a cubic and an area under it

Calculus

(a)

2 marks

The roots are 1, 5 and $a$ .

1 + 5 + a = 9 \Rightarrow a = 3

(b)

2 marks

The curve is above the $x$ -axis between 1 and 3.

\int_{1}^{3} (x^{3} - 9 x^{2} + 23 x - 15) d x = 4

Question 10

7 marks

Hat prices with a discount after the first 15

Functions

(a)

1 mark

Fifteen hats at $p$ euros each.

15 p

(b)

3 marks

The remaining $n - 15$ hats cost $p - 5$ each.

C (n) = 15 p + (p - 5) (n - 15)

(c)

3 marks

(i) The total is the mean price times the number of hats.

(ii) Substitute $n = 100$ in the cost equation.

25.73 \times 100 = 2573 EUR 2573 = 15 p + (p - 5) (85) \Rightarrow 2573 = 100 p - 425 \Rightarrow p = 29.98

Question 11

5 marks

Wind chill with a percentage error

Functions

(a)

2 marks

Substitute $v = 13$ .

W = - 34.1 - 7.33 ln 13 \approx - 52.9

(b)

3 marks

The approximate value $- 52.9$ is within 6% of the exact value $E$ , so $- 52.9 = E (1 \pm 0.06)$ .
The wind chill index is between

E = \frac{- 52.90}{1.06} \approx - 49.9 or E = \frac{- 52.90}{0.94} \approx - 56.3 - 56.3 (minimum) and - 49.9 (maximum)

Question 12

9 marks

A bag whose length is three times its width: the maximum volume and the limit M

Calculus

(a)

2 marks

Set the derivative to zero.

690 w - 36 w^{2} = 0 \Rightarrow w = \frac{690}{36} \approx 19.17 cm

(b)

4 marks

Integrate $\frac{d V}{d w} = 690 w - 36 w^{2}$ .
The volume is zero when the width is zero, so $c = 0$ .
Substitute $w = 19.17$ .

V = 345 w^{2} - 12 w^{3} + c V = 345 (19.17)^{2} - 12 (19.17)^{3} \approx 42246 \approx 42200 cm^{3}

(c)

3 marks

The length is $3 w = 57.5$ cm and the volume is width times length times height.
The sum of the width, length and height is $M$ .

h = \frac{42 246}{19.17 \times 57.5} \approx 38.3 cm M = 19.17 + 57.5 + 38.3 \approx 115 cm

Question 13

9 marks

A boat off course: the cosine rule and bearings

Geometry and trigonometry

(a)

7 marks

(i) In triangle $D B C$ we have $D B = 2$ , $B C = 13$ and $D B C = 17 0^{\circ}$ . Use the cosine rule.

(ii) Use the sine rule to find the angle $B D C$ .

Moving from the due east direction at $D$ (bearing $09 0^{\circ}$ ) round towards the south by this angle gives

D C^{2} = 2^{2} + 1 3^{2} - 2 (2) (13) cos 17 0^{\circ} \Rightarrow D C \approx 14.97 \approx 15.0 km \frac{sin B D C}{13} = \frac{sin 17 0 ^{\circ}}{14.97} \Rightarrow B D C \approx 8.6 7^{\circ} bearing = 90 + 8.67 \approx 098. 7^{\circ}

(b)

2 marks

D E = 13 km

Our analysis

Where the marks are

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

Calculus20 marks · 25%
Q8, Q9, Q12
Functions19 marks · 24%
Q5, Q10, Q11
Geometry and trigonometry17 marks · 21%
Q4, Q6, Q13
Statistics and probability16 marks · 20%
Q1, Q2, Q7
Number and algebra8 marks · 10%
Q3, Q4

Mentor insights

Traps to watch for

01
Symmetry · Q5
The two points at the same height are symmetric about the axis of symmetry.
02
Power rule · Q8
$\frac{d}{d x} x^{- n} = - n x^{- n - 1}$ .
03
Piecewise cost · Q10
The first 15 hats cost $p$ each, the rest cost $p - 5$ each.
04
Maximum · Q12
Set $\frac{d V}{d w} = 0$ .
05
Angle at B · Q13
The bearing of $C$ from $B$ is $10 0^{\circ}$ , and $D$ is due west of $B$ (bearing $27 0^{\circ}$ ), so the angle $D B C$ is $27 0^{\circ} - 10 0^{\circ} = 17 0^{\circ}$ .

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