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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: 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 November 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

5 marks

A geometric sequence 2, 6, 18

Number and algebra

(a)

1 mark

Divide consecutive terms.

r = \frac{6}{2} = 3

(b)

2 marks

Use $u_{8} = u_{1} r^{7}$ .

u_{8} = 2 \times 3^{7} = 4374

(c)

2 marks

Use the sum formula.

S_{10} = \frac{2 ( 3 ^{10} - 1 )}{3 - 1} = 3^{10} - 1 = 59048

Question 2

6 marks

A digital caliper: smallest values and percentage error

Geometry and trigonometry

(a)

4 marks

(i) The smallest diameter that rounds to 8.4 is

(ii) The radius is $\frac{8.35}{2} = 4.175$ . Use the volume of a sphere.

8.35 cm V = \frac{4}{3} π (4.175)^{3} \approx 305 cm^{3}

(b)

2 marks

Use the percentage error formula.

\frac{8.4 - 8.38}{8.38} \times 100 \approx 0.239%

Question 3

6 marks

Two delivery companies: a linear charge and a comparison

Functions

(a)

2 marks

Substitute $x = 5.1$ .

T (5.1) = 0.80 (5.1) + 1.50 = $5.58

(b)

2 marks

Read the graph: the line starts at $(0, 2)$ and passes through $(1, 3)$ , so the intercept is 2 and the gradient is 1.
(i) $a = 1$
(ii) $b = 2$

(c)

2 marks

TurboEats has the smaller gradient (0.8 compared with 1), so it costs less per kilometre. It also has the smaller fixed amount ($1.50 compared with $2). So $T (x) < G (x)$ for every $x \geq 0$ .

Question 4

7 marks

A two-sample t-test on fuel efficiency

Statistics and probability

(a)

1 mark

One assumption: the distances per litre for both fuel types come from normal populations with equal variances (and the days are independent).

(b)

2 marks

(i) The null hypothesis is that there is no difference in the mean distance per litre.

(ii) The alternative hypothesis, since Jacob is testing for further travel using premium.

H_{0} : μ_{s t an d a r d} = μ_{p r e mi u m} H_{1} : μ_{s t an d a r d} < μ_{p r e mi u m}

(c)

4 marks

(i) Run a pooled two-sample $t$ -test on the calculator.

(ii) The $p$ -value is less than the significance level, $0.0264 < 0.05$ , so reject $H_{0}$ . There is sufficient evidence at the 5% level that Jacob's car travels further per litre on premium fuel than on standard fuel.

p \approx 0.0264

Question 5

6 marks

Savings with monthly deposits: the interest rate and the interest earned

Number and algebra

(a)

3 marks

Enter the values into the finance solver, with $N = 10 \times 12 = 120$ payments and 12 compounding periods per year.

r \approx 7.15

(b)

3 marks

Total deposits: the initial $10000$ plus $120$ payments of $800$ .
Interest is what is left of the final amount.

10000 + 120 \times 800 = 106000 160000 - 106000 = $54 000

Question 6

8 marks

A triangle with sides 10, 8 and 15, and the area left after a sector

Geometry and trigonometry

(a)

3 marks

Use the cosine rule for the angle at $A$ , opposite side $B C = 8$ .

cos A = \frac{1 5 ^{2} + 1 0 ^{2} - 8 ^{2}}{2 ( 15 ) ( 10 )} = \frac{261}{300} = 0.87 B A C \approx 29. 5^{\circ} (0.516 rad)

(b)

5 marks

The area of the triangle.
The sector with radius $A B = 10$ and angle $0.5156$ radians.
Subtract.

\frac{1}{2} (15) (10) sin 29.5 4^{\circ} \approx 36.98 \frac{1}{2} (10)^{2} (0.5156) \approx 25.78 36.98 - 25.78 \approx 11.2 cm^{2}

Question 7

5 marks

Reaction times: a normal distribution and the slowest 15%

Statistics and probability

(a)

2 marks

$T \sim N (5, 1^{2})$ .

P (T < 4) \approx 0.159

(b)

3 marks

(i) Shade the right tail of the curve, beyond the point to the right of 5 that cuts off 15% of the area.

(ii) Use the inverse normal with an area of $0.85$ to the left.

P (T < t) = 0.85 \Rightarrow t \approx 6.04 seconds

Question 8

8 marks

A phone losing value: exponential decay with a floor

Functions

(a)

2 marks

Put $t = 0$ .

V (0) = 2000 + 30000 = 32000 PHP

(b)

4 marks

A 40% decrease leaves 60% of the value.
Set $V (10) = 19200$ .

0.6 \times 32000 = 19200 2000 + 30000 e^{- 10 k} = 19200 \Rightarrow e^{- 10 k} = \frac{17 200}{30 000} k = - \frac{1}{10} ln \frac{17200}{30000} \approx 0.0556

(c)

2 marks

(i) The graph of $V$ has a horizontal asymptote at $V = 2000$ , so $V$ gets close to 2000 but stays above it.

(ii) In reality, the phone could become obsolete when newer technology arrives, or it could stop working, and then its value could fall below 2000.

Question 9

7 marks

Guessing on a quiz show and choosing between two prizes

Statistics and probability

(a)

4 marks

Let $X \sim B (10, 0.25)$ be the number of correct answers.
(i) Exactly five.
(ii) At least three.

P (X = 5) = (5 10) (0.25)^{5} (0.75)^{5} \approx 0.0584 P (X \geq 3) = 1 - P (X \leq 2) \approx 0.474

(b)

2 marks

Heads wins $$500 000$ with probability $0.3$ and tails wins nothing with probability $0.7$ .

E = 0.3 \times 500000 + 0.7 \times 0 = $150 000

(c)

1 mark

Option B, because the expected winnings of $$150 000$ are much greater than the guaranteed $$50 000$ . (A person who does not want to risk getting nothing could reasonably take Option A instead, because it is a single toss and the chance of getting nothing is $0.7$ .)

Question 10

7 marks

Draining a reservoir into a pond: inverse variation

Functions

(a)

2 marks

Inverse variation means $V = \frac{k}{t}$ . Use $V = 300$ at $t = 5$ .

300 = \frac{k}{5} \Rightarrow k = 1500 V = \frac{1500}{t}

(b)

2 marks

The total volume is 300, shared equally when each has 150.

\frac{1500}{t} = 150 \Rightarrow t = 10

(c)

2 marks

The graph of $y = P (t)$ starts at $(5, 0)$ and increases, passing through about $(10, 150)$ . It is a smooth curve that is concave down and approaches 300 for large $t$ .

(d)

1 mark

The pond holds whatever has left the reservoir.

P (t) = 300 - \frac{1500}{t}

Question 11

8 marks

Minimising a cost-like function and the number of chefs

Calculus

(a)

6 marks

(i) Write $B (x) = 20 + 32 x^{- 1} + 2 x$ and differentiate.

(ii) Set the derivative to zero.

This is in the domain $x \geq 1$ and there is no local maximum, so it gives the minimum.

B^{'} (x) = - 32 x^{- 2} + 2 = - \frac{32}{x ^{2}} + 2 - \frac{32}{x ^{2}} + 2 = 0 \Rightarrow x^{2} = 16 \Rightarrow x = 4 B (4) = 20 + 8 + 8 = 36

(b)

2 marks

Solve $T (k) < 40$ .
So $2 < k < 8$ , and $k$ is a whole number.

20 + \frac{32}{k} + 2 k < 40 \Rightarrow 2 k^{2} - 20 k + 32 < 0 \Rightarrow (k - 2) (k - 8) < 0 k = 3, 4, 5, 6, 7

Question 12

7 marks

A Voronoi diagram for three police stations

Geometry and trigonometry

(a)

2 marks

(i) Every point in the shaded region is closer to $B$ than to any other police station.

(ii) The point $D$ is equidistant from $A$ , $B$ and $C$ .

(b)

5 marks

The edge between $A$ and $B$ is the perpendicular bisector of $A B$ and has gradient 2, so the line $A B$ has gradient $- \frac{1}{2}$ . It passes through $A (3, 5)$ .
The edge between $C$ and $B$ has gradient $- 1$ , so the line $C B$ has gradient 1. It passes through $C (10, 6)$ .
$B$ is where these two lines meet.
Check: the midpoint of $A B$ is $(5, 4)$ , which is on $y = 2 x - 6$ , and the midpoint of $B C$ is $(8.5, 4.5)$ , which is on $y = - x + 13$ .

y - 5 = - \frac{1}{2} (x - 3) \Rightarrow y = - 0.5 x + 6.5 y - 6 = 1 (x - 10) \Rightarrow y = x - 4 - 0.5 x + 6.5 = x - 4 \Rightarrow 1.5 x = 10.5 \Rightarrow x = 7, y = 3 B (7, 3)

Our analysis

Where the marks are

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

Functions21 marks · 26%
Q3, Q8, Q10
Geometry and trigonometry21 marks · 26%
Q2, Q6, Q12
Statistics and probability19 marks · 24%
Q4, Q7, Q9
Number and algebra11 marks · 14%
Q1, Q5
Calculus8 marks · 10%
Q11

Mentor insights

Traps to watch for

01
Pooled t-test · Q4
Assumes both populations are normal with the same variance.
02
Cosine rule · Q6
$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ .
03
Decay · Q8
$V = 2000 + 30000 e^{- k t}$ levels off at 2000.
04
Binomial · Q9
Ten independent guesses, each correct with probability $\frac{1}{4}$ .
05
Minimum · Q11
Set the derivative to 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.

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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: 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 November 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

5 marks

A geometric sequence 2, 6, 18

Number and algebra

(a)

1 mark

Divide consecutive terms.

r = \frac{6}{2} = 3

(b)

2 marks

Use $u_{8} = u_{1} r^{7}$ .

u_{8} = 2 \times 3^{7} = 4374

(c)

2 marks

Use the sum formula.

S_{10} = \frac{2 ( 3 ^{10} - 1 )}{3 - 1} = 3^{10} - 1 = 59048

Question 2

6 marks

A digital caliper: smallest values and percentage error

Geometry and trigonometry

(a)

4 marks

(i) The smallest diameter that rounds to 8.4 is

(ii) The radius is $\frac{8.35}{2} = 4.175$ . Use the volume of a sphere.

8.35 cm V = \frac{4}{3} π (4.175)^{3} \approx 305 cm^{3}

(b)

2 marks

Use the percentage error formula.

\frac{8.4 - 8.38}{8.38} \times 100 \approx 0.239%

Question 3

6 marks

Two delivery companies: a linear charge and a comparison

Functions

(a)

2 marks

Substitute $x = 5.1$ .

T (5.1) = 0.80 (5.1) + 1.50 = $5.58

(b)

2 marks

Read the graph: the line starts at $(0, 2)$ and passes through $(1, 3)$ , so the intercept is 2 and the gradient is 1.
(i) $a = 1$
(ii) $b = 2$

(c)

2 marks

TurboEats has the smaller gradient (0.8 compared with 1), so it costs less per kilometre. It also has the smaller fixed amount ($1.50 compared with $2). So $T (x) < G (x)$ for every $x \geq 0$ .

Question 4

7 marks

A two-sample t-test on fuel efficiency

Statistics and probability

(a)

1 mark

One assumption: the distances per litre for both fuel types come from normal populations with equal variances (and the days are independent).

(b)

2 marks

(i) The null hypothesis is that there is no difference in the mean distance per litre.

(ii) The alternative hypothesis, since Jacob is testing for further travel using premium.

H_{0} : μ_{s t an d a r d} = μ_{p r e mi u m} H_{1} : μ_{s t an d a r d} < μ_{p r e mi u m}

(c)

4 marks

(i) Run a pooled two-sample $t$ -test on the calculator.

p \approx 0.0264

Question 5

6 marks

Savings with monthly deposits: the interest rate and the interest earned

Number and algebra

(a)

3 marks

Enter the values into the finance solver, with $N = 10 \times 12 = 120$ payments and 12 compounding periods per year.

r \approx 7.15

(b)

3 marks

Total deposits: the initial $10000$ plus $120$ payments of $800$ .
Interest is what is left of the final amount.

10000 + 120 \times 800 = 106000 160000 - 106000 = $54 000

Question 6

8 marks

A triangle with sides 10, 8 and 15, and the area left after a sector

Geometry and trigonometry

(a)

3 marks

Use the cosine rule for the angle at $A$ , opposite side $B C = 8$ .

cos A = \frac{1 5 ^{2} + 1 0 ^{2} - 8 ^{2}}{2 ( 15 ) ( 10 )} = \frac{261}{300} = 0.87 B A C \approx 29. 5^{\circ} (0.516 rad)

(b)

5 marks

The area of the triangle.
The sector with radius $A B = 10$ and angle $0.5156$ radians.
Subtract.

\frac{1}{2} (15) (10) sin 29.5 4^{\circ} \approx 36.98 \frac{1}{2} (10)^{2} (0.5156) \approx 25.78 36.98 - 25.78 \approx 11.2 cm^{2}

Question 7

5 marks

Reaction times: a normal distribution and the slowest 15%

Statistics and probability

(a)

2 marks

$T \sim N (5, 1^{2})$ .

P (T < 4) \approx 0.159

(b)

3 marks

(i) Shade the right tail of the curve, beyond the point to the right of 5 that cuts off 15% of the area.

(ii) Use the inverse normal with an area of $0.85$ to the left.

P (T < t) = 0.85 \Rightarrow t \approx 6.04 seconds

Question 8

8 marks

A phone losing value: exponential decay with a floor

Functions

(a)

2 marks

Put $t = 0$ .

V (0) = 2000 + 30000 = 32000 PHP

(b)

4 marks

A 40% decrease leaves 60% of the value.
Set $V (10) = 19200$ .

0.6 \times 32000 = 19200 2000 + 30000 e^{- 10 k} = 19200 \Rightarrow e^{- 10 k} = \frac{17 200}{30 000} k = - \frac{1}{10} ln \frac{17200}{30000} \approx 0.0556

(c)

2 marks

(i) The graph of $V$ has a horizontal asymptote at $V = 2000$ , so $V$ gets close to 2000 but stays above it.

(ii) In reality, the phone could become obsolete when newer technology arrives, or it could stop working, and then its value could fall below 2000.

Question 9

7 marks

Guessing on a quiz show and choosing between two prizes

Statistics and probability

(a)

4 marks

Let $X \sim B (10, 0.25)$ be the number of correct answers.
(i) Exactly five.
(ii) At least three.

P (X = 5) = (5 10) (0.25)^{5} (0.75)^{5} \approx 0.0584 P (X \geq 3) = 1 - P (X \leq 2) \approx 0.474

(b)

2 marks

Heads wins $$500 000$ with probability $0.3$ and tails wins nothing with probability $0.7$ .

E = 0.3 \times 500000 + 0.7 \times 0 = $150 000

(c)

1 mark

Question 10

7 marks

Draining a reservoir into a pond: inverse variation

Functions

(a)

2 marks

Inverse variation means $V = \frac{k}{t}$ . Use $V = 300$ at $t = 5$ .

300 = \frac{k}{5} \Rightarrow k = 1500 V = \frac{1500}{t}

(b)

2 marks

The total volume is 300, shared equally when each has 150.

\frac{1500}{t} = 150 \Rightarrow t = 10

(c)

2 marks

The graph of $y = P (t)$ starts at $(5, 0)$ and increases, passing through about $(10, 150)$ . It is a smooth curve that is concave down and approaches 300 for large $t$ .

(d)

1 mark

The pond holds whatever has left the reservoir.

P (t) = 300 - \frac{1500}{t}

Question 11

8 marks

Minimising a cost-like function and the number of chefs

Calculus

(a)

6 marks

(i) Write $B (x) = 20 + 32 x^{- 1} + 2 x$ and differentiate.

(ii) Set the derivative to zero.

This is in the domain $x \geq 1$ and there is no local maximum, so it gives the minimum.

B^{'} (x) = - 32 x^{- 2} + 2 = - \frac{32}{x ^{2}} + 2 - \frac{32}{x ^{2}} + 2 = 0 \Rightarrow x^{2} = 16 \Rightarrow x = 4 B (4) = 20 + 8 + 8 = 36

(b)

2 marks

Solve $T (k) < 40$ .
So $2 < k < 8$ , and $k$ is a whole number.

20 + \frac{32}{k} + 2 k < 40 \Rightarrow 2 k^{2} - 20 k + 32 < 0 \Rightarrow (k - 2) (k - 8) < 0 k = 3, 4, 5, 6, 7

Question 12

7 marks

A Voronoi diagram for three police stations

Geometry and trigonometry

(a)

2 marks

(i) Every point in the shaded region is closer to $B$ than to any other police station.

(ii) The point $D$ is equidistant from $A$ , $B$ and $C$ .

(b)

5 marks

The edge between $A$ and $B$ is the perpendicular bisector of $A B$ and has gradient 2, so the line $A B$ has gradient $- \frac{1}{2}$ . It passes through $A (3, 5)$ .
The edge between $C$ and $B$ has gradient $- 1$ , so the line $C B$ has gradient 1. It passes through $C (10, 6)$ .
$B$ is where these two lines meet.
Check: the midpoint of $A B$ is $(5, 4)$ , which is on $y = 2 x - 6$ , and the midpoint of $B C$ is $(8.5, 4.5)$ , which is on $y = - x + 13$ .

y - 5 = - \frac{1}{2} (x - 3) \Rightarrow y = - 0.5 x + 6.5 y - 6 = 1 (x - 10) \Rightarrow y = x - 4 - 0.5 x + 6.5 = x - 4 \Rightarrow 1.5 x = 10.5 \Rightarrow x = 7, y = 3 B (7, 3)

Our analysis

Where the marks are

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

Functions21 marks · 26%
Q3, Q8, Q10
Geometry and trigonometry21 marks · 26%
Q2, Q6, Q12
Statistics and probability19 marks · 24%
Q4, Q7, Q9
Number and algebra11 marks · 14%
Q1, Q5
Calculus8 marks · 10%
Q11

Mentor insights

Traps to watch for

01
Pooled t-test · Q4
Assumes both populations are normal with the same variance.
02
Cosine rule · Q6
$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ .
03
Decay · Q8
$V = 2000 + 30000 e^{- k t}$ levels off at 2000.
04
Binomial · Q9
Ten independent guesses, each correct with probability $\frac{1}{4}$ .
05
Minimum · Q11
Set the derivative to 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