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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

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 1–5 · 80 marks

Question 1

15 marks

Marathon training: percentage error, arithmetic and geometric sequences

Number and algebra

(a)

3 marks

(i) To the nearest kilometre.

(ii) Percentage error of the rounded value.

42 km \frac{42 - 42.19}{42.19} \times 100 \approx 0.450%

(b)

3 marks

Rolando's distances form an arithmetic sequence with $u_{1} = 6$ and $d = 2$ .
He runs 42 km in week 19.

6 + 2 (n - 1) = 42 \Rightarrow 2 n + 4 = 42 \Rightarrow n = 19

(c)

2 marks

Use the sum formula with $n = 10$ .

S_{10} = \frac{10}{2} [2 (6) + 9 (2)] = 5 \times 30 = 150 km

(d)

4 marks

Davi's distances are geometric with $u_{1} = 3$ and $r = 1.2$ .
Check by table: $n = 15$ gives $3 (1.2)^{14} \approx 38.5$ and $n = 16$ gives $3 (1.2)^{15} \approx 46.2$ .

3 (1.2)^{n - 1} > 42 \Rightarrow 1. 2^{n - 1} > 14 n = 16

(e)

3 marks

The total after $n$ weeks.
Check: after 13 weeks the total is about 145.5, and after 14 weeks it is about 177.6.

\frac{3 ( 1. 2 ^{n} - 1 )}{0.2} > 150 \Rightarrow 1. 2^{n} > 11 n = 14

Question 2

14 marks

Name popularity and rank correlation, a wooden letter N and a t-test

Statistics and probability

(a)

4 marks

Use the calculator with the two lists of ranks (popularity position and alphabet position).
This is in the range 0.600 to 0.799, so the correlation is strong and positive.

r_{s} \approx 0.648

(b)

3 marks

Rectangle minus two cut-out triangles: the rectangle is $13 \times 14$ and each triangle has base 5 and height 7.

13 \times 14 - 2 \times \frac{1}{2} \times 7 \times 5 = 182 - 35 = 147 cm^{2}

(c)

1 mark

The volume is the front area times the depth of 4 cm.

147 \times 4 = 588 cm^{3}

(d)

6 marks

(i) The null hypothesis is that the population means are equal, and the alternative is that they are different.

(ii) Run a pooled two-sample $t$ -test.

(iii) Since $0.102 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that the mean weights of letters from the two machines are different.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} \neq = μ_{B} p \approx 0.102

Question 3

15 marks

A triangular farm, a path and a loan

Geometry and trigonometry

(a)

6 marks

Use the cosine rule for the angle between the sides 1600 and 1500, opposite the side 1400.
The area of the triangle.

cos θ = \frac{160 0 ^{2} + 150 0 ^{2} - 140 0 ^{2}}{2 ( 1600 ) ( 1500 )} \Rightarrow θ \approx 53.5 8^{\circ} \frac{1}{2} (1600) (1500) sin 53.5 8^{\circ} \approx 965579 m^{2} 9.66 \times 1 0^{5} m^{2}

(b)

2 marks

Use Pythagoras with 300 east and 200 north.

d = 30 0^{2} + 20 0^{2} \approx 361 m

(c)

2 marks

The path has area length times width, and costs $$120$ per square metre.

c = 360.555 \times 2 \times 120 \approx $86 533

(d)

5 marks

(i) She pays $60$ payments of $$1750$ .

(ii) Use the finance solver with $N = 60$ , $P V = 86533$ , $P M T = - 1750$ , $F V = 0$ and $P / Y = C / Y = 12$ .

60 \times 1750 - 86533 = 105000 - 86533 = $18 467 r \approx 7.89%

Question 4

18 marks

Tides on an island road: a sine model and when the road is open

Functions

(a)

2 marks

Use the maximum $1.26$ and the minimum $- 0.26$ .

a = \frac{1.26 - ( - 0.26 )}{2} = 0.76

(b)

2 marks

The midline.

d = \frac{1.26 + ( - 0.26 )}{2} = 0.5

(c)

3 marks

The maximum is at 03:36, which is $t = 3.6$ , and the minimum at 10:48, which is $t = 10.8$ . That is half a period.

period = 2 (10.8 - 3.6) = 14.4 hours

(d)

3 marks

In degrees, $b = \frac{360}{14.4} = 25$ . Check: the maximum is at $t = 3.6$ , and $25 \times 3.6 = 9 0^{\circ}$ , where sine is 1.

h (t) = 0.76 sin (25 t) + 0.5

(e)

2 marks

02:30 is $t = 2.5$ .

h (2.5) = 0.76 sin (62. 5^{\circ}) + 0.5 \approx 1.17 m

(f)

1 mark

When $h (t) < 0$ the water is below the road level, so the road is clear of water and the island can be reached by road.

(g)

5 marks

Solve $h (t) = 0$ for $0 \leq t \leq 24$ .
The road is open between $t = 8.85$ and $t = 12.75$ , and between $t = 23.25$ and midnight, $t = 24$ .

0.76 sin (25 t) = - 0.5 \Rightarrow sin (25 t) = - 0.658 t \approx 8.85, t \approx 12.75, t \approx 23.25 (12.7544 - 8.8456) + (24 - 23.2456) \approx 3.909 + 0.754 \approx 4.66 hours

Question 5

18 marks

A rollercoaster section: area, a maximum point, a trapezium estimate and a cost function

Calculus

(a)

4 marks

(i) The area between the curve and the $x$ -axis.

(ii) Evaluate on the calculator.

\int_{0}^{60} (0.001 x^{3} - 0.12 x^{2} + 4.05 x + 3) d x 2070 m^{2}

(b)

4 marks

(i) Find the local maximum of $f$ on the calculator, or solve $f^{'} (x) = 0$ .

(ii) The angle of elevation of $M$ from $O$ .

M \approx (24.2, 44.9) θ = tan^{- 1} (\frac{44.9}{24.2}) \approx 61. 7^{\circ}

(c)

3 marks

Use three trapezia, each of width 20.

\frac{1}{2} (20) (3 + 44) + \frac{1}{2} (20) (44 + 37) + \frac{1}{2} (20) (37 + 30) = 470 + 810 + 670 = 1950 m^{2}

(d)

5 marks

Integrate the rate of change.
Use $C (500) = 1750$ .

C (a) = \int (- 0.00292 a + 4.71) d a = - 0.00146 a^{2} + 4.71 a + c 1750 = - 0.00146 (500)^{2} + 4.71 (500) + c = - 365 + 2355 + c \Rightarrow c = - 240 C (a) = - 0.00146 a^{2} + 4.71 a - 240

(e)

2 marks

The total area of the three boards is 1950 m $^{2}$ .

C (1950) = - 0.00146 (1950)^{2} + 4.71 (1950) - 240 \approx $3390

Our analysis

Where the marks are

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

Functions18 marks · 23%
Q4
Calculus18 marks · 23%
Q5
Number and algebra15 marks · 19%
Q1
Geometry and trigonometry15 marks · 19%
Q3
Statistics and probability14 marks · 18%
Q2

Mentor insights

Traps to watch for

01
Arithmetic · Q1
$u_{n} = u_{1} + (n - 1) d$ and $S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d]$ .
02
Spearman · Q2
Rank correlation uses the ranks of the two variables.
03
Cosine rule and area · Q3
Find an angle with the cosine rule, then use $\frac{1}{2} ab sin C$ .
04
Amplitude and midline · Q4
$a = \frac{max - min}{2}$ and $d = \frac{max + min}{2}$ .
05
Area · Q5
The definite integral of the height over the horizontal span.

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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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

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 1–5 · 80 marks

Question 1

15 marks

Marathon training: percentage error, arithmetic and geometric sequences

Number and algebra

(a)

3 marks

(i) To the nearest kilometre.

(ii) Percentage error of the rounded value.

42 km \frac{42 - 42.19}{42.19} \times 100 \approx 0.450%

(b)

3 marks

Rolando's distances form an arithmetic sequence with $u_{1} = 6$ and $d = 2$ .
He runs 42 km in week 19.

6 + 2 (n - 1) = 42 \Rightarrow 2 n + 4 = 42 \Rightarrow n = 19

(c)

2 marks

Use the sum formula with $n = 10$ .

S_{10} = \frac{10}{2} [2 (6) + 9 (2)] = 5 \times 30 = 150 km

(d)

4 marks

Davi's distances are geometric with $u_{1} = 3$ and $r = 1.2$ .
Check by table: $n = 15$ gives $3 (1.2)^{14} \approx 38.5$ and $n = 16$ gives $3 (1.2)^{15} \approx 46.2$ .

3 (1.2)^{n - 1} > 42 \Rightarrow 1. 2^{n - 1} > 14 n = 16

(e)

3 marks

The total after $n$ weeks.
Check: after 13 weeks the total is about 145.5, and after 14 weeks it is about 177.6.

\frac{3 ( 1. 2 ^{n} - 1 )}{0.2} > 150 \Rightarrow 1. 2^{n} > 11 n = 14

Question 2

14 marks

Name popularity and rank correlation, a wooden letter N and a t-test

Statistics and probability

(a)

4 marks

Use the calculator with the two lists of ranks (popularity position and alphabet position).
This is in the range 0.600 to 0.799, so the correlation is strong and positive.

r_{s} \approx 0.648

(b)

3 marks

Rectangle minus two cut-out triangles: the rectangle is $13 \times 14$ and each triangle has base 5 and height 7.

13 \times 14 - 2 \times \frac{1}{2} \times 7 \times 5 = 182 - 35 = 147 cm^{2}

(c)

1 mark

The volume is the front area times the depth of 4 cm.

147 \times 4 = 588 cm^{3}

(d)

6 marks

(i) The null hypothesis is that the population means are equal, and the alternative is that they are different.

(ii) Run a pooled two-sample $t$ -test.

(iii) Since $0.102 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that the mean weights of letters from the two machines are different.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} \neq = μ_{B} p \approx 0.102

Question 3

15 marks

A triangular farm, a path and a loan

Geometry and trigonometry

(a)

6 marks

Use the cosine rule for the angle between the sides 1600 and 1500, opposite the side 1400.
The area of the triangle.

cos θ = \frac{160 0 ^{2} + 150 0 ^{2} - 140 0 ^{2}}{2 ( 1600 ) ( 1500 )} \Rightarrow θ \approx 53.5 8^{\circ} \frac{1}{2} (1600) (1500) sin 53.5 8^{\circ} \approx 965579 m^{2} 9.66 \times 1 0^{5} m^{2}

(b)

2 marks

Use Pythagoras with 300 east and 200 north.

d = 30 0^{2} + 20 0^{2} \approx 361 m

(c)

2 marks

The path has area length times width, and costs $$120$ per square metre.

c = 360.555 \times 2 \times 120 \approx $86 533

(d)

5 marks

(i) She pays $60$ payments of $$1750$ .

(ii) Use the finance solver with $N = 60$ , $P V = 86533$ , $P M T = - 1750$ , $F V = 0$ and $P / Y = C / Y = 12$ .

60 \times 1750 - 86533 = 105000 - 86533 = $18 467 r \approx 7.89%

Question 4

18 marks

Tides on an island road: a sine model and when the road is open

Functions

(a)

2 marks

Use the maximum $1.26$ and the minimum $- 0.26$ .

a = \frac{1.26 - ( - 0.26 )}{2} = 0.76

(b)

2 marks

The midline.

d = \frac{1.26 + ( - 0.26 )}{2} = 0.5

(c)

3 marks

The maximum is at 03:36, which is $t = 3.6$ , and the minimum at 10:48, which is $t = 10.8$ . That is half a period.

period = 2 (10.8 - 3.6) = 14.4 hours

(d)

3 marks

In degrees, $b = \frac{360}{14.4} = 25$ . Check: the maximum is at $t = 3.6$ , and $25 \times 3.6 = 9 0^{\circ}$ , where sine is 1.

h (t) = 0.76 sin (25 t) + 0.5

(e)

2 marks

02:30 is $t = 2.5$ .

h (2.5) = 0.76 sin (62. 5^{\circ}) + 0.5 \approx 1.17 m

(f)

1 mark

When $h (t) < 0$ the water is below the road level, so the road is clear of water and the island can be reached by road.

(g)

5 marks

Solve $h (t) = 0$ for $0 \leq t \leq 24$ .
The road is open between $t = 8.85$ and $t = 12.75$ , and between $t = 23.25$ and midnight, $t = 24$ .

0.76 sin (25 t) = - 0.5 \Rightarrow sin (25 t) = - 0.658 t \approx 8.85, t \approx 12.75, t \approx 23.25 (12.7544 - 8.8456) + (24 - 23.2456) \approx 3.909 + 0.754 \approx 4.66 hours

Question 5

18 marks

A rollercoaster section: area, a maximum point, a trapezium estimate and a cost function

Calculus

(a)

4 marks

(i) The area between the curve and the $x$ -axis.

(ii) Evaluate on the calculator.

\int_{0}^{60} (0.001 x^{3} - 0.12 x^{2} + 4.05 x + 3) d x 2070 m^{2}

(b)

4 marks

(i) Find the local maximum of $f$ on the calculator, or solve $f^{'} (x) = 0$ .

(ii) The angle of elevation of $M$ from $O$ .

M \approx (24.2, 44.9) θ = tan^{- 1} (\frac{44.9}{24.2}) \approx 61. 7^{\circ}

(c)

3 marks

Use three trapezia, each of width 20.

\frac{1}{2} (20) (3 + 44) + \frac{1}{2} (20) (44 + 37) + \frac{1}{2} (20) (37 + 30) = 470 + 810 + 670 = 1950 m^{2}

(d)

5 marks

Integrate the rate of change.
Use $C (500) = 1750$ .

C (a) = \int (- 0.00292 a + 4.71) d a = - 0.00146 a^{2} + 4.71 a + c 1750 = - 0.00146 (500)^{2} + 4.71 (500) + c = - 365 + 2355 + c \Rightarrow c = - 240 C (a) = - 0.00146 a^{2} + 4.71 a - 240

(e)

2 marks

The total area of the three boards is 1950 m $^{2}$ .

C (1950) = - 0.00146 (1950)^{2} + 4.71 (1950) - 240 \approx $3390

Our analysis

Where the marks are

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

Functions18 marks · 23%
Q4
Calculus18 marks · 23%
Q5
Number and algebra15 marks · 19%
Q1
Geometry and trigonometry15 marks · 19%
Q3
Statistics and probability14 marks · 18%
Q2

Mentor insights

Traps to watch for

01
Arithmetic · Q1
$u_{n} = u_{1} + (n - 1) d$ and $S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d]$ .
02
Spearman · Q2
Rank correlation uses the ranks of the two variables.
03
Cosine rule and area · Q3
Find an angle with the cosine rule, then use $\frac{1}{2} ab sin C$ .
04
Amplitude and midline · Q4
$a = \frac{max - min}{2}$ and $d = \frac{max + min}{2}$ .
05
Area · Q5
The definite integral of the height over the horizontal span.

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