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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

7 marks

River depth data: a line through the origin and a regression line

Statistics and probability

(a)

3 marks

(i) Compute the means on the calculator.

(ii) The gradient of the line through the origin and the mean point.

\overset{x}{ˉ} = \frac{1650}{7} \approx 236, \overset{y}{ˉ} = \frac{234}{7} \approx 33.4 m = \frac{33.43}{235.71} \approx 0.142 \Rightarrow y = 0.142 x

(b)

2 marks

Find the regression line of $y$ on $x$ on the calculator.

a \approx 0.126, b \approx 3.84

(c)

2 marks

Substitute $x = 350$ .

y = 0.12552 (350) + 3.84278 \approx 47.8 cm

Question 2

5 marks

A boat, a port and a lighthouse: bearings and the sine rule

Geometry and trigonometry

(a)

1 mark

The boat sails east (bearing 090°) and the lighthouse is on bearing 056° from $P$ . The angle between east and the direction 056° is $9 0^{\circ} - 5 6^{\circ} = 3 4^{\circ}$ .

L P B = 3 4^{\circ}

(b)

4 marks

At $B$ the lighthouse is on bearing 290°, which is $2 0^{\circ}$ north of west (since west is 270°). The line $B P$ points west, so
Use the sine rule in triangle $P B L$ , in which $L B$ is opposite the angle at $P$ and $P L = 5$ is opposite the angle at $B$ .

L B P = 29 0^{\circ} - 27 0^{\circ} = 2 0^{\circ} \frac{L B}{sin 3 4 ^{\circ}} = \frac{5}{sin 2 0 ^{\circ}} \Rightarrow L B = \frac{5 sin 3 4 ^{\circ}}{sin 2 0 ^{\circ}} \approx 8.17 km

Question 3

5 marks

A double angle identity and a trigonometric equation

Functions

(a)

2 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .

3 cos 2 x + 11 sin x = 3 (1 - 2 sin^{2} x) + 11 sin x = 3 + 11 sin x - 6 sin^{2} x

(b)

3 marks

Use part (a) in $3 cos 2 x + 11 sin x - 6 = 0$ .
Factorise (or use the quadratic formula).
Since $sin x \neq = \frac{3}{2}$ , we have $sin x = \frac{1}{3}$ .

3 + 11 sin x - 6 sin^{2} x - 6 = 0 \Rightarrow 6 sin^{2} x - 11 sin x + 3 = 0 (3 sin x - 1) (2 sin x - 3) = 0 x \approx 19. 5^{\circ} or x \approx 160. 5^{\circ}

Question 4

6 marks

A quiz median and finding two scores from the mean and range

Statistics and probability

(a)

1 mark

Aiden's score is below 10.5 and Brett's is above 10.5, so one new value is on each side of the median. The middle values stay the same, so the median is still 10.5.

(b)

5 marks

The total for the 28 pupils is $10.5 \times 28 = 294$ and the total for all 30 is $10.6 \times 30 = 318$ .
Aiden has the lowest score and Brett the highest, so the range is $B - A = 14$ .
Add the two equations.
Aiden scored 5 and Brett scored 19.

A + B = 318 - 294 = 24 2 B = 38 \Rightarrow B = 19, A = 24 - 19 = 5

Question 5

6 marks

A sector and a circle with related perimeter and area

Geometry and trigonometry

(a)

3 marks

The circumference of the circle is $2 π (8) = 16 π$ , so the perimeter of the sector is $1.5 \times 16 π = 24 π$ .

2 r + r θ = 24 π \Rightarrow r (2 + θ) = 24 π r = \frac{24 π}{2 + θ}

(b)

3 marks

Equate the areas and substitute for $r$ .
Simplify.
Solve this quadratic: $θ^{2} + (4 - 4.5 π) θ + 4 = 0$ . Its roots are about $0.411$ and $9.73$ . The angle is acute, so $θ < \frac{π}{2}$ .

\frac{1}{2} (\frac{24 π}{2 + θ})^{2} θ = 64 π 288 π^{2} θ = 64 π (2 + θ)^{2} \Rightarrow (2 + θ)^{2} = 4.5 π θ θ \approx 0.411

Question 6

7 marks

Normal heights of two sunflower varieties

Statistics and probability

(a)

3 marks

The height $180$ cm is exceeded by 25% of large sunflowers, so it is the upper quartile: $\frac{180 - m}{s} = z$ with $P (Z > z) = 0.25$ .

z = 0.6745 \Rightarrow m + 0.6745 s = 180 A \approx 0.674

(b)

4 marks

The giant sunflowers have mean $m + 35$ and standard deviation $2 s$ . Since 98% exceed 180 cm, 2% are below, so the standardised value of 180 is $- 2.0537$ .
Subtract this from the equation in part (a), $m + 0.6745 s = 180$ .

\frac{180 - ( m + 35 )}{2 s} = - 2.0537 \Rightarrow m - 4.1075 s = 145 (0.6745 + 4.1075) s = 180 - 145 = 35 \Rightarrow s \approx 7.32 m = 180 - 0.6745 (7.319) \approx 175

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

12 marks

Simple interest against compound interest

Number and algebra

(a)

4 marks

(i) $0.315%$ of $$5000$ .

(ii) Twelve months of interest.

(iii) After $n$ years the total is the initial deposit plus $n$ lots of the yearly interest.

(iv) Put $n = 5$ .

0.00315 \times 5000 = $15.75 12 \times 15.75 = $189 5000 + 189 n 5000 + 189 (5) = 5000 + 945 = $5945

(b)

3 marks

(i) Use compound interest for 5 years.

(ii) The interest is the increase.

5000 (1.03)^{5} \approx $5796 5796 - 5000 = $796

(c)

5 marks

(i) The amount after $n$ years.

(ii) Find the smallest $n$ with $5000 (1.03)^{n} > 5000 + 189 n$ . Use a table or graph on the calculator.

Jess has more money after

5000 (1.03)^{n} n = 16 : 8023.5 < 8024, n = 17 : 8264.2 > 8213 17 years

Question 8

15 marks

An ice cream cone with a hemisphere: volume, surface area and minimum wafer

Geometry and trigonometry

(a)

7 marks

(i) The total volume is the cone plus the hemisphere, with $r = 3$ and $h = 8$ .

(ii) The slant height of the cone.

The curved surface area of the wafer.

V = \frac{1}{3} π (3)^{2} (8) + \frac{2}{3} π (3)^{3} = 24 π + 18 π = 42 π \approx 132 cm^{3} l = 3^{2} + 8^{2} = 73 \approx 8.544 S = π r l = π (3) (8.544) \approx 80.5 cm^{2}

(b)

3 marks

Set the total volume to 120.
Multiply by 3 and solve for $h$ .

\frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = 120 π r^{2} h + 2 π r^{3} = 360 \Rightarrow h = \frac{360 - 2 π r ^{3}}{π r ^{2}}

(c)

2 marks

The slant height is $l = r^{2} + h^{2}$ .
Substitute for $h$ .

S = π r l = π r r^{2} + h^{2} S = π r r^{2} + (\frac{360 - 2 π r ^{3}}{π r ^{2}})^{2}

(d)

3 marks

Find the minimum of $S$ on the calculator.

S_{min} \approx 44.4 cm^{2} when r \approx 3.67 cm

Question 9

17 marks

A damped motion: maximum, minimum, total distance and a geometric series of peaks

Calculus

(a)

5 marks

(i) Find the first maximum of $s (t) = 2^{1 - t /5} sin \frac{2 π t}{3}$ on the calculator. It is at $t \approx 0.718$ .

(ii) The velocity is $\frac{d s}{d t}$ . The graph is steepest at $t = 0$ and the oscillations get smaller after that, so the largest speed is at $t = 0$ .

s_{ma x} \approx 1.81 cm v (0) = \frac{d s}{d t}_{t = 0} = 2 \cdot \frac{2 π}{3} cos 0 = \frac{4 π}{3} \approx 4.19 cm/s

(b)

3 marks

(i) The first minimum is at $t \approx 2.218$ .

(ii) Substitute $t = 3.5$ .

s_{min} \approx - 1.47 s (3.5) = 2^{1 - 0.7} sin \frac{7 π}{3} \approx 1.07

(c)

3 marks

In the first 3.5 seconds the particle goes out to $1.806$ and back to $O$ , then out to $1.467$ on the other side and back to $O$ , and then onwards to $1.066$ .

2 (1.806) + 2 (1.467) + 1.066 \approx 7.61 cm

(d)

1 mark

The sine factor is zero when $\frac{2 π t}{3} = π$ .

T = 1.5

(e)

5 marks

(i) The largest distances in each interval are $u_{1} \approx 1.806$ , $u_{2} \approx 1.467$ and $u_{3} \approx 1.192$ . The ratio.

Check: $\frac{u _{3}}{u _{2}} = \frac{1.192}{1.467} \approx 0.812$ . This is because the factor $2^{- t /5}$ shrinks by the same ratio $2^{- 1.5/5} = 2^{- 0.3}$ over every interval of length $T = 1.5$ .

(ii) In each interval the particle travels out and back, covering $2 u_{k}$ .

r = \frac{u _{2}}{u _{1}} = \frac{1.467}{1.806} \approx 0.812 Total = 2 S_{\infty} = \frac{2 ( 1.806 )}{1 - 0.8123} \approx 19.2 cm

Our analysis

Where the marks are

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

Geometry and trigonometry26 marks · 33%
Q2, Q5, Q8
Statistics and probability20 marks · 25%
Q1, Q4, Q6
Calculus17 marks · 21%
Q9
Number and algebra12 marks · 15%
Q7
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Mean point · Q1
A line through $(0, 0)$ and $(\overset{x}{ˉ}, \overset{y}{ˉ})$ has gradient $\frac{y ˉ}{x ˉ}$ .
02
Standard normal · Q6
$P (Z > z) = 0.25$ gives $z \approx 0.6745$ . $P (Z < z) = 0.02$ gives $z \approx - 2.0537$ .
03
Simple interest · Q7
A fixed amount is added each period, so the total follows an arithmetic pattern.
04
Volumes · Q8
Cone $\frac{1}{3} π r^{2} h$ and hemisphere $\frac{2}{3} π r^{3}$ .
05
Velocity · Q9
$v = \frac{d s}{d t}$ .

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

7 marks

River depth data: a line through the origin and a regression line

Statistics and probability

(a)

3 marks

(i) Compute the means on the calculator.

(ii) The gradient of the line through the origin and the mean point.

\overset{x}{ˉ} = \frac{1650}{7} \approx 236, \overset{y}{ˉ} = \frac{234}{7} \approx 33.4 m = \frac{33.43}{235.71} \approx 0.142 \Rightarrow y = 0.142 x

(b)

2 marks

Find the regression line of $y$ on $x$ on the calculator.

a \approx 0.126, b \approx 3.84

(c)

2 marks

Substitute $x = 350$ .

y = 0.12552 (350) + 3.84278 \approx 47.8 cm

Question 2

5 marks

A boat, a port and a lighthouse: bearings and the sine rule

Geometry and trigonometry

(a)

1 mark

The boat sails east (bearing 090°) and the lighthouse is on bearing 056° from $P$ . The angle between east and the direction 056° is $9 0^{\circ} - 5 6^{\circ} = 3 4^{\circ}$ .

L P B = 3 4^{\circ}

(b)

4 marks

At $B$ the lighthouse is on bearing 290°, which is $2 0^{\circ}$ north of west (since west is 270°). The line $B P$ points west, so
Use the sine rule in triangle $P B L$ , in which $L B$ is opposite the angle at $P$ and $P L = 5$ is opposite the angle at $B$ .

L B P = 29 0^{\circ} - 27 0^{\circ} = 2 0^{\circ} \frac{L B}{sin 3 4 ^{\circ}} = \frac{5}{sin 2 0 ^{\circ}} \Rightarrow L B = \frac{5 sin 3 4 ^{\circ}}{sin 2 0 ^{\circ}} \approx 8.17 km

Question 3

5 marks

A double angle identity and a trigonometric equation

Functions

(a)

2 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .

3 cos 2 x + 11 sin x = 3 (1 - 2 sin^{2} x) + 11 sin x = 3 + 11 sin x - 6 sin^{2} x

(b)

3 marks

Use part (a) in $3 cos 2 x + 11 sin x - 6 = 0$ .
Factorise (or use the quadratic formula).
Since $sin x \neq = \frac{3}{2}$ , we have $sin x = \frac{1}{3}$ .

3 + 11 sin x - 6 sin^{2} x - 6 = 0 \Rightarrow 6 sin^{2} x - 11 sin x + 3 = 0 (3 sin x - 1) (2 sin x - 3) = 0 x \approx 19. 5^{\circ} or x \approx 160. 5^{\circ}

Question 4

6 marks

A quiz median and finding two scores from the mean and range

Statistics and probability

(a)

1 mark

Aiden's score is below 10.5 and Brett's is above 10.5, so one new value is on each side of the median. The middle values stay the same, so the median is still 10.5.

(b)

5 marks

The total for the 28 pupils is $10.5 \times 28 = 294$ and the total for all 30 is $10.6 \times 30 = 318$ .
Aiden has the lowest score and Brett the highest, so the range is $B - A = 14$ .
Add the two equations.
Aiden scored 5 and Brett scored 19.

A + B = 318 - 294 = 24 2 B = 38 \Rightarrow B = 19, A = 24 - 19 = 5

Question 5

6 marks

A sector and a circle with related perimeter and area

Geometry and trigonometry

(a)

3 marks

The circumference of the circle is $2 π (8) = 16 π$ , so the perimeter of the sector is $1.5 \times 16 π = 24 π$ .

2 r + r θ = 24 π \Rightarrow r (2 + θ) = 24 π r = \frac{24 π}{2 + θ}

(b)

3 marks

Equate the areas and substitute for $r$ .
Simplify.
Solve this quadratic: $θ^{2} + (4 - 4.5 π) θ + 4 = 0$ . Its roots are about $0.411$ and $9.73$ . The angle is acute, so $θ < \frac{π}{2}$ .

\frac{1}{2} (\frac{24 π}{2 + θ})^{2} θ = 64 π 288 π^{2} θ = 64 π (2 + θ)^{2} \Rightarrow (2 + θ)^{2} = 4.5 π θ θ \approx 0.411

Question 6

7 marks

Normal heights of two sunflower varieties

Statistics and probability

(a)

3 marks

The height $180$ cm is exceeded by 25% of large sunflowers, so it is the upper quartile: $\frac{180 - m}{s} = z$ with $P (Z > z) = 0.25$ .

z = 0.6745 \Rightarrow m + 0.6745 s = 180 A \approx 0.674

(b)

4 marks

The giant sunflowers have mean $m + 35$ and standard deviation $2 s$ . Since 98% exceed 180 cm, 2% are below, so the standardised value of 180 is $- 2.0537$ .
Subtract this from the equation in part (a), $m + 0.6745 s = 180$ .

\frac{180 - ( m + 35 )}{2 s} = - 2.0537 \Rightarrow m - 4.1075 s = 145 (0.6745 + 4.1075) s = 180 - 145 = 35 \Rightarrow s \approx 7.32 m = 180 - 0.6745 (7.319) \approx 175

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

12 marks

Simple interest against compound interest

Number and algebra

(a)

4 marks

(i) $0.315%$ of $$5000$ .

(ii) Twelve months of interest.

(iii) After $n$ years the total is the initial deposit plus $n$ lots of the yearly interest.

(iv) Put $n = 5$ .

0.00315 \times 5000 = $15.75 12 \times 15.75 = $189 5000 + 189 n 5000 + 189 (5) = 5000 + 945 = $5945

(b)

3 marks

(i) Use compound interest for 5 years.

(ii) The interest is the increase.

5000 (1.03)^{5} \approx $5796 5796 - 5000 = $796

(c)

5 marks

(i) The amount after $n$ years.

(ii) Find the smallest $n$ with $5000 (1.03)^{n} > 5000 + 189 n$ . Use a table or graph on the calculator.

Jess has more money after

5000 (1.03)^{n} n = 16 : 8023.5 < 8024, n = 17 : 8264.2 > 8213 17 years

Question 8

15 marks

An ice cream cone with a hemisphere: volume, surface area and minimum wafer

Geometry and trigonometry

(a)

7 marks

(i) The total volume is the cone plus the hemisphere, with $r = 3$ and $h = 8$ .

(ii) The slant height of the cone.

The curved surface area of the wafer.

V = \frac{1}{3} π (3)^{2} (8) + \frac{2}{3} π (3)^{3} = 24 π + 18 π = 42 π \approx 132 cm^{3} l = 3^{2} + 8^{2} = 73 \approx 8.544 S = π r l = π (3) (8.544) \approx 80.5 cm^{2}

(b)

3 marks

Set the total volume to 120.
Multiply by 3 and solve for $h$ .

\frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = 120 π r^{2} h + 2 π r^{3} = 360 \Rightarrow h = \frac{360 - 2 π r ^{3}}{π r ^{2}}

(c)

2 marks

The slant height is $l = r^{2} + h^{2}$ .
Substitute for $h$ .

S = π r l = π r r^{2} + h^{2} S = π r r^{2} + (\frac{360 - 2 π r ^{3}}{π r ^{2}})^{2}

(d)

3 marks

Find the minimum of $S$ on the calculator.

S_{min} \approx 44.4 cm^{2} when r \approx 3.67 cm

Question 9

17 marks

A damped motion: maximum, minimum, total distance and a geometric series of peaks

Calculus

(a)

5 marks

(i) Find the first maximum of $s (t) = 2^{1 - t /5} sin \frac{2 π t}{3}$ on the calculator. It is at $t \approx 0.718$ .

(ii) The velocity is $\frac{d s}{d t}$ . The graph is steepest at $t = 0$ and the oscillations get smaller after that, so the largest speed is at $t = 0$ .

s_{ma x} \approx 1.81 cm v (0) = \frac{d s}{d t}_{t = 0} = 2 \cdot \frac{2 π}{3} cos 0 = \frac{4 π}{3} \approx 4.19 cm/s

(b)

3 marks

(i) The first minimum is at $t \approx 2.218$ .

(ii) Substitute $t = 3.5$ .

s_{min} \approx - 1.47 s (3.5) = 2^{1 - 0.7} sin \frac{7 π}{3} \approx 1.07

(c)

3 marks

In the first 3.5 seconds the particle goes out to $1.806$ and back to $O$ , then out to $1.467$ on the other side and back to $O$ , and then onwards to $1.066$ .

2 (1.806) + 2 (1.467) + 1.066 \approx 7.61 cm

(d)

1 mark

The sine factor is zero when $\frac{2 π t}{3} = π$ .

T = 1.5

(e)

5 marks

(i) The largest distances in each interval are $u_{1} \approx 1.806$ , $u_{2} \approx 1.467$ and $u_{3} \approx 1.192$ . The ratio.

(ii) In each interval the particle travels out and back, covering $2 u_{k}$ .

r = \frac{u _{2}}{u _{1}} = \frac{1.467}{1.806} \approx 0.812 Total = 2 S_{\infty} = \frac{2 ( 1.806 )}{1 - 0.8123} \approx 19.2 cm

Our analysis

Where the marks are

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

Geometry and trigonometry26 marks · 33%
Q2, Q5, Q8
Statistics and probability20 marks · 25%
Q1, Q4, Q6
Calculus17 marks · 21%
Q9
Number and algebra12 marks · 15%
Q7
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Mean point · Q1
A line through $(0, 0)$ and $(\overset{x}{ˉ}, \overset{y}{ˉ})$ has gradient $\frac{y ˉ}{x ˉ}$ .
02
Standard normal · Q6
$P (Z > z) = 0.25$ gives $z \approx 0.6745$ . $P (Z < z) = 0.02$ gives $z \approx - 2.0537$ .
03
Simple interest · Q7
A fixed amount is added each period, so the total follows an arithmetic pattern.
04
Volumes · Q8
Cone $\frac{1}{3} π r^{2} h$ and hemisphere $\frac{2}{3} π r^{3}$ .
05
Velocity · Q9
$v = \frac{d s}{d t}$ .

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