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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2TZ3

Session: May 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 May 2025 TZ2 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 · 34 marks

Question 1

4 marks

Play time and sleep time: a regression line and an estimate

Statistics and probability

(a)

2 marks

Enter the six pairs.

a \approx 1.98, b \approx 40.2

(b)

2 marks

Substitute $x = 20$ into $y = 1.9765 x + 40.229$ .

y = 1.9765 (20) + 40.229 \approx 79.8 hours

Question 2

6 marks

A square-based pyramid in 3D: height, diagonal and edge angle

Geometry and trigonometry

(a)

2 marks

With $V (1, 7, 0)$ and $X (- 3, 4, 2)$ .

V X = (1 + 3)^{2} + (7 - 4)^{2} + (0 - 2)^{2} = 29 \approx 5.39

(b)

2 marks

The diagonal of a square with side 5.

A C = 5^{2} + 5^{2} = 52 \approx 7.07 cm

(c)

2 marks

$X$ is the centre of the base, so $X C = \frac{5 2}{2}$ and the triangle $V X C$ has a right angle at $X$ .

tan θ = \frac{29}{5 2 /2} = \frac{5.385}{3.536} \Rightarrow θ \approx 56. 7^{\circ}

Question 3

6 marks

Where a function decreases and is concave up from its derivative

Calculus

(a)

3 marks

Solve $f^{'} (x) = 4 + 2 x - 3 e^{x} = 0$ with the calculator. The two roots are $x \approx - 1.736$ and $x \approx 0.518$ .
A graph of $f^{'}$ shows it is below the axis outside these roots.

x < - 1.74 and x > 0.518

(b)

3 marks

Differentiate $f^{'} (x)$ .
Concave up needs $f^{''} (x) > 0$ .

f^{''} (x) = 2 - 3 e^{x} 2 - 3 e^{x} > 0 \Rightarrow e^{x} < \frac{2}{3} \Rightarrow x < ln \frac{2}{3} \approx - 0.405

Question 4

5 marks

Finding k from a binomial coefficient

Number and algebra

5 marks

The term in $x^{6}$ .
Set the coefficient equal to $8.4 \times 1 0^{6}$ .

(6 10) (2 x)^{6} k^{4} = 210 \times 64 k^{4} x^{6} = 13440 k^{4} x^{6} 13440 k^{4} = 8400000 \Rightarrow k^{4} = 625 k = \pm 5

Question 5

7 marks

A particle with velocity e^(-sin t) cos 2t: speed, distance and acceleration

Calculus

(a)

2 marks

Sketch $v (t) = e^{- s i n t} cos 2 t$ on $0 \leq t \leq 5$ . The most extreme value of $v$ is the minimum at $t \approx 4.71 (\frac{3 π}{2})$ , where $v = - e$ .

maximum speed = e \approx 2.72 m s^{- 1}

(b)

2 marks

The distance travelled is the area under the graph of $∣ v ∣$ .

d = \int_{0}^{5} ∣ v (t) ∣ d t \approx 3.85 m

(c)

3 marks

The velocity is zero when $cos 2 t = 0$ : $t = \frac{π}{4}, \frac{3 π}{4}, \dots$ , and the sign of $v$ changes at each one. The second change of direction is at $t = \frac{3 π}{4} \approx 2.356$ .
Differentiate $v$ with the product rule.
At $t = \frac{3 π}{4}$ we have $cos 2 t = 0$ and $sin 2 t = - 1$ .

a = v^{'} (t) = e^{- s i n t} (- cos t cos 2 t - 2 sin 2 t) a = e^{- s i n (3 π /4)} (0 + 2) = 2 e^{- 1/ 2} \approx 0.986 m s^{- 2}

Question 6

6 marks

A circle sector split into a segment and a triangle in the ratio 3 : 5

Number and algebra

6 marks

The segment $P$ has area 12.8, so the triangle $Q$ has area $\frac{5}{3} \times 12.8 = 21.33$ and the sector has area $12.8 + 21.33 = 34.13$ .
Divide the second equation by the first.
Solve for $θ$ with the calculator (in radians).
Substitute into the sector area.

\frac{1}{2} r^{2} θ = 34.13, \frac{1}{2} r^{2} sin θ = 21.33 \frac{sin θ}{θ} = \frac{21.33}{34.13} = 0.625 θ \approx 1.599 \frac{1}{2} r^{2} (1.599) = 34.13 \Rightarrow r^{2} \approx 42.7 r \approx 6.53 cm

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A lighthouse, a Jet Ski and two rescue boats

Geometry and trigonometry

(a)

4 marks

Use the cosine rule in triangle $J C L$ with $C J = 5.5$ , $C L = 8$ .

J L^{2} = 5. 5^{2} + 8^{2} - 2 (5.5) (8) cos 2 8^{\circ} \Rightarrow J L \approx 4.07 km

(b)

3 marks

In the right triangle from the top of the lighthouse (60 m up), $tan 0.9 4^{\circ} = \frac{60}{B L}$ .

B L = \frac{60}{tan 0.9 4 ^{\circ}} \approx 3657 m \approx 3.66 km

(c)

2 marks

The bearing from $B$ to $L$ is $12 1^{\circ}$ . The bearing of the return direction is $18 0^{\circ}$ more.

121 + 180 = 30 1^{\circ}

(d)

2 marks

The time is distance over speed.

t = \frac{3.657}{48} hours = 0.0762 hours \approx 4.57 minutes

(e)

4 marks

The coastguard station $C$ is due west of $L$ (distance 8 km), and $B$ is on the bearing $30 1^{\circ}$ from $L$ , so the angle at $L$ in triangle $C L B$ is $30 1^{\circ} - 27 0^{\circ} = 3 1^{\circ}$ . Use the cosine rule for the distance $C B$ .
The coastguard boat's time at 55 km/h.
Since $4.57 < 5.69$ , the lighthouse rescue boat reaches the Jet Ski first.

C B^{2} = 8^{2} + 3.65 7^{2} - 2 (8) (3.657) cos 3 1^{\circ} \Rightarrow C B \approx 5.217 km t = \frac{5.217}{55} \times 60 \approx 5.69 minutes

Question 8

17 marks

Apple weights: normal probabilities, premium apples in boxes and a second orchard

Statistics and probability

(a)

2 marks

$W \sim N (175, 8^{2})$ .

P (W < 170) \approx 0.266

(b)

2 marks

Use the inverse normal with 20% in the upper tail.

w \approx 181.7 grams

(c)

2 marks

Premium apples weigh between 170 and 185 grams.

P (170 < W < 185) \approx 0.628 = 62.8%

(d)

3 marks

Let $X$ be the number of premium apples in a box, $X \sim B (40, 0.6284)$ .

P (X \geq 30) \approx 0.0739

(e)

2 marks

Let $Y$ be the number of boxes (out of 10) with at least 30 premium apples, $Y \sim B (10, 0.0739)$ .

P (Y = 4) \approx 0.00394

(f)

6 marks

Of all the apples, 82% are premium, so 18% are not. The percentage below 170 is twice the percentage above 185, so $P (M < 170) = 0.12$ and $P (M > 185) = 0.06$ .
The z-scores are $z = - 1.175$ and $z = 1.555$ .
Subtract the first from the second: $\frac{15}{s} = 2.730$ , so $s = 5.495$ . Then $m = 170 + 1.175 s$ .

\frac{170 - m}{s} = - 1.175, \frac{185 - m}{s} = 1.555 m \approx 176.5 \approx 176 grams

Question 9

14 marks

A bouncing tennis ball: geometric sequences and an arithmetic sequence of distances

Number and algebra

(a)

3 marks

Ball 1 starts from 10 m. After the 5th bounce the height is $10 (\frac{2}{3})^{5}$ .

10 (\frac{2}{3})^{5} = \frac{320}{243} \approx 1.32 m

(b)

3 marks

Before the 5th bounce the ball has fallen 10 m and then made four up-and-down trips.

10 + 2 [10 (\frac{2}{3}) + 10 (\frac{2}{3})^{2} + 10 (\frac{2}{3})^{3} + 10 (\frac{2}{3})^{4}] \approx 42.1 m

(c)

3 marks

The ball falls $x$ and then makes infinitely many up-and-down trips. The first rebound height is $\frac{2}{3} x$ , and the ratio is $\frac{2}{3}$ .

d = x + 2 \cdot \frac{\frac{2}{3} x}{1 - \frac{2}{3}} = x + 2 \cdot 2 x = 5 x

(d)

1 mark

With $x = 10$ .

d_{1} = 50

(e)

4 marks

Ball 2 has $d_{2} = 5 \times 9.56 = 47.8$ , so the common difference is $47.8 - 50 = - 2.2$ .
The first ball is

d_{n} = 50 + (n - 1) (- 2.2) < 25 \Rightarrow n - 1 > 11.36 \Rightarrow n > 12.36 Ball 13

Our analysis

Where the marks are

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

Number and algebra25 marks · 31%
Q4, Q6, Q9
Geometry and trigonometry21 marks · 26%
Q2, Q7
Statistics and probability21 marks · 26%
Q1, Q8
Calculus13 marks · 16%
Q3, Q5

Mentor insights

Traps to watch for

01
Distance between points · Q2
Use $Δ x^{2} + Δ y^{2} + Δ z^{2}$ .
02
Maximum speed · Q5
The speed is $∣ v ∣$ , so look at the largest value of $∣ v ∣$ , including a negative minimum of $v$ .
03
Angle at C · Q7
$C$ sees the lighthouse due east, and the Jet Ski on a bearing of $06 2^{\circ}$ , so $J C L = 9 0^{\circ} - 6 2^{\circ} = 2 8^{\circ}$ .
04
Normal and binomial · Q8
Premium apples in a box of 40 follow a binomial distribution with the probability from the normal model.
05
Geometric sequence · Q9
The maximum height after each bounce is $\frac{2}{3}$ of the one before.

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2TZ3

Session: May 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 May 2025 TZ2 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 · 34 marks

Question 1

4 marks

Play time and sleep time: a regression line and an estimate

Statistics and probability

(a)

2 marks

Enter the six pairs.

a \approx 1.98, b \approx 40.2

(b)

2 marks

Substitute $x = 20$ into $y = 1.9765 x + 40.229$ .

y = 1.9765 (20) + 40.229 \approx 79.8 hours

Question 2

6 marks

A square-based pyramid in 3D: height, diagonal and edge angle

Geometry and trigonometry

(a)

2 marks

With $V (1, 7, 0)$ and $X (- 3, 4, 2)$ .

V X = (1 + 3)^{2} + (7 - 4)^{2} + (0 - 2)^{2} = 29 \approx 5.39

(b)

2 marks

The diagonal of a square with side 5.

A C = 5^{2} + 5^{2} = 52 \approx 7.07 cm

(c)

2 marks

$X$ is the centre of the base, so $X C = \frac{5 2}{2}$ and the triangle $V X C$ has a right angle at $X$ .

tan θ = \frac{29}{5 2 /2} = \frac{5.385}{3.536} \Rightarrow θ \approx 56. 7^{\circ}

Question 3

6 marks

Where a function decreases and is concave up from its derivative

Calculus

(a)

3 marks

Solve $f^{'} (x) = 4 + 2 x - 3 e^{x} = 0$ with the calculator. The two roots are $x \approx - 1.736$ and $x \approx 0.518$ .
A graph of $f^{'}$ shows it is below the axis outside these roots.

x < - 1.74 and x > 0.518

(b)

3 marks

Differentiate $f^{'} (x)$ .
Concave up needs $f^{''} (x) > 0$ .

f^{''} (x) = 2 - 3 e^{x} 2 - 3 e^{x} > 0 \Rightarrow e^{x} < \frac{2}{3} \Rightarrow x < ln \frac{2}{3} \approx - 0.405

Question 4

5 marks

Finding k from a binomial coefficient

Number and algebra

5 marks

The term in $x^{6}$ .
Set the coefficient equal to $8.4 \times 1 0^{6}$ .

(6 10) (2 x)^{6} k^{4} = 210 \times 64 k^{4} x^{6} = 13440 k^{4} x^{6} 13440 k^{4} = 8400000 \Rightarrow k^{4} = 625 k = \pm 5

Question 5

7 marks

A particle with velocity e^(-sin t) cos 2t: speed, distance and acceleration

Calculus

(a)

2 marks

Sketch $v (t) = e^{- s i n t} cos 2 t$ on $0 \leq t \leq 5$ . The most extreme value of $v$ is the minimum at $t \approx 4.71 (\frac{3 π}{2})$ , where $v = - e$ .

maximum speed = e \approx 2.72 m s^{- 1}

(b)

2 marks

The distance travelled is the area under the graph of $∣ v ∣$ .

d = \int_{0}^{5} ∣ v (t) ∣ d t \approx 3.85 m

(c)

3 marks

The velocity is zero when $cos 2 t = 0$ : $t = \frac{π}{4}, \frac{3 π}{4}, \dots$ , and the sign of $v$ changes at each one. The second change of direction is at $t = \frac{3 π}{4} \approx 2.356$ .
Differentiate $v$ with the product rule.
At $t = \frac{3 π}{4}$ we have $cos 2 t = 0$ and $sin 2 t = - 1$ .

a = v^{'} (t) = e^{- s i n t} (- cos t cos 2 t - 2 sin 2 t) a = e^{- s i n (3 π /4)} (0 + 2) = 2 e^{- 1/ 2} \approx 0.986 m s^{- 2}

Question 6

6 marks

A circle sector split into a segment and a triangle in the ratio 3 : 5

Number and algebra

6 marks

The segment $P$ has area 12.8, so the triangle $Q$ has area $\frac{5}{3} \times 12.8 = 21.33$ and the sector has area $12.8 + 21.33 = 34.13$ .
Divide the second equation by the first.
Solve for $θ$ with the calculator (in radians).
Substitute into the sector area.

\frac{1}{2} r^{2} θ = 34.13, \frac{1}{2} r^{2} sin θ = 21.33 \frac{sin θ}{θ} = \frac{21.33}{34.13} = 0.625 θ \approx 1.599 \frac{1}{2} r^{2} (1.599) = 34.13 \Rightarrow r^{2} \approx 42.7 r \approx 6.53 cm

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A lighthouse, a Jet Ski and two rescue boats

Geometry and trigonometry

(a)

4 marks

Use the cosine rule in triangle $J C L$ with $C J = 5.5$ , $C L = 8$ .

J L^{2} = 5. 5^{2} + 8^{2} - 2 (5.5) (8) cos 2 8^{\circ} \Rightarrow J L \approx 4.07 km

(b)

3 marks

In the right triangle from the top of the lighthouse (60 m up), $tan 0.9 4^{\circ} = \frac{60}{B L}$ .

B L = \frac{60}{tan 0.9 4 ^{\circ}} \approx 3657 m \approx 3.66 km

(c)

2 marks

The bearing from $B$ to $L$ is $12 1^{\circ}$ . The bearing of the return direction is $18 0^{\circ}$ more.

121 + 180 = 30 1^{\circ}

(d)

2 marks

The time is distance over speed.

t = \frac{3.657}{48} hours = 0.0762 hours \approx 4.57 minutes

(e)

4 marks

The coastguard station $C$ is due west of $L$ (distance 8 km), and $B$ is on the bearing $30 1^{\circ}$ from $L$ , so the angle at $L$ in triangle $C L B$ is $30 1^{\circ} - 27 0^{\circ} = 3 1^{\circ}$ . Use the cosine rule for the distance $C B$ .
The coastguard boat's time at 55 km/h.
Since $4.57 < 5.69$ , the lighthouse rescue boat reaches the Jet Ski first.

C B^{2} = 8^{2} + 3.65 7^{2} - 2 (8) (3.657) cos 3 1^{\circ} \Rightarrow C B \approx 5.217 km t = \frac{5.217}{55} \times 60 \approx 5.69 minutes

Question 8

17 marks

Apple weights: normal probabilities, premium apples in boxes and a second orchard

Statistics and probability

(a)

2 marks

$W \sim N (175, 8^{2})$ .

P (W < 170) \approx 0.266

(b)

2 marks

Use the inverse normal with 20% in the upper tail.

w \approx 181.7 grams

(c)

2 marks

Premium apples weigh between 170 and 185 grams.

P (170 < W < 185) \approx 0.628 = 62.8%

(d)

3 marks

Let $X$ be the number of premium apples in a box, $X \sim B (40, 0.6284)$ .

P (X \geq 30) \approx 0.0739

(e)

2 marks

Let $Y$ be the number of boxes (out of 10) with at least 30 premium apples, $Y \sim B (10, 0.0739)$ .

P (Y = 4) \approx 0.00394

(f)

6 marks

Of all the apples, 82% are premium, so 18% are not. The percentage below 170 is twice the percentage above 185, so $P (M < 170) = 0.12$ and $P (M > 185) = 0.06$ .
The z-scores are $z = - 1.175$ and $z = 1.555$ .
Subtract the first from the second: $\frac{15}{s} = 2.730$ , so $s = 5.495$ . Then $m = 170 + 1.175 s$ .

\frac{170 - m}{s} = - 1.175, \frac{185 - m}{s} = 1.555 m \approx 176.5 \approx 176 grams

Question 9

14 marks

A bouncing tennis ball: geometric sequences and an arithmetic sequence of distances

Number and algebra

(a)

3 marks

Ball 1 starts from 10 m. After the 5th bounce the height is $10 (\frac{2}{3})^{5}$ .

10 (\frac{2}{3})^{5} = \frac{320}{243} \approx 1.32 m

(b)

3 marks

Before the 5th bounce the ball has fallen 10 m and then made four up-and-down trips.

10 + 2 [10 (\frac{2}{3}) + 10 (\frac{2}{3})^{2} + 10 (\frac{2}{3})^{3} + 10 (\frac{2}{3})^{4}] \approx 42.1 m

(c)

3 marks

The ball falls $x$ and then makes infinitely many up-and-down trips. The first rebound height is $\frac{2}{3} x$ , and the ratio is $\frac{2}{3}$ .

d = x + 2 \cdot \frac{\frac{2}{3} x}{1 - \frac{2}{3}} = x + 2 \cdot 2 x = 5 x

(d)

1 mark

With $x = 10$ .

d_{1} = 50

(e)

4 marks

Ball 2 has $d_{2} = 5 \times 9.56 = 47.8$ , so the common difference is $47.8 - 50 = - 2.2$ .
The first ball is

d_{n} = 50 + (n - 1) (- 2.2) < 25 \Rightarrow n - 1 > 11.36 \Rightarrow n > 12.36 Ball 13

Our analysis

Where the marks are

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

Number and algebra25 marks · 31%
Q4, Q6, Q9
Geometry and trigonometry21 marks · 26%
Q2, Q7
Statistics and probability21 marks · 26%
Q1, Q8
Calculus13 marks · 16%
Q3, Q5

Mentor insights

Traps to watch for

01
Distance between points · Q2
Use $Δ x^{2} + Δ y^{2} + Δ z^{2}$ .
02
Maximum speed · Q5
The speed is $∣ v ∣$ , so look at the largest value of $∣ v ∣$ , including a negative minimum of $v$ .
03
Angle at C · Q7
$C$ sees the lighthouse due east, and the Jet Ski on a bearing of $06 2^{\circ}$ , so $J C L = 9 0^{\circ} - 6 2^{\circ} = 2 8^{\circ}$ .
04
Normal and binomial · Q8
Premium apples in a box of 40 follow a binomial distribution with the probability from the normal model.
05
Geometric sequence · Q9
The maximum height after each bounce is $\frac{2}{3}$ of the one before.

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