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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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.

Questions

Questions 1–2 · 55 marks

Question 1

28 marks

Comparing two schools: samples, t-tests, correlation and school value added

Statistics and probability

(a)(i)

1 mark

A larger sample is more likely to be representative of the school, and gives results with less sampling error.

(a)(ii)

1 mark

A larger sample means more interviews, so it takes more time and costs more.

(b)

2 marks

Enter the six results from school A.

s_{n - 1} \approx 7.60

(c)

1 mark

The standard deviation of 7.60 is only from a sample of six students. It is an estimate, and the spread of results in the whole school could be bigger than the spread in school B even though this sample shows less.

(d)(i)

1 mark

A pooled $t$ -test needs the two population variances to be equal.

(d)(ii)

2 marks

The sample standard deviations are 7.60 and 7.66, which are very close, so it is plausible that the population variances are equal. Augustin can use a pooled $t$ -test.

(e)(i)

2 marks

Let $μ_{A}$ and $μ_{B}$ be the population mean results of the two schools. Augustin believed school B was higher.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} < μ_{B}

(e)(ii)

2 marks

Use a one-tailed pooled two-sample $t$ -test with means 51.7 and 60, standard deviations 7.60 and 7.66 and sample sizes 6 and 6.

p \approx 0.0445

(e)(iii)

2 marks

Since $0.0445 < 0.05$ , reject $H_{0}$ . There is significant evidence that the population mean result of school B is higher than that of school A.

(f)(i)

4 marks

$H_{0}$ : $ρ = 0$ and $H_{1}$ : $ρ \neq = 0$ , where $ρ$ is the population correlation between entry exam and final exam results.
Since $0.876 > 0.576$ , reject $H_{0}$ . There is significant evidence of a correlation between the entry exam and final exam results.

(f)(ii)

1 mark

If the requirements (such as both variables being normally distributed) were not met, Augustin could use Spearman's rank correlation coefficient.

(g)

1 mark

Every extra mark in the entry exam increases the model's predicted final exam result by 0.37 marks.

(h)(i)

2 marks

The model predicts for student 10003 (with entry result 38.7):
The school value added is $y - \overset{y}{^}$ .

\overset{y}{^} = 0.37 (38.7) + 37.6 = 51.919 q = 52.5 - 51.919 = 0.581 \approx 0.6

(h)(ii)

4 marks

Let $μ_{A}$ and $μ_{B}$ be the population mean school value added.
Run a one-tailed pooled two-sample $t$ -test on the two lists of values.
Since $0.0213 < 0.05$ , reject $H_{0}$ . There is evidence that school A has a higher mean school value added than school B.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} > μ_{B} p \approx 0.0213

(i)

2 marks

School A can claim it is doing better because its students have a higher value added, as the test in part (h) shows. School B can claim it is doing better because its students have a higher mean final exam result, as the test in part (e) shows.

Question 2

27 marks

Distances and directions between cities: a flat model and a spherical model

Geometry and trigonometry

(a)(i)

2 marks

Moscow is 6000 km north of Nairobi, and Bogota is 12 500 km west of Nairobi. Use Pythagoras.

BM = 1250 0^{2} + 600 0^{2} \approx 13865 km

(a)(ii)

3 marks

From Bogota, Moscow is 12 500 km east and 6000 km north.
The bearing is measured from north, so the bearing of Moscow from Bogota is

tan ϕ = \frac{12500}{6000} \Rightarrow ϕ \approx 64. 4^{\circ} 64. 4^{\circ}

(b)(i)

2 marks

With $p = (0, 0, 6)$ and $n = (0, 6, 0)$ , the scalar product is $p \cdot n = 0$ .

cos θ = \frac{0}{6 \times 6} = 0 \Rightarrow θ = 9 0^{\circ} = \frac{π}{2}

(b)(ii)

2 marks

The arc length is $r θ$ with radius 6000 km and angle $\frac{π}{2}$ .

6000 \times \frac{π}{2} = 3000 π km

(c)(i)

2 marks

Take the vector product of $a = (6, 0, 0)$ and $p = (0, 0, 6)$ .

a \times p = 0 \cdot 6 - 0 \cdot 0 0 \cdot 0 - 6 \cdot 6 0 = 0 - 36 0

(c)(ii)

3 marks

Similarly, with $n = (0, 6, 0)$ .
The two vectors have length 36 each. The scalar product is zero, so they are perpendicular.

a \times n = 0036 cos θ = \frac{( a \times p ) \cdot ( a \times n )}{36 \times 36} = \frac{0}{1296} = 0 \Rightarrow θ = 9 0^{\circ}

(d)

2 marks

The shortest distance is the arc $r θ$ with $r = 6$ (thousand km) and arc length 6.

6 θ = 6 \Rightarrow θ = 1 radian = 57. 3^{\circ}

(e)

5 marks

With $b = (6 sin 12 0^{\circ}, 6 cos 12 0^{\circ}, 0)$ and $m = (0, 6 cos θ, 6 sin θ)$ , use the scalar product to find the angle $B O M$ at the centre.
The arc length is $r θ$ with $r = 6$ thousand km.

cos B O M = \frac{b \cdot m}{36} = \frac{36 cos 12 0 ^{\circ} cos 57. 3 ^{\circ}}{36} = cos 12 0^{\circ} cos 57. 3^{\circ} \approx - 0.2701 B O M \approx 105.6 7^{\circ} \approx 1.844 radians 6 \times 1.8443 \approx 11.07 thousand km \approx 11100 km

(f)

6 marks

Using the method of part (c), find the vector products $b \times m$ and $b \times p$ .
The angle between them is given by the scalar product, with $∣ b \times p ∣ = 36$ and $∣ b \times m ∣ \approx 34.66$ .
The bearing of Moscow from Bogota on the sphere is about $29. 1^{\circ}$ (compared with $64. 4^{\circ}$ on the flat model).

b \times m = 36 cos 12 0^{\circ} sin θ - 36 sin 12 0^{\circ} sin θ 36 sin 12 0^{\circ} cos θ \approx - 15.15 - 26.23 16.84 b \times p = 36 cos 12 0^{\circ} - 36 sin 12 0^{\circ} 0 \approx - 18 - 31.18 0 cos ϕ = \frac{( - 15.15 ) ( - 18 ) + ( - 26.23 ) ( - 31.18 )}{36 \times 34.66} = \frac{1090.5}{1247.8} \approx 0.874 ϕ \approx 29. 1^{\circ}

Our analysis

Where the marks are

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

Statistics and probability28 marks · 51%
Q1
Geometry and trigonometry27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Pooled t-test · Q1
It assumes the two populations are normal and have the same variance, which is judged from the sample standard deviations.
02
Correlation test · Q1
$H_{0}$ : $ρ = 0$ . Reject when $∣ r ∣$ is above the critical value.
03
Value added · Q1
$y - \overset{y}{^}$ is how far a student's result is above or below the model's prediction from the entry exam.
04
Flat model · Q2
Use right-angled triangles. The bearing is measured clockwise from north.
05
Sphere · Q2
The distance along the surface is $r θ$ where $θ$ is the angle at the centre in radians.

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 HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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.

Questions

Questions 1–2 · 55 marks

Question 1

28 marks

Comparing two schools: samples, t-tests, correlation and school value added

Statistics and probability

(a)(i)

1 mark

A larger sample is more likely to be representative of the school, and gives results with less sampling error.

(a)(ii)

1 mark

A larger sample means more interviews, so it takes more time and costs more.

(b)

2 marks

Enter the six results from school A.

s_{n - 1} \approx 7.60

(c)

1 mark

(d)(i)

1 mark

A pooled $t$ -test needs the two population variances to be equal.

(d)(ii)

2 marks

The sample standard deviations are 7.60 and 7.66, which are very close, so it is plausible that the population variances are equal. Augustin can use a pooled $t$ -test.

(e)(i)

2 marks

Let $μ_{A}$ and $μ_{B}$ be the population mean results of the two schools. Augustin believed school B was higher.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} < μ_{B}

(e)(ii)

2 marks

Use a one-tailed pooled two-sample $t$ -test with means 51.7 and 60, standard deviations 7.60 and 7.66 and sample sizes 6 and 6.

p \approx 0.0445

(e)(iii)

2 marks

Since $0.0445 < 0.05$ , reject $H_{0}$ . There is significant evidence that the population mean result of school B is higher than that of school A.

(f)(i)

4 marks

$H_{0}$ : $ρ = 0$ and $H_{1}$ : $ρ \neq = 0$ , where $ρ$ is the population correlation between entry exam and final exam results.
Since $0.876 > 0.576$ , reject $H_{0}$ . There is significant evidence of a correlation between the entry exam and final exam results.

(f)(ii)

1 mark

If the requirements (such as both variables being normally distributed) were not met, Augustin could use Spearman's rank correlation coefficient.

(g)

1 mark

Every extra mark in the entry exam increases the model's predicted final exam result by 0.37 marks.

(h)(i)

2 marks

The model predicts for student 10003 (with entry result 38.7):
The school value added is $y - \overset{y}{^}$ .

\overset{y}{^} = 0.37 (38.7) + 37.6 = 51.919 q = 52.5 - 51.919 = 0.581 \approx 0.6

(h)(ii)

4 marks

Let $μ_{A}$ and $μ_{B}$ be the population mean school value added.
Run a one-tailed pooled two-sample $t$ -test on the two lists of values.
Since $0.0213 < 0.05$ , reject $H_{0}$ . There is evidence that school A has a higher mean school value added than school B.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} > μ_{B} p \approx 0.0213

(i)

2 marks

Question 2

27 marks

Distances and directions between cities: a flat model and a spherical model

Geometry and trigonometry

(a)(i)

2 marks

Moscow is 6000 km north of Nairobi, and Bogota is 12 500 km west of Nairobi. Use Pythagoras.

BM = 1250 0^{2} + 600 0^{2} \approx 13865 km

(a)(ii)

3 marks

From Bogota, Moscow is 12 500 km east and 6000 km north.
The bearing is measured from north, so the bearing of Moscow from Bogota is

tan ϕ = \frac{12500}{6000} \Rightarrow ϕ \approx 64. 4^{\circ} 64. 4^{\circ}

(b)(i)

2 marks

With $p = (0, 0, 6)$ and $n = (0, 6, 0)$ , the scalar product is $p \cdot n = 0$ .

cos θ = \frac{0}{6 \times 6} = 0 \Rightarrow θ = 9 0^{\circ} = \frac{π}{2}

(b)(ii)

2 marks

The arc length is $r θ$ with radius 6000 km and angle $\frac{π}{2}$ .

6000 \times \frac{π}{2} = 3000 π km

(c)(i)

2 marks

Take the vector product of $a = (6, 0, 0)$ and $p = (0, 0, 6)$ .

a \times p = 0 \cdot 6 - 0 \cdot 0 0 \cdot 0 - 6 \cdot 6 0 = 0 - 36 0

(c)(ii)

3 marks

Similarly, with $n = (0, 6, 0)$ .
The two vectors have length 36 each. The scalar product is zero, so they are perpendicular.

a \times n = 0036 cos θ = \frac{( a \times p ) \cdot ( a \times n )}{36 \times 36} = \frac{0}{1296} = 0 \Rightarrow θ = 9 0^{\circ}

(d)

2 marks

The shortest distance is the arc $r θ$ with $r = 6$ (thousand km) and arc length 6.

6 θ = 6 \Rightarrow θ = 1 radian = 57. 3^{\circ}

(e)

5 marks

With $b = (6 sin 12 0^{\circ}, 6 cos 12 0^{\circ}, 0)$ and $m = (0, 6 cos θ, 6 sin θ)$ , use the scalar product to find the angle $B O M$ at the centre.
The arc length is $r θ$ with $r = 6$ thousand km.

cos B O M = \frac{b \cdot m}{36} = \frac{36 cos 12 0 ^{\circ} cos 57. 3 ^{\circ}}{36} = cos 12 0^{\circ} cos 57. 3^{\circ} \approx - 0.2701 B O M \approx 105.6 7^{\circ} \approx 1.844 radians 6 \times 1.8443 \approx 11.07 thousand km \approx 11100 km

(f)

6 marks

Using the method of part (c), find the vector products $b \times m$ and $b \times p$ .
The angle between them is given by the scalar product, with $∣ b \times p ∣ = 36$ and $∣ b \times m ∣ \approx 34.66$ .
The bearing of Moscow from Bogota on the sphere is about $29. 1^{\circ}$ (compared with $64. 4^{\circ}$ on the flat model).

b \times m = 36 cos 12 0^{\circ} sin θ - 36 sin 12 0^{\circ} sin θ 36 sin 12 0^{\circ} cos θ \approx - 15.15 - 26.23 16.84 b \times p = 36 cos 12 0^{\circ} - 36 sin 12 0^{\circ} 0 \approx - 18 - 31.18 0 cos ϕ = \frac{( - 15.15 ) ( - 18 ) + ( - 26.23 ) ( - 31.18 )}{36 \times 34.66} = \frac{1090.5}{1247.8} \approx 0.874 ϕ \approx 29. 1^{\circ}

Our analysis

Where the marks are

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

Statistics and probability28 marks · 51%
Q1
Geometry and trigonometry27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Pooled t-test · Q1
It assumes the two populations are normal and have the same variance, which is judged from the sample standard deviations.
02
Correlation test · Q1
$H_{0}$ : $ρ = 0$ . Reject when $∣ r ∣$ is above the critical value.
03
Value added · Q1
$y - \overset{y}{^}$ is how far a student's result is above or below the model's prediction from the entry exam.
04
Flat model · Q2
Use right-angled triangles. The bearing is measured clockwise from north.
05
Sphere · Q2
The distance along the surface is $r θ$ where $θ$ is the angle at the centre in radians.

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