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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Session: November 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 Global paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

15 marks

A jogging track on circular arcs: arc length, track area and concrete

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 50 m long.

B M = 25 m

(b)

4 marks

Triangle BFM has a right angle at M, with $B M = 25$ and $F M = 20$ .

B F = 2 5^{2} + 2 0^{2} = 1025 \approx 32.0 m tan B F M = \frac{25}{20} \Rightarrow B F M = 51. 3^{\circ}

(c)

3 marks

The angle $A F B = 2 \times 51.3 4^{\circ} = 102.6 8^{\circ} = 1.792$ radians and the radius is $B F = 32.02$ .

arc A B = 32.02 \times 1.792 \approx 57.4 m

(d)

4 marks

The outer edge is an arc of radius $32.02 + 2 = 34.02$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.792) (34.0 2^{2} - 32.0 2^{2}) = \frac{1}{2} (1.792) (132.1) \approx 118 m^{2}

(e)

3 marks

The depth is 12 cm $= 0.12$ m.

V = 118.3 \times 0.12 \approx 14.2 m^{3}

Question 2

18 marks

Student heights: grouped mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $140 \leq h < 160$ is 150. Use the midpoints $150, 165, 175, 185, 200$ .

\overset{ˉ}{h} = \frac{150 ( 11 ) + 165 ( 51 ) + 175 ( 68 ) + 185 ( 47 ) + 200 ( 23 )}{200} = \frac{35260}{200} \approx 176 cm

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 168, Q_{3} = 183 \Rightarrow IQR = 15 cm

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
Laszlo's height is $204 < 205.5$ , so his height is not an outlier.

183 + 1.5 (15) = 205.5

(d)

2 marks

The null hypothesis says that the heights follow this normal distribution.

H_{0} : the heights can be modelled by N (176, 13. 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (176, 13. 5^{2})$ for 200 students.
Run a goodness of fit test with 5 classes: $ν = 4$ .
Since $0.0166 < 0.05$ , reject $H_{0}$ : there is sufficient evidence that the heights of these students are not well modelled by $N (176, 13. 5^{2})$ .

a = 200 P (170 \leq h < 180) = 57.6, b = 200 P (h \geq 190) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 3

16 marks

A house loan: payments, a faster repayment and the savings

Number and algebra

(a)

5 marks

The down payment is 15%, so the loan is 85% of the price.
Use a finance solver with $N = 360$ , $I = 4%$ , PV $= 242250$ , FV $= 0$ and 12 payments a year.

0.85 \times 285000 = 242250 USD PMT = 1156.54 USD

(b)

2 marks

Total paid over 360 months.

360 \times 1156.54 \approx 416354 USD

(c)

2 marks

Solve for the number of payments with PMT $= 1300$ .

N = 291.7 \Rightarrow 292 payments

(d)

4 marks

After 291 payments of 1300 the balance, after interest for the last month, is the final payment.

final payment \approx 874.81 USD

(e)

3 marks

Total paid with the faster plan.
Savings compared with the original total.

291 (1300) + 874.81 = 379 174.81 416354 - 379175 \approx 37179 USD

Question 4

12 marks

A plane taking off: a logistic height model and a gradient regulation

Calculus

(a)

2 marks

The model is $h (x) = \frac{10}{1 + 150 e ^{- 0.07 x}} - 0.06$ .
It is the height of the nose of the plane above the runway when the plane is just starting to leave the runway (about 6 metres).

h (0) = \frac{10}{151} - 0.06 \approx 0.00623 km

(b)

2 marks

As $x \to \infty$ , $e^{- 0.07 x} \to 0$ .
It is the maximum height (in km) that the nose of the plane approaches but does not reach.

y = 10 - 0.06 = 9.94

(c)

4 marks

Differentiate with the chain rule.

h^{'} (x) = - 10 (1 + 150 e^{- 0.07 x})^{- 2} (150 e^{- 0.07 x}) (- 0.07) = \frac{105 e ^{- 0.07 x}}{( 1 + 150 e ^{- 0.07 x} ) ^{2}}

(d)

4 marks

Find the maximum value of $h^{'} (x)$ on a graph.
The greatest gradient is $0.175 < 0.2$ , so the plane is following the safety regulation.

x = 71.6, h^{'} (71.6) \approx 0.175

Question 5

18 marks

Boat routes between islands: adjacency matrix, Eulerian trails and travelling salesman bounds

Number and algebra

(a)

1 mark

From the adjacency matrix in part (b), the routes are A-D, B-C, B-D, C-D, C-E and D-E.
Sketch: five vertices A, B, C, D, E with exactly these six edges.

(b)

3 marks

The adjacency matrix has rows and columns A to E.
Cube the matrix. The missing entry is
The entry for B to C in $P^{3}$ is 5.

P = 0001000110010111110100110 a = 6 5 ways

(c)

2 marks

The vertices with an odd number of routes are A (1) and C (3), so the trail starts at one and ends at the other.

A \to D \to E \to C \to D \to B \to C

(d)

3 marks

The trail costs $10 + 30 + 45 + 50 + 25 + 20 = 180$ USD and finishes at C. The cheapest way from C back to A is C-B-D-A.

20 + 25 + 10 = 55 180 + 55 = 235 USD

(e)

2 marks

Adding a route between A and C makes every vertex even, so the trip can start and end at the same port.
To keep the total at 235 USD the new route must cost the cheapest return, 55 USD.

A to C, 55 USD

(f)

7 marks

Nearest neighbour from A using the distance table.
Delete A. The minimum spanning tree of B, C, D, E uses BC (30), CD (45) and DE (55).
Add the two shortest edges at A: AD (50) and AE (60).

A \to D (50) \to C (45) \to B (30) \to E (120) \to A (60) 50 + 45 + 30 + 120 + 60 = 305 km 30 + 45 + 55 = 130 130 + 50 + 60 = 240 km

Question 6

14 marks

A line transformed by a matrix and a decomposition into rotation, enlargement and reflection

Geometry and trigonometry

(a)

1 mark

The line is $r = (- 1 1) + λ (2 - 1)$ .

x = - 1 + 2 λ, y = 1 - λ

(b)

4 marks

Multiply $T = (17 7 - 1)$ by a general point of $L_{1}$ .

(17 7 - 1) (- 1 + 2 λ 1 - λ) = (6 - 5 λ - 8 + 15 λ) L_{2} : r = (6 - 8) + λ (- 5 15)

(c)

2 marks

The rotation by $\frac{π}{4}$ anticlockwise and the enlargement by $52$ .

(i) (\frac{1}{2} \frac{1}{2} - \frac{1}{2} \frac{1}{2}), (ii) (520052)

(d)

1 mark

The reflection in the line $y = x tan α$ .

R = (cos 2 α sin 2 α sin 2 α - cos 2 α)

(e)

6 marks

The rotation comes first, then the enlargement, then the reflection, so $T = R X$ where $X$ is the enlargement times the rotation.
Then $R = T X^{- 1}$ .
Compare with $R$ from (d).

X = (520052) (\frac{1}{2} \frac{1}{2} - \frac{1}{2} \frac{1}{2}) = (55 - 5 5) R = \frac{1}{50} (17 7 - 1) (5 - 5 55) = (- \frac{3}{5} \frac{4}{5} \frac{4}{5} \frac{3}{5}) cos 2 α = - \frac{3}{5}, sin 2 α = \frac{4}{5} \Rightarrow 2 α = 2.214 \Rightarrow α = 1.11

Question 7

17 marks

Weights of adults in a city: binomial counts, a sum of normals and a one-tailed test

Statistics and probability

(a)

4 marks

First find $P (X > 85)$ for $X \sim N (72, 1 0^{2})$ .
Let $Y \sim B (10, 0.0968)$ be the number above 85 kg.

P (X > 85) = 0.0968 P (Y > 3) = P (Y \geq 4) \approx 0.0114

(b)

3 marks

The total weight $W$ of 8 adults is normal with mean $8 \times 72$ and variance $8 \times 1 0^{2}$ .

W \sim N (576, 800)

(c)

2 marks

Use the inverse normal with $P (W > w) = 0.01$ .

w = 642 kg

(d)

8 marks

Assumptions: Laetitia's clients are a random sample of the population, and each client's choice is independent of the others with the same probability $p$ of going abroad.
Under $H_{0}$ the number abroad is $Q \sim B (200, 0.42)$ .
Since $0.0085 < 0.05$ , reject $H_{0}$ : there is evidence that the proportion going abroad is less than 42%, so Laetitia's belief is reasonable.

H_{0} : p = 0.42, H_{1} : p < 0.42 P (Q \leq 67) \approx 0.00850

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 probability35 marks · 32%
Q2, Q7
Number and algebra34 marks · 31%
Q3, Q5
Geometry and trigonometry29 marks · 26%
Q1, Q6
Calculus12 marks · 11%
Q4

Mentor insights

Traps to watch for

01
Radians for arc length · Q1
The angle at F must be in radians in $s = r θ$ and in the sector area formula.
02
Midpoints for the mean · Q2
Use the middle of each class with its frequency.
03
Down payment first · Q3
The loan is the price minus the 15% down payment.
04
Eulerian trail needs two odd vertices · Q5
Start and end at the two vertices with an odd number of routes.
05
Variances add for a sum · Q7
The total of 8 independent weights has mean $8 \times 72$ and variance $8 \times 1 0^{2}$ .

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 HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Session: November 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 Global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

15 marks

A jogging track on circular arcs: arc length, track area and concrete

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 50 m long.

B M = 25 m

(b)

4 marks

Triangle BFM has a right angle at M, with $B M = 25$ and $F M = 20$ .

B F = 2 5^{2} + 2 0^{2} = 1025 \approx 32.0 m tan B F M = \frac{25}{20} \Rightarrow B F M = 51. 3^{\circ}

(c)

3 marks

The angle $A F B = 2 \times 51.3 4^{\circ} = 102.6 8^{\circ} = 1.792$ radians and the radius is $B F = 32.02$ .

arc A B = 32.02 \times 1.792 \approx 57.4 m

(d)

4 marks

The outer edge is an arc of radius $32.02 + 2 = 34.02$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.792) (34.0 2^{2} - 32.0 2^{2}) = \frac{1}{2} (1.792) (132.1) \approx 118 m^{2}

(e)

3 marks

The depth is 12 cm $= 0.12$ m.

V = 118.3 \times 0.12 \approx 14.2 m^{3}

Question 2

18 marks

Student heights: grouped mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $140 \leq h < 160$ is 150. Use the midpoints $150, 165, 175, 185, 200$ .

\overset{ˉ}{h} = \frac{150 ( 11 ) + 165 ( 51 ) + 175 ( 68 ) + 185 ( 47 ) + 200 ( 23 )}{200} = \frac{35260}{200} \approx 176 cm

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 168, Q_{3} = 183 \Rightarrow IQR = 15 cm

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
Laszlo's height is $204 < 205.5$ , so his height is not an outlier.

183 + 1.5 (15) = 205.5

(d)

2 marks

The null hypothesis says that the heights follow this normal distribution.

H_{0} : the heights can be modelled by N (176, 13. 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (176, 13. 5^{2})$ for 200 students.
Run a goodness of fit test with 5 classes: $ν = 4$ .
Since $0.0166 < 0.05$ , reject $H_{0}$ : there is sufficient evidence that the heights of these students are not well modelled by $N (176, 13. 5^{2})$ .

a = 200 P (170 \leq h < 180) = 57.6, b = 200 P (h \geq 190) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 3

16 marks

A house loan: payments, a faster repayment and the savings

Number and algebra

(a)

5 marks

The down payment is 15%, so the loan is 85% of the price.
Use a finance solver with $N = 360$ , $I = 4%$ , PV $= 242250$ , FV $= 0$ and 12 payments a year.

0.85 \times 285000 = 242250 USD PMT = 1156.54 USD

(b)

2 marks

Total paid over 360 months.

360 \times 1156.54 \approx 416354 USD

(c)

2 marks

Solve for the number of payments with PMT $= 1300$ .

N = 291.7 \Rightarrow 292 payments

(d)

4 marks

After 291 payments of 1300 the balance, after interest for the last month, is the final payment.

final payment \approx 874.81 USD

(e)

3 marks

Total paid with the faster plan.
Savings compared with the original total.

291 (1300) + 874.81 = 379 174.81 416354 - 379175 \approx 37179 USD

Question 4

12 marks

A plane taking off: a logistic height model and a gradient regulation

Calculus

(a)

2 marks

The model is $h (x) = \frac{10}{1 + 150 e ^{- 0.07 x}} - 0.06$ .
It is the height of the nose of the plane above the runway when the plane is just starting to leave the runway (about 6 metres).

h (0) = \frac{10}{151} - 0.06 \approx 0.00623 km

(b)

2 marks

As $x \to \infty$ , $e^{- 0.07 x} \to 0$ .
It is the maximum height (in km) that the nose of the plane approaches but does not reach.

y = 10 - 0.06 = 9.94

(c)

4 marks

Differentiate with the chain rule.

h^{'} (x) = - 10 (1 + 150 e^{- 0.07 x})^{- 2} (150 e^{- 0.07 x}) (- 0.07) = \frac{105 e ^{- 0.07 x}}{( 1 + 150 e ^{- 0.07 x} ) ^{2}}

(d)

4 marks

Find the maximum value of $h^{'} (x)$ on a graph.
The greatest gradient is $0.175 < 0.2$ , so the plane is following the safety regulation.

x = 71.6, h^{'} (71.6) \approx 0.175

Question 5

18 marks

Boat routes between islands: adjacency matrix, Eulerian trails and travelling salesman bounds

Number and algebra

(a)

1 mark

From the adjacency matrix in part (b), the routes are A-D, B-C, B-D, C-D, C-E and D-E.
Sketch: five vertices A, B, C, D, E with exactly these six edges.

(b)

3 marks

The adjacency matrix has rows and columns A to E.
Cube the matrix. The missing entry is
The entry for B to C in $P^{3}$ is 5.

P = 0001000110010111110100110 a = 6 5 ways

(c)

2 marks

The vertices with an odd number of routes are A (1) and C (3), so the trail starts at one and ends at the other.

A \to D \to E \to C \to D \to B \to C

(d)

3 marks

The trail costs $10 + 30 + 45 + 50 + 25 + 20 = 180$ USD and finishes at C. The cheapest way from C back to A is C-B-D-A.

20 + 25 + 10 = 55 180 + 55 = 235 USD

(e)

2 marks

Adding a route between A and C makes every vertex even, so the trip can start and end at the same port.
To keep the total at 235 USD the new route must cost the cheapest return, 55 USD.

A to C, 55 USD

(f)

7 marks

Nearest neighbour from A using the distance table.
Delete A. The minimum spanning tree of B, C, D, E uses BC (30), CD (45) and DE (55).
Add the two shortest edges at A: AD (50) and AE (60).

A \to D (50) \to C (45) \to B (30) \to E (120) \to A (60) 50 + 45 + 30 + 120 + 60 = 305 km 30 + 45 + 55 = 130 130 + 50 + 60 = 240 km

Question 6

14 marks

A line transformed by a matrix and a decomposition into rotation, enlargement and reflection

Geometry and trigonometry

(a)

1 mark

The line is $r = (- 1 1) + λ (2 - 1)$ .

x = - 1 + 2 λ, y = 1 - λ

(b)

4 marks

Multiply $T = (17 7 - 1)$ by a general point of $L_{1}$ .

(17 7 - 1) (- 1 + 2 λ 1 - λ) = (6 - 5 λ - 8 + 15 λ) L_{2} : r = (6 - 8) + λ (- 5 15)

(c)

2 marks

The rotation by $\frac{π}{4}$ anticlockwise and the enlargement by $52$ .

(i) (\frac{1}{2} \frac{1}{2} - \frac{1}{2} \frac{1}{2}), (ii) (520052)

(d)

1 mark

The reflection in the line $y = x tan α$ .

R = (cos 2 α sin 2 α sin 2 α - cos 2 α)

(e)

6 marks

The rotation comes first, then the enlargement, then the reflection, so $T = R X$ where $X$ is the enlargement times the rotation.
Then $R = T X^{- 1}$ .
Compare with $R$ from (d).

X = (520052) (\frac{1}{2} \frac{1}{2} - \frac{1}{2} \frac{1}{2}) = (55 - 5 5) R = \frac{1}{50} (17 7 - 1) (5 - 5 55) = (- \frac{3}{5} \frac{4}{5} \frac{4}{5} \frac{3}{5}) cos 2 α = - \frac{3}{5}, sin 2 α = \frac{4}{5} \Rightarrow 2 α = 2.214 \Rightarrow α = 1.11

Question 7

17 marks

Weights of adults in a city: binomial counts, a sum of normals and a one-tailed test

Statistics and probability

(a)

4 marks

First find $P (X > 85)$ for $X \sim N (72, 1 0^{2})$ .
Let $Y \sim B (10, 0.0968)$ be the number above 85 kg.

P (X > 85) = 0.0968 P (Y > 3) = P (Y \geq 4) \approx 0.0114

(b)

3 marks

The total weight $W$ of 8 adults is normal with mean $8 \times 72$ and variance $8 \times 1 0^{2}$ .

W \sim N (576, 800)

(c)

2 marks

Use the inverse normal with $P (W > w) = 0.01$ .

w = 642 kg

(d)

8 marks

Assumptions: Laetitia's clients are a random sample of the population, and each client's choice is independent of the others with the same probability $p$ of going abroad.
Under $H_{0}$ the number abroad is $Q \sim B (200, 0.42)$ .
Since $0.0085 < 0.05$ , reject $H_{0}$ : there is evidence that the proportion going abroad is less than 42%, so Laetitia's belief is reasonable.

H_{0} : p = 0.42, H_{1} : p < 0.42 P (Q \leq 67) \approx 0.00850

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 probability35 marks · 32%
Q2, Q7
Number and algebra34 marks · 31%
Q3, Q5
Geometry and trigonometry29 marks · 26%
Q1, Q6
Calculus12 marks · 11%
Q4

Mentor insights

Traps to watch for

01
Radians for arc length · Q1
The angle at F must be in radians in $s = r θ$ and in the sector area formula.
02
Midpoints for the mean · Q2
Use the middle of each class with its frequency.
03
Down payment first · Q3
The loan is the price minus the 15% down payment.
04
Eulerian trail needs two odd vertices · Q5
Start and end at the two vertices with an odd number of routes.
05
Variances add for a sum · Q7
The total of 8 independent weights has mean $8 \times 72$ and variance $8 \times 1 0^{2}$ .

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