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IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: November 2024
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 2024 TZ1 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

13 marks

A restaurant floor plan: Voronoi diagram, a bisector and a two-sample t-test

Geometry and trigonometry
Statistics and probability

(a)

1 mark

The perpendicular bisector of $P Q$ is $y = 4$ , which is horizontal, so $P Q$ is vertical, and $Q$ is the reflection of $P (2, 2)$ in $y = 4$ .

Q (2, 6)

(b)

6 marks

(i) The midpoint of $R (11, 0)$ and $S (12, 6)$ .

(ii) The gradient of $R S$ .

The perpendicular gradient is $- \frac{1}{6}$ . The line through the midpoint.

(\frac{11 + 12}{2}, \frac{0 + 6}{2}) = (11.5, 3) \frac{6 - 0}{12 - 11} = 6 y - 3 = - \frac{1}{6} (x - 11.5) \Rightarrow y = - \frac{1}{6} x + \frac{59}{12}

(c)

1 mark

The point $(6, 8)$ is closest to site $Q (2, 6)$ , since it is above the line $y = 4$ and far from $R$ and $S$ . Using nearest-neighbour interpolation, the waiting time is that of $Q$ .

4 minutes

(d)

1 mark

She is testing whether customers spend more at the weekend, which means weekday spending is less than weekend spending.

H_{1} : μ_{1} < μ_{2}

(e)

2 marks

(i) She takes the first 12 orders on chosen days, which is convenience sampling.

(ii) One disadvantage: the sample might not be representative of all customers (for example, early orders may not be typical).

(f)

2 marks

Since $0.0897 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that customers spend more at the weekend than on weekdays.

Question 2

16 marks

The Scheveningen Ferris wheel: radius, arc length, a sine model and revolutions

Geometry and trigonometry

(a)

2 marks

(i) The diameter is $45 - 8 = 37$ m.

(ii) The circumference.

r = \frac{37}{2} = 18.5 m 2 π (18.5) \approx 116 m

(b)

3 marks

A rotation of $1 0^{\circ}$ is $\frac{10}{360}$ of the circumference.

\frac{10}{360} \times 2 π (18.5) = \frac{37 π}{36} \approx 3.23 m

(c)

5 marks

(i) One revolution takes 15 minutes, so in degrees $b = \frac{360}{15}$ .

(ii) The midline is halfway between the lowest and highest points.

(iii) The amplitude is the radius, $a = 18.5$ .

b = 24 d = \frac{45 + 8}{2} = 26.5 h (t) = 18.5 sin (24 t) + 26.5

(d)

3 marks

Solve $h (t) = 33$ for $0 \leq t \leq 15$ .
Use the calculator.

18.5 sin (24 t) + 26.5 = 33 t \approx 0.857 and t \approx 6.64 minutes

(e)

3 marks

The wheel makes 4 revolutions per hour, so $4 \times 8 = 32$ revolutions per day.

32 \times 3000 = 96000 = 9.6 \times 1 0^{4}

Question 3

14 marks

Happiness scores of expatriate countries: quartiles, outliers and Spearman's rank

Statistics and probability

(a)

4 marks

Order the scores: $6.0, 6.2, 6.5, 6.6, 6.7, 7.0, 7.2, 7.2, 7.4, 7.5$ .
(i) The upper quartile.
(ii) The lower quartile is $Q_{1} = 6.5$ .

Q_{3} = 7.2 IQR = 7.2 - 6.5 = 0.7

(b)

3 marks

Find the upper boundary for outliers.
Switzerland's score is $7.5 < 8.25$ , so it is not an outlier.

Q_{3} + 1.5 \times IQR = 7.2 + 1.05 = 8.25

(c)

3 marks

(i) New Zealand and Australia both score 7.2, which would be ranks 3 and 4, so they share the rank 3.5.

(ii) Spain's score of 6.5 is the 8th highest.

(iii) Australia has the other tied 7.2.

a = 3.5 b = 8 c = 3.5

(d)

3 marks

(i) Use the calculator on the ranks 1 to 10 of the countries and the happiness ranks.

(ii) France's score would change from 6.7 to 6.9, but it would still be below 7.0 (Ireland) and above 6.6 (UAE), so its rank stays at 6 and the ranks do not change.

r_{s} \approx 0.164

(e)

1 mark

Jose's conclusion is not appropriate, because $r_{s} \approx 0.164$ is close to zero. It shows only a weak correlation between the ranks, so we cannot say that happy countries are favourite expatriate countries.

Question 4

20 marks

A dog food label and container: trapezium rule, an integral and the minimum material

Calculus

(a)

3 marks

Use five intervals of width $h = 1$ .

A \approx \frac{1}{2} \times 1 [6 + 4.3 + 2 (1.7 + 1.3 + 3.0 + 4.7)] = \frac{1}{2} (10.3 + 21.4) \approx 15.9 cm^{2}

(b)

4 marks

(i) The area under the curve from $x = 0$ to $x = 5$ .

(ii) Integrate.

\int_{0}^{5} (- \frac{x ^{3}}{3} + 3 x^{2} - 7 x + 6) d x [- \frac{x ^{4}}{12} + x^{3} - \frac{7 x ^{2}}{2} + 6 x]_{0}^{5} = - \frac{625}{12} + 125 - \frac{175}{2} + 30 \frac{185}{12} \approx 15.4 cm^{2}

(c)

1 mark

The volume of a cylinder is $π r^{2} h$ .

π r^{2} h = 400

(d)

4 marks

The total surface area is two ends plus the curved surface.
From part (c), $h = \frac{400}{π r ^{2}}$ , so $2 π r h = \frac{800}{r}$ .
So $k = 800$ .

A = 2 π r^{2} + 2 π r h A = 2 π r^{2} + \frac{800}{r}

(e)

5 marks

(i) Differentiate $A = 2 π r^{2} + 800 r^{- 1}$ .

(ii) Set the derivative to zero.

\frac{d A}{d r} = 4 π r - \frac{800}{r ^{2}} 4 π r^{3} = 800 \Rightarrow r = 3 \frac{200}{π} \approx 3.99 cm

(f)

3 marks

The height of this container.
The circumference.
The label is 5 cm wide and 7 cm high. Its height of 7 cm is less than the container's height of 7.99 cm, and its width of 5 cm is less than the circumference of 25.1 cm, so the label fits on the curved surface.

h = \frac{400}{π ( 3.9929 ) ^{2}} \approx 7.99 cm C = 2 π (3.9929) \approx 25.1 cm

Question 5

17 marks

A tent with sides of 3 m and edges of 4.2 m: lengths, volume and a rope

Geometry and trigonometry

(a)

2 marks

Triangle $A O T$ is right-angled at $O$ , with $O A = 3$ and $A T = 4.2$ .

O T = 4. 2^{2} - 3^{2} = 8.64 \approx 2.94 m

(b)

5 marks

(i) The base is six equilateral triangles of side 3.

(ii) The volume of a pyramid.

Area = 6 \times \frac{1}{2} (3) (3) sin 6 0^{\circ} = \frac{27 3}{2} \approx 23.4 m^{2} V = \frac{1}{3} (23.383) (2.9394) \approx 22.9 m^{3}

(c)

2 marks

In the right triangle $A O T$ .

cos O A T = \frac{3}{4.2} \Rightarrow O A T \approx 44. 4^{\circ}

(d)

4 marks

Triangle $A P Q$ has $A QP = 3 0^{\circ}$ and $P Q = 3.5$ . We need the angle at $A$ , between $A Q$ and $A T$ . $A$ and $B$ are opposite corners of the base, so $Q$ , $A$ , $O$ and $B$ all lie on one line, and the angle at $A$ is the supplement of $O A T$ from (c).
The angle at $P$ .
Use the sine rule.

Q A P = 18 0^{\circ} - 44.4 2^{\circ} \approx 135. 6^{\circ} A P Q = 18 0^{\circ} - 135. 6^{\circ} - 3 0^{\circ} \approx 14. 4^{\circ} \frac{A Q}{sin 14. 4 ^{\circ}} = \frac{3.5}{sin 135. 6 ^{\circ}} \Rightarrow A Q \approx 1.25 m

(e)

4 marks

$R$ is on $[P Q]$ , so in triangle $A QR$ we know $A Q \approx 1.245$ , $A R = 0.8$ and the angle $3 0^{\circ}$ at $Q$ . Use the cosine rule with $QR$ unknown.
Both roots are shorter than the rope (3.5 m), so both positions of $R$ are possible.

0. 8^{2} = 1.24 5^{2} + Q R^{2} - 2 (1.245) (QR) cos 3 0^{\circ} Q R^{2} - 2.156 QR + 0.910 = 0 QR \approx 0.576 m or QR \approx 1.58 m

Mentor tip

Read the base diagram, not just the words: the 2D view shows $A$ and $B$ on either side of $O$ . Taking them as neighbouring corners gives the wrong angle at $A$ .

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 trigonometry41 marks · 51%
Q1, Q2, Q5
Calculus20 marks · 25%
Q4
Statistics and probability19 marks · 24%
Q1, Q3

Mentor insights

Traps to watch for

01
Voronoi · Q1
The edge between two sites is the perpendicular bisector of the line joining them, and a point takes the value of its nearest site.
02
Radius · Q2
Half the difference between the highest and lowest points.
03
Outlier test · Q3
A value above $Q_{3} + 1.5 \times IQR$ is an outlier.
04
Trapezium rule · Q4
$A \approx \frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + \dots + y_{n - 1})]$ .
05
Pythagoras · Q5
In the right triangle $O A T$ , $O T^{2} = A T^{2} - O A^{2}$ .

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 AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: November 2024
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 2024 TZ1 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

13 marks

A restaurant floor plan: Voronoi diagram, a bisector and a two-sample t-test

Geometry and trigonometry
Statistics and probability

(a)

1 mark

The perpendicular bisector of $P Q$ is $y = 4$ , which is horizontal, so $P Q$ is vertical, and $Q$ is the reflection of $P (2, 2)$ in $y = 4$ .

Q (2, 6)

(b)

6 marks

(i) The midpoint of $R (11, 0)$ and $S (12, 6)$ .

(ii) The gradient of $R S$ .

The perpendicular gradient is $- \frac{1}{6}$ . The line through the midpoint.

(\frac{11 + 12}{2}, \frac{0 + 6}{2}) = (11.5, 3) \frac{6 - 0}{12 - 11} = 6 y - 3 = - \frac{1}{6} (x - 11.5) \Rightarrow y = - \frac{1}{6} x + \frac{59}{12}

(c)

1 mark

The point $(6, 8)$ is closest to site $Q (2, 6)$ , since it is above the line $y = 4$ and far from $R$ and $S$ . Using nearest-neighbour interpolation, the waiting time is that of $Q$ .

4 minutes

(d)

1 mark

She is testing whether customers spend more at the weekend, which means weekday spending is less than weekend spending.

H_{1} : μ_{1} < μ_{2}

(e)

2 marks

(i) She takes the first 12 orders on chosen days, which is convenience sampling.

(ii) One disadvantage: the sample might not be representative of all customers (for example, early orders may not be typical).

(f)

2 marks

Since $0.0897 > 0.05$ , do not reject $H_{0}$ . There is not enough evidence at the 5% level that customers spend more at the weekend than on weekdays.

Question 2

16 marks

The Scheveningen Ferris wheel: radius, arc length, a sine model and revolutions

Geometry and trigonometry

(a)

2 marks

(i) The diameter is $45 - 8 = 37$ m.

(ii) The circumference.

r = \frac{37}{2} = 18.5 m 2 π (18.5) \approx 116 m

(b)

3 marks

A rotation of $1 0^{\circ}$ is $\frac{10}{360}$ of the circumference.

\frac{10}{360} \times 2 π (18.5) = \frac{37 π}{36} \approx 3.23 m

(c)

5 marks

(i) One revolution takes 15 minutes, so in degrees $b = \frac{360}{15}$ .

(ii) The midline is halfway between the lowest and highest points.

(iii) The amplitude is the radius, $a = 18.5$ .

b = 24 d = \frac{45 + 8}{2} = 26.5 h (t) = 18.5 sin (24 t) + 26.5

(d)

3 marks

Solve $h (t) = 33$ for $0 \leq t \leq 15$ .
Use the calculator.

18.5 sin (24 t) + 26.5 = 33 t \approx 0.857 and t \approx 6.64 minutes

(e)

3 marks

The wheel makes 4 revolutions per hour, so $4 \times 8 = 32$ revolutions per day.

32 \times 3000 = 96000 = 9.6 \times 1 0^{4}

Question 3

14 marks

Happiness scores of expatriate countries: quartiles, outliers and Spearman's rank

Statistics and probability

(a)

4 marks

Order the scores: $6.0, 6.2, 6.5, 6.6, 6.7, 7.0, 7.2, 7.2, 7.4, 7.5$ .
(i) The upper quartile.
(ii) The lower quartile is $Q_{1} = 6.5$ .

Q_{3} = 7.2 IQR = 7.2 - 6.5 = 0.7

(b)

3 marks

Find the upper boundary for outliers.
Switzerland's score is $7.5 < 8.25$ , so it is not an outlier.

Q_{3} + 1.5 \times IQR = 7.2 + 1.05 = 8.25

(c)

3 marks

(i) New Zealand and Australia both score 7.2, which would be ranks 3 and 4, so they share the rank 3.5.

(ii) Spain's score of 6.5 is the 8th highest.

(iii) Australia has the other tied 7.2.

a = 3.5 b = 8 c = 3.5

(d)

3 marks

(i) Use the calculator on the ranks 1 to 10 of the countries and the happiness ranks.

(ii) France's score would change from 6.7 to 6.9, but it would still be below 7.0 (Ireland) and above 6.6 (UAE), so its rank stays at 6 and the ranks do not change.

r_{s} \approx 0.164

(e)

1 mark

Question 4

20 marks

A dog food label and container: trapezium rule, an integral and the minimum material

Calculus

(a)

3 marks

Use five intervals of width $h = 1$ .

A \approx \frac{1}{2} \times 1 [6 + 4.3 + 2 (1.7 + 1.3 + 3.0 + 4.7)] = \frac{1}{2} (10.3 + 21.4) \approx 15.9 cm^{2}

(b)

4 marks

(i) The area under the curve from $x = 0$ to $x = 5$ .

(ii) Integrate.

\int_{0}^{5} (- \frac{x ^{3}}{3} + 3 x^{2} - 7 x + 6) d x [- \frac{x ^{4}}{12} + x^{3} - \frac{7 x ^{2}}{2} + 6 x]_{0}^{5} = - \frac{625}{12} + 125 - \frac{175}{2} + 30 \frac{185}{12} \approx 15.4 cm^{2}

(c)

1 mark

The volume of a cylinder is $π r^{2} h$ .

π r^{2} h = 400

(d)

4 marks

The total surface area is two ends plus the curved surface.
From part (c), $h = \frac{400}{π r ^{2}}$ , so $2 π r h = \frac{800}{r}$ .
So $k = 800$ .

A = 2 π r^{2} + 2 π r h A = 2 π r^{2} + \frac{800}{r}

(e)

5 marks

(i) Differentiate $A = 2 π r^{2} + 800 r^{- 1}$ .

(ii) Set the derivative to zero.

\frac{d A}{d r} = 4 π r - \frac{800}{r ^{2}} 4 π r^{3} = 800 \Rightarrow r = 3 \frac{200}{π} \approx 3.99 cm

(f)

3 marks

The height of this container.
The circumference.
The label is 5 cm wide and 7 cm high. Its height of 7 cm is less than the container's height of 7.99 cm, and its width of 5 cm is less than the circumference of 25.1 cm, so the label fits on the curved surface.

h = \frac{400}{π ( 3.9929 ) ^{2}} \approx 7.99 cm C = 2 π (3.9929) \approx 25.1 cm

Question 5

17 marks

A tent with sides of 3 m and edges of 4.2 m: lengths, volume and a rope

Geometry and trigonometry

(a)

2 marks

Triangle $A O T$ is right-angled at $O$ , with $O A = 3$ and $A T = 4.2$ .

O T = 4. 2^{2} - 3^{2} = 8.64 \approx 2.94 m

(b)

5 marks

(i) The base is six equilateral triangles of side 3.

(ii) The volume of a pyramid.

Area = 6 \times \frac{1}{2} (3) (3) sin 6 0^{\circ} = \frac{27 3}{2} \approx 23.4 m^{2} V = \frac{1}{3} (23.383) (2.9394) \approx 22.9 m^{3}

(c)

2 marks

In the right triangle $A O T$ .

cos O A T = \frac{3}{4.2} \Rightarrow O A T \approx 44. 4^{\circ}

(d)

4 marks

Triangle $A P Q$ has $A QP = 3 0^{\circ}$ and $P Q = 3.5$ . We need the angle at $A$ , between $A Q$ and $A T$ . $A$ and $B$ are opposite corners of the base, so $Q$ , $A$ , $O$ and $B$ all lie on one line, and the angle at $A$ is the supplement of $O A T$ from (c).
The angle at $P$ .
Use the sine rule.

Q A P = 18 0^{\circ} - 44.4 2^{\circ} \approx 135. 6^{\circ} A P Q = 18 0^{\circ} - 135. 6^{\circ} - 3 0^{\circ} \approx 14. 4^{\circ} \frac{A Q}{sin 14. 4 ^{\circ}} = \frac{3.5}{sin 135. 6 ^{\circ}} \Rightarrow A Q \approx 1.25 m

(e)

4 marks

$R$ is on $[P Q]$ , so in triangle $A QR$ we know $A Q \approx 1.245$ , $A R = 0.8$ and the angle $3 0^{\circ}$ at $Q$ . Use the cosine rule with $QR$ unknown.
Both roots are shorter than the rope (3.5 m), so both positions of $R$ are possible.

0. 8^{2} = 1.24 5^{2} + Q R^{2} - 2 (1.245) (QR) cos 3 0^{\circ} Q R^{2} - 2.156 QR + 0.910 = 0 QR \approx 0.576 m or QR \approx 1.58 m

Mentor tip

Read the base diagram, not just the words: the 2D view shows $A$ and $B$ on either side of $O$ . Taking them as neighbouring corners gives the wrong angle at $A$ .

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 trigonometry41 marks · 51%
Q1, Q2, Q5
Calculus20 marks · 25%
Q4
Statistics and probability19 marks · 24%
Q1, Q3

Mentor insights

Traps to watch for

01
Voronoi · Q1
The edge between two sites is the perpendicular bisector of the line joining them, and a point takes the value of its nearest site.
02
Radius · Q2
Half the difference between the highest and lowest points.
03
Outlier test · Q3
A value above $Q_{3} + 1.5 \times IQR$ is an outlier.
04
Trapezium rule · Q4
$A \approx \frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + \dots + y_{n - 1})]$ .
05
Pythagoras · Q5
In the right triangle $O A T$ , $O T^{2} = A T^{2} - O A^{2}$ .

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