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IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2024 TZ2 paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

16 marks

Absences at a primary school: a Venn diagram and a model of influenza cases

Statistics and probability

(a)

3 marks

The region for all three illnesses has 1. The overlaps of exactly two have $H \cap I$ : 4, $H \cap C$ : 2 and $I \cap C$ : 3. The regions for one illness only have $H$ : 35, $I$ : 20 and $C$ : 5.
Sketch: a rectangle for the 500 students, with three overlapping circles labelled $H$ , $I$ and $C$ containing these numbers.

(b)

2 marks

Add the students in every circle.
The rest of the 500 had none of the three illnesses.

35 + 4 + 20 + 3 + 1 + 2 + 5 = 70 500 - 70 = 430

(c)

4 marks

(i) The headlice circle has $35 + 4 + 2 + 1 = 42$ students.

(ii) Of those 42, the ones who also had influenza are $4 + 1 = 5$ .

P (H) = \frac{42}{500} = 0.084 P (I ∣ H) = \frac{5}{42} \approx 0.119

(d)

2 marks

At the start of the school year $t = 0$ .

n (0) = 250 - 240 (2)^{0} = 10

(e)

2 marks

Substitute $t = 10$ and $n = 130$ .

130 = 250 - 240 (2)^{10 k} \Rightarrow 2^{10 k} = 0.5 \Rightarrow 10 k = - 1 k = - 0.1

(f)

2 marks

Solve $200 = 250 - 240 (2)^{- 0.1 t}$ .
So it takes about 23 days.

2^{- 0.1 t} = \frac{50}{240} \Rightarrow t \approx 22.6

(g)

1 mark

With $k < 0$ , $2^{k t}$ tends to 0, so $n (t)$ approaches but never exceeds 250. The model can never give 300 students, so it is not appropriate.

Question 2

14 marks

Three attractions in an amusement park: distances, a bearing and the fountain

Geometry and trigonometry

(a)

1 mark

$B (12, 20)$ and $C (12, 0)$ are on the same vertical line.

B C = 20 m

(b)

2 marks

Use Pythagoras with $A (0, 16)$ and $B (12, 20)$ .

A B = 1 2^{2} + 4^{2} = 160 \approx 12.6 m

(c)

3 marks

From $B$ to $A$ the walk goes 12 m west and 4 m south. The acute angle with the west direction (below the horizontal) is
West is the bearing $27 0^{\circ}$ , and the direction is rotated towards south, so subtract.

tan^{- 1} \frac{4}{12} \approx 18. 4^{\circ} 270 - 18.4 \approx 25 2^{\circ}

(d)

8 marks

(i) The gradient of $[A C]$ with $A (0, 16)$ and $C (12, 0)$ .

(ii) The midpoint of $[A C]$ .

(iii) The fountain is on the perpendicular bisector of $[A C]$ , which has gradient $\frac{3}{4}$ and passes through $(6, 8)$ .

It is also on the perpendicular bisector of $[B C]$ . Since $[B C]$ is vertical with midpoint $(12, 10)$ , that bisector is the horizontal line $y = 10$ .

The fountain is at

m_{A C} = \frac{0 - 16}{12 - 0} = - \frac{4}{3} (6, 8) y - 8 = \frac{3}{4} (x - 6) 10 - 8 = \frac{3}{4} (x - 6) \Rightarrow x = 6 + \frac{8}{3} = \frac{26}{3} (\frac{26}{3}, 10) \approx (8.67, 10)

Question 3

15 marks

A water slide: the trapezoidal rule, an exact area and the percentage error

Calculus

(a)

5 marks

Evaluate $f (x) = - \frac{1}{4} x^{2} + 2 x$ at $x = 0, 1, 2, 3, 4$ .
Use the trapezoidal rule with width 1.

0, 1.75, 3, 3.75, 4 A \approx \frac{1}{2} [0 + 4 + 2 (1.75 + 3 + 3.75)] = \frac{1}{2} (4 + 17) = 10.5 m^{2}

(b)

3 marks

Integrate term by term.

\int (- \frac{1}{4} x^{2} + 2 x) d x = - \frac{x ^{3}}{12} + x^{2} + c

(c)

4 marks

The area for $0 \leq x \leq 4$ is
For $4 \leq x \leq 5$ the height is constant at 4, a rectangle of area $1 \times 4 = 4$ .
For $5 \leq x \leq 12$ the height falls in a straight line from 4 to 0, a triangle of area $\frac{1}{2} \times 7 \times 4 = 14$ .

[- \frac{x ^{3}}{12} + x^{2}]_{0}^{4} = - \frac{64}{12} + 16 = \frac{32}{3} Area = \frac{32}{3} + 4 + 14 = \frac{86}{3} \approx 28.67 m^{2}

(d)

3 marks

The approximate total area uses 10.5 in place of $\frac{32}{3}$ .

10.5 + 4 + 14 = 28.5 \frac{28.6667 - 28.5}{28.6667} \times 100 \approx 0.581%

Question 4

19 marks

Heights of Dutch women: normal probabilities, a goodness of fit test and a t-test

Statistics and probability

(a)

4 marks

$X \sim N (170.7, 6. 3^{2})$ .
(i) $P (X < 160) \approx 0.0447$
(ii) $P (160 < X < 170) \approx 0.411$

0.0447 and 0.411

(b)

2 marks

Use the inverse normal with 27% in the upper tail.

h \approx 174.6 cm \approx 1.75 m

(c)

2 marks

(i) $a = 200 \times P (160 < X \leq 170)$ .

(ii) $b = 200 \times P (170 < X \leq 180)$ . The four expected frequencies add to 200, so $b = 200 - 8.943 - 82.21 - 13.99$ .

a = 200 (0.41105) \approx 82.21 b \approx 94.86

(d)

1 mark

There are 4 classes and no parameters were estimated from the data (the mean and standard deviation are given).

ν = 4 - 1 = 3

(e)

4 marks

Compute the statistic.
Since $1.80 < 7.815$ (and $p = 0.616 > 0.05$ ), do not reject $H_{0}$ . There is not enough evidence against the heights being normally distributed with mean 170.7 cm and standard deviation 6.3 cm.

χ^{2} = \frac{( 6 - 8.943 ) ^{2}}{8.943} + \frac{( 85 - 82.21 ) ^{2}}{82.21} + \frac{( 92 - 94.86 ) ^{2}}{94.86} + \frac{( 17 - 13.99 ) ^{2}}{13.99} \approx 1.80, p \approx 0.616

(f)

2 marks

Let $μ_{L}$ and $μ_{D}$ be the population mean heights of Latvian and Dutch women.

H_{0} : μ_{L} = μ_{D}, H_{1} : μ_{L} > μ_{D}

(g)

4 marks

Run a one-tailed pooled two-sample $t$ -test on the two samples.
Since $0.673 > 0.05$ do not reject $H_{0}$ . There is not enough evidence that Latvian women are taller than Dutch women on average.

t \approx - 0.456, p \approx 0.673

Question 5

16 marks

A skip as a trapezoidal prism: volumes and the maximum volume

Geometry and trigonometry

(a)

9 marks

(i) The top edge is 11 m, so the sloping edge reaches 1 m across, with height $4^{2} - 1^{2} = 15$ .

(ii) The height is 3.2 m, so the sloping edge reaches $4^{2} - 3. 2^{2} = 2.4$ m across and the top is 12.4 m.

(iii) With $θ = 6 0^{\circ}$ the height is $4 sin 6 0^{\circ}$ and the extra top length is $4 cos 6 0^{\circ} = 2$ .

V = 3 \times \frac{1}{2} (10 + 11) 15 \approx 122 m^{3} V = 3 \times \frac{1}{2} (10 + 12.4) (3.2) \approx 108 m^{3} V = 3 \times \frac{1}{2} (10 + 12) (4 sin 6 0^{\circ}) \approx 114 m^{3}

(b)

2 marks

The volume is the cross-section area times the width 3.

V = 3 \times \frac{1}{2} (10 + 10 + 4 cos θ) (4 sin θ) = 6 sin θ (20 + 4 cos θ) V = 24 sin θ (5 + cos θ)

(c)

1 mark

If $θ = 0$ the sloping edge would lie flat along the ground, so the skip would have no height and zero volume, and the garbage would fall out.

(d)

4 marks

(i) Sketch: from $(0, 0)$ the graph of $V$ rises to a maximum of about 122 near $θ = 7 9^{\circ}$ and then falls a little to 120 at $9 0^{\circ}$ . The horizontal axis runs from $0^{\circ}$ to $9 0^{\circ}$ .

(ii) Find the maximum on the calculator.

θ \approx 79. 3^{\circ}, V_{m a x} \approx 122 m^{3}

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 · 44%
Q1, Q4
Geometry and trigonometry30 marks · 38%
Q2, Q5
Calculus15 marks · 19%
Q3

Mentor insights

Traps to watch for

01
Venn diagram · Q1
Fill in the centre first (all three), then the overlaps of two, then the single regions.
02
Perpendicular bisectors · Q2
A point equally far from two points lies on the perpendicular bisector of the segment joining them.
03
Trapezoidal rule · Q3
With strips of width 1, area $\approx \frac{1}{2} [y_{0} + y_{n} + 2 (middle values)]$ .
04
Expected frequency · Q4
Expected frequency $= 200 \times$ the probability of the class under the model.
05
Cross-section · Q5
A trapezoid with parallel sides 10 and $10 + 4 cos θ$ and height $4 sin θ$ .

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 SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2024 TZ2 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

16 marks

Absences at a primary school: a Venn diagram and a model of influenza cases

Statistics and probability

(a)

3 marks

The region for all three illnesses has 1. The overlaps of exactly two have $H \cap I$ : 4, $H \cap C$ : 2 and $I \cap C$ : 3. The regions for one illness only have $H$ : 35, $I$ : 20 and $C$ : 5.
Sketch: a rectangle for the 500 students, with three overlapping circles labelled $H$ , $I$ and $C$ containing these numbers.

(b)

2 marks

Add the students in every circle.
The rest of the 500 had none of the three illnesses.

35 + 4 + 20 + 3 + 1 + 2 + 5 = 70 500 - 70 = 430

(c)

4 marks

(i) The headlice circle has $35 + 4 + 2 + 1 = 42$ students.

(ii) Of those 42, the ones who also had influenza are $4 + 1 = 5$ .

P (H) = \frac{42}{500} = 0.084 P (I ∣ H) = \frac{5}{42} \approx 0.119

(d)

2 marks

At the start of the school year $t = 0$ .

n (0) = 250 - 240 (2)^{0} = 10

(e)

2 marks

Substitute $t = 10$ and $n = 130$ .

130 = 250 - 240 (2)^{10 k} \Rightarrow 2^{10 k} = 0.5 \Rightarrow 10 k = - 1 k = - 0.1

(f)

2 marks

Solve $200 = 250 - 240 (2)^{- 0.1 t}$ .
So it takes about 23 days.

2^{- 0.1 t} = \frac{50}{240} \Rightarrow t \approx 22.6

(g)

1 mark

With $k < 0$ , $2^{k t}$ tends to 0, so $n (t)$ approaches but never exceeds 250. The model can never give 300 students, so it is not appropriate.

Question 2

14 marks

Three attractions in an amusement park: distances, a bearing and the fountain

Geometry and trigonometry

(a)

1 mark

$B (12, 20)$ and $C (12, 0)$ are on the same vertical line.

B C = 20 m

(b)

2 marks

Use Pythagoras with $A (0, 16)$ and $B (12, 20)$ .

A B = 1 2^{2} + 4^{2} = 160 \approx 12.6 m

(c)

3 marks

From $B$ to $A$ the walk goes 12 m west and 4 m south. The acute angle with the west direction (below the horizontal) is
West is the bearing $27 0^{\circ}$ , and the direction is rotated towards south, so subtract.

tan^{- 1} \frac{4}{12} \approx 18. 4^{\circ} 270 - 18.4 \approx 25 2^{\circ}

(d)

8 marks

(i) The gradient of $[A C]$ with $A (0, 16)$ and $C (12, 0)$ .

(ii) The midpoint of $[A C]$ .

(iii) The fountain is on the perpendicular bisector of $[A C]$ , which has gradient $\frac{3}{4}$ and passes through $(6, 8)$ .

It is also on the perpendicular bisector of $[B C]$ . Since $[B C]$ is vertical with midpoint $(12, 10)$ , that bisector is the horizontal line $y = 10$ .

The fountain is at

m_{A C} = \frac{0 - 16}{12 - 0} = - \frac{4}{3} (6, 8) y - 8 = \frac{3}{4} (x - 6) 10 - 8 = \frac{3}{4} (x - 6) \Rightarrow x = 6 + \frac{8}{3} = \frac{26}{3} (\frac{26}{3}, 10) \approx (8.67, 10)

Question 3

15 marks

A water slide: the trapezoidal rule, an exact area and the percentage error

Calculus

(a)

5 marks

Evaluate $f (x) = - \frac{1}{4} x^{2} + 2 x$ at $x = 0, 1, 2, 3, 4$ .
Use the trapezoidal rule with width 1.

0, 1.75, 3, 3.75, 4 A \approx \frac{1}{2} [0 + 4 + 2 (1.75 + 3 + 3.75)] = \frac{1}{2} (4 + 17) = 10.5 m^{2}

(b)

3 marks

Integrate term by term.

\int (- \frac{1}{4} x^{2} + 2 x) d x = - \frac{x ^{3}}{12} + x^{2} + c

(c)

4 marks

The area for $0 \leq x \leq 4$ is
For $4 \leq x \leq 5$ the height is constant at 4, a rectangle of area $1 \times 4 = 4$ .
For $5 \leq x \leq 12$ the height falls in a straight line from 4 to 0, a triangle of area $\frac{1}{2} \times 7 \times 4 = 14$ .

[- \frac{x ^{3}}{12} + x^{2}]_{0}^{4} = - \frac{64}{12} + 16 = \frac{32}{3} Area = \frac{32}{3} + 4 + 14 = \frac{86}{3} \approx 28.67 m^{2}

(d)

3 marks

The approximate total area uses 10.5 in place of $\frac{32}{3}$ .

10.5 + 4 + 14 = 28.5 \frac{28.6667 - 28.5}{28.6667} \times 100 \approx 0.581%

Question 4

19 marks

Heights of Dutch women: normal probabilities, a goodness of fit test and a t-test

Statistics and probability

(a)

4 marks

$X \sim N (170.7, 6. 3^{2})$ .
(i) $P (X < 160) \approx 0.0447$
(ii) $P (160 < X < 170) \approx 0.411$

0.0447 and 0.411

(b)

2 marks

Use the inverse normal with 27% in the upper tail.

h \approx 174.6 cm \approx 1.75 m

(c)

2 marks

(i) $a = 200 \times P (160 < X \leq 170)$ .

(ii) $b = 200 \times P (170 < X \leq 180)$ . The four expected frequencies add to 200, so $b = 200 - 8.943 - 82.21 - 13.99$ .

a = 200 (0.41105) \approx 82.21 b \approx 94.86

(d)

1 mark

There are 4 classes and no parameters were estimated from the data (the mean and standard deviation are given).

ν = 4 - 1 = 3

(e)

4 marks

Compute the statistic.
Since $1.80 < 7.815$ (and $p = 0.616 > 0.05$ ), do not reject $H_{0}$ . There is not enough evidence against the heights being normally distributed with mean 170.7 cm and standard deviation 6.3 cm.

χ^{2} = \frac{( 6 - 8.943 ) ^{2}}{8.943} + \frac{( 85 - 82.21 ) ^{2}}{82.21} + \frac{( 92 - 94.86 ) ^{2}}{94.86} + \frac{( 17 - 13.99 ) ^{2}}{13.99} \approx 1.80, p \approx 0.616

(f)

2 marks

Let $μ_{L}$ and $μ_{D}$ be the population mean heights of Latvian and Dutch women.

H_{0} : μ_{L} = μ_{D}, H_{1} : μ_{L} > μ_{D}

(g)

4 marks

Run a one-tailed pooled two-sample $t$ -test on the two samples.
Since $0.673 > 0.05$ do not reject $H_{0}$ . There is not enough evidence that Latvian women are taller than Dutch women on average.

t \approx - 0.456, p \approx 0.673

Question 5

16 marks

A skip as a trapezoidal prism: volumes and the maximum volume

Geometry and trigonometry

(a)

9 marks

(i) The top edge is 11 m, so the sloping edge reaches 1 m across, with height $4^{2} - 1^{2} = 15$ .

(ii) The height is 3.2 m, so the sloping edge reaches $4^{2} - 3. 2^{2} = 2.4$ m across and the top is 12.4 m.

(iii) With $θ = 6 0^{\circ}$ the height is $4 sin 6 0^{\circ}$ and the extra top length is $4 cos 6 0^{\circ} = 2$ .

V = 3 \times \frac{1}{2} (10 + 11) 15 \approx 122 m^{3} V = 3 \times \frac{1}{2} (10 + 12.4) (3.2) \approx 108 m^{3} V = 3 \times \frac{1}{2} (10 + 12) (4 sin 6 0^{\circ}) \approx 114 m^{3}

(b)

2 marks

The volume is the cross-section area times the width 3.

V = 3 \times \frac{1}{2} (10 + 10 + 4 cos θ) (4 sin θ) = 6 sin θ (20 + 4 cos θ) V = 24 sin θ (5 + cos θ)

(c)

1 mark

If $θ = 0$ the sloping edge would lie flat along the ground, so the skip would have no height and zero volume, and the garbage would fall out.

(d)

4 marks

(ii) Find the maximum on the calculator.

θ \approx 79. 3^{\circ}, V_{m a x} \approx 122 m^{3}

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 · 44%
Q1, Q4
Geometry and trigonometry30 marks · 38%
Q2, Q5
Calculus15 marks · 19%
Q3

Mentor insights

Traps to watch for

01
Venn diagram · Q1
Fill in the centre first (all three), then the overlaps of two, then the single regions.
02
Perpendicular bisectors · Q2
A point equally far from two points lies on the perpendicular bisector of the segment joining them.
03
Trapezoidal rule · Q3
With strips of width 1, area $\approx \frac{1}{2} [y_{0} + y_{n} + 2 (middle values)]$ .
04
Expected frequency · Q4
Expected frequency $= 200 \times$ the probability of the class under the model.
05
Cross-section · Q5
A trapezoid with parallel sides 10 and $10 + 4 cos θ$ and height $4 sin θ$ .

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