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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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

17 marks

Reaction times: cumulative frequency graph and grouped data

Statistics and probability

(a)

4 marks

Read the graph at cumulative frequency 20, 10 and 30 (the total is 40).

median \approx 0.58 s IQR = 0.70 - 0.42 = 0.28 s

(b)

2 marks

From the graph, about 9 teenagers have a reaction time of 0.4 s or less.

40 - 9 = 31

(c)

2 marks

The 90th percentile is at cumulative frequency $0.9 \times 40 = 36$ .

0.8 s

(d)

2 marks

The cumulative frequency at 0.4 is 9, so $3 + a = 9$ . The total is 40.

a = 6, b = 40 - (3 + 6 + 13 + 14) = 4

(e)

1 mark

The modal class has the highest frequency, 14.

0.6 < t \leq 0.8

(f)

2 marks

Use the midpoints $0.1, 0.3, 0.5, 0.7, 0.9$ with frequencies $3, 6, 13, 14, 4$ .

\overset{ˉ}{t} = \frac{0.3 + 1.8 + 6.5 + 9.8 + 3.6}{40} = 0.55 s

(g)

4 marks

Moving values from the lowest class to the second class replaces their midpoint 0.1 with 0.3, so the total of the times goes up and the estimated mean increases.
The median is in the class $0.4 < t \leq 0.6$ , which is above both classes involved. Moving values between the two lower classes does not change the middle of the data, so the median stays the same.

Question 2

13 marks

Dog food: arithmetic and geometric sequences in context

Number and algebra

(a)

4 marks

The terms are $u_{3} = 115.5$ and $u_{8} = 108$ . Five days pass between them.
So the dog is fed 1.5 cups per day.
Go back two days from day 3.

5 d = 108 - 115.5 = - 7.5 \Rightarrow d = - 1.5 u_{1} = 115.5 + 2 (1.5) = 118.5 cups

(b)

2 marks

The bag is empty when $u_{n} = 0$ .
One bag lasts 80 days.

118.5 - 1.5 (n - 1) = 0 \Rightarrow n = 80

(c)

3 marks

The spending in year $n$ (with 2021 as $n = 1$ ) is $625 \times 1.06 4^{n - 1}$ .

625 \times 1.06 4^{4} = 801.03 \approx $801

(d)

3 marks

Sum of a geometric series.
It is the total Scott expects to spend on dog food over the 10 years from 2021 to 2030 inclusive, about $8394.

n = 1 \sum 10 625 (1.064)^{n - 1} = 625 \cdot \frac{1.06 4 ^{10} - 1}{0.064} \approx 8394

(e)

1 mark

The model is not realistic for a long time: a constant yearly growth rate of 6.4% would make the cost very large over many years, and inflation, the dog's age and its appetite will not stay the same.

Question 3

18 marks

Voronoi diagram of four supermarkets

Geometry and trigonometry

(a)

2 marks

Midpoint of $B (6, 0)$ and $D (2, 2)$ .

(\frac{6 + 2}{2}, \frac{0 + 2}{2}) = (4, 1)

(b)

4 marks

Gradient of BD:
The perpendicular bisector has gradient 2 and passes through $(4, 1)$ .

\frac{2 - 0}{2 - 6} = - \frac{1}{2} y - 1 = 2 (x - 4) \Rightarrow y = 2 x - 7

(c)

3 marks

X lies on both $y = 2 - x$ and $y = 2 x - 7$ .
So $X = (3, - 1)$ .

2 - x = 2 x - 7 \Rightarrow x = 3, y = - 1

(d)

2 marks

Use the distance formula with $Y (- 1, 3)$ and $Z (7, 7)$ .

Y Z = (7 + 1)^{2} + (7 - 3)^{2} = 80 = 45

(e)

4 marks

Find XZ and use the cosine rule in triangle XYZ.

X Z = (7 - 3)^{2} + (7 + 1)^{2} = 80 cos X Y Z = \frac{80 + 32 - 80}{2 80 32} = 0.3162 X Y Z = 71. 6^{\circ}

(f)

2 marks

Use $\frac{1}{2} ab sin C$ with the two sides next to the angle at Y.

Area = \frac{1}{2} 8032 sin 71.5 7^{\circ} = 24 km^{2}

(g)

1 mark

A larger cell only means D is the closest shop to more land. It does not mean more people live there, and people choose where to shop for reasons other than distance.

Question 4

17 marks

A Ferris wheel: cosine model, domain, range and height above a statue

Functions

(a)

6 marks

The amplitude is the radius of the wheel. The period is 40 s. The centre is the lowest height plus the radius.

a = \frac{61.8}{2} = 30.9 b = \frac{360}{40} = 9 (degrees; \frac{π}{20} in radians) d = 30.9 + 2.7 = 33.6

(b)

2 marks

The ride lasts 12 minutes at 1.5 revolutions per minute.

12 \times 1.5 = 18 revolutions

(c)

3 marks

The ride lasts $12 \times 60 = 720$ seconds. The lowest chair height is 2.7 m and the highest is $2.7 + 61.8 = 64.5$ m.

0 \leq t \leq 720, 2.7 \leq h \leq 64.5

(d)

3 marks

Solve $h (t) = 16.7$ in the first revolution.
The chair is higher than the statue between these times.

t = 6.316 and t = 33.684 33.684 - 6.316 \approx 27.4 s

(e)

3 marks

A taller platform lifts the whole wheel but does not change its size or its speed, so only the centre line $d$ changes.
The parameter $d$ changes to 34.3.

d + 30.9 = 65.2 \Rightarrow d = 34.3

Question 5

15 marks

A cafe's profit: differentiation, maximum and a break-even point

Calculus

(a)

3 marks

Expand $C = \frac{k ^{2} x}{10} - \frac{3 x ^{3}}{1000}$ and differentiate.

\frac{d C}{d x} = \frac{k ^{2}}{10} - \frac{9 x ^{2}}{1000}

(b)

4 marks

Set the derivative to zero.
Substitute back.

\frac{9 x ^{2}}{1000} = \frac{k ^{2}}{10} \Rightarrow x^{2} = \frac{100 k ^{2}}{9} \Rightarrow x = \frac{10 k}{3} C_{m a x} = \frac{k}{3} (k^{2} - \frac{3}{100} \cdot \frac{100 k ^{2}}{9}) = \frac{k}{3} \cdot \frac{2 k ^{2}}{3} = \frac{2}{9} k^{3}

(c)

3 marks

Use $x = 20$ and $C = 426$ .
Then the profit-maximising quantity is

2 (k^{2} - 12) = 426 \Rightarrow k^{2} = 225 \Rightarrow k = 15 x = \frac{10 ( 15 )}{3} = 50 litres

(d)

3 marks

The maximum is $C = \frac{2}{9} (15)^{3} = 750$ . The profit is 0 at $x = 0$ and where $225 = 0.03 x^{2}$ , which gives $x = 86.6$ .
Sketch: a cubic arch in the first quadrant that starts at the origin, rises to the maximum at $(50, 750)$ and comes back to the axis at $(86.6, 0)$ .

(e)

2 marks

The cafe does not lose money while $C \geq 0$ , so the most coffee is at the right-hand intercept.

x = 7500 \approx 86.6 litres

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 trigonometry18 marks · 23%
Q3
Functions17 marks · 21%
Q4
Statistics and probability17 marks · 21%
Q1
Calculus15 marks · 19%
Q5
Number and algebra13 marks · 16%
Q2

Mentor insights

Traps to watch for

01
Median and quartiles come from the cumulative frequency curve · Q1
With 40 teenagers, the median is at the 20th value, the lower quartile at the 10th and the upper quartile at the 30th.
02
Which sequence is which · Q2
The food left falls by the same amount each day, so it is arithmetic. The yearly spending grows by a fixed percentage, so it is geometric.
03
Boundaries are perpendicular bisectors · Q3
A Voronoi edge between two sites is the perpendicular bisector of the segment joining them. Find the midpoint and the negative reciprocal gradient.
04
Starts at the bottom · Q4
The chair is at its lowest point at $t = 0$ , which is why the model is $- a cos (b t) + d$ .
05
Use the given profit to find k · Q5
Substitute $x = 20$ and $C = 426$ . The equation for $k^{2}$ is simple, and $k$ is positive.

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 2022
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 2022 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

17 marks

Reaction times: cumulative frequency graph and grouped data

Statistics and probability

(a)

4 marks

Read the graph at cumulative frequency 20, 10 and 30 (the total is 40).

median \approx 0.58 s IQR = 0.70 - 0.42 = 0.28 s

(b)

2 marks

From the graph, about 9 teenagers have a reaction time of 0.4 s or less.

40 - 9 = 31

(c)

2 marks

The 90th percentile is at cumulative frequency $0.9 \times 40 = 36$ .

0.8 s

(d)

2 marks

The cumulative frequency at 0.4 is 9, so $3 + a = 9$ . The total is 40.

a = 6, b = 40 - (3 + 6 + 13 + 14) = 4

(e)

1 mark

The modal class has the highest frequency, 14.

0.6 < t \leq 0.8

(f)

2 marks

Use the midpoints $0.1, 0.3, 0.5, 0.7, 0.9$ with frequencies $3, 6, 13, 14, 4$ .

\overset{ˉ}{t} = \frac{0.3 + 1.8 + 6.5 + 9.8 + 3.6}{40} = 0.55 s

(g)

4 marks

Moving values from the lowest class to the second class replaces their midpoint 0.1 with 0.3, so the total of the times goes up and the estimated mean increases.
The median is in the class $0.4 < t \leq 0.6$ , which is above both classes involved. Moving values between the two lower classes does not change the middle of the data, so the median stays the same.

Question 2

13 marks

Dog food: arithmetic and geometric sequences in context

Number and algebra

(a)

4 marks

The terms are $u_{3} = 115.5$ and $u_{8} = 108$ . Five days pass between them.
So the dog is fed 1.5 cups per day.
Go back two days from day 3.

5 d = 108 - 115.5 = - 7.5 \Rightarrow d = - 1.5 u_{1} = 115.5 + 2 (1.5) = 118.5 cups

(b)

2 marks

The bag is empty when $u_{n} = 0$ .
One bag lasts 80 days.

118.5 - 1.5 (n - 1) = 0 \Rightarrow n = 80

(c)

3 marks

The spending in year $n$ (with 2021 as $n = 1$ ) is $625 \times 1.06 4^{n - 1}$ .

625 \times 1.06 4^{4} = 801.03 \approx $801

(d)

3 marks

Sum of a geometric series.
It is the total Scott expects to spend on dog food over the 10 years from 2021 to 2030 inclusive, about $8394.

n = 1 \sum 10 625 (1.064)^{n - 1} = 625 \cdot \frac{1.06 4 ^{10} - 1}{0.064} \approx 8394

(e)

1 mark

The model is not realistic for a long time: a constant yearly growth rate of 6.4% would make the cost very large over many years, and inflation, the dog's age and its appetite will not stay the same.

Question 3

18 marks

Voronoi diagram of four supermarkets

Geometry and trigonometry

(a)

2 marks

Midpoint of $B (6, 0)$ and $D (2, 2)$ .

(\frac{6 + 2}{2}, \frac{0 + 2}{2}) = (4, 1)

(b)

4 marks

Gradient of BD:
The perpendicular bisector has gradient 2 and passes through $(4, 1)$ .

\frac{2 - 0}{2 - 6} = - \frac{1}{2} y - 1 = 2 (x - 4) \Rightarrow y = 2 x - 7

(c)

3 marks

X lies on both $y = 2 - x$ and $y = 2 x - 7$ .
So $X = (3, - 1)$ .

2 - x = 2 x - 7 \Rightarrow x = 3, y = - 1

(d)

2 marks

Use the distance formula with $Y (- 1, 3)$ and $Z (7, 7)$ .

Y Z = (7 + 1)^{2} + (7 - 3)^{2} = 80 = 45

(e)

4 marks

Find XZ and use the cosine rule in triangle XYZ.

X Z = (7 - 3)^{2} + (7 + 1)^{2} = 80 cos X Y Z = \frac{80 + 32 - 80}{2 80 32} = 0.3162 X Y Z = 71. 6^{\circ}

(f)

2 marks

Use $\frac{1}{2} ab sin C$ with the two sides next to the angle at Y.

Area = \frac{1}{2} 8032 sin 71.5 7^{\circ} = 24 km^{2}

(g)

1 mark

A larger cell only means D is the closest shop to more land. It does not mean more people live there, and people choose where to shop for reasons other than distance.

Question 4

17 marks

A Ferris wheel: cosine model, domain, range and height above a statue

Functions

(a)

6 marks

The amplitude is the radius of the wheel. The period is 40 s. The centre is the lowest height plus the radius.

a = \frac{61.8}{2} = 30.9 b = \frac{360}{40} = 9 (degrees; \frac{π}{20} in radians) d = 30.9 + 2.7 = 33.6

(b)

2 marks

The ride lasts 12 minutes at 1.5 revolutions per minute.

12 \times 1.5 = 18 revolutions

(c)

3 marks

The ride lasts $12 \times 60 = 720$ seconds. The lowest chair height is 2.7 m and the highest is $2.7 + 61.8 = 64.5$ m.

0 \leq t \leq 720, 2.7 \leq h \leq 64.5

(d)

3 marks

Solve $h (t) = 16.7$ in the first revolution.
The chair is higher than the statue between these times.

t = 6.316 and t = 33.684 33.684 - 6.316 \approx 27.4 s

(e)

3 marks

A taller platform lifts the whole wheel but does not change its size or its speed, so only the centre line $d$ changes.
The parameter $d$ changes to 34.3.

d + 30.9 = 65.2 \Rightarrow d = 34.3

Question 5

15 marks

A cafe's profit: differentiation, maximum and a break-even point

Calculus

(a)

3 marks

Expand $C = \frac{k ^{2} x}{10} - \frac{3 x ^{3}}{1000}$ and differentiate.

\frac{d C}{d x} = \frac{k ^{2}}{10} - \frac{9 x ^{2}}{1000}

(b)

4 marks

Set the derivative to zero.
Substitute back.

\frac{9 x ^{2}}{1000} = \frac{k ^{2}}{10} \Rightarrow x^{2} = \frac{100 k ^{2}}{9} \Rightarrow x = \frac{10 k}{3} C_{m a x} = \frac{k}{3} (k^{2} - \frac{3}{100} \cdot \frac{100 k ^{2}}{9}) = \frac{k}{3} \cdot \frac{2 k ^{2}}{3} = \frac{2}{9} k^{3}

(c)

3 marks

Use $x = 20$ and $C = 426$ .
Then the profit-maximising quantity is

2 (k^{2} - 12) = 426 \Rightarrow k^{2} = 225 \Rightarrow k = 15 x = \frac{10 ( 15 )}{3} = 50 litres

(d)

3 marks

The maximum is $C = \frac{2}{9} (15)^{3} = 750$ . The profit is 0 at $x = 0$ and where $225 = 0.03 x^{2}$ , which gives $x = 86.6$ .
Sketch: a cubic arch in the first quadrant that starts at the origin, rises to the maximum at $(50, 750)$ and comes back to the axis at $(86.6, 0)$ .

(e)

2 marks

The cafe does not lose money while $C \geq 0$ , so the most coffee is at the right-hand intercept.

x = 7500 \approx 86.6 litres

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 trigonometry18 marks · 23%
Q3
Functions17 marks · 21%
Q4
Statistics and probability17 marks · 21%
Q1
Calculus15 marks · 19%
Q5
Number and algebra13 marks · 16%
Q2

Mentor insights

Traps to watch for

01
Median and quartiles come from the cumulative frequency curve · Q1
With 40 teenagers, the median is at the 20th value, the lower quartile at the 10th and the upper quartile at the 30th.
02
Which sequence is which · Q2
The food left falls by the same amount each day, so it is arithmetic. The yearly spending grows by a fixed percentage, so it is geometric.
03
Boundaries are perpendicular bisectors · Q3
A Voronoi edge between two sites is the perpendicular bisector of the segment joining them. Find the midpoint and the negative reciprocal gradient.
04
Starts at the bottom · Q4
The chair is at its lowest point at $t = 0$ , which is why the model is $- a cos (b t) + d$ .
05
Use the given profit to find k · Q5
Substitute $x = 20$ and $C = 426$ . The equation for $k^{2}$ is simple, and $k$ is positive.

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