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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 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

15 marks

Hours of daylight: choosing a cosine model

Functions

(a)

1 mark

Daylight hours repeat every year, and a quadratic cannot repeat. It has only one turning point.

(b)

4 marks

The maximum is 16 and the minimum is 8, and the curve goes from the minimum to the maximum in 6 months.

amplitude = \frac{16 - 8}{2} = 4, period = 12, y = \frac{16 + 8}{2} = 12

(c)

3 marks

The curve starts at its minimum at $t = 0$ , so it is an upside-down cosine with $a = - 4$ , $b = 30$ and $d = 12$ .
In radians this is $- 4 cos (\frac{π t}{6}) + 12$ .

f (t) = - 4 cos (30 t) + 12

(d)

4 marks

Solve $f (t) = 10.5$ .
The solutions in the first year are $t = 2.266$ and $t = 12 - 2.266 = 9.734$ .

- 4 cos (30 t) + 12 = 10.5 \Rightarrow cos (30 t) = \frac{3}{8} time above 10.5 hours = 9.734 - 2.266 \approx 7.47 months

(e)

3 marks

The true maximum is $16 + \frac{14}{60} = 16.233$ hours.

\frac{∣16 - 16.233∣}{16.233} \times 100 \approx 1.44%

Question 2

19 marks

A path of tiles: arithmetic sequences, area and cost

Number and algebra

(a)

2 marks

Both $w_{n}$ and $l_{n}$ increase by 10 each time.

a = 30, b = 40

(b)

3 marks

Arithmetic sequences with common difference 10.

w_{n} = 20 + 10 (n - 1) = 10 n + 10 l_{n} = 30 + 10 (n - 1) = 10 n + 20

(c)

3 marks

Set $l_{n} = 740$ .
So the number of tiles is $2 \times 72 = 144$ .

10 n + 20 = 740 \Rightarrow n = 72 pairs w_{72} = 10 (72) + 10 = 730

(d)

3 marks

Each tile is $10 \times 20 = 200$ cm $^{2}$ .

144 \times 200 = 28800 = 2.88 \times 1 0^{4} cm^{2}

(e)

3 marks

One square metre is $100 \times 100 = 10000$ cm $^{2}$ , which holds $\frac{10000}{200} = 50$ tiles, so 10 packs of 5.

\frac{24.50}{10} = $2.45 per pack

(f)

3 marks

Eddie wants 8% more tiles than the 144 he needs.
He must order 32 packs.

1.08 \times 144 = 155.52 tiles \Rightarrow \frac{155.52}{5} = 31.1 packs

(g)

2 marks

Packs plus the fixed delivery cost.

35 + 32 (2.45) = 113.4 \approx $113

Question 3

16 marks

Eurovision scores: quartiles, outliers and rank correlation

Statistics and probability

(a)

4 marks

Order the scores: 302, 305, 331, 334, 364, 370, 472, 498. With 8 values the lower half is the first four and the upper half is the last four.

Q_{1} = \frac{305 + 331}{2} = 318, Q_{3} = \frac{370 + 472}{2} = 421 IQR = 421 - 318 = 103

(b)

3 marks

An outlier lies above $Q_{3} + 1.5 \times IQR$ .
The Netherlands scored $498 < 575.5$ , so it is not an outlier.

421 + 1.5 (103) = 575.5

(c)

2 marks

The correlation coefficient $r = 0.249$ is very close to 0, so the linear relationship is too weak.
No, a regression line would not be appropriate for predicting a country's score.

(d)

3 marks

Rank the populations from largest to smallest: Russia 145 (1), Italy 60 (2), Netherlands 17 (3), then Sweden and Azerbaijan both 10, then Switzerland 9, Norway 5 and North Macedonia 2.

a = 6, b = 4.5, c = 4.5

(e)

3 marks

Spearman's coefficient is the product-moment coefficient of the two sets of ranks.
Interpretation: a moderately strong positive association between population size and Eurovision score. Larger countries tend to rank higher.

r_{s} = 0.683

(f)

1 mark

The scores are ranked, and 478 is still the highest score. The ranks do not change, so $r_{s}$ is unchanged.

Question 4

15 marks

A goat in a square field: segment, reachable area and a rate of eating

Geometry and trigonometry
Calculus

(a)

8 marks

In the right triangle with hypotenuse 4.5 and adjacent side 4:
Sector area, with $θ = 0.9518$ rad:
Triangle area, with chord $A B = 2 4. 5^{2} - 4^{2} = 4.123$ :
The segment is the sector minus the triangle.

cos \frac{1}{2} A O B = \frac{4}{4.5} \Rightarrow A O B = 54. 5^{\circ} (0.952 rad) \frac{1}{2} (4.5)^{2} (0.9518) = 9.637 m^{2} \frac{1}{2} (4.123) (4) = 8.246 m^{2} 9.637 - 8.246 = 1.39 m^{2}

(b)

5 marks

Area of the full circle:
The goat cannot reach the four segments outside the square.

π (4.5)^{2} = 63.6 m^{2} 63.62 - 4 (1.390) = 58.1 m^{2}

(c)

2 marks

The rate is $\frac{d V}{d t} = 0.3 t e^{- t}$ . Differentiate and set to zero.

0.3 e^{- t} (1 - t) = 0 \Rightarrow t = 1 hour

Question 5

15 marks

Overbooking a flight: binomial model and expected gain

Statistics and probability

(a)

3 marks

$T \sim B (74, 0.9)$ , and more than 72 passengers means 73 or 74.

P (T \geq 73) = 1 - P (T \leq 72) \approx 0.00379

(b)(i)

2 marks

The expected value of a binomial is $n p$ with $n = 72$ and $p = 0.9$ .

E (T) = 72 \times 0.9 = 64.8

(b)(ii)

2 marks

Need $0.9 n \leq 72$ .
So at most 80 tickets.

n \leq 80

(c)

8 marks

With 74 tickets sold the income is $74 \times 150 = 11100$ .
Expected compensation:
Expected income with 74 tickets sold:
Compare with 72 tickets sold, which gives $72 \times 150 = 10800$ .

P (T = 73) = 0.003380, P (T = 74) = 0.000411 300 (0.003380) + 600 (0.000411) = 1.2607 11100 - 1.2607 = 11098.74 11098.74 - 10800 \approx $299 increase

Mentor tip

A different route: compare the changes. Two extra tickets bring $+ 300$ most of the time, $0$ when 73 arrive, and $- 300$ when 74 arrive.

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 probability31 marks · 39%
Q3, Q5
Number and algebra19 marks · 24%
Q2
Functions15 marks · 19%
Q1
Geometry and trigonometry13 marks · 16%
Q4
Calculus2 marks · 3%
Q4

Mentor insights

Traps to watch for

01
Sine and cosine in degrees · Q1
The period is 12 months, so $b = \frac{360}{12} = 30$ in degrees mode. Check your calculator mode before solving.
02
Both width and length grow by 10 each pair · Q2
Each extra pair adds 10 cm to both, so both are arithmetic with common difference 10.
03
Tied values share a rank · Q3
Populations of 10 million appear twice. They occupy ranks 4 and 5, so each gets $4.5$ , which is the mean of the two ranks.
04
Find the half-angle first · Q4
The 4 is the perpendicular distance from the centre to the chord. The half-angle $α$ has $cos α = \frac{4}{4.5}$ .
05
Compensation applies to the extra passengers only · Q5
If 73 arrive, one cannot board and costs $300; if 74 arrive, two cannot board and cost $600.

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 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

15 marks

Hours of daylight: choosing a cosine model

Functions

(a)

1 mark

Daylight hours repeat every year, and a quadratic cannot repeat. It has only one turning point.

(b)

4 marks

The maximum is 16 and the minimum is 8, and the curve goes from the minimum to the maximum in 6 months.

amplitude = \frac{16 - 8}{2} = 4, period = 12, y = \frac{16 + 8}{2} = 12

(c)

3 marks

The curve starts at its minimum at $t = 0$ , so it is an upside-down cosine with $a = - 4$ , $b = 30$ and $d = 12$ .
In radians this is $- 4 cos (\frac{π t}{6}) + 12$ .

f (t) = - 4 cos (30 t) + 12

(d)

4 marks

Solve $f (t) = 10.5$ .
The solutions in the first year are $t = 2.266$ and $t = 12 - 2.266 = 9.734$ .

- 4 cos (30 t) + 12 = 10.5 \Rightarrow cos (30 t) = \frac{3}{8} time above 10.5 hours = 9.734 - 2.266 \approx 7.47 months

(e)

3 marks

The true maximum is $16 + \frac{14}{60} = 16.233$ hours.

\frac{∣16 - 16.233∣}{16.233} \times 100 \approx 1.44%

Question 2

19 marks

A path of tiles: arithmetic sequences, area and cost

Number and algebra

(a)

2 marks

Both $w_{n}$ and $l_{n}$ increase by 10 each time.

a = 30, b = 40

(b)

3 marks

Arithmetic sequences with common difference 10.

w_{n} = 20 + 10 (n - 1) = 10 n + 10 l_{n} = 30 + 10 (n - 1) = 10 n + 20

(c)

3 marks

Set $l_{n} = 740$ .
So the number of tiles is $2 \times 72 = 144$ .

10 n + 20 = 740 \Rightarrow n = 72 pairs w_{72} = 10 (72) + 10 = 730

(d)

3 marks

Each tile is $10 \times 20 = 200$ cm $^{2}$ .

144 \times 200 = 28800 = 2.88 \times 1 0^{4} cm^{2}

(e)

3 marks

One square metre is $100 \times 100 = 10000$ cm $^{2}$ , which holds $\frac{10000}{200} = 50$ tiles, so 10 packs of 5.

\frac{24.50}{10} = $2.45 per pack

(f)

3 marks

Eddie wants 8% more tiles than the 144 he needs.
He must order 32 packs.

1.08 \times 144 = 155.52 tiles \Rightarrow \frac{155.52}{5} = 31.1 packs

(g)

2 marks

Packs plus the fixed delivery cost.

35 + 32 (2.45) = 113.4 \approx $113

Question 3

16 marks

Eurovision scores: quartiles, outliers and rank correlation

Statistics and probability

(a)

4 marks

Order the scores: 302, 305, 331, 334, 364, 370, 472, 498. With 8 values the lower half is the first four and the upper half is the last four.

Q_{1} = \frac{305 + 331}{2} = 318, Q_{3} = \frac{370 + 472}{2} = 421 IQR = 421 - 318 = 103

(b)

3 marks

An outlier lies above $Q_{3} + 1.5 \times IQR$ .
The Netherlands scored $498 < 575.5$ , so it is not an outlier.

421 + 1.5 (103) = 575.5

(c)

2 marks

The correlation coefficient $r = 0.249$ is very close to 0, so the linear relationship is too weak.
No, a regression line would not be appropriate for predicting a country's score.

(d)

3 marks

Rank the populations from largest to smallest: Russia 145 (1), Italy 60 (2), Netherlands 17 (3), then Sweden and Azerbaijan both 10, then Switzerland 9, Norway 5 and North Macedonia 2.

a = 6, b = 4.5, c = 4.5

(e)

3 marks

Spearman's coefficient is the product-moment coefficient of the two sets of ranks.
Interpretation: a moderately strong positive association between population size and Eurovision score. Larger countries tend to rank higher.

r_{s} = 0.683

(f)

1 mark

The scores are ranked, and 478 is still the highest score. The ranks do not change, so $r_{s}$ is unchanged.

Question 4

15 marks

A goat in a square field: segment, reachable area and a rate of eating

Geometry and trigonometry
Calculus

(a)

8 marks

In the right triangle with hypotenuse 4.5 and adjacent side 4:
Sector area, with $θ = 0.9518$ rad:
Triangle area, with chord $A B = 2 4. 5^{2} - 4^{2} = 4.123$ :
The segment is the sector minus the triangle.

cos \frac{1}{2} A O B = \frac{4}{4.5} \Rightarrow A O B = 54. 5^{\circ} (0.952 rad) \frac{1}{2} (4.5)^{2} (0.9518) = 9.637 m^{2} \frac{1}{2} (4.123) (4) = 8.246 m^{2} 9.637 - 8.246 = 1.39 m^{2}

(b)

5 marks

Area of the full circle:
The goat cannot reach the four segments outside the square.

π (4.5)^{2} = 63.6 m^{2} 63.62 - 4 (1.390) = 58.1 m^{2}

(c)

2 marks

The rate is $\frac{d V}{d t} = 0.3 t e^{- t}$ . Differentiate and set to zero.

0.3 e^{- t} (1 - t) = 0 \Rightarrow t = 1 hour

Question 5

15 marks

Overbooking a flight: binomial model and expected gain

Statistics and probability

(a)

3 marks

$T \sim B (74, 0.9)$ , and more than 72 passengers means 73 or 74.

P (T \geq 73) = 1 - P (T \leq 72) \approx 0.00379

(b)(i)

2 marks

The expected value of a binomial is $n p$ with $n = 72$ and $p = 0.9$ .

E (T) = 72 \times 0.9 = 64.8

(b)(ii)

2 marks

Need $0.9 n \leq 72$ .
So at most 80 tickets.

n \leq 80

(c)

8 marks

With 74 tickets sold the income is $74 \times 150 = 11100$ .
Expected compensation:
Expected income with 74 tickets sold:
Compare with 72 tickets sold, which gives $72 \times 150 = 10800$ .

P (T = 73) = 0.003380, P (T = 74) = 0.000411 300 (0.003380) + 600 (0.000411) = 1.2607 11100 - 1.2607 = 11098.74 11098.74 - 10800 \approx $299 increase

Mentor tip

A different route: compare the changes. Two extra tickets bring $+ 300$ most of the time, $0$ when 73 arrive, and $- 300$ when 74 arrive.

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 probability31 marks · 39%
Q3, Q5
Number and algebra19 marks · 24%
Q2
Functions15 marks · 19%
Q1
Geometry and trigonometry13 marks · 16%
Q4
Calculus2 marks · 3%
Q4

Mentor insights

Traps to watch for

01
Sine and cosine in degrees · Q1
The period is 12 months, so $b = \frac{360}{12} = 30$ in degrees mode. Check your calculator mode before solving.
02
Both width and length grow by 10 each pair · Q2
Each extra pair adds 10 cm to both, so both are arithmetic with common difference 10.
03
Tied values share a rank · Q3
Populations of 10 million appear twice. They occupy ranks 4 and 5, so each gets $4.5$ , which is the mean of the two ranks.
04
Find the half-angle first · Q4
The 4 is the perpendicular distance from the centre to the chord. The half-angle $α$ has $cos α = \frac{4}{4.5}$ .
05
Compensation applies to the extra passengers only · Q5
If 73 arrive, one cannot board and costs $300; if 74 arrive, two cannot board and cost $600.

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