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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global paper, which your school or the IB can provide.

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Questions

Questions 1–5 · 80 marks

Question 1

17 marks

Time spent in a library: grouped data, box plot and outliers

Statistics and probability

(a)

1 mark

Time can take any value in an interval, not just separate values.

continuous

(b)

2 marks

The frequencies add to 160.

k = 160 - (50 + 62 + 14 + 8) = 26

(c)

2 marks

The highest frequency is 62, in the class $20 \leq T < 40$ , whose middle value is

30

(d)

2 marks

Use the midpoints $10, 30, 50, 70, 90$ .

\overset{ˉ}{T} = \frac{10 ( 50 ) + 30 ( 62 ) + 50 ( 26 ) + 70 ( 14 ) + 90 ( 8 )}{160} = \frac{5360}{160} = 33.5

(e)

1 mark

The most people who could have spent 35 minutes or less is everyone in the two classes below 40 minutes.

50 + 62 = 112

(f)

2 marks

At least 60 minutes means the last two classes.

P (T \geq 60) = \frac{14 + 8}{160} = \frac{11}{80} = 0.1375

(g)

1 mark

Read the line inside the box.

median = 26 minutes

(h)

2 marks

From the diagram $Q_{1} = 16$ and $Q_{3} = 50$ .

IQR = 50 - 16 = 34

(i)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
The longest time is 92 minutes. Since $92 < 101$ , it is not an outlier.

50 + 1.5 (34) = 101

(j)

1 mark

The median is closer to the lower quartile and the upper whisker is much longer than the lower one, so the data are skewed to the right. A normal distribution is symmetric.

Question 2

17 marks

A park shaped as a quadrilateral and a redesign with a circular arc

Geometry and trigonometry

(a)

4 marks

The angles in triangle BDC add to $18 0^{\circ}$ .
Use the sine rule with BD opposite the $12 0^{\circ}$ angle.

B D C = 180 - 120 - 41 = 1 9^{\circ} \frac{B D}{sin 12 0 ^{\circ}} = \frac{40}{sin 1 9 ^{\circ}} \Rightarrow B D = 106 m

(b)

3 marks

Use the cosine rule in the isosceles triangle ABD.

cos B A D = \frac{8 5 ^{2} + 8 5 ^{2} - 106.4 0 ^{2}}{2 ( 85 ) ( 85 )} \Rightarrow B A D = 77.49 5^{\circ}

(c)

3 marks

Area of triangle ABD with $B A D = 7 7^{\circ}$ .

\frac{1}{2} (85) (85) sin 7 7^{\circ} = 3520 m^{2}

(d)

1 mark

E is on the arc whose centre is A and which passes through B, so $A E = A B$ .

85 m

(e)

3 marks

The angle is $7 7^{\circ} = 1.344$ radians. The arc length is

85 \times 1.344 = 114.2 m perimeter = 85 + 85 + 114.2 = 284 m

(f)

3 marks

The shaded region is the sector minus triangle ABD.

\frac{1}{2} (85)^{2} (1.344) - 3520 = 4855 - 3520 \approx 1330 m^{2}

Question 3

13 marks

Restaurants and a Voronoi diagram

Geometry and trigonometry

(a)

2 marks

The point $(2, 7)$ lies in the cell of B.
With A closed, compare distances from $(0, 1)$ : F is 5 away, B is $34 \approx 5.8$ away and D is $65 \approx 8.1$ away.

(i) B, (ii) F

(b)

2 marks

C $(7, 8)$ and D $(7, 5)$ have a vertical segment between them, so the bisector is horizontal through the midpoint $(7, 6.5)$ .

y = 6.5

(c)

5 marks

Midpoint of B $(3, 6)$ and C $(7, 8)$ is $(5, 7)$ . The gradient of BC is $\frac{8 - 6}{7 - 3} = \frac{1}{2}$ , so the bisector has gradient $- 2$ .

y - 7 = - 2 (x - 5) \Rightarrow y = - 2 x + 17

(d)

4 marks

The point equidistant from B, C and D is where the two bisectors meet.
Distance to D $(7, 5)$ :

- 2 x + 17 = 6.5 \Rightarrow x = 5.25, (5.25, 6.5) (7 - 5.25)^{2} + (5 - 6.5)^{2} = 5.3125 \approx 2.31 km

Question 4

16 marks

School survey: Venn diagram, conditional probability and a binomial model

Statistics and probability

(a)

4 marks

Both dog and cat is 14%. Dog only is $68 - 14$ , cat only is $36 - 14$ , and the rest have neither.

m = 54, n = 14, p = 22, q = 10

(b)

1 mark

Add all the regions except "neither".

54 + 14 + 22 = 90%

(c)

3 marks

Dog but no cat is the dog-only region. Given no cat, the sample space is dog-only plus neither.

P (dog, no cat) = 0.54, P (dog ∣ no cat) = \frac{0.54}{0.64} = \frac{27}{32} \approx 0.844

(d)

7 marks

Let $X \sim B (10, 0.68)$ be the number of captains with a dog.
Exactly nine in succession means DDDDDDDDDN or NDDDDDDDDD.

P (X = 5) \approx 0.123, P (X > 3) \approx 0.984 2 (0.68)^{9} (0.32) \approx 0.0199

(e)

1 mark

John picks 10 students from the survey without putting them back, so the choices are not independent and the probability of a dog changes from pick to pick.

Question 5

17 marks

Linda's field: maximum, area, trapezoidal error and a square building

Calculus

(a)

5 marks

The northern boundary is $f (x) = - \frac{x ^{2}}{50} + 2 x + 30$ .
The furthest north point has $f^{'} (x) = 0$ .

f^{'} (x) = - \frac{x}{25} + 2 x = 50, f (50) = - 50 + 100 + 30 = 80, (50, 80)

(b)

4 marks

The area under the northern boundary from $x = 0$ to $x = 70$ :

A = \int_{0}^{70} (- \frac{x ^{2}}{50} + 2 x + 30) d x A = [- \frac{x ^{3}}{150} + x^{2} + 30 x]_{0}^{70} = 4713 m^{2}

(c)

3 marks

The trapezoidal estimate was 11.4 m $^{2}$ too small.
Linda can reduce the error by using more intervals, so that each trapezoid is narrower.

\frac{11.4}{4713.3} \times 100 \approx 0.242%

(d)

5 marks

Let E be at $x = e$ . The side of the square is $70 - e$ and F is at height $f (e)$ , so $f (e) = 70 - e$ .
The root in the range $0 \leq e \leq 70$ :

- \frac{e ^{2}}{50} + 2 e + 30 = 70 - e \Rightarrow e^{2} - 150 e + 2000 = 0 e = 14.8 area = (70 - 14.79)^{2} \approx 3050 m^{2}

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 probability33 marks · 41%
Q1, Q4
Geometry and trigonometry30 marks · 38%
Q2, Q3
Calculus17 marks · 21%
Q5

Mentor insights

Traps to watch for

01
Estimate the mean from midpoints · Q1
Use the middle of each class with its frequency.
02
Sine rule in triangle BDC · Q2
The angle at D is $18 0^{\circ} - 12 0^{\circ} - 4 1^{\circ}$ , and BD faces the $12 0^{\circ}$ angle.
03
Read the cell · Q3
A point belongs to the restaurant whose cell it lies in. If a site is removed, the point goes to the next nearest site.
04
Fill the Venn diagram from the overlap outwards · Q4
Start with the 14% who have both, then subtract it from each total.
05
Percentage error uses the true area · Q5
Divide the underestimate by the exact area.

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

Session: November 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 November 2022 Global 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

Time spent in a library: grouped data, box plot and outliers

Statistics and probability

(a)

1 mark

Time can take any value in an interval, not just separate values.

continuous

(b)

2 marks

The frequencies add to 160.

k = 160 - (50 + 62 + 14 + 8) = 26

(c)

2 marks

The highest frequency is 62, in the class $20 \leq T < 40$ , whose middle value is

30

(d)

2 marks

Use the midpoints $10, 30, 50, 70, 90$ .

\overset{ˉ}{T} = \frac{10 ( 50 ) + 30 ( 62 ) + 50 ( 26 ) + 70 ( 14 ) + 90 ( 8 )}{160} = \frac{5360}{160} = 33.5

(e)

1 mark

The most people who could have spent 35 minutes or less is everyone in the two classes below 40 minutes.

50 + 62 = 112

(f)

2 marks

At least 60 minutes means the last two classes.

P (T \geq 60) = \frac{14 + 8}{160} = \frac{11}{80} = 0.1375

(g)

1 mark

Read the line inside the box.

median = 26 minutes

(h)

2 marks

From the diagram $Q_{1} = 16$ and $Q_{3} = 50$ .

IQR = 50 - 16 = 34

(i)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
The longest time is 92 minutes. Since $92 < 101$ , it is not an outlier.

50 + 1.5 (34) = 101

(j)

1 mark

The median is closer to the lower quartile and the upper whisker is much longer than the lower one, so the data are skewed to the right. A normal distribution is symmetric.

Question 2

17 marks

A park shaped as a quadrilateral and a redesign with a circular arc

Geometry and trigonometry

(a)

4 marks

The angles in triangle BDC add to $18 0^{\circ}$ .
Use the sine rule with BD opposite the $12 0^{\circ}$ angle.

B D C = 180 - 120 - 41 = 1 9^{\circ} \frac{B D}{sin 12 0 ^{\circ}} = \frac{40}{sin 1 9 ^{\circ}} \Rightarrow B D = 106 m

(b)

3 marks

Use the cosine rule in the isosceles triangle ABD.

cos B A D = \frac{8 5 ^{2} + 8 5 ^{2} - 106.4 0 ^{2}}{2 ( 85 ) ( 85 )} \Rightarrow B A D = 77.49 5^{\circ}

(c)

3 marks

Area of triangle ABD with $B A D = 7 7^{\circ}$ .

\frac{1}{2} (85) (85) sin 7 7^{\circ} = 3520 m^{2}

(d)

1 mark

E is on the arc whose centre is A and which passes through B, so $A E = A B$ .

85 m

(e)

3 marks

The angle is $7 7^{\circ} = 1.344$ radians. The arc length is

85 \times 1.344 = 114.2 m perimeter = 85 + 85 + 114.2 = 284 m

(f)

3 marks

The shaded region is the sector minus triangle ABD.

\frac{1}{2} (85)^{2} (1.344) - 3520 = 4855 - 3520 \approx 1330 m^{2}

Question 3

13 marks

Restaurants and a Voronoi diagram

Geometry and trigonometry

(a)

2 marks

The point $(2, 7)$ lies in the cell of B.
With A closed, compare distances from $(0, 1)$ : F is 5 away, B is $34 \approx 5.8$ away and D is $65 \approx 8.1$ away.

(i) B, (ii) F

(b)

2 marks

C $(7, 8)$ and D $(7, 5)$ have a vertical segment between them, so the bisector is horizontal through the midpoint $(7, 6.5)$ .

y = 6.5

(c)

5 marks

Midpoint of B $(3, 6)$ and C $(7, 8)$ is $(5, 7)$ . The gradient of BC is $\frac{8 - 6}{7 - 3} = \frac{1}{2}$ , so the bisector has gradient $- 2$ .

y - 7 = - 2 (x - 5) \Rightarrow y = - 2 x + 17

(d)

4 marks

The point equidistant from B, C and D is where the two bisectors meet.
Distance to D $(7, 5)$ :

- 2 x + 17 = 6.5 \Rightarrow x = 5.25, (5.25, 6.5) (7 - 5.25)^{2} + (5 - 6.5)^{2} = 5.3125 \approx 2.31 km

Question 4

16 marks

School survey: Venn diagram, conditional probability and a binomial model

Statistics and probability

(a)

4 marks

Both dog and cat is 14%. Dog only is $68 - 14$ , cat only is $36 - 14$ , and the rest have neither.

m = 54, n = 14, p = 22, q = 10

(b)

1 mark

Add all the regions except "neither".

54 + 14 + 22 = 90%

(c)

3 marks

Dog but no cat is the dog-only region. Given no cat, the sample space is dog-only plus neither.

P (dog, no cat) = 0.54, P (dog ∣ no cat) = \frac{0.54}{0.64} = \frac{27}{32} \approx 0.844

(d)

7 marks

Let $X \sim B (10, 0.68)$ be the number of captains with a dog.
Exactly nine in succession means DDDDDDDDDN or NDDDDDDDDD.

P (X = 5) \approx 0.123, P (X > 3) \approx 0.984 2 (0.68)^{9} (0.32) \approx 0.0199

(e)

1 mark

John picks 10 students from the survey without putting them back, so the choices are not independent and the probability of a dog changes from pick to pick.

Question 5

17 marks

Linda's field: maximum, area, trapezoidal error and a square building

Calculus

(a)

5 marks

The northern boundary is $f (x) = - \frac{x ^{2}}{50} + 2 x + 30$ .
The furthest north point has $f^{'} (x) = 0$ .

f^{'} (x) = - \frac{x}{25} + 2 x = 50, f (50) = - 50 + 100 + 30 = 80, (50, 80)

(b)

4 marks

The area under the northern boundary from $x = 0$ to $x = 70$ :

A = \int_{0}^{70} (- \frac{x ^{2}}{50} + 2 x + 30) d x A = [- \frac{x ^{3}}{150} + x^{2} + 30 x]_{0}^{70} = 4713 m^{2}

(c)

3 marks

The trapezoidal estimate was 11.4 m $^{2}$ too small.
Linda can reduce the error by using more intervals, so that each trapezoid is narrower.

\frac{11.4}{4713.3} \times 100 \approx 0.242%

(d)

5 marks

Let E be at $x = e$ . The side of the square is $70 - e$ and F is at height $f (e)$ , so $f (e) = 70 - e$ .
The root in the range $0 \leq e \leq 70$ :

- \frac{e ^{2}}{50} + 2 e + 30 = 70 - e \Rightarrow e^{2} - 150 e + 2000 = 0 e = 14.8 area = (70 - 14.79)^{2} \approx 3050 m^{2}

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 probability33 marks · 41%
Q1, Q4
Geometry and trigonometry30 marks · 38%
Q2, Q3
Calculus17 marks · 21%
Q5

Mentor insights

Traps to watch for

01
Estimate the mean from midpoints · Q1
Use the middle of each class with its frequency.
02
Sine rule in triangle BDC · Q2
The angle at D is $18 0^{\circ} - 12 0^{\circ} - 4 1^{\circ}$ , and BD faces the $12 0^{\circ}$ angle.
03
Read the cell · Q3
A point belongs to the restaurant whose cell it lies in. If a site is removed, the point goes to the next nearest site.
04
Fill the Venn diagram from the overlap outwards · Q4
Start with the 14% who have both, then subtract it from each total.
05
Percentage error uses the true area · Q5
Divide the underestimate by the exact area.

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