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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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 1–2 · 55 marks

Question 1

28 marks

Modelling a computer virus: regression, exponential growth and a logistic model

Calculus

(a)(i)

2 marks

Use linear regression of $Q$ on $t$ with the seven data points.

Q = 3094 t - 54042

(a)(ii)

1 mark

Read the correlation coefficient from the same regression.

r = 0.755

(a)(iii)

1 mark

The scatter of the data is strongly curved and grows by a larger factor every five days, so the data are not from a bivariate normal distribution. A hypothesis test on $r$ would not be valid.

(b)(i)

4 marks

Separate the variables.

\frac{Q ^{'}}{Q} = β N \Rightarrow \int \frac{d Q}{Q} = \int β N d t ln Q = β N t + c \Rightarrow Q = C e^{β N t}

(b)(ii)

2 marks

The solution is exponential, so fit an exponential regression $Q = a e^{b t}$ to the data.

Q (t) = 1.15 e^{0.292 t}

(b)(iii)

1 mark

Read the coefficient of determination from the exponential regression.

R^{2} = 0.999

(b)(iv)

2 marks

The exponential model has $R^{2} = 0.999$ whereas the linear model has $r^{2} \approx 0.570$ . Because $R^{2}$ is much closer to 1, the exponential model fits the data much better.

(b)(v)

1 mark

For large $t$ the exponential model grows without limit, which would predict more infected computers than there are computers in the city.

(c)

2 marks

Doubling means $e^{0.292 t} = 2$ .

t = \frac{ln 2}{0.292} \approx 2.37 days

(d)

3 marks

The growth rate is $β N = 0.292$ , so for city X:
This is bigger than $9.64 \times 1 0^{- 8}$ for city Y, so the virus is spreading more easily in city X.

β = \frac{0.292055}{2.6 \times 1 0 ^{6}} = 1.12 \times 1 0^{- 7}

(e)

2 marks

Use the formula with $t = 15$ and $t = 30$ .

a = \frac{Q ( 20 ) - Q ( 10 )}{10} = \frac{403 - 20}{10} = 38.3 b = \frac{Q ( 35 ) - Q ( 25 )}{10} = \frac{32667 - 1806}{10} = 3086.1

(f)(i)

5 marks

Work out $\frac{Q ^{'}}{Q}$ for $t = 15, 20, 25, 30, 35$ and regress it on $Q$ .
Compare with $\frac{Q ^{'}}{Q} = k - \frac{k}{L} Q$ : the intercept is $k$ and the gradient is $- \frac{k}{L}$ .

\frac{Q ^{'}}{Q} = 0.4223 - 2.556 \times 1 0^{- 6} Q k \approx 0.422, L = \frac{0.4223}{2.556 \times 1 0 ^{- 6}} \approx 165000

(f)(ii)

2 marks

As $t \to \infty$ the logistic solution tends to $L$ , the eventual number of infected computers.

\frac{165205}{2600000} \times 100 \approx 6.35%

Question 2

27 marks

Planning a new town: Voronoi diagrams and a sewer network

Geometry and trigonometry
Number and algebra

(a)

1 mark

It is reasonable if each town is small compared with the distances between towns, so its extent does not matter when comparing distances.

(b)

1 mark

The four towns are at the corners of a square, so the regions meet along the two perpendicular bisectors through the centre.
Sketch: the square split into four quarters by the lines $x = 20$ and $y = 20$ , each town in its own quarter.

(c)(i)

4 marks

The midpoint of $I (30, 20)$ and $F (40, 40)$ is $(35, 30)$ . The gradient of IF is $\frac{40 - 20}{40 - 30} = 2$ , so the bisector has gradient $- \frac{1}{2}$ .

y - 30 = - \frac{1}{2} (x - 35) \Rightarrow y = - \frac{1}{2} x + \frac{95}{2}

(c)(ii)

4 marks

Find the bisectors of IH and IG, then intersect them with the other lines.
Bisector of IH: midpoint $(15, 10)$ and gradient $- \frac{3}{2}$ , so $y = - \frac{3}{2} x + 32.5$ . Bisector of IG: midpoint $(35, 10)$ and gradient $\frac{1}{2}$ , so $y = \frac{1}{2} x - 7.5$ .
Edison, Hamilton and Isaacopolis: the bisector of IE with $y = 20$ (the bisector of EH).
Hamilton, Gaussville and Isaacopolis: the bisectors of IH and IG meet where
The other two vertices are $(\frac{25}{3}, 20)$ and $(20, 2.5)$ .

\frac{3}{2} x + 7.5 = 20 \Rightarrow x = \frac{25}{3} - \frac{3}{2} x + 32.5 = \frac{1}{2} x - 7.5 \Rightarrow x = 20, y = 2.5

(c)(iii)

2 marks

Sketch: Isaacopolis occupies a cell in the middle-right of the square, bordered by the vertices $(20, 37.5)$ , $(\frac{25}{3}, 20)$ and $(20, 2.5)$ and running out to the right-hand edge. Edison, Fermitown, Gaussville and Hamilton keep the remaining four regions.

(d)

4 marks

The car drives along a line $y = c$ . It crosses the Isaacopolis district between the bisector of IE and the bisector of IF.
The distance inside Isaacopolis's district:
It takes 30% of the 40 km, which is 12 km.
The destination is 33 km north of Gaussville.

x_{I E} = \frac{2 c - 15}{3}, x_{I F} = 95 - 2 c 95 - 2 c - \frac{2 c - 15}{3} = 100 - \frac{8 c}{3} 100 - \frac{8 c}{3} = 12 \Rightarrow c = 33

(e)(i)

4 marks

The point farthest from the nearest town must be a vertex. Compare the distances from the vertices to their nearest towns.
The best location is $(\frac{25}{3}, 20) \approx (8.33, 20)$ , 21.7 km from the three nearest towns.

(20, 37.5) : 2 0^{2} + 2. 5^{2} \approx 20.2 (\frac{25}{3}, 20) : \frac{65}{3} \approx 21.7 (20, 2.5) : 2 0^{2} + 2. 5^{2} \approx 20.2

(e)(ii)

1 mark

This ignores population density and travel: the dump would be close to the western towns, and the decision only takes distance to the nearest town into account.

(f)(i)

3 marks

Read the matrix as a graph with the order E, F, G, H, I, T. The connections between different places are E-G, E-H, F-T, G-I and H-T. This is a path I - G - E - H - T - F.
Pollution starts at T. Day 1: H and F. Day 2: E. Day 3: G. Day 4: I.
Isaacopolis is the last town to be polluted. Check with powers of $M$ : the last row of $M^{3}$ still has a 0 for the T to I entry, and $M^{4}$ has none.

(f)(ii)

1 mark

It takes 4 days for the pollution to reach I.

(f)(iii)

2 marks

Ignore the loops. The number of connections at each place are E 2, F 1, G 2, H 2, I 1, T 2. The only places with an odd number of connections are F and I.
A route that uses every sewer once must start at one odd place and end at the other: start at F and end at I (or the reverse).

Mentor tip

Check with the path: F, T, H, E, G, I uses each sewer exactly once.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus28 marks · 51%
Q1
Geometry and trigonometry21 marks · 38%
Q2
Number and algebra6 marks · 11%
Q2

Mentor insights

Traps to watch for

01
Hypothesis tests on r need the right conditions · Q1
A test on $r$ assumes the data come from a random sample of a bivariate normal population. Time-series data growing by a constant percentage is not like that.
02
Compare R-squared fairly · Q1
$R^{2}$ of the exponential model (0.999) is compared with $r^{2}$ of the linear model ( $0.75 5^{2} \approx 0.570$ ), or $R$ with $r$ .
03
Estimating a derivative from the table · Q1
The central difference $\frac{Q ( t + 5 ) - Q ( t - 5 )}{10}$ needs data 5 days either side, so it exists only for $t = 15$ to $35$ .
04
Logistic: divide by Q · Q1
Plot $\frac{Q ^{'}}{Q}$ against $Q$ . The line is $k - \frac{k}{L} Q$ , so the intercept is $k$ and the gradient is $- \frac{k}{L}$ .
05
Edges are perpendicular bisectors · Q2
Each edge of a Voronoi cell lies on the perpendicular bisector of two neighbouring sites. A vertex is equidistant from three sites.

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 HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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 1–2 · 55 marks

Question 1

28 marks

Modelling a computer virus: regression, exponential growth and a logistic model

Calculus

(a)(i)

2 marks

Use linear regression of $Q$ on $t$ with the seven data points.

Q = 3094 t - 54042

(a)(ii)

1 mark

Read the correlation coefficient from the same regression.

r = 0.755

(a)(iii)

1 mark

The scatter of the data is strongly curved and grows by a larger factor every five days, so the data are not from a bivariate normal distribution. A hypothesis test on $r$ would not be valid.

(b)(i)

4 marks

Separate the variables.

\frac{Q ^{'}}{Q} = β N \Rightarrow \int \frac{d Q}{Q} = \int β N d t ln Q = β N t + c \Rightarrow Q = C e^{β N t}

(b)(ii)

2 marks

The solution is exponential, so fit an exponential regression $Q = a e^{b t}$ to the data.

Q (t) = 1.15 e^{0.292 t}

(b)(iii)

1 mark

Read the coefficient of determination from the exponential regression.

R^{2} = 0.999

(b)(iv)

2 marks

The exponential model has $R^{2} = 0.999$ whereas the linear model has $r^{2} \approx 0.570$ . Because $R^{2}$ is much closer to 1, the exponential model fits the data much better.

(b)(v)

1 mark

For large $t$ the exponential model grows without limit, which would predict more infected computers than there are computers in the city.

(c)

2 marks

Doubling means $e^{0.292 t} = 2$ .

t = \frac{ln 2}{0.292} \approx 2.37 days

(d)

3 marks

The growth rate is $β N = 0.292$ , so for city X:
This is bigger than $9.64 \times 1 0^{- 8}$ for city Y, so the virus is spreading more easily in city X.

β = \frac{0.292055}{2.6 \times 1 0 ^{6}} = 1.12 \times 1 0^{- 7}

(e)

2 marks

Use the formula with $t = 15$ and $t = 30$ .

a = \frac{Q ( 20 ) - Q ( 10 )}{10} = \frac{403 - 20}{10} = 38.3 b = \frac{Q ( 35 ) - Q ( 25 )}{10} = \frac{32667 - 1806}{10} = 3086.1

(f)(i)

5 marks

Work out $\frac{Q ^{'}}{Q}$ for $t = 15, 20, 25, 30, 35$ and regress it on $Q$ .
Compare with $\frac{Q ^{'}}{Q} = k - \frac{k}{L} Q$ : the intercept is $k$ and the gradient is $- \frac{k}{L}$ .

\frac{Q ^{'}}{Q} = 0.4223 - 2.556 \times 1 0^{- 6} Q k \approx 0.422, L = \frac{0.4223}{2.556 \times 1 0 ^{- 6}} \approx 165000

(f)(ii)

2 marks

As $t \to \infty$ the logistic solution tends to $L$ , the eventual number of infected computers.

\frac{165205}{2600000} \times 100 \approx 6.35%

Question 2

27 marks

Planning a new town: Voronoi diagrams and a sewer network

Geometry and trigonometry
Number and algebra

(a)

1 mark

It is reasonable if each town is small compared with the distances between towns, so its extent does not matter when comparing distances.

(b)

1 mark

The four towns are at the corners of a square, so the regions meet along the two perpendicular bisectors through the centre.
Sketch: the square split into four quarters by the lines $x = 20$ and $y = 20$ , each town in its own quarter.

(c)(i)

4 marks

The midpoint of $I (30, 20)$ and $F (40, 40)$ is $(35, 30)$ . The gradient of IF is $\frac{40 - 20}{40 - 30} = 2$ , so the bisector has gradient $- \frac{1}{2}$ .

y - 30 = - \frac{1}{2} (x - 35) \Rightarrow y = - \frac{1}{2} x + \frac{95}{2}

(c)(ii)

4 marks

Find the bisectors of IH and IG, then intersect them with the other lines.
Bisector of IH: midpoint $(15, 10)$ and gradient $- \frac{3}{2}$ , so $y = - \frac{3}{2} x + 32.5$ . Bisector of IG: midpoint $(35, 10)$ and gradient $\frac{1}{2}$ , so $y = \frac{1}{2} x - 7.5$ .
Edison, Hamilton and Isaacopolis: the bisector of IE with $y = 20$ (the bisector of EH).
Hamilton, Gaussville and Isaacopolis: the bisectors of IH and IG meet where
The other two vertices are $(\frac{25}{3}, 20)$ and $(20, 2.5)$ .

\frac{3}{2} x + 7.5 = 20 \Rightarrow x = \frac{25}{3} - \frac{3}{2} x + 32.5 = \frac{1}{2} x - 7.5 \Rightarrow x = 20, y = 2.5

(c)(iii)

2 marks

(d)

4 marks

The car drives along a line $y = c$ . It crosses the Isaacopolis district between the bisector of IE and the bisector of IF.
The distance inside Isaacopolis's district:
It takes 30% of the 40 km, which is 12 km.
The destination is 33 km north of Gaussville.

x_{I E} = \frac{2 c - 15}{3}, x_{I F} = 95 - 2 c 95 - 2 c - \frac{2 c - 15}{3} = 100 - \frac{8 c}{3} 100 - \frac{8 c}{3} = 12 \Rightarrow c = 33

(e)(i)

4 marks

The point farthest from the nearest town must be a vertex. Compare the distances from the vertices to their nearest towns.
The best location is $(\frac{25}{3}, 20) \approx (8.33, 20)$ , 21.7 km from the three nearest towns.

(20, 37.5) : 2 0^{2} + 2. 5^{2} \approx 20.2 (\frac{25}{3}, 20) : \frac{65}{3} \approx 21.7 (20, 2.5) : 2 0^{2} + 2. 5^{2} \approx 20.2

(e)(ii)

1 mark

This ignores population density and travel: the dump would be close to the western towns, and the decision only takes distance to the nearest town into account.

(f)(i)

3 marks

Read the matrix as a graph with the order E, F, G, H, I, T. The connections between different places are E-G, E-H, F-T, G-I and H-T. This is a path I - G - E - H - T - F.
Pollution starts at T. Day 1: H and F. Day 2: E. Day 3: G. Day 4: I.
Isaacopolis is the last town to be polluted. Check with powers of $M$ : the last row of $M^{3}$ still has a 0 for the T to I entry, and $M^{4}$ has none.

(f)(ii)

1 mark

It takes 4 days for the pollution to reach I.

(f)(iii)

2 marks

Ignore the loops. The number of connections at each place are E 2, F 1, G 2, H 2, I 1, T 2. The only places with an odd number of connections are F and I.
A route that uses every sewer once must start at one odd place and end at the other: start at F and end at I (or the reverse).

Mentor tip

Check with the path: F, T, H, E, G, I uses each sewer exactly once.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus28 marks · 51%
Q1
Geometry and trigonometry21 marks · 38%
Q2
Number and algebra6 marks · 11%
Q2

Mentor insights

Traps to watch for

01
Hypothesis tests on r need the right conditions · Q1
A test on $r$ assumes the data come from a random sample of a bivariate normal population. Time-series data growing by a constant percentage is not like that.
02
Compare R-squared fairly · Q1
$R^{2}$ of the exponential model (0.999) is compared with $r^{2}$ of the linear model ( $0.75 5^{2} \approx 0.570$ ), or $R$ with $r$ .
03
Estimating a derivative from the table · Q1
The central difference $\frac{Q ( t + 5 ) - Q ( t - 5 )}{10}$ needs data 5 days either side, so it exists only for $t = 15$ to $35$ .
04
Logistic: divide by Q · Q1
Plot $\frac{Q ^{'}}{Q}$ against $Q$ . The line is $k - \frac{k}{L} Q$ , so the intercept is $k$ and the gradient is $- \frac{k}{L}$ .
05
Edges are perpendicular bisectors · Q2
Each edge of a Voronoi cell lies on the perpendicular bisector of two neighbouring sites. A vertex is equidistant from three sites.

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