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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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Questions

Questions 1–2 · 55 marks

Question 1

26 marks

The fastest way from home to school over a field and rough ground

Calculus

(a)

6 marks

The straight line from H to S has length $4^{2} + 1. 2^{2} = 4.176$ km. It crosses the boundary 0.4 km down, which is $\frac{0.4}{1.2} = \frac{1}{3}$ of the way.
Time = distance over speed.

field: \frac{4.176}{3} = 1.392 km, rough: \frac{2 ( 4.176 )}{3} = 2.784 km \frac{1.392}{15} + \frac{2.784}{5} = 0.0928 + 0.5568 = 0.6496 h \approx 39 minutes

(b)(i)

3 marks

Run from H to P: $H P = 0. 4^{2} + x^{2}$ at 15 km/h. Walk from P to S: S is $4 - x$ east of P and $0.8$ further south, so $P S = 0. 8^{2} + (4 - x)^{2}$ at 5 km/h.

T (x) = \frac{0. 4 ^{2} + x ^{2}}{15} + \frac{0. 8 ^{2} + ( 4 - x ) ^{2}}{5}

(b)(ii)

2 marks

Sketch for $0 \leq x \leq 4$ : the graph starts at $T (0) \approx 0.84$ hours, falls steadily to a minimum of about $0.42$ hours at $x \approx 3.7$ and then rises slightly to $T (4) \approx 0.43$ hours.

(b)(iii)

1 mark

Find the minimum of $T (x)$ on a graph.

x = 3.72 km

(b)(iv)

2 marks

The minimum time is
Compared with the straight line (39.0 minutes), the saving is

T (3.72) = 0.419 h = 25.1 minutes 39.0 - 25.1 \approx 14 minutes

(c)(i)

3 marks

Differentiate each term with the chain rule.

T^{'} (x) = \frac{x}{15 0.16 + x ^{2}} - \frac{4 - x}{5 0.64 + ( 4 - x ) ^{2}}

(c)(ii)

1 mark

Set $T^{'} (x) = 0$ and multiply by 15.

\frac{x}{0.16 + x ^{2}} = \frac{3 ( 4 - x )}{0.64 + ( 4 - x ) ^{2}}

(c)(iii)

2 marks

In triangle HPM, $cos H P M = \frac{x}{0.16 + x ^{2}}$ . In triangle SPN, $cos S P N = \frac{4 - x}{0.64 + ( 4 - x ) ^{2}}$ .
Dividing the two sides of the equation in (c)(ii):

\frac{cos H P M}{cos S P N} = 3 = \frac{15}{5} = \frac{speed over field}{speed over rough ground}

(d)

6 marks

Let $θ$ be the angle that HJ and KS make with the boundary, and $φ$ the angle that JK makes. The rule at J gives
Let $M J = a$ . Then $tan θ = \frac{0.4}{a}$ . The horizontal distances add to 4 km: $a$ for HJ, $\frac{0.6}{tan φ}$ for JK, and $\frac{a}{2}$ for KS (the drop is 0.2, half of 0.4, along the parallel line).
Solve these equations together on the calculator.

\frac{cos θ}{cos φ} = \frac{15}{5} = 3 \Rightarrow cos φ = \frac{cos θ}{3} a + \frac{a}{2} + \frac{0.6}{tan φ} = 4 θ = 0.157 rad, M J = a \approx 2.53 km

Question 2

29 marks

Exam marks: paired t-test, correlation, regression and Kendall's tau

Statistics and probability

(a)(i)

1 mark

The standard deviation of the population is not known and the sample is small ( $n = 8$ ), so a $t$ -test is used.

(a)(ii)

1 mark

Pierre decided before the exams that he wanted to know if the means differ, in either direction. He had no reason to expect one language to be higher, so the test is two-tailed.

(b)(i)

1 mark

The null hypothesis says the mean marks are equal; the alternative says they differ.

H_{0} : μ_{F} = μ_{G}, H_{1} : μ_{F} \neq = μ_{G}

(b)(ii)

2 marks

Run a paired $t$ -test on the eight differences (French minus German), with $ν = 7$ .

p \approx 0.153

(b)(iii)

1 mark

The $p$ -value is the probability of getting a mean difference as extreme as, or more extreme than, the observed one, if $H_{0}$ is true.

(b)(iv)

2 marks

Compare the $p$ -value with 0.05.
Do not reject $H_{0}$ : there is not enough evidence of a difference between the mean French and German marks.

0.153 > 0.05

(c)(i)

1 mark

Pierre expects a positive correlation, so the alternative is one-sided.

H_{0} : ρ = 0, H_{1} : ρ > 0

(c)(ii)

4 marks

Run a one-tailed test for correlation.
Since $0.00286 < 0.05$ , reject $H_{0}$ : there is significant evidence of a positive correlation between the French and German marks, so students who do well in one tend to do well in the other.

r = 0.864, p = 0.00286

(d)(i)

3 marks

Use the regression line of German marks on French marks.
Put $F = 58$ .

G = 0.737 F + 10.2 G = 0.7375 (58) + 10.24 \approx 53

(d)(ii)

2 marks

Use the regression line of French on German.
Set $F = 71$ and solve for $G$ .

F = 1.011 G + 4.04 71 = 1.011 G + 4.04 \Rightarrow G \approx 66

(e)(i)

1 mark

With no ties $C + D = \frac{n ( n - 1 )}{2}$ and both $C$ and $D$ are at least 0, so $∣ C - D ∣ \leq C + D$ .

- 1 \leq τ = \frac{C - D}{n ( n - 1 ) /2} \leq 1

(e)(ii)

1 mark

For P1 $(53, 41)$ and P2 $(76, 70)$ :
The product is positive, so the pair is concordant.

(76 - 53) (70 - 41) = 23 \times 29 > 0

(e)(iii)

4 marks

There are $\frac{6 \times 5}{2} = 15$ pairs. Check each pair $((x_{i} - x_{j}) (y_{i} - y_{j}))$ .
Concordant pairs: P1P2, P1P4, P1P5, P1P6, P2P3, P2P4, P2P5, P3P5 and P4P6, so $C = 9$ .
Discordant pairs: P1P3, P2P6, P3P4, P3P6, P4P5 and P5P6, so $D = 6$ .

τ = \frac{9 - 6}{15} = 0.2

(f)(i)

1 mark

Angela is testing whether there is an association between the marks.

H_{0} : there is no association between mathematics and history marks, H_{1} : there is an association

(f)(ii)

2 marks

The critical region is $∣ τ ∣ \geq 0.733$ . The value $τ = 0.2$ is not in the critical region.
Do not reject $H_{0}$ : there is insufficient evidence of an association between the mathematics and history marks.

(g)

2 marks

I disagree. Multiplying all the history marks by the same positive number does not change which pairs are concordant or discordant, so $τ$ would not change. Scaling is not needed.

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 probability29 marks · 53%
Q2
Calculus26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Straight line is not fastest · Q1
Moving faster over the field means it can pay to run further in the field and walk less on the rough ground.
02
Minimum from the derivative · Q1
Differentiate $T (x)$ and set it to zero, or use a graph.
03
Paired data · Q2
Each student sat both examinations, so use a paired test on the differences.
04
Regression direction · Q2
To predict German from French use the regression of German on French. To predict from German use French on German.
05
Kendall's tau · Q2
Count the concordant ( $C$ ) and discordant ( $D$ ) pairs among the $\frac{n ( n - 1 )}{2}$ pairs.

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 2023
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 2023 TZ2 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

26 marks

The fastest way from home to school over a field and rough ground

Calculus

(a)

6 marks

The straight line from H to S has length $4^{2} + 1. 2^{2} = 4.176$ km. It crosses the boundary 0.4 km down, which is $\frac{0.4}{1.2} = \frac{1}{3}$ of the way.
Time = distance over speed.

field: \frac{4.176}{3} = 1.392 km, rough: \frac{2 ( 4.176 )}{3} = 2.784 km \frac{1.392}{15} + \frac{2.784}{5} = 0.0928 + 0.5568 = 0.6496 h \approx 39 minutes

(b)(i)

3 marks

Run from H to P: $H P = 0. 4^{2} + x^{2}$ at 15 km/h. Walk from P to S: S is $4 - x$ east of P and $0.8$ further south, so $P S = 0. 8^{2} + (4 - x)^{2}$ at 5 km/h.

T (x) = \frac{0. 4 ^{2} + x ^{2}}{15} + \frac{0. 8 ^{2} + ( 4 - x ) ^{2}}{5}

(b)(ii)

2 marks

(b)(iii)

1 mark

Find the minimum of $T (x)$ on a graph.

x = 3.72 km

(b)(iv)

2 marks

The minimum time is
Compared with the straight line (39.0 minutes), the saving is

T (3.72) = 0.419 h = 25.1 minutes 39.0 - 25.1 \approx 14 minutes

(c)(i)

3 marks

Differentiate each term with the chain rule.

T^{'} (x) = \frac{x}{15 0.16 + x ^{2}} - \frac{4 - x}{5 0.64 + ( 4 - x ) ^{2}}

(c)(ii)

1 mark

Set $T^{'} (x) = 0$ and multiply by 15.

\frac{x}{0.16 + x ^{2}} = \frac{3 ( 4 - x )}{0.64 + ( 4 - x ) ^{2}}

(c)(iii)

2 marks

In triangle HPM, $cos H P M = \frac{x}{0.16 + x ^{2}}$ . In triangle SPN, $cos S P N = \frac{4 - x}{0.64 + ( 4 - x ) ^{2}}$ .
Dividing the two sides of the equation in (c)(ii):

\frac{cos H P M}{cos S P N} = 3 = \frac{15}{5} = \frac{speed over field}{speed over rough ground}

(d)

6 marks

Let $θ$ be the angle that HJ and KS make with the boundary, and $φ$ the angle that JK makes. The rule at J gives
Let $M J = a$ . Then $tan θ = \frac{0.4}{a}$ . The horizontal distances add to 4 km: $a$ for HJ, $\frac{0.6}{tan φ}$ for JK, and $\frac{a}{2}$ for KS (the drop is 0.2, half of 0.4, along the parallel line).
Solve these equations together on the calculator.

\frac{cos θ}{cos φ} = \frac{15}{5} = 3 \Rightarrow cos φ = \frac{cos θ}{3} a + \frac{a}{2} + \frac{0.6}{tan φ} = 4 θ = 0.157 rad, M J = a \approx 2.53 km

Question 2

29 marks

Exam marks: paired t-test, correlation, regression and Kendall's tau

Statistics and probability

(a)(i)

1 mark

The standard deviation of the population is not known and the sample is small ( $n = 8$ ), so a $t$ -test is used.

(a)(ii)

1 mark

Pierre decided before the exams that he wanted to know if the means differ, in either direction. He had no reason to expect one language to be higher, so the test is two-tailed.

(b)(i)

1 mark

The null hypothesis says the mean marks are equal; the alternative says they differ.

H_{0} : μ_{F} = μ_{G}, H_{1} : μ_{F} \neq = μ_{G}

(b)(ii)

2 marks

Run a paired $t$ -test on the eight differences (French minus German), with $ν = 7$ .

p \approx 0.153

(b)(iii)

1 mark

The $p$ -value is the probability of getting a mean difference as extreme as, or more extreme than, the observed one, if $H_{0}$ is true.

(b)(iv)

2 marks

Compare the $p$ -value with 0.05.
Do not reject $H_{0}$ : there is not enough evidence of a difference between the mean French and German marks.

0.153 > 0.05

(c)(i)

1 mark

Pierre expects a positive correlation, so the alternative is one-sided.

H_{0} : ρ = 0, H_{1} : ρ > 0

(c)(ii)

4 marks

Run a one-tailed test for correlation.
Since $0.00286 < 0.05$ , reject $H_{0}$ : there is significant evidence of a positive correlation between the French and German marks, so students who do well in one tend to do well in the other.

r = 0.864, p = 0.00286

(d)(i)

3 marks

Use the regression line of German marks on French marks.
Put $F = 58$ .

G = 0.737 F + 10.2 G = 0.7375 (58) + 10.24 \approx 53

(d)(ii)

2 marks

Use the regression line of French on German.
Set $F = 71$ and solve for $G$ .

F = 1.011 G + 4.04 71 = 1.011 G + 4.04 \Rightarrow G \approx 66

(e)(i)

1 mark

With no ties $C + D = \frac{n ( n - 1 )}{2}$ and both $C$ and $D$ are at least 0, so $∣ C - D ∣ \leq C + D$ .

- 1 \leq τ = \frac{C - D}{n ( n - 1 ) /2} \leq 1

(e)(ii)

1 mark

For P1 $(53, 41)$ and P2 $(76, 70)$ :
The product is positive, so the pair is concordant.

(76 - 53) (70 - 41) = 23 \times 29 > 0

(e)(iii)

4 marks

There are $\frac{6 \times 5}{2} = 15$ pairs. Check each pair $((x_{i} - x_{j}) (y_{i} - y_{j}))$ .
Concordant pairs: P1P2, P1P4, P1P5, P1P6, P2P3, P2P4, P2P5, P3P5 and P4P6, so $C = 9$ .
Discordant pairs: P1P3, P2P6, P3P4, P3P6, P4P5 and P5P6, so $D = 6$ .

τ = \frac{9 - 6}{15} = 0.2

(f)(i)

1 mark

Angela is testing whether there is an association between the marks.

H_{0} : there is no association between mathematics and history marks, H_{1} : there is an association

(f)(ii)

2 marks

The critical region is $∣ τ ∣ \geq 0.733$ . The value $τ = 0.2$ is not in the critical region.
Do not reject $H_{0}$ : there is insufficient evidence of an association between the mathematics and history marks.

(g)

2 marks

I disagree. Multiplying all the history marks by the same positive number does not change which pairs are concordant or discordant, so $τ$ would not change. Scaling is not needed.

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 probability29 marks · 53%
Q2
Calculus26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Straight line is not fastest · Q1
Moving faster over the field means it can pay to run further in the field and walk less on the rough ground.
02
Minimum from the derivative · Q1
Differentiate $T (x)$ and set it to zero, or use a graph.
03
Paired data · Q2
Each student sat both examinations, so use a paired test on the differences.
04
Regression direction · Q2
To predict German from French use the regression of German on French. To predict from German use French on German.
05
Kendall's tau · Q2
Count the concordant ( $C$ ) and discordant ( $D$ ) pairs among the $\frac{n ( n - 1 )}{2}$ pairs.

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