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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

17 marks

Points in a park: sine rule, area and an arc (with an angle of 10 degrees)

Geometry and trigonometry

(a)

3 marks

Use the sine rule in triangle AOB.
The angle at A is obtuse, so angle B is acute.

\frac{sin A B O}{25.9} = \frac{sin 1 0 ^{\circ}}{6.36} \Rightarrow sin A B O = 0.7072 A B O = 45. 0^{\circ}

(b)

4 marks

The angle at A is $180 - 10 - 45 = 12 5^{\circ}$ .

Area = \frac{1}{2} (25.9) (6.36) sin 12 5^{\circ} \approx 67.5 m^{2}

(c)

3 marks

Cosine rule in triangle AKB with $A K = 12$ , $A B = 6.36$ and angle $4 5^{\circ}$ at A.

K B^{2} = 1 2^{2} + 6.3 6^{2} - 2 (12) (6.36) cos 4 5^{\circ} \Rightarrow K B \approx 8.75 m

(d)

4 marks

In triangle OKX the angle at X is $180 - 53.8 - 51.1 = 75. 1^{\circ}$ . Use the sine rule for $O X$ , opposite the angle at K.
Khemil is 22.2 m from the football (KX), and Odette is 21.4 m. Odette is closer.

\frac{O X}{sin 51. 1 ^{\circ}} = \frac{22.2}{sin 53. 8 ^{\circ}} \Rightarrow O X \approx 21.4 m

(e)

3 marks

$K A X = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ and the radius is 12 m.

arc K X = 12 \times \frac{135 π}{180} = 9 π \approx 28.3 m

Question 2

15 marks

Pendulums made each day, and a swinging ball with decreasing distances

Number and algebra

(a)

3 marks

Arithmetic sequence with $u_{1} = 10$ and $d = 6$ .

u_{12} = 10 + 11 (6) = 76

(b)

2 marks

Sum of the first 15 terms.

S_{15} = \frac{15}{2} (2 (10) + 14 (6)) = 780

(c)

3 marks

With common difference $x$ .
The minimum integer is

S_{15} = \frac{15}{2} (20 + 14 x) \geq 1000 \Rightarrow 150 + 105 x \geq 1000 \Rightarrow x \geq 8.1 x = 9

(d)

3 marks

Geometric sequence with $u_{1} = 17.1$ and $r = 0.95$ .

u_{5} = 17.1 (0.95)^{4} \approx 13.9 cm

(e)

2 marks

Sum of the first 16 terms.

S_{16} = \frac{17.1 ( 1 - 0.9 5 ^{16} )}{1 - 0.95} \approx 191 cm

(f)

2 marks

The distance before recording is the swing before the first recorded swing, which was 95% of it.

\frac{17.1}{0.95} = 18 cm

Question 3

15 marks

Two bacteria experiments and filling a container

Number and algebra

(a)

2 marks

Put $t = 0$ .
It is the number of bacteria at the start of the experiment.

P (0) = 1200 \times k^{0} = 1200

(b)

2 marks

Use $P (3) = 18750$ .

1200 k^{3} = 18750 \Rightarrow k^{3} = 15.625 \Rightarrow k = 2.5

(c)

2 marks

1 hour 30 minutes is $t = 1.5$ .

P (1.5) = 1200 (2.5)^{1.5} \approx 4743

(d)

2 marks

Set $P (t) = S (t)$ .

1200 (2.5)^{t} = 5000 (1.65)^{t} \Rightarrow t = \frac{ln ( 25/6 )}{ln ( 2.5/1.65 )} \approx 3.43

(e)

4 marks

Set $S (t) = 19000$ .
That is 2 hours and $0.6659 \times 60 = 39.95$ minutes.

1.6 5^{t} = 3.8 \Rightarrow t = \frac{ln 3.8}{ln 1.65} = 2.6659 hours m = 40

(f)

3 marks

The number of bacteria that fit in the container is the container volume divided by the volume of one bacterium.
Solve $S (t) = 2.1 \times 1 0^{13}$ .

\frac{2.1 \times 1 0 ^{- 5}}{1 \times 1 0 ^{- 18}} = 2.1 \times 1 0^{13} 5000 (1.65)^{t} = 2.1 \times 1 0^{13} \Rightarrow t = \frac{ln ( 4.2 \times 1 0 ^{9} )}{ln 1.65} \approx 44.2 hours

Question 4

17 marks

A remedy claimed to cure 82%: binomial probabilities and a two-sample t-test

Statistics and probability

(a)

3 marks

Let $X$ be the number cured: $X \sim B (115, 0.82)$ .

P (X = 90) = (90 115) (0.82)^{90} (0.18)^{25} \approx 0.0535

(b)

2 marks

At least 95 is the complement of at most 94.

P (X \geq 95) = 1 - P (X \leq 94) \approx 0.491

(c)

2 marks

The variance of a binomial is $n p (1 - p)$ .

115 (0.82) (0.18) \approx 16.97

(d)

3 marks

Find the smallest $n$ with $P (X \geq n) < 0.3$ .

P (X \geq 97) = 0.3033 > 0.3, P (X \geq 98) = 0.2218 < 0.3 n = 98

(e)

2 marks

The new remedy is expected to have the shorter mean recovery time.

H_{0} : μ_{N} = μ_{O}, H_{1} : μ_{N} < μ_{O}

(f)

2 marks

Enter both samples into a pooled one-tailed two-sample $t$ -test (the variances are assumed equal).

p \approx 0.062

(g)

2 marks

Compare with the 10% significance level.
Reject $H_{0}$ : there is evidence at the 10% level that the mean recovery time with the new remedy is less than with the old remedy.

0.062 < 0.10

(h)

1 mark

The $p$ -value is the probability of getting a difference in sample means at least as large as the one observed (in favour of the new remedy), if the two population means were really equal.

Question 5

16 marks

A garden surrounded by a path: area model and maximum

Calculus

(a)

5 marks

The park has length $x$ and width $\frac{1200}{x}$ . The garden has length $x - 3$ and width $\frac{1200}{x} - 4$ .

A = (x - 3) (\frac{1200}{x} - 4) = 1200 - 4 x - \frac{3600}{x} + 12 = 1212 - 4 x - \frac{3600}{x}

(b)

4 marks

Set $A = 800$ .
The width is $\frac{1200}{x}$ .

1212 - 4 x - \frac{3600}{x} = 800 \Rightarrow 4 x^{2} - 412 x + 3600 = 0 \Rightarrow x^{2} - 103 x + 900 = 0 x = 93.4 or x = 9.64 93.4 m by 12.9 m or 9.64 m by 124.5 m

(c)

3 marks

Differentiate $A = 1212 - 4 x - 3600 x^{- 1}$ .

\frac{d A}{d x} = - 4 + \frac{3600}{x ^{2}}

(d)

2 marks

Set the derivative to zero.
This is a maximum because $\frac{d ^{2} A}{d x ^{2}} = - \frac{7200}{x ^{3}} < 0$ .

\frac{3600}{x ^{2}} = 4 \Rightarrow x^{2} = 900 \Rightarrow x = 30

(e)

2 marks

Substitute $x = 30$ .

A = 1212 - 120 - 120 = 972 m^{2}

Our analysis

Where the marks are

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

Number and algebra30 marks · 38%
Q2, Q3
Geometry and trigonometry17 marks · 21%
Q1
Statistics and probability17 marks · 21%
Q4
Calculus16 marks · 20%
Q5

Mentor insights

Traps to watch for

01
The obtuse angle · Q1
Sine can give an acute angle, but the angle at A is obtuse, which settles which triangle is meant.
02
Arithmetic then geometric · Q2
The daily production grows by a constant amount (arithmetic). The swing distances shrink by a constant factor (geometric).
03
Initial value · Q3
At $t = 0$ the exponent is 0, so $P (0)$ is the starting population.
04
Binomial for the number cured · Q4
$X \sim B (115, 0.82)$ if the claim is true.
05
Subtract the path from both sides · Q5
The path takes $2 \times 1.5 = 3$ m from the length and $2 \times 2 = 4$ m from the width of the park.

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 2023
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 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

17 marks

Points in a park: sine rule, area and an arc (with an angle of 10 degrees)

Geometry and trigonometry

(a)

3 marks

Use the sine rule in triangle AOB.
The angle at A is obtuse, so angle B is acute.

\frac{sin A B O}{25.9} = \frac{sin 1 0 ^{\circ}}{6.36} \Rightarrow sin A B O = 0.7072 A B O = 45. 0^{\circ}

(b)

4 marks

The angle at A is $180 - 10 - 45 = 12 5^{\circ}$ .

Area = \frac{1}{2} (25.9) (6.36) sin 12 5^{\circ} \approx 67.5 m^{2}

(c)

3 marks

Cosine rule in triangle AKB with $A K = 12$ , $A B = 6.36$ and angle $4 5^{\circ}$ at A.

K B^{2} = 1 2^{2} + 6.3 6^{2} - 2 (12) (6.36) cos 4 5^{\circ} \Rightarrow K B \approx 8.75 m

(d)

4 marks

In triangle OKX the angle at X is $180 - 53.8 - 51.1 = 75. 1^{\circ}$ . Use the sine rule for $O X$ , opposite the angle at K.
Khemil is 22.2 m from the football (KX), and Odette is 21.4 m. Odette is closer.

\frac{O X}{sin 51. 1 ^{\circ}} = \frac{22.2}{sin 53. 8 ^{\circ}} \Rightarrow O X \approx 21.4 m

(e)

3 marks

$K A X = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ and the radius is 12 m.

arc K X = 12 \times \frac{135 π}{180} = 9 π \approx 28.3 m

Question 2

15 marks

Pendulums made each day, and a swinging ball with decreasing distances

Number and algebra

(a)

3 marks

Arithmetic sequence with $u_{1} = 10$ and $d = 6$ .

u_{12} = 10 + 11 (6) = 76

(b)

2 marks

Sum of the first 15 terms.

S_{15} = \frac{15}{2} (2 (10) + 14 (6)) = 780

(c)

3 marks

With common difference $x$ .
The minimum integer is

S_{15} = \frac{15}{2} (20 + 14 x) \geq 1000 \Rightarrow 150 + 105 x \geq 1000 \Rightarrow x \geq 8.1 x = 9

(d)

3 marks

Geometric sequence with $u_{1} = 17.1$ and $r = 0.95$ .

u_{5} = 17.1 (0.95)^{4} \approx 13.9 cm

(e)

2 marks

Sum of the first 16 terms.

S_{16} = \frac{17.1 ( 1 - 0.9 5 ^{16} )}{1 - 0.95} \approx 191 cm

(f)

2 marks

The distance before recording is the swing before the first recorded swing, which was 95% of it.

\frac{17.1}{0.95} = 18 cm

Question 3

15 marks

Two bacteria experiments and filling a container

Number and algebra

(a)

2 marks

Put $t = 0$ .
It is the number of bacteria at the start of the experiment.

P (0) = 1200 \times k^{0} = 1200

(b)

2 marks

Use $P (3) = 18750$ .

1200 k^{3} = 18750 \Rightarrow k^{3} = 15.625 \Rightarrow k = 2.5

(c)

2 marks

1 hour 30 minutes is $t = 1.5$ .

P (1.5) = 1200 (2.5)^{1.5} \approx 4743

(d)

2 marks

Set $P (t) = S (t)$ .

1200 (2.5)^{t} = 5000 (1.65)^{t} \Rightarrow t = \frac{ln ( 25/6 )}{ln ( 2.5/1.65 )} \approx 3.43

(e)

4 marks

Set $S (t) = 19000$ .
That is 2 hours and $0.6659 \times 60 = 39.95$ minutes.

1.6 5^{t} = 3.8 \Rightarrow t = \frac{ln 3.8}{ln 1.65} = 2.6659 hours m = 40

(f)

3 marks

The number of bacteria that fit in the container is the container volume divided by the volume of one bacterium.
Solve $S (t) = 2.1 \times 1 0^{13}$ .

\frac{2.1 \times 1 0 ^{- 5}}{1 \times 1 0 ^{- 18}} = 2.1 \times 1 0^{13} 5000 (1.65)^{t} = 2.1 \times 1 0^{13} \Rightarrow t = \frac{ln ( 4.2 \times 1 0 ^{9} )}{ln 1.65} \approx 44.2 hours

Question 4

17 marks

A remedy claimed to cure 82%: binomial probabilities and a two-sample t-test

Statistics and probability

(a)

3 marks

Let $X$ be the number cured: $X \sim B (115, 0.82)$ .

P (X = 90) = (90 115) (0.82)^{90} (0.18)^{25} \approx 0.0535

(b)

2 marks

At least 95 is the complement of at most 94.

P (X \geq 95) = 1 - P (X \leq 94) \approx 0.491

(c)

2 marks

The variance of a binomial is $n p (1 - p)$ .

115 (0.82) (0.18) \approx 16.97

(d)

3 marks

Find the smallest $n$ with $P (X \geq n) < 0.3$ .

P (X \geq 97) = 0.3033 > 0.3, P (X \geq 98) = 0.2218 < 0.3 n = 98

(e)

2 marks

The new remedy is expected to have the shorter mean recovery time.

H_{0} : μ_{N} = μ_{O}, H_{1} : μ_{N} < μ_{O}

(f)

2 marks

Enter both samples into a pooled one-tailed two-sample $t$ -test (the variances are assumed equal).

p \approx 0.062

(g)

2 marks

Compare with the 10% significance level.
Reject $H_{0}$ : there is evidence at the 10% level that the mean recovery time with the new remedy is less than with the old remedy.

0.062 < 0.10

(h)

1 mark

The $p$ -value is the probability of getting a difference in sample means at least as large as the one observed (in favour of the new remedy), if the two population means were really equal.

Question 5

16 marks

A garden surrounded by a path: area model and maximum

Calculus

(a)

5 marks

The park has length $x$ and width $\frac{1200}{x}$ . The garden has length $x - 3$ and width $\frac{1200}{x} - 4$ .

A = (x - 3) (\frac{1200}{x} - 4) = 1200 - 4 x - \frac{3600}{x} + 12 = 1212 - 4 x - \frac{3600}{x}

(b)

4 marks

Set $A = 800$ .
The width is $\frac{1200}{x}$ .

1212 - 4 x - \frac{3600}{x} = 800 \Rightarrow 4 x^{2} - 412 x + 3600 = 0 \Rightarrow x^{2} - 103 x + 900 = 0 x = 93.4 or x = 9.64 93.4 m by 12.9 m or 9.64 m by 124.5 m

(c)

3 marks

Differentiate $A = 1212 - 4 x - 3600 x^{- 1}$ .

\frac{d A}{d x} = - 4 + \frac{3600}{x ^{2}}

(d)

2 marks

Set the derivative to zero.
This is a maximum because $\frac{d ^{2} A}{d x ^{2}} = - \frac{7200}{x ^{3}} < 0$ .

\frac{3600}{x ^{2}} = 4 \Rightarrow x^{2} = 900 \Rightarrow x = 30

(e)

2 marks

Substitute $x = 30$ .

A = 1212 - 120 - 120 = 972 m^{2}

Our analysis

Where the marks are

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

Number and algebra30 marks · 38%
Q2, Q3
Geometry and trigonometry17 marks · 21%
Q1
Statistics and probability17 marks · 21%
Q4
Calculus16 marks · 20%
Q5

Mentor insights

Traps to watch for

01
The obtuse angle · Q1
Sine can give an acute angle, but the angle at A is obtuse, which settles which triangle is meant.
02
Arithmetic then geometric · Q2
The daily production grows by a constant amount (arithmetic). The swing distances shrink by a constant factor (geometric).
03
Initial value · Q3
At $t = 0$ the exponent is 0, so $P (0)$ is the starting population.
04
Binomial for the number cured · Q4
$X \sim B (115, 0.82)$ if the claim is true.
05
Subtract the path from both sides · Q5
The path takes $2 \times 1.5 = 3$ m from the length and $2 \times 2 = 4$ m from the width of the park.

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