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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: November 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 November 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

12 marks

A model aeroplane wing cross-section: trapezoidal rule and percentage error

Calculus

(a)

3 marks

Substitute $x = 25, 50, 75$ into $y = 2 x - \frac{x}{5} + 1$ .

a = 10 - 5 + 1 = 6, b = 5.14, c = 3.32

(b)

3 marks

Four intervals of width 25.

A \approx \frac{25}{2} [1 + 1 + 2 (6 + 5.1421 + 3.3205)] \approx 387 cm^{2}

(c)

4 marks

The exact area is the integral of the curve.
Evaluate with the calculator.

\int_{0}^{100} (2 x - \frac{x}{5} + 1) d x A \approx 433.3 cm^{2}

(d)

2 marks

Percentage error of Gabriel's estimate.

\frac{∣386.6 - 433.3∣}{433.3} \times 100 \approx 10.8%

Question 2

15 marks

A jogging track on circular arcs: arc length, track area and concrete

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 50 m long.

B M = 25 m

(b)

4 marks

Triangle BFM has a right angle at M, with $B M = 25$ and $F M = 20$ .

B F = 2 5^{2} + 2 0^{2} = 1025 \approx 32.0 m tan B F M = \frac{25}{20} \Rightarrow B F M = 51. 3^{\circ}

(c)

3 marks

The angle $A F B = 2 \times 51.3 4^{\circ} = 102.6 8^{\circ} = 1.792$ radians and the radius is $B F = 32.02$ .

arc A B = 32.02 \times 1.792 \approx 57.4 m

(d)

4 marks

The outer edge is an arc of radius $32.02 + 2 = 34.02$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.792) (34.0 2^{2} - 32.0 2^{2}) = \frac{1}{2} (1.792) (132.1) \approx 118 m^{2}

(e)

3 marks

The depth is 12 cm $= 0.12$ m.

V = 118.3 \times 0.12 \approx 14.2 m^{3}

Question 3

18 marks

Student heights: grouped mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $140 \leq h < 160$ is 150. Use the midpoints $150, 165, 175, 185, 200$ .

\overset{ˉ}{h} = \frac{150 ( 11 ) + 165 ( 51 ) + 175 ( 68 ) + 185 ( 47 ) + 200 ( 23 )}{200} = \frac{35260}{200} \approx 176 cm

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 168, Q_{3} = 183 \Rightarrow IQR = 15 cm

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
Laszlo's height is $204 < 205.5$ , so his height is not an outlier.

183 + 1.5 (15) = 205.5

(d)

2 marks

The null hypothesis says that the heights follow this normal distribution.

H_{0} : the heights can be modelled by N (176, 13. 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (176, 13. 5^{2})$ for 200 students.
Run a goodness of fit test with 5 classes: $ν = 4$ .
Since $0.0166 < 0.05$ , reject $H_{0}$ : there is sufficient evidence that the heights of these students are not well modelled by $N (176, 13. 5^{2})$ .

a = 200 P (170 \leq h < 180) = 57.6, b = 200 P (h \geq 190) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 4

16 marks

A house loan: payments, a faster repayment and the savings

Number and algebra

(a)

5 marks

The down payment is 15%, so the loan is 85% of the price.
Use a finance solver with $N = 360$ , $I = 4%$ , PV $= 242250$ , FV $= 0$ and 12 payments a year.

0.85 \times 285000 = 242250 USD PMT = 1156.54 USD

(b)

2 marks

Total paid over 360 months.

360 \times 1156.54 \approx 416354 USD

(c)

2 marks

Solve for the number of payments with PMT $= 1300$ .

N = 291.7 \Rightarrow 292 payments

(d)

4 marks

After 291 payments of 1300 the balance, after interest for the last month, is the final payment.

final payment \approx 874.81 USD

(e)

3 marks

Total paid with the faster plan.
Savings compared with the original total.

291 (1300) + 874.81 = 379 174.81 416354 - 379175 \approx 37179 USD

Question 5

19 marks

Pricing phone cases: demand, income, profit and a supplier discount

Functions
Calculus

(a)

2 marks

Substitute $x = 16$ into $n = 20000 - 1000 x$ .

n = 20000 - 16000 = 4000 cases

(b)

2 marks

Income is the number sold times the price.

4000 \times 16 = 64000 EUR

(c)

1 mark

At a price $x$ , the number sold is $20000 - 1000 x$ .

R (x) = x (20000 - 1000 x) = 20000 x - 1000 x^{2}

(d)

3 marks

Costs are fixed costs 10 000 plus $10 \times 4000$ for the cases.

profit = 64000 - (10000 + 40000) = 14000 EUR

(e)

2 marks

Profit is income minus fixed and per-case costs.

P (x) = x (20000 - 1000 x) - 10000 - 10 (20000 - 1000 x) P (x) = 20000 x - 1000 x^{2} - 10000 - 200000 + 10000 x = - 1000 x^{2} + 30000 x - 210000

(f)

4 marks

Differentiate.
At the salesperson's price $x = 16$ the derivative is not zero.
The maximum is where $P^{'} (x) = 0$ , at $x = 15$ , which is not 16. So 16 EUR does not maximise the profit.

P^{'} (x) = - 2000 x + 30000 P^{'} (16) = - 2000 (16) + 30000 = - 2000 \neq = 0

(g)

3 marks

With $n = 20000 - 1000 x$ the cost per case is
The new profit is income minus fixed costs minus the cost of the cases.

C = 10 - 0.0001 (20000 - 1000 x) = 8 + 0.1 x P (x) = x (20000 - 1000 x) - 10000 - (8 + 0.1 x) (20000 - 1000 x) P (x) = - 900 x^{2} + 26000 x - 170000

(h)

2 marks

The maximum is at the vertex.

x = \frac{26000}{1800} \approx 14.4 EUR

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 probability18 marks · 23%
Q3
Calculus18 marks · 23%
Q1, Q5
Number and algebra16 marks · 20%
Q4
Geometry and trigonometry15 marks · 19%
Q2
Functions13 marks · 16%
Q5

Mentor insights

Traps to watch for

01
Trapezoidal rule · Q1
With equal strips of width $h = 25$ the area is $\frac{h}{2} [first + last + 2 (middle)]$ .
02
Radians for arc length · Q2
The angle at F must be in radians in $s = r θ$ and in the sector area formula.
03
Midpoints for the mean · Q3
Use the middle of each class with its frequency.
04
Down payment first · Q4
The loan is the price minus the 15% down payment.
05
Profit is income minus all costs · Q5
Fixed monthly costs plus the cost per item times the number sold.

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: November 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 November 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

12 marks

A model aeroplane wing cross-section: trapezoidal rule and percentage error

Calculus

(a)

3 marks

Substitute $x = 25, 50, 75$ into $y = 2 x - \frac{x}{5} + 1$ .

a = 10 - 5 + 1 = 6, b = 5.14, c = 3.32

(b)

3 marks

Four intervals of width 25.

A \approx \frac{25}{2} [1 + 1 + 2 (6 + 5.1421 + 3.3205)] \approx 387 cm^{2}

(c)

4 marks

The exact area is the integral of the curve.
Evaluate with the calculator.

\int_{0}^{100} (2 x - \frac{x}{5} + 1) d x A \approx 433.3 cm^{2}

(d)

2 marks

Percentage error of Gabriel's estimate.

\frac{∣386.6 - 433.3∣}{433.3} \times 100 \approx 10.8%

Question 2

15 marks

A jogging track on circular arcs: arc length, track area and concrete

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 50 m long.

B M = 25 m

(b)

4 marks

Triangle BFM has a right angle at M, with $B M = 25$ and $F M = 20$ .

B F = 2 5^{2} + 2 0^{2} = 1025 \approx 32.0 m tan B F M = \frac{25}{20} \Rightarrow B F M = 51. 3^{\circ}

(c)

3 marks

The angle $A F B = 2 \times 51.3 4^{\circ} = 102.6 8^{\circ} = 1.792$ radians and the radius is $B F = 32.02$ .

arc A B = 32.02 \times 1.792 \approx 57.4 m

(d)

4 marks

The outer edge is an arc of radius $32.02 + 2 = 34.02$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.792) (34.0 2^{2} - 32.0 2^{2}) = \frac{1}{2} (1.792) (132.1) \approx 118 m^{2}

(e)

3 marks

The depth is 12 cm $= 0.12$ m.

V = 118.3 \times 0.12 \approx 14.2 m^{3}

Question 3

18 marks

Student heights: grouped mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $140 \leq h < 160$ is 150. Use the midpoints $150, 165, 175, 185, 200$ .

\overset{ˉ}{h} = \frac{150 ( 11 ) + 165 ( 51 ) + 175 ( 68 ) + 185 ( 47 ) + 200 ( 23 )}{200} = \frac{35260}{200} \approx 176 cm

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 168, Q_{3} = 183 \Rightarrow IQR = 15 cm

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
Laszlo's height is $204 < 205.5$ , so his height is not an outlier.

183 + 1.5 (15) = 205.5

(d)

2 marks

The null hypothesis says that the heights follow this normal distribution.

H_{0} : the heights can be modelled by N (176, 13. 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (176, 13. 5^{2})$ for 200 students.
Run a goodness of fit test with 5 classes: $ν = 4$ .
Since $0.0166 < 0.05$ , reject $H_{0}$ : there is sufficient evidence that the heights of these students are not well modelled by $N (176, 13. 5^{2})$ .

a = 200 P (170 \leq h < 180) = 57.6, b = 200 P (h \geq 190) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 4

16 marks

A house loan: payments, a faster repayment and the savings

Number and algebra

(a)

5 marks

The down payment is 15%, so the loan is 85% of the price.
Use a finance solver with $N = 360$ , $I = 4%$ , PV $= 242250$ , FV $= 0$ and 12 payments a year.

0.85 \times 285000 = 242250 USD PMT = 1156.54 USD

(b)

2 marks

Total paid over 360 months.

360 \times 1156.54 \approx 416354 USD

(c)

2 marks

Solve for the number of payments with PMT $= 1300$ .

N = 291.7 \Rightarrow 292 payments

(d)

4 marks

After 291 payments of 1300 the balance, after interest for the last month, is the final payment.

final payment \approx 874.81 USD

(e)

3 marks

Total paid with the faster plan.
Savings compared with the original total.

291 (1300) + 874.81 = 379 174.81 416354 - 379175 \approx 37179 USD

Question 5

19 marks

Pricing phone cases: demand, income, profit and a supplier discount

Functions
Calculus

(a)

2 marks

Substitute $x = 16$ into $n = 20000 - 1000 x$ .

n = 20000 - 16000 = 4000 cases

(b)

2 marks

Income is the number sold times the price.

4000 \times 16 = 64000 EUR

(c)

1 mark

At a price $x$ , the number sold is $20000 - 1000 x$ .

R (x) = x (20000 - 1000 x) = 20000 x - 1000 x^{2}

(d)

3 marks

Costs are fixed costs 10 000 plus $10 \times 4000$ for the cases.

profit = 64000 - (10000 + 40000) = 14000 EUR

(e)

2 marks

Profit is income minus fixed and per-case costs.

P (x) = x (20000 - 1000 x) - 10000 - 10 (20000 - 1000 x) P (x) = 20000 x - 1000 x^{2} - 10000 - 200000 + 10000 x = - 1000 x^{2} + 30000 x - 210000

(f)

4 marks

Differentiate.
At the salesperson's price $x = 16$ the derivative is not zero.
The maximum is where $P^{'} (x) = 0$ , at $x = 15$ , which is not 16. So 16 EUR does not maximise the profit.

P^{'} (x) = - 2000 x + 30000 P^{'} (16) = - 2000 (16) + 30000 = - 2000 \neq = 0

(g)

3 marks

With $n = 20000 - 1000 x$ the cost per case is
The new profit is income minus fixed costs minus the cost of the cases.

C = 10 - 0.0001 (20000 - 1000 x) = 8 + 0.1 x P (x) = x (20000 - 1000 x) - 10000 - (8 + 0.1 x) (20000 - 1000 x) P (x) = - 900 x^{2} + 26000 x - 170000

(h)

2 marks

The maximum is at the vertex.

x = \frac{26000}{1800} \approx 14.4 EUR

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 probability18 marks · 23%
Q3
Calculus18 marks · 23%
Q1, Q5
Number and algebra16 marks · 20%
Q4
Geometry and trigonometry15 marks · 19%
Q2
Functions13 marks · 16%
Q5

Mentor insights

Traps to watch for

01
Trapezoidal rule · Q1
With equal strips of width $h = 25$ the area is $\frac{h}{2} [first + last + 2 (middle)]$ .
02
Radians for arc length · Q2
The angle at F must be in radians in $s = r θ$ and in the sector area formula.
03
Midpoints for the mean · Q3
Use the middle of each class with its frequency.
04
Down payment first · Q4
The loan is the price minus the 15% down payment.
05
Profit is income minus all costs · Q5
Fixed monthly costs plus the cost per item times the number sold.

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