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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 TZ1 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 propeller blade cross-section: table of values, trapezoidal rule and percentage error

Calculus

(a)

3 marks

Substitute $x = 16, 32, 48$ into $y = 4 x - \frac{x}{2} + 1$ .

a = 9, b = 7.63, c = 4.71

(b)

3 marks

Four intervals of width 16.

A \approx \frac{16}{2} [1 + 1 + 2 (9 + 7.627 + 4.713)] = 8 (44.68) \approx 357 mm^{2}

(c)

4 marks

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

\int_{0}^{64} (4 x - \frac{x}{2} + 1) d x A \approx 405.3 mm^{2}

(d)

2 marks

Percentage error of Jan's estimate.

\frac{∣357.4 - 405.3∣}{405.3} \times 100 \approx 11.8%

Question 2

15 marks

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

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 90 m long.

B M = 45 m

(b)

4 marks

Triangle BEM has a right angle at M, with $B M = 45$ and $E M = 32$ .

B E = 4 5^{2} + 3 2^{2} = 3049 \approx 55.2 m tan B E M = \frac{45}{32} \Rightarrow B E M = 54. 6^{\circ}

(c)

3 marks

The angle $A E B = 2 \times 54.5 8^{\circ} = 109.1 7^{\circ} = 1.905$ radians and the radius is 55.22.

arc A B = 55.22 \times 1.905 \approx 105 m

(d)

4 marks

The outer edge is an arc of radius $55.22 + 4 = 59.22$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.905) (59.2 2^{2} - 55.2 2^{2}) = \frac{1}{2} (1.905) (457.8) \approx 436 m^{2}

(e)

3 marks

The depth is 15 cm $= 0.15$ m.

V = 436.1 \times 0.15 \approx 65.4 m^{3}

Question 3

18 marks

Family movie running times: mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $70 \leq t < 80$ is 75. Use the midpoints $75, 82.5, 87.5, 92.5, 100$ .

\overset{ˉ}{t} = \frac{75 ( 11 ) + 82.5 ( 51 ) + 87.5 ( 68 ) + 92.5 ( 47 ) + 100 ( 23 )}{200} = 88.15 minutes

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 84, Q_{3} = 91.5 \Rightarrow IQR = 7.5 minutes

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
"Star Feud" lasts $100 < 102.75$ minutes, so it is not an outlier.

91.5 + 1.5 (7.5) = 102.75

(d)

2 marks

The null hypothesis says that the running times follow this normal distribution.

H_{0} : the running times can be modelled by N (88, 6.7 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (88, 6.7 5^{2})$ for 200 movies.
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 running times are not well modelled by $N (88, 6.7 5^{2})$ .

a = 200 P (85 \leq t < 90) = 57.6, b = 200 P (t \geq 95) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 4

16 marks

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

Number and algebra

(a)

5 marks

The down payment is 25%, so the loan is 75% of the price.
Use a finance solver with $N = 60$ , $I = 4.5%$ , PV $= 213750$ , FV $= 0$ and 12 payments a year.

0.75 \times 285000 = 213750 ZAR PMT = 3984.95 ZAR

(b)

2 marks

Total paid over 60 months.

60 \times 3984.95 \approx 239097 ZAR

(c)

2 marks

Solve for the number of payments with PMT $= 4600$ .

N = 51.15 \Rightarrow 52 payments

(d)

4 marks

After 51 payments of 4600 the balance, after interest for the last month, is the final payment.

final payment \approx 706.80 ZAR

(e)

3 marks

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

51 (4600) + 706.80 = 235 306.80 239097 - 235307 \approx 3790 ZAR

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 $h = 16$ the area is $\frac{h}{2} [first + last + 2 (middle)]$ .
02
Radians for arc length · Q2
The angle at E 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 25% 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 TZ1 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 propeller blade cross-section: table of values, trapezoidal rule and percentage error

Calculus

(a)

3 marks

Substitute $x = 16, 32, 48$ into $y = 4 x - \frac{x}{2} + 1$ .

a = 9, b = 7.63, c = 4.71

(b)

3 marks

Four intervals of width 16.

A \approx \frac{16}{2} [1 + 1 + 2 (9 + 7.627 + 4.713)] = 8 (44.68) \approx 357 mm^{2}

(c)

4 marks

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

\int_{0}^{64} (4 x - \frac{x}{2} + 1) d x A \approx 405.3 mm^{2}

(d)

2 marks

Percentage error of Jan's estimate.

\frac{∣357.4 - 405.3∣}{405.3} \times 100 \approx 11.8%

Question 2

15 marks

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

Geometry and trigonometry

(a)

1 mark

M is the midpoint of [AB], which is 90 m long.

B M = 45 m

(b)

4 marks

Triangle BEM has a right angle at M, with $B M = 45$ and $E M = 32$ .

B E = 4 5^{2} + 3 2^{2} = 3049 \approx 55.2 m tan B E M = \frac{45}{32} \Rightarrow B E M = 54. 6^{\circ}

(c)

3 marks

The angle $A E B = 2 \times 54.5 8^{\circ} = 109.1 7^{\circ} = 1.905$ radians and the radius is 55.22.

arc A B = 55.22 \times 1.905 \approx 105 m

(d)

4 marks

The outer edge is an arc of radius $55.22 + 4 = 59.22$ with the same angle. The area of the track is the difference between the two sectors.

A = \frac{1}{2} (1.905) (59.2 2^{2} - 55.2 2^{2}) = \frac{1}{2} (1.905) (457.8) \approx 436 m^{2}

(e)

3 marks

The depth is 15 cm $= 0.15$ m.

V = 436.1 \times 0.15 \approx 65.4 m^{3}

Question 3

18 marks

Family movie running times: mean, quartiles, outlier and a chi-squared fit

Statistics and probability

(a)

3 marks

The mid-interval value of $70 \leq t < 80$ is 75. Use the midpoints $75, 82.5, 87.5, 92.5, 100$ .

\overset{ˉ}{t} = \frac{75 ( 11 ) + 82.5 ( 51 ) + 87.5 ( 68 ) + 92.5 ( 47 ) + 100 ( 23 )}{200} = 88.15 minutes

(b)

2 marks

Read the curve at cumulative frequency 50 and 150.

Q_{1} = 84, Q_{3} = 91.5 \Rightarrow IQR = 7.5 minutes

(c)

3 marks

The upper fence is $Q_{3} + 1.5 \times IQR$ .
"Star Feud" lasts $100 < 102.75$ minutes, so it is not an outlier.

91.5 + 1.5 (7.5) = 102.75

(d)

2 marks

The null hypothesis says that the running times follow this normal distribution.

H_{0} : the running times can be modelled by N (88, 6.7 5^{2}), H_{1} : they cannot

(e)

8 marks

Expected frequencies from $N (88, 6.7 5^{2})$ for 200 movies.
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 running times are not well modelled by $N (88, 6.7 5^{2})$ .

a = 200 P (85 \leq t < 90) = 57.6, b = 200 P (t \geq 95) = 30.0 χ^{2} = 12.1, p = 0.0166

Question 4

16 marks

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

Number and algebra

(a)

5 marks

The down payment is 25%, so the loan is 75% of the price.
Use a finance solver with $N = 60$ , $I = 4.5%$ , PV $= 213750$ , FV $= 0$ and 12 payments a year.

0.75 \times 285000 = 213750 ZAR PMT = 3984.95 ZAR

(b)

2 marks

Total paid over 60 months.

60 \times 3984.95 \approx 239097 ZAR

(c)

2 marks

Solve for the number of payments with PMT $= 4600$ .

N = 51.15 \Rightarrow 52 payments

(d)

4 marks

After 51 payments of 4600 the balance, after interest for the last month, is the final payment.

final payment \approx 706.80 ZAR

(e)

3 marks

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

51 (4600) + 706.80 = 235 306.80 239097 - 235307 \approx 3790 ZAR

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 $h = 16$ the area is $\frac{h}{2} [first + last + 2 (middle)]$ .
02
Radians for arc length · Q2
The angle at E 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 25% 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