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IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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

18 marks

Reading times, outliers, regression and Spearman's rank correlation

Statistics and probability

(a)

1 mark

Jason set a target of 10 students from each class, so he used quota sampling.

(b)

1 mark

The median is the line inside the box.

10 hours

(c)

2 marks

The interquartile range is the upper quartile minus the lower quartile: $Q_{3} = 15$ and $Q_{1} = 7$ .

15 - 7 = 8

(d)

4 marks

A value is an outlier if it is above $Q_{3} + 1.5 \times IQR$ .
Since $25 < 27$ , Mackenzie's time is not an outlier, so Jason is correct.

15 + 1.5 \times 8 = 27

(e)

1 mark

As the time to read the book increases, the exam percentage goes down: a negative correlation.

(f)

2 marks

Substitute $x = 1.5$ into the equation.

y = - 1.54 (1.5) + 98.8 = 96.49 \approx 96.5%

(g)

2 marks

It's not reliable, because 1.5 hours is outside the range of times in the data, so this is extrapolation.

(h)

2 marks

Rank the number of pages with 1 as the largest, and the rating with 1 as the best.

pages: A 1, B 3, C 5, D 2, E 6, F 8, G 4, H 7 rating: A 1, B 2, C 3, D 4, E 5, F 6, G 7, H 8

(i)

3 marks

(i) The correlation of the two sets of ranks:

(ii) There is a moderate to strong positive association: books with more pages tend to have a better top-50 rating.

r_{s} \approx 0.714

Question 2

17 marks

Clock hands: angles, arcs and where two hands meet

Geometry and trigonometry

(a)

2 marks

At 4:00 the hour hand points at 4, and each hour on the clock is $3 0^{\circ}$ . The minute hand points at 12.

4 \times 3 0^{\circ} = 12 0^{\circ}

(b)

3 marks

Apply the cosine rule in triangle AOB.

A B^{2} = 1 0^{2} + 6^{2} - 2 (10) (6) cos 12 0^{\circ} = 136 + 60 = 196 \Rightarrow A B = 14 cm

(c)

2 marks

The minute hand turns $6^{\circ}$ each minute.

θ = 13 \times 6^{\circ} = 7 8^{\circ}

(d)

2 marks

The arc is a fraction $\frac{78}{360}$ of the circumference.

\frac{78}{360} \times 2 π (10) \approx 13.6 cm

(e)

2 marks

The sector is the same fraction of the circle's area.

\frac{78}{360} \times π (10)^{2} \approx 68.1 cm^{2}

(f)

1 mark

At 6:00 the minute hand points straight up. The bottom of the clock is 3 cm above the shelf and the clock face is 20 cm across.

3 + 20 = 23 cm

(g)

2 marks

Substitute $θ = 16 0^{\circ}$ into the model, with the calculator in degree mode.

h (160) = 10 cos 16 0^{\circ} + 13 \approx 3.60 cm

(h)

1 mark

The amplitude is the number multiplying the cosine.

10

(i)

2 marks

The tips are at the same height when $h (θ) = g (θ)$ .
The first solution is $θ \approx 166. 2^{\circ}$ , but there the minute hand is on the right of the clock and the hour hand is on the left, so they are not at the same point. The next solution has both on the same side.

10 cos θ + 13 = - 10 cos \frac{θ}{12} + 13 \Rightarrow cos θ = - cos \frac{θ}{12} k \approx 19 6^{\circ}

Question 3

19 marks

Savings and a loan: compound interest and amortisation

Number and algebra

(a)

3 marks

Two years is 8 quarters, each at $1.6%$ interest.

37000 (1 + \frac{0.064}{4})^{8} \approx 42010 AUD

(b)

4 marks

Solve $37000 (1.016)^{n} > 50000$ for the number of quarters $n$ , on the calculator or by logs.
Interest is only added at the end of each quarter, so convert to months.

n > \frac{ln ( 50000/37000 )}{ln 1.016} \approx 18.97 \Rightarrow n = 19 quarters m = 19 \times 3 = 57

(c)

1 mark

The deposit is 25% of 200 000, so the loan is the remaining 75%.

0.75 \times 200000 = 150000 AUD

(d)

5 marks

(i) Over 10 years there are 120 payments of 1700, so the interest is what Daisy pays beyond the loan.

(ii) Use the calculator's finance solver: $N = 120$ , $PV = 150000$ , $PMT = - 1700$ , $FV = 0$ , payments and compounding 12 per year.

120 \times 1700 - 150000 = 54000 AUD r \approx 6.46%

(e)

3 marks

After 5 years she has made 60 payments. Use the same solver with $N = 60$ , $I = 6.4578%$ , $PV = 150000$ , $PMT = - 1700$ , and read off the future value, which is the balance still owed.

final payment \approx 86973 AUD

(f)

3 marks

Without the early payoff she would have paid $120 \times 1700 = 204000$ . With it, she pays 60 payments plus the final amount.

204000 - (60 \times 1700 + 86973) = 204000 - 188973 = 15027 AUD

Question 4

13 marks

Do bicycle stopping distances fit a normal model? A goodness-of-fit test

Statistics and probability

(a)

3 marks

With $X \sim N (6.76, 0.1 2^{2})$ , use the normal cumulative distribution.

P (X < 6.5) \approx 0.0151, P (X > 7) \approx 0.0228

(b)

3 marks

Multiply the interval probabilities by 1000.

1000 P (6.5 < X < 6.75) \approx 451.7, 1000 P (6.75 < X < 7) \approx 510.5

(c)

2 marks

State the model as the null hypothesis.

H_{0} : the stopping distances follow N (6.76, 0.1 2^{2}), H_{1} : they do not

(d)

3 marks

The expected counts are about 15.1, 451.7, 510.5 and 22.8. Run the $χ^{2}$ goodness-of-fit test on the calculator with these and the observed 12, 428, 527, 33. There are 4 classes, so $df = 3$ .

p \approx 0.073

Mentor tip

With the expected counts rounded as above, $p = 0.0727$ . With unrounded ones it is 0.0706. Both are above 0.05, so the conclusion is the same.

(e)

2 marks

Compare with the 5% level.
So there is insufficient evidence to reject $H_{0}$ : the normal model is a reasonable fit.

p > 0.05

Question 5

13 marks

A tunnel: maximum height, trapezoidal rule and exact area

Calculus

(a)

6 marks

(i) Differentiate each term.

(ii) At the maximum the derivative is zero.

\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x x (- 0.3 x + 1.6) = 0 \Rightarrow x = \frac{16}{3} y = - 0.1 (\frac{16}{3})^{3} + 0.8 (\frac{16}{3})^{2} \approx 7.59 m

(b)

3 marks

With three intervals, $h = 2$ . The heights at $x = 2, 4, 6, 8$ are $2.4$ , $6.4$ , $7.2$ and $0$ .

A \approx \frac{1}{2} (2) [(2.4 + 0) + 2 (6.4 + 7.2)] = 29.6 m^{2}

(c)

4 marks

(i) The area is a definite integral between the two walls.

(ii) Evaluate the integral.

A = \int_{2}^{8} (- 0.1 x^{3} + 0.8 x^{2}) d x A = [- \frac{x ^{4}}{40} + \frac{0.8 x ^{3}}{3}]_{2}^{8} = 32.4 m^{2}

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 probability31 marks · 39%
Q1, Q4
Number and algebra19 marks · 24%
Q3
Geometry and trigonometry17 marks · 21%
Q2
Calculus13 marks · 16%
Q5

Mentor insights

Traps to watch for

01
Extrapolation is not reliable · Q1
A regression line is only reliable inside the range of its data. In (g) say that 1.5 hours lies outside that range.
02
Equal heights are not the same point · Q2
Two hand tips can be at the same height while on opposite sides of the clock. The tips meet only when they are also on the same side, which is why the first height match at about $16 6^{\circ}$ is rejected.
03
Quarterly means four periods a year · Q3
With interest compounded quarterly the rate per period is $\frac{6.4}{4} %$ and the number of periods is four times the years. The answer in months must then be a multiple of 3.
04
Expected counts, then compare · Q4
The test needs expected frequencies from the normal model (1000 times each probability). The degrees of freedom are the number of classes minus 1, since the mean and standard deviation were given, not estimated.
05
Use the unrounded stationary point · Q5
The maximum is at $x = \frac{16}{3}$ . Rounding to 5.33 before substituting changes the third figure of the height.

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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Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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

18 marks

Reading times, outliers, regression and Spearman's rank correlation

Statistics and probability

(a)

1 mark

Jason set a target of 10 students from each class, so he used quota sampling.

(b)

1 mark

The median is the line inside the box.

10 hours

(c)

2 marks

The interquartile range is the upper quartile minus the lower quartile: $Q_{3} = 15$ and $Q_{1} = 7$ .

15 - 7 = 8

(d)

4 marks

A value is an outlier if it is above $Q_{3} + 1.5 \times IQR$ .
Since $25 < 27$ , Mackenzie's time is not an outlier, so Jason is correct.

15 + 1.5 \times 8 = 27

(e)

1 mark

As the time to read the book increases, the exam percentage goes down: a negative correlation.

(f)

2 marks

Substitute $x = 1.5$ into the equation.

y = - 1.54 (1.5) + 98.8 = 96.49 \approx 96.5%

(g)

2 marks

It's not reliable, because 1.5 hours is outside the range of times in the data, so this is extrapolation.

(h)

2 marks

Rank the number of pages with 1 as the largest, and the rating with 1 as the best.

pages: A 1, B 3, C 5, D 2, E 6, F 8, G 4, H 7 rating: A 1, B 2, C 3, D 4, E 5, F 6, G 7, H 8

(i)

3 marks

(i) The correlation of the two sets of ranks:

(ii) There is a moderate to strong positive association: books with more pages tend to have a better top-50 rating.

r_{s} \approx 0.714

Question 2

17 marks

Clock hands: angles, arcs and where two hands meet

Geometry and trigonometry

(a)

2 marks

At 4:00 the hour hand points at 4, and each hour on the clock is $3 0^{\circ}$ . The minute hand points at 12.

4 \times 3 0^{\circ} = 12 0^{\circ}

(b)

3 marks

Apply the cosine rule in triangle AOB.

A B^{2} = 1 0^{2} + 6^{2} - 2 (10) (6) cos 12 0^{\circ} = 136 + 60 = 196 \Rightarrow A B = 14 cm

(c)

2 marks

The minute hand turns $6^{\circ}$ each minute.

θ = 13 \times 6^{\circ} = 7 8^{\circ}

(d)

2 marks

The arc is a fraction $\frac{78}{360}$ of the circumference.

\frac{78}{360} \times 2 π (10) \approx 13.6 cm

(e)

2 marks

The sector is the same fraction of the circle's area.

\frac{78}{360} \times π (10)^{2} \approx 68.1 cm^{2}

(f)

1 mark

At 6:00 the minute hand points straight up. The bottom of the clock is 3 cm above the shelf and the clock face is 20 cm across.

3 + 20 = 23 cm

(g)

2 marks

Substitute $θ = 16 0^{\circ}$ into the model, with the calculator in degree mode.

h (160) = 10 cos 16 0^{\circ} + 13 \approx 3.60 cm

(h)

1 mark

The amplitude is the number multiplying the cosine.

10

(i)

2 marks

The tips are at the same height when $h (θ) = g (θ)$ .
The first solution is $θ \approx 166. 2^{\circ}$ , but there the minute hand is on the right of the clock and the hour hand is on the left, so they are not at the same point. The next solution has both on the same side.

10 cos θ + 13 = - 10 cos \frac{θ}{12} + 13 \Rightarrow cos θ = - cos \frac{θ}{12} k \approx 19 6^{\circ}

Question 3

19 marks

Savings and a loan: compound interest and amortisation

Number and algebra

(a)

3 marks

Two years is 8 quarters, each at $1.6%$ interest.

37000 (1 + \frac{0.064}{4})^{8} \approx 42010 AUD

(b)

4 marks

Solve $37000 (1.016)^{n} > 50000$ for the number of quarters $n$ , on the calculator or by logs.
Interest is only added at the end of each quarter, so convert to months.

n > \frac{ln ( 50000/37000 )}{ln 1.016} \approx 18.97 \Rightarrow n = 19 quarters m = 19 \times 3 = 57

(c)

1 mark

The deposit is 25% of 200 000, so the loan is the remaining 75%.

0.75 \times 200000 = 150000 AUD

(d)

5 marks

(i) Over 10 years there are 120 payments of 1700, so the interest is what Daisy pays beyond the loan.

(ii) Use the calculator's finance solver: $N = 120$ , $PV = 150000$ , $PMT = - 1700$ , $FV = 0$ , payments and compounding 12 per year.

120 \times 1700 - 150000 = 54000 AUD r \approx 6.46%

(e)

3 marks

After 5 years she has made 60 payments. Use the same solver with $N = 60$ , $I = 6.4578%$ , $PV = 150000$ , $PMT = - 1700$ , and read off the future value, which is the balance still owed.

final payment \approx 86973 AUD

(f)

3 marks

Without the early payoff she would have paid $120 \times 1700 = 204000$ . With it, she pays 60 payments plus the final amount.

204000 - (60 \times 1700 + 86973) = 204000 - 188973 = 15027 AUD

Question 4

13 marks

Do bicycle stopping distances fit a normal model? A goodness-of-fit test

Statistics and probability

(a)

3 marks

With $X \sim N (6.76, 0.1 2^{2})$ , use the normal cumulative distribution.

P (X < 6.5) \approx 0.0151, P (X > 7) \approx 0.0228

(b)

3 marks

Multiply the interval probabilities by 1000.

1000 P (6.5 < X < 6.75) \approx 451.7, 1000 P (6.75 < X < 7) \approx 510.5

(c)

2 marks

State the model as the null hypothesis.

H_{0} : the stopping distances follow N (6.76, 0.1 2^{2}), H_{1} : they do not

(d)

3 marks

The expected counts are about 15.1, 451.7, 510.5 and 22.8. Run the $χ^{2}$ goodness-of-fit test on the calculator with these and the observed 12, 428, 527, 33. There are 4 classes, so $df = 3$ .

p \approx 0.073

Mentor tip

With the expected counts rounded as above, $p = 0.0727$ . With unrounded ones it is 0.0706. Both are above 0.05, so the conclusion is the same.

(e)

2 marks

Compare with the 5% level.
So there is insufficient evidence to reject $H_{0}$ : the normal model is a reasonable fit.

p > 0.05

Question 5

13 marks

A tunnel: maximum height, trapezoidal rule and exact area

Calculus

(a)

6 marks

(i) Differentiate each term.

(ii) At the maximum the derivative is zero.

\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x x (- 0.3 x + 1.6) = 0 \Rightarrow x = \frac{16}{3} y = - 0.1 (\frac{16}{3})^{3} + 0.8 (\frac{16}{3})^{2} \approx 7.59 m

(b)

3 marks

With three intervals, $h = 2$ . The heights at $x = 2, 4, 6, 8$ are $2.4$ , $6.4$ , $7.2$ and $0$ .

A \approx \frac{1}{2} (2) [(2.4 + 0) + 2 (6.4 + 7.2)] = 29.6 m^{2}

(c)

4 marks

(i) The area is a definite integral between the two walls.

(ii) Evaluate the integral.

A = \int_{2}^{8} (- 0.1 x^{3} + 0.8 x^{2}) d x A = [- \frac{x ^{4}}{40} + \frac{0.8 x ^{3}}{3}]_{2}^{8} = 32.4 m^{2}

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 probability31 marks · 39%
Q1, Q4
Number and algebra19 marks · 24%
Q3
Geometry and trigonometry17 marks · 21%
Q2
Calculus13 marks · 16%
Q5

Mentor insights

Traps to watch for

01
Extrapolation is not reliable · Q1
A regression line is only reliable inside the range of its data. In (g) say that 1.5 hours lies outside that range.
02
Equal heights are not the same point · Q2
Two hand tips can be at the same height while on opposite sides of the clock. The tips meet only when they are also on the same side, which is why the first height match at about $16 6^{\circ}$ is rejected.
03
Quarterly means four periods a year · Q3
With interest compounded quarterly the rate per period is $\frac{6.4}{4} %$ and the number of periods is four times the years. The answer in months must then be a multiple of 3.
04
Expected counts, then compare · Q4
The test needs expected frequencies from the normal model (1000 times each probability). The degrees of freedom are the number of classes minus 1, since the mean and standard deviation were given, not estimated.
05
Use the unrounded stationary point · Q5
The maximum is at $x = \frac{16}{3}$ . Rounding to 5.33 before substituting changes the third figure of the height.

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