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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

6 marks

Ship and cliff: angle, horizontal distance and standard form

Geometry and trigonometry

(a)

2 marks

In the right triangle KBS, the side opposite S is 218 and the hypotenuse is 1200.

sin θ = \frac{218}{1200} \Rightarrow θ = 10. 5^{\circ}

(b)

2 marks

Use Pythagoras for the horizontal side BS.

B S = 120 0^{2} - 21 8^{2} = 1180 m

(c)

2 marks

Move the decimal point three places.

1.18 \times 1 0^{3} m

Question 2

7 marks

Trees planted each month: arithmetic sequence

Number and algebra

(a)

3 marks

The 15th term is $u_{1} + 14 d$ .

u_{15} = 85 + 14 (30) = 505

(b)

2 marks

Sum of an arithmetic series.

S_{15} = \frac{15}{2} (85 + 505) = 4425

(c)

2 marks

Mean per month is the total over 15 months divided by 15.

\frac{4425}{15} = 295

Question 3

5 marks

Javelin height: a quadratic model

Functions

(a)

1 mark

At the moment of throwing, $t = 0$ .

h (0) = 1.8 m

(b)

2 marks

Use the axis of symmetry.

t = - \frac{10.8}{2 ( - 3.6 )} = 1.5 s

(c)

2 marks

Solve $h (t) = 0$ and reject the negative root.

- 3.6 t^{2} + 10.8 t + 1.8 = 0 \Rightarrow t = 3.16 s

Question 4

5 marks

Is favourite berry independent of income? A chi-squared test

Statistics and probability

(a)

1 mark

The alternative hypothesis is that they are not independent.

H_{0} : a person’s favourite berry is independent of their income level

(b)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence ( $ν = 4$ ).

χ^{2} = 2.27

(c)

2 marks

Compare the statistic with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 10% level that favourite berry depends on income level.

2.27 < 7.78

Mentor tip

The $p$ -value, about 0.687, is also much bigger than 0.10.

Question 5

6 marks

Coffee cooling: evaluating, asymptote and an inverse value

Functions

(a)

2 marks

Substitute $t = 16$ .

T (16) = 71 e^{- 0.0514 (16)} + 23 = 54.2^{\circ} C

(b)

1 mark

The exponential term tends to 0 as $t \to \infty$ .

y = 23

(c)

1 mark

The coffee ends at the temperature of the room.

23^{\circ} C

(d)

2 marks

$T^{- 1} (50) = k$ means $T (k) = 50$ .

71 e^{- 0.0514 k} + 23 = 50 \Rightarrow k = \frac{ln ( 71/27 )}{0.0514} = 18.8 minutes

Question 6

5 marks

Comparing two classes: a two-tailed t-test

Statistics and probability

(a)

1 mark

The test is two-tailed.

H_{1} : μ_{1} \neq = μ_{2}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test.

p = 0.97652

(c)

2 marks

Compare the p-value with 0.05.
Annabelle is correct: the $p$ -value is far larger than the significance level, so there is insufficient evidence to reject the null hypothesis.

0.97652 > 0.05

Question 7

7 marks

Savings with monthly deposits: finding the number of months

Number and algebra

(a)

4 marks

The monthly rate is $\frac{7.5}{12} %$ and each period is one month. Use a finance solver with PV = $- 800$ , PMT = $- 500$ , $I = 7.5$ , $P / Y = C / Y = 12$ and FV = 10000, and solve for $N$ .
Check: after 17 months the balance is below 10000, and after 18 it is above.

N = 17.2 \Rightarrow k = 18 800 (1.00625)^{18} + 500 \cdot \frac{1.0062 5 ^{18} - 1}{0.00625} = 10389

(b)

3 marks

The total paid in is $800 + 18 \times 500 = 9800$ .

interest = 10389 - 9800 \approx 589 EUR

Question 8

5 marks

Lap times: a normal probability and a time that 8.6 laps exceed

Statistics and probability

(a)

2 marks

Use $T \sim N (59, 3^{2})$ .

P (T < 55) = 0.0912

(b)

3 marks

The proportion of laps longer than $t$ is $\frac{8.6}{20} = 0.43$ .
Use the inverse normal.

P (T > t) = 0.43 \Rightarrow P (T < t) = 0.57 t = 59.5 s

Question 9

7 marks

A bottle-throwing game: total of two games and a fair game

Statistics and probability

(a)

4 marks

List the ways to get a total of two bottles over two games: 1 and 1, 0 and 2, 2 and 0.

P (total 2) = 0.4 \times 0.4 + 0.5 \times 0.1 + 0.1 \times 0.5 = 0.16 + 0.05 + 0.05 = 0.26

(b)

3 marks

The points are $k$ with probability 0.1, $4$ with 0.4 and $- 8$ with 0.5. A fair game has an expected value of 0.

0.1 k + 0.4 (4) + 0.5 (- 8) = 0 \Rightarrow 0.1 k - 2.4 = 0 k = 24

Question 10

6 marks

Star magnitude and brightness: comparing two stars

Functions

(a)

2 marks

Put $b = 0.0525$ into the equation.

m = 1 - 2.5 lo g_{10} (0.0525) = 4.20

(b)

2 marks

Set $m = 7$ and solve for $b$ .

7 = 1 - 2.5 lo g_{10} b \Rightarrow lo g_{10} b = - 2.4 \Rightarrow b = 0.00398

(c)

2 marks

Divide the brightness of Acubens by that of Ceres.
Acubens is about 13 times brighter than Ceres.

\frac{0.0525}{0.00398} \approx 13.2

Question 11

7 marks

Kristi's training runs: a geometric sequence of positions

Number and algebra

(a)

2 marks

The points are $2$ , then $B$ , then $4.5$ . In a geometric sequence $r = \frac{u _{3}}{u _{2}} = \frac{u _{2}}{u _{1}}$ , so $u_{2}^{2} = 2 (4.5) = 9$ and $u_{2} = \pm 3$ .
She runs west on day 2, so $u_{2} = - 3$ .

r = \frac{- 3}{2} = - 1.5, \frac{4.5}{- 3} = - 1.5

(b)

2 marks

The 6th term is $u_{1} r^{5}$ .

u_{6} = 2 (- 1.5)^{5} = - 15.1875 F = (- 15.19, 0)

(c)

3 marks

The distance each day is $∣ u_{n} ∣ = 2 (1.5)^{n - 1}$ . Sum the first 7 terms.

n = 1 \sum 7 2 (1.5)^{n - 1} = 2 \cdot \frac{1. 5 ^{7} - 1}{0.5} = 64.3 km

Question 12

6 marks

A cat in an exercise wheel: a sine model for the height of a point

Functions

(a)

1 mark

At $t = 0$ , $sin (0) = 0$ , so $h (0) = c$ .

c = 78

(b)

3 marks

The amplitude is the maximum minus the central height.
The first maximum of a sine curve is a quarter of a period in, so the period is $4 \times 4 = 16$ s.

a = 143 - 78 = 65 b = \frac{2 π}{16} = \frac{π}{8} \approx 0.393

(c)

1 mark

The minimum height is the centre minus the amplitude.

78 - 65 = 13 cm

(d)

1 mark

Twice as long means a period of 32 s.

b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

Question 13

8 marks

Cost per hour of hiring a boat: integrating a rate

Calculus

(a)

2 marks

The rate is $\frac{d P}{d t} = 20 - \frac{980}{t ^{2}}$ . Set it to zero.

20 = \frac{980}{t ^{2}} \Rightarrow t^{2} = 49 \Rightarrow h = 7

(b)

6 marks

Integrate.
Use $P (5) = 328$ .
For Yvonne at 7 hours:

P = \int (20 - 980 t^{- 2}) d t = 20 t + \frac{980}{t} + c 100 + 196 + c = 328 \Rightarrow c = 32 P (7) = 20 (7) + \frac{980}{7} + 32 = 140 + 140 + 32 = 312

Our analysis

Where the marks are

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

Functions23 marks · 29%
Q3, Q5, Q10, Q12
Statistics and probability22 marks · 28%
Q4, Q6, Q8, Q9
Number and algebra21 marks · 26%
Q2, Q7, Q11
Calculus8 marks · 10%
Q13
Geometry and trigonometry6 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Common difference · Q2
The number planted rises by 30 each month, so $d = 30$ .
02
Deposits at the end of each month · Q7
The first deposit earns interest for $k - 1$ months. The solver should be set to end-of-period payments.
03
Two ways to knock down two bottles in total · Q9
Total of 2 bottles can come from (1, 1), (0, 2) or (2, 0).
04
Signed positions · Q11
Running west gives a negative $x$ -coordinate, so the ratio is negative. Distances are the absolute values.
05
Constant of integration · Q13
Use Sandra's 5 hours at NOK 328 to find the constant.

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.

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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

6 marks

Ship and cliff: angle, horizontal distance and standard form

Geometry and trigonometry

(a)

2 marks

In the right triangle KBS, the side opposite S is 218 and the hypotenuse is 1200.

sin θ = \frac{218}{1200} \Rightarrow θ = 10. 5^{\circ}

(b)

2 marks

Use Pythagoras for the horizontal side BS.

B S = 120 0^{2} - 21 8^{2} = 1180 m

(c)

2 marks

Move the decimal point three places.

1.18 \times 1 0^{3} m

Question 2

7 marks

Trees planted each month: arithmetic sequence

Number and algebra

(a)

3 marks

The 15th term is $u_{1} + 14 d$ .

u_{15} = 85 + 14 (30) = 505

(b)

2 marks

Sum of an arithmetic series.

S_{15} = \frac{15}{2} (85 + 505) = 4425

(c)

2 marks

Mean per month is the total over 15 months divided by 15.

\frac{4425}{15} = 295

Question 3

5 marks

Javelin height: a quadratic model

Functions

(a)

1 mark

At the moment of throwing, $t = 0$ .

h (0) = 1.8 m

(b)

2 marks

Use the axis of symmetry.

t = - \frac{10.8}{2 ( - 3.6 )} = 1.5 s

(c)

2 marks

Solve $h (t) = 0$ and reject the negative root.

- 3.6 t^{2} + 10.8 t + 1.8 = 0 \Rightarrow t = 3.16 s

Question 4

5 marks

Is favourite berry independent of income? A chi-squared test

Statistics and probability

(a)

1 mark

The alternative hypothesis is that they are not independent.

H_{0} : a person’s favourite berry is independent of their income level

(b)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence ( $ν = 4$ ).

χ^{2} = 2.27

(c)

2 marks

Compare the statistic with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 10% level that favourite berry depends on income level.

2.27 < 7.78

Mentor tip

The $p$ -value, about 0.687, is also much bigger than 0.10.

Question 5

6 marks

Coffee cooling: evaluating, asymptote and an inverse value

Functions

(a)

2 marks

Substitute $t = 16$ .

T (16) = 71 e^{- 0.0514 (16)} + 23 = 54.2^{\circ} C

(b)

1 mark

The exponential term tends to 0 as $t \to \infty$ .

y = 23

(c)

1 mark

The coffee ends at the temperature of the room.

23^{\circ} C

(d)

2 marks

$T^{- 1} (50) = k$ means $T (k) = 50$ .

71 e^{- 0.0514 k} + 23 = 50 \Rightarrow k = \frac{ln ( 71/27 )}{0.0514} = 18.8 minutes

Question 6

5 marks

Comparing two classes: a two-tailed t-test

Statistics and probability

(a)

1 mark

The test is two-tailed.

H_{1} : μ_{1} \neq = μ_{2}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test.

p = 0.97652

(c)

2 marks

Compare the p-value with 0.05.
Annabelle is correct: the $p$ -value is far larger than the significance level, so there is insufficient evidence to reject the null hypothesis.

0.97652 > 0.05

Question 7

7 marks

Savings with monthly deposits: finding the number of months

Number and algebra

(a)

4 marks

The monthly rate is $\frac{7.5}{12} %$ and each period is one month. Use a finance solver with PV = $- 800$ , PMT = $- 500$ , $I = 7.5$ , $P / Y = C / Y = 12$ and FV = 10000, and solve for $N$ .
Check: after 17 months the balance is below 10000, and after 18 it is above.

N = 17.2 \Rightarrow k = 18 800 (1.00625)^{18} + 500 \cdot \frac{1.0062 5 ^{18} - 1}{0.00625} = 10389

(b)

3 marks

The total paid in is $800 + 18 \times 500 = 9800$ .

interest = 10389 - 9800 \approx 589 EUR

Question 8

5 marks

Lap times: a normal probability and a time that 8.6 laps exceed

Statistics and probability

(a)

2 marks

Use $T \sim N (59, 3^{2})$ .

P (T < 55) = 0.0912

(b)

3 marks

The proportion of laps longer than $t$ is $\frac{8.6}{20} = 0.43$ .
Use the inverse normal.

P (T > t) = 0.43 \Rightarrow P (T < t) = 0.57 t = 59.5 s

Question 9

7 marks

A bottle-throwing game: total of two games and a fair game

Statistics and probability

(a)

4 marks

List the ways to get a total of two bottles over two games: 1 and 1, 0 and 2, 2 and 0.

P (total 2) = 0.4 \times 0.4 + 0.5 \times 0.1 + 0.1 \times 0.5 = 0.16 + 0.05 + 0.05 = 0.26

(b)

3 marks

The points are $k$ with probability 0.1, $4$ with 0.4 and $- 8$ with 0.5. A fair game has an expected value of 0.

0.1 k + 0.4 (4) + 0.5 (- 8) = 0 \Rightarrow 0.1 k - 2.4 = 0 k = 24

Question 10

6 marks

Star magnitude and brightness: comparing two stars

Functions

(a)

2 marks

Put $b = 0.0525$ into the equation.

m = 1 - 2.5 lo g_{10} (0.0525) = 4.20

(b)

2 marks

Set $m = 7$ and solve for $b$ .

7 = 1 - 2.5 lo g_{10} b \Rightarrow lo g_{10} b = - 2.4 \Rightarrow b = 0.00398

(c)

2 marks

Divide the brightness of Acubens by that of Ceres.
Acubens is about 13 times brighter than Ceres.

\frac{0.0525}{0.00398} \approx 13.2

Question 11

7 marks

Kristi's training runs: a geometric sequence of positions

Number and algebra

(a)

2 marks

The points are $2$ , then $B$ , then $4.5$ . In a geometric sequence $r = \frac{u _{3}}{u _{2}} = \frac{u _{2}}{u _{1}}$ , so $u_{2}^{2} = 2 (4.5) = 9$ and $u_{2} = \pm 3$ .
She runs west on day 2, so $u_{2} = - 3$ .

r = \frac{- 3}{2} = - 1.5, \frac{4.5}{- 3} = - 1.5

(b)

2 marks

The 6th term is $u_{1} r^{5}$ .

u_{6} = 2 (- 1.5)^{5} = - 15.1875 F = (- 15.19, 0)

(c)

3 marks

The distance each day is $∣ u_{n} ∣ = 2 (1.5)^{n - 1}$ . Sum the first 7 terms.

n = 1 \sum 7 2 (1.5)^{n - 1} = 2 \cdot \frac{1. 5 ^{7} - 1}{0.5} = 64.3 km

Question 12

6 marks

A cat in an exercise wheel: a sine model for the height of a point

Functions

(a)

1 mark

At $t = 0$ , $sin (0) = 0$ , so $h (0) = c$ .

c = 78

(b)

3 marks

The amplitude is the maximum minus the central height.
The first maximum of a sine curve is a quarter of a period in, so the period is $4 \times 4 = 16$ s.

a = 143 - 78 = 65 b = \frac{2 π}{16} = \frac{π}{8} \approx 0.393

(c)

1 mark

The minimum height is the centre minus the amplitude.

78 - 65 = 13 cm

(d)

1 mark

Twice as long means a period of 32 s.

b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

Question 13

8 marks

Cost per hour of hiring a boat: integrating a rate

Calculus

(a)

2 marks

The rate is $\frac{d P}{d t} = 20 - \frac{980}{t ^{2}}$ . Set it to zero.

20 = \frac{980}{t ^{2}} \Rightarrow t^{2} = 49 \Rightarrow h = 7

(b)

6 marks

Integrate.
Use $P (5) = 328$ .
For Yvonne at 7 hours:

P = \int (20 - 980 t^{- 2}) d t = 20 t + \frac{980}{t} + c 100 + 196 + c = 328 \Rightarrow c = 32 P (7) = 20 (7) + \frac{980}{7} + 32 = 140 + 140 + 32 = 312

Our analysis

Where the marks are

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

Functions23 marks · 29%
Q3, Q5, Q10, Q12
Statistics and probability22 marks · 28%
Q4, Q6, Q8, Q9
Number and algebra21 marks · 26%
Q2, Q7, Q11
Calculus8 marks · 10%
Q13
Geometry and trigonometry6 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Common difference · Q2
The number planted rises by 30 each month, so $d = 30$ .
02
Deposits at the end of each month · Q7
The first deposit earns interest for $k - 1$ months. The solver should be set to end-of-period payments.
03
Two ways to knock down two bottles in total · Q9
Total of 2 bottles can come from (1, 1), (0, 2) or (2, 0).
04
Signed positions · Q11
Running west gives a negative $x$ -coordinate, so the ratio is negative. Distances are the absolute values.
05
Constant of integration · Q13
Use Sandra's 5 hours at NOK 328 to find the constant.

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