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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 May 2023 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

6 marks

Decathlon: Spearman's rank correlation for two events

Statistics and probability

(a)

2 marks

Rank the high jumps with the highest as 1: 2.11 (1), 2.08 (2), 2.05 (3), 2.02 twice (4.5 each), 1.99 (6), 1.96 (7), 1.90 (8), 1.87 (9), 1.84 (10).

Germany 1, France 2, Estonia 10, Canada 4.5, Netherlands 3, Ukraine 4.5, Algeria 8, Austria 9, Grenada 6, Japan 7

(b)

2 marks

Spearman's coefficient is the correlation between the two sets of ranks.

r_{s} = 0.541

(c)

2 marks

A value of 0.541 lies in the range 0.400 to 0.599, so the correlation is moderate.
Athletes who rank higher (better) in the long jump tend, to a moderate extent, to rank higher in the high jump.

Question 2

6 marks

A hot-air balloon seen from two points

Geometry and trigonometry

(a)

2 marks

The angle of elevation of H from A is $4 0^{\circ}$ , so at A the angle between the downward vertical AB and AH is $9 0^{\circ} + 4 0^{\circ} = 13 0^{\circ}$ . At B the angle between the upward vertical BA and BH is $9 0^{\circ} - 5 7^{\circ} = 3 3^{\circ}$ .

A H B = 180 - 130 - 33 = 1 7^{\circ}

(b)

3 marks

Sine rule in triangle ABH with $A B = 156$ .

\frac{B H}{sin 13 0 ^{\circ}} = \frac{156}{sin 1 7 ^{\circ}} \Rightarrow B H \approx 409 m

(c)

1 mark

The angle of depression from H to B decreases (it gets smaller) as the horizontal distance increases, because the balloon is at a constant height.

Question 3

5 marks

Savings with a deposit at the end of each month

Number and algebra

(a)

3 marks

1 January 2024 is 24 months after 1 January 2022. Monthly rate $\frac{4}{12} %$ .

A = 1000 (1 + \frac{0.04}{12})^{24} + 100 \cdot \frac{( 1 + \frac{0.04}{12} ) ^{24} - 1}{0.04/12} = $3577.43

(b)

2 marks

Test the balance after 36 and 37 months.
So it takes 37 complete months.

A_{36} = $4945.43 < 5000, A_{37} = $5061.91 > 5000

Question 4

6 marks

Memory scores: a one-tailed two-sample t-test

Statistics and probability

(a)

2 marks

The alternative hypothesis is one-sided because Carys believes bilingual people will score higher.

H_{0} : μ_{b} = μ_{m}, H_{1} : μ_{b} > μ_{m}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test (10 and 8 values, so $ν = 16$ ).

p \approx 0.119

(c)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level that the mean memory score of bilingual people is higher than that of monolingual people.

0.119 > 0.05

Question 5

5 marks

A tangent, a perpendicular line and whether it is the normal

Functions

(a)

1 mark

L2 has gradient $- \frac{1}{2}$ and L1 is perpendicular to it, so the gradients multiply to $- 1$ .

m_{1} = 2

(b)

2 marks

L1 passes through $P (3, - 1)$ .

y + 1 = 2 (x - 3) \Rightarrow y = 2 x - 7

(c)

2 marks

Put $x = 3$ into the equation of L2.
L2 does not pass through P, so it is not the normal to $f$ at P.

y = - \frac{1}{2} (3) - \frac{5}{2} = - 4 \neq = - 1

Question 6

5 marks

Stopping distance and speed: direct proportion to the square

Number and algebra

(a)

2 marks

Use $v = 50$ and $d = 12.3$ .

12.3 = k (50)^{2} \Rightarrow k = 0.00492 \Rightarrow d = 0.00492 v^{2}

(b)

2 marks

Set $d = 33$ and solve for $v$ .

33 = 0.00492 v^{2} \Rightarrow v = \frac{33}{0.00492} \approx 81.9 km h^{- 1}

(c)

1 mark

The state of the road surface (for example wet or dry) would affect the stopping distance. Other possible answers are the mass of the car, the condition of the tyres or the slope of the road.

Question 7

6 marks

An open box from a sheet of cardboard: maximum volume

Calculus

(a)

1 mark

The side of each square cannot be more than half of the shorter side, which is 30 cm.

k = 15

(b)

2 marks

Find the maximum of $V (x) = (48 - 2 x) (30 - 2 x) x$ on $0 < x < 15$ with a graph or by differentiating.

x = 6 cm

(c)

3 marks

The remaining cardboard is $(48 - 4)$ by $(30 - 4)$ , that is 44 cm by 26 cm.
Find the maximum on $0 < x < 13$ .

V (x) = (44 - 2 x) (26 - 2 x) x x = 5.28, V_{m a x} \approx 2730 cm^{3}

Question 8

7 marks

Password entropy: a logarithmic relationship

Functions

(a)

2 marks

Substitute $G = 5000$ .

0.301 p = lo g_{10} 5000 = 3.699 \Rightarrow p = 12.3 bits

(b)

1 mark

Undo the logarithm.

G (p) = 1 0^{0.301 p}

(c)

3 marks

Substitute $p = 28$ .

G = 1 0^{0.301 \times 28} = 1 0^{8.428} \approx 2.68 \times 1 0^{8}

(d)

1 mark

The point $(0, 1)$ means a password with entropy 0 bits can be decoded in 1 guess. It is completely predictable.

Question 9

8 marks

A hill profile: trapezoidal rule, exact area and percentage error

Calculus

(a)

2 marks

Trapezoidal rule with $h = 10$ and the five heights.

A \approx \frac{10}{2} [0 + 0 + 2 (3 + 8 + 9)] = 200 cm^{2}

(b)

4 marks

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

A = \int_{0}^{40} (0.04 x^{2} - 0.001 x^{3}) d x A = [\frac{0.04 x ^{3}}{3} - \frac{0.001 x ^{4}}{4}]_{0}^{40} = 853.33 - 640 = 213.33 cm^{2}

(c)

2 marks

Percentage error compares the trapezoidal estimate with the exact area.

\frac{∣200 - 213.33∣}{213.33} \times 100 = 6.25%

Question 10

7 marks

Baseball: a strike and a ball against the wall

Functions

(a)

4 marks

The gradient of the path is $- 0.045$ .
The ball is 1.25 m high at the front of the plate and about $1.25 - 0.0194 = 1.2307$ m at the back. The back height is between 0.53 m and 1.24 m, so at some point over the plate the ball is in the strike zone: it is a strike.

a = - 0.045 \times 0.43 = - 0.01935 (\approx - 0.0194)

(b)

3 marks

Find the height at the wall, $d = 96$ .
Since $8.34 > 5$ , the ball is above the 5 m wall when it reaches it, so it goes over the wall.

h (96) = - 0.01 (96)^{2} + 1.04 (96) + 0.66 = 8.34 m

Question 11

7 marks

A park post: bounds on a length and a safety angle

Geometry and trigonometry

(a)

2 marks

The length of AC is given to the nearest tenth of a centimetre.

lower bound 14.55 cm, upper bound 14.65 cm

(b)

3 marks

In triangle ABC the line AH to the midpoint of BC is perpendicular to BC, and $B H = 10$ . Use Pythagoras.

A H = A C^{2} - 1 0^{2} upper bound 14.6 5^{2} - 100 = 10.7 cm, lower bound 14.5 5^{2} - 100 = 10.6 cm

(c)

2 marks

The angle $α$ between the slant height and the base satisfies $cos α = \frac{10}{A H}$ . It is largest when $A H$ is largest.
Even the largest possible angle, $20. 9^{\circ}$ , is less than $2 2^{\circ}$ , so the post is safe.

α_{m a x} = arccos \frac{10}{10.706} = 20. 9^{\circ}

Question 12

7 marks

Car speeds: the standard deviation, speeding tickets and an asymmetric interval

Statistics and probability

(a)

2 marks

The speed 75.7 is two standard deviations above the mean 67.3.

2 σ = 75.7 - 67.3 = 8.4 \Rightarrow σ = 4.2

(b)

2 marks

Use $X \sim N (67.3, 4. 2^{2})$ .

P (X > 72) = 0.132

(c)

3 marks

A central region with 82% leaves 9% in each tail, so it would be $67.3 \pm 1.341 (4.2) = 67.3 \pm 5.63$ .
The speed 74 km/h is in the 82% interval from $p$ to $q$ but is above 72.9. So the interval is not this symmetric one: it is not symmetrical about the mean.

(61.7, 72.9)

Question 13

5 marks

A boat's return bearing

Number and algebra

5 marks

Take the starting point as the origin. After 8 km on $31 5^{\circ}$ and 6 km on $04 5^{\circ}$ :
The boat is 1.414 km west and 9.899 km north of the start, so to return it must go east 1.414 km and south 9.899 km.
The direction is south and slightly east, so the bearing is

east = - 8 sin 4 5^{\circ} + 6 sin 4 5^{\circ} = - 1.414, north = 8 cos 4 5^{\circ} + 6 cos 4 5^{\circ} = 9.899 tan α = \frac{1.414}{9.899} \Rightarrow α = 8.1 3^{\circ} 18 0^{\circ} - 8.1 3^{\circ} \approx 17 2^{\circ}

Our analysis

Where the marks are

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

Functions19 marks · 24%
Q5, Q8, Q10
Statistics and probability19 marks · 24%
Q1, Q4, Q12
Number and algebra15 marks · 19%
Q3, Q6, Q13
Calculus14 marks · 18%
Q7, Q9
Geometry and trigonometry13 marks · 16%
Q2, Q11

Mentor insights

Traps to watch for

01
Undo the log · Q8
$lo g_{10} G = 0.301 p$ means $G = 1 0^{0.301 p}$ .
02
Exact area uses the equation · Q9
Integrate the given cubic from 0 to 40.
03
Height change over the plate · Q10
The ball is over the plate for 0.43 m of horizontal distance, so the height changes by gradient times 0.43.
04
Bounds for a length to the nearest tenth · Q11
A length given as 14.6 to the nearest 0.1 lies between 14.55 and 14.65.
05
Two standard deviations · Q12
The speed 75.7 is $2 σ$ above 67.3, so $σ$ is half the difference.

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

Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 May 2023 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

6 marks

Decathlon: Spearman's rank correlation for two events

Statistics and probability

(a)

2 marks

Rank the high jumps with the highest as 1: 2.11 (1), 2.08 (2), 2.05 (3), 2.02 twice (4.5 each), 1.99 (6), 1.96 (7), 1.90 (8), 1.87 (9), 1.84 (10).

Germany 1, France 2, Estonia 10, Canada 4.5, Netherlands 3, Ukraine 4.5, Algeria 8, Austria 9, Grenada 6, Japan 7

(b)

2 marks

Spearman's coefficient is the correlation between the two sets of ranks.

r_{s} = 0.541

(c)

2 marks

A value of 0.541 lies in the range 0.400 to 0.599, so the correlation is moderate.
Athletes who rank higher (better) in the long jump tend, to a moderate extent, to rank higher in the high jump.

Question 2

6 marks

A hot-air balloon seen from two points

Geometry and trigonometry

(a)

2 marks

A H B = 180 - 130 - 33 = 1 7^{\circ}

(b)

3 marks

Sine rule in triangle ABH with $A B = 156$ .

\frac{B H}{sin 13 0 ^{\circ}} = \frac{156}{sin 1 7 ^{\circ}} \Rightarrow B H \approx 409 m

(c)

1 mark

The angle of depression from H to B decreases (it gets smaller) as the horizontal distance increases, because the balloon is at a constant height.

Question 3

5 marks

Savings with a deposit at the end of each month

Number and algebra

(a)

3 marks

1 January 2024 is 24 months after 1 January 2022. Monthly rate $\frac{4}{12} %$ .

A = 1000 (1 + \frac{0.04}{12})^{24} + 100 \cdot \frac{( 1 + \frac{0.04}{12} ) ^{24} - 1}{0.04/12} = $3577.43

(b)

2 marks

Test the balance after 36 and 37 months.
So it takes 37 complete months.

A_{36} = $4945.43 < 5000, A_{37} = $5061.91 > 5000

Question 4

6 marks

Memory scores: a one-tailed two-sample t-test

Statistics and probability

(a)

2 marks

The alternative hypothesis is one-sided because Carys believes bilingual people will score higher.

H_{0} : μ_{b} = μ_{m}, H_{1} : μ_{b} > μ_{m}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test (10 and 8 values, so $ν = 16$ ).

p \approx 0.119

(c)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level that the mean memory score of bilingual people is higher than that of monolingual people.

0.119 > 0.05

Question 5

5 marks

A tangent, a perpendicular line and whether it is the normal

Functions

(a)

1 mark

L2 has gradient $- \frac{1}{2}$ and L1 is perpendicular to it, so the gradients multiply to $- 1$ .

m_{1} = 2

(b)

2 marks

L1 passes through $P (3, - 1)$ .

y + 1 = 2 (x - 3) \Rightarrow y = 2 x - 7

(c)

2 marks

Put $x = 3$ into the equation of L2.
L2 does not pass through P, so it is not the normal to $f$ at P.

y = - \frac{1}{2} (3) - \frac{5}{2} = - 4 \neq = - 1

Question 6

5 marks

Stopping distance and speed: direct proportion to the square

Number and algebra

(a)

2 marks

Use $v = 50$ and $d = 12.3$ .

12.3 = k (50)^{2} \Rightarrow k = 0.00492 \Rightarrow d = 0.00492 v^{2}

(b)

2 marks

Set $d = 33$ and solve for $v$ .

33 = 0.00492 v^{2} \Rightarrow v = \frac{33}{0.00492} \approx 81.9 km h^{- 1}

(c)

1 mark

The state of the road surface (for example wet or dry) would affect the stopping distance. Other possible answers are the mass of the car, the condition of the tyres or the slope of the road.

Question 7

6 marks

An open box from a sheet of cardboard: maximum volume

Calculus

(a)

1 mark

The side of each square cannot be more than half of the shorter side, which is 30 cm.

k = 15

(b)

2 marks

Find the maximum of $V (x) = (48 - 2 x) (30 - 2 x) x$ on $0 < x < 15$ with a graph or by differentiating.

x = 6 cm

(c)

3 marks

The remaining cardboard is $(48 - 4)$ by $(30 - 4)$ , that is 44 cm by 26 cm.
Find the maximum on $0 < x < 13$ .

V (x) = (44 - 2 x) (26 - 2 x) x x = 5.28, V_{m a x} \approx 2730 cm^{3}

Question 8

7 marks

Password entropy: a logarithmic relationship

Functions

(a)

2 marks

Substitute $G = 5000$ .

0.301 p = lo g_{10} 5000 = 3.699 \Rightarrow p = 12.3 bits

(b)

1 mark

Undo the logarithm.

G (p) = 1 0^{0.301 p}

(c)

3 marks

Substitute $p = 28$ .

G = 1 0^{0.301 \times 28} = 1 0^{8.428} \approx 2.68 \times 1 0^{8}

(d)

1 mark

The point $(0, 1)$ means a password with entropy 0 bits can be decoded in 1 guess. It is completely predictable.

Question 9

8 marks

A hill profile: trapezoidal rule, exact area and percentage error

Calculus

(a)

2 marks

Trapezoidal rule with $h = 10$ and the five heights.

A \approx \frac{10}{2} [0 + 0 + 2 (3 + 8 + 9)] = 200 cm^{2}

(b)

4 marks

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

A = \int_{0}^{40} (0.04 x^{2} - 0.001 x^{3}) d x A = [\frac{0.04 x ^{3}}{3} - \frac{0.001 x ^{4}}{4}]_{0}^{40} = 853.33 - 640 = 213.33 cm^{2}

(c)

2 marks

Percentage error compares the trapezoidal estimate with the exact area.

\frac{∣200 - 213.33∣}{213.33} \times 100 = 6.25%

Question 10

7 marks

Baseball: a strike and a ball against the wall

Functions

(a)

4 marks

The gradient of the path is $- 0.045$ .
The ball is 1.25 m high at the front of the plate and about $1.25 - 0.0194 = 1.2307$ m at the back. The back height is between 0.53 m and 1.24 m, so at some point over the plate the ball is in the strike zone: it is a strike.

a = - 0.045 \times 0.43 = - 0.01935 (\approx - 0.0194)

(b)

3 marks

Find the height at the wall, $d = 96$ .
Since $8.34 > 5$ , the ball is above the 5 m wall when it reaches it, so it goes over the wall.

h (96) = - 0.01 (96)^{2} + 1.04 (96) + 0.66 = 8.34 m

Question 11

7 marks

A park post: bounds on a length and a safety angle

Geometry and trigonometry

(a)

2 marks

The length of AC is given to the nearest tenth of a centimetre.

lower bound 14.55 cm, upper bound 14.65 cm

(b)

3 marks

In triangle ABC the line AH to the midpoint of BC is perpendicular to BC, and $B H = 10$ . Use Pythagoras.

A H = A C^{2} - 1 0^{2} upper bound 14.6 5^{2} - 100 = 10.7 cm, lower bound 14.5 5^{2} - 100 = 10.6 cm

(c)

2 marks

The angle $α$ between the slant height and the base satisfies $cos α = \frac{10}{A H}$ . It is largest when $A H$ is largest.
Even the largest possible angle, $20. 9^{\circ}$ , is less than $2 2^{\circ}$ , so the post is safe.

α_{m a x} = arccos \frac{10}{10.706} = 20. 9^{\circ}

Question 12

7 marks

Car speeds: the standard deviation, speeding tickets and an asymmetric interval

Statistics and probability

(a)

2 marks

The speed 75.7 is two standard deviations above the mean 67.3.

2 σ = 75.7 - 67.3 = 8.4 \Rightarrow σ = 4.2

(b)

2 marks

Use $X \sim N (67.3, 4. 2^{2})$ .

P (X > 72) = 0.132

(c)

3 marks

A central region with 82% leaves 9% in each tail, so it would be $67.3 \pm 1.341 (4.2) = 67.3 \pm 5.63$ .
The speed 74 km/h is in the 82% interval from $p$ to $q$ but is above 72.9. So the interval is not this symmetric one: it is not symmetrical about the mean.

(61.7, 72.9)

Question 13

5 marks

A boat's return bearing

Number and algebra

5 marks

Take the starting point as the origin. After 8 km on $31 5^{\circ}$ and 6 km on $04 5^{\circ}$ :
The boat is 1.414 km west and 9.899 km north of the start, so to return it must go east 1.414 km and south 9.899 km.
The direction is south and slightly east, so the bearing is

east = - 8 sin 4 5^{\circ} + 6 sin 4 5^{\circ} = - 1.414, north = 8 cos 4 5^{\circ} + 6 cos 4 5^{\circ} = 9.899 tan α = \frac{1.414}{9.899} \Rightarrow α = 8.1 3^{\circ} 18 0^{\circ} - 8.1 3^{\circ} \approx 17 2^{\circ}

Our analysis

Where the marks are

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

Functions19 marks · 24%
Q5, Q8, Q10
Statistics and probability19 marks · 24%
Q1, Q4, Q12
Number and algebra15 marks · 19%
Q3, Q6, Q13
Calculus14 marks · 18%
Q7, Q9
Geometry and trigonometry13 marks · 16%
Q2, Q11

Mentor insights

Traps to watch for

01
Undo the log · Q8
$lo g_{10} G = 0.301 p$ means $G = 1 0^{0.301 p}$ .
02
Exact area uses the equation · Q9
Integrate the given cubic from 0 to 40.
03
Height change over the plate · Q10
The ball is over the plate for 0.43 m of horizontal distance, so the height changes by gradient times 0.43.
04
Bounds for a length to the nearest tenth · Q11
A length given as 14.6 to the nearest 0.1 lies between 14.55 and 14.65.
05
Two standard deviations · Q12
The speed 75.7 is $2 σ$ above 67.3, so $σ$ is half the difference.

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