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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 TZ2 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

5 marks

Percentage error and standard form for a bridge's steel

Number and algebra

(a)

2 marks

Percentage error.

\frac{∣53632000 - 55625000∣}{55625000} \times 100 \approx 3.58%

(b)

3 marks

Five identical bridges.

5 \times 55625000 = 278125000 \approx 278000000 kg 2.78 \times 1 0^{8} kg

Question 2

6 marks

Savings with quarterly interest or a phone that loses value

Number and algebra

(a)

3 marks

Compound interest with 20 quarters.

A = 520 (1 + \frac{0.012}{4})^{20} = $552.11

(b)

3 marks

The phone's value falls from $520 to $30 in 5 years.

520 (1 - r)^{5} = 30 \Rightarrow 1 - r = (\frac{30}{520})^{1/5} r = 0.435 (43.5%)

Question 3

7 marks

Times to solve a problem: quartiles from a cumulative frequency graph and an outlier

Statistics and probability

(a)

4 marks

Read the graph at cumulative frequency 100, 50 and 150.

median = 38 s, Q_{1} = 32 s, Q_{3} = 42 s IQR = 42 - 32 = 10 s

(b)

3 marks

Find the lower fence.
Since $14 < 17$ , Cedric's time is an outlier.

32 - 1.5 (10) = 17

Question 4

6 marks

Age and 100 m time: ranks, Spearman's coefficient and interpretation

Statistics and probability

(a)

2 marks

Rank the ages with the oldest as 1, and the times with the slowest as 1.

Age rank: 7, 6, 3, 5, 4, 2, 8, 1 Time rank: 3.5, 2, 3.5, 6, 7, 8, 1, 5

(b)

2 marks

Spearman's coefficient is the correlation of the ranks.

r_{s} = - 0.671

(c)

1 mark

The coefficient is negative: older athletes in this sample tend to have faster (shorter) 100 m times.

(d)

1 mark

The relationship between age and time is unlikely to be linear, and an outlier (such as 17.0 s for the youngest athlete) would affect Pearson's coefficient.

Question 5

4 marks

Test grades: a frequency table with an unknown frequency

Statistics and probability

(a)

1 mark

Add the frequencies.

1 + 4 + 7 + 9 + p + 9 + 4 = 34 + p

(b)

3 marks

$\sum f x = 1 + 8 + 21 + 36 + 5 p + 54 + 28 = 148 + 5 p$ .

\frac{148 + 5 p}{34 + p} = 4.5 \Rightarrow 148 + 5 p = 153 + 4.5 p \Rightarrow p = 10

Question 6

6 marks

Restaurant food quality: conditional probability and a chi-squared test

Statistics and probability

(a)

2 marks

Among the 232 breakfast items, 101 are perfect.

P (not perfect ∣ breakfast) = 1 - \frac{101}{232} = \frac{131}{232} \approx 0.565

(b)

2 marks

Run a $χ^{2}$ test for independence on the $3 \times 3$ table of observed values, with $ν = 4$ .

χ^{2} \approx 11.0

(c)

2 marks

Compare with the critical value.
Reject $H_{0}$ : there is significant evidence at the 5% level that the quality of the food depends on the meal.

11.0 > 9.488

Question 7

6 marks

Delivery regions for four cafes: a Voronoi diagram

Geometry and trigonometry

(a)

3 marks

The gradient of CD is $\frac{1}{3}$ , so the perpendicular bisector has gradient $- 3$ . It passes through the midpoint $(5.5, 1.5)$ .

y - 1.5 = - 3 (x - 5.5) \Rightarrow y = - 3 x + 18

(b)

1 mark

The boundary between A and D is the perpendicular bisector of AD, a vertical line that crosses the $x$ -axis at 2.5. Draw it up to the point where it meets the other boundaries (not into the region of B).

(c)

2 marks

The office is where the bisectors of CD and BC meet.

- 3 x + 18 = \frac{2 x - 1.5}{3} \Rightarrow - 9 x + 54 = 2 x - 1.5 \Rightarrow x = 5.05 y = - 3 (5.045) + 18 = 2.86, (5.05, 2.86)

Question 8

5 marks

An ice cream scoop and the cone it sits on

Geometry and trigonometry

(a)

3 marks

Volume of the sphere of radius 3.4.
Four fifths of it is not inside the cone.

V = \frac{4}{3} π (3.4)^{3} = 164.6 cm^{3} \frac{4}{5} (164.6) \approx 132 cm^{3}

(b)

2 marks

The outside of the cone is its curved surface, with $r = 3$ and slant height $l = 11$ .

A = π r l = π (3) (11) \approx 104 cm^{2}

Question 9

6 marks

Mango seed lengths: normal probabilities and inverse normal

Statistics and probability

(a)

2 marks

Use $X \sim N (4, 0.2 5^{2})$ .

P (X < 3.7) = 0.115

(b)

2 marks

30% above $k$ means 70% below.

k = 4 + 0.25 (0.5244) \approx 4.13

(c)

2 marks

The interval $4 \pm m$ holds 60%, leaving 20% in each tail, so $P (d < 4 + m) = 0.8$ .

m = 0.25 \times 0.8416 \approx 0.210

Question 10

8 marks

A basketball's height: quadratic model, landing, a catch and its limitations

Functions

(a)

2 marks

The maximum of a quadratic is at $t = - \frac{b}{2 a}$ .

t = \frac{8.75}{2 ( 4.75 )} = 0.921 s

(b)

2 marks

The ball hits the ground when $h (t) = 0$ . Reject the negative root.

- 4.75 t^{2} + 8.75 t + 1.5 = 0 \Rightarrow t = 2 s

(c)

2 marks

Set $h (t) = 1.2$ .
The other root is negative, so it is not used.

- 4.75 t^{2} + 8.75 t + 0.3 = 0 \Rightarrow t = 1.88 s

(d)

2 marks

The model ignores air resistance and spin. It also gives negative heights after the ball would have hit the ground, so it is valid only until the ball lands.

Question 11

7 marks

f(x) = 3x^2 - 5/x: a graph for negative x, equal values and an asymptote

Functions

(a)

4 marks

For $x < 0$ the function is positive: $f (- 5) = 75 + 1 = 76$ , and $f (x) \to + \infty$ as $x \to 0^{-}$ . In between there is a local minimum, found with the calculator.
Sketch: start at about $(- 5, 76)$ , curve down to a smooth minimum at about $(- 0.94, 7.97)$ and then rise steeply towards the $y$ -axis without touching it.

x = - 0.941

(b)

2 marks

Find the intersections of $y = f (x)$ with $y = 20$ .

x = - 2.45, x = - 0.252, x = 2.70

(c)

1 mark

The function is undefined at $x = 0$ and tends to infinity there.

x = 0

Question 12

5 marks

A target game: a probability table and whether the game is fair

Statistics and probability

(a)

2 marks

All probabilities add to 1.

0.15 + 0.2 + k + 0.16 + 2 k + 0.25 = 1 \Rightarrow 0.76 + 3 k = 1 \Rightarrow k = 0.08

(b)

3 marks

The probabilities are $0.15, 0.2, 0.08, 0.16, 0.16, 0.25$ .
The expected gain is not zero, so the game is not fair (the player loses $0.12 per game on average).

E (gain) = - 4 (0.15) - 3 (0.2) - 1 (0.08) + 0 (0.16) + 1 (0.16) + 4 (0.25) = - 0.12

Question 13

9 marks

The cross-section of a dam: trapezoidal rule and exact area

Calculus

(a)

2 marks

Use the values at $x = 0, 1, 2, 3$ with $h = 1$ .

A \approx \frac{1}{2} (1) [3 + 42 + 2 (8 + 19)] = 49.5 m^{2}

(b)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 4$ .
Use $f (0) = 3$ .

f (x) = x^{3} + 4 x + c f (x) = x^{3} + 4 x + 3

(c)

3 marks

The area under the curve from 0 to 3:
The remaining part is a rectangle $1 \times 42$ and a triangle with base 2 and height 42.

\int_{0}^{3} (x^{3} + 4 x + 3) d x = 47.25 42 (1) + \frac{1}{2} (42) (2) = 84 total area = 47.25 + 84 \approx 131 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 probability34 marks · 43%
Q3, Q4, Q5, Q6, Q9, Q12
Functions15 marks · 19%
Q10, Q11
Number and algebra11 marks · 14%
Q1, Q2
Geometry and trigonometry11 marks · 14%
Q7, Q8
Calculus9 marks · 11%
Q13

Mentor insights

Traps to watch for

01
Quarterly compounding · Q2
The rate per quarter is $\frac{1.2}{4} %$ and there are $5 \times 4 = 20$ quarters.
02
Quartiles at 25%, 50% and 75% · Q3
With 200 students read the graph at cumulative frequency 50, 100 and 150.
03
Use the quadratic formula for the time · Q10
Reject the negative root because time starts at 0.
04
Negative x is a different branch · Q11
For $x < 0$ the term $- \frac{5}{x}$ is positive, so the graph comes down from large positive values near $x = 0$ .
05
Trapezoidal rule with three intervals · Q13
The step is $h = 1$ with ordinates at $x = 0, 1, 2, 3$ .

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 TZ2 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

5 marks

Percentage error and standard form for a bridge's steel

Number and algebra

(a)

2 marks

Percentage error.

\frac{∣53632000 - 55625000∣}{55625000} \times 100 \approx 3.58%

(b)

3 marks

Five identical bridges.

5 \times 55625000 = 278125000 \approx 278000000 kg 2.78 \times 1 0^{8} kg

Question 2

6 marks

Savings with quarterly interest or a phone that loses value

Number and algebra

(a)

3 marks

Compound interest with 20 quarters.

A = 520 (1 + \frac{0.012}{4})^{20} = $552.11

(b)

3 marks

The phone's value falls from $520 to $30 in 5 years.

520 (1 - r)^{5} = 30 \Rightarrow 1 - r = (\frac{30}{520})^{1/5} r = 0.435 (43.5%)

Question 3

7 marks

Times to solve a problem: quartiles from a cumulative frequency graph and an outlier

Statistics and probability

(a)

4 marks

Read the graph at cumulative frequency 100, 50 and 150.

median = 38 s, Q_{1} = 32 s, Q_{3} = 42 s IQR = 42 - 32 = 10 s

(b)

3 marks

Find the lower fence.
Since $14 < 17$ , Cedric's time is an outlier.

32 - 1.5 (10) = 17

Question 4

6 marks

Age and 100 m time: ranks, Spearman's coefficient and interpretation

Statistics and probability

(a)

2 marks

Rank the ages with the oldest as 1, and the times with the slowest as 1.

Age rank: 7, 6, 3, 5, 4, 2, 8, 1 Time rank: 3.5, 2, 3.5, 6, 7, 8, 1, 5

(b)

2 marks

Spearman's coefficient is the correlation of the ranks.

r_{s} = - 0.671

(c)

1 mark

The coefficient is negative: older athletes in this sample tend to have faster (shorter) 100 m times.

(d)

1 mark

The relationship between age and time is unlikely to be linear, and an outlier (such as 17.0 s for the youngest athlete) would affect Pearson's coefficient.

Question 5

4 marks

Test grades: a frequency table with an unknown frequency

Statistics and probability

(a)

1 mark

Add the frequencies.

1 + 4 + 7 + 9 + p + 9 + 4 = 34 + p

(b)

3 marks

$\sum f x = 1 + 8 + 21 + 36 + 5 p + 54 + 28 = 148 + 5 p$ .

\frac{148 + 5 p}{34 + p} = 4.5 \Rightarrow 148 + 5 p = 153 + 4.5 p \Rightarrow p = 10

Question 6

6 marks

Restaurant food quality: conditional probability and a chi-squared test

Statistics and probability

(a)

2 marks

Among the 232 breakfast items, 101 are perfect.

P (not perfect ∣ breakfast) = 1 - \frac{101}{232} = \frac{131}{232} \approx 0.565

(b)

2 marks

Run a $χ^{2}$ test for independence on the $3 \times 3$ table of observed values, with $ν = 4$ .

χ^{2} \approx 11.0

(c)

2 marks

Compare with the critical value.
Reject $H_{0}$ : there is significant evidence at the 5% level that the quality of the food depends on the meal.

11.0 > 9.488

Question 7

6 marks

Delivery regions for four cafes: a Voronoi diagram

Geometry and trigonometry

(a)

3 marks

The gradient of CD is $\frac{1}{3}$ , so the perpendicular bisector has gradient $- 3$ . It passes through the midpoint $(5.5, 1.5)$ .

y - 1.5 = - 3 (x - 5.5) \Rightarrow y = - 3 x + 18

(b)

1 mark

(c)

2 marks

The office is where the bisectors of CD and BC meet.

- 3 x + 18 = \frac{2 x - 1.5}{3} \Rightarrow - 9 x + 54 = 2 x - 1.5 \Rightarrow x = 5.05 y = - 3 (5.045) + 18 = 2.86, (5.05, 2.86)

Question 8

5 marks

An ice cream scoop and the cone it sits on

Geometry and trigonometry

(a)

3 marks

Volume of the sphere of radius 3.4.
Four fifths of it is not inside the cone.

V = \frac{4}{3} π (3.4)^{3} = 164.6 cm^{3} \frac{4}{5} (164.6) \approx 132 cm^{3}

(b)

2 marks

The outside of the cone is its curved surface, with $r = 3$ and slant height $l = 11$ .

A = π r l = π (3) (11) \approx 104 cm^{2}

Question 9

6 marks

Mango seed lengths: normal probabilities and inverse normal

Statistics and probability

(a)

2 marks

Use $X \sim N (4, 0.2 5^{2})$ .

P (X < 3.7) = 0.115

(b)

2 marks

30% above $k$ means 70% below.

k = 4 + 0.25 (0.5244) \approx 4.13

(c)

2 marks

The interval $4 \pm m$ holds 60%, leaving 20% in each tail, so $P (d < 4 + m) = 0.8$ .

m = 0.25 \times 0.8416 \approx 0.210

Question 10

8 marks

A basketball's height: quadratic model, landing, a catch and its limitations

Functions

(a)

2 marks

The maximum of a quadratic is at $t = - \frac{b}{2 a}$ .

t = \frac{8.75}{2 ( 4.75 )} = 0.921 s

(b)

2 marks

The ball hits the ground when $h (t) = 0$ . Reject the negative root.

- 4.75 t^{2} + 8.75 t + 1.5 = 0 \Rightarrow t = 2 s

(c)

2 marks

Set $h (t) = 1.2$ .
The other root is negative, so it is not used.

- 4.75 t^{2} + 8.75 t + 0.3 = 0 \Rightarrow t = 1.88 s

(d)

2 marks

The model ignores air resistance and spin. It also gives negative heights after the ball would have hit the ground, so it is valid only until the ball lands.

Question 11

7 marks

f(x) = 3x^2 - 5/x: a graph for negative x, equal values and an asymptote

Functions

(a)

4 marks

For $x < 0$ the function is positive: $f (- 5) = 75 + 1 = 76$ , and $f (x) \to + \infty$ as $x \to 0^{-}$ . In between there is a local minimum, found with the calculator.
Sketch: start at about $(- 5, 76)$ , curve down to a smooth minimum at about $(- 0.94, 7.97)$ and then rise steeply towards the $y$ -axis without touching it.

x = - 0.941

(b)

2 marks

Find the intersections of $y = f (x)$ with $y = 20$ .

x = - 2.45, x = - 0.252, x = 2.70

(c)

1 mark

The function is undefined at $x = 0$ and tends to infinity there.

x = 0

Question 12

5 marks

A target game: a probability table and whether the game is fair

Statistics and probability

(a)

2 marks

All probabilities add to 1.

0.15 + 0.2 + k + 0.16 + 2 k + 0.25 = 1 \Rightarrow 0.76 + 3 k = 1 \Rightarrow k = 0.08

(b)

3 marks

The probabilities are $0.15, 0.2, 0.08, 0.16, 0.16, 0.25$ .
The expected gain is not zero, so the game is not fair (the player loses $0.12 per game on average).

E (gain) = - 4 (0.15) - 3 (0.2) - 1 (0.08) + 0 (0.16) + 1 (0.16) + 4 (0.25) = - 0.12

Question 13

9 marks

The cross-section of a dam: trapezoidal rule and exact area

Calculus

(a)

2 marks

Use the values at $x = 0, 1, 2, 3$ with $h = 1$ .

A \approx \frac{1}{2} (1) [3 + 42 + 2 (8 + 19)] = 49.5 m^{2}

(b)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 4$ .
Use $f (0) = 3$ .

f (x) = x^{3} + 4 x + c f (x) = x^{3} + 4 x + 3

(c)

3 marks

The area under the curve from 0 to 3:
The remaining part is a rectangle $1 \times 42$ and a triangle with base 2 and height 42.

\int_{0}^{3} (x^{3} + 4 x + 3) d x = 47.25 42 (1) + \frac{1}{2} (42) (2) = 84 total area = 47.25 + 84 \approx 131 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 probability34 marks · 43%
Q3, Q4, Q5, Q6, Q9, Q12
Functions15 marks · 19%
Q10, Q11
Number and algebra11 marks · 14%
Q1, Q2
Geometry and trigonometry11 marks · 14%
Q7, Q8
Calculus9 marks · 11%
Q13

Mentor insights

Traps to watch for

01
Quarterly compounding · Q2
The rate per quarter is $\frac{1.2}{4} %$ and there are $5 \times 4 = 20$ quarters.
02
Quartiles at 25%, 50% and 75% · Q3
With 200 students read the graph at cumulative frequency 50, 100 and 150.
03
Use the quadratic formula for the time · Q10
Reject the negative root because time starts at 0.
04
Negative x is a different branch · Q11
For $x < 0$ the term $- \frac{5}{x}$ is positive, so the graph comes down from large positive values near $x = 0$ .
05
Trapezoidal rule with three intervals · Q13
The step is $h = 1$ with ordinates at $x = 0, 1, 2, 3$ .

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