Loading…

Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Heights of basketball players: mean, median, mode, range and a rounded height

Statistics and probability

(a)

6 marks

(i) Add the six heights and divide by 6.

(ii) In order the heights are $1.60, 1.67, 1.68, 2.19, 2.31, 2.31$ . The median is the mean of the middle two.

(iii) The most common height.

(iv) The largest minus the smallest.

\overset{x}{ˉ} = \frac{1.67 + 1.60 + 1.68 + 2.31 + 2.31 + 2.19}{6} = \frac{11.76}{6} = 1.96 m \frac{1.68 + 2.19}{2} = 1.935 \approx 1.94 m 2.31 m 2.31 - 1.60 = 0.71 m

(b)

1 mark

A height of 1.98 m to the nearest centimetre means the true height is at least 1.975 m.

1.975 m

Question 2

6 marks

A Venn diagram of two cafes: counts, probability and mutually exclusive events

Statistics and probability

(a)

1 mark

Add every region of the Venn diagram: 10 (Thai only), 13 (both), 8 (Indian only) and 2 (neither).

10 + 13 + 8 + 2 = 33

(b)

1 mark

Those who have not eaten at the Indian cafe are in the Thai-only region or outside both circles.

10 + 2 = 12

(c)

1 mark

13 of the 33 students are in both circles.

\frac{13}{33} \approx 0.394

(d)

1 mark

$T \cup I$ is everyone who has eaten at one or both cafes, which is everyone except the 2 who have eaten at neither.

P (T \cup I) = \frac{31}{33} \approx 0.939

(e)

2 marks

$P (T \cap I) = \frac{13}{33} \neq = 0$ , because 13 students have eaten at both cafes. The events can both happen, so they are not mutually exclusive.

Question 3

7 marks

A new car's monthly sales as a geometric sequence

Number and algebra

(a)

2 marks

The twelfth term of a geometric sequence with $u_{1} = 40$ and $r = 1.1$ .

u_{12} = 40 (1.1)^{11} \approx 114

(b)

5 marks

(i) The total for 2025 is the sum of the first 12 terms.

(ii) The total for 2026 is months 13 to 24, so $S_{24} - S_{12}$ .

S_{12} = \frac{40 ( 1. 1 ^{12} - 1 )}{1.1 - 1} \approx 855 S_{24} = \frac{40 ( 1. 1 ^{24} - 1 )}{0.1} \approx 3539.9 S_{24} - S_{12} \approx 3539.9 - 855.4 \approx 2685

Question 4

7 marks

A phone charging: an exponential model for the percentage of power

Functions

(a)

2 marks

At 07:00, $t = 0$ .

P = 100 - 60 a^{0} = 100 - 60 = 40%

(b)

2 marks

At 08:00, $t = 1$ and $P = 75$ .

75 = 100 - 60 a^{- 1} \Rightarrow \frac{60}{a} = 25 \Rightarrow a = 2.4

(c)

2 marks

Plot $P = 100 - 60 (2.4)^{- t}$ .
Sketch: an increasing curve that is concave down. It starts at $(0, 40)$ , passes through $(1, 75)$ and flattens out as it approaches 100, getting to about 100 by $t = 5$ .

(d)

1 mark

The term $60 a^{- t}$ is always positive, so $P$ is always less than 100 and the graph has a horizontal asymptote at $P = 100$ .

Question 5

6 marks

A savings account and then an annuity: finding the rate and the balance

Number and algebra

(a)

3 marks

Compound growth over 96 months.

100000 (1 + \frac{I}{1200})^{96} = 150000 \Rightarrow (1 + \frac{I}{1200})^{96} = 1.5 I = 1200 (1. 5^{1/96} - 1) \approx 5.08

(b)

3 marks

Use the finance solver with $N = 120$ , $I = 6.1%$ , $P V = 150000$ , $P M T = - 1000$ and 12 payments a year.
Check by iterating the monthly rule $B \to B (1 + 0.061/12) - 1000$ one hundred and twenty times, which gives $110 866.71.

F V \approx $110 867

Question 6

6 marks

Perpendicular bisectors of a triangle and its Voronoi diagram

Geometry and trigonometry

(a)

2 marks

$B (9, 6)$ and $C (11, 2)$ have midpoint $(10, 4)$ and gradient $\frac{2 - 6}{11 - 9} = - 2$ . The bisector has gradient $\frac{1}{2}$ .

y - 4 = \frac{1}{2} (x - 10) \Rightarrow y = 0.5 x - 1

(b)

2 marks

Solve $0.5 x - 1 = - 3 x + 23$ .

3.5 x = 24 \Rightarrow x = 6.857, y = 0.5 (6.857) - 1 = 2.429 V \approx (6.857, 2.429)

(c)

2 marks

The Voronoi diagram has three edges, all starting at $V$ and going off to infinity, one on each bisector.
The edge between $A$ and $B$ is on $y = - 3 x + 23$ , going up and to the left of $V$ .
The edge between $B$ and $C$ is on $y = 0.5 x - 1$ , going up and to the right of $V$ .
The edge between $A$ and $C$ is on the bisector $y = 4 x - 25$ (midpoint $(7, 3)$ and gradient 4), going down and to the left of $V$ .
These directions are where each pair of sites are the two nearest. Because angle $B$ in the triangle is obtuse, $V$ lies outside triangle $A B C$ .

Question 7

6 marks

River depth modelled by a sine curve

Functions

(a)

1 mark

The midline is 4, the highest depth is 7, so the amplitude is

a = 7 - 4 = 3

(b)

2 marks

The highest depth is at 15:00 ( $t = 3$ ) and the lowest is at 21:00 ( $t = 9$ ). These are half a period apart, so the period is 12 hours.

\frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(c)

3 marks

Solve $3 sin (\frac{π t}{6}) + 4 = 3$ , so $sin (\frac{π t}{6}) = - \frac{1}{3}$ .
The first solution after $t = 0$ is in the third quadrant of the sine curve: $\frac{π t}{6} = π + sin^{- 1} \frac{1}{3}$ .
This is 18:39 (6:39 pm).

t = 6.649 hours = 6 h 39 min

Question 8

6 marks

Celsius and Fahrenheit: converting a formula and a mean and standard deviation

Functions
Statistics and probability

(a)

3 marks

(i) Rearrange $F = 1.8 C + 32$ .

(ii) Put $F = 77$ .

C = \frac{F - 32}{1.8} = \frac{5}{9} (F - 32) C = \frac{77 - 32}{1.8} = 25

(b)

3 marks

(i) The mean transforms like every value.

(ii) Adding 32 shifts every value and leaves the spread unchanged. The standard deviation is only multiplied by 1.8.

1.8 (17) + 32 = 62.6^{\circ} F 1.8 \times 9 = 16.2^{\circ} F

Question 9

8 marks

A cube of sugar dissolving: a linear model and a cubic volume model

Calculus

(a)

3 marks

(i) Each side drops by 0.2 mm every second, for 20 seconds.

(ii) The volume of a cube with side 6 mm.

10 - 0.2 \times 20 = 6 mm 6^{3} = 216 mm^{3}

(b)

2 marks

Differentiate $V (t) = 1000 - 60 t + 1.2 t^{2} - 0.008 t^{3}$ .

V^{'} (t) = - 60 + 2.4 t - 0.024 t^{2}

(c)

2 marks

Substitute $t = 20$ .

V^{'} (20) = - 60 + 2.4 (20) - 0.024 (20)^{2} = - 60 + 48 - 9.6 = - 21.6 mm^{3} s^{- 1}

(d)

1 mark

The temperature of the coffee falls as time passes, so the sugar dissolves more slowly. The surface of the block exposed to the coffee also changes as it shrinks, or the cube could break up into smaller pieces. Any of these would stop the side length decreasing at a steady rate.

Question 10

8 marks

Big cats: mass against shoulder height, and energy needs

Functions

(a)

4 marks

(i) $m = a h^{3}$ , and for the cheetah $64 = a (0.8)^{3}$ .

(ii) Substitute $h = 0.75$ .

a = \frac{64}{0.512} = 125 \Rightarrow m = 125 h^{3} m = 125 (0.75)^{3} \approx 52.7 kg

(b)

4 marks

Find the shoulder height of the lion from its mass.
Since $E = c h^{2}$ , the ratio of the lion's to the cheetah's energy needs is $(\frac{h _{L}}{h _{C}})^{2}$ .

220 = 125 h^{3} \Rightarrow h = (1.76)^{1/3} \approx 1.207 m k = (\frac{1.2074}{0.8})^{2} \approx 2.28

Question 11

5 marks

The arc BQ of a semicircle containing a rectangle

Number and algebra

5 marks

The radius is the length $O B$ .
The angle $B O P$ is in the right triangle with sides 5 and 8.
The semicircle angle is $π$ , so the angle $B O Q$ is the rest.
Arc length with radians.

r = 5^{2} + 8^{2} = 89 \approx 9.434 B O P = tan^{- 1} \frac{5}{8} \approx 0.5586 B O Q = π - 0.5586 \approx 2.583 B Q = r θ = 89 (2.583) \approx 24.4 m

Question 12

8 marks

A spinner game for a library: expected winnings and a fair game

Statistics and probability

(a)

3 marks

(i) Winning is positive and losing is negative.

(ii) In the long run a player loses an average of $2.80 per game, so the game is not fair (it favours the library).

E (X) = 5 (0.4) + (- 5) (0.2) + (- 8) (0.1) + (- 10) (0.3) = 2 - 1 - 0.8 - 3 = - 2.80

(b)

5 marks

(i) In game $k$ the prize is $5 + (k - 1)$ dollars. The game is fair when $E (X) = 0$ .

(ii) For each of the first seven games the prize is less than $12, so $E (X) < 0$ and a player loses on average, and the eighth game is fair (expected value 0). So over the 8 games Emily plays, Zac expects to make money.

(4 + k) (0.4) - 1 - 0.8 - 3 = 0 \Rightarrow (4 + k) (0.4) = 4.8 \Rightarrow 4 + k = 12 k = 8

Our analysis

Where the marks are

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

Functions24 marks · 30%
Q4, Q7, Q8, Q10
Statistics and probability24 marks · 30%
Q1, Q2, Q8, Q12
Number and algebra18 marks · 23%
Q3, Q5, Q11
Calculus8 marks · 10%
Q9
Geometry and trigonometry6 marks · 8%
Q6

Mentor insights

Traps to watch for

01
Rounding · Q1
A height given to the nearest centimetre can be up to 0.5 cm lower than the stated value.
02
Month and year · Q3
December 2025 is month 12. The 2026 total is the sum of months 13 to 24.
03
Volume of a cube · Q9
$V = s^{3}$ where $s$ is the side length.
04
Direct proportion to a power · Q10
$m = a h^{3}$ with a constant found from the data.
05
Expected value · Q12
$E (X) = \sum x P (X = x)$ . A game is fair if $E (X) = 0$ .

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 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Heights of basketball players: mean, median, mode, range and a rounded height

Statistics and probability

(a)

6 marks

(i) Add the six heights and divide by 6.

(ii) In order the heights are $1.60, 1.67, 1.68, 2.19, 2.31, 2.31$ . The median is the mean of the middle two.

(iii) The most common height.

(iv) The largest minus the smallest.

\overset{x}{ˉ} = \frac{1.67 + 1.60 + 1.68 + 2.31 + 2.31 + 2.19}{6} = \frac{11.76}{6} = 1.96 m \frac{1.68 + 2.19}{2} = 1.935 \approx 1.94 m 2.31 m 2.31 - 1.60 = 0.71 m

(b)

1 mark

A height of 1.98 m to the nearest centimetre means the true height is at least 1.975 m.

1.975 m

Question 2

6 marks

A Venn diagram of two cafes: counts, probability and mutually exclusive events

Statistics and probability

(a)

1 mark

Add every region of the Venn diagram: 10 (Thai only), 13 (both), 8 (Indian only) and 2 (neither).

10 + 13 + 8 + 2 = 33

(b)

1 mark

Those who have not eaten at the Indian cafe are in the Thai-only region or outside both circles.

10 + 2 = 12

(c)

1 mark

13 of the 33 students are in both circles.

\frac{13}{33} \approx 0.394

(d)

1 mark

$T \cup I$ is everyone who has eaten at one or both cafes, which is everyone except the 2 who have eaten at neither.

P (T \cup I) = \frac{31}{33} \approx 0.939

(e)

2 marks

$P (T \cap I) = \frac{13}{33} \neq = 0$ , because 13 students have eaten at both cafes. The events can both happen, so they are not mutually exclusive.

Question 3

7 marks

A new car's monthly sales as a geometric sequence

Number and algebra

(a)

2 marks

The twelfth term of a geometric sequence with $u_{1} = 40$ and $r = 1.1$ .

u_{12} = 40 (1.1)^{11} \approx 114

(b)

5 marks

(i) The total for 2025 is the sum of the first 12 terms.

(ii) The total for 2026 is months 13 to 24, so $S_{24} - S_{12}$ .

S_{12} = \frac{40 ( 1. 1 ^{12} - 1 )}{1.1 - 1} \approx 855 S_{24} = \frac{40 ( 1. 1 ^{24} - 1 )}{0.1} \approx 3539.9 S_{24} - S_{12} \approx 3539.9 - 855.4 \approx 2685

Question 4

7 marks

A phone charging: an exponential model for the percentage of power

Functions

(a)

2 marks

At 07:00, $t = 0$ .

P = 100 - 60 a^{0} = 100 - 60 = 40%

(b)

2 marks

At 08:00, $t = 1$ and $P = 75$ .

75 = 100 - 60 a^{- 1} \Rightarrow \frac{60}{a} = 25 \Rightarrow a = 2.4

(c)

2 marks

Plot $P = 100 - 60 (2.4)^{- t}$ .
Sketch: an increasing curve that is concave down. It starts at $(0, 40)$ , passes through $(1, 75)$ and flattens out as it approaches 100, getting to about 100 by $t = 5$ .

(d)

1 mark

The term $60 a^{- t}$ is always positive, so $P$ is always less than 100 and the graph has a horizontal asymptote at $P = 100$ .

Question 5

6 marks

A savings account and then an annuity: finding the rate and the balance

Number and algebra

(a)

3 marks

Compound growth over 96 months.

100000 (1 + \frac{I}{1200})^{96} = 150000 \Rightarrow (1 + \frac{I}{1200})^{96} = 1.5 I = 1200 (1. 5^{1/96} - 1) \approx 5.08

(b)

3 marks

Use the finance solver with $N = 120$ , $I = 6.1%$ , $P V = 150000$ , $P M T = - 1000$ and 12 payments a year.
Check by iterating the monthly rule $B \to B (1 + 0.061/12) - 1000$ one hundred and twenty times, which gives $110 866.71.

F V \approx $110 867

Question 6

6 marks

Perpendicular bisectors of a triangle and its Voronoi diagram

Geometry and trigonometry

(a)

2 marks

$B (9, 6)$ and $C (11, 2)$ have midpoint $(10, 4)$ and gradient $\frac{2 - 6}{11 - 9} = - 2$ . The bisector has gradient $\frac{1}{2}$ .

y - 4 = \frac{1}{2} (x - 10) \Rightarrow y = 0.5 x - 1

(b)

2 marks

Solve $0.5 x - 1 = - 3 x + 23$ .

3.5 x = 24 \Rightarrow x = 6.857, y = 0.5 (6.857) - 1 = 2.429 V \approx (6.857, 2.429)

(c)

2 marks

The Voronoi diagram has three edges, all starting at $V$ and going off to infinity, one on each bisector.
The edge between $A$ and $B$ is on $y = - 3 x + 23$ , going up and to the left of $V$ .
The edge between $B$ and $C$ is on $y = 0.5 x - 1$ , going up and to the right of $V$ .
The edge between $A$ and $C$ is on the bisector $y = 4 x - 25$ (midpoint $(7, 3)$ and gradient 4), going down and to the left of $V$ .
These directions are where each pair of sites are the two nearest. Because angle $B$ in the triangle is obtuse, $V$ lies outside triangle $A B C$ .

Question 7

6 marks

River depth modelled by a sine curve

Functions

(a)

1 mark

The midline is 4, the highest depth is 7, so the amplitude is

a = 7 - 4 = 3

(b)

2 marks

The highest depth is at 15:00 ( $t = 3$ ) and the lowest is at 21:00 ( $t = 9$ ). These are half a period apart, so the period is 12 hours.

\frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(c)

3 marks

Solve $3 sin (\frac{π t}{6}) + 4 = 3$ , so $sin (\frac{π t}{6}) = - \frac{1}{3}$ .
The first solution after $t = 0$ is in the third quadrant of the sine curve: $\frac{π t}{6} = π + sin^{- 1} \frac{1}{3}$ .
This is 18:39 (6:39 pm).

t = 6.649 hours = 6 h 39 min

Question 8

6 marks

Celsius and Fahrenheit: converting a formula and a mean and standard deviation

Functions
Statistics and probability

(a)

3 marks

(i) Rearrange $F = 1.8 C + 32$ .

(ii) Put $F = 77$ .

C = \frac{F - 32}{1.8} = \frac{5}{9} (F - 32) C = \frac{77 - 32}{1.8} = 25

(b)

3 marks

(i) The mean transforms like every value.

(ii) Adding 32 shifts every value and leaves the spread unchanged. The standard deviation is only multiplied by 1.8.

1.8 (17) + 32 = 62.6^{\circ} F 1.8 \times 9 = 16.2^{\circ} F

Question 9

8 marks

A cube of sugar dissolving: a linear model and a cubic volume model

Calculus

(a)

3 marks

(i) Each side drops by 0.2 mm every second, for 20 seconds.

(ii) The volume of a cube with side 6 mm.

10 - 0.2 \times 20 = 6 mm 6^{3} = 216 mm^{3}

(b)

2 marks

Differentiate $V (t) = 1000 - 60 t + 1.2 t^{2} - 0.008 t^{3}$ .

V^{'} (t) = - 60 + 2.4 t - 0.024 t^{2}

(c)

2 marks

Substitute $t = 20$ .

V^{'} (20) = - 60 + 2.4 (20) - 0.024 (20)^{2} = - 60 + 48 - 9.6 = - 21.6 mm^{3} s^{- 1}

(d)

1 mark

Question 10

8 marks

Big cats: mass against shoulder height, and energy needs

Functions

(a)

4 marks

(i) $m = a h^{3}$ , and for the cheetah $64 = a (0.8)^{3}$ .

(ii) Substitute $h = 0.75$ .

a = \frac{64}{0.512} = 125 \Rightarrow m = 125 h^{3} m = 125 (0.75)^{3} \approx 52.7 kg

(b)

4 marks

Find the shoulder height of the lion from its mass.
Since $E = c h^{2}$ , the ratio of the lion's to the cheetah's energy needs is $(\frac{h _{L}}{h _{C}})^{2}$ .

220 = 125 h^{3} \Rightarrow h = (1.76)^{1/3} \approx 1.207 m k = (\frac{1.2074}{0.8})^{2} \approx 2.28

Question 11

5 marks

The arc BQ of a semicircle containing a rectangle

Number and algebra

5 marks

The radius is the length $O B$ .
The angle $B O P$ is in the right triangle with sides 5 and 8.
The semicircle angle is $π$ , so the angle $B O Q$ is the rest.
Arc length with radians.

r = 5^{2} + 8^{2} = 89 \approx 9.434 B O P = tan^{- 1} \frac{5}{8} \approx 0.5586 B O Q = π - 0.5586 \approx 2.583 B Q = r θ = 89 (2.583) \approx 24.4 m

Question 12

8 marks

A spinner game for a library: expected winnings and a fair game

Statistics and probability

(a)

3 marks

(i) Winning is positive and losing is negative.

(ii) In the long run a player loses an average of $2.80 per game, so the game is not fair (it favours the library).

E (X) = 5 (0.4) + (- 5) (0.2) + (- 8) (0.1) + (- 10) (0.3) = 2 - 1 - 0.8 - 3 = - 2.80

(b)

5 marks

(i) In game $k$ the prize is $5 + (k - 1)$ dollars. The game is fair when $E (X) = 0$ .

(4 + k) (0.4) - 1 - 0.8 - 3 = 0 \Rightarrow (4 + k) (0.4) = 4.8 \Rightarrow 4 + k = 12 k = 8

Our analysis

Where the marks are

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

Functions24 marks · 30%
Q4, Q7, Q8, Q10
Statistics and probability24 marks · 30%
Q1, Q2, Q8, Q12
Number and algebra18 marks · 23%
Q3, Q5, Q11
Calculus8 marks · 10%
Q9
Geometry and trigonometry6 marks · 8%
Q6

Mentor insights

Traps to watch for

01
Rounding · Q1
A height given to the nearest centimetre can be up to 0.5 cm lower than the stated value.
02
Month and year · Q3
December 2025 is month 12. The 2026 total is the sum of months 13 to 24.
03
Volume of a cube · Q9
$V = s^{3}$ where $s$ is the side length.
04
Direct proportion to a power · Q10
$m = a h^{3}$ with a constant found from the data.
05
Expected value · Q12
$E (X) = \sum x P (X = x)$ . A game is fair if $E (X) = 0$ .

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