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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: November 2023
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 November 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Cycling time against hill gradient: regression and its limits

Statistics and probability

(a)

4 marks

Linear regression of $T$ on $G$ with the five points.
A value of $r$ this close to 1 shows a very strong positive linear correlation between gradient and time.

T = 0.799 G + 2.14 r = 0.996

(b)

2 marks

Substitute $G = 17$ .

T = 0.7988 (17) + 2.1397 \approx 15.7 minutes

(c)

1 mark

The line of $T$ on $G$ is meant for predicting time from gradient. To estimate a gradient from a time use the line of $G$ on $T$ . Also 22 minutes is outside the data range, so it would be extrapolation.

Question 2

6 marks

The Great Pyramid: bounds on a conversion factor and a volume

Number and algebra

(a)

2 marks

A value of 0.52 correct to two decimal places lies between 0.515 and 0.525.

lower bound 0.515 m, upper bound 0.525 m

(b)

4 marks

The smallest volume uses 1 REC $= 0.515$ m.
Volume of a square-based pyramid.

base = 440 \times 0.515 = 226.6 m, height = 280 \times 0.515 = 144.2 m V = \frac{1}{3} (226.6)^{2} (144.2) \approx 2.47 \times 1 0^{6} m^{3}

Question 3

6 marks

g(x) = 8/x + x^2/2: asymptote, derivative and where it increases

Functions
Calculus

(a)

1 mark

The term $\frac{8}{x}$ is not defined at $x = 0$ and tends to infinity there.

x = 0

(b)

3 marks

Write $g (x) = 8 x^{- 1} + \frac{1}{2} x^{2}$ and differentiate.

g^{'} (x) = - 8 x^{- 2} + x = - \frac{8}{x ^{2}} + x

(c)

2 marks

$g^{'} (x) = 0$ when $x = \frac{8}{x ^{2}}$ , so $x^{3} = 8$ and $x = 2$ . The derivative is positive for $x > 2$ .

x > 2 (the interval (2, \infty))

Question 4

8 marks

Voronoi diagrams of farmhouses: a bisector and the best place for a turbine

Geometry and trigonometry

(a)

1 mark

Read the point where the three boundaries meet on Diagram 1.

(4, 8)

(b)

3 marks

Midpoint of $A (0, 3)$ and $C (8, 13)$ is $(4, 8)$ . The gradient of AC is $\frac{13 - 3}{8 - 0} = \frac{5}{4}$ , so the bisector has gradient $- \frac{4}{5}$ .

y - 8 = - \frac{4}{5} (x - 4) \Rightarrow y = - 0.8 x + 11.2

(c)

4 marks

Find the distance from each vertex to its nearest farmhouses.
The vertex with the largest distance is $(\frac{9}{11}, \frac{116}{11})$ , so the turbine goes there, and the distance to the nearest farmhouse is

(4, 1) : to A 4^{2} + 2^{2} = 20 = 4.47 (11, 8) : to B or D 34 = 5.83 (\frac{9}{11}, \frac{116}{11}) : to A (\frac{9}{11})^{2} + (\frac{116}{11} - 3)^{2} = 7.59 7.59 km

Question 5

6 marks

A boat's journey with bearings: an angle and the return distance

Geometry and trigonometry

(a)

2 marks

At B the north line gives alternate angles: $18 0^{\circ} - 11 2^{\circ} = 6 8^{\circ}$ on one side and $4 1^{\circ}$ on the other.

A B C = 68 + 41 = 10 9^{\circ}

(b)

4 marks

C is due east of A, so the angle at A between AB (bearing $11 2^{\circ}$ ) and AC (bearing $09 0^{\circ}$ ) is $2 2^{\circ}$ . The angles in the triangle give
Use the sine rule with AC opposite the angle at B.

A C B = 180 - 109 - 22 = 4 9^{\circ} \frac{A C}{sin 10 9 ^{\circ}} = \frac{100}{sin 4 9 ^{\circ}} \Rightarrow A C \approx 125 km

Question 6

4 marks

An inverse value and the domain of an inverse

Functions

(a)

2 marks

Solve $h (x) = 1$ for $h (x) = \frac{2}{x - 1} + \frac{1}{2}$ .

\frac{2}{x - 1} = \frac{1}{2} \Rightarrow x - 1 = 4 \Rightarrow x = 17 h^{- 1} (1) = 17

(b)

2 marks

For $x > 1$ the fraction $\frac{2}{x - 1}$ takes every positive value, so $h (x) > \frac{1}{2}$ . The domain of $h^{- 1}$ is the range of $h$ .

x > \frac{1}{2}

Question 7

9 marks

A quadratic model through three points and a fence

Calculus

(a)

1 mark

The point $(3, 2.36)$ is on the curve.

2.36 = 9 a + 3 b + c

(b)

3 marks

Do the same for the other points.
Solve the three equations.

5 = 100 a + 10 b + c, 7.16 = 289 a + 17 b + c a = - 0.00490, b = 0.441, c = 1.08 f (x) = - 0.00490 x^{2} + 0.441 x + 1.08

(c)

3 marks

Evaluate the model at the fence, 80 m away.
Since $5.00 > 4$ , the ball is above the 4 m fence when it reaches it, so it goes over.

f (80) = - 0.0049 (6400) + 0.441 (80) + 1.08 = 5.00

(d)

2 marks

Solve $f (x) = 0$ and take the positive root.

- 0.0049 x^{2} + 0.441 x + 1.08 = 0 \Rightarrow x \approx 92.4 m

Question 8

7 marks

A road modelled by a cubic and a bridge normal to it (fire station at A)

Calculus

(a)

2 marks

Use the calculator to find the gradient of $r (x) = (x + 1)^{3} + 2 x^{2} - 4 x$ at $x = - 1$ .

r^{'} (- 1) = - 8

(b)

2 marks

The normal gradient is $\frac{1}{8}$ . It passes through $A (- 1, 6)$ .

y - 6 = \frac{1}{8} (x + 1)

(c)

3 marks

Find where the normal meets the road: solve $r (x) = 6 + \frac{x + 1}{8}$ . The solution other than $x = - 1$ on the lake side is $x = - 5.02$ .
The length of the bridge is the distance AB.

B \approx (- 5.02, 5.50) A B = (- 1 + 5.02)^{2} + (6 - 5.50)^{2} \approx 4.05 km

Question 9

8 marks

Two classes' history grades: box plots and a t-test on satisfaction scores

Statistics and probability

(a)

2 marks

From the diagram, Class A has $Q_{1} = 72$ and $Q_{3} = 85$ (IQR 13), and Class B has $Q_{1} = 67$ and $Q_{3} = 86$ (IQR 19).
Class B has a lower quartile below 70, so more than 25% of Class B scored under 70, while fewer than 25% of Class A did. So "a higher percentage of Class B received a grade less than 70" must be true.
The IQR of Class A (13) is less than that of Class B (19), so the last statement must be true.

(b)

2 marks

The null hypothesis says the population mean satisfaction is the same in the two classes.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} \neq = μ_{B}

(c)

2 marks

Enter the two samples into a two-sample $t$ -test.

p \approx 0.111

(d)

2 marks

Compare $p$ with the 5% level.
Do not reject $H_{0}$ : there is insufficient evidence of a difference in satisfaction between the classes.

0.111 > 0.05

Mentor tip

The unpooled version of the test gives $p = 0.097$ , which leads to the same conclusion.

Question 10

7 marks

A coin and a die game: a tree diagram and a conditional probability

Statistics and probability

(a)

2 marks

The missing probabilities: tails $\frac{1}{2}$ ; on tails the die gives a win with probability $\frac{2}{6} = \frac{1}{3}$ and no win with probability $\frac{2}{3}$ .

(b)

2 marks

No prize needs tails and then not a five or six.

P (no win) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

(c)

3 marks

Use conditional probability. Winning happens with heads ( $\frac{1}{2}$ ) or tails then five or six ( $\frac{1}{2} \times \frac{1}{3}$ ).

P (heads ∣ win) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}

Question 11

6 marks

Toktam's car and the weather: binomial probabilities over a week

Statistics and probability

(a)

2 marks

Let $X \sim B (5, 0.65)$ be the number of mornings the car starts.

P (X = 2) = (2 5) (0.65)^{2} (0.35)^{3} \approx 0.181

(b)

4 marks

The probability of taking a taxi on a morning is the car failing and it raining.
No taxi on any of the five mornings:

P (taxi) = 0.35 \times 0.45 = 0.1575, P (no taxi) = 0.8425 0.842 5^{5} \approx 0.424

Question 12

6 marks

Social media followers and album sales: Spearman's rank correlation test

Statistics and probability

(a)

1 mark

Rank the albums sold, with the largest as 1.

2, 4, 8, 3, 5, 6, 7, 1

(b)

2 marks

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

r_{s} = 0.595

(c)

1 mark

Thurston believes that more followers means more albums sold.

H_{1} : in the population there is a positive monotonic relationship between followers and albums sold

(d)

2 marks

Compare $r_{s}$ with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence of a positive monotonic relationship.

0.595 < 0.643

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%
Q1, Q9, Q10, Q11, Q12
Calculus21 marks · 26%
Q3, Q7, Q8
Geometry and trigonometry14 marks · 18%
Q4, Q5
Number and algebra6 marks · 8%
Q2
Functions5 marks · 6%
Q3, Q6

Mentor insights

Traps to watch for

01
Regress T on G only to predict T · Q1
The line of $T$ on $G$ is for estimating time from gradient, not the other way round.
02
Bisector of AC · Q4
Use the midpoint and the negative reciprocal gradient.
03
Three points give three equations · Q7
Substitute each point into $f (x) = a x^{2} + b x + c$ and solve the system.
04
Normal gradient · Q8
The bridge is perpendicular to the road at A, so its gradient is $- \frac{1}{r ^{'} ( - 1 )}$ .
05
Read the box plot carefully · Q9
It gives quartiles, medians and range, not normality, and not exact numbers of students above a score.

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: November 2023
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 November 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Cycling time against hill gradient: regression and its limits

Statistics and probability

(a)

4 marks

Linear regression of $T$ on $G$ with the five points.
A value of $r$ this close to 1 shows a very strong positive linear correlation between gradient and time.

T = 0.799 G + 2.14 r = 0.996

(b)

2 marks

Substitute $G = 17$ .

T = 0.7988 (17) + 2.1397 \approx 15.7 minutes

(c)

1 mark

Question 2

6 marks

The Great Pyramid: bounds on a conversion factor and a volume

Number and algebra

(a)

2 marks

A value of 0.52 correct to two decimal places lies between 0.515 and 0.525.

lower bound 0.515 m, upper bound 0.525 m

(b)

4 marks

The smallest volume uses 1 REC $= 0.515$ m.
Volume of a square-based pyramid.

base = 440 \times 0.515 = 226.6 m, height = 280 \times 0.515 = 144.2 m V = \frac{1}{3} (226.6)^{2} (144.2) \approx 2.47 \times 1 0^{6} m^{3}

Question 3

6 marks

g(x) = 8/x + x^2/2: asymptote, derivative and where it increases

Functions
Calculus

(a)

1 mark

The term $\frac{8}{x}$ is not defined at $x = 0$ and tends to infinity there.

x = 0

(b)

3 marks

Write $g (x) = 8 x^{- 1} + \frac{1}{2} x^{2}$ and differentiate.

g^{'} (x) = - 8 x^{- 2} + x = - \frac{8}{x ^{2}} + x

(c)

2 marks

$g^{'} (x) = 0$ when $x = \frac{8}{x ^{2}}$ , so $x^{3} = 8$ and $x = 2$ . The derivative is positive for $x > 2$ .

x > 2 (the interval (2, \infty))

Question 4

8 marks

Voronoi diagrams of farmhouses: a bisector and the best place for a turbine

Geometry and trigonometry

(a)

1 mark

Read the point where the three boundaries meet on Diagram 1.

(4, 8)

(b)

3 marks

Midpoint of $A (0, 3)$ and $C (8, 13)$ is $(4, 8)$ . The gradient of AC is $\frac{13 - 3}{8 - 0} = \frac{5}{4}$ , so the bisector has gradient $- \frac{4}{5}$ .

y - 8 = - \frac{4}{5} (x - 4) \Rightarrow y = - 0.8 x + 11.2

(c)

4 marks

Find the distance from each vertex to its nearest farmhouses.
The vertex with the largest distance is $(\frac{9}{11}, \frac{116}{11})$ , so the turbine goes there, and the distance to the nearest farmhouse is

(4, 1) : to A 4^{2} + 2^{2} = 20 = 4.47 (11, 8) : to B or D 34 = 5.83 (\frac{9}{11}, \frac{116}{11}) : to A (\frac{9}{11})^{2} + (\frac{116}{11} - 3)^{2} = 7.59 7.59 km

Question 5

6 marks

A boat's journey with bearings: an angle and the return distance

Geometry and trigonometry

(a)

2 marks

At B the north line gives alternate angles: $18 0^{\circ} - 11 2^{\circ} = 6 8^{\circ}$ on one side and $4 1^{\circ}$ on the other.

A B C = 68 + 41 = 10 9^{\circ}

(b)

4 marks

C is due east of A, so the angle at A between AB (bearing $11 2^{\circ}$ ) and AC (bearing $09 0^{\circ}$ ) is $2 2^{\circ}$ . The angles in the triangle give
Use the sine rule with AC opposite the angle at B.

A C B = 180 - 109 - 22 = 4 9^{\circ} \frac{A C}{sin 10 9 ^{\circ}} = \frac{100}{sin 4 9 ^{\circ}} \Rightarrow A C \approx 125 km

Question 6

4 marks

An inverse value and the domain of an inverse

Functions

(a)

2 marks

Solve $h (x) = 1$ for $h (x) = \frac{2}{x - 1} + \frac{1}{2}$ .

\frac{2}{x - 1} = \frac{1}{2} \Rightarrow x - 1 = 4 \Rightarrow x = 17 h^{- 1} (1) = 17

(b)

2 marks

For $x > 1$ the fraction $\frac{2}{x - 1}$ takes every positive value, so $h (x) > \frac{1}{2}$ . The domain of $h^{- 1}$ is the range of $h$ .

x > \frac{1}{2}

Question 7

9 marks

A quadratic model through three points and a fence

Calculus

(a)

1 mark

The point $(3, 2.36)$ is on the curve.

2.36 = 9 a + 3 b + c

(b)

3 marks

Do the same for the other points.
Solve the three equations.

5 = 100 a + 10 b + c, 7.16 = 289 a + 17 b + c a = - 0.00490, b = 0.441, c = 1.08 f (x) = - 0.00490 x^{2} + 0.441 x + 1.08

(c)

3 marks

Evaluate the model at the fence, 80 m away.
Since $5.00 > 4$ , the ball is above the 4 m fence when it reaches it, so it goes over.

f (80) = - 0.0049 (6400) + 0.441 (80) + 1.08 = 5.00

(d)

2 marks

Solve $f (x) = 0$ and take the positive root.

- 0.0049 x^{2} + 0.441 x + 1.08 = 0 \Rightarrow x \approx 92.4 m

Question 8

7 marks

A road modelled by a cubic and a bridge normal to it (fire station at A)

Calculus

(a)

2 marks

Use the calculator to find the gradient of $r (x) = (x + 1)^{3} + 2 x^{2} - 4 x$ at $x = - 1$ .

r^{'} (- 1) = - 8

(b)

2 marks

The normal gradient is $\frac{1}{8}$ . It passes through $A (- 1, 6)$ .

y - 6 = \frac{1}{8} (x + 1)

(c)

3 marks

Find where the normal meets the road: solve $r (x) = 6 + \frac{x + 1}{8}$ . The solution other than $x = - 1$ on the lake side is $x = - 5.02$ .
The length of the bridge is the distance AB.

B \approx (- 5.02, 5.50) A B = (- 1 + 5.02)^{2} + (6 - 5.50)^{2} \approx 4.05 km

Question 9

8 marks

Two classes' history grades: box plots and a t-test on satisfaction scores

Statistics and probability

(a)

2 marks

From the diagram, Class A has $Q_{1} = 72$ and $Q_{3} = 85$ (IQR 13), and Class B has $Q_{1} = 67$ and $Q_{3} = 86$ (IQR 19).
Class B has a lower quartile below 70, so more than 25% of Class B scored under 70, while fewer than 25% of Class A did. So "a higher percentage of Class B received a grade less than 70" must be true.
The IQR of Class A (13) is less than that of Class B (19), so the last statement must be true.

(b)

2 marks

The null hypothesis says the population mean satisfaction is the same in the two classes.

H_{0} : μ_{A} = μ_{B}, H_{1} : μ_{A} \neq = μ_{B}

(c)

2 marks

Enter the two samples into a two-sample $t$ -test.

p \approx 0.111

(d)

2 marks

Compare $p$ with the 5% level.
Do not reject $H_{0}$ : there is insufficient evidence of a difference in satisfaction between the classes.

0.111 > 0.05

Mentor tip

The unpooled version of the test gives $p = 0.097$ , which leads to the same conclusion.

Question 10

7 marks

A coin and a die game: a tree diagram and a conditional probability

Statistics and probability

(a)

2 marks

The missing probabilities: tails $\frac{1}{2}$ ; on tails the die gives a win with probability $\frac{2}{6} = \frac{1}{3}$ and no win with probability $\frac{2}{3}$ .

(b)

2 marks

No prize needs tails and then not a five or six.

P (no win) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

(c)

3 marks

Use conditional probability. Winning happens with heads ( $\frac{1}{2}$ ) or tails then five or six ( $\frac{1}{2} \times \frac{1}{3}$ ).

P (heads ∣ win) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}

Question 11

6 marks

Toktam's car and the weather: binomial probabilities over a week

Statistics and probability

(a)

2 marks

Let $X \sim B (5, 0.65)$ be the number of mornings the car starts.

P (X = 2) = (2 5) (0.65)^{2} (0.35)^{3} \approx 0.181

(b)

4 marks

The probability of taking a taxi on a morning is the car failing and it raining.
No taxi on any of the five mornings:

P (taxi) = 0.35 \times 0.45 = 0.1575, P (no taxi) = 0.8425 0.842 5^{5} \approx 0.424

Question 12

6 marks

Social media followers and album sales: Spearman's rank correlation test

Statistics and probability

(a)

1 mark

Rank the albums sold, with the largest as 1.

2, 4, 8, 3, 5, 6, 7, 1

(b)

2 marks

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

r_{s} = 0.595

(c)

1 mark

Thurston believes that more followers means more albums sold.

H_{1} : in the population there is a positive monotonic relationship between followers and albums sold

(d)

2 marks

Compare $r_{s}$ with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence of a positive monotonic relationship.

0.595 < 0.643

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%
Q1, Q9, Q10, Q11, Q12
Calculus21 marks · 26%
Q3, Q7, Q8
Geometry and trigonometry14 marks · 18%
Q4, Q5
Number and algebra6 marks · 8%
Q2
Functions5 marks · 6%
Q3, Q6

Mentor insights

Traps to watch for

01
Regress T on G only to predict T · Q1
The line of $T$ on $G$ is for estimating time from gradient, not the other way round.
02
Bisector of AC · Q4
Use the midpoint and the negative reciprocal gradient.
03
Three points give three equations · Q7
Substitute each point into $f (x) = a x^{2} + b x + c$ and solve the system.
04
Normal gradient · Q8
The bridge is perpendicular to the road at A, so its gradient is $- \frac{1}{r ^{'} ( - 1 )}$ .
05
Read the box plot carefully · Q9
It gives quartiles, medians and range, not normality, and not exact numbers of students above a score.

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