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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 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

Walking 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 the gradient and the time.

T = 0.552 G + 6.36 r = 0.994

(b)

2 marks

Substitute $G = 13$ .

T = 0.5521 (13) + 6.357 \approx 13.5 minutes

(c)

1 mark

The regression line of $T$ on $G$ is meant for predicting a time from a gradient. To estimate a gradient from a time you should use the line of $G$ on $T$ . Also 22 minutes is outside the range of the data, 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

f(x) = 16/x + x^2/8: asymptote, derivative and where it increases

Functions
Calculus

(a)

1 mark

The term $\frac{16}{x}$ is not defined at $x = 0$ and tends to infinity there.

x = 0

(b)

3 marks

Write $f (x) = 16 x^{- 1} + \frac{1}{8} x^{2}$ and differentiate.

f^{'} (x) = - 16 x^{- 2} + \frac{1}{4} x = - \frac{16}{x ^{2}} + \frac{x}{4}

(c)

2 marks

$f$ increases when $f^{'} (x) > 0$ . The derivative is zero at $\frac{x}{4} = \frac{16}{x ^{2}}$ , that is $x^{3} = 64$ , $x = 4$ , and it is positive for $x > 4$ .

x > 4 (the interval (4, \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

Ron's boat journey: bearings and the sine rule

Geometry and trigonometry

(a)

2 marks

At S the north line gives alternate angles: $18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ on one side and $3 8^{\circ}$ on the other.

R S T = 60 + 38 = 9 8^{\circ}

(b)

4 marks

T is due east of R, so the angle at R between RS (bearing $12 0^{\circ}$ ) and RT is $3 0^{\circ}$ . The angle at T is
Use the sine rule with RT opposite the angle at S.

R T S = 180 - 98 - 30 = 5 2^{\circ} \frac{R T}{sin 9 8 ^{\circ}} = \frac{150}{sin 5 2 ^{\circ}} \Rightarrow R T \approx 189 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 discus path: a quadratic through three points and a wall

Calculus

(a)

1 mark

The point $(3, 2.82)$ is on the curve.

2.82 = 9 a + 3 b + c

(b)

3 marks

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

4.25 = 36 a + 6 b + c, 5.30 = 81 a + 9 b + c a = - 0.0211, b = 0.667, c = 1.01 f (x) = - 0.0211 x^{2} + 0.667 x + 1.01

(c)

3 marks

Evaluate the model at the wall, 30 m away.
Since $2.01 > 1.8$ , the discus is above the 1.8 m wall when it reaches it, so it goes over.

f (30) = - 0.02111 (900) + 0.66667 (30) + 1.01 = 2.01

(d)

2 marks

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

- 0.0211 x^{2} + 0.667 x + 1.01 = 0 \Rightarrow x \approx 33.0 m

Question 8

7 marks

A road modelled by a cubic and a bridge normal to it

Calculus

(a)

2 marks

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

r^{'} (- 2) = - 14

(b)

2 marks

The normal gradient is $\frac{1}{14}$ . It passes through $H (- 2, 16)$ .

y - 16 = \frac{1}{14} (x + 2)

(c)

3 marks

Find where the normal meets the road: solve $r (x) = 16 + \frac{x + 2}{14}$ and take the point G on the lake side.
The length of the bridge is the distance HG.

G \approx (- 7.54, 15.60) H G = (- 2 + 7.54)^{2} + (16 - 15.60)^{2} \approx 5.55 km

Question 9

8 marks

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

Statistics and probability

(a)

2 marks

From the box and whisker diagram, the statements that must be true are:
A higher percentage of students in Class A received a grade less than 70 than in Class B, and the interquartile range for Class B is less than the interquartile range for Class A.

(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 0.095

(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 two classes.

0.095 > 0.05

Mentor tip

The unpooled version of the test gives $p = 0.080$ , which leads to the same conclusion.

Question 10

7 marks

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

Statistics and probability

(a)

2 marks

The die gives a win with probability $\frac{2}{6} = \frac{1}{3}$ and no win with probability $\frac{2}{3}$ . After no win the coin gives a win with probability $\frac{1}{2}$ and no win with probability $\frac{1}{2}$ .

(b)

2 marks

No prize needs a miss on the die and tails on the coin.

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

(c)

3 marks

Winning happens with a five or six ( $\frac{1}{3}$ ), or a miss then heads ( $\frac{2}{3} \times \frac{1}{2}$ ).

P (five or six ∣ win) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}

Question 11

6 marks

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

Statistics and probability

(a)

2 marks

Let $X \sim B (5, 0.72)$ be the number of mornings the car starts.

P (X = 3) = (3 5) (0.72)^{3} (0.28)^{2} \approx 0.293

(b)

4 marks

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

P (bus) = 0.28 \times 0.42 = 0.1176, P (no bus) = 0.8824 0.882 4^{5} \approx 0.535

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 the time from the 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 $f (x) = a x^{2} + b x + c$ .
04
Normal gradient · Q8
The bridge is perpendicular to the road at H, so its gradient is $- \frac{1}{r ^{'} ( - 2 )}$ .
05
Read what a box plot can tell you · Q9
It shows medians, quartiles and range, not whether data are normal, and not counts of students above a score unless the box plot makes it certain.

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 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

Walking 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 the gradient and the time.

T = 0.552 G + 6.36 r = 0.994

(b)

2 marks

Substitute $G = 13$ .

T = 0.5521 (13) + 6.357 \approx 13.5 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

f(x) = 16/x + x^2/8: asymptote, derivative and where it increases

Functions
Calculus

(a)

1 mark

The term $\frac{16}{x}$ is not defined at $x = 0$ and tends to infinity there.

x = 0

(b)

3 marks

Write $f (x) = 16 x^{- 1} + \frac{1}{8} x^{2}$ and differentiate.

f^{'} (x) = - 16 x^{- 2} + \frac{1}{4} x = - \frac{16}{x ^{2}} + \frac{x}{4}

(c)

2 marks

$f$ increases when $f^{'} (x) > 0$ . The derivative is zero at $\frac{x}{4} = \frac{16}{x ^{2}}$ , that is $x^{3} = 64$ , $x = 4$ , and it is positive for $x > 4$ .

x > 4 (the interval (4, \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

Ron's boat journey: bearings and the sine rule

Geometry and trigonometry

(a)

2 marks

At S the north line gives alternate angles: $18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ on one side and $3 8^{\circ}$ on the other.

R S T = 60 + 38 = 9 8^{\circ}

(b)

4 marks

T is due east of R, so the angle at R between RS (bearing $12 0^{\circ}$ ) and RT is $3 0^{\circ}$ . The angle at T is
Use the sine rule with RT opposite the angle at S.

R T S = 180 - 98 - 30 = 5 2^{\circ} \frac{R T}{sin 9 8 ^{\circ}} = \frac{150}{sin 5 2 ^{\circ}} \Rightarrow R T \approx 189 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 discus path: a quadratic through three points and a wall

Calculus

(a)

1 mark

The point $(3, 2.82)$ is on the curve.

2.82 = 9 a + 3 b + c

(b)

3 marks

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

4.25 = 36 a + 6 b + c, 5.30 = 81 a + 9 b + c a = - 0.0211, b = 0.667, c = 1.01 f (x) = - 0.0211 x^{2} + 0.667 x + 1.01

(c)

3 marks

Evaluate the model at the wall, 30 m away.
Since $2.01 > 1.8$ , the discus is above the 1.8 m wall when it reaches it, so it goes over.

f (30) = - 0.02111 (900) + 0.66667 (30) + 1.01 = 2.01

(d)

2 marks

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

- 0.0211 x^{2} + 0.667 x + 1.01 = 0 \Rightarrow x \approx 33.0 m

Question 8

7 marks

A road modelled by a cubic and a bridge normal to it

Calculus

(a)

2 marks

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

r^{'} (- 2) = - 14

(b)

2 marks

The normal gradient is $\frac{1}{14}$ . It passes through $H (- 2, 16)$ .

y - 16 = \frac{1}{14} (x + 2)

(c)

3 marks

Find where the normal meets the road: solve $r (x) = 16 + \frac{x + 2}{14}$ and take the point G on the lake side.
The length of the bridge is the distance HG.

G \approx (- 7.54, 15.60) H G = (- 2 + 7.54)^{2} + (16 - 15.60)^{2} \approx 5.55 km

Question 9

8 marks

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

Statistics and probability

(a)

2 marks

From the box and whisker diagram, the statements that must be true are:
A higher percentage of students in Class A received a grade less than 70 than in Class B, and the interquartile range for Class B is less than the interquartile range for Class A.

(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 0.095

(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 two classes.

0.095 > 0.05

Mentor tip

The unpooled version of the test gives $p = 0.080$ , which leads to the same conclusion.

Question 10

7 marks

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

Statistics and probability

(a)

2 marks

(b)

2 marks

No prize needs a miss on the die and tails on the coin.

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

(c)

3 marks

Winning happens with a five or six ( $\frac{1}{3}$ ), or a miss then heads ( $\frac{2}{3} \times \frac{1}{2}$ ).

P (five or six ∣ win) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}

Question 11

6 marks

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

Statistics and probability

(a)

2 marks

Let $X \sim B (5, 0.72)$ be the number of mornings the car starts.

P (X = 3) = (3 5) (0.72)^{3} (0.28)^{2} \approx 0.293

(b)

4 marks

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

P (bus) = 0.28 \times 0.42 = 0.1176, P (no bus) = 0.8824 0.882 4^{5} \approx 0.535

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 the time from the 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 $f (x) = a x^{2} + b x + c$ .
04
Normal gradient · Q8
The bridge is perpendicular to the road at H, so its gradient is $- \frac{1}{r ^{'} ( - 2 )}$ .
05
Read what a box plot can tell you · Q9
It shows medians, quartiles and range, not whether data are normal, and not counts of students above a score unless the box plot makes it certain.

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