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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

7 marks

Rabbit ear lengths: quartiles and comparing two distributions

Statistics and probability

(a)

3 marks

Read the Spanish box plot: the lower quartile is 68, the median is 75 and the upper quartile is 84.

median = 75 mm IQR = 84 - 68 = 16 mm

(b)

4 marks

(i) The Australian median (80) is bigger than the Spanish median (75), so the typical rabbit in Australia has longer ears than the typical rabbit in Spain.

(ii) The Spanish IQR (16) is bigger than the Australian IQR (11), so the ear lengths of the middle half of the rabbits vary more in Spain than in Australia.

Question 2

7 marks

A car losing value: 15% then 11% a year

Number and algebra

(a)

2 marks

A 15% loss leaves 85%.

0.85 \times 35000 = $29 750

(b)

2 marks

Nine further years of 11% loss after the first.

29750 \times 0.8 9^{9} \approx $10 423

(c)

3 marks

10% of the original price is $3500. Solve for the number of years.
The value after 19 years is $3250 and after 18 years is $3652, so the first time below $3500 is

29750 \times 0.8 9^{n - 1} < 3500 \Rightarrow n - 1 > 18.36 n = 20

Question 3

5 marks

A cone-shaped monument seen at an angle: radius and volume

Geometry and trigonometry

(a)

3 marks

Use the right triangle from Oliver's eye to the vertex.

tan 5 8^{\circ} = \frac{18.2}{5 + r} \Rightarrow 5 + r = 11.373 r \approx 6.37 m

(b)

2 marks

Volume of a cone.

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (6.3726)^{2} (20) \approx 851 m^{3}

Question 4

4 marks

A normal variable: standard deviations above the mean

Statistics and probability

(a)

2 marks

$1.5$ standard deviations above the mean is $X = 10 + 3 = 13$ .

P (X > 13) = P (Z > 1.5) \approx 0.0668

(b)

2 marks

Use the inverse normal with $P (Z < k) = 0.9$ .

k \approx 1.28

Question 5

6 marks

A particle with sinusoidal velocity: direction change, displacement and distance

Calculus

(a)

2 marks

Solve $v = 2 sin (0.5 t) + 0.3 t - 2 = 0$ on a graph or with the solver.

t \approx 1.69 s

(b)

2 marks

Displacement increases while $v > 0$ . The zeros of $v$ in $[0, 10]$ are $t \approx 1.687$ and $t \approx 6.119$ , and $v$ is positive between them.

1.69 < t < 6.12

(c)

2 marks

The displacement at $t = 10$ is the definite integral of $v$ .
The particle ends up 2.13 m behind $O$ .

\int_{0}^{10} (2 sin (0.5 t) + 0.3 t - 2) d t \approx - 2.13 m

Question 6

7 marks

Two test scores: correlation, extrapolation and predicting x from y

Statistics and probability

(a)

2 marks

Enter the ten pairs in the calculator.

r \approx 0.901

(b)

1 mark

The Test A scores in the data run from 52 to 100, and 10 is far below this range. Using the line there is extrapolation, so the estimate of 27 is not reliable.

(c)

4 marks

(i) The line of $y$ on $x$ is for predicting $y$ from a given $x$ . It is the wrong line for predicting $x$ from $y$ .

(ii) Use the regression line of $x$ on $y$ from the same data.

Substitute $y = 90$ .

x = 0.987 y - 3.22 x = 0.987124 (90) - 3.21971 \approx 85.6 \approx 86

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

12 marks

An exponential function and its inverse, then a reflection and translation

Functions

(a)

3 marks

Start with $y = \frac{3}{2} e^{x - 2}$ and swap $x$ and $y$ .
Take natural logarithms.

x = \frac{3}{2} e^{y - 2} \Rightarrow e^{y - 2} = \frac{2 x}{3} y - 2 = ln \frac{2 x}{3} \Rightarrow f^{- 1} (x) = 2 + ln (\frac{2 x}{3})

(b)

3 marks

The graphs meet on $y = x$ , so solve $f (x) = x$ .
$P$ and $Q$ are $(0.2645, 0.2645)$ and $(2.518, 2.518)$ , so the distance is

x \approx 0.2645 and x \approx 2.518 P Q = 2 (2.518 - 0.2645) \approx 3.19

(c)

3 marks

(i) Reflect in the $x$ -axis to get $- f (x)$ , then translate up 5.

(ii) The domain of $g$ is the same as that of $f$ .

g (x) = 5 - \frac{3}{2} e^{x - 2} 0 \leq x \leq 4

(d)

3 marks

Set $f (x) = g (x)$ .

\frac{3}{2} e^{x - 2} = 5 - \frac{3}{2} e^{x - 2} \Rightarrow 3 e^{x - 2} = 5 \Rightarrow e^{x - 2} = \frac{5}{3} x = 2 + ln \frac{5}{3}

Question 8

16 marks

Tide heights at two islands: a sine model and fitting a second model

Calculus

(a)

3 marks

The first low tide and first high tide of $H (t)$ are at $t \approx 5.138$ and $t \approx 11.262$ .

11.262 - 5.138 = 6.124 hours = 6 h 7.4 min m = 7

(b)

2 marks

Solve $H (t) = 1$ either side of the low tide at $t = 5.138$ .

t \approx 3.569 and t \approx 6.707 6.707 - 3.569 \approx 3.14 hours

(c)

2 marks

Differentiate and evaluate at $t = 13$ .

H^{'} (t) = 1.63 (0.513) cos (0.513 (t - 8.20)) \Rightarrow H^{'} (13) \approx - 0.651 m h^{- 1}

(d)

7 marks

Half the range gives $a$ and the midpoint gives $d$ .
Low tide at 02:41 and high tide at 09:02 are half a period apart.
The curve is at its midline, rising, halfway between the low tide and the high tide.

a = \frac{2.74 - 0.40}{2} = 1.17, d = \frac{2.74 + 0.40}{2} = 1.57 9 \frac{2}{60} - 2 \frac{41}{60} = 6.35 h \Rightarrow period = 12.7 h b = \frac{2 π}{12.7} \approx 0.495 c = \frac{2.6833 + 9.0333}{2} \approx 5.86

(e)

2 marks

Plot $H (t)$ and $h (t)$ and find the first intersection.

T \approx 4.16

Question 9

16 marks

A triangular field between a wall and a hedge: the ambiguous case and an area

Geometry and trigonometry

(a)

8 marks

(i) Use the sine rule in triangle $O A B$ with $O A = 50$ , $A B = 25$ and $A O B = 2 8^{\circ}$ .

The angle at $A$ is what is left.

(ii) Area $= \frac{1}{2} (O A) (A B) sin O A B$ .

\frac{sin A B O}{50} = \frac{sin 2 8 ^{\circ}}{25} \Rightarrow sin A B O = 0.9390 A B O \approx 69.8 7^{\circ} or 110.1 3^{\circ} O A B = 180 - 28 - 69.87 = 82. 1^{\circ} or 180 - 28 - 110.13 = 41. 9^{\circ} \frac{1}{2} (50) (25) sin 82.1 3^{\circ} \approx 619 m^{2} or \frac{1}{2} (50) (25) sin 41.8 7^{\circ} \approx 417 m^{2}

(b)

5 marks

Use the cosine rule in triangle $O C D$ (with $C D = 10$ and angle $2 8^{\circ}$ at $O$ ).
The area of the small triangle is 60.
Substitute for $x y$ in the cosine rule.

1 0^{2} = x^{2} + y^{2} - 2 x y cos 2 8^{\circ} \frac{1}{2} x y sin 2 8^{\circ} = 60 \Rightarrow x y = \frac{120}{sin 2 8 ^{\circ}} x^{2} + y^{2} = 100 + 2 cos 2 8^{\circ} \times \frac{120}{sin 2 8 ^{\circ}} = 100 + \frac{240}{tan 2 8 ^{\circ}}

(c)

3 marks

Replace $y$ by $\frac{120}{x sin 2 8 ^{\circ}}$ in the equation from (b).
Solve with the calculator.

x^{2} + (\frac{120}{x sin 2 8 ^{\circ}})^{2} = 100 + \frac{240}{tan 2 8 ^{\circ}} O C \approx 13.1 m or O C \approx 19.5 m

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q5, Q8
Geometry and trigonometry21 marks · 26%
Q3, Q9
Statistics and probability18 marks · 23%
Q1, Q4, Q6
Functions12 marks · 15%
Q7
Number and algebra7 marks · 9%
Q2

Mentor insights

Traps to watch for

01
Compare centre and spread · Q1
Use the median for the centre and the interquartile range for the spread, and say what each means in context.
02
First year is different · Q2
The first-year factor is 0.85, the later years use 0.89.
03
Inverse function · Q7
Swap $x$ and $y$ and make $y$ the subject.
04
Tide times · Q8
Low and high tide are half a period apart, so the period is twice the gap.
05
Ambiguous case · Q9
Sine rule with two sides and a non-included angle can give an acute or an obtuse angle.

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

7 marks

Rabbit ear lengths: quartiles and comparing two distributions

Statistics and probability

(a)

3 marks

Read the Spanish box plot: the lower quartile is 68, the median is 75 and the upper quartile is 84.

median = 75 mm IQR = 84 - 68 = 16 mm

(b)

4 marks

(i) The Australian median (80) is bigger than the Spanish median (75), so the typical rabbit in Australia has longer ears than the typical rabbit in Spain.

(ii) The Spanish IQR (16) is bigger than the Australian IQR (11), so the ear lengths of the middle half of the rabbits vary more in Spain than in Australia.

Question 2

7 marks

A car losing value: 15% then 11% a year

Number and algebra

(a)

2 marks

A 15% loss leaves 85%.

0.85 \times 35000 = $29 750

(b)

2 marks

Nine further years of 11% loss after the first.

29750 \times 0.8 9^{9} \approx $10 423

(c)

3 marks

10% of the original price is $3500. Solve for the number of years.
The value after 19 years is $3250 and after 18 years is $3652, so the first time below $3500 is

29750 \times 0.8 9^{n - 1} < 3500 \Rightarrow n - 1 > 18.36 n = 20

Question 3

5 marks

A cone-shaped monument seen at an angle: radius and volume

Geometry and trigonometry

(a)

3 marks

Use the right triangle from Oliver's eye to the vertex.

tan 5 8^{\circ} = \frac{18.2}{5 + r} \Rightarrow 5 + r = 11.373 r \approx 6.37 m

(b)

2 marks

Volume of a cone.

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (6.3726)^{2} (20) \approx 851 m^{3}

Question 4

4 marks

A normal variable: standard deviations above the mean

Statistics and probability

(a)

2 marks

$1.5$ standard deviations above the mean is $X = 10 + 3 = 13$ .

P (X > 13) = P (Z > 1.5) \approx 0.0668

(b)

2 marks

Use the inverse normal with $P (Z < k) = 0.9$ .

k \approx 1.28

Question 5

6 marks

A particle with sinusoidal velocity: direction change, displacement and distance

Calculus

(a)

2 marks

Solve $v = 2 sin (0.5 t) + 0.3 t - 2 = 0$ on a graph or with the solver.

t \approx 1.69 s

(b)

2 marks

Displacement increases while $v > 0$ . The zeros of $v$ in $[0, 10]$ are $t \approx 1.687$ and $t \approx 6.119$ , and $v$ is positive between them.

1.69 < t < 6.12

(c)

2 marks

The displacement at $t = 10$ is the definite integral of $v$ .
The particle ends up 2.13 m behind $O$ .

\int_{0}^{10} (2 sin (0.5 t) + 0.3 t - 2) d t \approx - 2.13 m

Question 6

7 marks

Two test scores: correlation, extrapolation and predicting x from y

Statistics and probability

(a)

2 marks

Enter the ten pairs in the calculator.

r \approx 0.901

(b)

1 mark

The Test A scores in the data run from 52 to 100, and 10 is far below this range. Using the line there is extrapolation, so the estimate of 27 is not reliable.

(c)

4 marks

(i) The line of $y$ on $x$ is for predicting $y$ from a given $x$ . It is the wrong line for predicting $x$ from $y$ .

(ii) Use the regression line of $x$ on $y$ from the same data.

Substitute $y = 90$ .

x = 0.987 y - 3.22 x = 0.987124 (90) - 3.21971 \approx 85.6 \approx 86

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

12 marks

An exponential function and its inverse, then a reflection and translation

Functions

(a)

3 marks

Start with $y = \frac{3}{2} e^{x - 2}$ and swap $x$ and $y$ .
Take natural logarithms.

x = \frac{3}{2} e^{y - 2} \Rightarrow e^{y - 2} = \frac{2 x}{3} y - 2 = ln \frac{2 x}{3} \Rightarrow f^{- 1} (x) = 2 + ln (\frac{2 x}{3})

(b)

3 marks

The graphs meet on $y = x$ , so solve $f (x) = x$ .
$P$ and $Q$ are $(0.2645, 0.2645)$ and $(2.518, 2.518)$ , so the distance is

x \approx 0.2645 and x \approx 2.518 P Q = 2 (2.518 - 0.2645) \approx 3.19

(c)

3 marks

(i) Reflect in the $x$ -axis to get $- f (x)$ , then translate up 5.

(ii) The domain of $g$ is the same as that of $f$ .

g (x) = 5 - \frac{3}{2} e^{x - 2} 0 \leq x \leq 4

(d)

3 marks

Set $f (x) = g (x)$ .

\frac{3}{2} e^{x - 2} = 5 - \frac{3}{2} e^{x - 2} \Rightarrow 3 e^{x - 2} = 5 \Rightarrow e^{x - 2} = \frac{5}{3} x = 2 + ln \frac{5}{3}

Question 8

16 marks

Tide heights at two islands: a sine model and fitting a second model

Calculus

(a)

3 marks

The first low tide and first high tide of $H (t)$ are at $t \approx 5.138$ and $t \approx 11.262$ .

11.262 - 5.138 = 6.124 hours = 6 h 7.4 min m = 7

(b)

2 marks

Solve $H (t) = 1$ either side of the low tide at $t = 5.138$ .

t \approx 3.569 and t \approx 6.707 6.707 - 3.569 \approx 3.14 hours

(c)

2 marks

Differentiate and evaluate at $t = 13$ .

H^{'} (t) = 1.63 (0.513) cos (0.513 (t - 8.20)) \Rightarrow H^{'} (13) \approx - 0.651 m h^{- 1}

(d)

7 marks

Half the range gives $a$ and the midpoint gives $d$ .
Low tide at 02:41 and high tide at 09:02 are half a period apart.
The curve is at its midline, rising, halfway between the low tide and the high tide.

a = \frac{2.74 - 0.40}{2} = 1.17, d = \frac{2.74 + 0.40}{2} = 1.57 9 \frac{2}{60} - 2 \frac{41}{60} = 6.35 h \Rightarrow period = 12.7 h b = \frac{2 π}{12.7} \approx 0.495 c = \frac{2.6833 + 9.0333}{2} \approx 5.86

(e)

2 marks

Plot $H (t)$ and $h (t)$ and find the first intersection.

T \approx 4.16

Question 9

16 marks

A triangular field between a wall and a hedge: the ambiguous case and an area

Geometry and trigonometry

(a)

8 marks

(i) Use the sine rule in triangle $O A B$ with $O A = 50$ , $A B = 25$ and $A O B = 2 8^{\circ}$ .

The angle at $A$ is what is left.

(ii) Area $= \frac{1}{2} (O A) (A B) sin O A B$ .

\frac{sin A B O}{50} = \frac{sin 2 8 ^{\circ}}{25} \Rightarrow sin A B O = 0.9390 A B O \approx 69.8 7^{\circ} or 110.1 3^{\circ} O A B = 180 - 28 - 69.87 = 82. 1^{\circ} or 180 - 28 - 110.13 = 41. 9^{\circ} \frac{1}{2} (50) (25) sin 82.1 3^{\circ} \approx 619 m^{2} or \frac{1}{2} (50) (25) sin 41.8 7^{\circ} \approx 417 m^{2}

(b)

5 marks

Use the cosine rule in triangle $O C D$ (with $C D = 10$ and angle $2 8^{\circ}$ at $O$ ).
The area of the small triangle is 60.
Substitute for $x y$ in the cosine rule.

1 0^{2} = x^{2} + y^{2} - 2 x y cos 2 8^{\circ} \frac{1}{2} x y sin 2 8^{\circ} = 60 \Rightarrow x y = \frac{120}{sin 2 8 ^{\circ}} x^{2} + y^{2} = 100 + 2 cos 2 8^{\circ} \times \frac{120}{sin 2 8 ^{\circ}} = 100 + \frac{240}{tan 2 8 ^{\circ}}

(c)

3 marks

Replace $y$ by $\frac{120}{x sin 2 8 ^{\circ}}$ in the equation from (b).
Solve with the calculator.

x^{2} + (\frac{120}{x sin 2 8 ^{\circ}})^{2} = 100 + \frac{240}{tan 2 8 ^{\circ}} O C \approx 13.1 m or O C \approx 19.5 m

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q5, Q8
Geometry and trigonometry21 marks · 26%
Q3, Q9
Statistics and probability18 marks · 23%
Q1, Q4, Q6
Functions12 marks · 15%
Q7
Number and algebra7 marks · 9%
Q2

Mentor insights

Traps to watch for

01
Compare centre and spread · Q1
Use the median for the centre and the interquartile range for the spread, and say what each means in context.
02
First year is different · Q2
The first-year factor is 0.85, the later years use 0.89.
03
Inverse function · Q7
Swap $x$ and $y$ and make $y$ the subject.
04
Tide times · Q8
Low and high tide are half a period apart, so the period is twice the gap.
05
Ambiguous case · Q9
Sine rule with two sides and a non-included angle can give an acute or an obtuse angle.

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