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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 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 · 35 marks

Question 1

5 marks

Three cities: the sine rule with a bearing

Geometry and trigonometry

(a)

2 marks

The angle at J is $5 6^{\circ}$ and the angle at L is $2 5^{\circ}$ . The angles of triangle JLD add to $18 0^{\circ}$ .

J D L = 180 - 56 - 25 = 9 9^{\circ}

(b)

3 marks

Use the sine rule, with LD opposite the angle at J.

\frac{L D}{sin 5 6 ^{\circ}} = \frac{500}{sin 9 9 ^{\circ}} \Rightarrow L D \approx 420 km

Question 2

6 marks

A car that loses value and a savings account to pay for it

Number and algebra

(a)

1 mark

The yearly multiplier is $0.91$ .

1 - 0.91 = 9%

(b)

2 marks

On 1 January 2028, $t = 5$ .

C = 40000 (0.91)^{5} \approx $24 961

(c)

3 marks

Alvie's account after 5 years at 3% compounded yearly.
The extra money needed is the car's value minus the account balance.

15000 (1.03)^{5} = $17 389.11 M = 24961.29 - 17389.11 \approx $7572

Question 3

5 marks

A composite function of tan and a cubic: expression and sketch

Functions

(a)

2 marks

Substitute $g (x) = tan x$ into $f$ .

(f \circ g) (x) = 2 tan x - tan^{3} x

(b)

3 marks

The graph is an odd function through the origin.
The turning points on $- 1 \leq x \leq 1$ are
Sketch: starting at $(- 1, 0.66)$ the curve falls to the local minimum $(- 0.685, - 1.09)$ , rises through the origin to the local maximum $(0.685, 1.09)$ and then falls to $(1, - 0.66)$ .

local maximum (0.685, 1.09), local minimum (- 0.685, - 1.09)

Question 4

7 marks

Children in a park and ice creams: a model and a regression of x on y

Calculus
Statistics and probability

(a)

2 marks

Substitute $T = 25$ into the model.

y = - 0.6 (25)^{2} + 23 (25) + 110 = 310

(b)

3 marks

Regress $x$ on $y$ using the five data points.

x = 0.0935 y + 7.43

(c)

2 marks

Use the 310 children from part (a).
About 36 ice creams.

x = 0.0935 (310) + 7.43 \approx 36.4

Question 5

5 marks

Tube diameters: finding sigma from the interquartile range

Number and algebra

5 marks

The quartiles are the 25th and 75th percentiles. For $N (32, s^{2})$ they are $32 \pm 0.6745 s$ .

IQR = 2 (0.6745) s = 0.28 s = 0.208

Question 6

7 marks

Finding a and b from two binomial coefficients

Number and algebra

7 marks

The term in $x^{6}$ is $(2 n) (a x^{3})^{2} b^{n - 2}$ .
Divide the second by the first.
Then $a^{2} = \frac{16}{64} = \frac{1}{4}$ .

(2 8) a^{2} b^{6} = 28 a^{2} b^{6} = 448 \Rightarrow a^{2} b^{6} = 16 (2 10) a^{2} b^{8} = 45 a^{2} b^{8} = 2880 \Rightarrow a^{2} b^{8} = 64 b^{2} = 4 \Rightarrow b = 2 (b > 0) a = \frac{1}{2}, b = 2

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A cooling cup of tea: value, rate of change, limit and room temperature

Functions
Calculus

(a)

2 marks

Substitute $t = 0$ and $t = 3$ .

H (0) = 21 + 75 = 96^{\circ} C H (3) = 21 + 75 e^{- 0.24} \approx 80.0^{\circ} C

(b)

2 marks

Differentiate $H (t) = 21 + 75 e^{- 0.08 t}$ .

H^{'} (t) = - 6 e^{- 0.08 t} \Rightarrow H^{'} (3) = - 6 e^{- 0.24} \approx - 4.72

(c)

2 marks

Three minutes after being poured, the tea is cooling at a rate of about 4.72 degrees Celsius per minute.

(d)

3 marks

Solve $H (k) < 67$ .
The least integer is

21 + 75 e^{- 0.08 k} < 67 \Rightarrow e^{- 0.08 k} < \frac{46}{75} \Rightarrow k > \frac{ln ( 75/46 )}{0.08} = 6.11 k = 7

(e)

2 marks

As $t \to \infty$ the exponential term tends to 0.

21^{\circ} C

(f)

2 marks

$H^{'} (t) = - 6 e^{- 0.08 t} \to 0$ as $t \to \infty$ .

t \to \infty lim H^{'} (t) = 0

Question 8

17 marks

Survey on chocolate: two-way table, conditional probability and independence

Statistics and probability

(a)

2 marks

The children total 60, of whom 48 prefer milk chocolate, and the total surveyed is 160.

p = 60 - 48 = 12, q = 160 - 60 = 100

(b)

4 marks

Three adults chosen without replacement.

P = \frac{100}{160} \times \frac{99}{159} \times \frac{98}{158} \approx 0.241

(c)

4 marks

The number who prefer milk chocolate is $48 + x$ .
An adult who prefers milk chocolate:

P (A ∣ M) = \frac{x}{48 + x} = \frac{1}{3} \Rightarrow 3 x = 48 + x \Rightarrow x = 24 P (A \cap M) = \frac{24}{160} = 0.15

(d)

3 marks

Compare $P (A \cap M)$ with $P (A) P (M)$ , where $P (A) = \frac{100}{160}$ and $P (M) = \frac{72}{160}$ .
The events are not independent.

P (A) P (M) = 0.625 \times 0.45 = 0.281 \neq = 0.15 = P (A \cap M)

(e)

4 marks

The number who prefer dark chocolate is $12 + (100 - 24) = 88$ , so $P (D) = \frac{88}{160} = 0.55$ . Let $X \sim B (10, 0.55)$ .

P (X \geq 5) = 1 - P (X \leq 4) \approx 0.738

Question 9

15 marks

A rain gutter: angle, cross-section area and overflow

Geometry and trigonometry
Calculus

(a)

3 marks

The width of the sheet is the two vertical sides plus the arc AB.

45 = 10 + 10 + 12 θ \Rightarrow θ = \frac{25}{12} = 2.08

(b)

7 marks

Segment AHB: sector minus triangle.
The width of the gutter is the chord AB.
Add the rectangle $10 \times w$ below the arc.

\frac{1}{2} (12)^{2} (2.08) - \frac{1}{2} (12)^{2} sin (2.08) = 149.76 - 62.87 = 86.89 w = 2 (12) sin (1.04) = 20.70 area = 86.89 + 10 (20.70) \approx 294 cm^{2}

(c)

5 marks

The gutter holds $294 \times 600 \approx 176000 cm^{3}$ .
In any 60 seconds the rainwater is
The cosine term gives 0 since 60 seconds is exactly 12 periods of 5 seconds, so every 60-second period brings 180 000 cm $^{3}$ .
Since $180000 > 176000$ , the gutter overflowed.

\int_{0}^{60} (50 cos \frac{2 π t}{5} + 3000) d t = [\frac{250}{2 π} sin \frac{2 π t}{5} + 3000 t]_{0}^{60} = 180000 cm^{3}

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 probability22 marks · 28%
Q4, Q8
Number and algebra18 marks · 23%
Q2, Q5, Q6
Calculus18 marks · 23%
Q4, Q7, Q9
Geometry and trigonometry15 marks · 19%
Q1, Q9
Functions7 marks · 9%
Q3, Q7

Mentor insights

Traps to watch for

01
Regress x on y · Q4
The vendor wants to predict ice creams ( $x$ ) from the number of children ( $y$ ), so use the regression line of $x$ on $y$ .
02
Pick the right term · Q6
In $(a x^{3} + b)^{n}$ the $x^{6}$ term needs $(a x^{3})^{2}$ , which is the term with $r = 2$ .
03
Derivative as a rate · Q7
$H^{'} (3)$ is the rate of change of temperature in degrees per minute. A negative sign means cooling.
04
Without replacement · Q8
Choosing three people from the survey changes the denominator each time.
05
The sheet width fixes the arc · Q9
The 45 cm sheet is two vertical sides of 10 cm plus the curved arc, so the arc is 25 cm.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 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 · 35 marks

Question 1

5 marks

Three cities: the sine rule with a bearing

Geometry and trigonometry

(a)

2 marks

The angle at J is $5 6^{\circ}$ and the angle at L is $2 5^{\circ}$ . The angles of triangle JLD add to $18 0^{\circ}$ .

J D L = 180 - 56 - 25 = 9 9^{\circ}

(b)

3 marks

Use the sine rule, with LD opposite the angle at J.

\frac{L D}{sin 5 6 ^{\circ}} = \frac{500}{sin 9 9 ^{\circ}} \Rightarrow L D \approx 420 km

Question 2

6 marks

A car that loses value and a savings account to pay for it

Number and algebra

(a)

1 mark

The yearly multiplier is $0.91$ .

1 - 0.91 = 9%

(b)

2 marks

On 1 January 2028, $t = 5$ .

C = 40000 (0.91)^{5} \approx $24 961

(c)

3 marks

Alvie's account after 5 years at 3% compounded yearly.
The extra money needed is the car's value minus the account balance.

15000 (1.03)^{5} = $17 389.11 M = 24961.29 - 17389.11 \approx $7572

Question 3

5 marks

A composite function of tan and a cubic: expression and sketch

Functions

(a)

2 marks

Substitute $g (x) = tan x$ into $f$ .

(f \circ g) (x) = 2 tan x - tan^{3} x

(b)

3 marks

The graph is an odd function through the origin.
The turning points on $- 1 \leq x \leq 1$ are
Sketch: starting at $(- 1, 0.66)$ the curve falls to the local minimum $(- 0.685, - 1.09)$ , rises through the origin to the local maximum $(0.685, 1.09)$ and then falls to $(1, - 0.66)$ .

local maximum (0.685, 1.09), local minimum (- 0.685, - 1.09)

Question 4

7 marks

Children in a park and ice creams: a model and a regression of x on y

Calculus
Statistics and probability

(a)

2 marks

Substitute $T = 25$ into the model.

y = - 0.6 (25)^{2} + 23 (25) + 110 = 310

(b)

3 marks

Regress $x$ on $y$ using the five data points.

x = 0.0935 y + 7.43

(c)

2 marks

Use the 310 children from part (a).
About 36 ice creams.

x = 0.0935 (310) + 7.43 \approx 36.4

Question 5

5 marks

Tube diameters: finding sigma from the interquartile range

Number and algebra

5 marks

The quartiles are the 25th and 75th percentiles. For $N (32, s^{2})$ they are $32 \pm 0.6745 s$ .

IQR = 2 (0.6745) s = 0.28 s = 0.208

Question 6

7 marks

Finding a and b from two binomial coefficients

Number and algebra

7 marks

The term in $x^{6}$ is $(2 n) (a x^{3})^{2} b^{n - 2}$ .
Divide the second by the first.
Then $a^{2} = \frac{16}{64} = \frac{1}{4}$ .

(2 8) a^{2} b^{6} = 28 a^{2} b^{6} = 448 \Rightarrow a^{2} b^{6} = 16 (2 10) a^{2} b^{8} = 45 a^{2} b^{8} = 2880 \Rightarrow a^{2} b^{8} = 64 b^{2} = 4 \Rightarrow b = 2 (b > 0) a = \frac{1}{2}, b = 2

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A cooling cup of tea: value, rate of change, limit and room temperature

Functions
Calculus

(a)

2 marks

Substitute $t = 0$ and $t = 3$ .

H (0) = 21 + 75 = 96^{\circ} C H (3) = 21 + 75 e^{- 0.24} \approx 80.0^{\circ} C

(b)

2 marks

Differentiate $H (t) = 21 + 75 e^{- 0.08 t}$ .

H^{'} (t) = - 6 e^{- 0.08 t} \Rightarrow H^{'} (3) = - 6 e^{- 0.24} \approx - 4.72

(c)

2 marks

Three minutes after being poured, the tea is cooling at a rate of about 4.72 degrees Celsius per minute.

(d)

3 marks

Solve $H (k) < 67$ .
The least integer is

21 + 75 e^{- 0.08 k} < 67 \Rightarrow e^{- 0.08 k} < \frac{46}{75} \Rightarrow k > \frac{ln ( 75/46 )}{0.08} = 6.11 k = 7

(e)

2 marks

As $t \to \infty$ the exponential term tends to 0.

21^{\circ} C

(f)

2 marks

$H^{'} (t) = - 6 e^{- 0.08 t} \to 0$ as $t \to \infty$ .

t \to \infty lim H^{'} (t) = 0

Question 8

17 marks

Survey on chocolate: two-way table, conditional probability and independence

Statistics and probability

(a)

2 marks

The children total 60, of whom 48 prefer milk chocolate, and the total surveyed is 160.

p = 60 - 48 = 12, q = 160 - 60 = 100

(b)

4 marks

Three adults chosen without replacement.

P = \frac{100}{160} \times \frac{99}{159} \times \frac{98}{158} \approx 0.241

(c)

4 marks

The number who prefer milk chocolate is $48 + x$ .
An adult who prefers milk chocolate:

P (A ∣ M) = \frac{x}{48 + x} = \frac{1}{3} \Rightarrow 3 x = 48 + x \Rightarrow x = 24 P (A \cap M) = \frac{24}{160} = 0.15

(d)

3 marks

Compare $P (A \cap M)$ with $P (A) P (M)$ , where $P (A) = \frac{100}{160}$ and $P (M) = \frac{72}{160}$ .
The events are not independent.

P (A) P (M) = 0.625 \times 0.45 = 0.281 \neq = 0.15 = P (A \cap M)

(e)

4 marks

The number who prefer dark chocolate is $12 + (100 - 24) = 88$ , so $P (D) = \frac{88}{160} = 0.55$ . Let $X \sim B (10, 0.55)$ .

P (X \geq 5) = 1 - P (X \leq 4) \approx 0.738

Question 9

15 marks

A rain gutter: angle, cross-section area and overflow

Geometry and trigonometry
Calculus

(a)

3 marks

The width of the sheet is the two vertical sides plus the arc AB.

45 = 10 + 10 + 12 θ \Rightarrow θ = \frac{25}{12} = 2.08

(b)

7 marks

Segment AHB: sector minus triangle.
The width of the gutter is the chord AB.
Add the rectangle $10 \times w$ below the arc.

\frac{1}{2} (12)^{2} (2.08) - \frac{1}{2} (12)^{2} sin (2.08) = 149.76 - 62.87 = 86.89 w = 2 (12) sin (1.04) = 20.70 area = 86.89 + 10 (20.70) \approx 294 cm^{2}

(c)

5 marks

The gutter holds $294 \times 600 \approx 176000 cm^{3}$ .
In any 60 seconds the rainwater is
The cosine term gives 0 since 60 seconds is exactly 12 periods of 5 seconds, so every 60-second period brings 180 000 cm $^{3}$ .
Since $180000 > 176000$ , the gutter overflowed.

\int_{0}^{60} (50 cos \frac{2 π t}{5} + 3000) d t = [\frac{250}{2 π} sin \frac{2 π t}{5} + 3000 t]_{0}^{60} = 180000 cm^{3}

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 probability22 marks · 28%
Q4, Q8
Number and algebra18 marks · 23%
Q2, Q5, Q6
Calculus18 marks · 23%
Q4, Q7, Q9
Geometry and trigonometry15 marks · 19%
Q1, Q9
Functions7 marks · 9%
Q3, Q7

Mentor insights

Traps to watch for

01
Regress x on y · Q4
The vendor wants to predict ice creams ( $x$ ) from the number of children ( $y$ ), so use the regression line of $x$ on $y$ .
02
Pick the right term · Q6
In $(a x^{3} + b)^{n}$ the $x^{6}$ term needs $(a x^{3})^{2}$ , which is the term with $r = 2$ .
03
Derivative as a rate · Q7
$H^{'} (3)$ is the rate of change of temperature in degrees per minute. A negative sign means cooling.
04
Without replacement · Q8
Choosing three people from the survey changes the denominator each time.
05
The sheet width fixes the arc · Q9
The 45 cm sheet is two vertical sides of 10 cm plus the curved arc, so the arc is 25 cm.

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