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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

15 marks

Goals in a soccer tournament: a frequency table, the mean, comparing data and sampling without replacement

Statistics and probability

(a)

1 mark

Read the bar for 3 goals in a game from the bar graph.

k = 5

(b)

3 marks

(i) The frequencies are 3, 11, 7, 5, $p$ and 1 for 0 to 5 goals.

(ii) Solve for $p$ .

\frac{0 ( 3 ) + 1 ( 11 ) + 2 ( 7 ) + 3 ( 5 ) + 4 p + 5 ( 1 )}{3 + 11 + 7 + 5 + p + 1} = 2.2 \Rightarrow \frac{45 + 4 p}{27 + p} = 2.2 45 + 4 p = 59.4 + 2.2 p \Rightarrow 1.8 p = 14.4 \Rightarrow p = 8

(c)

3 marks

The 2024 data give a median of 2, a lower quartile of 1 and an upper quartile of 4, so the interquartile range is 3.
Two observations that support Paul: the median number of goals in a game is 2 in both years, and the interquartile range is 3 in both years (with quartiles 1 and 4).

(d)

2 marks

$F^{'}$ is scoring neither 0 nor exactly 1 goal, which means scoring at least 2 goals, or more than 1 goal.

B and C

(e)

2 marks

There are 35 games and one game with exactly 1 goal has been watched. Of the 34 left, 10 have exactly 1 goal (there were 11 at the start).

\frac{10}{34} = \frac{5}{17}

(f)

4 marks

The probability that the first game has 5 goals is $\frac{1}{35}$ . Then 0 goals in the second game has probability $\frac{3}{34}$ .

\frac{1}{35} \times \frac{3}{34} = \frac{3}{1190} \approx 0.00252

Question 2

19 marks

A Voronoi diagram for three fire stations

Geometry and trigonometry

(a)

6 marks

The stations are $B (16, 12)$ and $C (8, 13)$ .
(i) The midpoint of $[B C]$ .
(ii) The gradient of $B C$ is $\frac{13 - 12}{8 - 16} = - \frac{1}{8}$ , so the perpendicular bisector has gradient 8.
(iii) The edge between $B$ and $C$ .

(\frac{16 + 8}{2}, \frac{12 + 13}{2}) = (12, 12.5) y - 12.5 = 8 (x - 12) \Rightarrow y = 8 x - 83.5

(b)

2 marks

(i) The location $(14, 10)$ is nearest to $B$ : $B X = 2^{2} + 2^{2} = 2.83$ , compared with $A X = 4^{2} + 2^{2} = 4.47$ and $C X = 6^{2} + 3^{2} = 6.71$ . By the Voronoi diagram, station $B$ responds.

(ii) A different station might respond because the roads to $B$ are longer than the straight line distance, or station $B$ is already attending another call.

(c)

4 marks

(i) The point $D$ is $(11 \frac{1}{3}, 7 \frac{1}{6})$ . The distance from $B (16, 12)$ is

(ii) The distance from $A (18, 8)$ is $(6.667)^{2} + (0.8333)^{2} = 45.14 \approx 6.71855$ and the distance from $C (8, 13)$ is $(3.333)^{2} + (5.833)^{2} = 45.14 \approx 6.71855$ . All three stations are the same distance away, so any of the three could be expected to respond.

(16 - 11.333)^{2} + (12 - 7.1667)^{2} = 21.78 + 23.36 = 45.14 \approx 6.71855

(d)

7 marks

(i) The region for $A$ has corners at $(7.75, 0)$ , $(20, 0)$ , $(20, 11.5)$ and $D (11 \frac{1}{3}, 7 \frac{1}{6})$ . The edge $A B$ is $y = 0.5 x + 1.5$ , which at $x = 20$ gives $y = 11.5$ . The edge $A C$ goes from $(7.75, 0)$ to $D$ .

Use the shoelace formula with these four corners.

(ii) One unit is 2.5 km, so one square unit is $2.5 \times 2.5 = 6.25$ km $^{2}$ .

Area = \frac{1}{2} \sum (x_{i} y_{i + 1} - x_{i + 1} y_{i}) \approx 93.7 \approx 94 square units 94 \times 6.25 \approx 588 km^{2}

Question 3

17 marks

The population of Los Angeles: a quadratic model and a logistic model

Calculus

(a)

3 marks

Use quadratic regression on the pairs $(0, 102)$ , $(20, 577)$ , $(40, 1504)$ , $(60, 2479)$ , $(80, 2967)$ , $(100, 3685)$ , $(120, 3899)$ .

f (x) = - 0.0989 x^{2} + 45.9 x - 67.8

(b)

2 marks

The value 0.98843 is very close to 1, so the quadratic explains nearly all of the variation in the data. It supports Ethan's proposal.

(c)

1 mark

The constant term is negative, $c = - 67.8$ , but the population in 1900 (when $x = 0$ ) cannot be negative. (Also, the negative coefficient of $x^{2}$ means the model predicts the population will eventually fall back to zero.)

(d)

4 marks

(i) Differentiate.

This is the rate of change in 2010: the population is predicted to be growing by about 24.2 thousand (24 200) people per year.

(ii) From 2000 to 2020 the population grew from 3685 to 3899 thousand, which is an average of $\frac{214}{20} = 10.7$ thousand per year. The model predicts 24.2 thousand per year, which is more than twice as large. The growth is slowing down in the data, so this casts doubt on the validity of Ethan's model.

f^{'} (x) = - 0.1978 x + 45.9 \Rightarrow f^{'} (110) \approx 24.2

(e)

1 mark

The logistic model levels off at a limit of 4000 thousand, matching the slowing growth in the data, which suggests the population tends towards a carrying capacity.

(f)

6 marks

(i) Differentiate $g (x) = \frac{4000}{1 + 14 e ^{- 0.05 x}}$ with the chain rule.

(ii) The greatest growth rate is at the maximum of $g^{'} (x)$ .

Since $x$ is the number of years after 1900, the greatest growth occurred during the year 1952.

g^{'} (x) = 4000 \cdot \frac{14 ( 0.05 ) e ^{- 0.05 x}}{( 1 + 14 e ^{- 0.05 x} ) ^{2}} = \frac{2800 e ^{- 0.05 x}}{( 1 + 14 e ^{- 0.05 x} ) ^{2}} x \approx 52.78

Question 4

17 marks

Air traffic: an airplane, a flock of birds and a jet

Number and algebra

(a)

2 marks

The direction vector of the birds is $(- 0.40, - 0.35, 0.1)$ .

speed = 0. 4^{2} + 0.3 5^{2} + 0. 1^{2} = 0.2925 \approx 0.541 km per minute

(b)

2 marks

The airplane flies in a straight line at a constant height of 7 km. It is heading north-west, moving west at 1.4 km per minute and north at 1.65 km per minute.

(c)

6 marks

(i) The heights are equal when $7 = 6 + 0.1 t_{1}$ .

(ii) At $t_{1} = 10$ the positions are

The heights are the same only at this time, and at this time the $x$ and $y$ coordinates are different, so the birds and the airplane are in different places. They will not collide.

t_{1} = 10 minutes, that is 10 : 10 am airplane: (5 - 14, - 2 + 16.5, 7) = (- 9, 14.5, 7) birds: (- 11 - 4, 30 - 3.5, 6 + 1) = (- 15, 26.5, 7)

(d)

2 marks

At 10:20, $t_{1} = 20$ , so $t_{1} = t_{2} + 20$ .

r = 5 - 2 7 + (t_{2} + 20) - 1.4 1.65 0 = - 23 317 + t_{2} - 1.4 1.65 0

(e)

5 marks

The jet is at $(- 95 + 1.3 t_{2}, 32 + 1.45 t_{2}, 1.5 + 0.2 t_{2})$ and the airplane is at $(- 23 - 1.4 t_{2}, 31 + 1.65 t_{2}, 7)$ . The difference is
Find the minimum with the calculator.
Since $4.33 < 5$ , the jet and the airplane will break the law.

(- 72 + 2.7 t_{2}, 1 - 0.2 t_{2}, - 5.5 + 0.2 t_{2}) d = (2.7 t_{2} - 72)^{2} + (1 - 0.2 t_{2})^{2} + (0.2 t_{2} - 5.5)^{2} d_{m i n} \approx 4.33 km at t_{2} \approx 26.6

Question 5

16 marks

A triangular prism doorstop: lengths, volume, an angle and a price

Number and algebra

(a)

2 marks

$A (13, 1, 0)$ and $C (13, 25, 7)$ .

A C = 0^{2} + 2 4^{2} + 7^{2} = 25 cm

(b)

2 marks

$A B = 24$ (along the $y$ -axis) and $B C = 7$ (along the $z$ -axis).
By the converse of Pythagoras, triangle $A B C$ has a right angle at $B$ .

2 4^{2} + 7^{2} = 576 + 49 = 625 = 2 5^{2}

(c)

3 marks

The prism has width $C D = 13 - 4 = 9$ and the triangular face has area $\frac{1}{2} (24) (7) = 84$ .

V = 84 \times 9 = 756 cm^{3}

(d)

6 marks

(i) $θ$ is the angle at $A$ in triangle $A B C$ .

(ii) The width is still 9, so the new triangular face has area $\frac{625}{9}$ . Its angle at $A$ is $θ$ and its height is $A X tan θ$ .

tan θ = \frac{7}{24} \Rightarrow θ \approx 16. 3^{\circ} \frac{1}{2} (A X)^{2} tan θ = \frac{625}{9} \Rightarrow (A X)^{2} = \frac{1250}{9 tan θ} A X \approx 21.8 cm

(e)

3 marks

The material needed is 10% more than 625 cm $^{3}$ , at 0.025 USD per cubic centimetre, and the price is 20% more than the cost.
The price is about 20.63 USD.

625 \times 1.1 \times 0.025 \times 1.2 = 20.625

Question 6

14 marks

Virus particles and healthy skin cells: a differential equation

Calculus

(a)

3 marks

The rate is proportional to $\frac{h}{v}$ , so $\frac{d h}{d t} = k \frac{h}{v}$ . At $t = 0$ , $h = 1350$ and $v = 27$ .

- 5 = k \frac{1350}{27} = 50 k \Rightarrow k = - \frac{1}{10} \frac{d h}{d t} = - \frac{h}{10 v}

(b)

7 marks

Put $v = 27 + 0.3 t$ and separate the variables.
Integrate.
Use $h = 1350$ at $t = 0$ : $1350 = A (27)^{- 1/3} = \frac{A}{3}$ .

\int \frac{1}{h} d h = - \int \frac{1}{10 ( 27 + 0.3 t )} d t ln h = - \frac{1}{10 ( 0.3 )} ln (27 + 0.3 t) + c = - \frac{1}{3} ln (27 + 0.3 t) + c h = A (27 + 0.3 t)^{- 1/3} A = 4050 \Rightarrow h = 4050 (27 + 0.3 t)^{- 1/3}

(c)

4 marks

The number of virus particles is $1000 v$ (since $v$ is in thousands). We want $1000 v \geq 100 h$ .
Solve with the calculator.
To be at least 100 times, the answer is the next whole hour, 217 hours.

1000 (27 + 0.28 t) \geq 100 \times 4380 (27 + 0.28 t)^{- 0.36} t \approx 216.3 hours

Question 7

12 marks

A table designed by rotating a region about the y-axis

Calculus

(a)

4 marks

The curve is $y = \frac{2}{3} cos^{- 1} (\frac{1}{10 x})$ . The region between the curve and the line $y = 1$ for $0.1 \leq x \leq 1$ , together with the strip from $x = 0$ to $x = 0.1$ , has area equal to the area of the unit square minus the area under the curve.

area under the curve = \int_{0.1}^{1} \frac{2}{3} cos^{- 1} (\frac{1}{10 x}) d x \approx 0.7809 shaded area = 1 - 0.7809 \approx 0.219 m^{2}

(b)

6 marks

Rearrange for $x$ : $cos (\frac{3 y}{2}) = \frac{1}{10 x}$ , so $x = \frac{1}{10 cos \frac{3 y}{2}}$ . At $x = 1$ , $y = \frac{2}{3} cos^{- 1} (0.1) = 0.9804$ .
The top of the table is a cylinder of radius 1 and height $1 - 0.9804 = 0.0196$ .
The curved part, from $y = 0$ to $y = 0.9804$ , has radius $x$ .
The base is a cylinder of radius 0.3 and height 0.075.

V_{1} = π (1)^{2} (0.0196) V_{2} = π \int_{0}^{0.9804} (\frac{1}{10 cos \frac{3 y}{2}})^{2} d y V_{3} = π (0.3)^{2} (0.075) V = π (1)^{2} (0.0196) + π \int_{0}^{0.9804} (\frac{1}{10 cos \frac{3 y}{2}})^{2} d y + π (0.3)^{2} (0.075)

(c)

2 marks

Evaluate with the calculator.

V \approx 0.291 m^{3}

Our analysis

Where the marks are

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

Calculus43 marks · 39%
Q3, Q6, Q7
Number and algebra33 marks · 30%
Q4, Q5
Geometry and trigonometry19 marks · 17%
Q2
Statistics and probability15 marks · 14%
Q1

Mentor insights

Traps to watch for

01
Mean from a frequency table · Q1
$\overset{x}{ˉ} = \frac{\sum f x}{\sum f}$ .
02
Perpendicular bisector · Q2
The edge between two sites is the perpendicular bisector of the segment joining them.
03
Coefficient of determination · Q3
$R^{2}$ close to 1 means the model explains the data well.
04
Speed · Q4
The speed is the length of the direction vector.
05
Right angle · Q5
Check with Pythagoras or with a scalar product.

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 HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

15 marks

Goals in a soccer tournament: a frequency table, the mean, comparing data and sampling without replacement

Statistics and probability

(a)

1 mark

Read the bar for 3 goals in a game from the bar graph.

k = 5

(b)

3 marks

(i) The frequencies are 3, 11, 7, 5, $p$ and 1 for 0 to 5 goals.

(ii) Solve for $p$ .

\frac{0 ( 3 ) + 1 ( 11 ) + 2 ( 7 ) + 3 ( 5 ) + 4 p + 5 ( 1 )}{3 + 11 + 7 + 5 + p + 1} = 2.2 \Rightarrow \frac{45 + 4 p}{27 + p} = 2.2 45 + 4 p = 59.4 + 2.2 p \Rightarrow 1.8 p = 14.4 \Rightarrow p = 8

(c)

3 marks

The 2024 data give a median of 2, a lower quartile of 1 and an upper quartile of 4, so the interquartile range is 3.
Two observations that support Paul: the median number of goals in a game is 2 in both years, and the interquartile range is 3 in both years (with quartiles 1 and 4).

(d)

2 marks

$F^{'}$ is scoring neither 0 nor exactly 1 goal, which means scoring at least 2 goals, or more than 1 goal.

B and C

(e)

2 marks

There are 35 games and one game with exactly 1 goal has been watched. Of the 34 left, 10 have exactly 1 goal (there were 11 at the start).

\frac{10}{34} = \frac{5}{17}

(f)

4 marks

The probability that the first game has 5 goals is $\frac{1}{35}$ . Then 0 goals in the second game has probability $\frac{3}{34}$ .

\frac{1}{35} \times \frac{3}{34} = \frac{3}{1190} \approx 0.00252

Question 2

19 marks

A Voronoi diagram for three fire stations

Geometry and trigonometry

(a)

6 marks

The stations are $B (16, 12)$ and $C (8, 13)$ .
(i) The midpoint of $[B C]$ .
(ii) The gradient of $B C$ is $\frac{13 - 12}{8 - 16} = - \frac{1}{8}$ , so the perpendicular bisector has gradient 8.
(iii) The edge between $B$ and $C$ .

(\frac{16 + 8}{2}, \frac{12 + 13}{2}) = (12, 12.5) y - 12.5 = 8 (x - 12) \Rightarrow y = 8 x - 83.5

(b)

2 marks

(ii) A different station might respond because the roads to $B$ are longer than the straight line distance, or station $B$ is already attending another call.

(c)

4 marks

(i) The point $D$ is $(11 \frac{1}{3}, 7 \frac{1}{6})$ . The distance from $B (16, 12)$ is

(16 - 11.333)^{2} + (12 - 7.1667)^{2} = 21.78 + 23.36 = 45.14 \approx 6.71855

(d)

7 marks

Use the shoelace formula with these four corners.

(ii) One unit is 2.5 km, so one square unit is $2.5 \times 2.5 = 6.25$ km $^{2}$ .

Area = \frac{1}{2} \sum (x_{i} y_{i + 1} - x_{i + 1} y_{i}) \approx 93.7 \approx 94 square units 94 \times 6.25 \approx 588 km^{2}

Question 3

17 marks

The population of Los Angeles: a quadratic model and a logistic model

Calculus

(a)

3 marks

Use quadratic regression on the pairs $(0, 102)$ , $(20, 577)$ , $(40, 1504)$ , $(60, 2479)$ , $(80, 2967)$ , $(100, 3685)$ , $(120, 3899)$ .

f (x) = - 0.0989 x^{2} + 45.9 x - 67.8

(b)

2 marks

The value 0.98843 is very close to 1, so the quadratic explains nearly all of the variation in the data. It supports Ethan's proposal.

(c)

1 mark

(d)

4 marks

(i) Differentiate.

This is the rate of change in 2010: the population is predicted to be growing by about 24.2 thousand (24 200) people per year.

f^{'} (x) = - 0.1978 x + 45.9 \Rightarrow f^{'} (110) \approx 24.2

(e)

1 mark

The logistic model levels off at a limit of 4000 thousand, matching the slowing growth in the data, which suggests the population tends towards a carrying capacity.

(f)

6 marks

(i) Differentiate $g (x) = \frac{4000}{1 + 14 e ^{- 0.05 x}}$ with the chain rule.

(ii) The greatest growth rate is at the maximum of $g^{'} (x)$ .

Since $x$ is the number of years after 1900, the greatest growth occurred during the year 1952.

g^{'} (x) = 4000 \cdot \frac{14 ( 0.05 ) e ^{- 0.05 x}}{( 1 + 14 e ^{- 0.05 x} ) ^{2}} = \frac{2800 e ^{- 0.05 x}}{( 1 + 14 e ^{- 0.05 x} ) ^{2}} x \approx 52.78

Question 4

17 marks

Air traffic: an airplane, a flock of birds and a jet

Number and algebra

(a)

2 marks

The direction vector of the birds is $(- 0.40, - 0.35, 0.1)$ .

speed = 0. 4^{2} + 0.3 5^{2} + 0. 1^{2} = 0.2925 \approx 0.541 km per minute

(b)

2 marks

The airplane flies in a straight line at a constant height of 7 km. It is heading north-west, moving west at 1.4 km per minute and north at 1.65 km per minute.

(c)

6 marks

(i) The heights are equal when $7 = 6 + 0.1 t_{1}$ .

(ii) At $t_{1} = 10$ the positions are

The heights are the same only at this time, and at this time the $x$ and $y$ coordinates are different, so the birds and the airplane are in different places. They will not collide.

t_{1} = 10 minutes, that is 10 : 10 am airplane: (5 - 14, - 2 + 16.5, 7) = (- 9, 14.5, 7) birds: (- 11 - 4, 30 - 3.5, 6 + 1) = (- 15, 26.5, 7)

(d)

2 marks

At 10:20, $t_{1} = 20$ , so $t_{1} = t_{2} + 20$ .

r = 5 - 2 7 + (t_{2} + 20) - 1.4 1.65 0 = - 23 317 + t_{2} - 1.4 1.65 0

(e)

5 marks

The jet is at $(- 95 + 1.3 t_{2}, 32 + 1.45 t_{2}, 1.5 + 0.2 t_{2})$ and the airplane is at $(- 23 - 1.4 t_{2}, 31 + 1.65 t_{2}, 7)$ . The difference is
Find the minimum with the calculator.
Since $4.33 < 5$ , the jet and the airplane will break the law.

(- 72 + 2.7 t_{2}, 1 - 0.2 t_{2}, - 5.5 + 0.2 t_{2}) d = (2.7 t_{2} - 72)^{2} + (1 - 0.2 t_{2})^{2} + (0.2 t_{2} - 5.5)^{2} d_{m i n} \approx 4.33 km at t_{2} \approx 26.6

Question 5

16 marks

A triangular prism doorstop: lengths, volume, an angle and a price

Number and algebra

(a)

2 marks

$A (13, 1, 0)$ and $C (13, 25, 7)$ .

A C = 0^{2} + 2 4^{2} + 7^{2} = 25 cm

(b)

2 marks

$A B = 24$ (along the $y$ -axis) and $B C = 7$ (along the $z$ -axis).
By the converse of Pythagoras, triangle $A B C$ has a right angle at $B$ .

2 4^{2} + 7^{2} = 576 + 49 = 625 = 2 5^{2}

(c)

3 marks

The prism has width $C D = 13 - 4 = 9$ and the triangular face has area $\frac{1}{2} (24) (7) = 84$ .

V = 84 \times 9 = 756 cm^{3}

(d)

6 marks

(i) $θ$ is the angle at $A$ in triangle $A B C$ .

(ii) The width is still 9, so the new triangular face has area $\frac{625}{9}$ . Its angle at $A$ is $θ$ and its height is $A X tan θ$ .

tan θ = \frac{7}{24} \Rightarrow θ \approx 16. 3^{\circ} \frac{1}{2} (A X)^{2} tan θ = \frac{625}{9} \Rightarrow (A X)^{2} = \frac{1250}{9 tan θ} A X \approx 21.8 cm

(e)

3 marks

The material needed is 10% more than 625 cm $^{3}$ , at 0.025 USD per cubic centimetre, and the price is 20% more than the cost.
The price is about 20.63 USD.

625 \times 1.1 \times 0.025 \times 1.2 = 20.625

Question 6

14 marks

Virus particles and healthy skin cells: a differential equation

Calculus

(a)

3 marks

The rate is proportional to $\frac{h}{v}$ , so $\frac{d h}{d t} = k \frac{h}{v}$ . At $t = 0$ , $h = 1350$ and $v = 27$ .

- 5 = k \frac{1350}{27} = 50 k \Rightarrow k = - \frac{1}{10} \frac{d h}{d t} = - \frac{h}{10 v}

(b)

7 marks

Put $v = 27 + 0.3 t$ and separate the variables.
Integrate.
Use $h = 1350$ at $t = 0$ : $1350 = A (27)^{- 1/3} = \frac{A}{3}$ .

\int \frac{1}{h} d h = - \int \frac{1}{10 ( 27 + 0.3 t )} d t ln h = - \frac{1}{10 ( 0.3 )} ln (27 + 0.3 t) + c = - \frac{1}{3} ln (27 + 0.3 t) + c h = A (27 + 0.3 t)^{- 1/3} A = 4050 \Rightarrow h = 4050 (27 + 0.3 t)^{- 1/3}

(c)

4 marks

The number of virus particles is $1000 v$ (since $v$ is in thousands). We want $1000 v \geq 100 h$ .
Solve with the calculator.
To be at least 100 times, the answer is the next whole hour, 217 hours.

1000 (27 + 0.28 t) \geq 100 \times 4380 (27 + 0.28 t)^{- 0.36} t \approx 216.3 hours

Question 7

12 marks

A table designed by rotating a region about the y-axis

Calculus

(a)

4 marks

area under the curve = \int_{0.1}^{1} \frac{2}{3} cos^{- 1} (\frac{1}{10 x}) d x \approx 0.7809 shaded area = 1 - 0.7809 \approx 0.219 m^{2}

(b)

6 marks

Rearrange for $x$ : $cos (\frac{3 y}{2}) = \frac{1}{10 x}$ , so $x = \frac{1}{10 cos \frac{3 y}{2}}$ . At $x = 1$ , $y = \frac{2}{3} cos^{- 1} (0.1) = 0.9804$ .
The top of the table is a cylinder of radius 1 and height $1 - 0.9804 = 0.0196$ .
The curved part, from $y = 0$ to $y = 0.9804$ , has radius $x$ .
The base is a cylinder of radius 0.3 and height 0.075.

V_{1} = π (1)^{2} (0.0196) V_{2} = π \int_{0}^{0.9804} (\frac{1}{10 cos \frac{3 y}{2}})^{2} d y V_{3} = π (0.3)^{2} (0.075) V = π (1)^{2} (0.0196) + π \int_{0}^{0.9804} (\frac{1}{10 cos \frac{3 y}{2}})^{2} d y + π (0.3)^{2} (0.075)

(c)

2 marks

Evaluate with the calculator.

V \approx 0.291 m^{3}

Our analysis

Where the marks are

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

Calculus43 marks · 39%
Q3, Q6, Q7
Number and algebra33 marks · 30%
Q4, Q5
Geometry and trigonometry19 marks · 17%
Q2
Statistics and probability15 marks · 14%
Q1

Mentor insights

Traps to watch for

01
Mean from a frequency table · Q1
$\overset{x}{ˉ} = \frac{\sum f x}{\sum f}$ .
02
Perpendicular bisector · Q2
The edge between two sites is the perpendicular bisector of the segment joining them.
03
Coefficient of determination · Q3
$R^{2}$ close to 1 means the model explains the data well.
04
Speed · Q4
The speed is the length of the direction vector.
05
Right angle · Q5
Check with Pythagoras or with a scalar product.

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