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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

16 marks

Thai cushions made of layers of triangular prisms

Number and algebra

(a)

2 marks

Read the number of prisms in the bottom layer from the diagrams.

7 for 4 layers, 9 for 5 layers

(b)

3 marks

(i) The difference between the 4-layer and 5-layer bottom layers is 2.

(ii) With 1 prism in the first cushion (1 layer), $u_{n} = 1 + 2 (n - 1)$ .

d = 2 u_{n} = 2 n - 1

(c)

3 marks

(i) The bottom layer of 9 layers.

(ii) The layers have $1, 3, 5, \dots, 17$ prisms. Their sum is

u_{9} = 2 (9) - 1 = 17 \frac{9}{2} (1 + 17) = 81

(d)

2 marks

The layers have $1, 3, 5, \dots, 2 n - 1$ prisms.

S_{n} = \frac{n}{2} (1 + (2 n - 1)) = n^{2}

(e)

1 mark

The cushion with 4 layers has $4^{2} = 16$ triangles in all, of which 6 are white.

16 - 6 = 10

(f)

2 marks

The black triangles per layer form an arithmetic sequence. The first layer has 1, and the fifth layer has $9 - 4 = 5$ , so the layers have $1, 2, 3, \dots, n$ black triangles.

black = 1 + 2 + 3 + \dots + n = \frac{n ( n + 1 )}{2}

(g)

3 marks

Add the black and the white triangles.

\frac{n ( n + 1 )}{2} + \frac{n ( n - 1 )}{2} = \frac{n [ ( n + 1 ) + ( n - 1 ) ]}{2} = \frac{2 n ^{2}}{2} = n^{2}

Question 2

12 marks

An exponential and a reciprocal function: intersections, areas and parallel tangents

Calculus

(a)

3 marks

Solve $1. 5^{x} = 6 - \frac{3}{x}$ .

x \approx 0.638 or x \approx 4.10

(b)

6 marks

(i) The shaded region is under the curve $y = f (x)$ between the intersection points.

(ii) Evaluate.

(iii) The region between the curves is under $g$ minus under $f$ .

\int_{0.638}^{4.10} 1. 5^{x} d x Area \approx 9.80 \int_{0.638}^{4.10} (6 - \frac{3}{x} - 1. 5^{x}) d x \approx 5.38

(c)

3 marks

The tangents are parallel when $f^{'} (k) = g^{'} (k)$ , where $f^{'} (x) = 1. 5^{x} ln 1.5$ and $g^{'} (x) = \frac{3}{x ^{2}}$ .
Solve with the calculator (find where the graphs of the two derivatives meet).

1. 5^{k} ln 1.5 = \frac{3}{k ^{2}} k \approx 1.86

Question 3

16 marks

A cover made from a cylinder and a hemisphere: minimising the plastic

Calculus

(a)

2 marks

The curved surface of the cylinder is $2 π r h$ and the curved surface of a hemisphere is $2 π r^{2}$ .

A = 2 π r^{2} + 2 π r h

(b)

2 marks

A cylinder is $π r^{2} h$ and a hemisphere is $\frac{2}{3} π r^{3}$ .

V = π r^{2} h + \frac{2 π r ^{3}}{3} = \frac{3 π r ^{2} h + 2 π r ^{3}}{3}

(c)

2 marks

Put $V = 10000$ and solve for $h$ .

3 π r^{2} h + 2 π r^{3} = 30000 \Rightarrow h = \frac{30000 - 2 π r ^{3}}{3 π r ^{2}}

(d)

2 marks

Substitute $h$ into $A = 2 π r^{2} + 2 π r h$ .

A = 2 π r^{2} + 2 π r \cdot \frac{30000 - 2 π r ^{3}}{3 π r ^{2}} = 2 π r^{2} + \frac{2 ( 30000 - 2 π r ^{3} )}{3 r} A = 2 π r^{2} + \frac{20000}{r} - \frac{4 π r ^{2}}{3} = \frac{2 π r ^{2}}{3} + \frac{20000}{r}

(e)

3 marks

Differentiate $A = \frac{2 π}{3} r^{2} + 20000 r^{- 1}$ .

\frac{d A}{d r} = \frac{4 π r}{3} - \frac{20000}{r ^{2}}

(f)

4 marks

Set the derivative to zero.
Substitute into $h = \frac{30000 - 2 π r ^{3}}{3 π r ^{2}}$ , using $2 π r^{3} = 30000$ .

\frac{4 π r}{3} = \frac{20000}{r ^{2}} \Rightarrow r^{3} = \frac{15000}{π} \Rightarrow r \approx 16.8 cm h = 0

(g)

1 mark

With $h = 0$ the cylinder disappears and the cover is just a hemisphere, so a hemisphere is the best shape for minimum plastic.

Question 4

13 marks

Cabling a wind farm and inspecting it: Prim's, nearest neighbour and bounds

Geometry and trigonometry

(a)

4 marks

Start at $A$ . The shortest edge from $A$ is $A E$ (0.86). Then from ${A, E}$ the shortest new edge is $E C$ (0.62). Then $C B$ (0.74), and then $C D$ (0.78).

A \to E \to C \to B \to D 0.86 + 0.62 + 0.74 + 0.78 = 3.00 km

(b)

2 marks

Delete $F$ , and the minimum spanning tree of the five turbines has length 3.00. Add the two shortest edges from $F$ : $F A = 0.96$ and $F E = 1.14$ .

lower bound = 3.00 + 0.96 + 1.14 = 5.10 km

(c)

3 marks

Start at $F$ . The nearest turbine is $A$ (0.96). From $A$ the nearest unvisited is $E$ (0.86), then $C$ (0.62), then $B$ (0.74), then $D$ (0.94), and then back to $F$ (2.24).

0.96 + 0.86 + 0.62 + 0.74 + 0.94 + 2.24 = 6.36 km

(d)

4 marks

The best (largest) lower bound in the table is 5.02 (from deleting $A$ or $C$ ) and the best (smallest) upper bound is 6.36 (from starting at $A$ , $B$ or $F$ ).
The inspections take $5 \times 12 = 60$ minutes, which is 1 hour. Travel time is distance divided by 28 km/h.

\frac{5.02}{28} + 1 \approx 1.18, \frac{6.36}{28} + 1 \approx 1.23 1.18 \leq T \leq 1.23

Question 5

15 marks

Cane beetles: a mean from grouped data, probability, a chi-squared test and a binomial test

Statistics and probability

(a)

1 mark

Add all the frequencies.

42 + 25 + 19 + 61 + 27 + 12 = 186

(b)

2 marks

The class midpoints are 11, 14 and 17. The frequencies are $103$ , $52$ and $31$ .

\overset{x}{ˉ} = \frac{11 ( 103 ) + 14 ( 52 ) + 17 ( 31 )}{186} \approx 12.8 mm

(c)

3 marks

There are $42 + 25 + 19 = 86$ females, and 42 of them are small. The beetles are chosen with replacement, so the two choices are independent.

(\frac{42}{86})^{2} = \frac{441}{1849} \approx 0.239

(d)

4 marks

$H_{0}$ : length category and sex are independent. $H_{1}$ : they are not independent. Run a $χ^{2}$ test for independence on the $2 \times 3$ table.
Since $0.127 > 0.05$ , do not reject $H_{0}$ . There is no significant evidence that length category and sex are related for these beetles.

p -value \approx 0.127

(e)

5 marks

$H_{0}$ : $ϕ = 0.45$ . $H_{1}$ : $ϕ > 0.45$ .
The number of males is $61 + 27 + 12 = 100$ . If $H_{0}$ is true, the number of males $X \sim B (186, 0.45)$ .
Since $0.0101 < 0.05$ , reject $H_{0}$ . There is significant evidence that more than 45% of cane beetles are male.

P (X \geq 100) \approx 0.0101

Question 6

21 marks

A share price model from a second-order differential equation

Calculus

(a)

5 marks

The matrix of the system is $(0 - 3 1 - 4)$ . Find its eigenvalues.

λ (λ + 4) + 3 = 0 \Rightarrow λ^{2} + 4 λ + 3 = 0 \Rightarrow (λ + 1) (λ + 3) = 0 λ = - 1 or - 3 x = A e^{- t} + B e^{- 3 t}

(b)

6 marks

(i) Use $x (0) = 0$ and $x^{'} (0) = - 1$ .

(ii) Sketch: the graph starts at $(0, 0)$ , dips to a minimum of about $- 0.19$ near $t = 0.55$ and then rises back towards the $t$ -axis, approaching $x = 0$ from below without crossing it.

A + B = 0, x^{'} (0) = - A - 3 B = - 1 B = \frac{1}{2}, A = - \frac{1}{2} \Rightarrow x = \frac{1}{2} e^{- 3 t} - \frac{1}{2} e^{- t}

(c)

4 marks

(i) The best time to buy is when the price is lowest, at the minimum of the graph. Solving $x^{'} (t) = 0$ gives $- \frac{3}{2} e^{- 3 t} + \frac{1}{2} e^{- t} = 0$ , so $e^{- 2 t} = \frac{1}{3}$ .

(ii) The price later tends to 0, so the most she can gain per share is the size of the minimum.

t = \frac{1}{2} ln 3 \approx 0.549 minutes x_{m i n} = \frac{1}{2} e^{- 1.648} - \frac{1}{2} e^{- 0.549} \approx - 0.192, profit \leq 0.192 dollars

(d)

6 marks

With $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 4 y - 3 x + x sin t$ , apply Euler's method from $t = 0$ , $x = 0$ , $y = - 1$ with step 0.1.
The first step gives $x_{1} = - 0.1$ and $y_{1} = - 0.6$ . After 10 steps, at $t = 1$ :

x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (- 4 y_{n} - 3 x_{n} + x_{n} sin t_{n}), t_{n + 1} = t_{n} + 0.1 x \approx - 0.172

Question 7

17 marks

Charging a phone: a Markov chain

Number and algebra

(a)

2 marks

The first column is the probability of charging today and of not charging today, given charged yesterday: 0.4 and 0.6.

a = 0.4, b = 0.6 M = (0.4 0.6 p 1 - p)

(b)

5 marks

(i) He charges it on each of the four days, with probability 0.4 each time.

(ii) With $p = 0.7$ , find $M^{4}$ times the starting vector $(1, 0)$ .

0. 4^{4} = 0.0256 M^{4} (10) = (0.5422 0.4578) P (charges on day 4) \approx 0.542

(c)

4 marks

Multiply $M$ by $(1 - 1)$ .
So $(1 - 1)$ is an eigenvector, for every $p$ , with eigenvalue

(0.4 0.6 p 1 - p) (1 - 1) = (0.4 - p p - 0.4) = (0.4 - p) (1 - 1) λ = 0.4 - p

(d)

4 marks

The steady state satisfies $M v = v$ with $x + y = 1$ . From the first row,
Use $x + y = 1$ .

0.4 x + p y = x \Rightarrow p y = 0.6 x \Rightarrow y = \frac{0.6 x}{p} x + \frac{0.6 x}{p} = 1 \Rightarrow x = \frac{p}{p + 0.6}

(e)

2 marks

Require $\frac{p}{p + 0.6} \geq 0.6$ .

p \geq 0.6 p + 0.36 \Rightarrow 0.4 p \geq 0.36 \Rightarrow p \geq 0.9 p_{m i n} = 0.9

Our analysis

Where the marks are

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

Calculus49 marks · 45%
Q2, Q3, Q6
Number and algebra33 marks · 30%
Q1, Q7
Statistics and probability15 marks · 14%
Q5
Geometry and trigonometry13 marks · 12%
Q4

Mentor insights

Traps to watch for

01
Arithmetic sequence · Q1
The number of prisms in the bottom layer goes up by the same amount each layer.
02
Surface and volume · Q3
The cover has a curved surface of a cylinder and of a hemisphere, with no base.
03
Grouped mean · Q5
Use the middle of each class.
04
Coupled system · Q6
Write $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 3 x - 4 y$ , which is $\frac{d}{d t} (x y) = (0 - 3 1 - 4) (x y)$ .
05
Transition matrix · Q7
The columns give the next-day probabilities from "charged" and from "not charged".

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.

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Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

16 marks

Thai cushions made of layers of triangular prisms

Number and algebra

(a)

2 marks

Read the number of prisms in the bottom layer from the diagrams.

7 for 4 layers, 9 for 5 layers

(b)

3 marks

(i) The difference between the 4-layer and 5-layer bottom layers is 2.

(ii) With 1 prism in the first cushion (1 layer), $u_{n} = 1 + 2 (n - 1)$ .

d = 2 u_{n} = 2 n - 1

(c)

3 marks

(i) The bottom layer of 9 layers.

(ii) The layers have $1, 3, 5, \dots, 17$ prisms. Their sum is

u_{9} = 2 (9) - 1 = 17 \frac{9}{2} (1 + 17) = 81

(d)

2 marks

The layers have $1, 3, 5, \dots, 2 n - 1$ prisms.

S_{n} = \frac{n}{2} (1 + (2 n - 1)) = n^{2}

(e)

1 mark

The cushion with 4 layers has $4^{2} = 16$ triangles in all, of which 6 are white.

16 - 6 = 10

(f)

2 marks

The black triangles per layer form an arithmetic sequence. The first layer has 1, and the fifth layer has $9 - 4 = 5$ , so the layers have $1, 2, 3, \dots, n$ black triangles.

black = 1 + 2 + 3 + \dots + n = \frac{n ( n + 1 )}{2}

(g)

3 marks

Add the black and the white triangles.

\frac{n ( n + 1 )}{2} + \frac{n ( n - 1 )}{2} = \frac{n [ ( n + 1 ) + ( n - 1 ) ]}{2} = \frac{2 n ^{2}}{2} = n^{2}

Question 2

12 marks

An exponential and a reciprocal function: intersections, areas and parallel tangents

Calculus

(a)

3 marks

Solve $1. 5^{x} = 6 - \frac{3}{x}$ .

x \approx 0.638 or x \approx 4.10

(b)

6 marks

(i) The shaded region is under the curve $y = f (x)$ between the intersection points.

(ii) Evaluate.

(iii) The region between the curves is under $g$ minus under $f$ .

\int_{0.638}^{4.10} 1. 5^{x} d x Area \approx 9.80 \int_{0.638}^{4.10} (6 - \frac{3}{x} - 1. 5^{x}) d x \approx 5.38

(c)

3 marks

The tangents are parallel when $f^{'} (k) = g^{'} (k)$ , where $f^{'} (x) = 1. 5^{x} ln 1.5$ and $g^{'} (x) = \frac{3}{x ^{2}}$ .
Solve with the calculator (find where the graphs of the two derivatives meet).

1. 5^{k} ln 1.5 = \frac{3}{k ^{2}} k \approx 1.86

Question 3

16 marks

A cover made from a cylinder and a hemisphere: minimising the plastic

Calculus

(a)

2 marks

The curved surface of the cylinder is $2 π r h$ and the curved surface of a hemisphere is $2 π r^{2}$ .

A = 2 π r^{2} + 2 π r h

(b)

2 marks

A cylinder is $π r^{2} h$ and a hemisphere is $\frac{2}{3} π r^{3}$ .

V = π r^{2} h + \frac{2 π r ^{3}}{3} = \frac{3 π r ^{2} h + 2 π r ^{3}}{3}

(c)

2 marks

Put $V = 10000$ and solve for $h$ .

3 π r^{2} h + 2 π r^{3} = 30000 \Rightarrow h = \frac{30000 - 2 π r ^{3}}{3 π r ^{2}}

(d)

2 marks

Substitute $h$ into $A = 2 π r^{2} + 2 π r h$ .

A = 2 π r^{2} + 2 π r \cdot \frac{30000 - 2 π r ^{3}}{3 π r ^{2}} = 2 π r^{2} + \frac{2 ( 30000 - 2 π r ^{3} )}{3 r} A = 2 π r^{2} + \frac{20000}{r} - \frac{4 π r ^{2}}{3} = \frac{2 π r ^{2}}{3} + \frac{20000}{r}

(e)

3 marks

Differentiate $A = \frac{2 π}{3} r^{2} + 20000 r^{- 1}$ .

\frac{d A}{d r} = \frac{4 π r}{3} - \frac{20000}{r ^{2}}

(f)

4 marks

Set the derivative to zero.
Substitute into $h = \frac{30000 - 2 π r ^{3}}{3 π r ^{2}}$ , using $2 π r^{3} = 30000$ .

\frac{4 π r}{3} = \frac{20000}{r ^{2}} \Rightarrow r^{3} = \frac{15000}{π} \Rightarrow r \approx 16.8 cm h = 0

(g)

1 mark

With $h = 0$ the cylinder disappears and the cover is just a hemisphere, so a hemisphere is the best shape for minimum plastic.

Question 4

13 marks

Cabling a wind farm and inspecting it: Prim's, nearest neighbour and bounds

Geometry and trigonometry

(a)

4 marks

Start at $A$ . The shortest edge from $A$ is $A E$ (0.86). Then from ${A, E}$ the shortest new edge is $E C$ (0.62). Then $C B$ (0.74), and then $C D$ (0.78).

A \to E \to C \to B \to D 0.86 + 0.62 + 0.74 + 0.78 = 3.00 km

(b)

2 marks

Delete $F$ , and the minimum spanning tree of the five turbines has length 3.00. Add the two shortest edges from $F$ : $F A = 0.96$ and $F E = 1.14$ .

lower bound = 3.00 + 0.96 + 1.14 = 5.10 km

(c)

3 marks

Start at $F$ . The nearest turbine is $A$ (0.96). From $A$ the nearest unvisited is $E$ (0.86), then $C$ (0.62), then $B$ (0.74), then $D$ (0.94), and then back to $F$ (2.24).

0.96 + 0.86 + 0.62 + 0.74 + 0.94 + 2.24 = 6.36 km

(d)

4 marks

The best (largest) lower bound in the table is 5.02 (from deleting $A$ or $C$ ) and the best (smallest) upper bound is 6.36 (from starting at $A$ , $B$ or $F$ ).
The inspections take $5 \times 12 = 60$ minutes, which is 1 hour. Travel time is distance divided by 28 km/h.

\frac{5.02}{28} + 1 \approx 1.18, \frac{6.36}{28} + 1 \approx 1.23 1.18 \leq T \leq 1.23

Question 5

15 marks

Cane beetles: a mean from grouped data, probability, a chi-squared test and a binomial test

Statistics and probability

(a)

1 mark

Add all the frequencies.

42 + 25 + 19 + 61 + 27 + 12 = 186

(b)

2 marks

The class midpoints are 11, 14 and 17. The frequencies are $103$ , $52$ and $31$ .

\overset{x}{ˉ} = \frac{11 ( 103 ) + 14 ( 52 ) + 17 ( 31 )}{186} \approx 12.8 mm

(c)

3 marks

There are $42 + 25 + 19 = 86$ females, and 42 of them are small. The beetles are chosen with replacement, so the two choices are independent.

(\frac{42}{86})^{2} = \frac{441}{1849} \approx 0.239

(d)

4 marks

$H_{0}$ : length category and sex are independent. $H_{1}$ : they are not independent. Run a $χ^{2}$ test for independence on the $2 \times 3$ table.
Since $0.127 > 0.05$ , do not reject $H_{0}$ . There is no significant evidence that length category and sex are related for these beetles.

p -value \approx 0.127

(e)

5 marks

$H_{0}$ : $ϕ = 0.45$ . $H_{1}$ : $ϕ > 0.45$ .
The number of males is $61 + 27 + 12 = 100$ . If $H_{0}$ is true, the number of males $X \sim B (186, 0.45)$ .
Since $0.0101 < 0.05$ , reject $H_{0}$ . There is significant evidence that more than 45% of cane beetles are male.

P (X \geq 100) \approx 0.0101

Question 6

21 marks

A share price model from a second-order differential equation

Calculus

(a)

5 marks

The matrix of the system is $(0 - 3 1 - 4)$ . Find its eigenvalues.

λ (λ + 4) + 3 = 0 \Rightarrow λ^{2} + 4 λ + 3 = 0 \Rightarrow (λ + 1) (λ + 3) = 0 λ = - 1 or - 3 x = A e^{- t} + B e^{- 3 t}

(b)

6 marks

(i) Use $x (0) = 0$ and $x^{'} (0) = - 1$ .

(ii) Sketch: the graph starts at $(0, 0)$ , dips to a minimum of about $- 0.19$ near $t = 0.55$ and then rises back towards the $t$ -axis, approaching $x = 0$ from below without crossing it.

A + B = 0, x^{'} (0) = - A - 3 B = - 1 B = \frac{1}{2}, A = - \frac{1}{2} \Rightarrow x = \frac{1}{2} e^{- 3 t} - \frac{1}{2} e^{- t}

(c)

4 marks

(i) The best time to buy is when the price is lowest, at the minimum of the graph. Solving $x^{'} (t) = 0$ gives $- \frac{3}{2} e^{- 3 t} + \frac{1}{2} e^{- t} = 0$ , so $e^{- 2 t} = \frac{1}{3}$ .

(ii) The price later tends to 0, so the most she can gain per share is the size of the minimum.

t = \frac{1}{2} ln 3 \approx 0.549 minutes x_{m i n} = \frac{1}{2} e^{- 1.648} - \frac{1}{2} e^{- 0.549} \approx - 0.192, profit \leq 0.192 dollars

(d)

6 marks

With $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 4 y - 3 x + x sin t$ , apply Euler's method from $t = 0$ , $x = 0$ , $y = - 1$ with step 0.1.
The first step gives $x_{1} = - 0.1$ and $y_{1} = - 0.6$ . After 10 steps, at $t = 1$ :

x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (- 4 y_{n} - 3 x_{n} + x_{n} sin t_{n}), t_{n + 1} = t_{n} + 0.1 x \approx - 0.172

Question 7

17 marks

Charging a phone: a Markov chain

Number and algebra

(a)

2 marks

The first column is the probability of charging today and of not charging today, given charged yesterday: 0.4 and 0.6.

a = 0.4, b = 0.6 M = (0.4 0.6 p 1 - p)

(b)

5 marks

(i) He charges it on each of the four days, with probability 0.4 each time.

(ii) With $p = 0.7$ , find $M^{4}$ times the starting vector $(1, 0)$ .

0. 4^{4} = 0.0256 M^{4} (10) = (0.5422 0.4578) P (charges on day 4) \approx 0.542

(c)

4 marks

Multiply $M$ by $(1 - 1)$ .
So $(1 - 1)$ is an eigenvector, for every $p$ , with eigenvalue

(0.4 0.6 p 1 - p) (1 - 1) = (0.4 - p p - 0.4) = (0.4 - p) (1 - 1) λ = 0.4 - p

(d)

4 marks

The steady state satisfies $M v = v$ with $x + y = 1$ . From the first row,
Use $x + y = 1$ .

0.4 x + p y = x \Rightarrow p y = 0.6 x \Rightarrow y = \frac{0.6 x}{p} x + \frac{0.6 x}{p} = 1 \Rightarrow x = \frac{p}{p + 0.6}

(e)

2 marks

Require $\frac{p}{p + 0.6} \geq 0.6$ .

p \geq 0.6 p + 0.36 \Rightarrow 0.4 p \geq 0.36 \Rightarrow p \geq 0.9 p_{m i n} = 0.9

Our analysis

Where the marks are

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

Calculus49 marks · 45%
Q2, Q3, Q6
Number and algebra33 marks · 30%
Q1, Q7
Statistics and probability15 marks · 14%
Q5
Geometry and trigonometry13 marks · 12%
Q4

Mentor insights

Traps to watch for

01
Arithmetic sequence · Q1
The number of prisms in the bottom layer goes up by the same amount each layer.
02
Surface and volume · Q3
The cover has a curved surface of a cylinder and of a hemisphere, with no base.
03
Grouped mean · Q5
Use the middle of each class.
04
Coupled system · Q6
Write $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 3 x - 4 y$ , which is $\frac{d}{d t} (x y) = (0 - 3 1 - 4) (x y)$ .
05
Transition matrix · Q7
The columns give the next-day probabilities from "charged" and from "not charged".

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