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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 8

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8

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

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Questions

Questions 1–8 · 110 marks

Question 1

15 marks

Delivery times: a normal distribution, a binomial count and delivery charges

Statistics and probability
Functions

(a)

1 mark

64 is the mean and the normal distribution is symmetric.

P (T < 64) = 0.5

(b)

2 marks

Use the normal CDF on the calculator.

P (44 < T < 64) \approx 0.452

(c)

4 marks

(i) Sketch: a bell-shaped curve centred on 64 with the area to the left of a value $k$ (below the mean) shaded.

(ii) Use the inverse normal with area 0.3, mean 64 and standard deviation 12.

k \approx 57.7

(d)

3 marks

Each package is below $k$ with probability 0.3. Let $X \sim B (5, 0.3)$ .

P (X = 2) = (2 5) (0.3)^{2} (0.7)^{3} \approx 0.309

(e)

2 marks

The charge is $4.50 plus $2.00 for each kilogram above 1.

C (x) = 4.5 + 2 (x - 1) = 2 x + 2.5

(f)

1 mark

Substitute $x = 5.3$ .

C (5.3) = 2 (5.3) + 2.5 = $13.10

(g)

2 marks

Set the charge equal to $7.20.

2 x + 2.5 = 7.2 \Rightarrow x = 2.35 kg

Question 2

12 marks

Three mountain summits: a vector, an angle and the sine rule

Number and algebra

(a)

3 marks

(i) Subtract coordinates, $B - A$ .

(ii) The distance is the length of the vector.

A B = 7.2 5.1 2.4 - 2.8 = 7.2 5.1 - 0.4 A B = 7. 2^{2} + 5. 1^{2} + 0. 4^{2} \approx 8.83 km

(b)

5 marks

The two vectors are $A B$ and $1.1 8.4 0.2$ , whose length is $1. 1^{2} + 8. 4^{2} + 0. 2^{2} \approx 8.474$ .

scalar product = 1.1 (7.2) + 8.4 (5.1) + 0.2 (- 0.4) = 7.92 + 42.84 - 0.08 = 50.68 cos θ = \frac{50.68}{8.832 \times 8.474} \Rightarrow θ \approx 47. 4^{\circ}

(c)

4 marks

This is the angle at $A$ in triangle $A B C$ (since $A C$ is parallel to the given vector). With the angle at $B$ equal to $55. 2^{\circ}$ , the angle at $C$ is
Use the sine rule for the side $A C$ opposite $B$ .

A C B = 180 - 47.38 - 55.2 \approx 77.4 2^{\circ} \frac{A C}{sin 55. 2 ^{\circ}} = \frac{8.832}{sin 77.4 2 ^{\circ}} \Rightarrow A C \approx 7.43 km

Question 3

13 marks

Smoothie sales and profit: a model n = 40000/x^2

Functions
Calculus

(a)

2 marks

Solve $\frac{40000}{x ^{2}} = 400$ for $x > 0$ .

x^{2} = 100 \Rightarrow x = 10 pesos

(b)

2 marks

(i) At $x = 50$ .

(ii) The income is the number sold times the price.

n = \frac{40000}{5 0 ^{2}} = 16 16 \times 50 = 800 pesos

(c)

9 marks

(i) Profit is the number sold times the profit per smoothie.

(ii) Differentiate $P = 40000 x^{- 1} - 800000 x^{- 2}$ .

(iii) Set the derivative to zero and multiply by $x^{3}$ .

(iv) Substitute $x = 40$ in the model for $n$ .

Since this is the only turning point and $P \to 0$ as $x \to \infty$ , this is the maximum profit.

P = \frac{40000}{x ^{2}} (x - 20) = \frac{40000}{x} - \frac{800000}{x ^{2}} \frac{d P}{d x} = - \frac{40000}{x ^{2}} + \frac{1600000}{x ^{3}} - 40000 x + 1600000 = 0 \Rightarrow x = 40 n = \frac{40000}{4 0 ^{2}} = 25

Question 4

12 marks

Two switches in a generator: a tree and a chi-squared test

Statistics and probability

(a)

2 marks

The switch $A$ fails with probability 0.1, so it does not fail with probability 0.9. The switch $B$ fails with probability 0.3, so it does not fail with probability 0.7.

a = 0.9, b = 0.3, c = 0.7

(b)

1 mark

Both switches fail.

0.1 \times 0.3 = 0.03

(c)

9 marks

State the hypotheses.
Find the probabilities of 0, 1 and 2 failures.
Multiply by 200 to get the expected frequencies.
All expected values are at least 5, so the test is valid. There are 3 categories, so $ν = 2$ .
Since $0.182 > 0.05$ there is not enough evidence to reject $H_{0}$ . The data are consistent with the manufacturer's claims.

H_{0} : the manufacturer’s claims are correct, H_{1} : they are not both correct P (0) = 0.9 \times 0.7 = 0.63, P (1) = 0.1 \times 0.7 + 0.9 \times 0.3 = 0.34, P (2) = 0.03 126, 68, 6 (observed 118, 72, 10) χ^{2} = \frac{( 118 - 126 ) ^{2}}{126} + \frac{( 72 - 68 ) ^{2}}{68} + \frac{( 10 - 6 ) ^{2}}{6} \approx 3.41, p \approx 0.182

Question 5

13 marks

A separable differential equation and its slope field

Calculus

(a)

2 marks

The gradient is $x e^{- 2 y}$ . It is zero along $x = 0$ and positive for every $x > 0$ , whatever $y$ is. Diagram C is the one with these features.

(b)

7 marks

Separate the variables and integrate both sides.
Use $y = 0$ when $x = 0$ .
Take the natural logarithm.

\int e^{2 y} d y = \int x d x \Rightarrow \frac{1}{2} e^{2 y} = \frac{1}{2} x^{2} + c \frac{1}{2} = c \Rightarrow e^{2 y} = x^{2} + 1 y = \frac{1}{2} ln (x^{2} + 1)

(c)

2 marks

Differentiate with the chain rule.

\frac{d y}{d x} = \frac{1}{2} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x = \frac{x}{x ^{2} + 1}

(d)

2 marks

From part (b), $e^{2 y} = x^{2} + 1$ , so $e^{- 2 y} = \frac{1}{x ^{2} + 1}$ .
The expression in (c) equals the right side of the differential equation, so $y = \frac{1}{2} ln (x^{2} + 1)$ is a solution.

x e^{- 2 y} = \frac{x}{x ^{2} + 1} = \frac{d y}{d x}

Question 6

13 marks

Spaceships and battleships: Poisson, expected score and the central limit theorem

Statistics and probability

(a)

5 marks

(i) In 2 minutes the spaceships have mean $2 \times 4.2 = 8.4$ .

(ii) The total of both kinds of target in one minute is Poisson with mean $4.2 + 2.3 = 6.5$ .

P (S \leq 10) \approx 0.774 P (S + B > 10) = 1 - P (S + B \leq 10) \approx 0.0668

(b)

3 marks

(i) The expected points are 3 per spaceship and 5 per battleship.

(ii) The variance of a Poisson variable is its mean.

E (T) = 3 (4.2) + 5 (2.3) = 24.1 Var (T) = 3^{2} (4.2) + 5^{2} (2.3) = 37.8 + 57.5 = 95.3

(c)

1 mark

A Poisson distribution has its mean equal to its variance, but $E (T) = 24.1$ and $Var (T) = 95.3$ are different.

(d)

4 marks

The mean score $\overset{ˉ}{T}$ over 60 minutes is approximately normal.
Find the probability it is above 25.

\overset{ˉ}{T} \approx N (24.1, \frac{95.3}{60}) P (\overset{ˉ}{T} > 25) \approx 0.238

Question 7

14 marks

A vase modelled by revolving a curve: volume, filling time and rate of rise

Calculus

(a)

5 marks

From $y = \frac{1}{2} x^{2} - 1$ get $x^{2} = 2 y + 2$ . The water up to height $h$ is a solid of revolution about the $y$ -axis.

V = π \int_{0}^{h} (2 y + 2) d y = π [y^{2} + 2 y]_{0}^{h} V = π (h^{2} + 2 h)

(b)

2 marks

The vase is full at $h = 15$ .
Divide by the filling rate.

V = π (1 5^{2} + 2 \times 15) = 255 π \approx 801.1 cm^{3} t = \frac{801.1}{20} \approx 40.1 s

(c)

7 marks

Differentiate the volume with respect to $h$ .
Use the chain rule with $\frac{d V}{d t} = 20$ .
Substitute $h = 10$ .

\frac{d V}{d h} = π (2 h + 2) \frac{d h}{d t} = \frac{d V / d t}{d V / d h} = \frac{20}{π ( 2 h + 2 )} \frac{d h}{d t} = \frac{20}{22 π} \approx 0.289 cm s^{- 1}

Question 8

18 marks

Walks on a graph, a bird on three posts and two birds

Number and algebra
Statistics and probability

(a)

2 marks

Each post is joined to itself (a loop) and $A$ to $B$ , $B$ to $C$ in both directions. With the rows and columns in the order $A, B, C$ :

M = 110101011

(b)

3 marks

Raise the matrix to the fifth power. The entry in the row $A$ and column $B$ is the number of walks of length 5 from $A$ to $B$ .

M^{5} = 111110111011101111 11 walks

(c)

3 marks

(i) Every flight has two equally likely choices, so a sequence of 5 flights has probability

(ii) From $A$ there are 11 walks of length 5 ending at $B$ , each with probability $\frac{1}{32}$ .

(\frac{1}{2})^{5} = \frac{1}{32} \frac{11}{32} \approx 0.344

(d)

3 marks

The states are the occupied pairs $A B$ , $A C$ , $B C$ . From $A B$ the bird on $B$ flies to the empty post $C$ , while the other bird goes to $B$ with probability 0.4 or stays with 0.6.
From $A C$ , each bird goes to $B$ with probability 0.5, and the new state is $B C$ if only $A$ 's bird moves ( $0.5 \times 0.5$ ) or $A B$ if only $C$ 's bird moves ( $0.5 \times 0.5$ ).
The state stays $A C$ if neither moves ( $0.5 \times 0.5$ ) or if both try to move to $B$ at the same time ( $0.5 \times 0.5$ ).

a = 0.4 b = 0.25 c = 0.25 + 0.25 = 0.5

(e)

4 marks

The transition matrix in the order $A B$ , $A C$ , $B C$ has columns (from) $A B$ : $(0, 0.6, 0.4)$ , $A C$ : $(0.25, 0.5, 0.25)$ and $B C$ : $(0.4, 0.6, 0)$ .
Start in the state $A B$ and take 5 steps.
The entry in the row $B C$ and the column $A B$ is the required probability.

T = 0 0.6 0.4 0.25 0.5 0.25 0.4 0.6 0 T^{5} \approx 0.2222 0.5455 0.2324 0.2273 0.5455 0.2273 0.2324 0.5455 0.2222 P (B C after 5 flights) \approx 0.232

(f)

3 marks

A high power of $T$ has all columns equal to the long-term proportions $(\frac{5}{22}, \frac{6}{11}, \frac{5}{22})$ for $A B$ , $A C$ , $B C$ .
Post $B$ is occupied in the states $A B$ and $B C$ , so it is occupied $\frac{5}{22} + \frac{5}{22} = \frac{5}{11}$ of the time. Posts $A$ and $C$ are each occupied $\frac{17}{22}$ of the time.
So post $B$ is the one that is sat on least, and the proportion of the time that it is free is

1 - \frac{5}{11} = \frac{6}{11} \approx 0.545

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 probability48 marks · 44%
Q1, Q4, Q6, Q8
Calculus36 marks · 33%
Q3, Q5, Q7
Number and algebra17 marks · 15%
Q2, Q8
Functions9 marks · 8%
Q1, Q3

Mentor insights

Traps to watch for

01
Normal and binomial · Q1
Part (c) turns the time distribution into a probability of 0.3, and part (d) counts how many of five packages are below $k$ .
02
Profit per item · Q3
Each smoothie earns $x - 20$ pesos, so profit is $n (x - 20)$ .
03
Separate the variables · Q5
$\frac{d y}{d x} = x e^{- 2 y}$ becomes $e^{2 y} d y = x d x$ .
04
Volume about the y-axis · Q7
$V = π \int x^{2} d y$ , so write $x^{2}$ in terms of $y$ .
05
Adjacency matrix powers · Q8
The $(i, j)$ entry of $M^{n}$ counts walks of length $n$ from vertex $i$ to vertex $j$ .

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 AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 8

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8

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.

Questions

Questions 1–8 · 110 marks

Question 1

15 marks

Delivery times: a normal distribution, a binomial count and delivery charges

Statistics and probability
Functions

(a)

1 mark

64 is the mean and the normal distribution is symmetric.

P (T < 64) = 0.5

(b)

2 marks

Use the normal CDF on the calculator.

P (44 < T < 64) \approx 0.452

(c)

4 marks

(i) Sketch: a bell-shaped curve centred on 64 with the area to the left of a value $k$ (below the mean) shaded.

(ii) Use the inverse normal with area 0.3, mean 64 and standard deviation 12.

k \approx 57.7

(d)

3 marks

Each package is below $k$ with probability 0.3. Let $X \sim B (5, 0.3)$ .

P (X = 2) = (2 5) (0.3)^{2} (0.7)^{3} \approx 0.309

(e)

2 marks

The charge is $4.50 plus $2.00 for each kilogram above 1.

C (x) = 4.5 + 2 (x - 1) = 2 x + 2.5

(f)

1 mark

Substitute $x = 5.3$ .

C (5.3) = 2 (5.3) + 2.5 = $13.10

(g)

2 marks

Set the charge equal to $7.20.

2 x + 2.5 = 7.2 \Rightarrow x = 2.35 kg

Question 2

12 marks

Three mountain summits: a vector, an angle and the sine rule

Number and algebra

(a)

3 marks

(i) Subtract coordinates, $B - A$ .

(ii) The distance is the length of the vector.

A B = 7.2 5.1 2.4 - 2.8 = 7.2 5.1 - 0.4 A B = 7. 2^{2} + 5. 1^{2} + 0. 4^{2} \approx 8.83 km

(b)

5 marks

The two vectors are $A B$ and $1.1 8.4 0.2$ , whose length is $1. 1^{2} + 8. 4^{2} + 0. 2^{2} \approx 8.474$ .

scalar product = 1.1 (7.2) + 8.4 (5.1) + 0.2 (- 0.4) = 7.92 + 42.84 - 0.08 = 50.68 cos θ = \frac{50.68}{8.832 \times 8.474} \Rightarrow θ \approx 47. 4^{\circ}

(c)

4 marks

This is the angle at $A$ in triangle $A B C$ (since $A C$ is parallel to the given vector). With the angle at $B$ equal to $55. 2^{\circ}$ , the angle at $C$ is
Use the sine rule for the side $A C$ opposite $B$ .

A C B = 180 - 47.38 - 55.2 \approx 77.4 2^{\circ} \frac{A C}{sin 55. 2 ^{\circ}} = \frac{8.832}{sin 77.4 2 ^{\circ}} \Rightarrow A C \approx 7.43 km

Question 3

13 marks

Smoothie sales and profit: a model n = 40000/x^2

Functions
Calculus

(a)

2 marks

Solve $\frac{40000}{x ^{2}} = 400$ for $x > 0$ .

x^{2} = 100 \Rightarrow x = 10 pesos

(b)

2 marks

(i) At $x = 50$ .

(ii) The income is the number sold times the price.

n = \frac{40000}{5 0 ^{2}} = 16 16 \times 50 = 800 pesos

(c)

9 marks

(i) Profit is the number sold times the profit per smoothie.

(ii) Differentiate $P = 40000 x^{- 1} - 800000 x^{- 2}$ .

(iii) Set the derivative to zero and multiply by $x^{3}$ .

(iv) Substitute $x = 40$ in the model for $n$ .

Since this is the only turning point and $P \to 0$ as $x \to \infty$ , this is the maximum profit.

P = \frac{40000}{x ^{2}} (x - 20) = \frac{40000}{x} - \frac{800000}{x ^{2}} \frac{d P}{d x} = - \frac{40000}{x ^{2}} + \frac{1600000}{x ^{3}} - 40000 x + 1600000 = 0 \Rightarrow x = 40 n = \frac{40000}{4 0 ^{2}} = 25

Question 4

12 marks

Two switches in a generator: a tree and a chi-squared test

Statistics and probability

(a)

2 marks

The switch $A$ fails with probability 0.1, so it does not fail with probability 0.9. The switch $B$ fails with probability 0.3, so it does not fail with probability 0.7.

a = 0.9, b = 0.3, c = 0.7

(b)

1 mark

Both switches fail.

0.1 \times 0.3 = 0.03

(c)

9 marks

State the hypotheses.
Find the probabilities of 0, 1 and 2 failures.
Multiply by 200 to get the expected frequencies.
All expected values are at least 5, so the test is valid. There are 3 categories, so $ν = 2$ .
Since $0.182 > 0.05$ there is not enough evidence to reject $H_{0}$ . The data are consistent with the manufacturer's claims.

H_{0} : the manufacturer’s claims are correct, H_{1} : they are not both correct P (0) = 0.9 \times 0.7 = 0.63, P (1) = 0.1 \times 0.7 + 0.9 \times 0.3 = 0.34, P (2) = 0.03 126, 68, 6 (observed 118, 72, 10) χ^{2} = \frac{( 118 - 126 ) ^{2}}{126} + \frac{( 72 - 68 ) ^{2}}{68} + \frac{( 10 - 6 ) ^{2}}{6} \approx 3.41, p \approx 0.182

Question 5

13 marks

A separable differential equation and its slope field

Calculus

(a)

2 marks

The gradient is $x e^{- 2 y}$ . It is zero along $x = 0$ and positive for every $x > 0$ , whatever $y$ is. Diagram C is the one with these features.

(b)

7 marks

Separate the variables and integrate both sides.
Use $y = 0$ when $x = 0$ .
Take the natural logarithm.

\int e^{2 y} d y = \int x d x \Rightarrow \frac{1}{2} e^{2 y} = \frac{1}{2} x^{2} + c \frac{1}{2} = c \Rightarrow e^{2 y} = x^{2} + 1 y = \frac{1}{2} ln (x^{2} + 1)

(c)

2 marks

Differentiate with the chain rule.

\frac{d y}{d x} = \frac{1}{2} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x = \frac{x}{x ^{2} + 1}

(d)

2 marks

From part (b), $e^{2 y} = x^{2} + 1$ , so $e^{- 2 y} = \frac{1}{x ^{2} + 1}$ .
The expression in (c) equals the right side of the differential equation, so $y = \frac{1}{2} ln (x^{2} + 1)$ is a solution.

x e^{- 2 y} = \frac{x}{x ^{2} + 1} = \frac{d y}{d x}

Question 6

13 marks

Spaceships and battleships: Poisson, expected score and the central limit theorem

Statistics and probability

(a)

5 marks

(i) In 2 minutes the spaceships have mean $2 \times 4.2 = 8.4$ .

(ii) The total of both kinds of target in one minute is Poisson with mean $4.2 + 2.3 = 6.5$ .

P (S \leq 10) \approx 0.774 P (S + B > 10) = 1 - P (S + B \leq 10) \approx 0.0668

(b)

3 marks

(i) The expected points are 3 per spaceship and 5 per battleship.

(ii) The variance of a Poisson variable is its mean.

E (T) = 3 (4.2) + 5 (2.3) = 24.1 Var (T) = 3^{2} (4.2) + 5^{2} (2.3) = 37.8 + 57.5 = 95.3

(c)

1 mark

A Poisson distribution has its mean equal to its variance, but $E (T) = 24.1$ and $Var (T) = 95.3$ are different.

(d)

4 marks

The mean score $\overset{ˉ}{T}$ over 60 minutes is approximately normal.
Find the probability it is above 25.

\overset{ˉ}{T} \approx N (24.1, \frac{95.3}{60}) P (\overset{ˉ}{T} > 25) \approx 0.238

Question 7

14 marks

A vase modelled by revolving a curve: volume, filling time and rate of rise

Calculus

(a)

5 marks

From $y = \frac{1}{2} x^{2} - 1$ get $x^{2} = 2 y + 2$ . The water up to height $h$ is a solid of revolution about the $y$ -axis.

V = π \int_{0}^{h} (2 y + 2) d y = π [y^{2} + 2 y]_{0}^{h} V = π (h^{2} + 2 h)

(b)

2 marks

The vase is full at $h = 15$ .
Divide by the filling rate.

V = π (1 5^{2} + 2 \times 15) = 255 π \approx 801.1 cm^{3} t = \frac{801.1}{20} \approx 40.1 s

(c)

7 marks

Differentiate the volume with respect to $h$ .
Use the chain rule with $\frac{d V}{d t} = 20$ .
Substitute $h = 10$ .

\frac{d V}{d h} = π (2 h + 2) \frac{d h}{d t} = \frac{d V / d t}{d V / d h} = \frac{20}{π ( 2 h + 2 )} \frac{d h}{d t} = \frac{20}{22 π} \approx 0.289 cm s^{- 1}

Question 8

18 marks

Walks on a graph, a bird on three posts and two birds

Number and algebra
Statistics and probability

(a)

2 marks

Each post is joined to itself (a loop) and $A$ to $B$ , $B$ to $C$ in both directions. With the rows and columns in the order $A, B, C$ :

M = 110101011

(b)

3 marks

Raise the matrix to the fifth power. The entry in the row $A$ and column $B$ is the number of walks of length 5 from $A$ to $B$ .

M^{5} = 111110111011101111 11 walks

(c)

3 marks

(i) Every flight has two equally likely choices, so a sequence of 5 flights has probability

(ii) From $A$ there are 11 walks of length 5 ending at $B$ , each with probability $\frac{1}{32}$ .

(\frac{1}{2})^{5} = \frac{1}{32} \frac{11}{32} \approx 0.344

(d)

3 marks

The states are the occupied pairs $A B$ , $A C$ , $B C$ . From $A B$ the bird on $B$ flies to the empty post $C$ , while the other bird goes to $B$ with probability 0.4 or stays with 0.6.
From $A C$ , each bird goes to $B$ with probability 0.5, and the new state is $B C$ if only $A$ 's bird moves ( $0.5 \times 0.5$ ) or $A B$ if only $C$ 's bird moves ( $0.5 \times 0.5$ ).
The state stays $A C$ if neither moves ( $0.5 \times 0.5$ ) or if both try to move to $B$ at the same time ( $0.5 \times 0.5$ ).

a = 0.4 b = 0.25 c = 0.25 + 0.25 = 0.5

(e)

4 marks

The transition matrix in the order $A B$ , $A C$ , $B C$ has columns (from) $A B$ : $(0, 0.6, 0.4)$ , $A C$ : $(0.25, 0.5, 0.25)$ and $B C$ : $(0.4, 0.6, 0)$ .
Start in the state $A B$ and take 5 steps.
The entry in the row $B C$ and the column $A B$ is the required probability.

T = 0 0.6 0.4 0.25 0.5 0.25 0.4 0.6 0 T^{5} \approx 0.2222 0.5455 0.2324 0.2273 0.5455 0.2273 0.2324 0.5455 0.2222 P (B C after 5 flights) \approx 0.232

(f)

3 marks

A high power of $T$ has all columns equal to the long-term proportions $(\frac{5}{22}, \frac{6}{11}, \frac{5}{22})$ for $A B$ , $A C$ , $B C$ .
Post $B$ is occupied in the states $A B$ and $B C$ , so it is occupied $\frac{5}{22} + \frac{5}{22} = \frac{5}{11}$ of the time. Posts $A$ and $C$ are each occupied $\frac{17}{22}$ of the time.
So post $B$ is the one that is sat on least, and the proportion of the time that it is free is

1 - \frac{5}{11} = \frac{6}{11} \approx 0.545

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 probability48 marks · 44%
Q1, Q4, Q6, Q8
Calculus36 marks · 33%
Q3, Q5, Q7
Number and algebra17 marks · 15%
Q2, Q8
Functions9 marks · 8%
Q1, Q3

Mentor insights

Traps to watch for

01
Normal and binomial · Q1
Part (c) turns the time distribution into a probability of 0.3, and part (d) counts how many of five packages are below $k$ .
02
Profit per item · Q3
Each smoothie earns $x - 20$ pesos, so profit is $n (x - 20)$ .
03
Separate the variables · Q5
$\frac{d y}{d x} = x e^{- 2 y}$ becomes $e^{2 y} d y = x d x$ .
04
Volume about the y-axis · Q7
$V = π \int x^{2} d y$ , so write $x^{2}$ in terms of $y$ .
05
Adjacency matrix powers · Q8
The $(i, j)$ entry of $M^{n}$ counts walks of length $n$ from vertex $i$ to vertex $j$ .

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