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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

18 marks

Two investment accounts: an arithmetic sequence of interest against compound interest

Number and algebra

(a)

3 marks

(i) The interest goes 200, 260, 320, so the common difference is 60.

(ii) The interest at the end of the 10th year.

d = 60 u_{10} = 200 + 9 (60) = 740 USD

(b)

3 marks

The total after $n$ years is the initial 5000 plus the sum of the first $n$ interest payments.

5000 + \frac{n}{2} [2 (200) + (n - 1) 60] = 5000 + \frac{n}{2} (340 + 60 n) 5000 + \frac{n ( 340 + 60 n )}{2}

(c)

2 marks

Put $n = 10$ .

5000 + \frac{10 ( 340 + 600 )}{2} = 5000 + 4700 = 9700 USD

(d)

4 marks

(i) 6.5% interest each year.

(ii) After 10 years Account 2 has

Since $9700 > 9386$ , Account 1 has more money than Account 2 after 10 years.

B = 1.065 5000 (1.065)^{10} \approx 9386 USD

(e)

3 marks

Find the first whole year with $5000 (1.065)^{m} > 5000 + \frac{m ( 340 + 60 m )}{2}$ . Use a graph or table on the calculator, where the crossing is at $m \approx 38.05$ .

m = 39

(f)

3 marks

After 39 years Account 2 has $5000 (1.065)^{39} \approx 58291$ .

interest = 58291 - 5000 \approx 53291 USD

Question 2

17 marks

Component weights: normal probabilities, binomial tests and a dispatch rule

Statistics and probability

(a)

2 marks

$W \sim N (2.5, 0.1 5^{2})$ .

P (W > 2.8) \approx 0.0228

(b)

4 marks

(i) Sketch a bell curve centred on 2.5 with the area to the right of $w$ shaded, where $w$ is a little to the left of the centre since 80% of the area is to its right.

(ii) Use the inverse normal with 20% below $w$ .

P (W < w) = 0.2 \Rightarrow w \approx 2.37

(c)

3 marks

(i) The probability of passing Test 1.

(ii) The expected number in a box of 200.

P (2.3 < W < 2.7) \approx 0.818 200 \times 0.8176 \approx 164

(d)

2 marks

The number passing Test 2 is $X \sim B (200, 0.95)$ .

P (X = 190) \approx 0.128

(e)

2 marks

At least 188.

P (X \geq 188) \approx 0.796

(f)

4 marks

The probability of passing both tests.
Let $Y$ be the number of the 12 components that pass both, $Y \sim B (12, 0.7767)$ .

0.8176 \times 0.95 \approx 0.7767 P (Y \geq 10) \approx 0.477

Question 3

20 marks

Drink containers: a cuboid and a cylinder, and extra material

Functions

(a)

1 mark

The height is three times the length of the base.

d = 3 l

(b)

3 marks

The volume is $l \times l \times d = 3 l^{3}$ .

3 l^{3} = 375 \Rightarrow l = 5, d = 15

(c)

3 marks

Two squares and four rectangles.

2 (5 \times 5) + 4 (5 \times 15) = 350 cm^{2}

(d)

2 marks

The volume of the cylinder is $π r^{2} h = 375$ .

h = \frac{375}{π r ^{2}}

(e)

2 marks

The total external surface area is $A = 2 π r^{2} + 2 π r h$ .

A = 2 π r^{2} + 2 π r \cdot \frac{375}{π r ^{2}} = 2 π r^{2} + \frac{750}{r}

(f)

3 marks

Differentiate.

\frac{d A}{d r} = 4 π r - \frac{750}{r ^{2}}

(g)

3 marks

(i) Set the derivative to zero.

(ii) Substitute $r = 3.908$ .

4 π r^{3} = 750 \Rightarrow r \approx 3.91 cm A = 2 π (3.908)^{2} + \frac{750}{3.908} \approx 288 cm^{2}

(h)

3 marks

The cuboid needs 10% more material and the cylinder needs 25% more.
Since $360 < 385$ , Kailash should choose the cylinder.

cuboid: 1.10 \times 350 = 385 cm^{2}, cylinder: 1.25 \times 287.87 \approx 360 cm^{2}

Question 4

13 marks

Tides at Albion harbour: a cosine model and when a boat can sail

Functions

(a)

1 mark

The tide repeats every 12 hours, so the period in degrees is $\frac{360}{b} = 12$ .

b = 30

(b)

2 marks

Substitute $t = 5$ (degrees).

h (5) = - 2.5 cos (15 0^{\circ}) + 4.5 \approx 6.67 m

(c)

3 marks

(i) The amplitude is 2.5 m.

(ii) The principal axis.

h = 4.5

(d)

3 marks

Solve $h (t) = 2.65$ after $t = 12$ .
That is 13 hours and 25 minutes after midnight.

- 2.5 cos (30 t^{\circ}) + 4.5 = 2.65 \Rightarrow t \approx 13.41 13 : 25

(e)

4 marks

The tide stays high enough until $h (t) = 2.65$ again, at $t \approx 22.59$ . The boat needs 15 minutes each way, which is 0.5 hours in all.

22.59 - 13.41 - 0.5 \approx 8.68 hours

Question 5

12 marks

Water flowing under a bridge and whether a boat fits

Calculus

(a)

3 marks

With strips of width 1.9 m.

A \approx \frac{1.9}{2} [0 + 0 + 2 (1.68 + 2.81 + 2.32)] \approx 12.9 m^{2}

(b)

2 marks

The volume per second is the area times the speed.

12.94 \times 0.3 \approx 3.88 m^{3} per second

(c)

1 mark

The deepest point of the river is 2.81 m, and the boat's lowest part is 1.6 m below the surface.

2.81 - 1.6 = 1.21 m

(d)

2 marks

The maximum of $y = - 0.15 x^{2} + 1.14 x + 0.9$ is at the vertex $x = \frac{1.14}{0.3} = 3.8$ .

y = - 0.15 (3.8)^{2} + 1.14 (3.8) + 0.9 \approx 3.07 m

(e)

4 marks

The boat is 5 m wide and travels down the centre of the river, so its top spans from $3.8 - 2.5 = 1.3$ to $3.8 + 2.5 = 6.3$ . The arch is lowest at the edges of the boat, so find its height at $x = 1.3$ (the same at $x = 6.3$ by symmetry).
The top of the boat is 2.35 m above the water, and $2.13 < 2.35$ , so the boat cannot pass under the bridge.

y (1.3) = - 0.15 (1.69) + 1.14 (1.3) + 0.9 \approx 2.13 m

Our analysis

Where the marks are

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

Functions33 marks · 41%
Q3, Q4
Number and algebra18 marks · 23%
Q1
Statistics and probability17 marks · 21%
Q2
Calculus12 marks · 15%
Q5

Mentor insights

Traps to watch for

01
Arithmetic series · Q1
The interest added each year grows by a constant amount.
02
Independent tests · Q2
The probability of passing both tests is the product of the two probabilities.
03
Surface area · Q3
The cuboid has two squares and four rectangles. The cylinder has two discs and a curved surface.
04
Height limit · Q4
Boats can only enter or leave when $h (t) \geq 2.65$ .
05
Trapezoidal rule · Q5
$Area \approx \frac{h}{2} [y_{0} + y_{n} + 2 (middle values)]$ .

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 SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

18 marks

Two investment accounts: an arithmetic sequence of interest against compound interest

Number and algebra

(a)

3 marks

(i) The interest goes 200, 260, 320, so the common difference is 60.

(ii) The interest at the end of the 10th year.

d = 60 u_{10} = 200 + 9 (60) = 740 USD

(b)

3 marks

The total after $n$ years is the initial 5000 plus the sum of the first $n$ interest payments.

5000 + \frac{n}{2} [2 (200) + (n - 1) 60] = 5000 + \frac{n}{2} (340 + 60 n) 5000 + \frac{n ( 340 + 60 n )}{2}

(c)

2 marks

Put $n = 10$ .

5000 + \frac{10 ( 340 + 600 )}{2} = 5000 + 4700 = 9700 USD

(d)

4 marks

(i) 6.5% interest each year.

(ii) After 10 years Account 2 has

Since $9700 > 9386$ , Account 1 has more money than Account 2 after 10 years.

B = 1.065 5000 (1.065)^{10} \approx 9386 USD

(e)

3 marks

Find the first whole year with $5000 (1.065)^{m} > 5000 + \frac{m ( 340 + 60 m )}{2}$ . Use a graph or table on the calculator, where the crossing is at $m \approx 38.05$ .

m = 39

(f)

3 marks

After 39 years Account 2 has $5000 (1.065)^{39} \approx 58291$ .

interest = 58291 - 5000 \approx 53291 USD

Question 2

17 marks

Component weights: normal probabilities, binomial tests and a dispatch rule

Statistics and probability

(a)

2 marks

$W \sim N (2.5, 0.1 5^{2})$ .

P (W > 2.8) \approx 0.0228

(b)

4 marks

(i) Sketch a bell curve centred on 2.5 with the area to the right of $w$ shaded, where $w$ is a little to the left of the centre since 80% of the area is to its right.

(ii) Use the inverse normal with 20% below $w$ .

P (W < w) = 0.2 \Rightarrow w \approx 2.37

(c)

3 marks

(i) The probability of passing Test 1.

(ii) The expected number in a box of 200.

P (2.3 < W < 2.7) \approx 0.818 200 \times 0.8176 \approx 164

(d)

2 marks

The number passing Test 2 is $X \sim B (200, 0.95)$ .

P (X = 190) \approx 0.128

(e)

2 marks

At least 188.

P (X \geq 188) \approx 0.796

(f)

4 marks

The probability of passing both tests.
Let $Y$ be the number of the 12 components that pass both, $Y \sim B (12, 0.7767)$ .

0.8176 \times 0.95 \approx 0.7767 P (Y \geq 10) \approx 0.477

Question 3

20 marks

Drink containers: a cuboid and a cylinder, and extra material

Functions

(a)

1 mark

The height is three times the length of the base.

d = 3 l

(b)

3 marks

The volume is $l \times l \times d = 3 l^{3}$ .

3 l^{3} = 375 \Rightarrow l = 5, d = 15

(c)

3 marks

Two squares and four rectangles.

2 (5 \times 5) + 4 (5 \times 15) = 350 cm^{2}

(d)

2 marks

The volume of the cylinder is $π r^{2} h = 375$ .

h = \frac{375}{π r ^{2}}

(e)

2 marks

The total external surface area is $A = 2 π r^{2} + 2 π r h$ .

A = 2 π r^{2} + 2 π r \cdot \frac{375}{π r ^{2}} = 2 π r^{2} + \frac{750}{r}

(f)

3 marks

Differentiate.

\frac{d A}{d r} = 4 π r - \frac{750}{r ^{2}}

(g)

3 marks

(i) Set the derivative to zero.

(ii) Substitute $r = 3.908$ .

4 π r^{3} = 750 \Rightarrow r \approx 3.91 cm A = 2 π (3.908)^{2} + \frac{750}{3.908} \approx 288 cm^{2}

(h)

3 marks

The cuboid needs 10% more material and the cylinder needs 25% more.
Since $360 < 385$ , Kailash should choose the cylinder.

cuboid: 1.10 \times 350 = 385 cm^{2}, cylinder: 1.25 \times 287.87 \approx 360 cm^{2}

Question 4

13 marks

Tides at Albion harbour: a cosine model and when a boat can sail

Functions

(a)

1 mark

The tide repeats every 12 hours, so the period in degrees is $\frac{360}{b} = 12$ .

b = 30

(b)

2 marks

Substitute $t = 5$ (degrees).

h (5) = - 2.5 cos (15 0^{\circ}) + 4.5 \approx 6.67 m

(c)

3 marks

(i) The amplitude is 2.5 m.

(ii) The principal axis.

h = 4.5

(d)

3 marks

Solve $h (t) = 2.65$ after $t = 12$ .
That is 13 hours and 25 minutes after midnight.

- 2.5 cos (30 t^{\circ}) + 4.5 = 2.65 \Rightarrow t \approx 13.41 13 : 25

(e)

4 marks

The tide stays high enough until $h (t) = 2.65$ again, at $t \approx 22.59$ . The boat needs 15 minutes each way, which is 0.5 hours in all.

22.59 - 13.41 - 0.5 \approx 8.68 hours

Question 5

12 marks

Water flowing under a bridge and whether a boat fits

Calculus

(a)

3 marks

With strips of width 1.9 m.

A \approx \frac{1.9}{2} [0 + 0 + 2 (1.68 + 2.81 + 2.32)] \approx 12.9 m^{2}

(b)

2 marks

The volume per second is the area times the speed.

12.94 \times 0.3 \approx 3.88 m^{3} per second

(c)

1 mark

The deepest point of the river is 2.81 m, and the boat's lowest part is 1.6 m below the surface.

2.81 - 1.6 = 1.21 m

(d)

2 marks

The maximum of $y = - 0.15 x^{2} + 1.14 x + 0.9$ is at the vertex $x = \frac{1.14}{0.3} = 3.8$ .

y = - 0.15 (3.8)^{2} + 1.14 (3.8) + 0.9 \approx 3.07 m

(e)

4 marks

The boat is 5 m wide and travels down the centre of the river, so its top spans from $3.8 - 2.5 = 1.3$ to $3.8 + 2.5 = 6.3$ . The arch is lowest at the edges of the boat, so find its height at $x = 1.3$ (the same at $x = 6.3$ by symmetry).
The top of the boat is 2.35 m above the water, and $2.13 < 2.35$ , so the boat cannot pass under the bridge.

y (1.3) = - 0.15 (1.69) + 1.14 (1.3) + 0.9 \approx 2.13 m

Our analysis

Where the marks are

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

Functions33 marks · 41%
Q3, Q4
Number and algebra18 marks · 23%
Q1
Statistics and probability17 marks · 21%
Q2
Calculus12 marks · 15%
Q5

Mentor insights

Traps to watch for

01
Arithmetic series · Q1
The interest added each year grows by a constant amount.
02
Independent tests · Q2
The probability of passing both tests is the product of the two probabilities.
03
Surface area · Q3
The cuboid has two squares and four rectangles. The cylinder has two discs and a curved surface.
04
Height limit · Q4
Boats can only enter or leave when $h (t) \geq 2.65$ .
05
Trapezoidal rule · Q5
$Area \approx \frac{h}{2} [y_{0} + y_{n} + 2 (middle values)]$ .

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