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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

5 marks

Windscreen wiper: arc length and cleared area

Geometry and trigonometry

(a)

2 marks

Arc length is $\frac{θ}{360} \times 2 π r$ with $r = 56$ .

\frac{140}{360} \times 2 π (56) \approx 137 cm

(b)

3 marks

The rubber blade runs from 10 cm to 56 cm along the arm, since the arm is 56 cm and the blade is 46 cm.

A = \frac{140}{360} π (5 6^{2} - 1 0^{2}) \approx 3710 cm^{2}

Question 2

6 marks

University applicants: table probabilities

Statistics and probability

(a)

1 mark

There are $17 + 25 = 42$ admitted out of 130.

P (admitted) = \frac{42}{130} = \frac{21}{65} \approx 0.323

(b)

2 marks

Only the 42 admitted applicants count, and 17 of them chose Arts.

P (Arts ∣ admitted) = \frac{17}{42} \approx 0.405

(c)

3 marks

There are $17 + 24 = 41$ Arts applicants.

P (both Arts) = \frac{41}{130} \times \frac{40}{129} = \frac{1640}{16770} \approx 0.0978

Question 3

8 marks

Security cameras: angle of depression and distances

Geometry and trigonometry

(a)

2 marks

The cameras are 3.1 m up and the register is 1.0 m up, so the vertical drop is $3.1 - 1.0 = 2.1$ m.

sin θ = \frac{2.1}{2.8} \Rightarrow θ = 48. 6^{\circ}

(b)

4 marks

Horizontal distance from Camera 1 to the register:
Camera 2 is $6.4 - 1.852 = 4.548$ m away horizontally and 2.1 m above the register.

2. 8^{2} - 2. 1^{2} = 1.852 m d = 4.54 8^{2} + 2. 1^{2} = 5.01 m

(c)

2 marks

Both cameras are at the same height, so the angle of depression is larger for the one that is horizontally closer.
Camera 1 is closer to the register (1.85 m against 4.55 m), so Camera 1 has the larger angle.

Mentor tip

This works because $tan (angle) = \frac{2.1}{horizontal distance}$ gets bigger as the horizontal distance shrinks.

Question 4

5 marks

pH of coffee: logarithms and what a lower pH means

Functions

(a)

2 marks

Put $C = 1.3 \times 1 0^{- 5}$ into the formula.

pH = - lo g_{10} (1.3 \times 1 0^{- 5}) = 4.89

(b)

3 marks

A concentration ten times larger is $1.3 \times 1 0^{- 4}$ .
A lower pH means more acidic, and $3.89 < 4.89$ , so the unknown liquid is more acidic than the coffee.

pH = - lo g_{10} (1.3 \times 1 0^{- 4}) = 3.89

Mentor tip

The graph of $y = - lo g_{10} x$ falls as $x$ rises, which shows the same thing.

Question 5

7 marks

Polygraph test: binomial probabilities

Statistics and probability

(a)

2 marks

Each of the 10 people passes with probability $0.8$ . Let $X$ be the number who pass: $X \sim B (10, 0.8)$ .

E (X) = 10 \times 0.8 = 8

(b)

2 marks

Four fail means six pass.

P (X = 6) = (6 10) (0.8)^{6} (0.2)^{4} \approx 0.0881

(c)

3 marks

Fewer than 7 pass is $X \leq 6$ .

P (X \leq 6) \approx 0.121

Mentor tip

Same answer from the failures: $Y \sim B (10, 0.2)$ and $P (Y \geq 4)$ .

Question 6

7 marks

Area under a line and a curve, and the area between them

Calculus

(a)

2 marks

The region is a trapezoid with parallel sides 7 and 4 and width 3.
Or by integration: $\int_{- 1}^{2} (6 - x) d x = 16.5$ .

A = \frac{1}{2} (7 + 4) (3) = 16.5

(b)

3 marks

Integral for the area under the curve between $x = - 1$ and $x = 2$ :
Evaluate.

A = \int_{- 1}^{2} (1.5 x^{2} - 2.5 x + 3) d x [0.5 x^{3} - 1.25 x^{2} + 3 x]_{- 1}^{2} = 5 - (- 4.75) = 9.75

(c)

2 marks

The area between the line and the curve is the difference of the two areas.

16.5 - 9.75 = 6.75

Question 7

5 marks

Test scores: outlier fences and comparing two box plots

Statistics and probability

(a)

3 marks

The interquartile range is $88 - 62 = 26$ , so $1.5 \times IQR = 39$ .
Since $25 > 23$ , a score of 25 is not an outlier.

lower fence = 62 - 39 = 23

(b)

2 marks

In favour of the researcher: the median score of the evening class is higher than that of the morning class.
Against: the evening scores are more spread out, with lower scores and a lower lower quartile, so the evening class is less consistent.

Question 8

6 marks

Reaction times while driving: a two-sample t-test

Statistics and probability

(a)

1 mark

The test compares the two means for a difference in either direction.

H_{1} : μ_{1} - μ_{2} \neq = 0

(b)

2 marks

Enter the two samples into a two-sample $t$ -test with pooled variance.

p \approx 0.0778

(c)

3 marks

Compare $p$ with the significance level.
Reject $H_{0}$ .
In context: there is significant evidence at the 10% level that the mean reaction time is different for drivers talking on a phone and drivers talking to a passenger.

0.0778 < 0.10

Question 9

8 marks

Graduate savings: fitting a cubic model

Functions

(a)

1 mark

The curve has one local maximum and one local minimum (and crosses the horizontal axis three times).

(b)

4 marks

At $t = 0$ the value is $- 5$ .
Substitute $t = 1, 2, 3$ into $S = a t^{3} + b t^{2} + c t - 5$ .
Solve the system.

d = - 5 a + b + c = 8, 8 a + 4 b + 2 c = 4, 27 a + 9 b + 3 c = 0 a = 2, b = - 12, c = 18

(c)

3 marks

The model is $S = 2 t^{3} - 12 t^{2} + 18 t - 5$ . Solve $S = 0$ .
$S$ is negative on $0 \leq t < 0.358$ and again on $1.832 < t < 3.810$ .

t \approx 0.358, 1.832, 3.810 time in debt = 0.358 + (3.810 - 1.832) \approx 2.34 years

Mentor tip

The graduate starts in debt ( $S = - 5$ ), so the first interval starts at $t = 0$ .

Question 10

5 marks

Fuji apples: normal probabilities and a lower boundary

Statistics and probability

(a)

2 marks

Use $X \sim N (163, 6.8 3^{2})$ .

P (172 < X < 183) = 0.0921

(b)

3 marks

First find the probability below 172.
The medium class has probability 0.68, so the probability below $k$ is
Use the inverse normal.

P (X < 172) = 0.9062 P (X < k) = 0.9062 - 0.68 = 0.2262 k = 157.9 \approx 158 g

Question 11

7 marks

Tangent gradient and parallel tangents

Calculus

(a)

2 marks

Write $f (x) = x^{2} - 3 x^{- 1}$ .

f^{'} (x) = 2 x + 3 x^{- 2}

(b)

2 marks

The gradient of the tangent at $x = 1$ is $f^{'} (1)$ .

f^{'} (1) = 2 + 3 = 5

(c)

3 marks

Parallel tangents have the same gradient, so solve $f^{'} (x) = 5$ .
There are three tangents with gradient 5, and one of them is L (at $x = 1$ ). So there are 2 other tangent lines parallel to L.

2 x + \frac{3}{x ^{2}} = 5 \Rightarrow x = - 0.686, 1, 2.19

Question 12

5 marks

Archway as a parabola: axis of symmetry and a crate

Functions

(a)

2 marks

The axis of symmetry of $y = a x^{2} + b x$ is $x = - \frac{b}{2 a}$ .

x = - \frac{4.48}{2 ( - 1.6 )} = 1.4

(b)

3 marks

A crate 1.6 m wide centred on $x = 1.4$ takes up $0.6 \leq x \leq 2.2$ .
The height of the arch at its two sides is
Since the arch is lowest at the edges of the crate and $2.112 > 2.0$ , the crate fits through.

y (0.6) = - 1.6 (0.36) + 4.48 (0.6) = 2.112

Question 13

6 marks

Saving for a real value of 4000: lump sum or yearly payments

Number and algebra

(a)

3 marks

The target in ten years' money is $4000 \times 1.0 2^{10} = 4875.98$ .
The minimum single investment is $3457.

P V = \frac{4875.98}{1.03 5 ^{10}} = 3456.67

(b)

3 marks

The $1000 grows to
The yearly payments must make up the rest.
So about $295 p er y e a r;$ 296 is the smallest whole-dollar payment that reaches the target.

1000 \times 1.03 5^{10} = 1410.60 x \cdot \frac{1.03 5 ^{10} - 1}{0.035} = 4875.98 - 1410.60 = 3465.38 x = \frac{3465.38}{11.7314} \approx 295.39

Mentor tip

Use a finance solver with N = 10, I = 3.5, PV = $- 1000$ , FV = 4875.98 and 1 payment per year to check.

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 probability29 marks · 36%
Q2, Q5, Q7, Q8, Q10
Functions18 marks · 23%
Q4, Q9, Q12
Calculus14 marks · 18%
Q6, Q11
Geometry and trigonometry13 marks · 16%
Q1, Q3
Number and algebra6 marks · 8%
Q13

Mentor insights

Traps to watch for

01
Angle of depression is measured from the horizontal · Q3
In the right triangle the 2.1 m drop is opposite the angle, and the 2.8 m sightline is the hypotenuse.
02
Pass versus fail · Q5
The 20% is the chance of failing. Passing has probability $0.8$ , so check which one the question asks about.
03
Area under a curve needs the integral sign and dx · Q6
Write the limits, the function and $d x$ . Using the word "y" is not enough when there are two equations.
04
Use the first point for d · Q9
The point $(0, - 5)$ gives $d = - 5$ immediately, which makes the other three equations simpler.
05
Solve f'(x) = gradient of L · Q11
The tangents parallel to L are where $f^{'} (x) = 5$ . One solution is $x = 1$ , the tangent L itself.

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 SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–13 · 80 marks

Question 1

5 marks

Windscreen wiper: arc length and cleared area

Geometry and trigonometry

(a)

2 marks

Arc length is $\frac{θ}{360} \times 2 π r$ with $r = 56$ .

\frac{140}{360} \times 2 π (56) \approx 137 cm

(b)

3 marks

The rubber blade runs from 10 cm to 56 cm along the arm, since the arm is 56 cm and the blade is 46 cm.

A = \frac{140}{360} π (5 6^{2} - 1 0^{2}) \approx 3710 cm^{2}

Question 2

6 marks

University applicants: table probabilities

Statistics and probability

(a)

1 mark

There are $17 + 25 = 42$ admitted out of 130.

P (admitted) = \frac{42}{130} = \frac{21}{65} \approx 0.323

(b)

2 marks

Only the 42 admitted applicants count, and 17 of them chose Arts.

P (Arts ∣ admitted) = \frac{17}{42} \approx 0.405

(c)

3 marks

There are $17 + 24 = 41$ Arts applicants.

P (both Arts) = \frac{41}{130} \times \frac{40}{129} = \frac{1640}{16770} \approx 0.0978

Question 3

8 marks

Security cameras: angle of depression and distances

Geometry and trigonometry

(a)

2 marks

The cameras are 3.1 m up and the register is 1.0 m up, so the vertical drop is $3.1 - 1.0 = 2.1$ m.

sin θ = \frac{2.1}{2.8} \Rightarrow θ = 48. 6^{\circ}

(b)

4 marks

Horizontal distance from Camera 1 to the register:
Camera 2 is $6.4 - 1.852 = 4.548$ m away horizontally and 2.1 m above the register.

2. 8^{2} - 2. 1^{2} = 1.852 m d = 4.54 8^{2} + 2. 1^{2} = 5.01 m

(c)

2 marks

Both cameras are at the same height, so the angle of depression is larger for the one that is horizontally closer.
Camera 1 is closer to the register (1.85 m against 4.55 m), so Camera 1 has the larger angle.

Mentor tip

This works because $tan (angle) = \frac{2.1}{horizontal distance}$ gets bigger as the horizontal distance shrinks.

Question 4

5 marks

pH of coffee: logarithms and what a lower pH means

Functions

(a)

2 marks

Put $C = 1.3 \times 1 0^{- 5}$ into the formula.

pH = - lo g_{10} (1.3 \times 1 0^{- 5}) = 4.89

(b)

3 marks

A concentration ten times larger is $1.3 \times 1 0^{- 4}$ .
A lower pH means more acidic, and $3.89 < 4.89$ , so the unknown liquid is more acidic than the coffee.

pH = - lo g_{10} (1.3 \times 1 0^{- 4}) = 3.89

Mentor tip

The graph of $y = - lo g_{10} x$ falls as $x$ rises, which shows the same thing.

Question 5

7 marks

Polygraph test: binomial probabilities

Statistics and probability

(a)

2 marks

Each of the 10 people passes with probability $0.8$ . Let $X$ be the number who pass: $X \sim B (10, 0.8)$ .

E (X) = 10 \times 0.8 = 8

(b)

2 marks

Four fail means six pass.

P (X = 6) = (6 10) (0.8)^{6} (0.2)^{4} \approx 0.0881

(c)

3 marks

Fewer than 7 pass is $X \leq 6$ .

P (X \leq 6) \approx 0.121

Mentor tip

Same answer from the failures: $Y \sim B (10, 0.2)$ and $P (Y \geq 4)$ .

Question 6

7 marks

Area under a line and a curve, and the area between them

Calculus

(a)

2 marks

The region is a trapezoid with parallel sides 7 and 4 and width 3.
Or by integration: $\int_{- 1}^{2} (6 - x) d x = 16.5$ .

A = \frac{1}{2} (7 + 4) (3) = 16.5

(b)

3 marks

Integral for the area under the curve between $x = - 1$ and $x = 2$ :
Evaluate.

A = \int_{- 1}^{2} (1.5 x^{2} - 2.5 x + 3) d x [0.5 x^{3} - 1.25 x^{2} + 3 x]_{- 1}^{2} = 5 - (- 4.75) = 9.75

(c)

2 marks

The area between the line and the curve is the difference of the two areas.

16.5 - 9.75 = 6.75

Question 7

5 marks

Test scores: outlier fences and comparing two box plots

Statistics and probability

(a)

3 marks

The interquartile range is $88 - 62 = 26$ , so $1.5 \times IQR = 39$ .
Since $25 > 23$ , a score of 25 is not an outlier.

lower fence = 62 - 39 = 23

(b)

2 marks

In favour of the researcher: the median score of the evening class is higher than that of the morning class.
Against: the evening scores are more spread out, with lower scores and a lower lower quartile, so the evening class is less consistent.

Question 8

6 marks

Reaction times while driving: a two-sample t-test

Statistics and probability

(a)

1 mark

The test compares the two means for a difference in either direction.

H_{1} : μ_{1} - μ_{2} \neq = 0

(b)

2 marks

Enter the two samples into a two-sample $t$ -test with pooled variance.

p \approx 0.0778

(c)

3 marks

Compare $p$ with the significance level.
Reject $H_{0}$ .
In context: there is significant evidence at the 10% level that the mean reaction time is different for drivers talking on a phone and drivers talking to a passenger.

0.0778 < 0.10

Question 9

8 marks

Graduate savings: fitting a cubic model

Functions

(a)

1 mark

The curve has one local maximum and one local minimum (and crosses the horizontal axis three times).

(b)

4 marks

At $t = 0$ the value is $- 5$ .
Substitute $t = 1, 2, 3$ into $S = a t^{3} + b t^{2} + c t - 5$ .
Solve the system.

d = - 5 a + b + c = 8, 8 a + 4 b + 2 c = 4, 27 a + 9 b + 3 c = 0 a = 2, b = - 12, c = 18

(c)

3 marks

The model is $S = 2 t^{3} - 12 t^{2} + 18 t - 5$ . Solve $S = 0$ .
$S$ is negative on $0 \leq t < 0.358$ and again on $1.832 < t < 3.810$ .

t \approx 0.358, 1.832, 3.810 time in debt = 0.358 + (3.810 - 1.832) \approx 2.34 years

Mentor tip

The graduate starts in debt ( $S = - 5$ ), so the first interval starts at $t = 0$ .

Question 10

5 marks

Fuji apples: normal probabilities and a lower boundary

Statistics and probability

(a)

2 marks

Use $X \sim N (163, 6.8 3^{2})$ .

P (172 < X < 183) = 0.0921

(b)

3 marks

First find the probability below 172.
The medium class has probability 0.68, so the probability below $k$ is
Use the inverse normal.

P (X < 172) = 0.9062 P (X < k) = 0.9062 - 0.68 = 0.2262 k = 157.9 \approx 158 g

Question 11

7 marks

Tangent gradient and parallel tangents

Calculus

(a)

2 marks

Write $f (x) = x^{2} - 3 x^{- 1}$ .

f^{'} (x) = 2 x + 3 x^{- 2}

(b)

2 marks

The gradient of the tangent at $x = 1$ is $f^{'} (1)$ .

f^{'} (1) = 2 + 3 = 5

(c)

3 marks

Parallel tangents have the same gradient, so solve $f^{'} (x) = 5$ .
There are three tangents with gradient 5, and one of them is L (at $x = 1$ ). So there are 2 other tangent lines parallel to L.

2 x + \frac{3}{x ^{2}} = 5 \Rightarrow x = - 0.686, 1, 2.19

Question 12

5 marks

Archway as a parabola: axis of symmetry and a crate

Functions

(a)

2 marks

The axis of symmetry of $y = a x^{2} + b x$ is $x = - \frac{b}{2 a}$ .

x = - \frac{4.48}{2 ( - 1.6 )} = 1.4

(b)

3 marks

A crate 1.6 m wide centred on $x = 1.4$ takes up $0.6 \leq x \leq 2.2$ .
The height of the arch at its two sides is
Since the arch is lowest at the edges of the crate and $2.112 > 2.0$ , the crate fits through.

y (0.6) = - 1.6 (0.36) + 4.48 (0.6) = 2.112

Question 13

6 marks

Saving for a real value of 4000: lump sum or yearly payments

Number and algebra

(a)

3 marks

The target in ten years' money is $4000 \times 1.0 2^{10} = 4875.98$ .
The minimum single investment is $3457.

P V = \frac{4875.98}{1.03 5 ^{10}} = 3456.67

(b)

3 marks

The $1000 grows to
The yearly payments must make up the rest.
So about $295 p er y e a r;$ 296 is the smallest whole-dollar payment that reaches the target.

1000 \times 1.03 5^{10} = 1410.60 x \cdot \frac{1.03 5 ^{10} - 1}{0.035} = 4875.98 - 1410.60 = 3465.38 x = \frac{3465.38}{11.7314} \approx 295.39

Mentor tip

Use a finance solver with N = 10, I = 3.5, PV = $- 1000$ , FV = 4875.98 and 1 payment per year to check.

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 probability29 marks · 36%
Q2, Q5, Q7, Q8, Q10
Functions18 marks · 23%
Q4, Q9, Q12
Calculus14 marks · 18%
Q6, Q11
Geometry and trigonometry13 marks · 16%
Q1, Q3
Number and algebra6 marks · 8%
Q13

Mentor insights

Traps to watch for

01
Angle of depression is measured from the horizontal · Q3
In the right triangle the 2.1 m drop is opposite the angle, and the 2.8 m sightline is the hypotenuse.
02
Pass versus fail · Q5
The 20% is the chance of failing. Passing has probability $0.8$ , so check which one the question asks about.
03
Area under a curve needs the integral sign and dx · Q6
Write the limits, the function and $d x$ . Using the word "y" is not enough when there are two equations.
04
Use the first point for d · Q9
The point $(0, - 5)$ gives $d = - 5$ immediately, which makes the other three equations simpler.
05
Solve f'(x) = gradient of L · Q11
The tangents parallel to L are where $f^{'} (x) = 5$ . One solution is $x = 1$ , the tangent L itself.

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