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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 TZ1 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

Area of the roof of a doghouse

Number and algebra

5 marks

The wall is 0.9 m high, so the triangle on top has height $1.35 - 0.9 = 0.45$ m and half-base $0.9 \div 2 = 0.45$ m.
Each roof rectangle is 0.6364 m by 0.9 m (the depth of the house).

slant length = 0.4 5^{2} + 0.4 5^{2} = 0.45 2 = 0.6364 m area to paint = 2 \times 0.6364 \times 0.9 = 1.15 m^{2}

Question 2

4 marks

Rope on a pole standing on level ground

Geometry and trigonometry

(a)

2 marks

Evaluate it.

A F = 3. 2^{2} + 4. 5^{2} + 5. 8^{2} A F = 8.01 m

(b)

2 marks

In the right triangle FOA the side opposite the angle at A is $F O = 5.8$ and the hypotenuse is $A F$ .

sin F A O = \frac{5.8}{8.008} \Rightarrow F A O = 46. 4^{\circ}

Question 3

5 marks

A baseball's height: quadratic model and domain

Functions

(a)

1 mark

At the instant of the hit $t = 0$ .

h (0) = 1.2 m

(b)

2 marks

The ball hits the ground when $h (t) = 0$ .

- 4.8 t^{2} + 21 t + 1.2 = 0 \Rightarrow t \approx 4.43 s (reject t < 0)

(c)

2 marks

Time starts at the hit and ends when the ball lands.

0 \leq t \leq 4.43

Question 4

7 marks

A road along the perpendicular bisector of two towns

Geometry and trigonometry

(a)

5 marks

Midpoint of A and B:
Gradient of AB and of the road:
Equation of the road:

(\frac{- 6 + 8}{2}, \frac{- 1 + 6}{2}) = (1, 2.5) m_{A B} = \frac{6 - ( - 1 )}{8 - ( - 6 )} = \frac{1}{2}, m_{road} = - 2 y - 2.5 = - 2 (x - 1) \Rightarrow y = - 2 x + 4.5

(b)

2 marks

Town C has the same $x$ -coordinate as A, which is $- 6$ .

y = - 2 (- 6) + 4.5 = 16.5

Question 5

5 marks

Concert tickets: forming and solving three equations

Number and algebra

(a)

3 marks

The total number of tickets, the total income and the ratio are:

x + y + z = 600, 15 x + 10 y + 12 z = 7816, x = 2 y

(b)

2 marks

Substitute $x = 2 y$ into the first two equations.
Eliminate $z = 600 - 3 y$ .

3 y + z = 600, 40 y + 12 z = 7816 40 y + 12 (600 - 3 y) = 7816 \Rightarrow 4 y = 616 \Rightarrow y = 154 x = 308 adult, y = 154 child, z = 138 student

Question 6

7 marks

A painting bounded by a cubic: trapezoidal rule and exact area

Calculus

(a)

3 marks

Trapezoidal rule with $h = 1$ and four ordinates.

A \approx \frac{1}{2} (1) [0.6 + 0 + 2 (1.2 + 1.2)] = 2.7 m^{2}

(b)

2 marks

Integrate the cubic from $- 1$ to $2$ .

A = \int_{- 1}^{2} \frac{- x ^{3} - 3 x ^{2} + 4 x + 12}{10} d x = \frac{1}{10} [- \frac{x ^{4}}{4} - x^{3} + 2 x^{2} + 12 x]_{- 1}^{2} A = \frac{1}{10} (20 - (- 9.25)) = 2.925 m^{2}

(c)

2 marks

The frame is a square with side 3, so its area is $3^{2} = 9$ .

9 - 2.925 = 6.075 m^{2}

Question 7

7 marks

Is the die fair? Chi-squared goodness of fit

Statistics and probability

(a)

1 mark

The null hypothesis says the die is fair, so each face has probability $\frac{1}{6}$ . The alternative says it is not fair.

H_{0} : P (each number) = \frac{1}{6}, H_{1} : the probabilities are not all \frac{1}{6}

(b)

1 mark

Six categories and no parameters estimated from the data.

ν = 6 - 1 = 5

(c)

1 mark

A fair die gives each face one sixth of the 60 rolls.

E = 60 \times \frac{1}{6} = 10

(d)

2 marks

All six expected frequencies are 10. Add up $\frac{( O - E ) ^{2}}{E}$ .
With $ν = 5$ the upper tail gives the p-value.

χ^{2} = \frac{4 + 9 + 16 + 25 + 4 + 4}{10} = 6.2 p \approx 0.287

(e)

2 marks

Compare $p = 0.287$ with the significance level $0.05$ .
There is insufficient evidence to reject $H_{0}$ : the data are consistent with the die being fair.

0.287 > 0.05

Mentor tip

Say "insufficient evidence to reject", not "the die is proved fair".

Question 8

6 marks

Bags of sugar: percentages, rejection and a cutoff value

Statistics and probability

(a)

1 mark

The normal distribution is symmetric about its mean of 500 g.

50%

(b)

2 marks

Use $X \sim N (500, 3^{2})$ .

P (X < 495) = 0.0478

(c)

3 marks

The top 2% of bags are rejected, so $P (X > k) = 0.02$ and $P (X < k) = 0.98$ .

k = 506.16 \approx 506 g

Question 9

7 marks

Derivative of 2/x + 3x^2 - 3 and the normal to the curve

Calculus

(a)

3 marks

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

f^{'} (x) = - 2 x^{- 2} + 6 x

(b)

4 marks

Gradient of the tangent at $x = 1$ :
The normal has gradient $- \frac{1}{4}$ and goes through $(1, 2)$ .

f^{'} (1) = - 2 + 6 = 4 y - 2 = - \frac{1}{4} (x - 1) \Rightarrow 4 y - 8 = - x + 1 x + 4 y - 9 = 0

Question 10

6 marks

Two socks from a drawer: tree diagram and conditional probability

Statistics and probability

(a)

1 mark

Seven socks: 3 brown and 4 black. If the first is brown, the second is brown with $\frac{2}{6}$ and black with $\frac{4}{6}$ . If the first is black, the second is brown with $\frac{3}{6}$ and black with $\frac{3}{6}$ .

(b)

2 marks

Multiply along the branches and add the two "same colour" outcomes.

P (same) = \frac{3}{7} \times \frac{2}{6} + \frac{4}{7} \times \frac{3}{6} = \frac{18}{42} = \frac{3}{7} \approx 0.429

(c)

3 marks

Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ .

P (brown, brown ∣ same) = \frac{\frac{3}{7} \times \frac{2}{6}}{\frac{3}{7}} = \frac{1}{3}

Question 11

8 marks

Earthquakes: Gutenberg-Richter equation and the wait between events

Number and algebra

(a)

2 marks

Put $N = 100$ and $M = 3$ into the equation.

lo g_{10} 100 = a - 3 \Rightarrow a = 5

(b)

2 marks

Undo the logarithm.

N = 1 0^{5 - M} = \frac{1 0 ^{5}}{1 0 ^{M}} \Rightarrow b = 100000

(c)

2 marks

$N = 1 0^{5 - M}$ is decreasing, so use the ends $M = 0$ and $M = 8$ .

1 0^{- 3} < N < 1 0^{5}

(d)

2 marks

At $M = 7.2$ :
The expected wait is the reciprocal.

N = 1 0^{5 - 7.2} = 0.00631 \frac{1}{0.00631} \approx 158 years

Question 12

6 marks

Rate of change of a company's profit

Calculus

(a)

2 marks

Evaluate the rate at $t = 2$ .
The rate is negative, so the profit is decreasing at $t = 2$ .

\frac{d P}{d t} = 3 (2)^{2} - 8 (2) = - 4

(b)

4 marks

Integrate the rate.
Use $P (1) = 4$ .

P (t) = t^{3} - 4 t^{2} + c 1 - 4 + c = 4 \Rightarrow c = 7 P (t) = t^{3} - 4 t^{2} + 7

Question 13

7 marks

Bouncing ball: geometric sequence and total distance

Number and algebra

(a)

2 marks

The bounce heights form a geometric sequence with common ratio $0.85$ .

h_{6} = 1.8 \times 0.8 5^{6} = 0.6789 \dots m = 67.9 cm \approx 68 cm

(b)

2 marks

Find the largest $n$ with $1.8 \times 0.8 5^{n} > 0.1$ .
So 17 bounces reach more than 10 cm.

1.8 \times 0.8 5^{17} = 0.114 m > 0.1, 1.8 \times 0.8 5^{18} = 0.0966 m < 0.1

(c)

3 marks

Up-and-down distances for bounces 1, 2 and 3 form a geometric series with first term $1.8 \times 0.85 = 1.53$ .
Add the first drop and double the bounce distances.

S_{3} = \frac{1.53 ( 1 - 0.8 5 ^{3} )}{1 - 0.85} = 3.935925 d = 1.8 + 2 (3.935925) = 9.67 m

Our analysis

Where the marks are

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

Number and algebra25 marks · 31%
Q1, Q5, Q11, Q13
Calculus20 marks · 25%
Q6, Q9, Q12
Statistics and probability19 marks · 24%
Q7, Q8, Q10
Geometry and trigonometry11 marks · 14%
Q2, Q4
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Midpoint and negative reciprocal · Q4
The perpendicular bisector passes through the midpoint of AB with gradient $- \frac{1}{m _{A B}}$ .
02
h is the strip width · Q6
The table has $x$ values 1 apart, so $h = 1$ in the trapezoidal rule.
03
Expected frequencies come from the null · Q7
If the die is fair every face is expected $60 \div 6 = 10$ times. Do not use the observed counts as expected ones.
04
Write 2/x as a power · Q9
$\frac{2}{x} = 2 x^{- 1}$ , so the derivative is $- 2 x^{- 2}$ .
05
Expected time between events is 1/N · Q11
If $N$ events happen per year on average, the expected wait between events is $\frac{1}{N}$ years.

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 TZ1 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

Area of the roof of a doghouse

Number and algebra

5 marks

The wall is 0.9 m high, so the triangle on top has height $1.35 - 0.9 = 0.45$ m and half-base $0.9 \div 2 = 0.45$ m.
Each roof rectangle is 0.6364 m by 0.9 m (the depth of the house).

slant length = 0.4 5^{2} + 0.4 5^{2} = 0.45 2 = 0.6364 m area to paint = 2 \times 0.6364 \times 0.9 = 1.15 m^{2}

Question 2

4 marks

Rope on a pole standing on level ground

Geometry and trigonometry

(a)

2 marks

Evaluate it.

A F = 3. 2^{2} + 4. 5^{2} + 5. 8^{2} A F = 8.01 m

(b)

2 marks

In the right triangle FOA the side opposite the angle at A is $F O = 5.8$ and the hypotenuse is $A F$ .

sin F A O = \frac{5.8}{8.008} \Rightarrow F A O = 46. 4^{\circ}

Question 3

5 marks

A baseball's height: quadratic model and domain

Functions

(a)

1 mark

At the instant of the hit $t = 0$ .

h (0) = 1.2 m

(b)

2 marks

The ball hits the ground when $h (t) = 0$ .

- 4.8 t^{2} + 21 t + 1.2 = 0 \Rightarrow t \approx 4.43 s (reject t < 0)

(c)

2 marks

Time starts at the hit and ends when the ball lands.

0 \leq t \leq 4.43

Question 4

7 marks

A road along the perpendicular bisector of two towns

Geometry and trigonometry

(a)

5 marks

Midpoint of A and B:
Gradient of AB and of the road:
Equation of the road:

(\frac{- 6 + 8}{2}, \frac{- 1 + 6}{2}) = (1, 2.5) m_{A B} = \frac{6 - ( - 1 )}{8 - ( - 6 )} = \frac{1}{2}, m_{road} = - 2 y - 2.5 = - 2 (x - 1) \Rightarrow y = - 2 x + 4.5

(b)

2 marks

Town C has the same $x$ -coordinate as A, which is $- 6$ .

y = - 2 (- 6) + 4.5 = 16.5

Question 5

5 marks

Concert tickets: forming and solving three equations

Number and algebra

(a)

3 marks

The total number of tickets, the total income and the ratio are:

x + y + z = 600, 15 x + 10 y + 12 z = 7816, x = 2 y

(b)

2 marks

Substitute $x = 2 y$ into the first two equations.
Eliminate $z = 600 - 3 y$ .

3 y + z = 600, 40 y + 12 z = 7816 40 y + 12 (600 - 3 y) = 7816 \Rightarrow 4 y = 616 \Rightarrow y = 154 x = 308 adult, y = 154 child, z = 138 student

Question 6

7 marks

A painting bounded by a cubic: trapezoidal rule and exact area

Calculus

(a)

3 marks

Trapezoidal rule with $h = 1$ and four ordinates.

A \approx \frac{1}{2} (1) [0.6 + 0 + 2 (1.2 + 1.2)] = 2.7 m^{2}

(b)

2 marks

Integrate the cubic from $- 1$ to $2$ .

A = \int_{- 1}^{2} \frac{- x ^{3} - 3 x ^{2} + 4 x + 12}{10} d x = \frac{1}{10} [- \frac{x ^{4}}{4} - x^{3} + 2 x^{2} + 12 x]_{- 1}^{2} A = \frac{1}{10} (20 - (- 9.25)) = 2.925 m^{2}

(c)

2 marks

The frame is a square with side 3, so its area is $3^{2} = 9$ .

9 - 2.925 = 6.075 m^{2}

Question 7

7 marks

Is the die fair? Chi-squared goodness of fit

Statistics and probability

(a)

1 mark

The null hypothesis says the die is fair, so each face has probability $\frac{1}{6}$ . The alternative says it is not fair.

H_{0} : P (each number) = \frac{1}{6}, H_{1} : the probabilities are not all \frac{1}{6}

(b)

1 mark

Six categories and no parameters estimated from the data.

ν = 6 - 1 = 5

(c)

1 mark

A fair die gives each face one sixth of the 60 rolls.

E = 60 \times \frac{1}{6} = 10

(d)

2 marks

All six expected frequencies are 10. Add up $\frac{( O - E ) ^{2}}{E}$ .
With $ν = 5$ the upper tail gives the p-value.

χ^{2} = \frac{4 + 9 + 16 + 25 + 4 + 4}{10} = 6.2 p \approx 0.287

(e)

2 marks

Compare $p = 0.287$ with the significance level $0.05$ .
There is insufficient evidence to reject $H_{0}$ : the data are consistent with the die being fair.

0.287 > 0.05

Mentor tip

Say "insufficient evidence to reject", not "the die is proved fair".

Question 8

6 marks

Bags of sugar: percentages, rejection and a cutoff value

Statistics and probability

(a)

1 mark

The normal distribution is symmetric about its mean of 500 g.

50%

(b)

2 marks

Use $X \sim N (500, 3^{2})$ .

P (X < 495) = 0.0478

(c)

3 marks

The top 2% of bags are rejected, so $P (X > k) = 0.02$ and $P (X < k) = 0.98$ .

k = 506.16 \approx 506 g

Question 9

7 marks

Derivative of 2/x + 3x^2 - 3 and the normal to the curve

Calculus

(a)

3 marks

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

f^{'} (x) = - 2 x^{- 2} + 6 x

(b)

4 marks

Gradient of the tangent at $x = 1$ :
The normal has gradient $- \frac{1}{4}$ and goes through $(1, 2)$ .

f^{'} (1) = - 2 + 6 = 4 y - 2 = - \frac{1}{4} (x - 1) \Rightarrow 4 y - 8 = - x + 1 x + 4 y - 9 = 0

Question 10

6 marks

Two socks from a drawer: tree diagram and conditional probability

Statistics and probability

(a)

1 mark

(b)

2 marks

Multiply along the branches and add the two "same colour" outcomes.

P (same) = \frac{3}{7} \times \frac{2}{6} + \frac{4}{7} \times \frac{3}{6} = \frac{18}{42} = \frac{3}{7} \approx 0.429

(c)

3 marks

Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ .

P (brown, brown ∣ same) = \frac{\frac{3}{7} \times \frac{2}{6}}{\frac{3}{7}} = \frac{1}{3}

Question 11

8 marks

Earthquakes: Gutenberg-Richter equation and the wait between events

Number and algebra

(a)

2 marks

Put $N = 100$ and $M = 3$ into the equation.

lo g_{10} 100 = a - 3 \Rightarrow a = 5

(b)

2 marks

Undo the logarithm.

N = 1 0^{5 - M} = \frac{1 0 ^{5}}{1 0 ^{M}} \Rightarrow b = 100000

(c)

2 marks

$N = 1 0^{5 - M}$ is decreasing, so use the ends $M = 0$ and $M = 8$ .

1 0^{- 3} < N < 1 0^{5}

(d)

2 marks

At $M = 7.2$ :
The expected wait is the reciprocal.

N = 1 0^{5 - 7.2} = 0.00631 \frac{1}{0.00631} \approx 158 years

Question 12

6 marks

Rate of change of a company's profit

Calculus

(a)

2 marks

Evaluate the rate at $t = 2$ .
The rate is negative, so the profit is decreasing at $t = 2$ .

\frac{d P}{d t} = 3 (2)^{2} - 8 (2) = - 4

(b)

4 marks

Integrate the rate.
Use $P (1) = 4$ .

P (t) = t^{3} - 4 t^{2} + c 1 - 4 + c = 4 \Rightarrow c = 7 P (t) = t^{3} - 4 t^{2} + 7

Question 13

7 marks

Bouncing ball: geometric sequence and total distance

Number and algebra

(a)

2 marks

The bounce heights form a geometric sequence with common ratio $0.85$ .

h_{6} = 1.8 \times 0.8 5^{6} = 0.6789 \dots m = 67.9 cm \approx 68 cm

(b)

2 marks

Find the largest $n$ with $1.8 \times 0.8 5^{n} > 0.1$ .
So 17 bounces reach more than 10 cm.

1.8 \times 0.8 5^{17} = 0.114 m > 0.1, 1.8 \times 0.8 5^{18} = 0.0966 m < 0.1

(c)

3 marks

Up-and-down distances for bounces 1, 2 and 3 form a geometric series with first term $1.8 \times 0.85 = 1.53$ .
Add the first drop and double the bounce distances.

S_{3} = \frac{1.53 ( 1 - 0.8 5 ^{3} )}{1 - 0.85} = 3.935925 d = 1.8 + 2 (3.935925) = 9.67 m

Our analysis

Where the marks are

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

Number and algebra25 marks · 31%
Q1, Q5, Q11, Q13
Calculus20 marks · 25%
Q6, Q9, Q12
Statistics and probability19 marks · 24%
Q7, Q8, Q10
Geometry and trigonometry11 marks · 14%
Q2, Q4
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Midpoint and negative reciprocal · Q4
The perpendicular bisector passes through the midpoint of AB with gradient $- \frac{1}{m _{A B}}$ .
02
h is the strip width · Q6
The table has $x$ values 1 apart, so $h = 1$ in the trapezoidal rule.
03
Expected frequencies come from the null · Q7
If the die is fair every face is expected $60 \div 6 = 10$ times. Do not use the observed counts as expected ones.
04
Write 2/x as a power · Q9
$\frac{2}{x} = 2 x^{- 1}$ , so the derivative is $- 2 x^{- 2}$ .
05
Expected time between events is 1/N · Q11
If $N$ events happen per year on average, the expected wait between events is $\frac{1}{N}$ years.

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