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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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

The amount of a drug in the body: an exponential decay model

Functions

(a)

3 marks

(i) At $t = 0$ the amount is the coefficient.

(ii) Each hour 85% remains, so 15% leaves.

D (0) = 23 mg 1 - 0.85 = 0.15 = 15%

(b)

2 marks

Put $t = 10$ into the model.

D (10) = 23 (0.85)^{10} \approx 4.53 mg

Question 2

5 marks

An inclined railway: distance, midpoint and height

Geometry and trigonometry

(a)

2 marks

Use the three-dimensional distance formula.

A B = (140 - 20)^{2} + (15 - 5)^{2} + (0 - 250)^{2} = 77000 \approx 277 m

(b)

2 marks

Average each coordinate.

M = (\frac{140 + 20}{2}, \frac{15 + 5}{2}, \frac{0 + 250}{2}) = (80, 10, 125)

(c)

1 mark

The height above the ground is the $z$ -coordinate.

125 m

Question 3

7 marks

Painting a storage box with a half-cylinder lid

Number and algebra

7 marks

The box faces to paint: two $90 \times 34$ sides, two $42 \times 34$ sides and the $90 \times 42$ base. The top is covered by the lid.
The lid has radius $\frac{42}{2} = 21$ cm. Its curved surface is half of a cylinder's side, and it has two semicircular ends that together make one circle.
Add the two parts.

2 (90 \times 34) + 2 (42 \times 34) + 90 \times 42 = 6120 + 2856 + 3780 = 12756 \frac{1}{2} \cdot 2 π (21) (90) + π (21)^{2} = 1890 π + 441 π \approx 5937.6 + 1385.4 12756 + 7323.0 \approx 20079 \approx 20100 cm^{2}

Question 4

7 marks

A swimmer and a helicopter: two angles of elevation

Geometry and trigonometry

(a)

1 mark

The horizontal line through H is parallel to the lake, so the angle of depression equals the angle of elevation from C.

2 5^{\circ}

(b)

2 marks

In the right-angled triangle HAC, $tan 2 5^{\circ} = \frac{380}{A C}$ .

A C = \frac{380}{tan 2 5 ^{\circ}} \approx 815 m

(c)

3 marks

In triangle HAB, $tan 4 0^{\circ} = \frac{380}{A B}$ , so $A B \approx 452.9$ m. Then take the difference.

B C = A C - A B \approx 814.9 - 452.9 \approx 362 m

(d)

1 mark

15 minutes is 0.25 hours.

\frac{362.05}{0.25} \approx 1450 m per hour

Question 5

5 marks

Reading a box plot and testing a claim about it

Statistics and probability

(a)

3 marks

Read the five values off the box plot: the left end of the whisker, the left edge of the box, and the line inside the box.

minimum = 2, Q_{1} = 6, median = 8

(b)

2 marks

The lower quartile is 6 and the upper quartile is 11. About 25% of employees took fewer than 6 days, and about 25% took more than 11, so the two percentages are about the same. A box plot doesn't show exact values near the quartiles, so Paul can't infer that one is smaller.
So Paul is not correct.

Question 6

7 marks

A perpendicular bisector and a bus stop on a road

Geometry and trigonometry

(a)

5 marks

The gradient of AB is $\frac{24 - 20}{14 - 2} = \frac{1}{3}$ , so the bisector's gradient is $- 3$ . The midpoint of AB is $(8, 22)$ .

y - 22 = - 3 (x - 8) \Rightarrow y = - 3 x + 46

(b)

2 marks

The road is $- x + y = 4$ , that is $y = x + 4$ . Set it equal to the bisector.

x + 4 = - 3 x + 46 \Rightarrow x = 10.5, y = 14.5 (10.5, 14.5)

Question 7

5 marks

Range of a function with a break, and one value of the inverse

Functions

(a)

3 marks

The end values are $f (- 7) = 2 - \frac{12}{- 2} = 8$ and $f (7) = 2 - 1 = 1$ . On $(- 5, 7]$ the function rises from $- \infty$ to 1. On $[- 7, - 5)$ it falls from $+ \infty$ to 8.

f (x) \leq 1 or f (x) \geq 8

(b)

2 marks

$f^{- 1} (0)$ is the value of $x$ with $f (x) = 0$ .

2 - \frac{12}{x + 5} = 0 \Rightarrow x + 5 = 6 \Rightarrow x = 1 f^{- 1} (0) = 1

Question 8

6 marks

Are chinchilla rabbits heavier than sable rabbits? A two-sample $t$ -test

Statistics and probability

(a)

2 marks

Let $μ_{c}$ and $μ_{s}$ be the population mean weights.

H_{0} : μ_{c} = μ_{s}, H_{1} : μ_{c} > μ_{s}

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances and the upper-tail alternative.

p \approx 0.0408

(c)

2 marks

Compare $p$ with the 5% level.
So reject $H_{0}$ : there is sufficient evidence that chinchilla rabbits are generally heavier than sable rabbits.

0.0408 < 0.05

Mentor tip

Use words in the conclusion, not just "reject".

Question 9

7 marks

A lawn bounded by an arc and a chord

Geometry and trigonometry

(a)

3 marks

The footpath goes along the arc, a fraction $\frac{135}{360}$ of the circumference.

arc = \frac{135}{360} \times 2 π (6) = \frac{135}{360} \times 12 π \approx 14.1 m

(b)

4 marks

The lawn is the sector minus the triangle.

\frac{135}{360} π (6)^{2} - \frac{1}{2} (6) (6) sin 13 5^{\circ} = 42.41 - 12.73 Area \approx 29.7 m^{2}

Question 10

6 marks

Savings accounts with half-yearly and quarterly compounding

Number and algebra

(a)

3 marks

Five years of half-yearly compounding is 10 periods at $\frac{2.75}{2} %$ each.

1500 (1 + \frac{0.0275}{2})^{10} \approx 1719.49 euros

(b)

3 marks

The target is $1.5 \times 1500 = 2250$ after 20 quarters at $\frac{r}{4} %$ each.

1500 (1 + \frac{r}{400})^{20} = 2250 \Rightarrow (1 + \frac{r}{400})^{20} = 1.5 r \approx 8.19

Question 11

6 marks

Testing a newspaper vendor's model: a goodness-of-fit test

Statistics and probability

(a)

1 mark

The total sold is 460 copies over 5 days.

\frac{460}{5} = 92

(b)

5 marks

(i) There are 5 days and no parameters estimated, so

(ii) The test statistic on the calculator, with expected counts of 92 each day:

So there is insufficient evidence to reject $H_{0}$ : the data are consistent with the vendor's model.

df = 5 - 1 = 4 χ^{2} \approx 8.54 < 9.49 (p \approx 0.0736 > 0.05)

Question 12

7 marks

A quadratic from its vertex and an intercept

Functions

(a)

1 mark

The axis of symmetry is $x = 0.5$ , which is halfway between $- 2$ and $p$ .

\frac{- 2 + p}{2} = 0.5 \Rightarrow p = 3

(b)

5 marks

Write the function in factorised form $f (x) = a (x + 2) (x - 3)$ and use the vertex $(0.5, - 12.5)$ .
Expand to read off $b$ and $c$ .

f (0.5) = a (2.5) (- 2.5) = - 6.25 a = - 12.5 \Rightarrow a = 2 f (x) = 2 (x^{2} - x - 6) = 2 x^{2} - 2 x - 12 a = 2, b = - 2, c = - 12

(c)

1 mark

The axis of symmetry passes through the vertex.

x = 0.5

Question 13

7 marks

Profit from a rate of change, and what happens after 30 cars

Calculus

(a)

5 marks

Integrate $\frac{d P}{d x} = - 1.6 x + 48$ term by term.
Use $P (15) = 260$ .

P (x) = - 0.8 x^{2} + 48 x + c 260 = - 0.8 (225) + 48 (15) + c = 540 + c \Rightarrow c = - 280 P (x) = - 0.8 x^{2} + 48 x - 280

(b)

2 marks

For $x > 30$ the rate $- 1.6 x + 48$ is negative, so profit decreases as production increases. Check: $P (30) = 440$ and $P (50) = 120$ .
So profit falls from 440 to 120 thousand dollars.

Our analysis

Where the marks are

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

Geometry and trigonometry26 marks · 33%
Q2, Q4, Q6, Q9
Functions17 marks · 21%
Q1, Q7, Q12
Statistics and probability17 marks · 21%
Q5, Q8, Q11
Number and algebra13 marks · 16%
Q3, Q10
Calculus7 marks · 9%
Q13

Mentor insights

Traps to watch for

01
Count only the faces that show · Q3
The lid sits exactly on top of the box, so the top face of the box is hidden. Paint is needed on the four sides, the base, the curved lid and its two semicircular ends.
02
Alternate angles · Q4
The angle of depression from H to C equals the angle of elevation from C to H, because they are alternate angles between parallel lines.
03
Equidistant means on the bisector · Q6
The point equally far from two schools lies on the perpendicular bisector of the segment joining them. The bus stop must also lie on the road, so it is where the two lines meet.
04
Segment area is sector minus triangle · Q9
The lawn lies between the arc and the chord. Subtract the triangle OAB from the sector OAB. Convert $13 5^{\circ}$ to $\frac{3 π}{4}$ for the arc length.
05
Use the symmetry · Q12
The vertex is on the axis of symmetry, halfway between the two $x$ -intercepts. That gives the missing intercept immediately.

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 2021
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 2021 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

The amount of a drug in the body: an exponential decay model

Functions

(a)

3 marks

(i) At $t = 0$ the amount is the coefficient.

(ii) Each hour 85% remains, so 15% leaves.

D (0) = 23 mg 1 - 0.85 = 0.15 = 15%

(b)

2 marks

Put $t = 10$ into the model.

D (10) = 23 (0.85)^{10} \approx 4.53 mg

Question 2

5 marks

An inclined railway: distance, midpoint and height

Geometry and trigonometry

(a)

2 marks

Use the three-dimensional distance formula.

A B = (140 - 20)^{2} + (15 - 5)^{2} + (0 - 250)^{2} = 77000 \approx 277 m

(b)

2 marks

Average each coordinate.

M = (\frac{140 + 20}{2}, \frac{15 + 5}{2}, \frac{0 + 250}{2}) = (80, 10, 125)

(c)

1 mark

The height above the ground is the $z$ -coordinate.

125 m

Question 3

7 marks

Painting a storage box with a half-cylinder lid

Number and algebra

7 marks

The box faces to paint: two $90 \times 34$ sides, two $42 \times 34$ sides and the $90 \times 42$ base. The top is covered by the lid.
The lid has radius $\frac{42}{2} = 21$ cm. Its curved surface is half of a cylinder's side, and it has two semicircular ends that together make one circle.
Add the two parts.

2 (90 \times 34) + 2 (42 \times 34) + 90 \times 42 = 6120 + 2856 + 3780 = 12756 \frac{1}{2} \cdot 2 π (21) (90) + π (21)^{2} = 1890 π + 441 π \approx 5937.6 + 1385.4 12756 + 7323.0 \approx 20079 \approx 20100 cm^{2}

Question 4

7 marks

A swimmer and a helicopter: two angles of elevation

Geometry and trigonometry

(a)

1 mark

The horizontal line through H is parallel to the lake, so the angle of depression equals the angle of elevation from C.

2 5^{\circ}

(b)

2 marks

In the right-angled triangle HAC, $tan 2 5^{\circ} = \frac{380}{A C}$ .

A C = \frac{380}{tan 2 5 ^{\circ}} \approx 815 m

(c)

3 marks

In triangle HAB, $tan 4 0^{\circ} = \frac{380}{A B}$ , so $A B \approx 452.9$ m. Then take the difference.

B C = A C - A B \approx 814.9 - 452.9 \approx 362 m

(d)

1 mark

15 minutes is 0.25 hours.

\frac{362.05}{0.25} \approx 1450 m per hour

Question 5

5 marks

Reading a box plot and testing a claim about it

Statistics and probability

(a)

3 marks

Read the five values off the box plot: the left end of the whisker, the left edge of the box, and the line inside the box.

minimum = 2, Q_{1} = 6, median = 8

(b)

2 marks

The lower quartile is 6 and the upper quartile is 11. About 25% of employees took fewer than 6 days, and about 25% took more than 11, so the two percentages are about the same. A box plot doesn't show exact values near the quartiles, so Paul can't infer that one is smaller.
So Paul is not correct.

Question 6

7 marks

A perpendicular bisector and a bus stop on a road

Geometry and trigonometry

(a)

5 marks

The gradient of AB is $\frac{24 - 20}{14 - 2} = \frac{1}{3}$ , so the bisector's gradient is $- 3$ . The midpoint of AB is $(8, 22)$ .

y - 22 = - 3 (x - 8) \Rightarrow y = - 3 x + 46

(b)

2 marks

The road is $- x + y = 4$ , that is $y = x + 4$ . Set it equal to the bisector.

x + 4 = - 3 x + 46 \Rightarrow x = 10.5, y = 14.5 (10.5, 14.5)

Question 7

5 marks

Range of a function with a break, and one value of the inverse

Functions

(a)

3 marks

The end values are $f (- 7) = 2 - \frac{12}{- 2} = 8$ and $f (7) = 2 - 1 = 1$ . On $(- 5, 7]$ the function rises from $- \infty$ to 1. On $[- 7, - 5)$ it falls from $+ \infty$ to 8.

f (x) \leq 1 or f (x) \geq 8

(b)

2 marks

$f^{- 1} (0)$ is the value of $x$ with $f (x) = 0$ .

2 - \frac{12}{x + 5} = 0 \Rightarrow x + 5 = 6 \Rightarrow x = 1 f^{- 1} (0) = 1

Question 8

6 marks

Are chinchilla rabbits heavier than sable rabbits? A two-sample $t$ -test

Statistics and probability

(a)

2 marks

Let $μ_{c}$ and $μ_{s}$ be the population mean weights.

H_{0} : μ_{c} = μ_{s}, H_{1} : μ_{c} > μ_{s}

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances and the upper-tail alternative.

p \approx 0.0408

(c)

2 marks

Compare $p$ with the 5% level.
So reject $H_{0}$ : there is sufficient evidence that chinchilla rabbits are generally heavier than sable rabbits.

0.0408 < 0.05

Mentor tip

Use words in the conclusion, not just "reject".

Question 9

7 marks

A lawn bounded by an arc and a chord

Geometry and trigonometry

(a)

3 marks

The footpath goes along the arc, a fraction $\frac{135}{360}$ of the circumference.

arc = \frac{135}{360} \times 2 π (6) = \frac{135}{360} \times 12 π \approx 14.1 m

(b)

4 marks

The lawn is the sector minus the triangle.

\frac{135}{360} π (6)^{2} - \frac{1}{2} (6) (6) sin 13 5^{\circ} = 42.41 - 12.73 Area \approx 29.7 m^{2}

Question 10

6 marks

Savings accounts with half-yearly and quarterly compounding

Number and algebra

(a)

3 marks

Five years of half-yearly compounding is 10 periods at $\frac{2.75}{2} %$ each.

1500 (1 + \frac{0.0275}{2})^{10} \approx 1719.49 euros

(b)

3 marks

The target is $1.5 \times 1500 = 2250$ after 20 quarters at $\frac{r}{4} %$ each.

1500 (1 + \frac{r}{400})^{20} = 2250 \Rightarrow (1 + \frac{r}{400})^{20} = 1.5 r \approx 8.19

Question 11

6 marks

Testing a newspaper vendor's model: a goodness-of-fit test

Statistics and probability

(a)

1 mark

The total sold is 460 copies over 5 days.

\frac{460}{5} = 92

(b)

5 marks

(i) There are 5 days and no parameters estimated, so

(ii) The test statistic on the calculator, with expected counts of 92 each day:

So there is insufficient evidence to reject $H_{0}$ : the data are consistent with the vendor's model.

df = 5 - 1 = 4 χ^{2} \approx 8.54 < 9.49 (p \approx 0.0736 > 0.05)

Question 12

7 marks

A quadratic from its vertex and an intercept

Functions

(a)

1 mark

The axis of symmetry is $x = 0.5$ , which is halfway between $- 2$ and $p$ .

\frac{- 2 + p}{2} = 0.5 \Rightarrow p = 3

(b)

5 marks

Write the function in factorised form $f (x) = a (x + 2) (x - 3)$ and use the vertex $(0.5, - 12.5)$ .
Expand to read off $b$ and $c$ .

f (0.5) = a (2.5) (- 2.5) = - 6.25 a = - 12.5 \Rightarrow a = 2 f (x) = 2 (x^{2} - x - 6) = 2 x^{2} - 2 x - 12 a = 2, b = - 2, c = - 12

(c)

1 mark

The axis of symmetry passes through the vertex.

x = 0.5

Question 13

7 marks

Profit from a rate of change, and what happens after 30 cars

Calculus

(a)

5 marks

Integrate $\frac{d P}{d x} = - 1.6 x + 48$ term by term.
Use $P (15) = 260$ .

P (x) = - 0.8 x^{2} + 48 x + c 260 = - 0.8 (225) + 48 (15) + c = 540 + c \Rightarrow c = - 280 P (x) = - 0.8 x^{2} + 48 x - 280

(b)

2 marks

For $x > 30$ the rate $- 1.6 x + 48$ is negative, so profit decreases as production increases. Check: $P (30) = 440$ and $P (50) = 120$ .
So profit falls from 440 to 120 thousand dollars.

Our analysis

Where the marks are

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

Geometry and trigonometry26 marks · 33%
Q2, Q4, Q6, Q9
Functions17 marks · 21%
Q1, Q7, Q12
Statistics and probability17 marks · 21%
Q5, Q8, Q11
Number and algebra13 marks · 16%
Q3, Q10
Calculus7 marks · 9%
Q13

Mentor insights

Traps to watch for

01
Count only the faces that show · Q3
The lid sits exactly on top of the box, so the top face of the box is hidden. Paint is needed on the four sides, the base, the curved lid and its two semicircular ends.
02
Alternate angles · Q4
The angle of depression from H to C equals the angle of elevation from C to H, because they are alternate angles between parallel lines.
03
Equidistant means on the bisector · Q6
The point equally far from two schools lies on the perpendicular bisector of the segment joining them. The bus stop must also lie on the road, so it is where the two lines meet.
04
Segment area is sector minus triangle · Q9
The lawn lies between the arc and the chord. Subtract the triangle OAB from the sector OAB. Convert $13 5^{\circ}$ to $\frac{3 π}{4}$ for the arc length.
05
Use the symmetry · Q12
The vertex is on the axis of symmetry, halfway between the two $x$ -intercepts. That gives the missing intercept immediately.

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