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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Session: November 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 November 2021 Global 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

6 marks

Age and 5000 m time: correlation, regression and a prediction

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.933

(b)

1 mark

Since $0.8 \leq ∣ r ∣ \leq 1$ , the correlation is strong (and positive: older runners tend to take longer).

(c)

1 mark

The regression line of $t$ on $x$ :

t \approx 0.228 x + 24.3

(d)

2 marks

Substitute $x = 57$ .

t = 0.2277 \times 57 + 24.3153 \approx 37.3 minutes

Question 2

5 marks

The Bermuda Triangle: an angle and an area from three sides

Geometry and trigonometry

(a)

3 marks

Use the cosine rule with $θ$ opposite the 1670 km side.

cos θ = \frac{166 0 ^{2} + 155 0 ^{2} - 167 0 ^{2}}{2 ( 1660 ) ( 1550 )} \approx 0.4620 \Rightarrow θ \approx 62. 6^{\circ}

(b)

2 marks

Use the area formula with the two sides on either side of $θ$ .

A = \frac{1}{2} (1660) (1550) sin 62.58 7^{\circ} \approx 1.14 \times 1 0^{6} km^{2}

Mentor tip

Keep the unrounded angle in the calculator. Using $6 3^{\circ}$ gives a slightly different area.

Question 3

4 marks

Mould growth: finding the rate constant of an exponential model

Functions

(a)

1 mark

At $t = 0$ , $P (0) = A e^{0} = A$ .

A = 112

(b)

3 marks

At $t = 5$ the area is 360.

112 e^{5 k} = 360 \Rightarrow e^{5 k} = \frac{360}{112} \Rightarrow k = \frac{1}{5} ln \frac{360}{112} \approx 0.234

Question 4

6 marks

A kite on coordinate axes: gradients and a perpendicular diagonal

Geometry and trigonometry

(a)

2 marks

The gradient of AC.

m_{A C} = \frac{6 - 0}{4 - 2} = 3

(b)

1 mark

BD is perpendicular to AC.

m_{B D} = - \frac{1}{3}

(c)

2 marks

Use the point B $(0, 4)$ with gradient $- \frac{1}{3}$ .

y = - \frac{1}{3} x + 4 \Rightarrow x + 3 y - 12 = 0

(d)

1 mark

D is on the $x$ -axis, so $y = 0$ .

x + 0 - 12 = 0 \Rightarrow x = 12

Question 5

7 marks

A tin can's height as a function of diameter, and its inverse

Functions

(a)

3 marks

$h$ is decreasing on its domain, so the largest value is at $x = 4$ and the smallest at $x = 14$ .

h (4) = \frac{640}{16} + 0.5 = 40.5, h (14) = \frac{640}{196} + 0.5 \approx 3.77 3.77 \leq h (x) \leq 40.5

(b)

4 marks

(i) Solve $h (x) = 10$ .

(ii) A tin that is 10 cm high has a diameter of about 8.21 cm.

(iii) The range of the inverse is the domain of $h$ .

\frac{640}{x ^{2}} + 0.5 = 10 \Rightarrow x^{2} = \frac{640}{9.5} \Rightarrow x \approx 8.21 4 \leq h^{- 1} (x) \leq 14

Question 6

5 marks

Carbon dioxide emissions: the trapezoidal rule and percentage error

Calculus

(a)

3 marks

Use the trapezoidal rule with width 0.4: half the width times (first + last + twice the middle values).

\frac{1}{2} (0.4) [30 + 50 + 2 (50 + 60 + 40 + 20)] = 0.2 (80 + 340) = 84 tonnes

(b)

2 marks

Percentage error compared with the real value, 72 tonnes.

\frac{84 - 72}{72} \times 100% \approx 16.7%

Question 7

6 marks

Fire stations and a Voronoi diagram: where to put a new station

Geometry and trigonometry

(a)

3 marks

The new station should be as far as possible from the nearest existing one, which happens at a vertex of the Voronoi diagram, P or Q. Compare the distance from each to its nearest stations.
P is farther from its nearest stations than Q is, so the new station should be at P.

AP = (10 - 6)^{2} + (6 - 2)^{2} = 32 \approx 5.66, BQ = (13 - 14)^{2} + (7 - 2)^{2} = 26 \approx 5.10

(b)

3 marks

(i) P is $(10, 6)$ and C is $(18, 6)$ . They have the same $y$ , so the bisector is vertical through the midpoint $(14, 6)$ .

(ii) The missing boundaries are the perpendicular bisectors of PC and PD, joined to the existing boundaries. The one for PD passes through the midpoint $(10.4, 8.8)$ and has gradient $- \frac{1}{7}$ , since PD has gradient 7.

x = 14 y - 8.8 = - \frac{1}{7} (x - 10.4)

Question 8

5 marks

A party hat from a circular sector

Geometry and trigonometry

(a)

3 marks

(i) The perimeter of the circular base of radius 6.5 cm is

(ii) This is the arc length of the sector of radius 18 cm and angle $θ$ .

2 π (6.5) = 13 π cm \frac{θ}{360} \times 2 π (18) = 13 π \Rightarrow θ = 13 0^{\circ}

(b)

2 marks

The outside of the hat is the curved surface, which is the area of the sector.

\frac{130}{360} π (18)^{2} = 117 π \approx 368 cm^{2}

Mentor tip

The same result comes from the curved area of a cone, $π r l = π (6.5) (18) = 117 π$ .

Question 9

8 marks

History exam scores: a cumulative frequency graph and grade boundaries

Statistics and probability

(a)

1 mark

The median is at the half-way cumulative frequency, $\frac{120}{2} = 60$ . Read across to the curve and down to the score axis.

median = 75

(b)

2 marks

The grades account for all 120 students.

6 + 13 + 26 + a + b = 120 \Rightarrow a = 75 - b

(c)

5 marks

(i) The mean grade is the weighted total of the grades divided by 120.

(ii) 28 students got grade 5, so the other $120 - 28 = 92$ scored lower. Read the graph at a cumulative frequency of 92.

\frac{6 ( 1 ) + 13 ( 2 ) + 26 ( 3 ) + 4 ( 75 - b ) + 5 b}{120} = 3.65 \Rightarrow \frac{410 + b}{120} = 3.65 410 + b = 438 \Rightarrow b = 28 minimum score for grade 5 = 84

Question 10

9 marks

A 30-year home loan: the payment, the balance and the interest paid

Number and algebra

(a)

3 marks

Use the finance solver with $N = 360$ , $I = 3.8%$ , $PV = 170000$ , $FV = 0$ , 12 payments and 12 compounding periods per year.

PMT \approx 792.13 AUD per month

(b)

6 marks

(i) After 10 years ( $N = 120$ ) the balance is the future value with the payment of $- 792.13$ .

(ii) They paid $120 \times 792.13 = 95 055.60$ . Of this, $170000 - 133019.94 = 36 980.06$ went to repaying the loan. The rest is interest.

balance \approx 133 019.94 AUD 95 055.60 - 36 980.06 = 58 075.54 AUD

Mentor tip

Using the unrounded payment gives 58 075.60. Both are fine to the cent, as long as you say which one you used.

Question 11

5 marks

Eye colour and hair colour from a tree diagram

Statistics and probability

(a)

1 mark

From "blue eyed", the branches add to 1, so $a$ is the chance of not fair-haired.

a = 1 - 0.58 = 0.42

(b)

1 mark

Not blue eyed (0.68) and then fair haired (probability $b$ ).

P (not blue and fair) = 0.68 b

(c)

3 marks

(i) Add the two ways to be fair haired, and set the total to 0.41.

(ii) The branches from "not blue eyed" add to 1.

0.32 \times 0.58 + 0.68 b = 0.41 \Rightarrow 0.68 b = 0.2244 \Rightarrow b = 0.33 c = 1 - 0.33 = 0.67

Question 12

6 marks

The cheapest open box: a derivative and what it means

Calculus

(a)

3 marks

Write the second term as $128 x^{- 1}$ and differentiate.

S (x) = x^{2} + 128 x^{- 1} \Rightarrow S^{'} (x) = 2 x - 128 x^{- 2} = 2 x - \frac{128}{x ^{2}}

(b)

3 marks

(i) Set the derivative to zero.

(ii) $x = 4$ cm is the side length of the square base that gives the smallest surface area for a box of volume 32 cm³.

2 x = \frac{128}{x ^{2}} \Rightarrow x^{3} = 64 \Rightarrow x = 4

Question 13

8 marks

A window with a quadratic top: finding the curve and the area

Functions

(a)

5 marks

(i) At $x = 0$ , $y = c$ , so

(ii) The end point $(8, 10)$ and the vertex $(4, 12)$ give two equations.

(iii) Solve the pair: the first gives $b = - 8 a$ , and the second becomes $16 a - 32 a = 2$ .

c = 10 64 a + 8 b + 10 = 10, 16 a + 4 b + 10 = 12 a = - \frac{1}{8}, b = 1 \Rightarrow y = - \frac{1}{8} x^{2} + x + 10

(b)

3 marks

The shaded region lies under the curve between $x = 0$ and $x = 8$ .

\int_{0}^{8} (- \frac{1}{8} x^{2} + x + 10) d x = [- \frac{x ^{3}}{24} + \frac{x ^{2}}{2} + 10 x]_{0}^{8} = \frac{272}{3} \approx 90.7 cm^{2}

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 trigonometry22 marks · 28%
Q2, Q4, Q7, Q8
Functions19 marks · 24%
Q3, Q5, Q13
Statistics and probability19 marks · 24%
Q1, Q9, Q11
Calculus11 marks · 14%
Q6, Q12
Number and algebra9 marks · 11%
Q10

Mentor insights

Traps to watch for

01
Describe the strength using the table · Q1
Compare $∣ r ∣$ with the bands in the question to say "strong", and say it is positive. Predicting for a 57-year-old is inside the data range, so it is an interpolation.
02
Interpret the inverse · Q5
The inverse takes a height and gives the diameter. So $h^{- 1} (10)$ is the diameter of a tin that is 10 cm high.
03
Use the total of 120 · Q9
Read the median at 60 on the cumulative axis. For the minimum score for grade 5, count down from 120 by the number of grade 5 students and read the graph there.
04
Use the finance solver correctly · Q10
PV and the payment must have opposite signs, and the payment is at the end of each month. Round the payment to the cent before using it, as banks do.
05
Use symmetry · Q13
The vertex $(4, 12)$ lies on the axis of symmetry halfway between the two end points $x = 0$ and $x = 8$ . That gives the vertex equation 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.

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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Session: November 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 November 2021 Global 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

6 marks

Age and 5000 m time: correlation, regression and a prediction

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.933

(b)

1 mark

Since $0.8 \leq ∣ r ∣ \leq 1$ , the correlation is strong (and positive: older runners tend to take longer).

(c)

1 mark

The regression line of $t$ on $x$ :

t \approx 0.228 x + 24.3

(d)

2 marks

Substitute $x = 57$ .

t = 0.2277 \times 57 + 24.3153 \approx 37.3 minutes

Question 2

5 marks

The Bermuda Triangle: an angle and an area from three sides

Geometry and trigonometry

(a)

3 marks

Use the cosine rule with $θ$ opposite the 1670 km side.

cos θ = \frac{166 0 ^{2} + 155 0 ^{2} - 167 0 ^{2}}{2 ( 1660 ) ( 1550 )} \approx 0.4620 \Rightarrow θ \approx 62. 6^{\circ}

(b)

2 marks

Use the area formula with the two sides on either side of $θ$ .

A = \frac{1}{2} (1660) (1550) sin 62.58 7^{\circ} \approx 1.14 \times 1 0^{6} km^{2}

Mentor tip

Keep the unrounded angle in the calculator. Using $6 3^{\circ}$ gives a slightly different area.

Question 3

4 marks

Mould growth: finding the rate constant of an exponential model

Functions

(a)

1 mark

At $t = 0$ , $P (0) = A e^{0} = A$ .

A = 112

(b)

3 marks

At $t = 5$ the area is 360.

112 e^{5 k} = 360 \Rightarrow e^{5 k} = \frac{360}{112} \Rightarrow k = \frac{1}{5} ln \frac{360}{112} \approx 0.234

Question 4

6 marks

A kite on coordinate axes: gradients and a perpendicular diagonal

Geometry and trigonometry

(a)

2 marks

The gradient of AC.

m_{A C} = \frac{6 - 0}{4 - 2} = 3

(b)

1 mark

BD is perpendicular to AC.

m_{B D} = - \frac{1}{3}

(c)

2 marks

Use the point B $(0, 4)$ with gradient $- \frac{1}{3}$ .

y = - \frac{1}{3} x + 4 \Rightarrow x + 3 y - 12 = 0

(d)

1 mark

D is on the $x$ -axis, so $y = 0$ .

x + 0 - 12 = 0 \Rightarrow x = 12

Question 5

7 marks

A tin can's height as a function of diameter, and its inverse

Functions

(a)

3 marks

$h$ is decreasing on its domain, so the largest value is at $x = 4$ and the smallest at $x = 14$ .

h (4) = \frac{640}{16} + 0.5 = 40.5, h (14) = \frac{640}{196} + 0.5 \approx 3.77 3.77 \leq h (x) \leq 40.5

(b)

4 marks

(i) Solve $h (x) = 10$ .

(ii) A tin that is 10 cm high has a diameter of about 8.21 cm.

(iii) The range of the inverse is the domain of $h$ .

\frac{640}{x ^{2}} + 0.5 = 10 \Rightarrow x^{2} = \frac{640}{9.5} \Rightarrow x \approx 8.21 4 \leq h^{- 1} (x) \leq 14

Question 6

5 marks

Carbon dioxide emissions: the trapezoidal rule and percentage error

Calculus

(a)

3 marks

Use the trapezoidal rule with width 0.4: half the width times (first + last + twice the middle values).

\frac{1}{2} (0.4) [30 + 50 + 2 (50 + 60 + 40 + 20)] = 0.2 (80 + 340) = 84 tonnes

(b)

2 marks

Percentage error compared with the real value, 72 tonnes.

\frac{84 - 72}{72} \times 100% \approx 16.7%

Question 7

6 marks

Fire stations and a Voronoi diagram: where to put a new station

Geometry and trigonometry

(a)

3 marks

The new station should be as far as possible from the nearest existing one, which happens at a vertex of the Voronoi diagram, P or Q. Compare the distance from each to its nearest stations.
P is farther from its nearest stations than Q is, so the new station should be at P.

AP = (10 - 6)^{2} + (6 - 2)^{2} = 32 \approx 5.66, BQ = (13 - 14)^{2} + (7 - 2)^{2} = 26 \approx 5.10

(b)

3 marks

(i) P is $(10, 6)$ and C is $(18, 6)$ . They have the same $y$ , so the bisector is vertical through the midpoint $(14, 6)$ .

x = 14 y - 8.8 = - \frac{1}{7} (x - 10.4)

Question 8

5 marks

A party hat from a circular sector

Geometry and trigonometry

(a)

3 marks

(i) The perimeter of the circular base of radius 6.5 cm is

(ii) This is the arc length of the sector of radius 18 cm and angle $θ$ .

2 π (6.5) = 13 π cm \frac{θ}{360} \times 2 π (18) = 13 π \Rightarrow θ = 13 0^{\circ}

(b)

2 marks

The outside of the hat is the curved surface, which is the area of the sector.

\frac{130}{360} π (18)^{2} = 117 π \approx 368 cm^{2}

Mentor tip

The same result comes from the curved area of a cone, $π r l = π (6.5) (18) = 117 π$ .

Question 9

8 marks

History exam scores: a cumulative frequency graph and grade boundaries

Statistics and probability

(a)

1 mark

The median is at the half-way cumulative frequency, $\frac{120}{2} = 60$ . Read across to the curve and down to the score axis.

median = 75

(b)

2 marks

The grades account for all 120 students.

6 + 13 + 26 + a + b = 120 \Rightarrow a = 75 - b

(c)

5 marks

(i) The mean grade is the weighted total of the grades divided by 120.

(ii) 28 students got grade 5, so the other $120 - 28 = 92$ scored lower. Read the graph at a cumulative frequency of 92.

\frac{6 ( 1 ) + 13 ( 2 ) + 26 ( 3 ) + 4 ( 75 - b ) + 5 b}{120} = 3.65 \Rightarrow \frac{410 + b}{120} = 3.65 410 + b = 438 \Rightarrow b = 28 minimum score for grade 5 = 84

Question 10

9 marks

A 30-year home loan: the payment, the balance and the interest paid

Number and algebra

(a)

3 marks

Use the finance solver with $N = 360$ , $I = 3.8%$ , $PV = 170000$ , $FV = 0$ , 12 payments and 12 compounding periods per year.

PMT \approx 792.13 AUD per month

(b)

6 marks

(i) After 10 years ( $N = 120$ ) the balance is the future value with the payment of $- 792.13$ .

(ii) They paid $120 \times 792.13 = 95 055.60$ . Of this, $170000 - 133019.94 = 36 980.06$ went to repaying the loan. The rest is interest.

balance \approx 133 019.94 AUD 95 055.60 - 36 980.06 = 58 075.54 AUD

Mentor tip

Using the unrounded payment gives 58 075.60. Both are fine to the cent, as long as you say which one you used.

Question 11

5 marks

Eye colour and hair colour from a tree diagram

Statistics and probability

(a)

1 mark

From "blue eyed", the branches add to 1, so $a$ is the chance of not fair-haired.

a = 1 - 0.58 = 0.42

(b)

1 mark

Not blue eyed (0.68) and then fair haired (probability $b$ ).

P (not blue and fair) = 0.68 b

(c)

3 marks

(i) Add the two ways to be fair haired, and set the total to 0.41.

(ii) The branches from "not blue eyed" add to 1.

0.32 \times 0.58 + 0.68 b = 0.41 \Rightarrow 0.68 b = 0.2244 \Rightarrow b = 0.33 c = 1 - 0.33 = 0.67

Question 12

6 marks

The cheapest open box: a derivative and what it means

Calculus

(a)

3 marks

Write the second term as $128 x^{- 1}$ and differentiate.

S (x) = x^{2} + 128 x^{- 1} \Rightarrow S^{'} (x) = 2 x - 128 x^{- 2} = 2 x - \frac{128}{x ^{2}}

(b)

3 marks

(i) Set the derivative to zero.

(ii) $x = 4$ cm is the side length of the square base that gives the smallest surface area for a box of volume 32 cm³.

2 x = \frac{128}{x ^{2}} \Rightarrow x^{3} = 64 \Rightarrow x = 4

Question 13

8 marks

A window with a quadratic top: finding the curve and the area

Functions

(a)

5 marks

(i) At $x = 0$ , $y = c$ , so

(ii) The end point $(8, 10)$ and the vertex $(4, 12)$ give two equations.

(iii) Solve the pair: the first gives $b = - 8 a$ , and the second becomes $16 a - 32 a = 2$ .

c = 10 64 a + 8 b + 10 = 10, 16 a + 4 b + 10 = 12 a = - \frac{1}{8}, b = 1 \Rightarrow y = - \frac{1}{8} x^{2} + x + 10

(b)

3 marks

The shaded region lies under the curve between $x = 0$ and $x = 8$ .

\int_{0}^{8} (- \frac{1}{8} x^{2} + x + 10) d x = [- \frac{x ^{3}}{24} + \frac{x ^{2}}{2} + 10 x]_{0}^{8} = \frac{272}{3} \approx 90.7 cm^{2}

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 trigonometry22 marks · 28%
Q2, Q4, Q7, Q8
Functions19 marks · 24%
Q3, Q5, Q13
Statistics and probability19 marks · 24%
Q1, Q9, Q11
Calculus11 marks · 14%
Q6, Q12
Number and algebra9 marks · 11%
Q10

Mentor insights

Traps to watch for

01
Describe the strength using the table · Q1
Compare $∣ r ∣$ with the bands in the question to say "strong", and say it is positive. Predicting for a 57-year-old is inside the data range, so it is an interpolation.
02
Interpret the inverse · Q5
The inverse takes a height and gives the diameter. So $h^{- 1} (10)$ is the diameter of a tin that is 10 cm high.
03
Use the total of 120 · Q9
Read the median at 60 on the cumulative axis. For the minimum score for grade 5, count down from 120 by the number of grade 5 students and read the graph there.
04
Use the finance solver correctly · Q10
PV and the payment must have opposite signs, and the payment is at the end of each month. Round the payment to the cent before using it, as banks do.
05
Use symmetry · Q13
The vertex $(4, 12)$ lies on the axis of symmetry halfway between the two end points $x = 0$ and $x = 8$ . That gives the vertex equation 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