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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 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

4 marks

A continued-fraction approximation of $π$

Number and algebra

(a)

2 marks

Work out the denominator first: $6 + \frac{13}{16} = \frac{109}{16}$ . The expression becomes $3 + \frac{16}{109}$ .

3 + \frac{16}{109} \approx 3.14679 \approx 3.1468

(b)

2 marks

Percentage error is the difference divided by the exact value, times 100.

\frac{∣3.1468 - π ∣}{π} \times 100 \approx 0.166%

Question 2

4 marks

Ten days of temperatures: mode, mean and standard deviation

Statistics and probability

(a)

1 mark

The mode is the most common value; 14 appears three times.

14

(b)

2 marks

Add the ten values and divide by ten.

\frac{14 + 15 + 14 + 11 + 10 + 9 + 14 + 15 + 16 + 13}{10} = 13.1

(c)

1 mark

Enter the data in the calculator's one-variable statistics.

σ \approx 2.21

Question 3

6 marks

Surface area of a hemisphere and the chocolate needed

Geometry and trigonometry

(a)

4 marks

The curved surface is half of a sphere's surface, and there is also the flat circle.

2 π (6)^{2} + π (6)^{2} = 3 π (36) = 108 π \approx 339 mm^{2}

(b)

2 marks

One gram covers 240 mm².

\frac{339.29}{240} \approx 1.41 g

Question 4

7 marks

Two petrol stations: a linear model, its inverse, and where one becomes cheaper

Functions

(a)

2 marks

Put $x = 40$ into $L (x) = 1.50 x - 5$ .

L (40) = 60 - 5 = $55

(b)

2 marks

Solve $1.50 x - 5 = 70$ .

x = \frac{75}{1.5} = 50 litres

(c)

3 marks

Erica's cost is $1.30 x$ . She is cheaper than Leon when $1.30 x < 1.50 x - 5$ .

5 < 0.2 x \Rightarrow x > 25 k = 25

Question 5

6 marks

Voronoi towers: nearest-tower regions and the edge between two towers

Geometry and trigonometry

(a)

1 mark

Every tower is identical, so the signal is strongest from the nearest one. Every point in the shaded region is closer to T4 than to any other tower.

(b)

2 marks

The edge $y = 3$ is the perpendicular bisector of T2 and T4, so T4 is the reflection of T2 in that line. T2 is 2 units above the line, so T4 is 2 units below it.

T_{4} = (- 9, 1)

(c)

3 marks

First find the gradient of the line joining T1 and T2.
The edge between them is perpendicular to this line, so its gradient is the negative reciprocal.

m = \frac{5 - 3}{- 9 - ( - 13 )} = \frac{1}{2} - 2

Question 6

5 marks

Are black goose eggs lighter than white goose eggs? A $t$ -test

Statistics and probability

(a)

1 mark

The null hypothesis says there is no difference between the population mean weights.

H_{0} : the mean weight of eggs from black geese equals the mean weight of eggs from white geese

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances, and the alternative that the black mean is lower.

p \approx 0.177

(c)

2 marks

Compare $p$ with the 10% level.
So there is insufficient evidence to reject $H_{0}$ . The result does not support Arriane's claim that black geese eggs are lighter.

0.177 > 0.10

Question 7

6 marks

Forgetting modelled by an exponential decay

Calculus

(a)

2 marks

After 1 day 50% remains, so $R (1) = 50$ .

100 e^{- p} = 50 \Rightarrow e^{- p} = \frac{1}{2} \Rightarrow p = ln 2 \approx 0.693

(b)

2 marks

36 hours is 1.5 days, since $t$ is in days.

R (1.5) = 100 e^{- 1.5 l n 2} \approx 35.4%

(c)

1 mark

An exponential function is always positive, so $R (t) > 0$ for every $t$ . The graph approaches the axis $R = 0$ but never reaches it.

(d)

1 mark

One valid limit: the model gives $R (0) = 100%$ , which is not realistic immediately at the start, or large values of $t$ are unrealistic because the model keeps predicting some retention forever.

Mentor tip

The limitation has to be about the domain $t \geq 0$ , not about people in general.

Question 8

8 marks

Two training plans: arithmetic against geometric growth

Number and algebra

(a)

5 marks

Charlie's distances are arithmetic, with $u_{1} = 500$ and $d = 100$ .
Daniella's are geometric with first term 500 and common ratio 1.02.

u_{20} = 500 + 19 (100) = 2400 m u_{20} = 500 (1.02)^{19} \approx 728 m

(b)

3 marks

Daniella is ahead on day $n$ when her distance exceeds Charlie's.
Solve on the calculator, by graphing both sides or tabulating values.

500 (1.02)^{n - 1} > 500 + 100 (n - 1) n > 184.2 \Rightarrow n = 185

Mentor tip

The first whole day above the crossing point is 185, not 184.

Question 9

5 marks

The largest area when every measurement is rounded

Number and algebra

5 marks

The sides can be at most 56.5 m and 82.5 m. The angle can be anywhere from $102. 5^{\circ}$ to $107. 5^{\circ}$ . Over that range $sin$ is decreasing (since the angle is past $9 0^{\circ}$ ), so the sine is largest at $102. 5^{\circ}$ .
To three significant figures that is about 2280 m².

Area = \frac{1}{2} (56.5) (82.5) sin 102. 5^{\circ} \approx 2275 m^{2}

Mentor tip

Check by trying both ends: $sin 107. 5^{\circ} \approx 0.954$ is smaller than $sin 102. 5^{\circ} \approx 0.976$ .

Question 10

7 marks

The larger of two dice: the distribution, a conditional and the mean

Statistics and probability

(a)

2 marks

The score is $t$ when at least one die shows $t$ and neither shows more. That happens in $2 t - 1$ of the 36 outcomes.

P (T = t) = \frac{2 t - 1}{36} for t = 1, \dots, 6 : \frac{1}{36}, \frac{3}{36}, \frac{5}{36}, \frac{7}{36}, \frac{9}{36}, \frac{11}{36}

(b)

3 marks

(i) "At least 3" has probability $\frac{5 + 7 + 9 + 11}{36}$ .

(ii) A conditional probability: divide by the probability of the condition.

P (T \geq 3) = \frac{32}{36} = \frac{8}{9} P (T = 6 ∣ T \geq 3) = \frac{11/36}{32/36} = \frac{11}{32} \approx 0.344

(c)

2 marks

Multiply each score by its probability and add.

E (T) = \frac{1 ( 1 ) + 2 ( 3 ) + 3 ( 5 ) + 4 ( 7 ) + 5 ( 9 ) + 6 ( 11 )}{36} = \frac{161}{36} \approx 4.47

Question 11

6 marks

A siren's sound intensity: inverse variation and the hearing limit

Functions

(a)

2 marks

Inverse variation with the square of the distance means $I = \frac{k}{d ^{2}}$ . Use $d = 1.5$ and $I = 4$ .

4 = \frac{k}{1. 5 ^{2}} \Rightarrow k = 9 \Rightarrow I = \frac{9}{d ^{2}}

(b)

2 marks

The curve is in the first quadrant only. It falls steeply from the top left (large $I$ for small $d$ ) through the point $(1.5, 4)$ , and then flattens out towards the $d$ -axis, approaching $I = 0$ without touching it. It has no $I$ -intercept.

(c)

2 marks

Scarlett can't hear the siren when the intensity is at most $1.5 \times 1 0^{- 6}$ .

\frac{9}{d ^{2}} \leq 1.5 \times 1 0^{- 6} \Rightarrow d^{2} \geq 6 \times 1 0^{6} d \geq 2450 m (to 3 sf)

Mentor tip

The exact boundary is $d = 6 \times 1 0^{6} \approx 2449.5$ m.

Question 12

8 marks

A gift box: area in terms of one variable, then minimised

Calculus

(a)

4 marks

The top is a rectangle of area $8 \times 6 = 48$ plus two right-angled triangles with areas $\frac{1}{2} \cdot 6 \cdot q = 3 q$ and $\frac{1}{2} \cdot 8 \cdot p = 4 p$ .
The two triangles are similar, so their sides are in proportion.
Replace $p$ in the area.

A = 3 q + 4 p + 48 \frac{p}{6} = \frac{8}{q} \Rightarrow pq = 48 \Rightarrow p = \frac{48}{q} A = 3 q + \frac{192}{q} + 48

(b)

2 marks

Differentiate, writing $\frac{192}{q} = 192 q^{- 1}$ .

\frac{d A}{d q} = 3 - \frac{192}{q ^{2}}

(c)

2 marks

For the minimum, set the derivative to zero.

3 - \frac{192}{q ^{2}} = 0 \Rightarrow q^{2} = 64 \Rightarrow q = 8 cm

Mentor tip

$q > 0$ is a length, so reject $- 8$ . The second derivative $\frac{384}{q ^{3}}$ is positive, so this is a minimum.

Question 13

8 marks

A golf ball's distance from a rate of change

Calculus

(a)

3 marks

Evaluate the rate of change at $θ = 5 0^{\circ}$ .
This is negative, so the graph of $l$ against $θ$ is decreasing at $θ = 5 0^{\circ}$ .

\frac{d l}{d θ} = - 0.2 (50) + 9 = - 1

(b)

5 marks

Integrate $- 0.2 θ + 9$ term by term.
Use $l (40) = 205.5$ to find $c$ .

l (θ) = - 0.1 θ^{2} + 9 θ + c 205.5 = - 0.1 (1600) + 9 (40) + c = 200 + c \Rightarrow c = 5.5 l (θ) = - 0.1 θ^{2} + 9 θ + 5.5

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q7, Q12, Q13
Number and algebra17 marks · 21%
Q1, Q8, Q9
Statistics and probability16 marks · 20%
Q2, Q6, Q10
Functions13 marks · 16%
Q4, Q11
Geometry and trigonometry12 marks · 15%
Q3, Q5

Mentor insights

Traps to watch for

01
Inverse means solve for $x$ · Q4
$L^{- 1} (70)$ asks how many litres cost $70. Set $L (x) = 70$ and solve.
02
Day 20 is 19 steps from day 1 · Q8
Day 20 means 19 increases from day 1, not 20. The $n^{th}$ term uses the exponent $n - 1$ .
03
Count the outcomes · Q10
For a score of $t$ , count how many of the 36 outcomes have a larger number equal to $t$ . That count is $2 t - 1$ .
04
Link $p$ and $q$ first · Q12
The area has two variables. A similar-triangles equation, $pq = 48$ , lets you write it in one variable before you differentiate.
05
Find the constant · Q13
After integrating $\frac{d l}{d θ}$ , use the known point $(40, 205.5)$ to find the constant of integration.

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 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

4 marks

A continued-fraction approximation of $π$

Number and algebra

(a)

2 marks

Work out the denominator first: $6 + \frac{13}{16} = \frac{109}{16}$ . The expression becomes $3 + \frac{16}{109}$ .

3 + \frac{16}{109} \approx 3.14679 \approx 3.1468

(b)

2 marks

Percentage error is the difference divided by the exact value, times 100.

\frac{∣3.1468 - π ∣}{π} \times 100 \approx 0.166%

Question 2

4 marks

Ten days of temperatures: mode, mean and standard deviation

Statistics and probability

(a)

1 mark

The mode is the most common value; 14 appears three times.

14

(b)

2 marks

Add the ten values and divide by ten.

\frac{14 + 15 + 14 + 11 + 10 + 9 + 14 + 15 + 16 + 13}{10} = 13.1

(c)

1 mark

Enter the data in the calculator's one-variable statistics.

σ \approx 2.21

Question 3

6 marks

Surface area of a hemisphere and the chocolate needed

Geometry and trigonometry

(a)

4 marks

The curved surface is half of a sphere's surface, and there is also the flat circle.

2 π (6)^{2} + π (6)^{2} = 3 π (36) = 108 π \approx 339 mm^{2}

(b)

2 marks

One gram covers 240 mm².

\frac{339.29}{240} \approx 1.41 g

Question 4

7 marks

Two petrol stations: a linear model, its inverse, and where one becomes cheaper

Functions

(a)

2 marks

Put $x = 40$ into $L (x) = 1.50 x - 5$ .

L (40) = 60 - 5 = $55

(b)

2 marks

Solve $1.50 x - 5 = 70$ .

x = \frac{75}{1.5} = 50 litres

(c)

3 marks

Erica's cost is $1.30 x$ . She is cheaper than Leon when $1.30 x < 1.50 x - 5$ .

5 < 0.2 x \Rightarrow x > 25 k = 25

Question 5

6 marks

Voronoi towers: nearest-tower regions and the edge between two towers

Geometry and trigonometry

(a)

1 mark

Every tower is identical, so the signal is strongest from the nearest one. Every point in the shaded region is closer to T4 than to any other tower.

(b)

2 marks

The edge $y = 3$ is the perpendicular bisector of T2 and T4, so T4 is the reflection of T2 in that line. T2 is 2 units above the line, so T4 is 2 units below it.

T_{4} = (- 9, 1)

(c)

3 marks

First find the gradient of the line joining T1 and T2.
The edge between them is perpendicular to this line, so its gradient is the negative reciprocal.

m = \frac{5 - 3}{- 9 - ( - 13 )} = \frac{1}{2} - 2

Question 6

5 marks

Are black goose eggs lighter than white goose eggs? A $t$ -test

Statistics and probability

(a)

1 mark

The null hypothesis says there is no difference between the population mean weights.

H_{0} : the mean weight of eggs from black geese equals the mean weight of eggs from white geese

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances, and the alternative that the black mean is lower.

p \approx 0.177

(c)

2 marks

Compare $p$ with the 10% level.
So there is insufficient evidence to reject $H_{0}$ . The result does not support Arriane's claim that black geese eggs are lighter.

0.177 > 0.10

Question 7

6 marks

Forgetting modelled by an exponential decay

Calculus

(a)

2 marks

After 1 day 50% remains, so $R (1) = 50$ .

100 e^{- p} = 50 \Rightarrow e^{- p} = \frac{1}{2} \Rightarrow p = ln 2 \approx 0.693

(b)

2 marks

36 hours is 1.5 days, since $t$ is in days.

R (1.5) = 100 e^{- 1.5 l n 2} \approx 35.4%

(c)

1 mark

An exponential function is always positive, so $R (t) > 0$ for every $t$ . The graph approaches the axis $R = 0$ but never reaches it.

(d)

1 mark

One valid limit: the model gives $R (0) = 100%$ , which is not realistic immediately at the start, or large values of $t$ are unrealistic because the model keeps predicting some retention forever.

Mentor tip

The limitation has to be about the domain $t \geq 0$ , not about people in general.

Question 8

8 marks

Two training plans: arithmetic against geometric growth

Number and algebra

(a)

5 marks

Charlie's distances are arithmetic, with $u_{1} = 500$ and $d = 100$ .
Daniella's are geometric with first term 500 and common ratio 1.02.

u_{20} = 500 + 19 (100) = 2400 m u_{20} = 500 (1.02)^{19} \approx 728 m

(b)

3 marks

Daniella is ahead on day $n$ when her distance exceeds Charlie's.
Solve on the calculator, by graphing both sides or tabulating values.

500 (1.02)^{n - 1} > 500 + 100 (n - 1) n > 184.2 \Rightarrow n = 185

Mentor tip

The first whole day above the crossing point is 185, not 184.

Question 9

5 marks

The largest area when every measurement is rounded

Number and algebra

5 marks

The sides can be at most 56.5 m and 82.5 m. The angle can be anywhere from $102. 5^{\circ}$ to $107. 5^{\circ}$ . Over that range $sin$ is decreasing (since the angle is past $9 0^{\circ}$ ), so the sine is largest at $102. 5^{\circ}$ .
To three significant figures that is about 2280 m².

Area = \frac{1}{2} (56.5) (82.5) sin 102. 5^{\circ} \approx 2275 m^{2}

Mentor tip

Check by trying both ends: $sin 107. 5^{\circ} \approx 0.954$ is smaller than $sin 102. 5^{\circ} \approx 0.976$ .

Question 10

7 marks

The larger of two dice: the distribution, a conditional and the mean

Statistics and probability

(a)

2 marks

The score is $t$ when at least one die shows $t$ and neither shows more. That happens in $2 t - 1$ of the 36 outcomes.

P (T = t) = \frac{2 t - 1}{36} for t = 1, \dots, 6 : \frac{1}{36}, \frac{3}{36}, \frac{5}{36}, \frac{7}{36}, \frac{9}{36}, \frac{11}{36}

(b)

3 marks

(i) "At least 3" has probability $\frac{5 + 7 + 9 + 11}{36}$ .

(ii) A conditional probability: divide by the probability of the condition.

P (T \geq 3) = \frac{32}{36} = \frac{8}{9} P (T = 6 ∣ T \geq 3) = \frac{11/36}{32/36} = \frac{11}{32} \approx 0.344

(c)

2 marks

Multiply each score by its probability and add.

E (T) = \frac{1 ( 1 ) + 2 ( 3 ) + 3 ( 5 ) + 4 ( 7 ) + 5 ( 9 ) + 6 ( 11 )}{36} = \frac{161}{36} \approx 4.47

Question 11

6 marks

A siren's sound intensity: inverse variation and the hearing limit

Functions

(a)

2 marks

Inverse variation with the square of the distance means $I = \frac{k}{d ^{2}}$ . Use $d = 1.5$ and $I = 4$ .

4 = \frac{k}{1. 5 ^{2}} \Rightarrow k = 9 \Rightarrow I = \frac{9}{d ^{2}}

(b)

2 marks

(c)

2 marks

Scarlett can't hear the siren when the intensity is at most $1.5 \times 1 0^{- 6}$ .

\frac{9}{d ^{2}} \leq 1.5 \times 1 0^{- 6} \Rightarrow d^{2} \geq 6 \times 1 0^{6} d \geq 2450 m (to 3 sf)

Mentor tip

The exact boundary is $d = 6 \times 1 0^{6} \approx 2449.5$ m.

Question 12

8 marks

A gift box: area in terms of one variable, then minimised

Calculus

(a)

4 marks

The top is a rectangle of area $8 \times 6 = 48$ plus two right-angled triangles with areas $\frac{1}{2} \cdot 6 \cdot q = 3 q$ and $\frac{1}{2} \cdot 8 \cdot p = 4 p$ .
The two triangles are similar, so their sides are in proportion.
Replace $p$ in the area.

A = 3 q + 4 p + 48 \frac{p}{6} = \frac{8}{q} \Rightarrow pq = 48 \Rightarrow p = \frac{48}{q} A = 3 q + \frac{192}{q} + 48

(b)

2 marks

Differentiate, writing $\frac{192}{q} = 192 q^{- 1}$ .

\frac{d A}{d q} = 3 - \frac{192}{q ^{2}}

(c)

2 marks

For the minimum, set the derivative to zero.

3 - \frac{192}{q ^{2}} = 0 \Rightarrow q^{2} = 64 \Rightarrow q = 8 cm

Mentor tip

$q > 0$ is a length, so reject $- 8$ . The second derivative $\frac{384}{q ^{3}}$ is positive, so this is a minimum.

Question 13

8 marks

A golf ball's distance from a rate of change

Calculus

(a)

3 marks

Evaluate the rate of change at $θ = 5 0^{\circ}$ .
This is negative, so the graph of $l$ against $θ$ is decreasing at $θ = 5 0^{\circ}$ .

\frac{d l}{d θ} = - 0.2 (50) + 9 = - 1

(b)

5 marks

Integrate $- 0.2 θ + 9$ term by term.
Use $l (40) = 205.5$ to find $c$ .

l (θ) = - 0.1 θ^{2} + 9 θ + c 205.5 = - 0.1 (1600) + 9 (40) + c = 200 + c \Rightarrow c = 5.5 l (θ) = - 0.1 θ^{2} + 9 θ + 5.5

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q7, Q12, Q13
Number and algebra17 marks · 21%
Q1, Q8, Q9
Statistics and probability16 marks · 20%
Q2, Q6, Q10
Functions13 marks · 16%
Q4, Q11
Geometry and trigonometry12 marks · 15%
Q3, Q5

Mentor insights

Traps to watch for

01
Inverse means solve for $x$ · Q4
$L^{- 1} (70)$ asks how many litres cost $70. Set $L (x) = 70$ and solve.
02
Day 20 is 19 steps from day 1 · Q8
Day 20 means 19 increases from day 1, not 20. The $n^{th}$ term uses the exponent $n - 1$ .
03
Count the outcomes · Q10
For a score of $t$ , count how many of the 36 outcomes have a larger number equal to $t$ . That count is $2 t - 1$ .
04
Link $p$ and $q$ first · Q12
The area has two variables. A similar-triangles equation, $pq = 48$ , lets you write it in one variable before you differentiate.
05
Find the constant · Q13
After integrating $\frac{d l}{d θ}$ , use the known point $(40, 205.5)$ to find the constant of integration.

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