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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: November 2024
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 2024 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

The steepest train track: labelling a right triangle and the height gained

Geometry and trigonometry

(a)

2 marks

Draw a right triangle with the track as the hypotenuse. Label the hypotenuse $1.52$ km, and label the angle at the bottom of the track $36. 1^{\circ}$ .

(b)

2 marks

The vertical height is opposite the angle of elevation.

height = 1.52 sin 36. 1^{\circ} \approx 0.896 km

Question 2

7 marks

Test scores from a frequency table: mean, standard deviation and changes of scale

Statistics and probability

(a)

3 marks

Enter the scores and the frequencies (total 240) in the calculator.

\overset{x}{ˉ} \approx 4.442, σ \approx 1.657

(b)

2 marks

Multiply both by 10.

\overset{x}{ˉ} \approx 44.4, σ \approx 16.6

(c)

2 marks

Adding 30 shifts the mean by 30 and leaves the spread alone.

\overset{x}{ˉ} \approx 74.4, σ \approx 16.6

Question 3

5 marks

Museum ticket prices from two group bookings, and a rounded count

Number and algebra

(a)

2 marks

A school group of 7 adults and 60 children paid $$832$ , and a family of 3 adults and 5 children paid $$108$ .

7 a + 60 c = 832, 3 a + 5 c = 108

(b)

2 marks

Solve the two equations on the calculator (or eliminate $c$ by multiplying the second by 12).
(i) An adult ticket costs $$16$ .
(ii) Substitute into $3 a + 5 c = 108$ .

36 a + 60 c = 1296 \Rightarrow 29 a = 464 \Rightarrow a = 16 48 + 5 c = 108 \Rightarrow c = 12 ($12 for a child)

(c)

1 mark

Rounding to the nearest thousand, the true number is at least $203000 - 500$ .

202500

Question 4

5 marks

A triangle with sides 39, 40 and 46: an angle and the area

Geometry and trigonometry

(a)

3 marks

The angle $T R S$ is opposite the side $S T = 40$ .

cos R = \frac{4 6 ^{2} + 3 9 ^{2} - 4 0 ^{2}}{2 ( 46 ) ( 39 )} = \frac{2037}{3588} \approx 0.5677 T R S \approx 55. 4^{\circ}

(b)

2 marks

Use the two sides next to the angle.

Area = \frac{1}{2} (46) (39) sin 55.4 1^{\circ} \approx 738 cm^{2}

Question 5

7 marks

Bicycle hire costs: meaning of a linear function and its inverse

Functions

(a)

2 marks

The number 60 is the cost per day of hiring the bicycle (CAD 60 each day).
The number 10 is the fixed charge for the helmet and repair kit (CAD 10).

(b)

2 marks

Substitute $d = 5$ .

C (5) = 60 (5) + 10 = 310 CAD

(c)

1 mark

The function is defined for $d \geq 3$ .

3 days

(d)

2 marks

$C^{- 1} (1270) = k$ means $C (k) = 1270$ .

60 k + 10 = 1270 \Rightarrow k = 21

Question 6

6 marks

Decay of radioactive carbon in a fossil

Functions

(a)

1 mark

Put $t = 0$ .

m (0) = 120 ng

(b)

2 marks

Put $t = 20000$ .

m (20000) = 120 e^{- 0.000121 \times 20000} \approx 10.7 ng

(c)

3 marks

More than half has decayed when $m (t) < 60$ .
So the smallest whole number of years is

120 e^{- 0.000121 t} < 60 \Rightarrow t > \frac{ln 2}{0.000121} \approx 5728.5 5729 years

Question 7

7 marks

Masses of Scottish Terriers: normal probabilities

Statistics and probability

(a)

2 marks

Use the calculator.

P (W > 7.2) \approx 0.922

(b)

3 marks

(i) $P (W < a) = 0.1$ , so use the inverse normal with area $0.1$ .

(ii) $P (W > b) = 0.1$ , so use the inverse normal with area $0.9$ to the left.

a \approx 7.36 b \approx 10.4

(c)

2 marks

First find $P (W < 7.2) = 1 - 0.92171 = 0.07829$ . The two dogs are independent.

P (both < 7.2) = (0.07829)^{2} \approx 0.00613

Question 8

5 marks

Simple interest of 4% on 18 000 dollars as an arithmetic sequence

Number and algebra

(a)

2 marks

The yearly interest is 4% of the original $$18 000$ .

d = 0.04 \times 18000 = $720

(b)

3 marks

The balance after $k$ complete years.

18000 + 720 k > 32000 \Rightarrow k > \frac{14 000}{720} \approx 19.44 k = 20

Question 9

7 marks

Driving test results: a binomial model

Statistics and probability

(a)

2 marks

(i) The mean.

(ii) The variance.

30 \times 0.6 = 18 30 \times 0.6 \times 0.4 = 7.2

(b)

4 marks

(i) Use the binomial probability.

(ii) Fewer than 12 means 11 or fewer.

P (X = 21) = (21 30) (0.6)^{21} (0.4)^{9} \approx 0.0823 P (X \leq 11) \approx 0.00830

(c)

1 mark

One assumption: the students' results are independent of each other (or the probability of passing is the same, 0.6, for every student).

Question 10

6 marks

An annuity fund of 300 000 dollars with monthly withdrawals

Number and algebra

(a)

3 marks

Enter $N = 96$ (8 years of monthly withdrawals), $I = 3.8%$ , $P V = 300000$ , $P M T = - 2800$ , $P / Y = C / Y = 12$ , and find $F V$ .

F V \approx $92 828

(b)

3 marks

The fund runs out when $F V = 0$ . Solve for $N$ with the same inputs.
So Humberto can make 131 withdrawals.

N = 131

Question 11

7 marks

A fair game with wins, draws and losses

Statistics and probability

(a)

2 marks

Add the probabilities.

0.35 + 0.4 + p = 1 \Rightarrow p = 0.25

(b)

2 marks

In 60 games, the expected number of losses is $60 p$ .

60 \times 0.25 = 15

(c)

3 marks

Set the expected score to zero.
Solve.

E (X) = 0.35 w + 0.4 (0) - 7 (0.25) = 0 0.35 w = 1.75 \Rightarrow w = 5

Question 12

7 marks

A curve with a reciprocal power: gradient and normal

Calculus

(a)

3 marks

Write $y = 5 x^{2} - 3 x^{- 3}$ and differentiate.

\frac{d y}{d x} = 10 x + 9 x^{- 4} = 10 x + \frac{9}{x ^{4}}

(b)

1 mark

Substitute $x = 1$ .

\frac{d y}{d x} = 10 + 9 = 19

(c)

3 marks

The point on the curve at $x = 1$ .
The gradient of the normal is $- \frac{1}{19}$ .

y = 5 - 3 = 2 \Rightarrow (1, 2) y - 2 = - \frac{1}{19} (x - 1) \Rightarrow y = - \frac{x}{19} + \frac{39}{19}

Question 13

7 marks

Bacteria that double every ten minutes: a geometric sequence

Number and algebra

(a)

3 marks

The sequence is geometric with $u_{1} = 1250$ and $r = 2$ .

u_{9} = 1250 \times 2^{8} = 320000 u_{k} = 1250 \times 2^{k - 1}

(b)

4 marks

(i) Solve $1250 \times 2^{n - 1} = 1.28 \times 1 0^{6}$ .

(ii) Every new bacterium is made by a division, so the number of divisions is the increase in the number of bacteria.

(Check with the sum: $1250 + 2500 + \dots + 1250 \times 2^{9} = 1250 (2^{10} - 1) = 1278750$ .)

2^{n - 1} = 1024 = 2^{10} \Rightarrow n - 1 = 10 \Rightarrow n = 11 1.28 \times 1 0^{6} - 1250 = 1278750 \approx 1279000

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 probability28 marks · 35%
Q2, Q7, Q9, Q11
Number and algebra23 marks · 29%
Q3, Q8, Q10, Q13
Functions13 marks · 16%
Q5, Q6
Geometry and trigonometry9 marks · 11%
Q1, Q4
Calculus7 marks · 9%
Q12

Mentor insights

Traps to watch for

01
Scaling · Q2
Multiplying all values by $k$ multiplies the mean and standard deviation by $k$ .
02
Gradient and intercept · Q5
In $C = 60 d + 10$ , 60 is the cost per day and 10 is the fixed charge.
03
Normal · Q7
$W \sim N (8.9, 1. 2^{2})$ .
04
Binomial · Q9
$X \sim B (30, 0.6)$ .
05
Total · Q11
The probabilities add to 1.

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

Timezone

TZ1TZ2

Session: November 2024
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 2024 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

The steepest train track: labelling a right triangle and the height gained

Geometry and trigonometry

(a)

2 marks

Draw a right triangle with the track as the hypotenuse. Label the hypotenuse $1.52$ km, and label the angle at the bottom of the track $36. 1^{\circ}$ .

(b)

2 marks

The vertical height is opposite the angle of elevation.

height = 1.52 sin 36. 1^{\circ} \approx 0.896 km

Question 2

7 marks

Test scores from a frequency table: mean, standard deviation and changes of scale

Statistics and probability

(a)

3 marks

Enter the scores and the frequencies (total 240) in the calculator.

\overset{x}{ˉ} \approx 4.442, σ \approx 1.657

(b)

2 marks

Multiply both by 10.

\overset{x}{ˉ} \approx 44.4, σ \approx 16.6

(c)

2 marks

Adding 30 shifts the mean by 30 and leaves the spread alone.

\overset{x}{ˉ} \approx 74.4, σ \approx 16.6

Question 3

5 marks

Museum ticket prices from two group bookings, and a rounded count

Number and algebra

(a)

2 marks

A school group of 7 adults and 60 children paid $$832$ , and a family of 3 adults and 5 children paid $$108$ .

7 a + 60 c = 832, 3 a + 5 c = 108

(b)

2 marks

Solve the two equations on the calculator (or eliminate $c$ by multiplying the second by 12).
(i) An adult ticket costs $$16$ .
(ii) Substitute into $3 a + 5 c = 108$ .

36 a + 60 c = 1296 \Rightarrow 29 a = 464 \Rightarrow a = 16 48 + 5 c = 108 \Rightarrow c = 12 ($12 for a child)

(c)

1 mark

Rounding to the nearest thousand, the true number is at least $203000 - 500$ .

202500

Question 4

5 marks

A triangle with sides 39, 40 and 46: an angle and the area

Geometry and trigonometry

(a)

3 marks

The angle $T R S$ is opposite the side $S T = 40$ .

cos R = \frac{4 6 ^{2} + 3 9 ^{2} - 4 0 ^{2}}{2 ( 46 ) ( 39 )} = \frac{2037}{3588} \approx 0.5677 T R S \approx 55. 4^{\circ}

(b)

2 marks

Use the two sides next to the angle.

Area = \frac{1}{2} (46) (39) sin 55.4 1^{\circ} \approx 738 cm^{2}

Question 5

7 marks

Bicycle hire costs: meaning of a linear function and its inverse

Functions

(a)

2 marks

The number 60 is the cost per day of hiring the bicycle (CAD 60 each day).
The number 10 is the fixed charge for the helmet and repair kit (CAD 10).

(b)

2 marks

Substitute $d = 5$ .

C (5) = 60 (5) + 10 = 310 CAD

(c)

1 mark

The function is defined for $d \geq 3$ .

3 days

(d)

2 marks

$C^{- 1} (1270) = k$ means $C (k) = 1270$ .

60 k + 10 = 1270 \Rightarrow k = 21

Question 6

6 marks

Decay of radioactive carbon in a fossil

Functions

(a)

1 mark

Put $t = 0$ .

m (0) = 120 ng

(b)

2 marks

Put $t = 20000$ .

m (20000) = 120 e^{- 0.000121 \times 20000} \approx 10.7 ng

(c)

3 marks

More than half has decayed when $m (t) < 60$ .
So the smallest whole number of years is

120 e^{- 0.000121 t} < 60 \Rightarrow t > \frac{ln 2}{0.000121} \approx 5728.5 5729 years

Question 7

7 marks

Masses of Scottish Terriers: normal probabilities

Statistics and probability

(a)

2 marks

Use the calculator.

P (W > 7.2) \approx 0.922

(b)

3 marks

(i) $P (W < a) = 0.1$ , so use the inverse normal with area $0.1$ .

(ii) $P (W > b) = 0.1$ , so use the inverse normal with area $0.9$ to the left.

a \approx 7.36 b \approx 10.4

(c)

2 marks

First find $P (W < 7.2) = 1 - 0.92171 = 0.07829$ . The two dogs are independent.

P (both < 7.2) = (0.07829)^{2} \approx 0.00613

Question 8

5 marks

Simple interest of 4% on 18 000 dollars as an arithmetic sequence

Number and algebra

(a)

2 marks

The yearly interest is 4% of the original $$18 000$ .

d = 0.04 \times 18000 = $720

(b)

3 marks

The balance after $k$ complete years.

18000 + 720 k > 32000 \Rightarrow k > \frac{14 000}{720} \approx 19.44 k = 20

Question 9

7 marks

Driving test results: a binomial model

Statistics and probability

(a)

2 marks

(i) The mean.

(ii) The variance.

30 \times 0.6 = 18 30 \times 0.6 \times 0.4 = 7.2

(b)

4 marks

(i) Use the binomial probability.

(ii) Fewer than 12 means 11 or fewer.

P (X = 21) = (21 30) (0.6)^{21} (0.4)^{9} \approx 0.0823 P (X \leq 11) \approx 0.00830

(c)

1 mark

One assumption: the students' results are independent of each other (or the probability of passing is the same, 0.6, for every student).

Question 10

6 marks

An annuity fund of 300 000 dollars with monthly withdrawals

Number and algebra

(a)

3 marks

Enter $N = 96$ (8 years of monthly withdrawals), $I = 3.8%$ , $P V = 300000$ , $P M T = - 2800$ , $P / Y = C / Y = 12$ , and find $F V$ .

F V \approx $92 828

(b)

3 marks

The fund runs out when $F V = 0$ . Solve for $N$ with the same inputs.
So Humberto can make 131 withdrawals.

N = 131

Question 11

7 marks

A fair game with wins, draws and losses

Statistics and probability

(a)

2 marks

Add the probabilities.

0.35 + 0.4 + p = 1 \Rightarrow p = 0.25

(b)

2 marks

In 60 games, the expected number of losses is $60 p$ .

60 \times 0.25 = 15

(c)

3 marks

Set the expected score to zero.
Solve.

E (X) = 0.35 w + 0.4 (0) - 7 (0.25) = 0 0.35 w = 1.75 \Rightarrow w = 5

Question 12

7 marks

A curve with a reciprocal power: gradient and normal

Calculus

(a)

3 marks

Write $y = 5 x^{2} - 3 x^{- 3}$ and differentiate.

\frac{d y}{d x} = 10 x + 9 x^{- 4} = 10 x + \frac{9}{x ^{4}}

(b)

1 mark

Substitute $x = 1$ .

\frac{d y}{d x} = 10 + 9 = 19

(c)

3 marks

The point on the curve at $x = 1$ .
The gradient of the normal is $- \frac{1}{19}$ .

y = 5 - 3 = 2 \Rightarrow (1, 2) y - 2 = - \frac{1}{19} (x - 1) \Rightarrow y = - \frac{x}{19} + \frac{39}{19}

Question 13

7 marks

Bacteria that double every ten minutes: a geometric sequence

Number and algebra

(a)

3 marks

The sequence is geometric with $u_{1} = 1250$ and $r = 2$ .

u_{9} = 1250 \times 2^{8} = 320000 u_{k} = 1250 \times 2^{k - 1}

(b)

4 marks

(i) Solve $1250 \times 2^{n - 1} = 1.28 \times 1 0^{6}$ .

(ii) Every new bacterium is made by a division, so the number of divisions is the increase in the number of bacteria.

(Check with the sum: $1250 + 2500 + \dots + 1250 \times 2^{9} = 1250 (2^{10} - 1) = 1278750$ .)

2^{n - 1} = 1024 = 2^{10} \Rightarrow n - 1 = 10 \Rightarrow n = 11 1.28 \times 1 0^{6} - 1250 = 1278750 \approx 1279000

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 probability28 marks · 35%
Q2, Q7, Q9, Q11
Number and algebra23 marks · 29%
Q3, Q8, Q10, Q13
Functions13 marks · 16%
Q5, Q6
Geometry and trigonometry9 marks · 11%
Q1, Q4
Calculus7 marks · 9%
Q12

Mentor insights

Traps to watch for

01
Scaling · Q2
Multiplying all values by $k$ multiplies the mean and standard deviation by $k$ .
02
Gradient and intercept · Q5
In $C = 60 d + 10$ , 60 is the cost per day and 10 is the fixed charge.
03
Normal · Q7
$W \sim N (8.9, 1. 2^{2})$ .
04
Binomial · Q9
$X \sim B (30, 0.6)$ .
05
Total · Q11
The probabilities add to 1.

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