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IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Prices of garden chairs: range, mean, standard deviation and a sale

Statistics and probability

(a)

2 marks

The largest price is $255 and the smallest is $49.

255 - 49 = $206

(b)

3 marks

Enter the ten prices into the calculator.
(i) The mean.
(ii) The standard deviation (of the whole set of ten prices).

\overset{x}{ˉ} \approx $137 σ \approx $74.5

(c)

2 marks

(i) The mean falls by $20.

(ii) The spread does not change.

137.1 - 20 \approx $117 σ \approx $74.5

Question 2

6 marks

A toy crane: the cosine rule and the sine rule

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $A B C$ with $A B = 25$ , $A C = 15$ and $B A C = 2 2^{\circ}$ .

B C^{2} = 1 5^{2} + 2 5^{2} - 2 (15) (25) cos 2 2^{\circ} \Rightarrow B C \approx 12.4 cm

(b)

3 marks

Use the sine rule.
The angle is acute, so the sine rule gives the angle directly.

\frac{sin A B C}{15} = \frac{sin 2 2 ^{\circ}}{12.43} sin A B C = 0.4520 \Rightarrow A B C \approx 26. 9^{\circ}

Question 3

5 marks

Pepper sizes and colours: a chi-squared test for independence

Statistics and probability

(a)

1 mark

$H_{0}$ : the size of a pepper is independent of its colour.
$H_{1}$ : the size of a pepper is not independent of its colour.

(b)

4 marks

(i) Enter the $3 \times 3$ table into the calculator and run a $χ^{2}$ test. The degrees of freedom are $(3 - 1) (3 - 1) = 4$ , and every expected count is at least 20.

(ii) Since $22.5 > 9.49$ (equivalently, the p-value $0.000156 < 0.05$ ), reject $H_{0}$ . There is sufficient evidence that the size of the peppers is not independent of their colour.

χ_{c a l c}^{2} \approx 22.5

Question 4

6 marks

Bricklayers: direct and inverse variation

Functions

(a)

3 marks

The area is proportional to the number of bricklayers, $n = k B$ .

60 = 5 k \Rightarrow k = 12 n = 12 \times 7 = 84 m^{2}

(b)

3 marks

The time is inversely proportional to the number of bricklayers, $t = \frac{c}{B}$ .

8 = \frac{c}{5} \Rightarrow c = 40 t = \frac{40}{12} = \frac{10}{3} \approx 3.33 hours (3 hours 20 minutes)

Question 5

9 marks

Savings, then a car loan: compound interest and repayments

Number and algebra

(a)

3 marks

Six years is 72 months.

3000 (1 + \frac{0.0125}{12})^{72} \approx $3233.53

(b)

2 marks

Solve $3000 (1 + \frac{0.0125}{12})^{N} = 3550$ for $N$ .

N \approx 161.7 \Rightarrow 162 months

(c)

1 mark

The car costs $22 000 and the partial payment is $3550.

22000 - 3550 = $18 450

(d)

3 marks

Use the finance solver with $N = 96$ , $I = 12.6%$ , $P V = 18450$ , $F V = 0$ , 12 payments a year.

P M T \approx $306

Question 6

6 marks

Flawed chocolates in a batch: binomial probabilities and expected income

Statistics and probability

(a)

4 marks

(i) Let $X \sim B (20, 0.04)$ .

(ii) More than two flawed.

P (X = 2) = (2 20) (0.04)^{2} (0.96)^{18} \approx 0.146 P (X > 2) = 1 - P (X \leq 2) \approx 0.0439

(b)

2 marks

The expected numbers are $20 (0.96)$ perfect and $20 (0.04)$ flawed.

50 (20 \times 0.96) + 15 (20 \times 0.04) = 960 + 12 = 972 pesos

Question 7

7 marks

The pH scale: a logarithm and its inverse

Functions

(a)

2 marks

Substitute.

pH = - lo g_{10} (0.0003) \approx 3.52

(b)

2 marks

Rearrange for the concentration.

[H^{+}] = 1 0^{- 6.6} \approx 2.51 \times 1 0^{- 7} moles per litre

(c)

3 marks

The concentrations are $1 0^{- 2}$ for the lemon and $1 0^{- 4.5}$ for the tomato. Divide.
The acid in a lemon is about 316 times stronger.

\frac{1 0 ^{- 2}}{1 0 ^{- 4.5}} = 1 0^{2.5} \approx 316

Question 8

7 marks

A coin and a die: final points, a table and the expected value

Statistics and probability

(a)

2 marks

A score of 7 only happens with a head and a 6.

\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

(b)

3 marks

A score of 1 needs tails and a 1. A score of 7 needs heads and a 6. Each of 2 to 6 can come from tails and that value, or heads and one less.
Check: the seven probabilities add to $\frac{1}{12} + 5 (\frac{1}{6}) + \frac{1}{12} = 1$ .

P (1) = \frac{1}{12}, P (7) = \frac{1}{12} P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

(c)

2 marks

Add each score times its probability.
Check: the die has mean 3.5 and heads adds 1 half of the time, so $3.5 + 0.5 = 4$ .

E = 1 (\frac{1}{12}) + (2 + 3 + 4 + 5 + 6) (\frac{1}{6}) + 7 (\frac{1}{12}) = \frac{8}{12} + \frac{20}{6} = 4

Question 9

6 marks

Two marathon training plans: arithmetic against geometric

Number and algebra

6 marks

Eefje's distance on day $n$ is $5 + 2 (n - 1)$ . She reaches 42.195 km when
So Eefje first reaches the marathon distance on day 20, when she runs $5 + 2 (19) = 43$ km.
Shumay's distance on day $n$ is $5 (1.13)^{n - 1}$ .
So Shumay first reaches it on day 19, when she runs $5 (1.13)^{18} \approx 45.1$ km. (On day 18 she runs 39.9 km, which is less than 42.195.)
Shumay is the first to run a marathon distance, on day 19 of her training.

5 + 2 (n - 1) \geq 42.195 \Rightarrow n \geq 19.6 5 (1.13)^{n - 1} \geq 42.195 \Rightarrow n - 1 \geq 17.45 \Rightarrow n \geq 18.5

Question 10

7 marks

Finding a and b from the gradient of a normal

Number and algebra

7 marks

The normal has gradient $- \frac{1}{3}$ , so the tangent at $T (2, 4)$ has gradient 3.
Differentiate $y = a x^{2} + b x - 10$ .
The point $(2, 4)$ is on the curve.
Subtract the first equation from the second.

\frac{d y}{d x} = 2 a x + b \Rightarrow 4 a + b = 3 4 = 4 a + 2 b - 10 \Rightarrow 4 a + 2 b = 14 b = 11, a = - 2

Question 11

6 marks

Triangle field against sector field: which is larger

Number and algebra

6 marks

Area of the sheep's field (triangle $A B C$ ).
The goat's sector has angle $28 2^{\circ}$ and radius 8.
The goat's field is larger, by

\frac{1}{2} (15) (21) sin 7 8^{\circ} \approx 154.06 m^{2} \frac{282}{360} π (8)^{2} \approx 157.50 m^{2} 157.50 - 154.06 \approx 3.44 m^{2}

Question 12

8 marks

A carpet made of a rectangle and two semicircles: perimeter and maximum area

Calculus

(a)

1 mark

The perimeter is two straight sides of length $y$ plus the circle formed by the two semicircular ends.

2 y + π x = 20

(b)

3 marks

The area is the rectangle $x y$ plus the two semicircles (one circle of radius $\frac{x}{2}$ ).
From (a), $y = 10 - \frac{π x}{2}$ .

A = x y + π (\frac{x}{2})^{2} = x y + \frac{π x ^{2}}{4} A = x (10 - \frac{π x}{2}) + \frac{π x ^{2}}{4} = 10 x - \frac{π x ^{2}}{2} + \frac{π x ^{2}}{4} A = 10 x - \frac{π x ^{2}}{4}

(c)

2 marks

Differentiate.

\frac{d A}{d x} = 10 - \frac{π x}{2}

(d)

2 marks

Set the derivative to zero.
The second derivative is $- \frac{π}{2} < 0$ , so this is a maximum.

10 - \frac{π x}{2} = 0 \Rightarrow x = \frac{20}{π}

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 algebra28 marks · 35%
Q5, Q9, Q10, Q11
Statistics and probability25 marks · 31%
Q1, Q3, Q6, Q8
Functions13 marks · 16%
Q4, Q7
Calculus8 marks · 10%
Q12
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Shifting data · Q1
Subtracting a constant from every value subtracts it from the mean and leaves the standard deviation unchanged.
02
Nominal rate · Q5
A nominal annual rate of 1.25% compounded monthly is $\frac{1.25}{12} %$ per month.
03
Log and exponential · Q7
$pH = - lo g_{10} [H^{+}]$ means $[H^{+}] = 1 0^{- pH}$ .
04
Two ways to a score · Q8
Tails scores the die value, heads scores the die value plus one. A score of 2 to 6 can come from either.
05
Two semicircles · Q12
Two semicircles of diameter $x$ together make a circle, with circumference $π x$ and area $\frac{π x ^{2}}{4}$ .

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 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–12 · 80 marks

Question 1

7 marks

Prices of garden chairs: range, mean, standard deviation and a sale

Statistics and probability

(a)

2 marks

The largest price is $255 and the smallest is $49.

255 - 49 = $206

(b)

3 marks

Enter the ten prices into the calculator.
(i) The mean.
(ii) The standard deviation (of the whole set of ten prices).

\overset{x}{ˉ} \approx $137 σ \approx $74.5

(c)

2 marks

(i) The mean falls by $20.

(ii) The spread does not change.

137.1 - 20 \approx $117 σ \approx $74.5

Question 2

6 marks

A toy crane: the cosine rule and the sine rule

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $A B C$ with $A B = 25$ , $A C = 15$ and $B A C = 2 2^{\circ}$ .

B C^{2} = 1 5^{2} + 2 5^{2} - 2 (15) (25) cos 2 2^{\circ} \Rightarrow B C \approx 12.4 cm

(b)

3 marks

Use the sine rule.
The angle is acute, so the sine rule gives the angle directly.

\frac{sin A B C}{15} = \frac{sin 2 2 ^{\circ}}{12.43} sin A B C = 0.4520 \Rightarrow A B C \approx 26. 9^{\circ}

Question 3

5 marks

Pepper sizes and colours: a chi-squared test for independence

Statistics and probability

(a)

1 mark

$H_{0}$ : the size of a pepper is independent of its colour.
$H_{1}$ : the size of a pepper is not independent of its colour.

(b)

4 marks

(i) Enter the $3 \times 3$ table into the calculator and run a $χ^{2}$ test. The degrees of freedom are $(3 - 1) (3 - 1) = 4$ , and every expected count is at least 20.

(ii) Since $22.5 > 9.49$ (equivalently, the p-value $0.000156 < 0.05$ ), reject $H_{0}$ . There is sufficient evidence that the size of the peppers is not independent of their colour.

χ_{c a l c}^{2} \approx 22.5

Question 4

6 marks

Bricklayers: direct and inverse variation

Functions

(a)

3 marks

The area is proportional to the number of bricklayers, $n = k B$ .

60 = 5 k \Rightarrow k = 12 n = 12 \times 7 = 84 m^{2}

(b)

3 marks

The time is inversely proportional to the number of bricklayers, $t = \frac{c}{B}$ .

8 = \frac{c}{5} \Rightarrow c = 40 t = \frac{40}{12} = \frac{10}{3} \approx 3.33 hours (3 hours 20 minutes)

Question 5

9 marks

Savings, then a car loan: compound interest and repayments

Number and algebra

(a)

3 marks

Six years is 72 months.

3000 (1 + \frac{0.0125}{12})^{72} \approx $3233.53

(b)

2 marks

Solve $3000 (1 + \frac{0.0125}{12})^{N} = 3550$ for $N$ .

N \approx 161.7 \Rightarrow 162 months

(c)

1 mark

The car costs $22 000 and the partial payment is $3550.

22000 - 3550 = $18 450

(d)

3 marks

Use the finance solver with $N = 96$ , $I = 12.6%$ , $P V = 18450$ , $F V = 0$ , 12 payments a year.

P M T \approx $306

Question 6

6 marks

Flawed chocolates in a batch: binomial probabilities and expected income

Statistics and probability

(a)

4 marks

(i) Let $X \sim B (20, 0.04)$ .

(ii) More than two flawed.

P (X = 2) = (2 20) (0.04)^{2} (0.96)^{18} \approx 0.146 P (X > 2) = 1 - P (X \leq 2) \approx 0.0439

(b)

2 marks

The expected numbers are $20 (0.96)$ perfect and $20 (0.04)$ flawed.

50 (20 \times 0.96) + 15 (20 \times 0.04) = 960 + 12 = 972 pesos

Question 7

7 marks

The pH scale: a logarithm and its inverse

Functions

(a)

2 marks

Substitute.

pH = - lo g_{10} (0.0003) \approx 3.52

(b)

2 marks

Rearrange for the concentration.

[H^{+}] = 1 0^{- 6.6} \approx 2.51 \times 1 0^{- 7} moles per litre

(c)

3 marks

The concentrations are $1 0^{- 2}$ for the lemon and $1 0^{- 4.5}$ for the tomato. Divide.
The acid in a lemon is about 316 times stronger.

\frac{1 0 ^{- 2}}{1 0 ^{- 4.5}} = 1 0^{2.5} \approx 316

Question 8

7 marks

A coin and a die: final points, a table and the expected value

Statistics and probability

(a)

2 marks

A score of 7 only happens with a head and a 6.

\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

(b)

3 marks

A score of 1 needs tails and a 1. A score of 7 needs heads and a 6. Each of 2 to 6 can come from tails and that value, or heads and one less.
Check: the seven probabilities add to $\frac{1}{12} + 5 (\frac{1}{6}) + \frac{1}{12} = 1$ .

P (1) = \frac{1}{12}, P (7) = \frac{1}{12} P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

(c)

2 marks

Add each score times its probability.
Check: the die has mean 3.5 and heads adds 1 half of the time, so $3.5 + 0.5 = 4$ .

E = 1 (\frac{1}{12}) + (2 + 3 + 4 + 5 + 6) (\frac{1}{6}) + 7 (\frac{1}{12}) = \frac{8}{12} + \frac{20}{6} = 4

Question 9

6 marks

Two marathon training plans: arithmetic against geometric

Number and algebra

6 marks

Eefje's distance on day $n$ is $5 + 2 (n - 1)$ . She reaches 42.195 km when
So Eefje first reaches the marathon distance on day 20, when she runs $5 + 2 (19) = 43$ km.
Shumay's distance on day $n$ is $5 (1.13)^{n - 1}$ .
So Shumay first reaches it on day 19, when she runs $5 (1.13)^{18} \approx 45.1$ km. (On day 18 she runs 39.9 km, which is less than 42.195.)
Shumay is the first to run a marathon distance, on day 19 of her training.

5 + 2 (n - 1) \geq 42.195 \Rightarrow n \geq 19.6 5 (1.13)^{n - 1} \geq 42.195 \Rightarrow n - 1 \geq 17.45 \Rightarrow n \geq 18.5

Question 10

7 marks

Finding a and b from the gradient of a normal

Number and algebra

7 marks

The normal has gradient $- \frac{1}{3}$ , so the tangent at $T (2, 4)$ has gradient 3.
Differentiate $y = a x^{2} + b x - 10$ .
The point $(2, 4)$ is on the curve.
Subtract the first equation from the second.

\frac{d y}{d x} = 2 a x + b \Rightarrow 4 a + b = 3 4 = 4 a + 2 b - 10 \Rightarrow 4 a + 2 b = 14 b = 11, a = - 2

Question 11

6 marks

Triangle field against sector field: which is larger

Number and algebra

6 marks

Area of the sheep's field (triangle $A B C$ ).
The goat's sector has angle $28 2^{\circ}$ and radius 8.
The goat's field is larger, by

\frac{1}{2} (15) (21) sin 7 8^{\circ} \approx 154.06 m^{2} \frac{282}{360} π (8)^{2} \approx 157.50 m^{2} 157.50 - 154.06 \approx 3.44 m^{2}

Question 12

8 marks

A carpet made of a rectangle and two semicircles: perimeter and maximum area

Calculus

(a)

1 mark

The perimeter is two straight sides of length $y$ plus the circle formed by the two semicircular ends.

2 y + π x = 20

(b)

3 marks

The area is the rectangle $x y$ plus the two semicircles (one circle of radius $\frac{x}{2}$ ).
From (a), $y = 10 - \frac{π x}{2}$ .

A = x y + π (\frac{x}{2})^{2} = x y + \frac{π x ^{2}}{4} A = x (10 - \frac{π x}{2}) + \frac{π x ^{2}}{4} = 10 x - \frac{π x ^{2}}{2} + \frac{π x ^{2}}{4} A = 10 x - \frac{π x ^{2}}{4}

(c)

2 marks

Differentiate.

\frac{d A}{d x} = 10 - \frac{π x}{2}

(d)

2 marks

Set the derivative to zero.
The second derivative is $- \frac{π}{2} < 0$ , so this is a maximum.

10 - \frac{π x}{2} = 0 \Rightarrow x = \frac{20}{π}

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 algebra28 marks · 35%
Q5, Q9, Q10, Q11
Statistics and probability25 marks · 31%
Q1, Q3, Q6, Q8
Functions13 marks · 16%
Q4, Q7
Calculus8 marks · 10%
Q12
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Shifting data · Q1
Subtracting a constant from every value subtracts it from the mean and leaves the standard deviation unchanged.
02
Nominal rate · Q5
A nominal annual rate of 1.25% compounded monthly is $\frac{1.25}{12} %$ per month.
03
Log and exponential · Q7
$pH = - lo g_{10} [H^{+}]$ means $[H^{+}] = 1 0^{- pH}$ .
04
Two ways to a score · Q8
Tails scores the die value, heads scores the die value plus one. A score of 2 to 6 can come from either.
05
Two semicircles · Q12
Two semicircles of diameter $x$ together make a circle, with circumference $π x$ and area $\frac{π x ^{2}}{4}$ .

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