Loading…

Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

5 marks

Exponential cooling model

Functions

(a)

2 marks

The start means $t = 0$ , and $e^{0} = 1$ .

T (0) = 18 + 72 e^{0} = 18 + 72 T (0) = 90^{\circ} C

Mentor tip

Any time a model asks for the value at the start, set time to zero first.

(b)

2 marks

Substitute $t = 15$ and evaluate on the GDC.

T (15) = 18 + 72 e^{- 15/20} = 18 + 72 e^{- 0.75} T (15) = 52.0103 \dots \approx 52.0^{\circ} C

Mentor tip

Sanity check: the answer must sit between the starting 90 and the room temperature.

(c)

1 mark

As $t$ grows, $e^{- t /20} \to 0$ , so $T$ settles on the constant term. That long-run value is the room temperature.

t \to \infty : T \to 18 + 0 18^{\circ} C

Mentor tip

The horizontal asymptote of a decaying exponential is the number added on the outside.

Question 2

7 marks

Cumulative frequency, missing frequencies and a mean

Statistics and probability

(a)

2 marks

Half of 52 is 26. Go across from a cumulative frequency of 26 to the curve, then down to the height axis.

\frac{52}{2} = 26 median \approx 170 cm

Mentor tip

Draw the lines on the graph so your reading is visible. A careful reading lands at about 170 cm.

(b)

5 marks

(i) From the graph, read the cumulative frequencies at 170, 175 and 180 cm: 25, 42 and 52. Each class frequency is the jump between neighbouring readings.

(ii) Use the midpoint of each class as its representative height.

Multiply each midpoint by its frequency, add, and divide by 52. Give four significant figures, as asked.

p = 42 - 25 = 17, q = 52 - 42 = 10 \sum f x = 3 (157.5) + 8 (162.5) + 14 (167.5) + 17 (172.5) + 10 (177.5) = 8825 \overset{x}{ˉ} = \frac{8825}{52} = 169.711 \dots \overset{x}{ˉ} \approx 169.7 cm

Another way. Put the midpoints in one GDC list and the frequencies in another, then run one-variable statistics.

Mentor tip

Check that 3 + 8 + 14 + p + q comes to 52 before you work out the mean. And watch the accuracy: 170 is not four significant figures.

Question 3

4 marks

Where a function is decreasing

Calculus

4 marks

A function decreases where its derivative is negative, so solve $x^{2} - x - 1 < 0$ .
Find the roots of the quadratic with the formula (or the GDC).
The parabola opens upwards, so it is negative between its roots.

x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 \pm 5}{2} x = - 0.618 \dots or x = 1.618 \dots f^{'} (x) < 0 for \frac{1 - 5}{2} < x < \frac{1 + 5}{2} - 0.618 < x < 1.62

Mentor tip

Say clearly that you need $f^{'} (x) < 0$ . Finding the roots alone is only half the story; the interval is the answer.

Question 4

6 marks

Ferris wheel height as a trig model

Geometry and trigonometry

(a)

4 marks

The centre height is the middle of the motion, so it gives $a$ .
The radius is the amplitude, so it gives $b$ . The minus sign already makes $P$ start at the bottom.
One turn takes 4 minutes and the angle is in degrees, so $c \times 4 = 360$ .

a = 23 b = 20 4 c = 360 \Rightarrow c = 90

Mentor tip

Check with $t = 0$ : $23 - 20 cos 0 = 3$ m, which is the lowest point. The degree sign means $360$ , not $2 π$ .

(b)

2 marks

A sine curve is a cosine curve shifted by a quarter period. Here a quarter period is 1 minute.
Or test a point: at $t = 0$ the height must be 3.

3 = 23 - 20 sin (90 d)^{\circ} \Rightarrow sin (90 d)^{\circ} = 1 90 d = 90 \Rightarrow d = 1

Mentor tip

Graph both versions on the GDC and see that they sit on top of each other; that is a quick check on $d$ .

Question 5

7 marks

Turning points and the area between a curve and a line

Calculus

(a)

2 marks

Use the GDC maximum and minimum tools on the graph of $g$ (radian mode).

A (- 0.700, 0.469) B (0.700, - 0.469)

Mentor tip

The function is odd, so $A$ and $B$ are reflections of each other through the origin; a nice check.

(b)

5 marks

Find the line through $O$ and $B$ : its gradient is the $y$ -coordinate of $B$ over its $x$ -coordinate.
Between $x = 0$ and $x_{B}$ , the line lies above the curve.
Integrate line minus curve over that interval on the GDC, using stored values for $x_{B}$ and the gradient.

m = \frac{- 0.468564 \dots}{0.700108 \dots} = - 0.669274 \dots, y = - 0.669274 \dots x Area = \int_{0}^{0.700108 \dots} (- 0.669274 \dots x - g (x)) d x (GDC) = 0.0406871 \dots \approx 0.0407

Another way. Find the area between the curve and the $x$ -axis from $0$ to $x_{B}$ (0.2047...) and take away the triangle under $[O B]$ (0.1640...).

Mentor tip

Store full-accuracy values in the GDC. Rounding $B$ to 0.7 before integrating can shift the third significant figure of such a small area.

Question 6

7 marks

Normal distribution with conditional probability

Statistics and probability

7 marks

Turn the first fact into a $z$ -score: $P (H < 171) = 0.6$ .
The second fact is conditional. Rewrite it as $P (H > 175) \div P (H > 171) = 0.35$ to get $P (H > 175)$ .
Turn that into a second $z$ -score, using $P (H < 175) = 0.86$ .
Solve the two linear equations in $μ$ and $σ$ .

\frac{171 - μ}{σ} = Φ^{- 1} (0.6) = 0.253347 \dots P (H > 175) = 0.35 \times 0.4 = 0.14 \frac{175 - μ}{σ} = Φ^{- 1} (0.86) = 1.080319 \dots σ = \frac{4}{1.080319 \dots - 0.253347 \dots} = 4.83692 \dots σ \approx 4.84, μ \approx 170 (169.774 \dots)

Mentor tip

The trap is using 0.35 directly as $P (H > 175)$ . When a random adult is picked from a subgroup, the probability is conditional, so multiply by 0.4 first.

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

16 marks

Sphere from a diameter, cosine rule and triangle area

Geometry and trigonometry

(a)

2 marks

The centre of a sphere is the midpoint of any diameter. Average each coordinate.

C = (\frac{- 3 + ( - 1 )}{2}, \frac{6 + 2}{2}, \frac{4 + ( - 2 )}{2}) C = (- 2, 4, 1)

(b)

2 marks

The radius is the distance from $C$ to $A$ (or to $B$ ). Use the 3D distance formula and show every square.

A C = (- 3 + 2)^{2} + (6 - 4)^{2} + (4 - 1)^{2} = 1 + 4 + 9 = 14

Mentor tip

For a show-that, write the numbers inside the root. Just writing $14$ proves nothing.

(c)

2 marks

Use $V = \frac{4}{3} π r^{3}$ with $r = 14$ .

V = \frac{4}{3} π (14)^{3} = \frac{56 14}{3} π V = 219.422 \dots \approx 219 units^{3}

(d)

7 marks

(i) $A B$ is a diameter, so $A B = 214$ . With all three sides of triangle $B A P$ known, use the cosine rule for the angle at $A$ .

The side opposite angle $A$ is $B P = 4$ .

(ii) Use area $= \frac{1}{2} ab sin C$ with the two sides next to that angle.

cos B \hat{A} P = \frac{5 ^{2} + ( 2 14 ) ^{2} - 4 ^{2}}{2 ( 5 ) ( 2 14 )} = \frac{65}{20 14} = 0.868599 \dots B \hat{A} P = 29.7037 \dots^{\circ} \approx 29. 7^{\circ} (0.518 rad) Area = \frac{1}{2} (5) (214) sin 29.7037 \dots^{\circ} Area = 9.27024 \dots \approx 9.27 units^{2}

Another way. For the area, Heron's formula with sides 4, 5 and $214$ gives the same result without the angle.

Mentor tip

Keep the unrounded angle in your GDC for part (ii) and again in part (e).

(e)

3 marks

(i) $C$ lies on $[A B]$ , so the angle $C \hat{A} P$ is the same angle you found in (d), and $A C = 14$ .

Use the cosine rule in triangle $C A P$ to find $C P$ .

(ii) Compare $C P$ with the radius: a point is inside the sphere when its distance from the centre is less than $14$ .

C P^{2} = 5^{2} + (14)^{2} - 2 (5) (14) cos 29.7037 \dots^{\circ} = 6.5 C P = 6.5 = 2.54950 \dots \approx 2.55 2.55 < 14 = 3.74 \Rightarrow P is inside the sphere

Another way. Since $cos C \hat{A} P = \frac{65}{20 14}$ exactly, the cosine rule gives $C P^{2} = 25 + 14 - 32.5 = 6.5$ with no rounding at all.

Mentor tip

For the conclusion, write both numbers and the inequality, then say what it means. A bare number does not show anything.

Question 8

13 marks

Compound interest against depreciation

Number and algebra

(a)

3 marks

Quarterly compounding at 4.8% a year means 1.2% per quarter.
Five years is 20 quarters.
Use the formula or the GDC finance solver (PV = -15000, I% = 4.8, C/Y = 4).

F V = 15000 (1 + \frac{4.8}{4 \times 100})^{4 \times 5} = 15000 (1.012)^{20} F V = 19041.51 \dots \approx $19 042

Mentor tip

Round to the nearest dollar, as asked, and make sure your final value is positive even if the solver shows a minus sign.

(b)

3 marks

Set the future value equal to 28000 and solve for the time, on the GDC or with logs.
Convert quarters to years and count on from the start of 2026.

15000 (1.012)^{4 x} = 28000 x = \frac{ln ( 28000/15000 )}{4 ln 1.012} = 13.0810 \dots years 2026 + 13.08 \dots \Rightarrow during 2039

Another way. Tabulate the balance: after 52 quarters (start of 2039) it is 27 891.88 dollars, still short; after 53 quarters it is 28 226.59 dollars.

Mentor tip

Take care counting years: 13 whole years from 1 January 2026 is 1 January 2039, so 13.08 years lands during 2039.

(c)

2 marks

One year of depreciation multiplies the value by $1 - \frac{r}{100}$ .

28000 (1 - \frac{r}{100}) = 26222 \frac{r}{100} = \frac{28000 - 26222}{28000} = 0.0635 r = 6.35

(d)

5 marks

Write the boat's value after $x$ years as $28000 (0.9365)^{x}$ .
Write the account balance after $x$ years as $15000 (1.012)^{4 x}$ .
Solve for where they meet, with the GDC intersect tool.
Convert the time into a calendar year.

15000 (1.012)^{4 x} = 28000 (0.9365)^{x} x = 5.50788 \dots (GDC) 2026 + 5.51 \Rightarrow during 2031

Another way. Tabulate both values every quarter year: at $x = 5.5$ the account (19 501) is still just below the boat (19 519), and by $x = 5.75$ it is ahead.

Mentor tip

Do not round 6.35% or 1.012 along the way, and check the year the same way as in (b): 5.5 years after the start of 2026 is mid-2031.

Question 9

15 marks

Binomial probabilities and an expected-value game

Statistics and probability

(a)

2 marks

The 12 yellow counters make up 0.4 of the bag, which gives the total.
The rest are shared equally between red and blue.

\frac{12}{N} = 0.4 \Rightarrow N = 30 red = \frac{30 - 12}{2} = 9

Another way. Set up $\frac{12}{2 x + 12} = 0.4$ with $x$ red counters and solve for $x$ .

Mentor tip

Start from the given facts and end at 9; do not use 9 to prove itself.

(b)

3 marks

The counter goes back each time, so the draws are independent: $Y \sim B (10, 0.4)$ for the number of yellows.
At least 6 is the complement of at most 5.

P (Y \geq 6) = 1 - P (Y \leq 5) (GDC) = 0.166238 \dots \approx 0.16624

Mentor tip

Five significant figures is asked for here, so 0.166 is not enough. Also be sure you use $P (Y \leq 5)$ , not $P (Y \leq 6)$ .

(c)

1 mark

The probability 0.1662 matches at least 6 yellows, so the gain is the yellow prize minus the entry fee.

A = 10 - 5 = 5

(d)

5 marks

(i) The probabilities must add to 1.

(ii) Write $E (X)$ in terms of $B$ and require $E (X) > 0$ .

Solve the inequality and take the next whole number up.

p = 1 - 2 (0.0473) - 0.1662 = 0.7392 E (X) = 35 (0.0473) + (B - 5) (0.0473) + 5 (0.1662) - 5 (0.7392) E (X) = 0.0473 B - 1.446 0.0473 B - 1.446 > 0 \Rightarrow B > 30.57 \dots B = 31

Mentor tip

The $- 5$ outcome carries a negative sign in the expected value. Dropping it is the most common slip here.

(e)

4 marks

Missing out on a prize in one game has probability $0.7392$ .
At least one prize in $n$ games is the complement of no prize in every game.
Solve the inequality with logs or a GDC table and round up.

1 - 0.739 2^{n} > 0.99 0.739 2^{n} < 0.01 \Rightarrow n > \frac{ln 0.01}{ln 0.7392} = 15.239 \dots n = 15 : 0.98925 \dots, n = 16 : 0.99205 \dots n = 16

Mentor tip

Dividing by the log of a number below 1 flips the inequality. A quick table of values for $n = 15$ and $16$ avoids that worry.

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 probability29 marks · 36%
Q2, Q6, Q9
Geometry and trigonometry22 marks · 28%
Q4, Q7
Number and algebra13 marks · 16%
Q8
Calculus11 marks · 14%
Q3, Q5
Functions5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
Treating a conditional probability as plain · Q6
The 0.35 applies only to adults already over 171 cm. Multiply by 0.4 to get the probability of being over 175 cm before using inverse normal.
02
Too few significant figures in the mean · Q2
The mean is asked for to four significant figures, so write 169.7, not 170.
03
Counting the calendar year wrongly · Q8
A time of 5.5 years after 1 January 2026 falls in 2031, not 2032. Count whole years from the start date with care.
04
Rounding coordinates before integrating · Q5
The area is tiny, so using 0.7 and -0.469 for B can change the answer. Keep stored GDC values throughout.
05
Forgetting the fee in the net gain · Q9
Net gain is prize minus the 5 dollar entry, and losing games count as -5 in the expected value.

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

5 marks

Exponential cooling model

Functions

(a)

2 marks

The start means $t = 0$ , and $e^{0} = 1$ .

T (0) = 18 + 72 e^{0} = 18 + 72 T (0) = 90^{\circ} C

Mentor tip

Any time a model asks for the value at the start, set time to zero first.

(b)

2 marks

Substitute $t = 15$ and evaluate on the GDC.

T (15) = 18 + 72 e^{- 15/20} = 18 + 72 e^{- 0.75} T (15) = 52.0103 \dots \approx 52.0^{\circ} C

Mentor tip

Sanity check: the answer must sit between the starting 90 and the room temperature.

(c)

1 mark

As $t$ grows, $e^{- t /20} \to 0$ , so $T$ settles on the constant term. That long-run value is the room temperature.

t \to \infty : T \to 18 + 0 18^{\circ} C

Mentor tip

The horizontal asymptote of a decaying exponential is the number added on the outside.

Question 2

7 marks

Cumulative frequency, missing frequencies and a mean

Statistics and probability

(a)

2 marks

Half of 52 is 26. Go across from a cumulative frequency of 26 to the curve, then down to the height axis.

\frac{52}{2} = 26 median \approx 170 cm

Mentor tip

Draw the lines on the graph so your reading is visible. A careful reading lands at about 170 cm.

(b)

5 marks

(i) From the graph, read the cumulative frequencies at 170, 175 and 180 cm: 25, 42 and 52. Each class frequency is the jump between neighbouring readings.

(ii) Use the midpoint of each class as its representative height.

Multiply each midpoint by its frequency, add, and divide by 52. Give four significant figures, as asked.

p = 42 - 25 = 17, q = 52 - 42 = 10 \sum f x = 3 (157.5) + 8 (162.5) + 14 (167.5) + 17 (172.5) + 10 (177.5) = 8825 \overset{x}{ˉ} = \frac{8825}{52} = 169.711 \dots \overset{x}{ˉ} \approx 169.7 cm

Another way. Put the midpoints in one GDC list and the frequencies in another, then run one-variable statistics.

Mentor tip

Check that 3 + 8 + 14 + p + q comes to 52 before you work out the mean. And watch the accuracy: 170 is not four significant figures.

Question 3

4 marks

Where a function is decreasing

Calculus

4 marks

A function decreases where its derivative is negative, so solve $x^{2} - x - 1 < 0$ .
Find the roots of the quadratic with the formula (or the GDC).
The parabola opens upwards, so it is negative between its roots.

x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 \pm 5}{2} x = - 0.618 \dots or x = 1.618 \dots f^{'} (x) < 0 for \frac{1 - 5}{2} < x < \frac{1 + 5}{2} - 0.618 < x < 1.62

Mentor tip

Say clearly that you need $f^{'} (x) < 0$ . Finding the roots alone is only half the story; the interval is the answer.

Question 4

6 marks

Ferris wheel height as a trig model

Geometry and trigonometry

(a)

4 marks

The centre height is the middle of the motion, so it gives $a$ .
The radius is the amplitude, so it gives $b$ . The minus sign already makes $P$ start at the bottom.
One turn takes 4 minutes and the angle is in degrees, so $c \times 4 = 360$ .

a = 23 b = 20 4 c = 360 \Rightarrow c = 90

Mentor tip

Check with $t = 0$ : $23 - 20 cos 0 = 3$ m, which is the lowest point. The degree sign means $360$ , not $2 π$ .

(b)

2 marks

A sine curve is a cosine curve shifted by a quarter period. Here a quarter period is 1 minute.
Or test a point: at $t = 0$ the height must be 3.

3 = 23 - 20 sin (90 d)^{\circ} \Rightarrow sin (90 d)^{\circ} = 1 90 d = 90 \Rightarrow d = 1

Mentor tip

Graph both versions on the GDC and see that they sit on top of each other; that is a quick check on $d$ .

Question 5

7 marks

Turning points and the area between a curve and a line

Calculus

(a)

2 marks

Use the GDC maximum and minimum tools on the graph of $g$ (radian mode).

A (- 0.700, 0.469) B (0.700, - 0.469)

Mentor tip

The function is odd, so $A$ and $B$ are reflections of each other through the origin; a nice check.

(b)

5 marks

Find the line through $O$ and $B$ : its gradient is the $y$ -coordinate of $B$ over its $x$ -coordinate.
Between $x = 0$ and $x_{B}$ , the line lies above the curve.
Integrate line minus curve over that interval on the GDC, using stored values for $x_{B}$ and the gradient.

m = \frac{- 0.468564 \dots}{0.700108 \dots} = - 0.669274 \dots, y = - 0.669274 \dots x Area = \int_{0}^{0.700108 \dots} (- 0.669274 \dots x - g (x)) d x (GDC) = 0.0406871 \dots \approx 0.0407

Another way. Find the area between the curve and the $x$ -axis from $0$ to $x_{B}$ (0.2047...) and take away the triangle under $[O B]$ (0.1640...).

Mentor tip

Store full-accuracy values in the GDC. Rounding $B$ to 0.7 before integrating can shift the third significant figure of such a small area.

Question 6

7 marks

Normal distribution with conditional probability

Statistics and probability

7 marks

Turn the first fact into a $z$ -score: $P (H < 171) = 0.6$ .
The second fact is conditional. Rewrite it as $P (H > 175) \div P (H > 171) = 0.35$ to get $P (H > 175)$ .
Turn that into a second $z$ -score, using $P (H < 175) = 0.86$ .
Solve the two linear equations in $μ$ and $σ$ .

\frac{171 - μ}{σ} = Φ^{- 1} (0.6) = 0.253347 \dots P (H > 175) = 0.35 \times 0.4 = 0.14 \frac{175 - μ}{σ} = Φ^{- 1} (0.86) = 1.080319 \dots σ = \frac{4}{1.080319 \dots - 0.253347 \dots} = 4.83692 \dots σ \approx 4.84, μ \approx 170 (169.774 \dots)

Mentor tip

The trap is using 0.35 directly as $P (H > 175)$ . When a random adult is picked from a subgroup, the probability is conditional, so multiply by 0.4 first.

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

16 marks

Sphere from a diameter, cosine rule and triangle area

Geometry and trigonometry

(a)

2 marks

The centre of a sphere is the midpoint of any diameter. Average each coordinate.

C = (\frac{- 3 + ( - 1 )}{2}, \frac{6 + 2}{2}, \frac{4 + ( - 2 )}{2}) C = (- 2, 4, 1)

(b)

2 marks

The radius is the distance from $C$ to $A$ (or to $B$ ). Use the 3D distance formula and show every square.

A C = (- 3 + 2)^{2} + (6 - 4)^{2} + (4 - 1)^{2} = 1 + 4 + 9 = 14

Mentor tip

For a show-that, write the numbers inside the root. Just writing $14$ proves nothing.

(c)

2 marks

Use $V = \frac{4}{3} π r^{3}$ with $r = 14$ .

V = \frac{4}{3} π (14)^{3} = \frac{56 14}{3} π V = 219.422 \dots \approx 219 units^{3}

(d)

7 marks

(i) $A B$ is a diameter, so $A B = 214$ . With all three sides of triangle $B A P$ known, use the cosine rule for the angle at $A$ .

The side opposite angle $A$ is $B P = 4$ .

(ii) Use area $= \frac{1}{2} ab sin C$ with the two sides next to that angle.

cos B \hat{A} P = \frac{5 ^{2} + ( 2 14 ) ^{2} - 4 ^{2}}{2 ( 5 ) ( 2 14 )} = \frac{65}{20 14} = 0.868599 \dots B \hat{A} P = 29.7037 \dots^{\circ} \approx 29. 7^{\circ} (0.518 rad) Area = \frac{1}{2} (5) (214) sin 29.7037 \dots^{\circ} Area = 9.27024 \dots \approx 9.27 units^{2}

Another way. For the area, Heron's formula with sides 4, 5 and $214$ gives the same result without the angle.

Mentor tip

Keep the unrounded angle in your GDC for part (ii) and again in part (e).

(e)

3 marks

(i) $C$ lies on $[A B]$ , so the angle $C \hat{A} P$ is the same angle you found in (d), and $A C = 14$ .

Use the cosine rule in triangle $C A P$ to find $C P$ .

(ii) Compare $C P$ with the radius: a point is inside the sphere when its distance from the centre is less than $14$ .

C P^{2} = 5^{2} + (14)^{2} - 2 (5) (14) cos 29.7037 \dots^{\circ} = 6.5 C P = 6.5 = 2.54950 \dots \approx 2.55 2.55 < 14 = 3.74 \Rightarrow P is inside the sphere

Another way. Since $cos C \hat{A} P = \frac{65}{20 14}$ exactly, the cosine rule gives $C P^{2} = 25 + 14 - 32.5 = 6.5$ with no rounding at all.

Mentor tip

For the conclusion, write both numbers and the inequality, then say what it means. A bare number does not show anything.

Question 8

13 marks

Compound interest against depreciation

Number and algebra

(a)

3 marks

Quarterly compounding at 4.8% a year means 1.2% per quarter.
Five years is 20 quarters.
Use the formula or the GDC finance solver (PV = -15000, I% = 4.8, C/Y = 4).

F V = 15000 (1 + \frac{4.8}{4 \times 100})^{4 \times 5} = 15000 (1.012)^{20} F V = 19041.51 \dots \approx $19 042

Mentor tip

Round to the nearest dollar, as asked, and make sure your final value is positive even if the solver shows a minus sign.

(b)

3 marks

Set the future value equal to 28000 and solve for the time, on the GDC or with logs.
Convert quarters to years and count on from the start of 2026.

15000 (1.012)^{4 x} = 28000 x = \frac{ln ( 28000/15000 )}{4 ln 1.012} = 13.0810 \dots years 2026 + 13.08 \dots \Rightarrow during 2039

Another way. Tabulate the balance: after 52 quarters (start of 2039) it is 27 891.88 dollars, still short; after 53 quarters it is 28 226.59 dollars.

Mentor tip

Take care counting years: 13 whole years from 1 January 2026 is 1 January 2039, so 13.08 years lands during 2039.

(c)

2 marks

One year of depreciation multiplies the value by $1 - \frac{r}{100}$ .

28000 (1 - \frac{r}{100}) = 26222 \frac{r}{100} = \frac{28000 - 26222}{28000} = 0.0635 r = 6.35

(d)

5 marks

Write the boat's value after $x$ years as $28000 (0.9365)^{x}$ .
Write the account balance after $x$ years as $15000 (1.012)^{4 x}$ .
Solve for where they meet, with the GDC intersect tool.
Convert the time into a calendar year.

15000 (1.012)^{4 x} = 28000 (0.9365)^{x} x = 5.50788 \dots (GDC) 2026 + 5.51 \Rightarrow during 2031

Another way. Tabulate both values every quarter year: at $x = 5.5$ the account (19 501) is still just below the boat (19 519), and by $x = 5.75$ it is ahead.

Mentor tip

Do not round 6.35% or 1.012 along the way, and check the year the same way as in (b): 5.5 years after the start of 2026 is mid-2031.

Question 9

15 marks

Binomial probabilities and an expected-value game

Statistics and probability

(a)

2 marks

The 12 yellow counters make up 0.4 of the bag, which gives the total.
The rest are shared equally between red and blue.

\frac{12}{N} = 0.4 \Rightarrow N = 30 red = \frac{30 - 12}{2} = 9

Another way. Set up $\frac{12}{2 x + 12} = 0.4$ with $x$ red counters and solve for $x$ .

Mentor tip

Start from the given facts and end at 9; do not use 9 to prove itself.

(b)

3 marks

The counter goes back each time, so the draws are independent: $Y \sim B (10, 0.4)$ for the number of yellows.
At least 6 is the complement of at most 5.

P (Y \geq 6) = 1 - P (Y \leq 5) (GDC) = 0.166238 \dots \approx 0.16624

Mentor tip

Five significant figures is asked for here, so 0.166 is not enough. Also be sure you use $P (Y \leq 5)$ , not $P (Y \leq 6)$ .

(c)

1 mark

The probability 0.1662 matches at least 6 yellows, so the gain is the yellow prize minus the entry fee.

A = 10 - 5 = 5

(d)

5 marks

(i) The probabilities must add to 1.

(ii) Write $E (X)$ in terms of $B$ and require $E (X) > 0$ .

Solve the inequality and take the next whole number up.

p = 1 - 2 (0.0473) - 0.1662 = 0.7392 E (X) = 35 (0.0473) + (B - 5) (0.0473) + 5 (0.1662) - 5 (0.7392) E (X) = 0.0473 B - 1.446 0.0473 B - 1.446 > 0 \Rightarrow B > 30.57 \dots B = 31

Mentor tip

The $- 5$ outcome carries a negative sign in the expected value. Dropping it is the most common slip here.

(e)

4 marks

Missing out on a prize in one game has probability $0.7392$ .
At least one prize in $n$ games is the complement of no prize in every game.
Solve the inequality with logs or a GDC table and round up.

1 - 0.739 2^{n} > 0.99 0.739 2^{n} < 0.01 \Rightarrow n > \frac{ln 0.01}{ln 0.7392} = 15.239 \dots n = 15 : 0.98925 \dots, n = 16 : 0.99205 \dots n = 16

Mentor tip

Dividing by the log of a number below 1 flips the inequality. A quick table of values for $n = 15$ and $16$ avoids that worry.

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 probability29 marks · 36%
Q2, Q6, Q9
Geometry and trigonometry22 marks · 28%
Q4, Q7
Number and algebra13 marks · 16%
Q8
Calculus11 marks · 14%
Q3, Q5
Functions5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
Treating a conditional probability as plain · Q6
The 0.35 applies only to adults already over 171 cm. Multiply by 0.4 to get the probability of being over 175 cm before using inverse normal.
02
Too few significant figures in the mean · Q2
The mean is asked for to four significant figures, so write 169.7, not 170.
03
Counting the calendar year wrongly · Q8
A time of 5.5 years after 1 January 2026 falls in 2031, not 2032. Count whole years from the start date with care.
04
Rounding coordinates before integrating · Q5
The area is tiny, so using 0.7 and -0.469 for B can change the answer. Keep stored GDC values throughout.
05
Forgetting the fee in the net gain · Q9
Net gain is prize minus the 5 dollar entry, and losing games count as -5 in the expected value.

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