HL only · Paper 3

IB Maths AI HL Paper 3: guide and practice questions

Paper 3 is the HL paper that tests problem-solving rather than recall: two long questions that each take one new idea and lead you through it. This page covers the format, how it is marked and the techniques that score, then gives you worked investigations and full-length practice.

By Pete Bromfield, former Head of Maths with IB examining experience (independent) · Updated 25 October 2026

Time1 hour 15 minutes
QuestionsTwo, both compulsory
CalculatorGDC needed
Weight20% of your HL grade

Try a worked investigation Full Paper 3s

What Paper 3 is

Paper 3 is sat by HL students only, after Papers 1 and 2. It has two compulsory extended-response questions, and you have the formula booklet and your GDC throughout. Each question starts with something short and concrete, then asks you to spot a pattern, make a conjecture, prove or justify it, and extend it. Any topic in the course can come up, including the SL core, and one question can join several: matrices with graph theory, say, or statistics with a model you build from data.

AI HL paper	Time	Calculator	Weight
Paper 1	2 hours	Calculator (GDC)	30%
Paper 2	2 hours	Calculator (GDC)	30%
Paper 3	1 hour 15 minutes	Calculator (GDC)	20%
Internal assessment	Coursework	–	20%

Our full Paper 3s are 55 marks over the two questions. From the first exams of the new courses in May 2029, the IB has announced HL Paper 3 will be 1 hour (what is changing). More on every paper: AI HL past papers · AI HL formula booklet guide.

How Paper 3 is marked

Paper 3 uses the same kinds of marks as Papers 1 and 2, but more of them reward reasoning, because the questions ask you to explain and prove. Knowing what each mark is for tells you what to write.

M: method
For a valid method, shown in your working. You can earn it even if your answer goes wrong later, so always write the method down.
A: accuracy
For a correct answer or step, usually only after the method mark before it. A bare answer with no working can score nothing.
R: reasoning
For explaining why: a justification in words, a reason a step is valid, a concluding sentence. Paper 3 has many of these.
AG: answer given
On "show that" parts the answer is printed, so the marks are for the working that reaches it. If you cannot show it, use the given result in the next part anyway.

An early slip is usually penalised once if you then use your wrong value correctly (follow through). More: how IB Maths mark schemes work · where AI HL students lose marks.

The techniques that score

Spot the pattern

Let the GDC generate many cases quickly (matrix powers, a sequence, a model's output for several inputs), then look down the column for what stays the same and what changes.

Conjecture, then justify

AI questions ask you to justify rather than to give a formal proof. Explain in words why the relationship must hold, then test it on a case you have not used yet.

Use your GDC

Regression (transform the data with logs first when it curves), solving equations, matrix work and graphs. Write down what you entered and what you got, not only the final answer.

Communicate

Give every answer in context, with units, and comment on the model: what it assumes, and what happens outside the data.

When you are stuck: read to the end of the question to see where it is heading, use any given result and move on to the next part; later parts are often easier than the one before. Timing: about half the time for each question.

Two worked mini-investigations

Short Paper 3-style questions, with the mark codes on each solution so you can see where the marks are. Try each part before you open its solution. Every answer was checked.

Investigation 1: Closed walks in a complete graph

Use your GDC
Spot the pattern
Explain and generalise

The complete graph $K_{4}$ has four vertices $A$ , $B$ , $C$ and $D$ , and every pair of vertices is joined by one edge. Let $M$ be its adjacency matrix. A walk may use the same vertex or edge more than once.

(a) Write down $M$ , and use your GDC to find $M^{2}$ , $M^{3}$ and $M^{4}$ . [3]

Worked solution (a)

M = (\begin{matrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{matrix})

M^{2} = (\begin{matrix} 3 & 2 & 2 & 2 \\ 2 & 3 & 2 & 2 \\ 2 & 2 & 3 & 2 \\ 2 & 2 & 2 & 3 \end{matrix})

M^{3} = (\begin{matrix} 6 & 7 & 7 & 7 \\ 7 & 6 & 7 & 7 \\ 7 & 7 & 6 & 7 \\ 7 & 7 & 7 & 6 \end{matrix})

M^{4} = (\begin{matrix} 21 & 20 & 20 & 20 \\ 20 & 21 & 20 & 20 \\ 20 & 20 & 21 & 20 \\ 20 & 20 & 20 & 21 \end{matrix})

A1 A1 A1

(b) State what the entry in row $A$ , column $A$ of $M^{n}$ counts. [1]

Worked solution (b)

The number of walks of length $n$ that start and end at $A$ (closed walks). A1

(c) Explain why there are $3^{n}$ walks of length $n$ that start at $A$ . [2]

Worked solution (c)

Every vertex has degree 3, so each of the $n$ steps can be made in 3 ways, giving $3 \times 3 \times \dots \times 3 = 3^{n}$ walks. R1 A1

Check: each row of $M^{4}$ adds up to $21 + 20 + 20 + 20 = 81 = 3^{4}$ .

(d) Let $a_{n}$ be the number of closed walks of length $n$ from $A$ . Explain why $a_{n + 1} = 3^{n} - a_{n}$ . [3]

Worked solution (d)

Take a closed walk of length $n + 1$ . Its first $n$ steps form a walk of length $n$ from $A$ that ends at a vertex other than $A$ , because the last step must move back to $A$ along an edge. M1

There are $3^{n} - a_{n}$ walks of length $n$ from $A$ that do not end at $A$ . A1 Each of them ends at a vertex joined to $A$ , so it extends to a closed walk in exactly one way. R1 So $a_{n + 1} = 3^{n} - a_{n}$ .

(e) Show that $a_{n} = \frac{3^{n} + 3 (- 1)^{n}}{4}$ gives $a_{1} = 0$ and satisfies the relation in (d), and check it against your answer to (a). [3]

Worked solution (e)

$a_{1} = \frac{3 - 3}{4} = 0$ , which is right: a walk of length 1 cannot return to $A$ . A1

$3^{n} - a_{n} = \frac{4 \times 3^{n} - 3^{n} - 3 (- 1)^{n}}{4} = \frac{3^{n + 1} + 3 (- 1)^{n + 1}}{4} = a_{n + 1}$ M1 A1

Check: $a_{2} = \frac{9 + 3}{4} = 3$ , $a_{3} = \frac{27 - 3}{4} = 6$ and $a_{4} = \frac{81 + 3}{4} = 21$ , the diagonal entries in (a).

(f) Conjecture a formula for the number of closed walks of length $n$ from one vertex of $K_{m}$ . Test it with your GDC for $K_{5}$ and $n = 4$ . [2]

Worked solution (f)

Replace the degree 3 by $m - 1$ and the number of vertices 4 by $m$ : $a_{n} = \frac{(m - 1)^{n} + (m - 1) (- 1)^{n}}{m}$ . A1

For $K_{5}$ and $n = 4$ : $\frac{4^{4} + 4}{5} = 52$ , and the GDC gives 52 on the diagonal of $M^{4}$ . A1

The argument in (d) works for any $m$ and gives $a_{n + 1} = (m - 1)^{n} - a_{n}$ , which this formula satisfies. So it is more than a guess.

What this shows: The GDC does the heavy arithmetic in (a), but the marks in (c) and (d) are for explaining why. Say what you are counting, in words, before you write a formula. 14 marks

Investigation 2: Cooling tea: straightening data to build a model

Use your GDC
Test the model
Communicate

A cup of tea is left in a room at $20^{\circ} C$ . Its temperature $T^{\circ} C$ is measured $t$ minutes later.

$t$ (minutes)	0	5	10	15	20	30
$T$ ( $^{\circ} C$ )	90.0	74.2	62.8	52.9	46.1	35.4

Newton's law of cooling suggests the model $T = 20 + A e^{- k t}$ , where $A$ and $k$ are positive constants.

(a) Show that, if the model is right, the points $(t, \ln (T - 20))$ lie on a straight line. State its gradient and $y$ -intercept in terms of $A$ and $k$ . [2]

Worked solution (a)

$T - 20 = A e^{- k t} \Rightarrow \ln (T - 20) = \ln A - k t$ . M1

This is $y = m x + c$ with $y = \ln (T - 20)$ and $x = t$ : a straight line with gradient $- k$ and $y$ -intercept $\ln A$ . A1

(b) Use your GDC to find the equation of the regression line of $\ln (T - 20)$ on $t$ . Hence estimate $A$ and $k$ . [4]

Worked solution (b)

Put $t$ in List 1, $T$ in List 2, and $\ln (List 2 - 20)$ in List 3: $4.248, 3.993, 3.757, 3.493, 3.262, 2.734$ . M1 A1

Linear regression of List 3 on List 1: $\ln (T - 20) = 4.25 - 0.0502 t$ . A1

So $k \approx 0.0502$ and $A \approx e^{4.251} \approx 70.2$ , giving $T = 20 + 70.2 e^{- 0.0502 t}$ . A1

Use the GDC's unrounded intercept: $e^{4.25}$ would give $70.1$ .

(c) The tea is pleasant to drink once it has cooled to $50^{\circ} C$ . Use your model to find when this happens. [2]

Worked solution (c)

$20 + 70.2 e^{- 0.0502 t} = 50 \Rightarrow t = \frac{1}{0.0502} \ln \frac{70.2}{30} \approx 16.9$ minutes. M1 A1 (Or solve it with the GDC's graph or solver.)

(d) Find the correlation coefficient for $\ln (T - 20)$ and $t$ , and comment on what it tells you about the model. [2]

Worked solution (d)

$r \approx - 0.9998$ . A1 This is very close to $- 1$ , so the transformed points lie very close to a straight line: strong evidence that the model fits these data. R1

It does not prove the model is right. It only shows the model fits the times that were measured.

(e) A student uses the GDC's exponential regression on $T$ against $t$ instead, and gets $T = 87.0 \times {0.9695}^{t}$ . Use both models to predict the temperature after one hour, and explain which model is more realistic. [3]

Worked solution (e)

Newton's law model: $20 + 70.2 e^{- 0.0502 \times 60}$ , about $23^{\circ} C$ . A1

The student's model: $87.0 \times {0.9695}^{60}$ , about $14^{\circ} C$ . A1

The tea cannot cool below the room's $20^{\circ} C$ . The student's model tends to 0 as $t$ grows, but $T = 20 + A e^{- k t}$ tends to 20, so Newton's law model is more realistic for large $t$ . R1

What this shows: Straightening data with a log is one of the most useful GDC moves in AI. The last part is typical too: a model that fits the data can still give nonsense outside it, and saying why earns the reasoning mark. 13 marks

Full-length practice questions

Two full-length questions in our Paper 3 style: Voronoi diagrams and network algorithms for a chain of shops; then a rumour spreading through a school, modelled with a differential equation and tested against data. Together they make a full paper: sit them in 1 hour 15 minutes, then check each part.

Question 1 · Planning a coffee-shop network

Topics: Voronoi, perpendicular bisectors, graph theory, MST · 28 marks

A coffee-chain owner has three existing shops in a city at:

$A = (4, 2), B = (10, 4), C = (8, 10)$ (units = km).

She wants to open a fourth shop and to plan a delivery network between all four locations. Travel times (min) along direct roads are given later in the question.

(a) [3 marks]

Find the equation of the perpendicular bisector of $[A B]$ in the form $a x + b y + c = 0$ .

Reveal worked solution

Midpoint of $A B$ : $M = (7, 3)$ . (M1)

Gradient of $A B$ : $m_{A B} = \frac{4 - 2}{10 - 4} = \frac{1}{3}$ ; perpendicular gradient $= - 3$ . (M1)

y - 3 = - 3 (x - 7) \Rightarrow 3 x + y - 24 = 0.

(A1)

(b) [3 marks]

Find the equation of the perpendicular bisector of $[B C]$ .

Reveal worked solution

Midpoint of $B C$ : $(9, 7)$ . Gradient of $B C$ : $\frac{10 - 4}{8 - 10} = - 3$ ; perpendicular gradient $= \frac{1}{3}$ . (M1 · M1)

y - 7 = \frac{1}{3} (x - 9) \Rightarrow x - 3 y + 12 = 0.

(A1)

Hence find the point $P$ equidistant from $A$ , $B$ and $C$ , and calculate its distance from $A$ .

Reveal worked solution

Solve simultaneously: $y = - 3 x + 24$ and $x - 3 y + 12 = 0$ .

x - 3 (- 3 x + 24) + 12 = 0 \Rightarrow 10 x - 60 = 0 \Rightarrow x = 6, y = 6.

(M1 · A1)

| P A | = \sqrt{2^{2} + 4^{2}} = \sqrt{20} \approx 4.47 km .

(A1)

(d) [3 marks]

Explain in Voronoi terms what point $P$ represents, and where the owner should place a fourth shop $D$ if she wants to maximise the size of shop $A$ 's Voronoi cell.

Reveal worked solution

Interpretation. $P$ is the Voronoi vertex where the cells of $A$ , $B$ , $C$ meet — the point equidistant from all three shops. (R1)

Placement of $D$ . To maximise $A$ 's cell, place $D$ as far from $A$ as possible and far from where the customer base concentrates. Practically: place $D$ on the far side of $B$ or $C$ (e.g. north-east of $C$ ), so its Voronoi cell steals area from $B$ and $C$ rather than from $A$ . (R1 · A1)

(e) [5 marks]

The owner places the fourth shop at $D = (10, 10)$ . Travel times (min) between the four shops are: $A B = 12, A C = 20, B C = 15, A D = 22, B D = 10, C D = 8$ . Use Prim's algorithm starting at vertex $A$ to find the minimum spanning tree (MST). State the edges chosen in order and the total weight.

Reveal worked solution

Step 1. From $A$ , cheapest edge: $A B = 12$ . Add $B$ . (M1)

Step 2. From ${A, B}$ : edges to $C$ ( $A C = 20, B C = 15$ ) and $D$ ( $A D = 22, B D = 10$ ). Cheapest: $B D = 10$ . Add $D$ .

Step 3. From ${A, B, D}$ : edges to $C$ ( $A C = 20, B C = 15, C D = 8$ ). Cheapest: $C D = 8$ . Add $C$ . (A1 · A1)

MST edges (in order):

A B, B D, D C

. Total weight =

12 + 10 + 8 = 30

min. (A1 · A1)

(f) [5 marks]

The owner now wants a daily circuit: start at $A$ , visit every shop exactly once, return to $A$ . Use the nearest-neighbour algorithm starting at $A$ to find an upper bound for the shortest circuit.

Reveal worked solution

Nearest-neighbour trace.

$A \to$ nearest unvisited $= B$ (12). Path: $A, B$ . (M1)
$B \to$ nearest unvisited $= D$ (10). Path: $A, B, D$ .
$D \to$ nearest unvisited $= C$ (8). Path: $A, B, D, C$ .
Return to $A$ : $C \to A = 20$ .

Total:

12 + 10 + 8 + 20 = 50

min. Upper bound

= 50

min. (A1 · A1 · A1 · A1)

(g) [6 marks]

Now use the deleted-vertex algorithm (deleting $A$ ) to find a lower bound for the shortest Hamiltonian cycle. Compare with (f) and comment on whether the nearest-neighbour tour is optimal.

Reveal worked solution

Step 1. Remove $A$ . Remaining vertices: ${B, C, D}$ with edges $B C = 15, B D = 10, C D = 8$ .

Step 2. MST of the remaining graph (Prim from $B$ ): $B D = 10$ , then $D C = 8$ . MST weight $= 18$ . (M1 · A1)

Step 3. Add the two shortest edges from $A$ : $A B = 12$ and $A C = 20$ (skipping $A D = 22$ ). (M1 · A1)

Lower bound

= 18 + 12 + 20 = 50

min. (A1)

Comment. Upper bound = lower bound = 50, so the nearest-neighbour tour $A \to B \to D \to C \to A$ is optimal. (R1)

Question 2 · Modelling a rumour spreading through a school

Topics: differential equations, logistic model, regression, R² · 27 marks

A rumour spreads through a school of $N = 800$ students. Let $x (t)$ be the number who know the rumour after $t$ hours. Assume the rate of spread is proportional both to the number who know and the number who don't:

\frac{d x}{d t} = k x (N - x), x (0) = 1.

Below is data recorded during one Wednesday morning.

$t$ (hours)	0	1	2	3	4	5
$x$	1	14	140	540	760	795

(a) [4 marks]

Solve the differential equation by separation of variables to show that $x (t) = \frac{N}{1 + A e^{- k N t}}$ for some constant $A$ you should identify in terms of $N$ and $x (0)$ .

Reveal worked solution

Partial fractions. $\frac{1}{x (N - x)} = \frac{1}{N} (\frac{1}{x} + \frac{1}{N - x}) .$ (M1)

\int \frac{d x}{x (N - x)} = \frac{1}{N} \ln \frac{x}{N - x} = k t + C .

Rearrange. $\frac{x}{N - x} = e^{k N t + N C} = B e^{k N t}$ for some $B > 0$ . (M1)

x = \frac{N}{1 + \frac{1}{B} e^{- k N t}} = \frac{N}{1 + A e^{- k N t}}, with A = \frac{N - x (0)}{x (0)} .

With $x (0) = 1$ , $A = N - 1 = 799$ . (A1 · AG1)

(b) [4 marks]

Use the data point $(t, x) = (3, 540)$ to estimate $k$ to 3 significant figures.

Reveal worked solution

540 = \frac{800}{1 + 799 e^{- 2400 k}} \Rightarrow 1 + 799 e^{- 2400 k} = \frac{800}{540} = 1. \overset{―}{481}

799 e^{- 2400 k} = 0.481 \Rightarrow e^{- 2400 k} = 6.023 \times 10^{- 4} .

(M1 · A1)

- 2400 k = \ln (6.023 \times 10^{- 4}) = - 7.416 \Rightarrow k \approx 3.09 \times 10^{- 3} .

(M1 · A1)

Using the model with your $k$ , predict $x (4)$ and compare with the observed value $760$ .

Reveal worked solution

x (4) = \frac{800}{1 + 799 \cdot e^{- 3.09 \times 10^{- 3} \cdot 800 \cdot 4}} = \frac{800}{1 + 799 \cdot e^{- 9.888}} \approx \frac{800}{1 + 0.0410} \approx 768.

(M1 · A1)

Model predicts $\approx 768$ ; observed $= 760$ . Error $= 8$ (about $1 %$ ) — very good fit. (A1)

(d) [3 marks]

At what time is the rumour spreading fastest? Interpret this in the context.

Reveal worked solution

Max spread rate. $\frac{d x}{d t} = k x (N - x)$ is maximised when $x = N / 2 = 400$ . (M1)

Solve $400 = \frac{800}{1 + 799 e^{- 2.472 t}} \Rightarrow e^{- 2.472 t} = \frac{1}{799} \Rightarrow t \approx 2.7$ hours. (A1)

Interpretation: the rumour spreads fastest around the mid-morning break, when about half the school knows and half don't. (R1)

(e) [5 marks]

A researcher proposes an alternative model: exponential growth $y (t) = e^{c t}$ . Using GDC regression on all 6 data points, the researcher obtains $c \approx 1.90$ and $R^{2} \approx 0.72$ . Compare the two models. Which is better? Justify quantitatively and conceptually.

Reveal worked solution

Quantitative comparison. The logistic model gave a $1 %$ residual at $t = 4$ (part c). Computing the logistic prediction at all six data points gives $R^{2} \approx 0.9998$ . Exponential regression: $R^{2} = 0.72$ — much lower. (M1 · A1)

Conceptual reasoning. Exponential growth $e^{c t}$ has no ceiling — it would predict $x (5) \approx e^{9.5} \approx 13,400$ , hugely overshooting the school population of $800$ . (R1)

The logistic model has the carrying-capacity $N = 800$ built in, reflecting the fact that once everyone knows, spread stops. (R1)

Conclusion: the logistic model is superior — higher $R^{2}$ and physically meaningful bound. (A1)

(f) [4 marks]

Two students, initially resistant, refuse to believe the rumour and never adopt it. Modify the differential equation and describe qualitatively how the long-run behaviour changes.

Reveal worked solution

Modified model. Effective susceptible population $= N - 2 = 798$ . New equation:

\frac{d x}{d t} = k x (798 - x), x (0) = 1.

(M1)

Long-run behaviour. As $t \to \infty$ , $x (t) \to 798$ , not $800$ . Two students remain permanently uninformed. (M1 · A1)

The shape (S-curve, mid-point at $x = 399$ ) is essentially unchanged; only the ceiling shifts down by 2. This is why real epidemic / diffusion models often add "immune" or "resistant" compartments. (R1)

(g) [4 marks]

Suggest one further real-world factor not in the model that would change the shape (not just the ceiling) of the curve, and describe how the modified equation would look.

Reveal worked solution

Example factor: forgetting / disproving. Some students who "knew" the rumour lose interest or find out it's false, so $x$ can decrease. Add a linear decay:

\frac{d x}{d t} = k x (N - x) - μ x

, where

μ > 0

is the forgetting rate. (M1 · A1)

Behaviour: still an S-curve initially, but instead of levelling at $N$ , the population settles at a steady state $x^{*} = N - \frac{μ}{k}$ , where new "learners" balance "forgetters". If $μ \geq k N$ , no epidemic ever takes off. This is the classical SIS model from epidemiology. (R1 · A1)

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Full Paper 3s to sit

Once the techniques feel natural, sit a whole Paper 3 against the clock: 1 hour 15 minutes, two questions, then mark it line by line.

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Full Paper 3s we wrote in the format of the current IB guide, each with a full markscheme (M, A and R marks) and worked solutions. Every answer is checked twice, the second time by an independent solver.

November 2026 Paper 3 1 hour 15 minutes · 55 marks · 2 questions · out 14 October 2026
May 2027 Paper 3 1 hour 15 minutes · 55 marks · 2 questions · out 18 April 2027
New-syllabus practice Paper 3 (first exams May 2029) 1 hour · 50 marks · 2 questions · out 4 October 2026

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AI HL Paper 3 practice questions · The two full-length questions on this page, with worked solutions.
Interactive Practice Paper 3 · Five questions you answer on screen, with instant feedback and a timed mode. No sign-up.
Paper 3 toolkit · The problem-solving moves for unfamiliar contexts, with tips for when you are stuck.
How to prepare for Paper 3 · A deeper guide for both HL courses, with a walkthrough and a section for teachers.
Paper 3 to IA · Two step-by-step investigations, with the IA criterion each step earns.

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Common questions

How long is IB Maths AI HL Paper 3?

1 hour 15 minutes, with a graphic display calculator (GDC) and the formula booklet. It is two compulsory extended-response problem-solving questions and counts for 20% of your HL grade.

Can I use a calculator in Paper 3?

Yes. Paper 3 is a technology-allowed paper, so you need your GDC. Use it to explore cases, check answers and do heavy calculations, but write down what you did.

What topics come up on AI HL Paper 3?

Any topic in the course can come up, including the SL core, and one question can join several: matrices with graph theory, say, or statistics with a model you build from data. Each question introduces an idea and leads you through it, so you are tested on using what you know in a new setting.

How do I revise for Paper 3?

Learn the four techniques on this page, then work through investigation-style questions slowly with the solutions. Once that feels natural, sit full Paper 3s timed and mark them against the markscheme.

Is Paper 3 changing?

For the new courses (first exams in May 2029), the IB has announced that HL Paper 3 will be 1 hour instead of 1 hour 15 minutes. Exams before then keep the current format.

Are these official IB questions?

No. Every question on this page is our own, written in the style of Paper 3. This site is independent and is not connected to the IB.

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What Paper 3 is

How Paper 3 is marked

M: method

A: accuracy

R: reasoning

AG: answer given

The techniques that score

Spot the pattern

Conjecture, then justify

Use your GDC

Communicate

Two worked mini-investigations

Investigation 1: Closed walks in a complete graph

Investigation 2: Cooling tea: straightening data to build a model

Full-length practice questions

Question 1 · Planning a coffee-shop network

Question 2 · Modelling a rumour spreading through a school

Full Paper 3s to sit

AI HL Prediction Paper 3s

AI HL Mock A Paper 3

More Paper 3 practice

Practise Paper 3 properly: free account

Common questions

More for AI HL