Paper 3: extended problem solving

Tackle the two long Paper 3 problems by reading the stem as it unfolds, building from simple cases to a general result, working with parameters and using "show that" parts to re-enter.

IB Mathematics: Applications and Interpretation
Exam skills
HL only

13 min readLast reviewed 15 Oct 2026

Introduction

Paper 3 is taken only by HL students. It has two compulsory extended problem-solving questions in one hour, and each question follows a single theme from start to finish. A question typically starts with routine mathematics from the course, applies it to a new situation, and ends with a generalisation, a result involving a parameter, or an interpretation of the context. The marks reward sustained reasoning and inquiry more than recall.

The contexts are often new to candidates, while the mathematics used in them comes from the course. Questions can draw on any topic, SL as well as HL: real Paper 3 questions have used the cosine rule, Voronoi diagrams and linear regression alongside differential equations and graph algorithms. What the paper adds is the length of the chain of reasoning, and the skill of using each part to answer the next.

How a Paper 3 question is built

In the sessions from 2021 to 2022, each Paper 3 had two questions of between 24 and 31 marks, and each question had between 10 and 18 parts worth 1 to 7 marks each. The two questions need not be the same length. With 55 marks in 60 minutes, the time budget is just under 1.1 minutes per mark, so a question worth 28 marks deserves about 30 minutes.

\frac{60}{55} \approx 1.09 min per mark 28 \times 1.09 \approx 30.5 min

Each question opens with a framing sentence of the form "This question explores …", which states the theme and often the goal. The context is then introduced in stages: a first block of information and a few parts, then new information ("The analyst now considers …"), then more parts. By the end the question has usually moved from a specific case to something general.

Stage	What it usually asks	Example of the kind of part
Opening	Routine course mathematics on the given data or diagram	Write down a Hamiltonian cycle; find a regression line; find a mean and variance
Development	Course techniques applied to the context, often across topics	Use an algorithm on a modified graph; solve a differential equation; set up a test
Generalisation	A result in terms of a parameter or for a general case	Find a bound "in terms of $x$ "; show that a general solution has a given form
Interpretation	What the result means; a criticism or a design decision	Find the least value of a parameter for which something changes; design a device; criticise the model

The arcs of real Paper 3 questions show the pattern. One question took a network of tourist attractions with one unknown travel time $x$ : it asked for a Hamiltonian cycle, a lower bound in terms of $x$ , an upper bound for a specific value of $x$ and then for all larger values, and finally the least $x$ for which one route would never be used. Another started from data on water draining from a cylinder, fitted a linear and then a quadratic regression model, then derived the draining rate from a physical law with a differential equation, and ended by designing a 15-minute water timer. A third used linear regression, an exponential model and then a logistic model to describe the spread of a computer virus.

Reading a long stem

A Paper 3 question can take three or four minutes to read in full, and that time is well spent. On the first read, the aim is to understand the situation and the goal, not to solve anything. On later reads, each block of information is read just before the parts that use it.

Several habits help.

Write a short summary of the definitions. A list such as " $t$ = days since first infection; $Q (t)$ = total infected; $N$ = 2.6 million computers" kept at the top of the page saves rereading and stops variables being mixed up.
Mark where new information arrives. A line across the page or a note in the margin ("new: second container") shows which parts each fact belongs to.
Keep the question's letters. Paper 3 markschemes penalise equations that switch between $y$ , $x$ , $Q$ and $t$ for the same quantities, and a missing or changed variable can cost the mark for an otherwise correct equation.
Check the domain. A model "for $0 \leq t \leq 60$ " or "while the tank contains water" limits every later answer.

Using earlier parts

Paper 3 is the paper on which "use the earlier parts" matters most. Most parts depend on something already found, and the question is designed so that each step is short when the previous results are used and long or impossible when they are not.

Specific cases first

When a question works through specific cases ( $x = 3$ , then $x > 3$ ; one container, then two; a 2 × 2 example, then $n \times n$ ), the cases are a guide to the general answer. Write each case out fully, with the same layout, so that the parts which change and the parts which stay fixed are easy to see. A general expression should always reduce to the earlier specific answers when their values are substituted, which makes a quick and reliable check.

Show that parts as re-entry points

Paper 3 questions use "show that" parts heavily, and many of them exist to let candidates continue. A part such as "show that the general solution is $h = \dots$ " gives every candidate the result needed for the rest of the question. If you cannot complete the derivation, take the given result and use it: the later parts are marked independently. Examiners' reports on Paper 3 make the point directly: the purpose of "show that" parts is often to let candidates re-enter a question.

Working with parameters

Parameters are a central feature of Paper 3. A part may ask for a bound "in terms of $x$ ", a value of $k$ for which a model changes behaviour, or a general solution containing an arbitrary constant. Answers to such parts are expressions, not numbers, and replacing the parameter by a convenient value loses the marks even when the arithmetic is right. Treat the parameter as a known number throughout: carry it through each step exactly as you would a value such as 3 or 0.005.

Parameters also lead naturally to the last kind of part: for which values of the parameter does something happen? These parts are usually answered by setting up an inequality or an equation in the parameter, and the answer often has a meaning in context (a threshold, a break-even point, a least or greatest value).

A Paper 3 question worked in full

The question that follows is written in the style of Paper 3. It uses SL and HL content together: the finance solver (subtopic 1.4), geometric series (1.3), the laws of logarithms (1.9) and interpretation of a parameter. The loan is hypothetical.

This question explores how the time taken to repay a loan depends on the size of the monthly repayment.

Sam borrows $12,000. Interest of 0.5% is charged on the amount owed at the end of each month, after which Sam makes a repayment of $ $M$ . Let $A_{n}$ be the amount owed, in dollars, immediately after the $n$ th repayment.

(a) In the case $M = 300$ , use technology to find the number of repayments needed to repay the loan. [2]

(b) Show that $A_{1} = 12 000 (1.005) - M$ . [1]

(d) Write down an expression for $A_{n}$ , in the style of part (c). [1]

(e) Hence show that $A_{n} = 12 000 (1.005)^{n} - 200 M ({1.005}^{n} - 1)$ . [3]

(f) Find the value of $M$ for which the loan is repaid in exactly 48 months. [2]

(g) Show that the loan is repaid after $n$ repayments when ${1.005}^{n} = \frac{M}{M - 60}$ . [3]

(h) Explain why the loan is never repaid if $M \leq 60$ , and interpret the value 60 in context. [2]

(i) Hence find an expression for $n$ in terms of $M$ , and use it to verify your answer to part (a). [3]

Show the markscheme

(a) $I % = 6, P V = 12 000, P M T = - 300, F V = 0, P / Y = 12, C / Y = 12$ (M1)

$N = 44.74 \dots$ , so 45 repayments A1

Note: Accept $I % = 0.5$ with $P / Y = C / Y = 1$ . Award A0 for 44, which leaves money owing.

(b) $A_{1} = 12 000 \times 1.005 - M$ (interest added, then the repayment subtracted) A1AG

$= 1.005 (12 000 (1.005) - M) - M = 12 000 (1.005)^{2} - 1.005 M - M$ A1

$= 12 000 (1.005)^{2} - M (1 + 1.005)$ AG

(d) $A_{n} = 12 000 (1.005)^{n} - M (1 + 1.005 + {1.005}^{2} + \dots + {1.005}^{n - 1})$ A1

(e) the bracket is a geometric series with $u_{1} = 1$ , $r = 1.005$ and $n$ terms M1

$1 + 1.005 + \dots + {1.005}^{n - 1} = \frac{{1.005}^{n} - 1}{1.005 - 1}$ A1

$= \frac{{1.005}^{n} - 1}{0.005} = 200 ({1.005}^{n} - 1)$ A1

so $A_{n} = 12 000 (1.005)^{n} - 200 M ({1.005}^{n} - 1)$ AG

(f) $12 000 (1.005)^{48} - 200 M ({1.005}^{48} - 1) = 0$ or TVM with $N = 48$ , $F V = 0$ (M1)

$M = $ 281.82$ $(281.820 \dots)$ A1

(g) $A_{n} = 0$ : $12 000 (1.005)^{n} = 200 M ({1.005}^{n}) - 200 M$ M1

$200 M = {1.005}^{n} (200 M - 12 000)$ A1

${1.005}^{n} = \frac{200 M}{200 M - 12 000} = \frac{M}{M - 60}$ A1AG

(h) if $M \leq 60$ , the right-hand side is negative or undefined, but ${1.005}^{n} > 0$ , so no $n$ satisfies the equation R1

60 is the interest charged in the first month (0.5% of $12,000); a repayment of $60 or less never reduces the amount owed A1

(i) $n = \frac{\ln (\frac{M}{M - 60})}{\ln 1.005}$ M1A1

$M = 300$ : $n = \frac{\ln 1.25}{\ln 1.005} = 44.74 \dots$ , which agrees with part (a) A1

Total [19]

The question follows the standard arc. Part (a) is a specific case solved with technology. Parts (b) to (e) build the general formula, starting from the first two months so that the pattern in part (d) can be seen. Parts (f) and (g) use the general formula, and parts (h) and (i) ask what the parameter means and connect the general result back to the specific case. Parts (b), (c), (e) and (g) are all "show that" parts: a candidate who cannot derive the formula in (e) can still use it in (f), and one who cannot do (g) can still answer (h) and (i) from the given equation.

The general result in part (i) also explains a feature of loans that the single case in part (a) hides. The graph of $n$ against $M$ has a vertical asymptote at $M = 60$ : as the repayment falls towards the monthly interest, the time to repay grows without limit.

n = ln(M/(M − 60)) ÷ ln 1.005

Doubling the repayment from $150 to $300 cuts the repayment time from about 102 months to about 45, more than halving it, and repayments near $60 take ever longer.

The graph of the general result shows why the question asks for a threshold in part (h). Doubling the monthly repayment from $150 to $300 more than halves the time to repay, because a larger repayment also reduces the interest charged in every later month. A graph like this is a natural way to interpret a general result if a final part asks you to comment on it.

Writing for a long problem

Paper 3 markschemes credit method and reasoning heavily, so the working should read as a short argument. Brief words between lines ("interest is added before the repayment", "the bracket is a geometric series with $r = 1.005$ ") make the reasoning visible and earn method marks even when an arithmetic slip spoils the answer. Examiners' reports note that a correct method is easier to credit when there is appropriate working or a running commentary.

When technology is used, describe the mathematics, not the buttons. "Solving $A_{48} = 0$ for $M$ " or "using exponential regression on $Q$ against $t$ " is the right register; "pressed solver" or a calculator command is not. Data or values entered into a calculator for a test should still be written down, so that follow-through can be given if the final value is wrong.

Answers to the final interpretation parts follow the same rules as on Paper 2: a reason and a conclusion, both in the words of the context. On Paper 3 these parts often ask about the validity of a model or a method: whether a hypothesis test on $r$ is appropriate, whether a model can be used for large values of time, whether a test measures what it claims to. Specific reasons (" $t$ is not a random variable, so a test on $r$ does not apply"; "the model predicts more infected computers than exist") earn the marks; general ones ("the model is not realistic") do not.

Managing the hour

Two questions in 60 minutes leaves little slack. A practical plan is to spend about 2 minutes reading both framing sentences and stems, then about 28 to 30 minutes on each question. Starting with the question whose theme feels more familiar is reasonable, but neither question should be allowed to take 40 minutes, because the opening parts of the second question are the most accessible marks remaining.

Within each question, the end of each block of information is a natural checkpoint. If the first block has taken more than half the time for the question, move on to the next block, which often begins with a fresh "show that" or a new piece of data that does not depend on the parts left unfinished.

Summary

Paper 3 (HL only) has two problem-solving questions of about 24–31 marks each, in 60 minutes for 55 marks: budget about 30 minutes per question.
Each question follows one theme, introduced in stages, and moves from routine course mathematics to a generalisation or an interpretation.
Questions can use any topic of the course, SL as well as HL.
Read the framing sentence for the goal; summarise definitions; mark where new information arrives; keep the question's variables.
Work specific cases in a consistent layout so the general pattern is visible, and check general results against them.
"Show that" parts and given results are re-entry points: use them even if you could not derive them.
Keep parameters as parameters; a number in place of an expression does not answer the part.
Explain reasoning in words between lines and write any data entered into technology, so method and follow-through marks can be awarded.

Self-check

To do:
I can describe the usual arc of a Paper 3 question from specific case to generalisation.
To do:
I can read a long stem and summarise its definitions and conditions before starting.
To do:
I can use a given "show that" result to continue a question after a part I could not do.
To do:
I can carry a parameter through a calculation and give an answer in terms of it.
To do:
I can check a general expression by substituting an earlier specific case.
To do:
I can find the value of a parameter at which a model changes behaviour, and interpret it.
To do:
I can compare two models using a stated measure such as $R^{2}$ .
To do:
I can plan the hour so that both questions get a fair share of the time.

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