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ISTQB K3 Practice: Chapter 4 Test Design Techniques

This is the narrowest and hardest drill on the site: only the questions that are both chapter 4 and K3. Every one of them gives you something to work on - a range, a set of rules, a state model, a set of acceptance criteria - and expects a derived answer rather than a recognised one.

Every question in this drill comes from chapter 4, Test Analysis and Design, which is 11 of the 40 questions on the real exam. It is narrowed further to K3 only.

The five K3 objectives in chapter 4

Every question in this drill applies one of these to supplied data.

  • FL-4.2.1 - Apply equivalence partitioning to derive test cases
  • FL-4.2.2 - Apply boundary value analysis to derive test cases
  • FL-4.2.3 - Apply decision table testing to derive test cases
  • FL-4.2.4 - Apply state transition testing to derive test cases
  • FL-4.5.3 - Use acceptance test-driven development (ATDD) to derive test cases

Most wrong answers here are the neighbouring concept. See the 2 concepts this drill separates before you start, or work the drill first and use them to check your misses.

Start the drill

Certified Tester Foundation Level (CTFL v4.0)

Practising Chapter 4 and K3 questions only: every question in this set matches that filter. There is no clock - a drill is shorter than a real paper and covers one topic, so timing it would tell you nothing. Work at your own pace and submit when you are ready; you get a full answer review when you finish.

Questions
0
Time limit
None
Pass mark
65%
Cost
Free

No timer on a drill - take as long as you need.

How this level is examined

Five of the syllabus's eight K3 objectives sit in chapter 4, and chapter 4 is already the heaviest chapter at 11 of 40 questions. That concentration is why this drill exists: it is the densest block of applied marks on the exam, and it is the one that most reliably separates candidates who have practised from candidates who have only read.

Work these with a pen. Equivalence partitioning and boundary value analysis are quick once the partitions are written down and error-prone when done in your head - the classic slip is treating a boundary and the value beyond it as one test rather than two, which silently halves your count. Decision tables need the layout right: conditions and actions form the rows, and each column is one rule. State transition questions need the start state identified before you walk anything, and the invalid transitions are usually where the marks are.

ATDD is the fifth objective here and the one candidates least expect to be applied. You can be handed a user story with acceptance criteria and asked which test case correctly derives from them - which is a different skill from describing what ATDD is.

If this drill goes badly but the chapter 4 drill goes fine, your problem is execution under time pressure rather than knowledge, and the fix is repetition on the four procedures rather than re-reading the chapter.

Concepts this drill separates

Separate these and the distractors stop working, because there is nothing plausible left for a wrong answer to be built from.

Equivalence Partitioning versus Boundary Value Analysis

Equivalence partitioning divides data into partitions whose members should be handled alike and takes one value from each; boundary value analysis tests the values at the edges of those partitions, and can only be used where the partition is ordered.

How to tell them apart: The two are used together, so read what the question asks you to count. A representative from inside each partition is partitioning. For boundaries, check first whether the syllabus's 2-value form (the boundary and its nearest neighbour outside) or the 3-value form (the boundary plus its neighbours on both sides) is being asked for - they give different counts.

Work this distinction in full →

Decision Table Testing versus State Transition Testing

Decision table testing derives tests from combinations of conditions that hold at one moment; state transition testing derives them from the sequence of events over time and the states they move the system through.

How to tell them apart: Ask whether order matters. If the same set of inputs in a different order would behave differently, it is a state model. If the outcome depends only on which conditions are true together, it is a decision table.

Work this distinction in full →

Worked examples at K3

These 5 come with the answer and the reasoning already shown, so you can see what this kind of question looks like before you start. They are drawn from the questions-and-answers page and are deliberately held out of the drill above, so the drill still tells you something you did not already know.

  1. 1. A tester wants to achieve 100% valid transition coverage. What is the MINIMUM number of test cases required for the state model below?

    K3

    Document approval state model:

    SubmitApproveRejectReviseDraftApprovedSubmittedRejected
    • A.4
    • B.2
    • C.1
    • D.3

    Correct answer: C

    Draft-Submit-Submitted-Reject-Rejected-Revise-Draft-Submit-Submitted-Approve-Approved traverses all four valid transitions in a single test case. The Revise loop returns to Draft, so no second case is needed.

  2. 2. Which equivalence partition is currently NOT covered by the existing test cases?

    K3
    Order Amount RangeExisting Test Value
    0-9950
    100-499150
    500-999750
    1000+
    • A.100–499
    • B.500–999
    • C.1000+
    • D.0–99

    Correct answer: C

    Each row pairs a partition with the test value that exercises it. The 0-99, 100-499 and 500-999 partitions each have a value (50, 150 and 750), but the 1000+ partition has no test value against it, so it is the uncovered partition.

  3. 3. An input field accepts integer ages from 18 to 65 inclusive. Using equivalence partitioning, how many partitions exist for this field?

    K3
    • A.Two
    • B.Three
    • C.Forty-eight
    • D.Four

    Correct answer: B

    One valid partition (18-65) and two invalid partitions (below 18, above 65). Equivalence partitioning covers the whole value domain, invalid values included.

  4. 4. A decision table has three independent binary conditions. How many rules does the table contain before any simplification?

    K3
    • A.Eight
    • B.Nine
    • C.Three
    • D.Six

    Correct answer: A

    Each of the three conditions has two outcomes, giving 2^3 = 8 combinations. Rules may then be collapsed where a condition is irrelevant.

  5. 5. What is the MINIMUM number of test cases for 100% statement coverage and for 100% branch coverage respectively?

    K3
    if (a > 0) { doSomething(); } finish();
    • A.2 and 2
    • B.1 and 2
    • C.2 and 1
    • D.1 and 1

    Correct answer: B

    One test with a > 0 executes every statement. Branch coverage additionally needs the false outcome, so it requires two.

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