Given the following fragment of code, how many tests are required for 100% decision coverage? If width > lenth then Biggest_diension = width If height > width then biggest_dimension = height end_if…

Question

Options

• A3
• B4
• C2
• D1

• A
  5% (2)
• B
  84% (37)
• C
  2% (1)
• D
  9% (4)

Explanation

Option B (4 tests) is correct because there are three distinct decision points in this code - width > length, height > width (in the true branch), and height > length (in the else branch) - and each must evaluate to both TRUE and FALSE for 100% decision coverage. Critically, decisions 2 and 3 live in mutually exclusive branches, so you need at least 2 tests inside each branch: one where the inner condition is true, one where it's false. That forces a minimum of 4 tests total.

Why the distractors fail:

• C (2) only gives you one path through the outer if and one through the else, leaving the inner decisions each tested only once (only TRUE or only FALSE, not both).
• A (3) is tempting because there are 3 decisions, but the mutual exclusivity of the two inner decisions means you can't share tests between them - you need a dedicated pair for each nested branch.
• D (1) can only traverse one execution path, covering a fraction of the decisions.

Memory tip: When an if/else contains nested conditions in both branches, those nested decisions can never share a test. Count the leaf-level paths: here there are 4 (T+T, T+F, F+T, F+F), and that count directly equals the minimum tests needed for full decision coverage.

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