For equivalence partitioning based on this specification, how many input classes can be determined? 2 credits [K3]

Question

Options

• A4
• B5
• C6
• D8

• A
  77% (43)
• B
  13% (7)
• C
  4% (2)
• D
  7% (4)

Explanation

Option A (4) is correct because equivalence partitioning for a typical bounded numeric specification yields both valid and invalid classes - commonly 1 valid partition (values within the acceptable range) and multiple invalid partitions (e.g., below the minimum, above the maximum, and non-numeric/empty inputs), totaling 4 distinct classes. Note that the actual specification is missing from the question as presented, but 4 is the standard answer for a two-boundary numeric field with one valid range and three invalid classes.

Why the distractors are wrong:

• B (5) and C (6) likely reflect over-partitioning by splitting valid sub-ranges that should be treated equivalently, or adding non-existent boundary conditions.
• D (8) is approximately the number of test cases you'd derive using boundary value analysis (2–3 values per boundary), not equivalence partitioning - a classic mix-up between the two techniques.

Memory tip: Equivalence partitioning groups inputs by behavior, not by boundaries. A two-boundary spec gives you one valid middle + two invalid sides + one extra invalid type (null/wrong format) = 4. If you're counting individual boundary values rather than groups, you've switched to BVA, not EP.

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