CTFL_SYLL_4.0 · Question #83
A functional requirement states that pressure values must always be positive and stay within a range of 100 to 200 pounds for a particular tank (100 and 200 are valid within the range). Which of the…
The correct answer is A. A series of tests where pressure is set to 98, 110, and 201 pounds. Equivalence partitioning divides inputs into three distinct classes here: below the valid range (< 100), within the valid range (100–200), and above the valid range (> 200) - and you need exactly one representative value from each. Option A (98, 110, 201) achieves this minimum…
Question
A functional requirement states that pressure values must always be positive and stay within a range of 100 to 200 pounds for a particular tank (100 and 200 are valid within the range). Which of the following is the minimum set of values that would achieve coverage using the equivalence partitioning technique?
Options
- AA series of tests where pressure is set to 98, 110, and 201 pounds
- BA series of tests where pressure is set to 98, 100, and 200 pounds
- CA series of tests where pressure is set to 100, 110, and 200 pounds
- DA series of tests where pressure is set to -1, 0, 175, and 297 pounds
How the community answered
(46 responses)- A72% (33)
- B4% (2)
- C15% (7)
- D9% (4)
Explanation
Equivalence partitioning divides inputs into three distinct classes here: below the valid range (< 100), within the valid range (100–200), and above the valid range (> 200) - and you need exactly one representative value from each. Option A (98, 110, 201) achieves this minimum coverage perfectly: 98 represents the "too low" partition, 110 represents the "valid" partition, and 201 represents the "too high" partition.
Option B (98, 100, 200) fails because both 100 and 200 fall in the valid partition - wasting a test - while the "above 200" partition goes completely untested. Option C (100, 110, 200) tests only the valid partition three times, missing both invalid partitions entirely. Option D (-1, 0, 175, 297) uses four values but puts both -1 and 0 in the same "below range" partition, violating the minimum requirement.
Memory tip: Think "one zone, one test" - sketch three zones (low / valid / high) and pick one value clearly inside each zone, not on the edges (that's boundary value analysis, a different technique).
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