CTFL 4.2.1: Use equivalence partitioning to derive test cases
K3Syllabus section 4.2.1
5 original PrepBench practice questions for learning objective 4.2.1 of the CTFL 4.0 syllabus. Try the examples below and check each answer against its syllabus reference.
Practice questions
Question 1
An input field accepts integer order quantities: Quantities from 1 to 100 are processed at the standard rate. Quantities from 101 to 500 are processed with a bulk surcharge. Any other quantity is rejected. Using equivalence partitioning, and including the invalid partitions, how many partitions must the test cases cover?
- 3
- 4
- 5
- 6
Show answer
- 3Three would treat the two valid ranges as a single partition.
- Correct answer: 4Two valid partitions and two invalid partitions.
- 5Five would split one of the four partitions without a rule requiring it.
- 6Six would add partitions the specification does not distinguish.
In equivalence partitioning the coverage items are the equivalence partitions, and 100% coverage requires test cases to exercise all identified partitions, including the invalid ones, at least once. The field has two valid partitions, 1 to 100 and 101 to 500, and two invalid partitions, quantities below 1 and quantities above 500 — four in total.
Syllabus section 4.2.1
Question 2
An online shop gives no discount for purchases below 50, a 5% discount from 50 up to and including 99.99, and 10% from 100 upwards. Which test set covers each of these three valid partitions exactly once?
- 25, 75, 150
- 50, 99.99, 100
- 0, 49.99, 50
- 75, 80, 95
Show answer
- Correct answer: 25, 75, 15025, 75, and 150 each land in a different partition, covering all three exactly once.
- 50, 99.99, 10050 and 99.99 both represent the middle partition, while the first partition is omitted.
- 0, 49.99, 500 and 49.99 are both in the first partition, so a partition is covered twice.
- 75, 80, 9575, 80, and 95 all fall in the 5% partition, leaving two partitions untested.
The three partitions are [0, 50), [50, 100), and [100, ∞). The values 25, 75, and 150 each fall into a different partition, achieving 100% partition coverage with three tests. A boundary value is still a member of one partition; the other sets repeat partitions and omit others.
Syllabus section 4.2.1
Question 3
An online shop validates the order quantity for one article: Quantity 1 to 9: processed at the single-item price Quantity 10 to 99: processed at the bulk price Quantity 100 or more: rejected with “too many” Quantity 0 or less: rejected with “invalid quantity” Which quantities exercise every equivalence partition exactly once?
- Quantities 1, 9, 10 and 99
- Quantities 5, 50 and 150
- Quantities 0, 5, 50 and 150
- Quantities 5, 20, 50 and 150
Show answer
- Quantities 1, 9, 10 and 99These are the boundary values of the two valid partitions; both invalid partitions stay uncovered.
- Quantities 5, 50 and 150This covers 1–9, 10–99 and 100 and above, but leaves the invalid partition 0 and below uncovered.
- Correct answer: Quantities 0, 5, 50 and 150One value from each of the four partitions, including both invalid ones.
- Quantities 5, 20, 50 and 15020 and 50 both lie in 10–99, so that partition is exercised twice while 0 and below stays uncovered.
The requirement defines four non-overlapping partitions: the valid partitions 1–9 and 10–99, and the invalid partitions 100 and above and 0 and below. In equivalence partitioning the coverage items are the partitions, and to achieve 100% coverage the test cases must exercise all identified partitions, including the invalid ones, at least once. Since all elements of a partition are expected to be processed in the same way, one value per partition is sufficient: 0, 5, 50 and 150.
Syllabus section 4.2.1
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