BCS CTFL4 Exam With Confidence Using Practice Dumps

This answer is correct because configuration management is a process of establishing and maintaining consistency of a product’s performance, functional, and physical attributes with its requirements, design, and operational information throughout its life. Configuration management helps ensure that all relevant project documentation and software items are uniquely identified in all their versions and therefore can be unambiguously referenced in test documentation. This supports testing by providing traceability, consistency, and control over the test artifacts and the software under test. References: : ISTQB Glossary of Testing Terms v4.0, : ISTQB Foundation Level Syllabus v4.0, Section 2.2.2.2

This answer is correct because equivalence partitioning is a test design technique that divides the input domain of a system or component into partitions of equivalent data, such that each partition is expected to produce the same output or behavior. Equivalence partitioning aims to reduce the number of test cases by selecting one representative value from each partition. In this case, the input domain of the TAS can be divided into four partitions based on the bonus rules: less than 50k€, between 50k€ and 80k€, between 80k€ and 120k€, and more than 120k€. The test cases in the answer belong to the same partition, which is between 80k€ and 120k€, and they are expected to produce the same output, which is a bonus of 15%. References: ISTQB Glossary of Testing Terms v4.0, ISTQB Foundation Level Syllabus v4.0, Section 2.3.2.1

To apply one-dimensional equivalence partitioning to the ID requirement, we need to consider each condition individually:

Length of the ID: Valid partitions (5-10 characters), Invalid partitions (less than 5, more than 10) = 3 partitions.

Type of characters: Valid partitions (alphanumeric), Invalid partitions (non-alphanumeric) = 2 partitions.

First character: Valid partitions (letter), Invalid partitions (non-letter) = 2 partitions. Adding these partitions, we get a total of 3 (length) + 2 (character type) + 2 (first character) = 7 partitions. Thus, the correct answer is A​​.