ISTQB Foundation Exam Preparation-K Level wise-Q 51 to 60

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ISTQB Foundation Exam Preparation�K Level wise: Q. 51 to 60

The following questions prepared according to different K-Levels are of great help in reviewing your skills before appearing in ISTQB Foundation Level exam as per the Latest Syllabus updated in 2010.

Set of 10 Questions (Q.51 to 60) with explanation in Favor of the correct answer

The Correct Answer has been Hi-lighted in Yellow.

K2- Q. 51: Which of the following is most likely to cause failure in the implementation of a test tool?

A. Underestimating the demand for a tool.
B. The purchase price of the tool.
C. No agreed requirements for the tool.
D. The cost of resources to implement and maintain the tool.

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K2- Q. 52: What benefits do static analysis tools have over test execution tools?

A. Static analysis tools find defects earlier in the life cycle.

B. Static analysis tools can be used before code is written.

C. Static analysis tools test that the delivered code meets business requirements.

D. Static analysis tools are particularly effective for regression testing.

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K2- Q. 53: Which of the following principles should be followed when introducing a test tool into an organization?
(i) Assessing organizational maturity to establish whether a tool will provide expected benefits.
(ii) Requiring a quick payback on the initial investment.
(iii) Including a requirement for the tool to be easy to use without having to train unskilled testers.
(iv) Identifying and agreeing requirements before evaluating test tools.

A. (i) and (ii)
B. (i) and (iv)
C. (ii) and (iii)
D. (iii) and (iv)

Explanation in favor of the correct answer

Assessing organizational maturity (i) is very important when deciding whether to introduce a test tool, as implementing a tool in an immature test organization with poor processes is unlikely to produce any benefits.

A quick return on the initial investment (ii) in a test tool is rare.

Having a requirement that a tool should be easy to use for untrained and unskilled testers (iii) is generally a false hope. This is comparable with expecting someone who has never driven a car to be able to drive safely and effectively.

Agreeing requirements before evaluating tools (iv) is essential. Not to do so would be comparable with building and testing a system without requirements.

In conclusion, (i) and (iv) are good principles to follow when introducing a tool and (ii) and (iii) are not.

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K2- Q. 54: How can test execution tools be of most benefit during exploratory testing?

A. They can record user actions so that defects are easier to recreate.

B. They can be used to perform the regression aspects of exploratory testing.

C. They can help to mitigate the risk of low test coverage.

D. They can use data-driven tests to increase the amount of exploratory testing performed.

Explanation in favor of the correct answer

Exploratory testing is used when it is unclear what the system is supposed to do. Therefore test execution tools are of little use because expected results cannot be predicted.

However, the record feature of a test execution tool can be used to log the actions performed so that defects can be recreated (option (A)) and rectified more easily.

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K2- Q. 55: Which of the following types of test tool are most likely to include traceability functions?
(i) Performance testing tool
(ii)Requirements management tool
(iii)Configuration management tool
(iv)Static analysis tool

A. (i) and (ii)
B. (i) and (iv)
C. (ii) and (iii)
D. (iii) and (iv)

Explanation in favor of the correct answer

Requirements management tools (ii) have traceability because they enable test conditions and subsequently test scripts and defects to be traced back to requirements. Configuration management tools (iii) also need to trace the appropriate version of a test script to the release or version of a system or module.

Performance monitoring tools (i) and static analysis tools (iv) are designed for specific objectives. Neither of these tools particularly need traceability functions.

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K3 Level Questions

K3- Q. 56: A system is designed to accept values of examination marks as follows:
Fail :0�39 inclusive
Pass :40�59 inclusive
Merit :60�79 inclusive
Distinction :80�100 inclusive
In which of the following sets of values are all values in different equivalence partitions?

A. 25, 40, 60, 75
B. 0, 45, 79, 87
C. 35, 40, 59, 69
D. 25, 39, 60, 81

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K3 – Q. 57: A washing machine has three temperature bands for different kinds of fabrics: fragile fabrics are washed at temperatures between 15 and 30 degrees Celsius; normal fabrics are washed at temperatures between 31 and 60 degrees Celsius; heavily soiled and tough fabrics are washed at temperatures between 61 and 100 degrees Celsius.
Which of the following contains only values that are in different equivalence partitions?

A. 15, 30, 60
B. 20, 35, 60
C. 25, 45, 75
D. 12, 35, 55

Explanation in favor of the correct answer

Answer (A) includes two values from the lower partition, answer (B) contains two values from the second partition, answer (D) contains one value that is invalid (out of range).

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K3 – Q. 58: Consider the following pseudo code:
1 Begin
2 Read Time
3 If Time < 12 Then
4 Print(Time, “am”)
5 Endif
6 If Time > 12 Then
7 Print(Time −12, “pm”)
8 Endif
9 If Time = 12 Then
10 Print (Time, “noon”)
11 Endif
12 End
How many test cases are needed to achieve 100 per cent decision coverage?

A. 1
B. 2
C. 3
D. 4

Explanation in favor of the correct answer

The three decisions are in sequence and the conditions are all mutually exclusive (if any one is true the others must be false). Hence a test case that makes the first decision true will make the second and third decisions false and so on.

So test case 1 (say Time = 6) would exercise the path True, False, False, test case 2 (say Time = 15) would exercise the path False, True, False. Test case 3 would have to be Time = 12. This combination achieves 100 per cent decision coverage because each decision has been exercised through its true and its false outcomes.

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K4 Level Questions

K4 – Q. 59: Consider the following pseudo code:
1 Begin
2 Read Time
3 If Time < 12 Then
4 Print(Time, “am”)
5 Endif
6 If Time > 12 Then
7 Print(Time −12, “pm”)
8 Endif
9 If Time = 12 Then
10 Print (Time, “noon”)
11 Endif
12 End
If the test cases Time = 11 and Time = 15 were input, what level of decision coverage would be achieved?

A. 100% or 6/6
B. 50% or 3/6
C. 67% or 4/6
D. 83% or 5/6

Explanation in favor of the correct answer

Test case 1 exercises the decision outcomes True, False, False

Test case 2 exercises the decision outcomes False, True, False

This leaves the True outcome of decision 3 not exercised.

Of the 6 possible decision outcomes, 5 have been exercised, so the decision coverage is 5/6 (about 83%).

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K4 – Q. 60: A software component has the code shown below:
Program BiggestA,
Biggest: Integer
Begin
Read A
Biggest = 10
While A > 0
Do
If A > Biggest
Then Biggest = A
Endif
Read A
Enddo
End
The component has exit criteria for component testing that include 100% statement coverage. Which of the following test cases will satisfy this criterion?

A. 0
B. 10, 0
C. 10, 5, 0
D. 10, 11, 0

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