CTFL-Foundation Practice Exam Questions and Answers

A.  

A variable is defined but is then not used

B.  

A variable is used in a calculation before it is defined

C.  

A variable has the wrong numeric value passed into it

D.  

A variable is used but not declared

A.  

Decision table testing

B.  

Exploratory testing

C.  

Statement testing

D.  

Error guessing

A.  

Exploratory testing can replace black box techniques when testing time is very limited

B.  

Test execution scheduling should give priority to experienced based testing

C.  

Specification based techniques can be used as a substitute for a poorly defined test basis

D.  

Experienced based techniques are systematic and produce detailed test documentation

A.  

Decision Coverage

B.  

Error Guessing

C.  

Statement Coverage

D.  

Equivalence Partitioning

A.  

Unit, integration, system and maintenance.

B.  

Functional, glass box, incremental and maintenance.

C.  

Component, integration, system and acceptance.

D.  

Unit, component, functional and alpha/beta.

A.  

Use case testing

B.  

Boundary value testing

C.  

Statement testing and coverage

D.  

State transition testing

A.  

The total number of tests planned

B.  

The total number of requirements to be tested

C.  

The failure rate in testing already completed

D.  

The number of testers used for test execution so far

A.  

Likelihood of failure = 1%, potential cost of impact = $1m.

B.  

Likelihood of failure = 10%, potential cost of impact = $500,000.

C.  

Likelihood of failure = 20%, potential cost of impact = $150,000.

D.  

Likelihood of failure = 5%, potential cost of impact = $500,000.

A.  

At the component test level only

B.  

At all test levels

C.  

At integration testing and system testing levels only

D.  

At all levels from integration testing to acceptance testing

A.  

(i) = S1 – S2 – S3 – S4 – S2 and (ii) = S4 – S2 – S4 – S4 – S4

B.  

(i) = S1 – S2 – S3 – S4 – S4 and (ii) = S2 –S4 – S4 – S4 – S2

C.  

(i) = S2 – S3 – S4 – S2 – S2 and (ii) = S4 – S2 – S4 – S4 – S4

D.  

(i) = S2 – S3 – S4 - S4 – S2 and (ii) = S2 –S3 – S4 –S4 – S4