Wednesday 29 April 2015

JUNE 2013 - PAPER III

7. The “PROJECT” operator of a relational algebra creates a new table that has always
(A) More columns than columns in original table
(B) More rows than original table
(C) Same number of rows as the original table
(D) Same number of columns as the original table

Ans:-C

Explanation:-
The Projection operator is also a unary operator. Whereas the selection operator chooses a subset of the rows of the relation, the projection operator chooses a subset of the columns. The Projection operation on a table, simply forms another table by copying specified columns, from the original table. Symbol of projection is π. Given a employee table having the columns(Emp-id, name, salary).
To select only the name of employee :- πname(employee)
To know only the salary :- πsalary(employee)
Selection operation:
It yields a horizontal subset of a given relation that is the subset of row should be selected with in the given relation for which a particular condition is satisfied. Sign of selection is σ. To see the details of those employees whose salary is greater than 10000, the selection operation will be used.
σsalary>10000(employee).
So, looking at the options given for the answer, the PROJECT operator cannot create a table having more columns than the original table. Same number of columns also ruled out. More rows than the original table is also not correct. So, it will be Same number of rows as the original table, but less number of columns than the original table. SO, the correct answer is C.


8. The employee information of an Organization is stored in the relation : Employee (name, sex, salary, deptname) Consider the following SQL query Select deptname from Employee Where sex = ‘M’ group by deptname having avg (salary) > {select avg (salary) from Employee} Output of the given query corresponds to
(A) Average salary of employee more than average salary of the organization.
(B) Average salary less than average salary of the organization.
(C) Average salary of employee equal to average salary of the organization.
(D) Average salary of male employees in a department is more than average salary of the organization

Ans:- D

Explanation:-
The question is very length and intimidating. But the answer is surprisingly simple. Rule out the wrong options first and you will always get the correct answer. There is a greater than symbol used in the query and we all know what it stands for. Options B and C says that Average salary less and equal. So both B and C are ruled out. Option A says Average salary of employee more than average salary of the organization. But in the query there is a where clause which is sex='M' and so the query is average salary of male employees is more than average salary of the organization. So the correct answer is option D.


9. For a database relation R(a, b, c, d) where the domains of a, b, c, d include only the atomic values. The functional dependency a → c, b → d holds in the following relation
(A) In 1NF not in 2NF
(B) In 2NF not in 3NF
(C) In 3NF
(D) In 1NF

Ans:-A

Explanation:-
A relation is in the first normal form if the domain of each attribute contains only atomic values. It means atomicity must be present in the relation. So, the question clearly mentions the same. The domains of a,b,c,d include only the atomic values and so 1NF holds but the question is whether 2NF holds or not. A relation will be in second normal form if it is in the first normal form and all the non primary key attribute must be fully functionally dependent on the primary key attribute. Although the functional dependency is given, no information is provided about the primary key here. So the correct answer is A.


Wednesday 22 April 2015

JUNE 2013 - PAPER III SOLUTION

Hi all,
Now that the application process for the NET exam has started, we need to buck up and get on with the preparations real serious. I will be solving the paper III questions in a detailed manner in the subsequent posts. As always, HAPPY STUDYING....

1. The Software Maturity Index (SMI) is defined as SMI = [Mf – (Fa + Fc + Fd)] / Mf Where
Mf = the number of modules in the current release.
Fa = the number of modules in the current release that have been added.
Fc = the number of modules in the current release that have been changed.
Fd = the number of modules in the current release that have been deleted.
The product begins to stabilize when
(A) SMI approaches 1
(B) SMI approaches 0
(C) SMI approaches –1
(D) None of the above

Ans:- A

Explanation:-
IEEE suggested a software maturity index(SMI) that provides an indication of the stability of a software product(based on changes that occur for each release of the product). SMI is calculated using the formula, which is mentioned in the question itself. Explanation of the formula is also part of the question. The product begins to stabilise when SMI approaches 1.0. SMI may also be used as a metric for planning software maintenance activities.


2. Match the following :
a. Watson Felix model      i. Failure intensity
b. Quick-Fix model      ii. Cost estimation
c. Putnam resource allocation model      iii. Project planning
d. Logarithmetic Poisson Model      iv. Maintenance
Codes :
     a b c d
(A) ii i iv iii
(B) i ii iv iii
(C) ii i iii iv
(D) ii iv iii i

Ans:-D

Explanation:-
Watson Felix model is used for cost estimation.
Putnam resource allocation model is used in project planning.
Quick-Fix model is used in software maintenance.
Logarithmetic Poisson model is used in failure intensity.
The option which matches all of these is D. So D is the correct answer.


3. __________ is a process model that removes defects before they can precipitate serious hazards.
(A) Incremental model
(B) Spiral model
(C) Cleanroom software engineering
(D) Agile model

Ans:- C

Explanation:-
Cleanroom software engineering(CSE) is a process model that removes defects before they can precipitate serious hazards.This is the exact definition of CSE and so the correct option is C.


4.Equivalence partitioning is a __________ method that divides the input domain of a program into classes of data from which test cases can be derived.
(A) White-box testing
(B) Black-box testing
(C) Orthogonal array testing
(D) Stress testing

Ans:- B

Explanation:-
What is Equivalence partitioning is given in the question. But we will have to know whether it belongs to what kind of testing, whether white-box, or black-box or the other two testing methods. Equivalence partitioning is a black-box testing method. Other examples are boundary-value analysis, classification trees, cause-effect graphs, etc are also black-box testing methods.


5. The following three golden rules :
(i) Place the user in control
(ii) Reduce the user’s memory load
(iii) Make the interface consistent
are for
(A) User satisfaction
(B) Good interface design
(C) Saving system’s resources
(D) None of these

Ans:- B

Explanation:-
I did not look for any textbook explanation for the above question. I went by my intuition and the answer is B. I hope all of you will agree and also know why.


6. Software safety is a __________ activity that focuses on the identification and assessment of potential hazards that may affect software negatively and cause an entire system to fail.
(A) Risk mitigation, monitoring and management
(B) Software quality assurance
(C) Software cost estimation
(D) Defect removal efficiency

Ans:- B

Explanation:-
Software safety and hazard analysis are software quality assurance activities that focus on the identification and assessment of potential hazards that may impact software negatively and cause an entire system to fail.