UGC NET Computer Science Computer Previous Year Questions (PYQs) – Page 4 of 111
Study Stuff for MCA Examinations
Solution
Software validation checks whether the software meets the requirements and expectations of the customer.
Four persons: P, Q, R and S are in police custody and one of them has committed a crime. They confess as follows:
A. Person P: Q did it.
B. Person Q: S did it.
C. Person R: I did not do it.
D. Person S: Q lied.
If exactly one of the statements is false, which of the following is the guilty person.
Solution
Check each case one by one.
Case 1: Suppose P is guilty.
P says: Q did it → False
Q says: S did it → False
R says: I did not do it → True
S says: Q lied → True
Here, two statements are false. So, P is not guilty.
Case 2: Suppose Q is guilty.
P says: Q did it → True
Q says: S did it → False
R says: I did not do it → True
S says: Q lied → True
Here, exactly one statement is false.
So, Q can be guilty.
Case 3: Suppose R is guilty.
P says: Q did it → False
Q says: S did it → False
R says: I did not do it → False
S says: Q lied → True
Here, three statements are false. So, R is not guilty.
Case 4: Suppose S is guilty.
P says: Q did it → False
Q says: S did it → True
R says: I did not do it → True
S says: Q lied → False
Here, two statements are false. So, S is not guilty.
Therefore, the guilty person is Q.
$F(A,B,C,D)=\sum(0,1,2,3,6,12,13,14,15)$
is/are
(A) $A'B' + AB + A'C'D'$
(B) $A'B' + AB + A'CD'$
(C) $A'B' + AB + BC'D'$
(D) $A'B' + AB + BCD'$
Choose the correct answer from the options given below:
Solution
$0,1,2,3,6,12,13,14,15$
$A'B'$ covers minterms $0,1,2,3$
$AB$ covers minterms $12,13,14,15$
Remaining minterm is $6$
Option (B):
$A'CD'$ covers minterms $2,6$
Both are present in the function.
So, option (B) is correct.
Option (D):
$BCD'$ covers minterms $6,14$
Both are present in the function.
So, option (D) is also correct.
Options (A) and (C) include minterm $4$, which is not present in the function.
Solution
A grammar is ambiguous if a string generated by it has more than one parse tree.
Equivalent definitions:
• More than one leftmost derivation
• More than one rightmost derivation
Thus the correct statements are:
B and C.
Solution
Morphological analysis, semantic analysis and pragmatic analysis are standard components of natural language understanding.
Meaning analysis is not a standard separate component because meaning is covered under semantic analysis.
Assertion A:
A Raster scan device is a CRT graphic device and can use a television monitor for display.
Reason R:
In Raster scan display the picture is composed of a series of dots. These dots are traced out as a series of horizontal lines. Television works in a similar fashion.
In the light of the above statements, choose the correct answer from the options given below:
Solution
Raster scan display forms an image using many small dots called pixels.
These pixels are scanned line by line from left to right and top to bottom.
Television display also works in a similar raster scanning manner.
So, Assertion A is true.
Reason R correctly explains why raster scan devices can use television monitors.
So, Reason R is also true and it is the correct explanation of Assertion A.
(A) Maximum speedup that a pipeline can provide is k theoretically.
(B) It is impossible to achieve maximum speedup k in k-segment pipeline.
(C) All segments in pipeline take same time in computation.
Choose the correct answer from the options given below:
Solution
So, statement A is true.
In practical cases, due to pipeline filling time, emptying time, delays and unequal stages, exact maximum speedup $k$ is not achieved.
So, statement B is true.
All pipeline segments need not take exactly the same computation time.
So, statement C is false.
Solution
Steps in page replacement:
No frame is free → D
Find a frame to replace → A
Free that frame → E
Update the page table → B
Use the freed frame for the new page → C
Solution
A contains numbers of the form:
$4n + 2$
So, numbers in A are:
$6, 10, 14, 18, 22, 26, 30, ...$
B contains multiples of 3:
$3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...$
Common elements are:
$6, 18, 30, 42, ...$
These can be written as:
$12n - 6$
Solution
In Round Robin scheduling, each process gets CPU for one time quantum.
Given:
Time quantum $=4$ ms
Number of processes $=20$
Maximum waiting time occurs for the last process in the ready queue.
Before the last process gets CPU, the other $19$ processes will execute.
So,
Maximum waiting time $=19 \times 4$
$=76$ ms
UGC NET Computer Science
Online Test Series, Information About Examination,
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UGC NET Computer Science
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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