Kerala PSC Previous Years Question Paper & Answer

Title : LECTURER IN COMPUTER SCIENCE COLLEGIATE EDUCATION
Question Code : A

Page:6


Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'LECTURER IN COMPUTER SCIENCE COLLEGIATE EDUCATION' And exam conducted in the year 2015. And Question paper code was '028/2015/OL'. Medium of question paper was in Malayalam or English . Booklet Alphacode was 'A'. Answer keys are given at the bottom, but we suggest you to try answering the questions yourself and compare the key along wih to check your performance. Because we would like you to do and practice by yourself.

page: 6 out of 12
Excerpt of Question Code: 028/2015/OL

A:-AB+C*DE--FG+~
B:-AB-C*DE--FG+"
C:-AB+CDE*--FG+"~
D:-AB+*¥CDE--FG+~
Correct Answer:- Option-A
Question48:-Merge sort uses
A:-Divide and Conguer strategy
B:-Backiracking approach
C:-Heuristic search
D:-Greedy approach
Correct Answer:- Option-A
Question49:-Linked lists are best suited
A:-for relatively permanent collections of data
B:-for the size of the structure and the data in the structure are constantly changing
C:-for both of above situations
D:-for none of above situations
Correct Answer:- Option-B
Question50:-The complexity of the average case of an algorithm is
A:-much more complicated to analyze than that of worst case
B:-much more simpler to analyze than that of worst case
C:-some times more complicated and some other times simpler than that of worst case
D:-none of the above
Correct Answer:- Option-A
Question51:-What is the minimum number of units of resource R required such that no deadlock will ever occur in an
operating system containing 5 user processes each requiring 2 units of R7
A5
B:-6
೧7
0:-8
Correct Answer:- Option-B
Question52:-The page miss ratio in a paged memory is 0.72. The time required to access a page in primary and secondary
memory are 10ns and 100ns, respectively. The average time required to access a page is:
A:-64.8
B:-35.2
C:-74.8
D:-45.2
Correct Answer:- Option-C
Question53:-The value of a counting semaphore at a particular instance of time is 5. #' P operations and 16 V operations
were completed on this ssmaphore subsequently. If the final value of the semaphore is 8, then *#* will be
‏1ھ‎
‎62
‎C:-16
‎0:13
‎Correct Answer:- Option-D
Question54:-Given the following disk queue:
95, 180, 34, 119, 11, 123, 62, 64
with read-write head initially at the track 50 and the tail track being at 199. If C-LOOK disk scheduling algorithm is followed,
then the total number of head movement required is:
A:-157
B:-146
C:-187
D:-236
Correct Answer:- Option-A
Question55:-NTFS stands for
A:-New Type File System
B:-Novel Technology File Service
C:-New Technology File System
D:-New Technology File Service
Correct Answer:- Option-C
Question56:-Shift-reduce parsers are

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