# ISRO 2007 COMPUTER SCIENCE PART – 1

1. The Boolean expression Y=(A+B¯+A¯B)C¯Y=(A+B¯+A¯B)C¯ is given by
1. AC¯
2. BC¯¯
3. C¯
4. AB

Explanation:Y=(A+B¯+A¯B)C¯

Y=(A+(B¯+A¯).(B¯+B))C¯

Y=(A+B¯+A¯)C¯

Y=((A+A¯)+B¯)C¯

Y=(1+B¯)C¯

Y=C¯

ans is A

3.The circuit shown  in the given figure  is

(2)Full subtractor

(3)shift register

4.When two numbers are added in excess-3 code and the sum is les than 9,then in order to get the correct answer it is necessary to

(a)subtract 0011 from the sum

(c)subtract 0110 from the sum

5 . The characteristic equation of an SR flip-flop is given by

1. Qn+1=S+RQnQn+1=S+RQn
2. Qn+1=RQ¯n+S¯QnQn+1=RQ¯n+S¯Qn
3. Qn+1=S¯+RQnQn+1=S¯+RQn
4. Qn+1=S+R¯Qn

6.A graph with “n” vertices and n-1 edges  that is not a tree is

(a)connected

(b)disconnected

(c)Euler

(d)a circuit

7.If a graph requires  k  different colors for its proper coloring,then the chromatic number of graph is

(a)1

(b)k

(c) k-1

(d)k/2

Explanation: The chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices share the same color.

(a)and written by CPU

(b) and written  by peripheral

(c)by peripheral and written by CPU

(d)by CPU and written by peripheral

10.The term aging refers to

(a)booting up  the priority  of the process in multi-level of queue without feedback

(b)gradually increasing the priority of jobs  that wait in the system for  a long time to remedy  infinite blocking

(c)keeping track of the following the page  has been in memory  for the purpose of LRU replacement

(d)letting job reside in memory for a certain amount of time  so that the number of pages required can be estimated accurately

11.Consider a set of  n tasks with known runtimes  r1 ,r2…………..rn to be run on a uniprocessor macine .Which of the following  processor scheduling algorithms  will result in maximum throughput ?

a)Round Robin

b)Shortest  job first

c)Highest response ratio next

d)first cum first served

Explanation:

Throughput means total number of tasks executed per unit time. Shortest Job First has maximum throughput because in this scheduling technique shortest jobs are executed first hence maximum number of tasks are completed.
Note: Highest-Response-Ratio-Next policy favors shorter jobs, but it also limits the waiting time of longer jobs.

12. Consider a job scheduling problem with 4 jobs J1,J2,J3J1,J2,J3 and J4J4 with corresponding deadlines: (d1,d2,d3,d4)=(4,2,4,2)(d1,d2,d3,d4)=(4,2,4,2). Which of the following is not a feasible schedule without violating any job schedule?

1. J2,J4,J1,J3
2. J4,J1,J2,J3
3. J4,J2,J1,J3
4. J4,J2,J3,J1

Explanation:

TO OPTIMIZE AND TO GET A FEASIBLE SOLUTION WE HAVE TO FINISH ALL THE JOBS WITH IN THEIR DEAD LINE.

SINCE THE DEADLINE OF J2 AND J4 ARE 2.

SO THEY MUST BE COMPLETED BY 2.SO WE CAN SCHEDULE ANY ONE

THAT IS J2 OR J4..

IN OPTION C,D J4 IS SCHEDULED FIRST AND THEN J2 AND

IN OPTION A, FIRST J2 THEN J4             SO WE CAN ABLE TO SCHEDULE OTHER 2 JOBS

NOW CONSIDER OPTION B

J4 IS SCHEDULED AND NEED TO BE COMPLETED ….BY 2  —NO PROBLEM

J1 IS SCHEDULED AND NEED TO BE COMPLETED ….BY 4     –NO PROBLEM

BUT NOW AS J2 SHOULD COME BEFORE TIME 2..BUT WE SCHEDULED J1….SO J2 CANT BE SCHEDULED

13. By using an eight bit optical encoder the degree of resolution that can be obtained is (approximately)

1. 1.8 degree
2. 3.4 degree
3. 2.8 degree
4. 1.4 degree

Explanation:

A digital optical encoder is a device that converts motion into a sequence of digital pulses.Resolution is defined as the smallest discernible quantity.
Hence its is 360/2^8

Ans is 1.4 degree.

14 The principle of locality of reference justifies the use of

a) virtual memory                             b) interrupts                                                                                            c) main memory                        d)cache memory

Explanation:

locality of reference, also known as the principle of locality, is a term for the phenomenon in which the same values, or related storage locations, are frequently accessed, depending on the memory access pattern.

There are two basic types of reference locality –

temporal and spatial locality.

15 . Consider the following pseudo-code

x:=1;
i:=1;
while (x <= 1000)
begin
x:=2^x;
i:=i+1;
end;

What is the value of i at the end of the pseudo-code?

1. 4
2. 5
3. 6
4. 7

Explanation:

st time … x = 2 and i = 2

2nd time …x=4 and i=3

3rd time…..x=16 and i = 4

4th time ……before entering the loop value of x < 1000 ….it is 16 so it will enter the loop

and make i = 5

5th time x>1000 so while loop doesn’t execute…..

SO the answer is i = 5

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1. B.2. D3 full adder4. B5. B6. D7.C8. C9 not clearly viewed10. B11. D.12 13 not clear14. D.

Sashi pls publish solution too
Thanx