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WIPRO papers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 |
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Wipro paper(System software)
July1997
PART A
1) abcD+abcd+aBCd+aBCD then the simplified function is (
Capital letters are copliments of corresponding letters A=compliment of a)
[a] a [b] ab [c] abc [d] a(bc)* [e] none (bc)*=compliment of bc
Ans: e
2) A 12 address lines maps to the memory of
[a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none
Ans: b
3) In a processor these are 120 instructions . Bits needed to impliment
this instructions
[a] 6 [b] 7 [c] 10 [d] none
Ans: b
4) In 8085 microprocessor READY signal does.which of the following
is incorrect statements
[a]It is input to the microprocessor
[b] It sequences the instructions
Ans : b
5) Return address will be returned by function to
[a] Pushes to the stack by call
Ans : a
6)
n=7623
{
temp=n/10;
result=temp*10+ result;
n=n/10
}
Ans : 3267
7) If AB then
F=F(G);
else BC then
F=G(G);
in this , for 75% times AB and 25% times BC then,is 10000 instructions
are there ,then the ratio of F to G
[a] 7500:2500 [b] 7500:625 [c] 7500:625 if a=b=c else 7500:2500
8) In a compiler there is 36 bit for a word and to store a character 8bits
are needed. IN this to store a character two words are appended
.Then for storing a K characters string, How many words are needed.
[a] 2k/9 [b] (2k+8)/9 [c] (k+8)/9 [d]
2*(k+8)/9 [e] none
Ans: a
9) C program code
int zap(int n)
{
if(n<=1)then zap=1;
else zap=zap(n3)+zap(n1);
}
then the call zap(6) gives the values of zap
[a] 8 [b] 9 [c] 6 [d] 12 [e] 15
Ans: b
PARTB
1) Virtual memory size depends on
[a] address lines [b] data bus
[c] disc space [d] a & c [e] none
Ans : a
2) Critical section is
[a]
[b] statements which are accessing shared resourses
Ans : b
3) load a
mul a
store t1
load b
mul b
store t2
mul t2
add t1
then the content in accumulator is
Ans : a**2+b**4
4) question (3) in old paper
5) q(4) in old paper
6) question (7) in old paper
7) q(9) in old paper