All the following questions are memory based and have been submitted by
candidates who appeared for the tests/interviews in the past.
Wipro paper (System software)
July 1997 -- Source: Roorkee
University Student
PART --A
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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
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2) A 12 address lines maps to the memory of
[a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none
Ans: b
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3) In a processor these are 120 instructions . Bits needed to implement
this instructions
[a] 6 [b] 7 [c] 10 [d] none
Ans: b
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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
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5) Return address will be returned by function to [a]
Pushes to the stack by call
Ans : a
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6) n=7623
{temp=n/10;
result=temp*10+ result; n=n/10
}
Ans : 3267
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7) If A>B then
F=F(G);
else B>C then
F=G(G);
in this , for 75% times A>B and 25% times B>C 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
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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
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9) C program code
int zap(int n)
{
if(n<=1)then zap=1;
--More--
else zap=zap(n-3)+zap(n-1);
}
then the call zap(6) gives the values of zap
[a] 8 [b] 9 [c] 6 [d] 12 [e] 15
Ans: b
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PART - B
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1) Virtual memory size depends on
[a] address lines [b] data bus [c]
disc space [d] a & c [e] none
Ans : a
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2) 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
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