Wipro Placement Paper
Company: Wipro Infotech
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] mone
(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 impliment
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; 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 —————————————————————
PART-B
——-
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 ——————————]
Wipro Placement Paper
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