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Placement Papers - Wipro

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Wipro Paper Pattern
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WIPRO Placement Paper

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 implement 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 :
Ans : Pushes to the stack by call


6) n=7623
{temp=n/10; result=temp*10+ result; n=n/10 }
Ans : 3267
 
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
 
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;
More
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
 
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) 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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