Technical Question Paper

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Technical Question Paper



 Wipro paper(System software)
 July-1997
 ------------
 
 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    ab  [c] abc  [d] a(bc)* [e] mone
   (bc)*=compliment of bc
 
   Ans:  e
 
 -------------------------------------
 2) A 12 address lines maps to the memory of
 
  [a] 1k bytes  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  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
    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    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  (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(n-3)+zap(n-1);
    }
    then the call zap(6) gives the values of zap
    [a] 8    9  [c] 6  [d]  12  [e] 15
 
   Ans: b
 ---------------------------------------------------------------
 
 
 PART-B
 -------
 1) Virtual memory size depends on
    [a] address lines    data bus
    [c] disc space       [d] a & c    [e] none
 
  Ans :  a
 -----------------------------------------------
 2) Critical section is
    [a]
    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
 ------------------------------

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