Wipro 
    InfoTech Placement Papers
          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 
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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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