Placement Papers: CMC Paper 1997

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CMC: Test Pattern Total 149 Question. 100 Aptitide (Verbal, Reasoning, Analytical) (no-veMarks)

49 Technical (-ve marks) (c, c + +, Java, Operating System Concepts, Data Communication & Networks, PL/SQL) the test is for 1hr: 30 Mins.

Technical test

  1. Unix OS is a

    1. multi-taks

    2. mutual task

    3. multiple processe

    4. a&c

  2. The TV Satilite Transmission is a

    1. Half-Duplex

    2. Full Duplex

    3. one-way duplex

    4. non of the above

  3. Which among is the OOPL

    1. c

    2. java

    3. Cobol

    4. PAscal

  4. what will be the output of main () { int a = 5, c; int ptr; ptr = &a; c = * ptr * a; printf ( “%d, %d” c, a); }

    1. 5, 5

    2. 25, 25

    3. 25, 5

    4. Error

  5. What can a Friend Keywoard do?

    1. Free the memory

    2. provide access to private & public variables

    3. provide access to Private, Protected, public variables.

    4. never allowed to acccess any constructor

  6. Abstract Class can be

    1. Accessed as Constructors

    2. used for hiding information of class

    3. class is Abstract class

    4. none of the above

  7. The present LAn-card Supports

    1. 32kbphs

    2. 64kbphs

    3. 128kbphs

    4. 256kbphs

  8. What is CDMA Technology?

    1. Electronic calculator

    2. Cellular phones

    3. Rocket

    4. Car

  9. What is GC in JAVA?

    1. De-allocates the memory

    2. De-allocates the objects & its reference

    3. Finalising of Memory

    4. non-of the above

  10. which among is thread class?

    1. sleep ()

    2. IsAliva ()

    3. IsLive ()

    4. none of the above

  11. main () { int x = 10, y = 5, p, q; p = x>9; q = x>3&&y! = 3; printf ( “p = %d q = %d” p, q); }ans: 1, 1.

  12. main () { int x = 11, y = 6, z; z = x = = 5 |! = 4; printf ( “z = %d” z); }ans: 1

  13. main () { int c = 0, d = 5, e = 10, a; a = c>1? d>1 | e>1? 100: 200: 300; printf ( “a = %d” a); }ans: 300

  14. main () { int i = -5, j = -2; junk (i, &j); printf ( “i = %d, j = %d” i, j); } junk (i, j) int i, * j { i = i * i; * j = * j * * j; } ans: 5, 4

  15. #define NO #define YES main () { int i = 5, j; if (i>5) j = YES; else j = NO; printf ( “%d” j); } ans: Error message

  16. main () { int a = 0xff; if (a<<4>>12) printf ( “leftist” ) else printf ( “rightist” ) } ans: Rightist

  17. main () { int i = + 1; while (~i) printf ( “vicious circles” ) }

ans: Continuous loop.