Placement Papers: Deshaw Placement Paper 2008

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This paper comprises of an Aptitude test and C-language test

Aptitude-20 questions (2 marks each)

C-language-20 questions (3 marks each)

APTITUDE TEST

20 QUESTIONS; 2 MARKS EACH

TIME-20 MINUTES

  1. ONE RECTANGULAR PLATE WITH LENGTH 8INCHES, BREADTH 11 INCHES AND 2 INCHES THICKNESS IS THERE. WHAT IS THE LENGTH OF THE CIRCULAR ROD WITH DIAMETER 8 INCHES AND EQUAL TO VOLUME OF RECTANGULAR PLATE? ANS: 3.5INCHES

  2. WHAT IS THE NUMBER OF ZEROS AT THE END OF THE PRODUCT OF THE NUMBERS FROM 1 TO 100

  3. in some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held? ans: 138

  4. one fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish. Ans: 1hr 12min

  5. in 8 * 8 chess board what is the total number of squares refer odel ans: 204

  6. falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.

  7. gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings

  8. 2 oranges, 3 bananas and 4 apples cost Rs. 15.3 ornages 2 bananas 1 apple costs Rs. 10. What is the cost of 3 oranges, 3 bananas and 3 apples. Ans Rs. 15.

  9. in 80 coins one coin is counterfiet what is minimum number of weighings to find out counterfiet coin

  10. in a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability that a randomly choosen employee is a male or female?

  11. statement: All green are blue are blue, all blue are white conclusion:

    1. some blue are green

    2. some white are green

    3. some green are not white

    4. all white are blue-he has given four choices like gre type

  12. all teachers are students. Some students are girls. This type of questions are there. We cant able to reproduce them.

Section B

  1. The function strcmp (str1, str2) returns

  2. int * x[] (); means

  3. #define PRINT (int) printf ( “int = %d” int); main () { int x, y, z; x = 03; y = -1; z = 01; PRINT (xx); z< ⇐ 3; PRINT (x); y>> = 3; PRINT (y); }

  4. struct list { int x; struct list * next; } * head; the struct head. x = 100 above is correct/wrong

  5. ‘-’ = 45 ‘/’ = 47 printfr (%d/n, ‘-’ ‘-’ ‘-’ ‘-’ ‘/’ ‘/’ ‘/’ ); o/p =?

  6. o/p =? int i; i = 1; i = i + 2 * i + +; printf (%d, i)

  7. { ch = ‘A’ while (ch ⇐ ‘F’ ) { switch (ch) { case ‘A’ Case ‘B’ Case ‘C’ Case ‘D’ Ch + +; continue; case ‘E’ Case ‘F’ Ch + +; } putchar (ch); } }

    1. ABCDEF

    2. EFG

    3. FG

    4. error

  8. FILE * fp1, * fp2; fp1 = fopen ( “one” “w” ) fp2 = fopen ( “one” “w” ) fputc ( ‘A’ fp1) fputc ( ‘B’ fp2) fclose (fp1) fclose (fp2)

  9. int a = 1; b = 2; c = 3; * pointer; pointer = &c; a = c/* pointer; b = c; printf ( “a = %d b = %d” a, b)

    1. a = 1 b = 3

    2. a = 3 b = 3

    3. 3 2

    4. error

  10. #include char * f () { char * s = malloc (8); strcpy (s, “goodbye” ) }

  11. main () { char * f (); printf ( “%c” * f () = ‘A’ ); } o/p =?

  12. int sum (n) int n; if (n<1) return n; else return (n + sum (n − 1) )

  13. when a function is recursively called all, automatic variables are

  14. #define MAN (x, y) (x) > (y) (x)? (y) { int i = 10; j = 5; k = 0; k = MAN (i + +, + + j) printf (%d %d %d %d, i, j, k) }

  15. a = 10; b = 5; c = 3; d = 3; if (a < b) && (c = d + + ) printf (%d %d %d %d a, b, c, d) else printf ( “%d %d %d %d a, b, c, d” )

  16. if the following program (my prog) main (int size of, char * arg[]) { while (size of arg) printf ( “%s” arg[--size of arg]) } is run from the command line as myprog jan feb mar apr what would be the o/p

    1. myprog jan, feb, mar, apr

    2. rev

    3. jan, feb, mar, apr

    4. error

  17. what is o/p main () { int i = 3; while (i--) { int i = 100 i--; printf ( “%d.” i); } }

    1. infinite loop

    2. error

    3. 99. 99. 99. 99

    4. 3. 22. 1.

  18. What is the o/p of the program

main () { int rows = 3, colums = 4; int a[rows][colums] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }}.

i = j = k = 99.

for (i = 0; i < rows; i + + )

for (j = 0; j < colums; j + + )

if (a[k][j] < k) k = a[i][j].

printf ( “%d\n” k).