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Globaledge Aptitude Reasoning Aptitude Questions And Answers 04 September 2012
GlobalEdge Aptitude Questions and Answers1.If the number 481 * 673 is completely divisible by 9, then the smallest whole number in place of * will be:A.2B.5C.6D.7E.None of theseAnswer:DEx:Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9.Therefore, x = 7.2.3897 x 999 = ?A.3883203B.3893103C.3639403D.3791203E.None of theseAnswer:BOption : BEX:3897 x 999 = 3897 x (1000 - 1)= 3897 x 1000 - 3897 x 1= 3897000 - 3897= 3893103.3.Sakshi can do a piece of work in 20 days.…
Globaledge Placement Paper 31 December 2010
Globaledge Placement Paper 31 January 2011
GLOBAL EDGE Placement Question Paper1) main(){ int arr[]={ 1,2,3,4 };int *ptr ;;;;ptr++ = arr;printf("%d,%d",ptr[2],arr[2]);return 0;}what is the output :a> compile time error :multiple termination statements for pointerb> lvalue required for ptrc> prints 3 3d> printd 4 3ans b: lvalue required for ptr;2) main(){char s[10];scanf ("%s",s);printf(s);}what is the output if input is abcd :a> prints abcdb> compiler errorc> prints abcd and 6 junk charactersd> printd sans a: prints abcd.3) main(){char c…
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