Global Placement Papers

Global placement papers in pdf , doc and text format to prepare for global company. Check following old global placement papers with solutions and test interview questions from year 2010 to 2015-16. Global is one of reputed company in india for your career. We can provide global aptitude syllabus for your help too. Know about global placement procedure and questions with aptitude test papers on this page. Best of luck for your interview in global company.
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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. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:A.15B.16C.18D.25Answer:B Ex:Ratio of times taken by Sakshi and Tanya = 125:100 = 5:4.Suppose...
Sample Test Paper Globaledge1)what is big-endian.a) MSB at lower address LSB at higher addressb) LSB at lower address MSB at higher addressc) memory mgmt techniqued) none of the aboveans:a2)what is Little-endian.a) MSB at lower address LSB at higher addressb) LSB at lower address MSB at higher addressc) memory mgmt techniqued) none of the aboveans:b3)8086 hasa)16 bit data bus ,16 bit address busb)16 bit data bus,32 bit address busc)8 bit data bus,16 bit address busd)8 bit data bus,8 bit address busans:a4) what is the scheduling algorithm used in general operating systems.a) FCFS algorithmb) Highest Priority First algorithmc) Round-Robin algorithmd) None of the aboveans:c5)Router is present ata)Physical...
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 = 255;printf ("%d",c);return 0;}what is the outputa> illegal character assignmentb> prints -1c> prints 2d> prints 255ans b: prints -1.4) main(){int i;for (i=7;i prints hello 7 timesb> prints hello 8...
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Sample Test Paper  Globaledge 1)what is big-endian.a) MSB at lower address LSB at higher addressb) LSB at lower address MSB at higher addressc) memory mgmt techniqued) none of the aboveans:a2)what is Little-endian.a) MSB at lower address LSB at higher addressb) LSB at lower address MSB at higher addressc) memory mgmt techniqued) none of the aboveans:b3)8086 hasa)16 bit data bus ,16 bit address busb)16 bit data bus,32 bit address busc)8 bit data bus,16 bit address busd)8 bit data bus,8 bit address busans:a4) what is the scheduling algorithm used in general operating systems.a) FCFS algorithmb) Highest Priority First algorithmc) Round-Robin algorithmd) None of the aboveans:c5)Router is present...
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GLOBAL EDGE  Placement Paper 5 20051)Remove the odd onesa)job scheduler b)long term scheduler c)medium term schedulerd)process term schedulerans:d2) long term scheduler is also known asa)cpu scheduler b)job scheduler c)middle term scheduler d)none of theseans:b3)which one is not in the process statea)ready b)run c)terminated d)none of theseans: d4)Switching the cpu to another process requires saving the state of the old process and loading the saved state is calleda)mode switch b)process switch c)context switch d)noneans:cGLOBAL EDGE Latest Fresher Engineer Placement Sample Question Paper 6 20055)which one is not in the compilation stagea)lexical analyserb)parserc)assemblerd)code generatorans:c Operating...
GLOBAL EDGE  Placement Question Paper 4 20051. Where is the MBR stored?1. maintained by OS2. MBR is in Boot.3 MBR is in First sector of HDD4. None of the above.2. Where is the partition table stored?1. BIOS2. CMOS Setup3. MBR4. stored per partition.3. Where is the boot record stored?1. BIOS2. CMOS Setup3. MBR4. stored per partition.4. How many primary partitons can be created?1. 12. 23. 34. 4ans : 4.5. What is the difference between primary & extended partion?1. Both are same2. Primary and extended are in logical partion3. primary cannot be subdivided but extended can be.4. extended cannot be subdivided but primary can be.ans 3.6. Can we create 2 primary dos partions?a)Yes b)No c)Depends...
 GLOBAL EDGE Placement Question Paper 2 2005main(){ 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 = 255;printf ("%d",c);return 0;}what is the outputa> illegal character assignmentb> prints -1c> prints 2d> prints 255ans b: prints -1.4>main(){int...
GLOBAL EDGE  Placement Question Paper 4 2005 1. Where is the MBR stored?1. maintained by OS2. MBR is in Boot.3 MBR is in First sector of HDD4. None of the above.2. Where is the partition table stored?1. BIOS2. CMOS Setup3. MBR4. stored per partition.3. Where is the boot record stored?1. BIOS2. CMOS Setup3. MBR4. stored per partition.4. How many primary partitons can be created?1. 12. 23. 34. 4ans : 4. 5. What is the difference between primary & extended partion?1. Both are same2. Primary and extended are in logical partion3. primary cannot be subdivided but extended can be.4. extended cannot be subdivided but primary can be.ans 3. 6. Can we create 2 primary dos partions?a)Yes b)No c)Depends...
GLOBAL EDGE Placement Question Paper 2 2005 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 3 ans 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 s ans a: prints abcd. 3> main(){char c = 255;printf ("%d",c);return 0;} what is the output a> illegal character assignmentb> prints -1c> prints 2d> prints 255 ans b: prints -1....
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