SIEMENS Placement Papers

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Whole Testpaper Cummins College of Engineering - 18 July 2007

Hi Friends,...............

Criteria : 55% or above , no live ATKT

Total students Appeared : 30

Selected After Aptitude :11

Selected Finally : 4

The selection procedure was as follows :

Aptitude Test Technical Interview HR Interview 1. Aptitude Test :

It had 2 sections – Total marks- 45. Time – 50 mins

a. Java --- 20 questions (.1 mark each) Cut off - 8

b. Quantitative Aptitude --- 25 questions (1 mark each) cut off -13

There was –ve marking of -0.25 for each wrong answer.

Java Questions were tricky. They stressed on inheritance ,Garbage collection ,C++ questions were also there….we had to find out the output of the programs given.

Quantitative was easy…..questions were good n difficulty level was moderate. I referred Abhijeet Guha book for Aptitude and R.S.Agarwal also. You have to be fast in this section so that u can give more time to Java.

2.Technical Round :

After the Aptitude Test….only 11 students were selected for this round. I was one of them. They asked me the following questions….

Q : Tel us something abt ur family background.

A : I told…..

Q : What is ur fav subject ?

A : Software Engineering and OS

Q: What is SDLC ?

A: I told in detail all the phases.

Q: What is the configuration of ur PC ?

A : 40GB Hard-disk ,512 MB RAM….and so on.

Q: If u have a 1 GB file….how will it accommodate in 512 MB RAM ?

A : Paging concept will come into picture. Then I explained him the paging concept.

Q: What is the difference in C++ and Java ?

A: I told all major differences.

Q: What is the difference between pass by value and pass by reference ? Which is better n why ?

A : I answered…..Pass by reference n value first n said pass by reference is better coz we save memory here as only the references have to b changed n final result is reflected at the original place of storage itself.

Q : U have any plans for higher studies ?

A : No sir……I want to work.

Q: What r ur strengths and weaknesses ?

A : I told……

Q: Do u have any problems in working in shifts ?

A : No problems sir.

Q: What is ur preffered Job location ?

A : Pune is my first choice……anywhere in India and I am ready to go abroad on a company project.

That was all. The Panel was friendly. My suggestion would be that be cool n composed while answering……be confident n dress smartly.

There were no eliminations after this round. All 11 students gave HR round also.

3.HR Round :

Here I was asked …..

1.What r ur hobbies ?

2.What do u know aboutt Siemens ?

3.What do u know aboutt SAP ?

4.What is ur fav subject ? I had to explain SDLC again.

5.How was ur day ?

6.Are u a hostellite ?

Ans--I said Yes.

7.What was the most memorable moment in the hostel?

Ans--I told that…

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Whole Testpaper Siemens-Worli - 18 December 2006

SIEMENS PAPER ON 18th DECEMBER AT SIEMENS-WORLI

25 qns tech n 25 qns quant. 0.25 negative markingcut off is 8 and 12 respectivelytime limit 35 min.Round -1technical1. A c++ prgm is given n find the output(a static function calls a nonstatic function)2. what is normalization( choices are all very similar so read the proper defn frm textbook.3. In UNIX how to delete a file? ans>> ,rm, command4 In UNIX how to concatenate or see the details of a file ? and is ,cat>>filename, cmd5. qn related to editors in UNIX ans is, vi editor,6. in java to ensure that no threads are being updated simultaneously monitors are used and also synchronised keyword is used.7. sm qn related to oracle window application.8. what is primary key?9. oracle is a rdbms

quantitaive1. qn related to time n work2. qn related to typing papers (related to time n work) ans is 75 or sm value in 70s.3. qn related to percentage4 qn related to profit n loss5.qn related to simple interst6. qn related to time, speed n distance where distance is miles per hr and time is in minutes.7. complete the series1,4,9,16,? ans is 25two more such series8. encryted word(each alphsbet replaced by its next)two of such ones9. 4+x^2= x^3? whats x? ans x=2

start quantitaive frm behind as its easy n one can finish it fast.

Round-2Technical cum HR.

qns asked.1. what is normalization?2. what r the types of normal forms no explaination was asked3. explain all projects done.rest all qns wer hr

Round-3Technical cum HR

Qns asked1. if u delete an entry frm table what can u do?ans rollback if delete is used and cant rollback if truncate is used they asked me if i knew the concept of recovery but i didnt know as it was not thr in syllabus

2. explain the difference between delete n truncateans delte is ddl nad truncate is dml with truncate the structure of database is lost

3. whats a foreign key or wats the referential intergrity constraint?

4 in MS Access wats created first the table or dtabase? its databaseREst all wer Hr qns

Round -3HRqns asked1. favourite subject in college.As I said datawarehousing next qn was difference between datbase n datawarehouse2. rest all wer hr qns

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Aptitude - General

THIS PAPER CONSISTS 6 PARTS. all are multiple choice q,s

1 general2 c/unix3 c++/motif 4 database 5 x-windows6 ms-windowswe have written q,s not acc. to each part.total 50. q,s. time is sufficient. if u have basic idea about all of the u can easily answer the

which of following operator can,t be overloaded. a)== b)++ c)?! d)<=

#includemain()

{printf("Hello World");}the program prints Hello World without changing main() the o/p should be initialization Hello World Destruct the changes should bea)iostream operator<<<,intialisation,<<b) c) d)none of the above

CDPATH shell variable is in(c-shell)a) b) c) d)

semaphore variable is different from ordinary variable by

swap(int x,y){int temp;temp=x;x=y;y=temp; }main(){int x=2;y=3;swap(x,y);}after calling swap ,what are yhe values x&y?

static variable will be visible ina)fn. in which they are definedb)module " " " "c)all the programd)none

unix system is a)multi processingb)multi processing ,multiuserc)multi processing ,multiuser,multitaskingd)multiuser,multitasking

x.25 protocol encapsulates the follwing layersa)networkb)datalinkc)physicald)all of the abovee)none of the above

TCP/IP can work ona)ethernetb)tokenringc)a&bd)none

A node has the ip address 138.50.10.7 and 138.50.10.9.But it is transmitting data from node1 to node2only. The reason may bea)a node cannot have more than one addressb)class A should have second octet differentc)classB " " " " "d)a,b,c

for an application which exceeds 64k the memory model should bea)mediumb)hugec)large d)none

The condition required for dead lock in unix sustem is

set-user-id is related to (in unix)

wrong statement about c++a)code removablyb)encapsulation of data and codec)program easy maintenanced)program runs faster

struct base {int a,b;base();int virtual function1(); }struct derv1:base{int b,c,d;derv1()int virtual function1();}struct derv2 : base {int a,e;}base::base(){a=2;b=3;

}derv1::derv1(){b=5;c=10;d=11;}base::function1() {return(100);}derv1::function1() {return(200);}main()base ba;derv1 d1,d2;printf("%d %d",d1.a,d1.b) o/p is a)a=2;b=3;b)a=3; b=2; c)a=5; b=10;d)none

for the above program answer the following q,smain()base da;derv1 d1;derv2 d2;printf("%d %d %d",da.function1(),d1.function1(),d2.function1());o/p isa)100,200,200;b)200,100,200;c)200,200,100;d)none

struct {int x;int y;}abc;you can not access x by the following 1)abc-->x;2)abc[0]-->x;abc.x;(abc)-->x;a)1,2,3 b)2&3c)1&2d)1,3,4

automatic variables are destroyed after fn. ends because a)stored in swapb)stored in stack and poped out after fn. returnsc)stored in data aread)stored in disk

variable DESTDIR in make program is accessed asa) $(DESTDIR)b) ${DESTDIR}c) DESTDIRd) DESTDIR

The keystroke mouse entrie are interpreted in ms windows asa) interruptb) messagec) eventd) none of the above

link between program and out side world (ms -win)a)device driver and hardware diskb)application and device driverc)application and hardware device d)none

after logout the process still runs in the background by giving the commanda)nohopb)

process dies out but still waitaa)exitb)wakeupc)zombied)steep

n dynamic memory allocation we usea) doubly linked listb) circularly linkedc) B treesd) L treese) none

To find the key of search the data structure is a)hask keyb)treesc)linked lists

d)records

data base-------------------------------- employ_code salary employ_code leave from to--------------------------------------1236 1500 1238 --- ---2371237 2000 1238 --- --- 1237 1238 2500 1237 --- 1237 -------------------------- --------------------------------------select employ_code,employ_data ,leavethe number of rows in the o/p a)18b)6c)7d)3

which is truea)bridge connects dissimiler LANand protocol insensitiveb)router " " " " "c)gateway " " " " "d)none of the above

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Whole Testpaper

SIEMENS PAPER

All are multiple choice questions.60 questions to be answered in 60 mins.Distribution of questions:---10 questions from data structures and some general topics.-----10 questions from Unix and C.--7Questions from Data base.

----Remaining from Windows(x windows,MS Windows etc..)The distribution is not exact.Only approximate.Totally there are six sections as below:

1.General2.Unix and c3.RDBMS4.C++/Object orientedTCP/IP6not remembered.The questions are as follows:RDBMS-----1.What is RDBMS...Def2.Two tables are given.In 1st table 2 columns are there.oneisEmployee no,second is salary.In second table 3 columns are there,one is employee no,second is date,3rd is salary.Select employee no,from table1,table 2.How many records it will contain?.(This is somewhat difficult).

3.What is transaction?TCP/IP:1.X.25 protocol belongs to which layer.2,Order all the 7 layers in sequence3,One node has 2 IP address but data goes through only one link.What is the reason?4,Router,Bridge,Gateway....Which one of these can not connecttwo different Lans and is protocol sensitive.5,Client sends server---reqest or demand or -----Choicesare given.Another section...1.main(argc,argv){if(argc<1)printf("error");elseexit(0);}If this program is compiled without giving any argument ,what it will print.2.What are the static variables...def3.What is Dynamic allocation ?4.Dead lock condition...What may be the condition for it.5.Semaphore variable?..def6.Most of the Windows operating systems are....choices are given like----a,maltitasking, multiuser, multi processingb,only.......&......c,only........&......like that7.Properties of object oriented,(just recall the name of the properties likeinheritance,....8.A program of C++,and two questions were based on the o/p of that program..................---------.................

Some more questions are not remembered.Here I am concatenating another mail from my friend.The number of questions from different areas as mentioned before is not correct.The correct numbers are given below.Sections...1, 5Q ...x windows2, 5Q .....MS Windows3, 7Q.....Databases4, 20 Q....General5, 20Q....UNIX/C6,5Q ....OOPSSome questions.....1,Sticky bit concept2,Bourne shell.,no hup concept3,Net work layers4,Ethernet,Token Ring5,Unix...(Ans is Multitasking,Multi user)6,Windows...(ans is pseudo multitasking)7,IP Addresses8,Normalisation(in data basese)9,SQL Queries10, semaphores

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SIEMENS Paper Whole Testpaper

All are multiple choice questions.60 questions to be answered in 60 mins.Distribution of questions:---10 questions from data structures and some general topics.-----10 questions from Unix and C.--7Questions from Data base.

----Remaining from Windows(x windows,MS Windows etc..)The distribution is not exact.Only approximate.Totally there are six sections as below:1.General

2.Unix and c3.RDBMS4.C++/Object orientedTCP/IP6not remembered.The questions are as follows:RDBMS-----1.What is RDBMS...Def2.Two tables are given.In 1st table 2 columns are there.oneisEmployee no,second is salary.In second table 3 columns are there,one is employee no,second is date,3rd is salary.Select employee no,from table1,table 2.How many records it will contain?.(This is somewhat difficult).

3.What is transaction?TCP/IP:1.X.25 protocol belongs to which layer.2,Order all the 7 layers in sequence3,One node has 2 IP address but data goes through only one link.What is the reason?4,Router,Bridge,Gateway....Which one of these can not connecttwo different Lans and is protocol sensitive.5,Client sends server---reqest or demand or -----Choicesare given.Another section...1.main(argc,argv){if(argc<1)printf("error");elseexit(0);}If this program is compiled without giving any argument ,what it will print.2.What are the static variables...def3.What is Dynamic allocation ?4.Dead lock condition...What may be the condition for it.5.Semaphore variable?..def6.Most of the Windows operating systems are....choices are given like----a,maltitasking, multiuser, multi processingb,only.......&......c,only........&......like that7.Properties of object oriented,(just recall the name of the properties likeinheritance,....8.A program of C++,and two questions were based on the o/p of that program..................---------.................

Some more questions are not remembered.Here I am concatenating another mail from my friend.The number of questions from different areas as mentioned before is not correct.The correct numbers are given below.Sections...1, 5Q ...x windows2, 5Q .....MS Windows3, 7Q.....Databases4, 20 Q....General5, 20Q....UNIX/C6,5Q ....OOPSSome questions.....1,Sticky bit concept2,Bourne shell.,no hup concept3,Net work layers4,Ethernet,Token Ring5,Unix...(Ans is Multitasking,Multi user)6,Windows...(ans is pseudo multitasking)7,IP Addresses8,Normalisation(in data basese)9,SQL Queries10, semaphores 

 

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SIEMENS Paper Whole Testpaper

SIEMENS INFO------------------------------------------------------------------------

THIS PAPER CONSISTS 6 PARTS. all are multiple choice q,s1)general2)c/unix3)c++/motif4)database

5)x-windows6)ms-windowswe have written q,s not acc. to each part.total 50. q,s. time issufficient.if u have basic idea about all of the u can easily answer the paper.paper------1)which of following operator can,t be overloaded.a)== b)++ c)?! d)<=2)#include<iostream.h>main(){printf("Hello World");}the program prints Hello World without changing main() the o/p shouldbeintialisationHello WorldDesructthe changes should bea)iostream operator<<(iostream os, char*s)os<<,intialisation,<<(Hello World)<<Destructb) c) d)none of the above3)CDPATH shell variable is in(c-shell)a) b) c) d)4) term stickily bit is related to a)kernelb)undeletable filec) d)none5)semaphore variable is different from ordinary variable by6)swap(int x,y){int temp;temp=x;x=y;y=temp;}main(){int x=2;y=3;swap(x,y);}after calling swap ,what are yhe values x&y?7) static variable will be visible ina)fn. in which they are definedb)module " " " "c)all the program

d)none8) Unix system isa)multi processingb)multi processing, multiuserc)multi processing ,multiuser,multitaskingd)multiuser,multitasking9)x.25 protocol encapsulates the follwing layersa)networkb)datalinkc)physicald)all of the abovee)none of the above10)TCP/IP can work ona)ethernetb)tokenringc)a&bd)none11)a node has the ip address 138.50.10.7 and 138.50.10.9.But it istransmitting data from node1 to node2only. The reason may bea)a node cannot have more than one addressb)class A should have second octet differentc)classB " " " " "d)a,b,c 12) the OSI layer from bottom to top13)for an application which exceeds 64k the memory model should bea)mediumb)hugec)larged)none14)the condition required for dead lock in unix sustem is15)set-user-id is related to (in unix)16) bourne shell hasa)history recordb)c)d)17)wrong statement about c++a)code removablyb)encapsulation of data and codec)program easy maintenanced)program runs faster18)struct base {int a,b;base();int virtual function1();}struct derv1:base{int b,c,d;

derv1()int virtual function1();}struct derv2 : base{int a,e;}base::base(){a=2;b=3;}derv1::derv1(){b=5;c=10;d=11;}base::function1(){return(100);}derv1::function1(){return(200);}main()base ba;derv1 d1,d2;printf("%d %d",d1.a,d1.b)o/p isa)a=2;b=3;b)a=3; b=2;c)a=5; b=10;d)none19) for the above program answer the following q,smain()base da;derv1 d1;derv2 d2;printf("%d %d %d",da.function1(),d1.function1(),d2.function1());o/p isa)100,200,200;b)200,100,200;c)200,200,100;d)none20)struct {int x;int y;}abc;you can not access x by the following1)abc-->x;

2)abc[0]-->x;abc.x;(abc)-->x;a)1,2,3b)2&3c)1&2d)1,3,421) automatic variables are destroyed after fn. ends becausea)stored in swapb)stored in stack and poped out after fn. returnsc)stored in data aread)stored in disk22) relation between x-application and x-server (x-win)23)UIL(user interface language) (x-win)24)which is right in ms-windowsa)application has single qvalue system has multiple qvalueb) " multiple " " single "c)" " " multiple "d)none25) widget in x-windows is26) gadget  in x_windows is27)variable DESTDIR in make program is accessed asa)$(DESTDIR)b)${DESTDIR}c)DESTDIRd)DESTDIR28)the keystroke mouse entrie are interpreted in ms windows asa)interruptb)messagec)eventd)none of the above29)link between program and out side world (ms -win)a)device driver and hardware diskb)application and device driverc)application and hardware deviced)none30)ms -windows isa)multitaskingb) c) d)31)dynimic scoping is32) after logout the process still runs in the background by givingthe commanda)nohopb)33)process dies out but still waitaa)exit

b)wakeupc)zombied)steep34)in dynamic memory allocation we usea)doubly linked listb)circularly linkedc)B treesd)L treese)none35)to find the key of search the data structure isa)hask keyb)treesc)linked listsd)records36)data base----------------------------------------------------------employ_code salary employ_code leave-------------------------------------------------------------from to

--------------------------------------1236 1500 1238 --- ---1237 2000 1238 --- ---1238 2500 1237 --------1237 --- ---1237 --- ---1237 --- -------------------------------------------------------------------select employ_code,employ_data ,leavethe number of rows in the o/pa)18b)6c)7d)337)DBMS38)read about SQL,db39)which is truea)bridge connects dissimiler LANand protocol insensitiveb)router " " " " "c)gateway " " " " "d)none of the above

40)read types of tree traversals.41)42)43) simple programs on pointers in c