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Poornam Info Vision Selection Process

Written test (Aptitude, C, LINUX, General Technical Awareness, Communication)
Interviews (Technical and HR rounds)

The recruitment process for Poornam Infovision consists

1. Written Test -Aptitude

2. Technical Interview

3. HR Interview

4. Essay Section

Poornam Info vision Written Test consists Aptitude and Technical Question

1 Written Test

Quantitative Aptitude : 10 Questions

Logical Deductions

Statement Problems etc.,

C section :10 Questions

Linux Section : 20 Questions

Technical Section : 20 Questions

Essay :200 words 1 Topic

Poornam Info Visions Previously asked Questions in Placements

1.APTITUDE SECTION: 10 QUESTIONS

I) In each of the following questions, a set of six statements is given, followed by four answer choices. Each of the answer choices has a combination of three statements from the given set of six statements. You are required to identify the answer choices in which the statements are logically related.
i) A.All cats are goats
B. All Goats are dogs.
C. No goats are cows
D. No goats are dogs
E. All Cows are dogs
F. All dogs are cows
a) FAB
b) ABE
c) AFB
d) ABF

ii) A.Some lids are nibs
B. All hooks are lids.
C. All hooks are nibs
D. No lid is a nib
E. No lid is a hook
F. No nib is hook
a) EFD
b)BCA
c)DEA
d)CDA

iii)A.All MBA’s are logical
B.Sudir is rational.
C.Sudhir is a logical MBA
D.Sudhir is a man
E.Some men are MBA’s
F.All men are rational.
a) DEC
b) EAF
c) BCF
d) FDB

iv) A.Competitive examinations are tough to pass.
B.There is heavy competition in any field.
C. No student can pass MAT
D. Very few students can pass MAT.
E. MAT is a competitive examination
F. MAT is tough to pass
a) AEF
b) ABC
c) DFB
d) CDE

v) A.All Pens are knives
B. All knives are spoons
C. No knives are pens
D. No knives are spoons.
E. All pens are spoons.
F. All spoons are pens.
a) ABE
b) ABF
c) AFE
d) DBE

II) Questions 5 and 6 are based on the following data:
In a class of 150 students, 40 passed in Social Studies, 90 passed in Science and 30 failed in both the subjects.
vi) How many students passed in at most one subject among Science and Social Science?
a) 100
b) 110
c) 140
d) 150

vii) How many students passed in Science but failed in Social Studies?
a) 70
b) 10
c) 90
d) 80

Poornam Info Visions C section Technical Questions

2. C SECTION: 10 QUESTIONS

i) What will be the output for the following program?
void main()
{
int const *p=5;
printf("%d",++(*p));
}
a) 5
b) 6
c) Run time error
d) Compiler error

ii) What will be the output for the following program?
main()
{
int c[]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf("%d",*c);
++q; }
for(j=0;j<5;j++) {
printf("%d",*p);
++p; }
}
a) 2 2 2 2 2 2 2 3 4 5
b) 2 2 2 2 2 2 3 4 6 5
c) 2 3 4 6 5 2 3 4 6 5
d) 2 3 4 6 5 3 4 5 6 7

iii) What will be the output for the following program?
main()
{
extern int i;
i=20;
printf("%d",i);
}
a) 20
b) Grabage value
c) Linker error
d) Compiler error

iv) What will be the output for the following program?
main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
a) 0 0 1 3 0
b) 0 0 1 3 2
c) 0 0 1 3 1
d) 1 1 2 3 1

v) What will be the output for the following program?
main()
{
char *p;
printf("%d %d",sizeof(*p),sizeof(p));
}
a) 1 1
b) 2 1
c) 1 2
d) 2 2

vi) What will be the output for the following program?
main()
{
int i=3;
switch(i)
{
default: printf("zero");
case 1: printf("one");
break;
case 2: printf("two");
break;
case 3: printf("three");
break;
}
}
a) zero
b) zero one two three
c) three
d) zero three

vii) What will be the output for the following program?
main()
{
printf("%x",-1<<4);
}
a) ffff
b) 0000
c) 0fff
d) fff0

viii) What will be the output for the following program?
main()
{
char String[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
a) Hello World
b) Linker error
c) Compiler error
d) Hello

ix) What will be the output for the following program?
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct y *P;
};
}
a) Blank screen
b) Linker error
c) Compiler error
d) Infinite execution

x) What will be the output for the following program?
#include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
a) 3 hello
b) 3hello
c) Linker error
d) Compiler error

3.LINUX SECTION: 20 QUESTIONS

i) You are trying to create four partitions on a linux system and you get an error while creating the fourth partition. Why?
a) Linux does not allow four partitions.
b) There is not enough disk space for the fourth partition.
c) The swap partition must be created first.
d) The last partition must be a swap partition.

ii) Which of the following is not a valid Linux command?
a) rm
b) mv
c) copy
d) more

iii) Which Linux command is used for renaming a file?
a) rm
b) mv
c) chmod
d) rename

Poornam Info Visions Technical part has 20 Questions

4. TECHNICAL SECTION: 20 QUESTIONS
i) What is the maximum decimal number that can be accomodated in a byte?
a) 128
b) 256
c) 255
d) 512

ii) The BIOS of a computer is stored in:
a) RAM
b) ROM
c) CACHE
d) HARD DISK

iii) Which of the folowing does not work over TCP?
a) HTTP
b) SMTP
c) FTP
d) None of the above

iv) What is the number of control lines for a 16 output decoder?
a) 2
b) 4
c) 16
d) 10

v) Which IP Addressing class allows for the largest number of hosts?
a) Class A
b) Class A and B
c) Class B
d) Class C

Poornam Info Vision Technical Interview Questions

1What will be the output of the following code?

void main ()
{ int i = 0 , a[3] ;
a[i] = i++;
printf (“%d",a[i]) ;
}
Ans: The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value.

2 Why doesn,t the following code give the desired result?

int x = 3000, y = 2000 ;
long int z = x * y ;
Ans: Here the multiplication is carried out between two ints x and y, and the result that would overflow would be truncated before being assigned to the variable z of type long int. However, to get the correct output, we should use an explicit cast to force long arithmetic as shown below:

long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.

3Why doesn,t the following statement work?

char str[ ] = "Hello" ;
strcat ( str, ,!, ) ;
Ans: The string function strcat( ) concatenates strings and not a character. The basic difference between a string and a character is that a string is a collection of characters, represented by an array of characters whereas a character is a single character. To make the above statement work writes the statement as shown below:
strcat ( str, "!" ) ;

4How do I know how many elements an array can hold?

Ans: The amount of memory an array can consume depends on the data type of an array. In DOS environment, the amount of memory an array can consume depends on the current memory model (i.e. Tiny, Small, Large, Huge, etc.). In general an array cannot consume more than 64 kb. Consider following program, which shows the maximum number of elements an array of type int, float and char can have in case of Small memory model.
main( )
{
int i[32767] ;
float f[16383] ;
char s[65535] ;
}

5How do I write code that reads data at memory location specified by segment and offset?

Ans: Use peekb( ) function. This function returns byte(s) read from specific segment and offset locations in memory. The following program illustrates use of this function. In this program from VDU memory we have read characters and its attributes of the first row. The information stored in file is then further read and displayed using peek( ) function.

#include
#include

main( )
{

char far *scr = 0xB8000000 ;
FILE *fp ;
int offset ;
char ch ;

if ( ( fp = fopen ( "scr.dat", "wb" ) ) == NULL )
{

printf ( "\nUnable to open file" ) ;
exit( ) ;

}

// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
fprintf ( fp, "%c", peekb ( scr, offset ) ) ;
fclose ( fp ) ;

if ( ( fp = fopen ( "scr.dat", "rb" ) ) == NULL )
{

printf ( "\nUnable to open file" ) ;
exit( ) ;

}

// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
{

fscanf ( fp, "%c", &ch ) ;
printf ( "%c", ch ) ;

}

fclose ( fp ) ;

}


Ans: Sometimes in a program we require to compare memory ranges containing strings. In such a situation we can use functions like memcmp( ) or memicmp( ). The basic difference between two functions is that memcmp( ) does a case-sensitive comparison whereas memicmp( ) ignores case of characters. Following program illustrates the use of both the functions.

#include

main( )
{
char *arr1 = "Kicit" ;
char *arr2 = "kicitNagpur" ;

int c ;

c = memcmp ( arr1, arr2, sizeof ( arr1 ) ) ;

if ( c == 0 )
printf ( "\nStrings arr1 and arr2 compared using memcmp are identical" ) ;

else
printf ( "\nStrings arr1 and arr2 compared using memcmp are not identical"
) ;

c = memicmp ( arr1, arr2, sizeof ( arr1 ) ) ;

if ( c == 0 )
printf ( "\nStrings arr1 and arr2 compared using memicmp are identical" )
;
else
printf ( "\nStrings arr1 and arr2 compared using memicmp are not
identical" ) ;
}


7 Fixed-size objects are more appropriate as compared to variable size data objects. Using variable-size data objects saves very little space. Variable size data objects usually have some overhead. Manipulation of fixed-size data objects is usually faster and easier. Use fixed size when maximum size is clearly bounded and close to average. And use variable-size data objects when a few of the data items are bigger than the average size. For example,

char *num[10] = { "One", "Two", "Three", "Four",
"Five", "Six", "Seven", "Eight", "Nine", "Ten" } ;

Instead of using the above, use

char num[10][6] = { "One", "Two", "Three", "Four",
"Five", "Six", "Seven", "Eight", "Nine", "Ten" } ;

The first form uses variable-size data objects. It allocates 10 pointers, which are pointing to 10 string constants of variable size. Assuming each pointer is of 4 bytes, it requires 90 bytes. On the other hand, the second form uses fixed size data objects. It allocates 10 arrays of 6 characters each. It requires only 60 bytes of space. So, the variable-size in this case does not offer any advantage over fixed size.

8The Spawnl( ) function...

DOS is a single tasking operating system, thus only one program runs at a time. The Spawnl( ) function provides us with the capability of starting the execution of one program from within another program. The first program is called the parent process and the second program that gets called from within the first program is called a child process. Once the second program starts execution, the first is put on hold until the second program completes execution. The first program is then restarted. The following program demonstrates use of spawnl( ) function.

/* Mult.c */

int main ( int argc, char* argv[ ] )
{
int a[3], i, ret ;
if ( argc < 3 || argc > 3 )
{
printf ( "Too many or Too few arguments..." ) ;
exit ( 0 ) ;
}

for ( i = 1 ; i < argc ; i++ )
a[i] = atoi ( argv[i] ) ;
ret = a[1] * a[2] ;
return ret ;
}

/* Spawn.c */
#include
#include

main( )
{
int val ;
val = spawnl ( P_WAIT, "C:\\Mult.exe", "3", "10",
"20", NULL ) ;
printf ( "\nReturned value is: %d", val ) ;
}

Here, there are two programs. The program ,Mult.exe, works as a child process whereas ,Spawn.exe, works as a parent process. On execution of ,Spawn.exe, it invokes ,Mult.exe, and passes the command-line arguments to it. ,Mult.exe, in turn on execution, calculates the product of 10 and 20 and returns the value to val in ,Spawn.exe,. In our call to spawnl( ) function, we have passed 6 parameters, P_WAIT as the mode of execution, path of ,.exe, file to run as child process, total number of arguments to be passed to the child process, list of command line arguments and NULL. P_WAIT will cause our application to freeze execution until the child process has completed its execution. This parameter needs to be passed as the default parameter if you are working under DOS. under other operating systems that support multitasking, this parameter can be P_NOWAIT or P_OVERLAY. P_NOWAIT will cause the parent process to execute along with the child process, P_OVERLAY will load the child process on top of the parent process in the memory.

9Are the following two statements identical?

char str[6] = "Kicit" ;
char *str = "Kicit" ;
Ans: No! Arrays are not pointers. An array is a single, pre-allocated chunk of contiguous elements (all of the same type), fixed in size and location. A pointer on the other hand, is a reference to any data element (of a particular type) located anywhere. A pointer must be assigned to point to space allocated elsewhere, but it can be reassigned any time. The array declaration char str[6] ; requests that space for 6 characters be set aside, to be known
by name str. In other words there is a location named str at which six characters are stored. The pointer declaration char *str ; on the other hand, requests a place that holds a pointer, to be known by the name str. This pointer can point almost anywhere to any char, to any contiguous array of chars, or nowhere.

10 Is the following code fragment correct?

const int x = 10 ;
int arr[x] ;
Ans: No! Here, the variable x is first declared as an int so memory is reserved for it. Then it is qualified by a const qualifier. Hence, const qualified object is not a constant fully. It is an object with read only attribute, and in C, an object associated with memory cannot be used in array dimensions

POORNAM INFO SELECTION PROCEDURE

The Poornam Info Vision Interview Process:

     If you are contemplating a career with Poornam, you would like to know more about our Interview process. This process is very thorough and personal, that provides us an objective evaluation of your skills, experiences and competencies. This is an opportunity for you and us to evaluate each other in depth. What are we looking for? What should you expect? Here is a brief explanation of our 5 step process.

Stage1
Your resumes are screened by a team of experts. Usually each member of the screening team reviews resumes individually and the resumes are graded as per a set of predetermined criterion. The criterion depends on the position you apply for. We usually contact you on phone or by mail, so that we get you know you better and can make a final decision on whether you get to the next stage or not. In this phase we try to review and assess your credentials, experience and background.

Stage2
This stage consists of a personal and technical interview conducted at the Poornam premises. If you have been invited over for an interview at Poornam, you know you,ve made the first cut. Then, based on your performance during the interview, the recruitment team will decide whether you get thorough to the next round. Before the interview, you might have to complete a short aptitude test, followed by an essay. The interview is to assess your suitability for employment with Poornam. We try to find out whether you will fit in with the rest of the Poornam family. We look for your technical skills, leadership potential, the aptitude for teamwork, exemplary involvement in extra- curricular activities and community achievements. The interview provides you with an opportunity to interact with people working here.

Stage3
Once you are thorough stage 2 you are shortlisted for a practical test. You will be given a 3-4 days to master a topic before the test. The 4-5 hour hands on test ( on the given topic), evaluates your ability to comprehend and grasp a new subject in a short time. You will be evaluated on your technical acumen, quality of execution and the output itself. The test could also be on your area of expertise, in which case, you do not have the advantage of preparation. It also means that the test will be tougher. Normally the test topic is given when you finish Step 2. Your test and the output will be evaluated by a team of experts. This might take a day or two.

Stage4
Once your practical test is deemed satisfactory by our team of experts, you will be shortlisted for a talk with our directors. Getting to this stage is tough and it takes a great deal of character to make it this far. This is a very informal session and is the final stage of evaluation for your recruitment to Poornam.