Easy Tek placement papers


Company Name : EASI-TECH

 WRITE THE OUTPUT FOR FOLLOWING PROGRAMS.

1) printf(“%d%xn”,ox2,12);

2) int a=10;
   int b=20;
   a=a^b;
   b=a^b;
   a=a^b;
   printf(“%d%dn”,a,b);

    3)enum  
{ELLIPSE,
TRIANGLE,
RECTANGLE,
SQUARE=100,
CIRCLE=5
}
        printf{“%d%d%d%dn”,TRIANGLE-RECTANGLE,SQUARE*CIRCLE-RECTANGLE};
4) define the following…
   a) pointer to a integer
   b) pointer to a char
   c) function pointer returning a pointer integer.
5) void a(void);
   main()
   {
   a();
   }
   void a(void)
   {
   char a=”HELLOW”;
   char *b=”HELLOW”;
   char c[10]=”HELLOW”;
   printf(“%s%s%sn”,a,b,c);
   printf(“%d%d%dn”,sizeof(a),sizeof(b),sizeof(c)));
   }
  6)  int a=15;
int b=16;
printf(“%d %d n”,a&b,a/b); (bitwise operators)
 7)   int a[5],*p;
      for(p=a;p<&a[5];p++)
{
*p=p-a;
printf(“%dn”,*p);
        }
8) sscanf(“xyz abc ABC 345” “% *[a-z A-Z]lf”,&a);
   printf(“lf”,a);
  9) Not clear
        fprintf(- – – – –
10) main()
    {
    int i=10;
    printf(“%d”,i);
    {
    int i=20;
    printf(“%d”,i);
    }
    printf(“%d”,i);
    }
 11)struct class
{
int i;
float a;
string[12];
}
sizeof(class)=?
12)   int *p;
i=10;
p=i;
printf(“%d”,*p);
13)  fact(5)
     int n;
     fact(n)
     {
      sum=n*fact(n-1);
     }
NEXT FINITE ELEMENT ANALYSYS QUESTIONS
1)
NAME THE EQUATION AND EXPLAIN THE TERMS
 A) M/I=F/Y=E/R.
 B) mx..+cx.+kx=F(t)         ans: sdof mass-spring-dashpot equn.
2)
WHAT ARE THE DIFFT.TYPES OF FINITE ELEMENTS.
3)
WHAT IS THE SIZE OF STIFFNESS MATRIX FOR A 4 NODED SHELL ELEMENT IN 3-D
SPACE.
4)
HOW MANY NODES ARE THERE FOR A BEAM ELEMENT ? WHY?
5)
WRITE STIFFNESS MATRIX OF TRUSS ELEMENT.

   6)WHAT ARE PLANE STRESS AND PLANE STRAIN ELEMENTS

  7)WHAT IS MENT BY STRESS SMOTHENING
  8)WHAT IS GAUSS POINT
  9)WHAT ARE CO AND C1 PROBLEMS

[OK 13Q IN C  AND 9Q IN FEM]

  4 15:17:23 1998

1) YOU SHOULD LEARN POINTERS(they may ask in I.W)
2)****SHOULD LEARN C++(this is very very IMP,Based
on this only I got job.One more thing is
CLASS,INHERATENCI&POLYMORPHISM is sufficient)
3)YOU SHOULD PREPARE ONE FEM PROJECT AND TELL THEM
 THAT AS YOUR OWN WORK(This will increase chance
to 99%)
                      EASI-TECH
                 WRITTEN TEST PATTERN
——————–
1] C       TEST 10Q 20MINUITS
2]F E M    TEST 19Q 20MINUITS
3]APTITUDE TEST 15Q 20MINUITS
—————————–
F E M TEST
———-
1)WHO USED THE TERM FINITE ELEMENT FOR THE FIRST TIME?
  a)
  b)
  c) CLOUGH
2)DERIVE THE JACOBIEN |J| FOR BEAM ELEMENT WITH STRAIN
  ENERGY?
  (ANS:PROMLEM IS NOT CORRECT PLEASE DO NOT ATTEMPT)
3)FOR AN ELEMENT SIGMA Ni=1,WHICH TYPE OF ELEMENT IT IS?
   a)
   b)NATURAL CO-ORDINATE
    (ANS 100% CORRECT)
4)TIMOSHENKO BEAM ELEMENT THEORY TO CONSIDER — – – -?
   a)
   b)
   c)SHEAR DEFORMATION
5)SHEAR LOCKING – – – – ?
  (ANS IS VERY STIFF ‘K’)
6)MEMBRANE LOCKING
  (ANS IS ARCH ELEMENT)
7)Ex(epsiolan x)=dU/dX,Ey=dV/dY,r(X,Y)=?(gama(x,y)=?)
  (ANS IS dU/dY+dV/dX )
8)K=integral B(TRANSPOSE)*D*B  FOR LARGE DEFORMATION
  WHICH MATRIX WILL GET EFFECTED?
  (ANS IS  D matrix )100%correct
9)FOR PLANE STRAIN  f(Ex,Ey,Ez,r(x,y) ) – – –
  (ANS IS    Ez=0)
10)SERENDIPITY ELEMENT IS
   (ANS 8 NODED ELEMENT)
 (The element which is having nodes only on boundary is
   called SERENDIPITY element)
11)IF THE ROTATION OF ELEMENT AND THE DISPLACEMENT ABOUT
   N-A  IS SAME THEN THE ORDER OF CONTINUITY
   (ANS IS  C1)
12)FRONTAL THEORY IS APPLIED FOR
  (Please refer any fem book)
13)MINDLINS THEORY IS APPLIED FOR
  c) this is the answer(Both Co&C1problems)
14)X=SigmaNi*Xi,U=sigmaNi*Ui WHICH TYPE OF ELEMENT
  (Refer book)
15)BEAM SUBJECTED TO UDL FIND THE MOMENTS AT THE
   2 NODES
16)Integral B(Transpose)*sigma(here sigma means stress)*dV
    REPRESENTS?
    (ANS IS INTERNAL LOAD VECTOR)
17)Integral ET(epsiolan transpose)*sigma*dV
    P=strain displacement vector
    Q=stress-strain deformation
    Find [K]
    (ANS    [K]=[P]T(p transpose)*[Q]*[P] )
18)
19)

REMEMBER ORDER MAY NOT BE CORRECT


                      APTITUDE TEST
 1)33 1/3 of     101 + 296 is
(ans 1200) check
 2)0.625= ? (ans 27/40)
 3)One ship goes along the stream direction 28 km and in opposite
   direction 13 km in 5 hrs for each direction.What is the velocity
   of stream?
(ans 1.5 kmph)
 4)Cubic root of 3375=?
(ans 15)
 5)2020201-565656=?
(ans 1454545)
 6)CHAIRS PROBLEM
5 chairs=9 tables,12 tables = 7 stools likethat- – –
(ans is 80Rs)
 7)One clock ringes 7 O’clock in 7 sec.In how many seconds it will
   ring 10 O’clock.
(ans 10.5 sec)
 8)One watch is showing 30 past 3 .What is the angle between minutes &
   hours hand?
(ans 75 degrees)
 9)The average of 4 consecutive even numbers is 27. What is the largest
   number?
(ans 30)
 10) 25 stations ,24 stations are inbetween- – – – –
     how many tickets should be required.
                (ans 25*24=600)PUZZLES TO PUZZLE YOU “S.DEVI”PROB 24
 11)One ball was dropped from 8ft height and every time it goes half
    of the height. How much distance it will travell before coming to
    rest.
                 (ans 24 approximately)
 12)Two trains are travelline at equilateral .Train A is travelling
    in the direction of earths spin.Other train B is travelling in
    opposite direction of earths spin.Which trains wheels will wear
    first?and why?
        (ans TRAIN B .Because of less centrifugal force.)
     

C QUESTIONS:WHAT IS THE OUT PUT FOR FOLLOWING PROGRAMMS
1)main()
{
char a[2];
*a[0]=7;
*a[1]=5;
printf(“%d”,&a[1]-a)
ANS:
   ans may be 1.(illegal initialization)
2)
#include
main(){
char a[]=”hellow”;
char *b=”hellow”;
char c[5]=”hellow”;
printf(“%s %s %s “,a,b,c);
printf(” “,sizeof(a),sizeof(b),sizeof(c));
}
(ans is     hellow,hellow,hellow
              6,2,5   )
3)
#include
main()

float value=10.00;
printf(“%g %0.2g %0.4g %f”,value,value,value,value)
}
(ans is 10,10,10,10.000000)
4)
#include
void function1;
int i-value=100;
main()
{
 i-value=50;
function1;
printf(“i-value in the function=”,i-value);
printf(“i-value after the function=”,i-value);
}
printf(“i-value at the end of main=”,i-value);
functioni()
i-value=25;

THIS IS ROUGH IDEA OF THE PROGRAM
   ANS ARE
1)i-value in the function=25;
2)i-value after the function=50;
3)i-value at the end of the main=100;

5)
  main()
{
funct(int n);
{
switch(n)
  case1:
        m=2;
  break;
  case2:
m=5;
break;
  case3:
m=7;
break;
  default:
m=0;
}
THIS IS ROUGH IDEA:
    (ANS:Out put is m=0)

   

From mepg9766@violet.iitm.ernet.in Tue Aug  4 15:19:07 1998

I am sending mainly c paper and some questions.Rao
also will send somethig.There are 15 c q’s all are discriptive.

1)fallacy
f()
{
int a;
void c;f2(&c,&a);
2)a=0;
b=(a=0)?2:3;
a) What will be the value of b? why
b) If in 1st stmt a=0 is replaced by -1, b=?
c) If in second stmt a=0 is replaced by -1, b=?
3)char *a[2]
int const *p;
int *const p;
struct new { int a;int b; *var[5] (struct new)
4)f()
{
int a=2;
f1(a++);
}
f1(int c)
{
printf(“%d”, c);
}
c=?
5)f1()
{
f(3);}
f(int t)
{
switch(t);
{
case 2: c=3;
case 3: c=4;
case 4: c=5;
case 5: c=6;
default: c=0;}
value of c?
6)Fallacy
int *f1()
{
int a=5;
return &a;
}
f()
int *b=f1()
int c=*b;
}
7)a)Function returning an int pointer
b)Function ptr returning an int ptr
c)Function ptr returning an array of integers
d)array of function ptr returning an array of integers
(See Scham series book)
8)fallacy
int a;
short b;
b=a;
9)Define function ?Explain about arguments?
10)C passes  By value or By reference?
11)Post processed code for
abc=1;
b=abc1; (1 or 2 blank lines are given)
strcpy(s,”abc”);
z=abc;
12)difference between my-strcpy and strcpy ?check
13)f()
{
int *b;
*b=2;
}
14)Function which gives a pointer to a binary trees const an integer value
at each code, return function of all the nodes in binary tree.(Study)Check
15)Calling refernce draw the diagram of function stack illustrating the
variables in the —–then were pushed on the stack at the point when
function f2 has been introduced
type def struct
{ double x,double y} point;
main( int argc, char *arg[3])
{double a;
int b,c;
f1(a,b);}
f1(double x, int y)
{
point p;
stack int n;
f2(p,x,y)}
f2(point p, double angle)
{ int i,j,k,int max)
}
_____________________________________________________________
1)Least no. when divide by [7 gives remainder 6,6gives 5,5 gives 4 and
soon     ans;419
2)What compilation do (ans source code to obj)
3)Artficial language is provided which of the language (Lisp) check
4)241 change its equivalent octal ?
5)for cube and sphere 3 views are similarly draw one such figure?
6)Write a program to exchange two variaables without temp
7)Fortran cannot have value by reference
8)4,6,8,__
9)success is to failure, joy is to
10)MEANING OF JOLLY?
11)opposite to essential?
12)”Raw” means
 13)To be good “Wrestler ” one should have?
14)”Command” opposite?
15)genuine opposite?
16)Two proverbs are goven
17)Sum of two consecutive nos is 55, larger one is?
18)A person goes 4/5 of his usual speed reaches 10min lateto his
destinaton, time taken?
19)80% pass in english, 70%pass in maths , 10%fail in both , 144 pass in
both . How many all appeared to the test?
20)To get a parabola if you cut a section of?
21)Bird is flying 120km/hr b/w B to R. two trians at B to R at 60 kmph
The distance trvelled by the bird before it is killed.Ans.120
22)meaning of inert
 Prepare well
for the interview. Mostly on graphics , geometry .Prepare questions like
(for interview)Prove  some of the angles in a triangle are 180.Angle in a
half circle is 90.How will you measure hight of building when you  are at
the top of the building and if you have stone with you.
Best of luck

Amdocs Placement Papers 2011

Paper and Interview Experience

Hi guyz!

On 27th September 2011 , Amdocs visited our college ,they selected 11 out of  around 150 students from BIT MESRA.

Here I m sharing my experience with  you people .I hope it will help you.
Plz listen their PPT carefully,It will help you a lot during interview. Although it was too boring for me . I dnt remember all the questions but I hav tried my best to share with u people.

There would be 6 subjects sections :
1. Quantitative aptitude: 20 questions, 25 mins

(Venn diagrams, DS, Some questions on average, 2 Logical Reasoning Caselets each comprising of 4-5 questions) 



2. Attention TO detail (Easiest Section of paper): 15 questions 15 mins

(Pattern matching, arithmetic operators substitution (if + is *), 2 Decision making questions each comprising of 4-5 questtions.) This section in sample paper comprise mostly of pattern matching questions which is not the case in actual paper and would follow the above pattern.



3. Technical passage: 10 questions, 15 mins

(This sub-section comprise of a technical passage (USB, Cookies, Computer N/w, File Organization, O.S etc) and is followed by 10 questions. Don’t commit the mistake of reading the whole passage as all the questions are direct, so just browse through the passage.)



4. Programming ability test (C / C++ / Java): 20 questions, 25 mins

I would recommend you to choose between C and C++ as Java was quite difficult and remember that based on your language selection in this test you would be asked question in technical interview.


5. SQL (Toughest section of paper): 15 questions, 15 mins

This is indeed the toughest section of the paper. There can be things which you must have not heard of earlier. Remember questions given in sample paper are quite different from what actually appears in paper

6. UNIX: 15 questions, 15 mins


Basic commands 
Tee, chmod (would always be asked), tail, head, cut, paste, sed, vi editor internal commands(i, I, a, A), grep, etc.

System calls
May or may not be asked.
Do remember meaning of special symbols such as $$,$!,$#,$*,$1 etc.



Questions asked are mainly based on:

1. Index (Clustered, Non Clustered)
2. Views (Updateble and read only views, How they are internally stored)
3. Where condtion involving columns that contain NUll values (so in this case you must remember how AND and OR operators work on NULL values)
4. INSERT statements to a table.
5. Some built in functions (Decode(), Substr(), Instr() )
6. Nested queries
7. Join operations (involves ANSI AND THETA notations)
8. Result of a Query (A table schema would be given and you would be required to determine what would be its output whether it contains any error)

9. Error Correction in a Query i.e what should be modified to make a query run succesfully.
This mainly comprise of errors in group by, having clause, joins (for e.g. Joining two tables without any condition although is correct but would not yield the result)



Written was consist of two section. 
It was a online test. Sectional Cut-off was there.

1 – Technical ( (C/C++/JAVA), sql,unix)
2 – Non-Technical (Data Interpretation (very easy),  Time Distance (2 or 3 ques), Train(1 ques), etc but simple ques were there in this section, And a big  Comprehension from HOW stuff works.com).

Non-Tech portion was very easy,cut off goes till 90% in this section.U can refer R.S. Agarwalfor this.For this portion u require atmost An half day to prepare.

Now lets talk abt tech portion, it was difficult at least for me.
I opted  for  C++. 

C++ all question were from Test Your C++ Skills,And I  had revised so ,it was easy for me.

SQL queries were from join,Subqueries,Clusters etc,This portion was tough for me,
Unix was consist of  Pipelining,Shell Scripting and other vi’ Commands.It was very -very difficult for me.

I attempted more questions in C++ in comparison to Unix, I haven’t studied Unix any time before, I read unix   just for 2 or 1 day before the visit of AMDOCS.

25 students could clear Written.

Now the Time for INTERVIEW.
It was consist of two rounds.

1 – Pure Tech(around 30-35 minute)
2 – HR round( around  35-40 minute)

GUYZ KEEP IN MIND YOUR  HR INTERVIEW(COMMUNICATION SKILL) WILL PLAY A BIG ROLE IN SELECTION.
Actually ,I was unaware of this fact till the announcement of the AM DOCS Result.

In tech round ,I was asked abt ( What is autometa theory,Give me an example of Autometa theory application,Then I was asked to draw a flow chart  and to write Code for that example).Then I was asked puzzles.I coulnt ans any puzzle.

This interview was not a good experience for me. Others were also asked for writing SQL queries SQL datatypes etc.Be honest with whatever u hav mentioned in ur CV.He can ask question from any subject.Even they checked our result cards during interview .Some students were asked question from the subjects which they have studied in their Graduation.

Now it’s the time for HR INTERVIEW.
It started with my INTRODUCTION(Tell me abt yourself).
He asked me in which area I m intresetd to work(development/Testing).I was asked abt my projects etc.s
Guyz , be ready wid ur reason in whichever area u want to work. 

Then asked abt location preference(GURGAON/PUNE).
I  had already offer from Infosys . they asked  me Y AMDOCS?
He asked many question. I was feeling Like I wont be selected, I gave  so many unsatisfactory answers in both the interviews.

But I don’t know what happened I was suddenly called for 3rd interview,None other than me appeared for 3rd interview, This interview was taken by their senior executive Manager.I was so scared .I felt like ,I am forced to sent in jail and  I didn’t know what mistake I had committed.

Well ,Finally ,I went  for 3rd interview, it again started wid my introduction.
This round went for 40-45 minutes.This round was  of 80% tech and 50% non tech questions.

He asked question from DATA STRUCTURE, SOFTWARE ENGINEERING,C++(MANY CONCEPTS like object oriented ,object based, virtual destrutor  etc).
He asked my area of intrest to work(DEVLOPMENT/TESTING). And Y and Y not?
He asked in which phase of SDLC I m interested to work.

I felt HR interview was not easy to satisfy the interviewe.

In 3rd round ,I could ans 80% and of tech ques. It was little bit satisfactory for me.

They started taking interview in morning at 10:15 am, and they declared results at  9:30 pm.

Finally I was one of those 11. THANX TO GOD!
ALL THE BEST.

SmartBear Software Placement Papers Liscensing Specialist

Company Profile : 


SmartBear Software provides tools for over one million software professionals to build, test, and monitor some of the best software applications and websites anywhere – on the desktop, mobile and in the cloud. Our users can be found worldwide, in small businesses, Fortune 100 companies, and government agencies. Learn more about the SmartBear Quality Anywhere Platform, our award-winning tools, or join our active user community at www.smartbear.com, on Facebook, or follow us on Twitter @smartbear.

Job Description :

SmartBear Software is looking for a highly motivated individual to join our growing team in the fast paced computer software environment. The Licensing Specialist will be responsible for license key generation and distribution within a defined time period with the ability to reconcile all licenses back to our CRM and financial systems.

Location : Beverly, MA (Greater Boston Area)


Desired Skills & Experience

Salesforce.com
Strong MS Office skills
Understanding of Software Licensing
Strong attention to detail
Individual must be self-directed and well organized with the ability to meet fixed service deadlines
Able to identify process improvements and system enhancements in a fast-paced changing environment
Experience:
Two year college degree in a business concentration
Database experience

Remedy Interactive Placement Papers


Remedy Interactive
Remedy Interactive is a global SaaS provider in the workplace risk management market. EHS, Risk Management and HR leaders at Fortune 500 companies around the world use Remedy Interactive’s enterprise software solutions to identify employee’s health and safety risks, and reduce costs associated with workers’ compensations claims, lost work days and productivity.

Tideway Systems Placement Papers

Tideway

Tideway Systems was a software company, founded in November 2002 by Richard Muirhead, that offered products for data centre optimization, virtualization management, application availability and standards compliance, targeted toward enterprise customers.

In 2007, Tideway’s flagship product Foundation v7.0 was released in what was the biggest upgrade ever to the product line. Through its ability to identify patterns, it automatically models and maps an IT infrastructure, connecting the technology layers in a way that is relevant to both the business and IT professionals – from business applications through virtualized environments to switches and the dependencies that exist in between.

Category : IT/ System Software

Role : Senior Software Developer

Linsyssoft Technologies placement papers

LinSysSoft TechnologiesLinSysSoft is a software solutions company focussed on system level software. For more than 5 years, we have delivered technical solutions that have enabled companies to successfully roll out their products in SAN, embedded and virtualization space. LinSysSoft has been recognized by industry for its ability to solve complex technical problems.

Aci Worldwide Placement paper

ACI Worldwide powers electronic payments for financial institutions, retailers and processors around the world with the broadest, most integrated suite of electronic payment software in the market. More than 75 billion times each year, ACI’s solutions process consumer payments. On an average day, ACI software manages more than US$12 trillion in wholesale payments. And for more than 150 payments organizations worldwide, ACI software ensures people and businesses don’t fall victim to financial crime. We are trusted globally based on our unrivaled understanding of payments and related processes. We have a definitive vision of how electronic payment systems will look in the future and we have the knowledge, scale and resources to deliver it. Since 1975, ACI has provided software solutions to the world’s innovators. We welcome the opportunity to do the same for you.

Interviewed in Pune (India) (Nov 2011 – took 2 days)
1.Round of technical interview
2.second round of technical interview
3.Manager Round
4.HR Round

Good knowledge of C++.
Proven experience (including university/school projects) in programming/testing.
You will be given a test related to the job you apply (C++, UML, testing tools and other)
Also other skills and abilities will be tested, such as: attention, focusing on details, adaptation to stress, making decisions, work ethics.

Texas instruments placment papers

For more than 80 years, Texas Instruments has used increasingly complex signal-processing technology – with advances ranging from the incremental to the revolutionary – to literally and repeatedly change the world.
Every day, our semiconductor innovations help 90,000 customers unlock possibilities for a smarter, safer, greener, healthier and more enjoyable world. Our focus on building a better future is ingrained in everything we do, from responsible semiconductor manufacturing, to employee care and active involvement with the communities we live in.
TIers are a tremendously diverse group, coming from every continent, embracing scores of different cultures and viewpoints, and speaking dozens of languages; yet we all share a passion for discovery. After all, innovation is what we do.
TI’s amazing past is a prologue to an even more incredible future. And in many ways, our story is just beginning.

Here are the details of 2011 Texas Instruments Placement Paper – I job in Texas Instruments – TI

Welcome to Texas Instrements Placement Paper 2011. Here you will find TI Paper Pattern and Download questions of Texas Instrements Placement Paper 2011 with Answers & Solutions.

2011 Texas Instrements Placement Papers:-

1. Can we declare a static function as virtual?
Ans: No. The virtual function mechanism is used on the specific object that determines which virtual function to call. Since the static functions are not any way related to objects, they cannot be declared as virtual.

2. Can user-defined object be declared as static data member of another class?
Ans: Yes. The following code shows how to initialize a user-defined object.
      #include
      class test
      {
      int i ;
      public :
      test ( int ii = 0 )
      {
      i = ii ;
      }
      } ;
      class sample
      {
      static test s ;
      } ;
      test sample::s ( 26 ) ;
Here we have initialized the object s by calling the one-argument constructor. We
can use the same convention to initialize the object by calling multiple-argument constructor.

3. What is forward referencing and when should it be used?
Ans: Consider the following program:
     class test
      {
      public :
      friend void fun ( sample, test ) ;
      } ;
      class sample
      {
      public :
      friend void fun ( sample, test ) ;
      } ;
      void fun ( sample s, test t )
      {
      // code
      }
      void main( )
      {
      sample s ;
      test t ;
      fun ( s, t ) ;
      }
This program would not compile. It gives an error that sample is undeclared identifier in the statement friend void fun ( sample, test ) ; of the class test. This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample. Forward referencing is generally required when we make a class or a function as a friend.

4. The istream_withassign class has been derived from the istream class and overloaded assignment operator has been added to it. The _withassign classes are much like their base classes except that they include overloaded assignment operators. Using these operators the objects of the _withassign classes can be copied. The istream, ostream, and iostream classes are made uncopyable by making their overloaded copy constructor and assignment operators private.

5. How do I write my own zero-argument manipulator that should work same as hex?
Ans: This is shown in following program.
      #include
      ostream& myhex ( ostream &o )
      {
      o.setf ( ios::hex) ;
      return o ;
      }
      void main( )
      {
      cout << endl << myhex << 2000 ;
      }

6.We all know that a const variable needs to be initialized at the time of declaration. Then how come the program given below runs properly even when we have not initialized p?
      #include
      void main( )
      {
      const char *p ;
      p = “A const pointer” ;
      cout << p ;
      }
Ans: The output of the above program is ‘A const pointer’. This is because in this program p is declared as ‘const char*’ which means that value stored at p will be constant and not p and so the program works properly

7. How do I refer to a name of class or function that is defined within a namespace?
Ans: There are two ways in which we can refer to a name of class or function that is defined within a namespace: Using scope resolution operator through the using keyword. This is shown in following example:

      namespace name1
     {
      class sample1
      {
      // code
      } ;
      }
      namespace name2
      {
      class sample2
      {
      // code
      } ;
      }
      using namespace name2 ;
      void main( )
      {
      name1::sample1 s1 ;
      sample2 s2 ;
      }
Here, class sample1 is referred using the scope resolution operator. On the other hand we can directly refer to class sample2 because of the statement using namespace name2 ; the using keyword declares all the names in the namespace to be in the current scope. So we can use the names without any qualifiers.

8. While overloading a binary operator can we provide default values?
Ans: No!. This is because even if we provide the default arguments to the parameters of the overloaded operator function we would end up using the binary operator incorrectly. This is explained in the following example:

      sample operator + ( sample a, sample b = sample (2, 3.5f ) )
      {
      }
      void main( )
      {
      sample s1, s2, s3 ;
      s3 = s1 + ; // error
      }

9.
 How do I carry out conversion of one object of user-defined type to another?
Ans: To perform conversion from one user-defined type to another we need to provide conversion function. Following program demonstrates how to provide such conversion function.
      class circle
      {
      private :
      int radius ;
      public:
      circle ( int r = 0 )
      {
      radius = r ;
      }
      } ;
      class rectangle
      {
      private :
      int length, breadth ;
      public :
      rectangle( int l, int b )
      {
      length = l ;
      breadth = b ;
      }
      operator circle( )
      {
      return circle ( length ) ;
      }
      } ;
      void main( )
      {
      rectangle r ( 20, 10 ) ;
      circle c;
      c = r ;
      }
Here, when the statement c = r ; is executed the compiler searches for an overloaded assignment operator in the class circle which accepts the object of type rectangle. Since there is no such overloaded assignment operator, the conversion operator function that converts the rectangle object to the circle object is searched in the rectangle class. We have provided such a conversion function in the rectangle class. This conversion operator function returns a circle object. By default conversion operators have the name and return type same as the object type to which it converts to. Here the type of the object is circle and hence the name of the operator function as well as the return type is circle.

10. How do I write code that allows to create only one instance of a class?
Ans: This is shown in following code snippet.

      #include
      class sample
      {
      static sample *ptr ;
      private:
      sample( )
      {
      }
      public:
      static sample* create( )
      {
      if ( ptr == NULL )
      ptr = new sample ;
      return ptr ;
      }
      } ;
      sample *sample::ptr = NULL ;
      void main( )
      {
      sample *a = sample::create( ) ;
      sample *b = sample::create( ) ;
      }
Here, the class sample contains a static data member ptr, which is a pointer
to the object of same class. The constructor is private which avoids us from creating objects outside the class. A static member function called create( ) is used to create an object of the class. In this function the condition is checked whether or not ptr is NULL, if it is then an object is created dynamically and its address collected in ptr is returned. If ptr is not NULL, then the same address is returned. Thus, in main( ) on execution of the first statement one object of sample gets created whereas on execution of second statement, b holds the address of the first object. Thus, whatever number of times you call create( ) function, only one object of sample class will be available.

11. How do I write code to add functions, which would work as get and put properties of a class?
Ans: This is shown in following code.
      #include
      class sample
      {
      int data ;
      public:
      __declspec ( property ( put = fun1, get = fun2 ) )
      int x ;
      void fun1 ( int i )
      {
      if ( i < 0 )
      data = 0 ;
      else
      data = i ;
      }
      int fun2( )
      {
      return data ;
      }
      } ;
      void main( )
      {
      sample a ;
      a.x = -99 ;
      cout << a.x ;
      }
Here, the function fun1( ) of class sample is used to set the given integer value into data, whereas fun2( ) returns the current value of data. To set these functions as properties of a class we have given the statement as shown below:
__declspec ( property ( put = fun1, get = fun2 )) int x ;

As a result, the statement a.x = -99 ; would cause fun1( ) to get called to set the value in data. On the other hand, the last statement would cause fun2( ) to get called to return the value of data.

12. How do I write code to make an object work like a 2-D array?
Ans: Take a look at the following program.
      #include
      class emp
      {
      public :
      int a[3][3] ;
      emp( )
      {
      int c = 1 ;
      for ( int i = 0 ; i <= 2 ; i++ )
      {
      for ( int j = 0 ; j <= 2 ; j++ )
      {
      a[i][j] = c ;
      c++ ;
      }
      }
      }
      int* operator[] ( int i )
      {
      return a[i] ;
      }
      } ;
      void main( )
      {
      emp e ;
      cout << e[0][1] ;
      }
The class emp has an overloaded operator [ ] function. It takes one argument an integer representing an array index and returns an int pointer. The statement cout << e[0][1] ; would get converted into a call to the overloaded [ ] function as e.operator[ ] ( 0 ). 0 would get collected in i. The function would return a[i] that represents the base address of the zeroeth row. Next the statement would get expanded as base address of zeroeth row[1] that can be further expanded as *( base address + 1 ). This gives us a value in zeroth row and first column. 13. What are formatting flags in ios class?
Ans: The ios class contains formatting flags that help users to format the stream data. Formatting flags are a set of enum definitions. There are two types of formatting flags:
      On/Off flags
      Flags that work in-group
The On/Off flags are turned on using the setf( ) function and are turned off using the unsetf( ) function. To set the On/Off flags, the one argument setf( ) function is used. The flags working in groups are set through the two-argument setf( ) function. For example, to left justify a string we can set the flag as,
      cout.setf ( ios::left ) ;
      cout << "KICIT Nagpur" ;
      To remove the left justification for subsequent output we can say,
      cout.unsetf ( ios::left ) ;
The flags that can be set/unset include skipws, showbase, showpoint,
uppercase, showpos, unitbuf and stdio. The flags that work in a group can have only one of these flags set at a time.

14. What is the purpose of ios::basefield in the following statement?
      cout.setf ( ios::hex, ios::basefield ) ;
Ans: This is an example of formatting flags that work in a group. There is a flag for each numbering system (base) like decimal, octal and hexadecimal. Collectively, these flags are referred to as basefield and are specified by ios::basefield flag. We can have only one of these flags on at a time. If we set the hex flag as setf ( ios::hex ) then we will set the hex bit but we won’t clear the dec bit resulting in undefined behavior. The solution is to call setf( ) as setf ( ios::hex, ios::basefield ). This call first clears all the bits and then sets the hex bit.

15. Can we get the value of ios format flags?
Ans: Yes! The ios::flags( ) member function gives the value format flags. This function takes no arguments and returns a long ( typedefed to fmtflags) that contains the current format flags.

16. Is there any function that can skip certain number of characters present in the input stream?
Ans: Yes! This can be done using cin::ignore( ) function. The prototype of this function is as shown below:
      istream& ignore ( int n = 1, int d =EOF ) ;
Sometimes it happens that some extra characters are left in the input stream while taking the input such as, the ?n? (Enter) character. This extra character is then passed to the next input and may pose problem.

To get rid of such extra characters the cin::ignore( ) function is used. This is equivalent to fflush ( stdin ) used in C language. This function ignores the first n characters (if present) in the input stream, stops if delimiter d is encountered.

17. Write a program that implements a date class containing day, month and year as data members. Implement assignment operator and copy constructor in this class.
Ans: This is shown in following program:
      #include
      class date
      {
      private :
      int day ;
      int month ;
      int year ;
      public :
      date ( int d = 0, int m = 0, int y = 0 )
      {
      day = d ;
      month = m ;
      year = y ;
      }
      // copy constructor
      date ( date &d )
      {
      day = d.day ;
      month = d.month ;
      year = d.year ;
      }
      // an overloaded assignment operator
      date operator = ( date d )
      {
      day = d.day ;
      month = d.month ;
      year = d.year ;
      return d ;
      }
      void display( )
      {
      cout << day << "/" << month << "/" << year ;
      }
      } ;
      void main( )
      {
      date d1 ( 25, 9, 1979 ) ;
      date d2 = d1 ;
      date d3 ;
      d3 = d2 ;
      d3.display( ) ;
      }

18. When should I use unitbuf flag?
Ans: The unit buffering (unitbuf) flag should be turned on when we want to ensure that each character is output as soon as it is inserted into an output stream. The same can be done using unbuffered output but unit buffering provides a better performance than the unbuffered output.

19.What are manipulators?
Ans: Manipulators are the instructions to the output stream to modify the output in various ways. The manipulators provide a clean and easy way for formatted output in comparison to the formatting flags of the ios class. When manipulators are used, the formatting instructions are inserted directly into the stream. Manipulators are of two types, those that take an argument and those that don?t.

20. What is the difference between the manipulator and setf( ) function?
Ans: The difference between the manipulator and setf( ) function are as follows:

The setf( ) function is used to set the flags of the ios but manipulators directly insert the formatting instructions into the stream. We can create user-defined manipulators but setf( ) function uses data members of ios class only. The flags put on through the setf( ) function can be put off through unsetf( ) function. Such flexibility is not available with manipulators.

21. How do I get the current position of the file pointer?
Ans: We can get the current position of the file pointer by using the tellp( ) member function of ostream class or tellg( ) member function of istream class. These functions return (in bytes) positions of put pointer and get pointer respectively.

22. What are put and get pointers?
Ans: These are the long integers associated with the streams. The value present in the put pointer specifies the byte number in the file from where next write would take place in the file. The get pointer specifies the byte number in the file from where the next reading should take place.

23. What do the nocreate and noreplace flag ensure when they are used for opening a file?
Ans: nocreate and noreplace are file-opening modes. A bit in the ios class defines these modes. The flag nocreate ensures that the file must exist before opening it. On the other hand the flag noreplace ensures that while opening a file for output it does not get overwritten with new one unless ate or app is set. When the app flag is set then whatever we write gets appended to the existing file. When ate flag is set we can start reading or writing at the end of existing file.

24. What is the limitation of cin while taking input for character array?
Ans: To understand this consider following statements,
      char str[5] ;
      cin >> str ;
While entering the value for str if we enter more than 5 characters then there is no provision in cin to check the array bounds. If the array overflows, it may be dangerous. This can be avoided by using get( ) function. For example, consider following statement,
      cin.get ( str, 5 ) ;
On executing this statement if we enter more than 5 characters, then get( ) takes only first five characters and ignores rest of the characters. Some more variations of get( ) are available, such as shown below:
      get ( ch ) ? Extracts one character only
      get ( str, n ) ? Extracts up to n characters into str
      get ( str, DELIM ) ? Extracts characters into array str until specified delimiter (such as ‘n’). Leaves delimiting character in stream.
      get ( str, n, DELIM ) ? Extracts characters into array str until n characters or DELIM character, leaving delimiting character in stream.

25. What is the purpose of istream class?
Ans: The istream class performs activities specific to input. It is derived from the ios class. The most commonly used member function of this class is the overloaded >> operator which can extract values of all basic types. We can extract even a string using this operator.

26. Would the following code work?
      #include
      void main( )
      {
      ostream o ;
      o << "Dream. Then make it happen!" ;
      }
Ans: No! This is because we cannot create an object of the ostream class since its constructor and copy constructor are declared private.

27. Can we use this pointer inside static member function?
Ans: No! The this pointer cannot be used inside a static member function. This is because a static member function is never called through an object.

28. What is strstream?
Ans: strstream is a type of input/output stream that works with the memory. It allows using section of the memory as a stream object. These streams provide the classes that can be used for storing the stream of bytes into memory. For example, we can store integers, floats and strings as a stream of bytes. There are several classes that implement this in-memory formatting. The class ostrstream derived from ostream is used when output is to be sent to memory, the class istrstream derived from istream is used when input is taken from memory and strstream class derived from iostream is used for memory objects that do both input and output.  Ans: When we want to retrieve the streams of bytes from memory we can use istrestream. The following example shows the use of istrstream class.
      #include
      void main( )
      {
      int age ;
      float salary ;
      char name[50] ;
      char str[] = “22 12004.50 K. Vishwanatth” ;
      istrstream s ( str ) ;
      s >> age >> salary >> name ;
      cout << age << endl << salary << endl << name ;
      cout << endl << s.rdbuf( ) ;
      }
Here, s is the object of the class istrstream. When we are creating the object s, the constructor of istrstream gets called that receives a pointer to the zero terminated character array str. The statement s >> age >> salary >> name ; extracts the age, salary and the name from the istrstream object s. However, while extracting the name, only the first word of name gets extracted. The balance is extracted using rdbuf( ).

29. When the constructor of a base class calls a virtual function, why doesn’t the override function of the derived class gets called?
Ans: While building an object of a derived class first the constructor of the base class and then the constructor of the derived class gets called. The object is said an immature object at the stage when the constructor of base class is called. This object will be called a matured object after the execution of the constructor of the derived class. Thus, if we call a virtual function when an object is still immature, obviously, the virtual function of the base class would get called. This is illustrated in the following example.
      #include
      class base
      {
      protected :
      int i ;
      public :
      base ( int ii = 0 )
      {
      i = ii ;
      show( ) ;
      }
      virtual void show( )
      {
      cout << "base's show( )" << endl ;
      }
      } ;
      class derived : public base
      {
      private :
      int j ;
      public :
      derived ( int ii, int jj = 0 ) : base ( ii )
      {
      j = jj ;
      show( ) ;
      }
      void show( )
      {
      cout << "derived's show( )" << endl ;
      }
      } ;

      void main( )
      {
      derived dobj ( 20, 5 ) ;
      }
      The output of this program would be:
      base’s show( )
      derived’s show( )

30. Can I have a reference as a data member of a class? If yes, then how do I initialise it?
Ans: Yes, we can have a reference as a data member of a class. A reference as a data member of a class is initialized in the initialization list of the constructor. This is shown in following program.
      #include
      class sample
      {
      private :
      int& i ;
      public :
      sample ( int& ii ) : i ( ii )
      {
      }
      void show( )
      {
      cout << i << endl ;
      }
      } ;
      void main( )
      {
      int j = 10 ;
      sample s ( j ) ;
      s.show( ) ;
      }
Here, i refers to a variable j allocated on the stack. A point to note here is that we cannot bind a reference to an object passed to the constructor as a value. If we do so, then the reference i would refer to the function parameter (i.e. parameter ii in the constructor), which would disappear as soon as the function returns, thereby creating a situation of dangling reference.

31. Why does the following code fail?
      #include
      class sample
      {
      private :
      char *str ;
      public :
      sample ( char *s )
      {
      strcpy ( str, s ) ;
      }
      ~sample( )
      {
      delete str ;
      }
      } ;
      void main( )
      {
      sample s1 ( “abc” ) ;
      }
Ans: Here, through the destructor we are trying to deal locate memory, which has been allocated statically. To remove an exception, add following statement to the constructor.
      sample ( char *s )
      {
      str = new char[strlen(s) + 1] ;
      strcpy ( str, s ) ;
      }
Here, first we have allocated memory of required size, which then would get deal located through the destructor.

32. assert( ) macro…
We can use a macro called assert( ) to test for conditions that should not occur in a code. This macro expands to an if statement. If test evaluates to 0, assert prints an error message and calls abort to abort the program.
      #include
      #include
      void main( )
      {
      int i ;
      cout << "nEnter an integer: " ;
      cin >> i ;
      assert ( i >= 0 ) ;
      cout << i << endl ;
      }

33. Why it is unsafe to deal locate the memory using free( ) if it has been
allocated using new?
Ans: This can be explained with the following example:
      #include
      class sample
      {
      int *p ;
      public :
      sample( )
      {
      p = new int ;
      }
      ~sample( )
      {
      delete p ;
      }
      } ;
      void main( )
      {
      sample *s1 = new sample ;
      free ( s1 ) ;
      sample *s2 = ( sample * ) malloc ( sizeof ( sample ) ) ;
      delete s2 ;
      }
The new operator allocates memory and calls the constructor. In the constructor we have allocated memory on heap, which is pointed to by p. If we release the object using the free( ) function the object would die but the memory allocated in the constructor would leak. This is because free( ) being a C library function does not call the destructor where we have deal located the memory.

As against this, if we allocate memory by calling malloc( ) the constructor would not get called. Hence p holds a garbage address. Now if the memory is deal located using delete, the destructor would get called where we have tried to release the memory pointed to by p. Since p contains garbage this may result in a runtime error.

34. Can we distribute function templates and class templates in object libraries?
Ans: No! We can compile a function template or a class template into object code (.obj file). The code that contains a call to the function template or the code that creates an object from a class template can get compiled. This is because the compiler merely checks whether the call matches the declaration (in case of function template) and whether the object definition matches class declaration (in case of class template). Since the function template and the class template definitions are not found, the compiler leaves it to the linker to restore this. However, during linking, linker doesn’t find the matching definitions for the function call or a matching definition for object creation. In short the expanded versions of templates are not found in the object library. Hence the linker reports error.

35. What is the difference between an inspector and a mutator ?
Ans: An inspector is a member function that returns information about an object’s state (information stored in object’s data members) without changing the object’s state. A mutator is a member function that changes the state of an object. In the class Stack given below we have defined a mutator and an inspector.
      class Stack
      {
      public :
      int pop( ) ;
      int getcount( ) ;
      }
In the above example, the function pop( ) removes top element of stack thereby changing the state of an object. So, the function pop( ) is a mutator. The function getcount( ) is an inspector because it simply counts the number of elements in the stack without changing the stack.

36. Namespaces:
The C++ language provides a single global namespace. This can cause problems with global name clashes. For instance, consider these two C++ header files:
      // file1.h
      float f ( float, int ) ;
      class sample { … } ;
      // file2.h
      class sample { … } ;
With these definitions, it is impossible to use both header files in a single program; the sample classes will clash. A namespace is a declarative region that attaches an additional identifier to any names declared inside it. The additional identifier thus avoids the possibility that a name will conflict with names declared elsewhere in the program. It is possible to use the same name in separate namespaces without conflict even if the names appear in the same translation unit. As long as they appear in separate namespaces, each name will be unique because of the addition of the namespace identifier. For example:
     // file1.h
      namespace file1
      {
      float f ( float, int ) ;
      class sample { … } ;
      }
      // file2.h
      namespace file2
      {
      class sample { … } ;
      }

Now the class names will not clash because they become file1::sample and file2::sample, respectively.

37. What would be the output of the following program?
      #include
      class user
      {
      int i ;
      float f ;
      char c ;
      public :
      void displaydata( )
      {
      cout << endl << i << endl << f << endl << c ;
      }
      } ;
      void main( )
      {
      cout << sizeof ( user ) ;
      user u1 ;
      cout << endl << sizeof ( u1 ) ;
      u1.displaydata( ) ;
      }
Ans: The output of this program would be,
      9 or 7
      9 or 7
      Garbage
      Garbage
      Garbage
Since the user class contains three elements, int, float and char its size would be 9 bytes (int-4, float-4, char-1) under Windows and 7 bytes (int-2, float-4, char-1) under DOS. Second output is again the same because u1 is an object of the class user. Finally three garbage values are printed out because i, f and c are not initialized anywhere in the program.

Note that if you run this program you may not get the answer shown here. This is because packing is done for an object in memory to increase the access efficiency. For example, under DOS, the object would be aligned on a 2-byte boundary. As a result, the size of the object would be reported as 6 bytes. Unlike this, Windows being a 32-bit OS the object would be aligned on a 4-byte boundary. Hence the size of the object would be reported as 12 bytes. To force the alignment on a 1-byte boundary, write the following statement before the class declaration.
     #pragma pack ( 1 )

38. Write a program that will convert an integer pointer to an integer and vice-versa.
Ans: The following program demonstrates this.
      #include
      void main( )
      {
      int i = 65000 ;
      int *iptr = reinterpret_cast ( i ) ;
      cout << endl << iptr ;
      iptr++ ;
      cout << endl << iptr ;
      i = reinterpret_cast ( iptr ) ;
      cout << endl << i ;
      i++ ;
      cout << endl << i ;
      }

39. What is a const_cast?
Ans. The const_cast is used to convert a const to a non-const. This is shown in the following
      program:
      #include
      void main( )
      {
      const int a = 0 ;
      int *ptr = ( int * ) &a ; //one way
      ptr = const_cast_ ( &a ) ; //better way
      }
Here, the address of the const variable a is assigned to the pointer to a non-const variable. The const_cast is also used when we want to change the data members of a class inside the const member functions. The following code snippet shows this:
      class sample
      {
      private:
      int data;
      public:
      void func( ) const
      {
      (const_cast (this))->data = 70 ;
      }
      } ;

40. What is forward referencing and when should it be used?
Ans: Forward referencing is generally required when we make a class or a function as a friend.
Consider following program:
      class test
      {
      public:
      friend void fun ( sample, test ) ;
      } ;
      class sample
      {
      public:
      friend void fun ( sample, test ) ;
      } ;
      void fun ( sample s, test t )
      {
      // code
      }
      void main( )
      {
      sample s ;
      test t ;
      fun ( s, t ) ;
      }
On compiling this program it gives error on the following statement of test class. It gives an error that sample is undeclared identifier. friend void fun ( sample, test );
This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample.
      class sample ;

41. How would you give an alternate name to a namespace?
Ans: An alternate name given to namespace is called a namespace-alias. namespace-alias is generally used to save the typing effort when the names of namespaces are very long or complex. The following syntax is used to give an alias to a namespace.
      namespace myname = my_old_very_long_name ;

42. Using a smart pointer can we iterate through a container?
Ans: Yes. A container is a collection of elements or objects. It helps to properly organize and store the data. Stacks, linked lists, arrays are examples of containers. Following program shows how to iterate through a container using a smart pointer.
      #include
      class smartpointer
      {
      private :
      int *p ; // ordinary pointer
      public :
      smartpointer ( int n )
      {
      p = new int [ n ] ;
      int *t = p ;
      for ( int i = 0 ; i <= 9 ; i++ )
      *t++ = i * i ;
      }
      int* operator ++ ( int )
      {
      return p++ ;
      }
      int operator * ( )
      {
      return *p ;
      }
      } ;
      void main( )
      {
      smartpointer sp ( 10 ) ;
      for ( int i = 0 ; i <= 9 ; i++ )
      cout << *sp++ << endl ;
      }
Here, sp is a smart pointer. When we say *sp, the operator * ( ) function gets called. It returns the integer being pointed to by p. When we say sp++ the operator ++ ( ) function gets called. It increments p to point to The next element in the array and then returns the address of this new location.

43. Can objects read and write themselves?
Ans: Yes! This can be explained with the help of following example:
      #include
      #include
      class employee
      {
      private :
      char name [ 20 ] ;
      int age ;
      float salary ;
      public :
      void getdata( )
      {
      cout << "Enter name, age and salary of employee : " ;
      cin >> name >> age >> salary ;
      }
      void store( )
      {
      ofstream file ;
      file.open ( “EMPLOYEE.DAT”, ios::app | ios::binary ) ;
      file.write ( ( char * ) this, sizeof ( *this ) ) ;
      file.close( ) ;
      }
      void retrieve ( int n )
      {
      ifstream file ;
      file.open ( “EMPLOYEE.DAT”, ios:

TERADATA Placement Papers

Teradata is the world’s leading analytic data solutions company focused on integrated data warehousing, big data analytics, and business applications that provide actionable business intelligence. It’s our passion and it’s all we do. We deliver award-winning, integrated, purpose built platforms based on the most powerful, scalable, and reliable technology platform in the industry. Our assets include:

Why FLOAD doesn’t supports NUSI?

Where as Mload supports NUSI.

Why MLOAD needs Work Tables?

what is the command to copy windows files (i386) without formatting OS?

what are the types of facts ? explain them?

What are cookies and how to test cookies

What are the different functions of Syntax phase, Scheduler?

Flipkart.com India Placement Papers

Flipkart.com: Online Shopping IndiaFlipKart – India’s leading portal for Online Shopping of Books, Mobile Phones, Digital Cameras, Laptops, Watches & Other Products at best price.

Flipkart is built by a group of passionate, innovative, relentless individuals, who are constantly striving to provide a wholesome online shopping experience to the customer.



Job Category :   IT
Job Location :  Gurgaon,Delhi
Company :  Flipkart India



Flipkart Placement Paper January 2012:-

1. A secret can be told only 2 persons in 5 minutes .the same person tells to 2 more persons and so on . How long will take to tell it to 768 persons ?
    
a) 47.5 min b)50 min c) 500 min d)49 min  
    
Ans: 47.5 min

2. When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The  average age of my family during my marriage is same as today. What is the age of Father when he died?
    
Ans: 70.

3. A son and father goes for boating in river upstream . After rowing for 1 mile son notices the hat of his father falling in the river. After 5 min. he tells his father that his hat has fallen. So they turn round and are able to pick the hat at the point from where they began boating after 5min. Tell the speed of river? 
    
Ans: 6 miles/hr 

4. There are three departments having students 64,58,24 .In an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only (No mixing of departments. Find the minimum number rooms required?
     
Ans : 73
     
5. Argentina had football team of 22 player of which captain is from Brazilian team and goalki from European team. For remaining player they have picked 6 from Argentinean and 14 from European. Now for a team of 11 they must have goalki and captain so out of 9 now they plan to select 3 from Argentinean and 6 from European. Find out number of methods available for it.
    
Ans : 160600

6. What causes “Thrashing” of a program : 

A.) The constant swapping of program due to page faults. 
B.) The inability of a program to get assess to a network resource. 
C) A near overflow / underflow of a variable. 
D) Assessing a memory area not allocated to the process. 

Ans. A

7. Turbo-C is a / an 

A) IDE and C compiler/linker. 
B) C-compiler/linker 
C) C . 
D) code generator. 

Ans. A

8. The path of creation of an executable is : 

A) coding, linking, compiling, parsing. 
B) coding, parsing, compiling, linking. 
C) coding, compiling, parsing, linking. 
D) coding, compiling, linking, parsing. 

Ans. B

9. The classic way of checking whether a mathematical expression has matched parenthesis will employ the following data structure : 

A) List. 
B) Directed Graph 
C) Threaded Binary tree. 
D) Stack. 

Ans. D

10. The fastest sorting algorithm for a Random set of numbers is: 

A) Quick sort 
B) Shell sort 
C) Bubble sort 
D) Double Bubble sort.