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Logical Reasoning Section

2010-03-06T22:42:00.000+05:30

Logical Reasoning Section
The logical reasoning section consists of some passages followed by 4 to 7 questions
per passage. The questions are such that they require ability to read fast and
comprehend. The questions asked in this section have three choices TRUE, FALSE,
CAN'T SAY. Some examples of questions are given below. Please note that these
passages are not the exact passages asked. The passages used a good deal of difficult
words, which have been removed in this reproduction. Also the passages appearing in
the actual paper are much lengthier.
1. My father has no brothers. He has three sisters who have two Childs each.
1> my grandfather has two sons (f)
2> three of my aunts have two sons (can't say)
3> my father is only child to his father (f)
4> I have six cousins from my mother side (f)
5> I have one uncle (f)
2. Ether injected into gallbladder to dissolve gallstones. This type one-day treatment is
enough for gallstones not for calcium stones. This method is alternative to surgery for
millions of people who are suffering from this disease.
1>-calcium stones can be cured in one day (f)
2> hundreds of people contains calcium stones (can't say)
3> surgery is the only treatment to calcium stones (t)
4> either will be injected into the gallbladder to cure the cholesterol based gall
stones (t).
3. Hacking is illegal entry into other computer. This is done mostly because of lack of
knowledge of computer networking with networks one machine can access to another
machine. Hacking go about without knowing that each network is accredited to use
network facility.
1> Hacking people never break the code of the company, which they work for (can't
say).
2> Hacking is the only vulnerability of the computers for the usage of the data. (f)
3> Hacking is done mostly due to the lack of computer knowledge (f).
4. Alpine tunnels are closed tunnels. In the past 30 yrs not even a single accident has
been recorded for there is one accident in the railroad system. Even in case of a fire
accident it is possible to shift the passengers into adjacent wagons and even the living
fire can be detected and extinguished with in the duration of 30 min.
1> no accident can occur in the closed tunnels (True)
2> fire is allowed to live for 30 min. (False)
3> All the care that travel in the tunnels will be carried by rail shutters. (True)
5. In the past helicopters are forced to ground or crash because of the formation of the
ice on the rotors and engines. A new electronic device has been developed which can
detect the water content in the atmosphere and warns the pilot if the temperature is
below freezing temp about the formation of the ice on the rotors and wings.
1> the electronic device can avoid formation of the ice on the wings (False).
2> There will be the malfunction of rotor & engine because of formation of ice (t)
3> The helicopters are to be crashed or down (t)
4> There is only one device that warns about the formation of ice (t).
6. In the survey conducted in Mumbai out of 63 newly married house wives not a single
house wife felt that the husbands should take equal part in the household work as they
felt they loose their power over their husbands. In spite of their careers they opt to do
the kitchen work themselves after coming back to home. The wives get half as much
leisure time as the husbands get at the weekends.
1> housewives want the husbands to take part equally in the household (f)
2> wives have half as much leisure time as the husbands have (f)
3> 39% of the men will work equally in the house in cleaning and washing
7. Copernicus is the intelligent. In the days of Copernicus the transport and technology
development was less & it took place weeks to communicate a message at that time.
Where in we can send it through satellite with in no time ----------. Even with these fast
developments it has become difficult to understand each other.
1> people were not intelligent during Copernicus days (f).
2> Transport facilities are very much improved in now a days (can't say)
3> Even with the fast developments of the technology we can't live happily.(can't
say).
4> We can understand the people very much with the development of
communication (f).
8. Senior managers warned the workers that because of the introductory of Japanese
industry in the car market. There is the threat to the workers. They also said that there
would be the reduction in the purchase of the sales of car in public. The interest rates of
the car will be increased with the loss in demand.
1> Japanese workers are taking over the jobs of Indian industry (false)
2> Managers said car interests would go down after seeing the raise in interest rates
(true)
3> Japanese investments are ceasing to end in the car industry (false)
4> people are very much interested to buy the cars (false)
9. In the totalitarian days, the words have very much devalued. In the present day, they
are becoming domestic that is the words will be much more devalued. In those days, the
words will be very much affected in political area. But at present, the words came very
cheap .we can say they come free at cost.
1> totalitarian society words are devalued. (False)
2> totalitarian will have to come much about words (True)
3> The art totalitarian society the words are used for the political speeches.
10. There should be copyright for all arts. The reel has come that all the arts has come
under one copyright society, they were use the money that come from the arts for the
developments. There may be a lot of money will come from the Tagore works. We
have to ask the benefactors from Tagore work to help for the development of his works.
1> Tagore works are come under this copyright rule. (F)
2> People are free to go to the because of the copyright rule. (Can’t say)
3> People gives to theater and collect the money for development. (Can’t say)
4> We have ask the Tagore residents to help for the developments of art. (Can’t say)
11. Senior manager in a big company said that new Japanese company invades in India
for transferring the cars from industrial and warned that jobs were under threat from
Japanese company. They stated that increasing competence would be coupled with an
inevitable down term in car market and recent rise in interest rate which has already hit
demand.
1> Manager issue their warning after a rise in interest rate
2> Manager told workers that Japanese workers are taking jobs away from Indian
workers
3> Manager said that more people want to buy new cars in future
4> Increasing rate of interest mean that Japanese firm will create into operate in the
country
12. Human existence is suspicious of arbitrary divide between concise and unconcise.
The concise world invades shape activity of the unconcise, while many of great activity
of humanity waking as whole or partially improved by dreams. Even it could be
ignored that dreams precede exceptional such a dichotomy could not be drawn as the
influence of dream on waking state would remain unclear. But as yet no company
rebuilt exists to record the substitute of dreaming.
1> Sleepy can be creative state
2> It is difficult to tell whether a sleeper is dream or not
3> If we know what babies would dream about before they are born we could show
that the concise and unconcise mind influence on one another
4> It is untrue claim that concise and unconcise world never impinge one another
13. Any one who has systematic exam phases will have perceived a profound although
not a prolific of asymmetry whether or not the exception is volitions and self-control of
spontaneous appeal to predict facial as symmetry as does the type of emotion portrayed.
Position cannot displace symmetric at left side regret of a negative emotion is more
common posed. Expression negation emotions are likely to be symmetric representation
and where as symmetric occurs relative left sided expression is more common.
1> Any angry person is more likely to have left sided expression than some one who
has smiling
2> An actor is likely to smile symmetric when acting
3> Delicious facial expression will always be as symmetrical
14. In the totalitarian days, the words have very much devalued. In the present day, they
are becoming domestic that is the words will be much more devalued. In those days,
the words will be very much affected in political area. But at present, the words came
very cheap; we can say they come free at cost.
1> Totalitarian society words are devalued
2> Totalitarian will have to come much about words
3> The art totalitarian society the words are used for the political speeches
15. There should be copyright for all arts. The rule has come that all the arts has come
under one copy right society, they were use the money that come from the arts for the
developments. There may be a lot of money will come from the Tagore works. We
have to ask the benifitters from Tagore work to help for the development of his works.
1> Tagore works are come under this copyright rule.
2> People give to theater and collect the money for development.
3> People are free to go because of the copy right rule.
4> We have asked the Tagore residents to help for the developments of art.
16. Disease x succeeds lung disease. It may also occur healthy persons the person looks
healthier from outside. The number of red blood cells in the blood increased and this
leads to thickness of the blood as a result of which blood lacks inside the vessels .the
person suffers heart attack. Possible solution is to remove a liter of blood or to control
the growth of red blood cells.
1> lung disease precedes the disease -- T
2> a person who gets heart attack always has disease -- F
3> people suffering from disease x look healthier -- C
4> one liter of blood must be removed to cure this -- T
17. A star fire engine has been designed to work only for unleaded Petrol. This crossfire
engine is designed for both 1500 and 1800 cc they internally and externally look alike
except for the difference in wheels ----,2000 cc is different from the above. Two in
terms of 3 Features --- ----- ---- on the bumper.
1> the engine runs only on the leaded petrol -- F
2> 1500 and 1800 cc look alike -- F
3> apart from cross fire engines they manufacture -- F
4> internal structure of the 2000cc is same as that of both 1500 and 1800 cc -- C
18. A weed killer to kill the weeds has been developed. The weed killer solution has to
be applied to the growing tips. It needs not to be applied to all the tips. the solution has
to be prepared in a can in one hand and the plants are taken in the other hand with a
glove and are immersed in the solution if we cannot immerse them in the solution. The
solution can be applied to the roots with a brush. it used without care it can pose danger
to other plants.
1> the solutions has to applied to growing leaves
2> to use the weed killer first take the plants in the can as then pour the solution into
it.
3> it is sufficient to apply it only to some roots
4>it effects to the other plants.
The following Campus Recruitment papers are quite important. Read the book
references mentioned therein and practice complete questions from the given page nos.
Campus Recruitment at NIT Bhopal 2003
1.Eight players have to be seated. G, J, L are in the varsity baseball team. H, M, O are
both in the varsity baseball team and the football team. K and N are in the varsity
baseball team as well as the basketball team. You have to seat them in such a way that
two players who play in two teams never sit adjacent to each other. Now answer the
following questions: -(There were four questions. I can’t recollect them but once u
make the initial arrangement it becomes pretty easy.)
2. You have to make seating arrangements for 8 people. Andy and Goldberg have to
leave early so they must be seated next to the door, which is on the right rear end of the
table. Cooper and D’ameto are enemies so they have to be seated as far as possible in
the table. Edmund and Farley always sit together. Burns always occupies the center
position.
Answer the following questions.
1.If Edmund and D’ameto never sit opposite Burns then in how many ways can the
arrangement be made? Etc.
3.A person needs to get himself affiliated to the Nambrian Consulate. He must be either
a US citizen or a neutralized person. He has to get the certificate from the Nambrian
Consulate, which is open only on Monday, Wednesday and Friday from 12 to 4
o’clock. For this he also needs to get himself vaccinated from the Alabama Clinic,
which is near the consulate and is open on Wednesday till 5 pm. He may also get the
vaccination from The Beryl Clinic, which is open on Mondays to Thursdays from 9-12
pm. And on Fridays from 4-5 pm. However it takes one hour to get to the clinic from
the Nambrian Consulate. Also he must show a notarized bank account of more than
$1000, which can be done from Monday to Friday from 9am to 3pm.
Campus Recruitment NIT Durgapur2003
six floor is given. there are 7 or 8 people .each floor contain two apartments. In each
apartment there were 2 or persons. We have to determine the person what floors they
are staying (REFERENCE 12 TH BARRON)
25 PERSONS ARE IN A COMMITTEE. THERE A,B,C,D,E,F ,G,H.A&C and
C&E,D&F &D&G cant go together From this there are easy questions # barron’ page
487 sec 6 analytical ability prob 8-12 abt 6ix American city…….
# barron’ page 435 sec 5 analytical ability problem 1-4 about city a ,b and c …….
#3 Toll in a high way is given .The peoples cant reminding the data. It ‘s also from
Barron Book (12th Barron edition).
Refer Barron for these problems. There were 3 of those. One of them we have given
below:
A project to consolidate the programs of a large university and a small college is set up.
It is agreed that the representatives work in small committees of 3, with 2 from the
large university. Faculty members of same subject area should represent no
committee. The professors were:
Large University: J(English),K(Maths.),L,(Natural Science)
Small college: M(Maths.),N(Latin), O,P(Both English)
Questions:
1. Which of the following represents a properly composed committee?
(KLN,KLM,JKL,JON,JKM) akln
2.Which of the following may serve with P?
(KM,KL,KO,JK,MN) bkl
3. Which of the following must be true?
a. if J serves,P must be assigned.
b. if J can't serve,M also can't be assigned
c. if J can't serve,L must serve.
4. If L is not available,which two must be available ?
MJ,OJ,NJ,NO,PJ

C through

2010-03-06T22:38:00.000+05:30

C Thorough Test
1. fn(int n,int p,int r)
{
static int a=p;
switch(n);
{
case4: a+ = a*r;
case3: a+ = a*r;
case2: a+ = a*r;
case1: a+ = a*r;
}
}
The aboue programme calculates
a.Compound interest for 1 to 4 years
b.Amount of Compound interest for 4 years
c.Simple interest for 1 year
d.Simple interest for 4 year
2. int a[50], *pa;
pa = &a[0];
In order to access 5th element, find the incorrect one
a.pa[5]
b.a[5]
c. *(*(pa+5))
d.*(a+5)
3. Regarding the variables select incorrect one
Ans. Auto variables select incorrect one
4. Write onest equivalent to the following two
x + sqrt(a);
return(x);
Choose one of the alternatives
a.printf("%d",sqrt(a));
b.return(sqrt(a));
c.printf("sqrt(a)");
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b.return(a*a*a);
5. Which of the following is not an infinite loop
a.while(i){....}
b.do{......} while(i);
c.for(ii)
d.#define TRUE 0
while(TRUE){....};
6. What does the following function print ?
func(int i)
{ if (i%2)
return 0;
else
return 1;
}
main()
{
int i=3;
i=func(i);
i=func(i);
printf("%d",i);
}
a.2 b.3 c.1 d.0
7.to10.th bit: Choose one of the alternatives for the following
a.float b.int c.char d.string
7.'9'
8."1e02"
9.10e03
10.15
11. Consider the following structure
struc num_name{int no;
char name[25];}
struc num_name n1[]={{12,"find"},{15,"matrix"},{8,"pick"}};
.........
.........
.........
printf("%d%d",n1[2],no,(*(n1+2)),no+1);
What does the above st. do ?
a. 8,9 b. 8,8 c. 9,9 d. 8,unpredictable value
12. for(i=0; i!=10; i+=2)
printf(".....");
How many time will it execute ?
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a.10 b.0 c.5 d.not even once
13. 20 bytes are allocated to a string s
s="ENIRANC B"
l = strlen(s);
then i = ?
a.20 b.22 c.21 d.8
14. y=5;
if ( x == 10)
else if ( x == 9)
else y=3;
then y = ?
a.8 b.7 c.5 d.0
15. Which is incorrect ?
a.a+=b;
b.a*=b;
c.a>=b;
d.a**=b;
16. operator for
a. not available b. ** c. ^ d. %
17. cond1? cond2 ? cond3 ? : exp2 : exp2 : exp3 :exp4
the equvalent set of statements to the above is
a. b. c. d.
18. valuue stored in a variable of type double is
a. can be less athan int or float or long
b. always > int
c. always < long
d. always > float
19. Thetre are 100 functions and first 50 are in file 1.c and rest
in file 2.c then they must be declared in file 1.c as
a. auto
b. global
c. static
d. external
20. struct out {
struct in{ char c;int d;}s1,*p1;
}a1, *pa1;
pa1 = &a1;
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a1.p1 = &s1;
Which is incorrect?
a. a1.p1->c
b. pa1->p1->c
c. a1->p1
d. a1.s1.c
21. if a=z then value a <<1 is ?
a. 3 b.4 c.2 d.1
22. #define prod(a,b) a*b
x = 2; y = 2;
prod(x+2,y+1) = ?
a. 6 b.12 c. 7 d.16
23. int sum = 1;
switch (2)
{
case 2: sum sum+2;
case 3: sum*=2; break;
defdault: sum = 0;
}
which is the value of sum ?
a. 2 b. 6 c. 1 d. 0
24. Which one of the following is invalid?
a.if(a==1) b. if(a != 3)
c. if(a25. int x = 5, *p;
p = &x;
printf("%d",++ *p); what is the output?
a. 2 b. 6 c.1 d. 0
26. unsigned int i = 10;
while (i>=0) {---------; i--}
How many times the loop is executed?
a.10 b.9 c. infinite d.11
27. pick the odd man out
a. malloc
b. calloc
c. realloc
d. free
28. pick the odd man out
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a. c = fgetc(stdin)
b. c= getch();
c. -----
d. gets(s);
29. Which is incorrect regarding a recursive function
a. does not return a value at all
b. calls itself
c. equals to a loop
d. has a termination condition
30. Write an expression for " either x lies in the
range of 10 to 5o or y = 0"
...........
31. x = 7; y= 0;
if(x ==6) y =7;
else y = 1;
what is the value of y?
32. Choose the incorrect one
auto variables within a function are
a. global
b. local
c. its type must be declared before using
d. -----
33. The delimitor for statements is
a. semicolon
b. colon
c. ---
d. ---

C programming technical questions for interviews

2010-02-16T07:57:00.000+05:30

All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. 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; }
}

Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. 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);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char

14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
 *p that is value at the location currently pointed by p will be taken
 ++*p the retrieved value will be incremented
 when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem

25. main()
{
41printf("%p",main);
}8Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

27) main()
{
clrscr();
}
clrscr();

Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28) enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

29) void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer

30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words the scope of the labels is limited to functions. The label 'here' is available in function fun() Hence it is not visible in function main.

33) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

36) #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100

39) main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40) #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

41) #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

42) #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.

46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.

2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

**49) main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
M O U S E \0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
M T R A C K \0
The third input starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

51) main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

52) main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

53) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

59) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71) #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant

72) #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

73) #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)

76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes

78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.

80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.

81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

82) # include
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83) # include
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85) #include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.

86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0

89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.

92) What are the following notations of defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation

93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:

100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

101) void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of time.

102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration.
So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.

104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);

}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.

105) #include
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.

106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !

107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10

108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.

110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.

111) void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.

112). What is the output of the program given below

main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);

}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.

Apply clock-wise rule to find the meaning of this definition.

116). What is the output for the program given below

typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s convenience.

117) typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}

Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.

Note
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

118) #ifdef something
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.

119) #if something == 0
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.

120). What is the output for the following program

main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3 integers each .

The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.

121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand this.

123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b

124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.

125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.

126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}

main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing "bye".

127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}

Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.

128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

129) void ( * abc( int, void ( *def) () ) ) ();

Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer variable.(b). a ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.

130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}

Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.

133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}

Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;

136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal

2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal

3. Same as 1.

137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}

Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}

Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.

139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.

140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.

141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.

142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.

143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is the character array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.

144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.

145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.

146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].

147) main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

148) main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

149) main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.

150) main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.

151) #include
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptrprintf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

152) Declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])( );

153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of type struct date is given. The compiler doesn’t have the definition of date structure (forward reference is not allowed in C in this case) so it issues an error.

154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition of ‘student’ but to have a variable of type struct date the definition of the structure is required.

155) There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will return EOF only when fread tries to read another record and fails reading EOF (and returning EOF). So it prints the last record again. After this only the condition feof(fp) becomes false, hence comes out of the while loop.

156) Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.

157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++ p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So p remains NULL where *p =0 may result in problem (may rise to runtime error “NULL pointer assignment” and terminate the program).

158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The check s != NULL is for error/exception handling and for that assert shouldn’t be used. A plain if and the corresponding remedy statement has to be given.

159) What is the hidden bug with the following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In assert, the experssion involves side-effects. So the behavior of the code becomes different in case of debug version and the release version thus leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.

160) void main()
{
int *i = 0x400; // i points to the address 400
*i = 0; // set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to some location whose value may not be available for modification. This type of pointer in which the non-availability of the implementation of the referenced location is known as 'incomplete type'.

161) #define assert(cond) if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())

void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file %s, line %d \n",#cond, __FILE__,__LINE__), abort()))

Note:
However this problem of “matching with nearest else” cannot be solved by the usual method of placing the if statement inside a block like this,
#define assert(cond) { \
if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}

162) Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be recursive without end.

163) Is the following code legal?
struct a
{
int x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows, the size of the pointer to a structure even before the size of the structure
is determined(as you know the pointer to any type is of same size). This type of structures is known as ‘self-referencing’ structure.

164) Is the following code legal?
typedef struct a
{
int x;
aType *b;
}aType
Answer:
No
Explanation:
The typename aType is not known at the point of declaring the structure (forward references are not made for typedefs).

165) Is the following code legal?
typedef struct a aType;
struct a
{
int x;
aType *b;
};
Answer:
Yes
Explanation:
The typename aType is known at the point of declaring the structure, because it is already typedefined.

166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
aType *b;
};
}
Answer:
No
Explanation:
When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This is known as ‘incomplete types’.

167) void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
printf(“sizeof (int *) = %d \n”, sizeof(int *));
printf(“sizeof (double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));
}
Answer :
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *) = 2
sizeof(struct unknown *) = 2
Explanation:
The pointer to any type is of same size.

168) char inputString[100] = {0};
To get string input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString) doesn't know the size of the string passed and so, if a very big input (here, more than 100 chars) the charactes will be written past the input string. When fgets is used with stdin performs the same operation as gets but is safe.

169) Which version do you prefer of the following two,
1) printf(“%s”,str); // or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any format characters like %d then it will result in a subtle bug.

170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.

171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

172) Is this code legal?
int *ptr;
ptr = (int *) 0x400; Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location 0x400.
173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to get stored.

174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.

175) main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u %u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

176) main()
{
extern int i;
{ int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int i;

177) Printf can be implemented by using __________ list.
Answer:
Variable length argument lists
178) char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

Ways to Make a Good Impressions

2010-02-02T23:58:00.000+05:30

Impressions are important: They leave an initial taste in people's mouths that can remain prevalent for the entire relationship. If you are paranoid about what kind of impression you make, run through these seven list items and see if you are consistent with them; if you are, then you will probably expose the best of yourself. If not, then work to meet these standards.
1. Dress: The absolute first impression you will make on someone will be through your clothing, because that is what is seen from a distance, and cannot change throughout your meeting. Make sure to dress according to the situation-don't over or under dress-and maintain within the limits of good taste. If you aren't sure if what you're wearing looks good, ask people for an honest opinion. One last thought: always, and I mean always, pull up your pants
2. Hygiene: Take a shower! Shave! Brush your teeth! You must be fully bathed and groomed before you meet with someone for the first time, because scruffy looking people generally don't seem as neat and mature. Pay attention to the little elements like breath: keep a pack of mint gum with you wherever you go, and periodically check to make sure you aren't killing bugs every time you breathe out. If you sweat heavily, keep a small stick of deodorant/anti-perspirant close, and if you notice you're stinking you can freshen up. People notice the minutiae!
3. Manners: At the table and with other people be civilized, polite and respectful: keep your elbows off of the table, open doors for people and address everyone-initially, at least-by their formal title. This will make an especially good impression on senior citizens, because you will prove that you aren't one of those "new fangled punks."
4. Speech: Have clean, clear diction and speak sans "like" or "you know." It is important to be articulate because that inspires a feeling of intelligence and education in the person you are meeting with. Always leave out profanity, and whatever you do, make sure to speak loud enough for all to hear, because conversationalists are easily agitated if you force them say "excuse me?" more than a few times.
5. Discretion: Choose what to share about yourself: forget to tell everyone about that time you went camping and ruptured your appendix, then fell face first into a pile of bug infested leaves-it is rude and will alienate you from the group. Try to withhold from conversations on personal subjects like religion or more disgusting topics like personal medical care. Before you speak, think about the possible impact of what you might say, then imagine its implications in the long run.
6. Humor: Humor can be your most powerful tool or your doom, because everyone has a slightly different sense of humor. What might be hilarious to you might seem disgusting to another, or vice versa. Try to withhold from any jokes that aren't family or dinner table friendly; you can tell those later.
7. Start and End with a Bang: Whoever you are meeting with will remember how you greet them, and then in what manner you left them. If you feel you have trouble with this, practice a few different phrases in the mirror, and introduce elements like: "pleased to meet you," or "honored to make your acquaintance." Ignore the antiquity of these phrases; it often makes them more memorable.

Making a good impression will set any relationship off on a good foot. If you are in a situation where you need to be judged at face value-such as a job interview or date-then make sure to go through this list and make sure you are within bounds of reason and good taste on all of your decisions.

Happiness In 7 Simple Tricks

2010-02-02T23:39:00.000+05:30

1. Take the others as they are. Don’t try to change people around you, because nobody ever changes, unless they feel like doing so. Instead of making your friends unhappy because they are not able to live up to your standards, try to adjust your way of looking at them, try to accept their way, and enjoy the good things you saw in them from the very beginning (before becoming friends, there was a kind of attraction between you, wasn’t it?)
2. Indulge yourself. I know people who wouldn’t forgive themselves for making even the smallest mistake. They think and re-think the moments, blaming themselves, getting angry and frustrated, thus not being able to enjoy the present moment. The frustration comes from the fact that they cannot travel back in time to do things right. If you cannot fix things anyway, what’s the use of accusing yourself and punishing you ever and ever again? Get over the past and live your life!
3. Smile. Research has shown that smiling and feeling good work both ways. If you feel good, you’ll smile, right? Well, our body is strange, it seems that it has the power to influence the mind: so, if you smile, you’ll immediately feel better. Besides, you’ll have a positive effect on people around you: seeing you smiling, they will tend to smile back to you, thus contributing to a good atmosphere.
4. Respect yourself. You are your most valuable asset, so make sure you take good care of your soul. Don’t do things which don’t fit into your life philosophy. Don’t give you reasons to hate yourself later on.
5. Respect the others. It is not always true that you get what you give, but if you give lack of respect, you’ll get it back for sure.
6. Live the moment. Don’t wait for the most proper time to do something you wish to. Do it now, because the proper moment never comes. There is always something more to wait for, so you’ll end up old or dead and burried before you even notice that your moment never came. Do you want to learn how to play the piano? Don’t wait until you’ll have more time. Make yourself some time, find a place in your schedule and go take piano classes. Do you want to write in your blog every day? Just sit down and write. You’ll see that time will be there for you when you need it.
7. Give. Whatever you can, be it knowledge, money, old clothes or even a phone call to a lonely person, just make sure you give something every day. This will make you feel like a good person. Helping is in the human nature, so why not following our giving impulse? There are many people who are in need for your help, that you’ll never be refused if you know how to look for those people.

How to Create Lock in a folder.

2010-02-02T23:27:00.000+05:30

How to Create a Lock folder.

Today i will tell u How to make a lock folder.
Plz follow the simple steps..
Suppose you want to lock the folder movies in d: which has the path D:\movies.
In the same drive create a text file and type
ren movies movies.{21EC2020-3AEA-1069-A2DD-08002B30309D}

Now save this text file as loc.bat
Create another text file and type in it
ren movies.{21EC2020-3AEA-1069-A2DD-08002B30309D} movies

Now save this text file as key.bat
Now you can see two batch files “loc and key”. Press loc and the folder games will change to control panel and you cannot view its contents. Press key and you will get back your original folder.
Try it out!!!!!!!

MS-DOS HISTORY

2010-02-02T23:25:00.000+05:30

The history of MS-DOS was developed by Seattle Computer Products to run on IBM's new PC.
Version

Date

Comments
1.0

1981

The original version of MS-DOS. This was a renamed version of QDOS which had been purchased by an upstart company called Microsoft.
1.25

1982

This added support for double-sided disks. Previously the disk had to be turned over to use the other side
2.0

1983

This added support for IBM's 10 MB hard disk, directories and double-density 5.25" floppy disks with capacities of 360 KB
2.11

1983

Support for foreign and extended characters was added.
3.0

1984

Support for high-density (1.2 MB) floppy disks and 32 MB hard disks was added.
3.1

1984

Network support was added.
3.3

1987

This release was written to take advantage of IBM's PS/2 computer range. It added support for high density 3.5" floppy disks, more than one partition on hard disks (allowing use of disks bigger than 32 MB) and code pages.
4.0

1988

This version provided XMS support, support for partitions on hard disks up to 2 GB and a graphical shell. It also contained a large number of bugs and many programs refused to run on it.
4.01

1989

The bugs in version 4.0 were fixed.
5.0

1991

This was a major upgrade. It allowed parts of DOS to load itself in the high memory area and certain device drivers and TSRs to run in the unused parts of the upper memory area between 640K and 1024K. This version also added support for IBM's new 2.88 MB floppy disks. An improved BASIC interpreter and text editor were included, as was a disk cache, an undelete utility and a hard-disk partition-table backup program. After the problems with MS-DOS 4, it also provided a utility to make programs think they were running on a different version of MS-DOS.
5.0a

1992/3

This was a minor bug fix which dealt with possibly catastrophic problems with UNDELETE and CHKDSK.
6.0

1993

This was a catch-up with Novell's DR-DOS 6. It added a disk-compression utility called DoubleSpace, a basic anti-virus program and a disk defragmenter. It also finally included a MOVE command, an improved backup program, MSBACKUP and multiple boot configurations. Memory management was also improved by the addition of MEMMAKER. A number of older utilities, such as JOIN and RECOVER were removed. The DOS Shell was released separately as Microsoft felt that there were too many disks.
6.2

1993

Extra security was built into DoubleSpace following complaints of data loss. A new disk checker, SCANDISK, was also introduced, as well as improvements to DISKCOPY and Smart Drive.
6.21

1993

Following legal action by Stac Electronics, Microsoft released this version which had DoubleSpace removed. It came with a voucher for an alternative disk compression program.
6.22

1994

Microsoft licenced a disk-compression package called Double Disk from VertiSoft Systems and renamed it DriveSpace, which was included in this version.
7.0

1995

This version is part of the original version of Windows 95. It provides support for long filenames when Windows is running, but removes a large number of utilities, some of which are on the Windows 95 CD in the \other\oldmsdos directory.
7.1

1997

This version is part of OEM Service Release 2 and later of Windows 95. The main change is support for FAT 32 hard disks, a more efficient and robust way of storing data on large drives.

Microsoft Code name
Interface Manager

Windows 1.x

Chicago

Windows 95

Version 4.00
Frosting

Windows 95 PlusPack!

Detroit

Windows 95 OSR 2

Memphis

Windows 98

Version 4.10

Georgia

Windows ME

Version 4.90

Computer Commands

2010-02-02T23:23:00.000+05:30

Control Panel
· CONTROL: opens the control panel window
· CONTROL ADMINTOOLS: opens the administrative tools
· CONTROL KEYBOARD: opens keyboard properties
· CONTROL COLOUR: opens display properties.Appearance tab
· CONTROL FOLDERS: opens folder options
· CONTROL FONTS: opens font policy management
· CONTROL INTERNATIONAL or INTL.CPL: opens Regional and Language option
· CONTROL MOUSE or MAIN.CPL: opens mouse properties
· CONTROL USERPASSWORDS: opens User Accounts editor
· CONTROL USERPASSWORDS2 or NETPLWIZ: User account access restrictions
· CONTROL PRINTERS: opens faxes and printers available
· APPWIZ.CPL: opens Add or Remove programs utility tool
· OPTIONALFEATURES: opens Add or Remove Windows component utility
· DESK.CPL: opens display properties. Themes tab
· HDWWIZ.CPL: opens add hardware wizard
· IRPROPS.CPL: infrared utility tool
· JOY.CP: opens game controllers settings
· MMSYS.CPL: opens Sound and Audio device Properties. Volume tab
· SYSDM.CPL: opens System properties
· TELEPHON.CPL: Opens phone and Modem options
· TIMEDATE.CPL: Date and Time properties
· WSCUI.CPL: opens Windows Security Center
· ACCESS.CPL: opens Accessibility Options
· WUAUCPL.CPL: opens Automatic Updates
· POWERCFG.CPL: opens Power Options Properties
· AZMAN.MSC: opens authorisation management utility tool
· CERTMGR.MSC: opens certificate management tool
· COMPMGMT.MSC: opens the Computer management tool
· COMEXP.MSC or DCOMCNFG: opens the Computer Services management tool
· DEVMGMT.MSC: opens Device Manager
· EVENTVWR or EVENTVWR.MSC: opens Event Viewer
· FSMGMT.MSC: opens Shared Folders
· NAPCLCFG.MSC: NAP Client configuration utility tool
· SERVICES.MSC: opens Service manager
· TASKSCHD.MSC or CONTROL SCHEDTASKS: opens Schedule Tasks manager
· GPEDIT.MSC: opens Group Policy utility tool
· LUSRMGR.MSC: opens Local Users and Groups
· SECPOL.MSC: opens local security settings
· CIADV.MSC: opens indexing service
· NTMSMGR.MSC: removable storage manager
· NTMSOPRQ.MSC: removable storage operator requests
· WMIMGMT.MSC: opens (WMI) Window Management Instrumentation
· PERFMON or PERFMON.MSC: opens the Performance monitor
· MMC: opens empty Console
· MDSCHED: opens memory diagnostics tools
· DXDIAG: opens DirectX diagnostics tools
· ODBCAD32: opens ODBC Data source Administrator
· REGEDIT or REGEDT32: opens Registry Editor
· DRWTSN32: opens Dr. Watson
· VERIFIER: opens Driver Verifier Manager
· CLICONFG: opens SQL Server Client Network Utility
· UTILMAN: opens Utility Manager
· COLORCPL: opens color management
· CREDWIZ: back up and recovery tool for user passwords
· MOBSYNC: opens Synchronization center
· MSCONFIG: opens System Configuration Utility
· SYSEDIT: opens System Configuration Editor (careful while using this command)
· SYSKEY: Windows Account Database Security management (careful while using this command)

Windows utility and applications
· EPLORER: Opens windows Explorer
· IEXPLORER: Opens Internet explorer
· WAB: opens Contacts
· CHARMAP: opens Character Map
· WRITE: opens WordPad
· NOTEPAD: opens Notepad
· CALC: opens Calculator
· CLIPBRD: opens Clipbook Viewer
· WINCHAT: opens Microsoft Chat Interface
· SOUNDRECORDER: opens sound recording tool
· DVDPLAY: run CD or DVD
· WMPLAYER: opens Windows Media Player
· MOVIEMK: Opens untitled Windows Movie Maker
· OSK: opens on-screen Keyboard
· MAGNIFY: opens Magnifier
· WINCAL: opens Calendar
· DIALER: opens phone Dialer
· EUDCEDIT: opens Private Character Editor
· NDVOL: opens the mixer volume
· RSTRUI : opens Tool System Restore (For Vista only)
· %WINDIR%\SYSTEM32\RESTORE\rstrui.exe: opens Tool System Restore (for XP only).
· MSINFO32: Opens the System Information
· MRT : launches the utility removal of malware.
· Taskmgr : Opens the Windows Task Manager
· CMD: opens a command prompt
· MIGWIZ: Opens the tool for transferring files and settings from Windows (Vista only)
· Migwiz.exe: Opens the tool for transferring files and settings from Windows (for XP only)
· SIDEBAR: Open the Windows (Vista only)
· Sigverif : Opens the tool for verification of signatures of files
· Winver : Opens the window for your Windows version
· FSQUIRT: Bluetooth Transfer Wizard
· IExpress opens the wizard for creating self-extracting archives. Tutorial HERE
· MBLCTR: opens the mobility center (Windows Vista only)
· MSRA : Opens the Windows Remote Assistance
· Mstsc : opens the tool connection Remote Desktop
· MSDT: opens the diagnostic tools and support Microsoft
· WERCON: opens the reporting tool and solutions to problems (for Vista only)
· WINDOWSANYTIMEUPGRADE: Enables the upgrade of Windows Vista
· WINWORD : opens Word (if installed)
· PRINTBRMUI : Opens migration wizard printer (Vista only)

Disk management
· DISKMGMT.MSC: opens disk management utility
· CLEANMGR: opens disk drive clean up utility
· DFRG.MSC: opens disk defragmenter
· CHKDSK: complete analysis of disk partition
· DISKPART: disk partitioning tool

Connection management
· IPCONFIG: list the configuration of IP addresses on your PC (for more information type IPCONFIG/? in the CMD menu)
· INETCPL.CPL: opens internet properties
· FIREWALL.CPL: opens windows firewall
· NETSETUP.CPL: opens network setup wizard

Miscellaneous commands
· JAVAWS: View the cover of JAVA software (if installed)
· AC3FILTER.CPL: Opens the properties AC3 Filter (if installed)
· FIREFOX: Mozilla launches Firefox (if installed)
· NETPROJ: allow or not connecting to a network projector (For Vista only)
· LOGOFF: closes the current session
· SHUTDOWN: shut down Windows
· SHUTDOWN-A: to interrupt Windows shutdown
· %WINDIR% or %SYSTEMROOT%: opens the Windows installation
· %PROGRAMFILES%: Opens the folder where you installed other programs (Program Files)
· %USERPROFILE%: opens the profile of the user currently logged
· %HOMEDRIVE%: opens the browser on the partition or the operating system is installed
· %HOMEPATH%: opens the currently logged user C: \ Documents and Settings \ [username]
· %TEMP%: opens the temporary folder
· VSP1CLN: deletes the cache for installation of the service pack 1 for Vista
· System File Checker (Requires Windows CD if the cache is not available):
o SFC / scannow: immediately scans all system files and repairs damaged files
o SFC / VERIFYONLY: scans only those files system
o SFC / Scanfil = "name and file path": scans the specified file, and repaired if damaged
o SFC / VERIFYFILE = "name and file path": Scans only the file specified
o SFC / scanonce: scans the system files on the next restart
o SFC / REVERT: return the initial configuration (For more information, type SFC /? In the command prompt CMD.

Ways to Save The Environment

2010-02-02T23:22:00.000+05:30

In Your Home – Conserve Energy
1. Clean or replace air filters on your air conditioning unit at least once a month.
2. If you have central air conditioning, do not close vents in unused rooms.
3. Lower the thermostat on your water heater to 120.
4. Wrap your water heater in an insulated blanket.
5. Turn down or shut off your water heater when you will be away for extended periods.
6. Turn off unneeded lights even when leaving a room for a short time.
7. Set your refrigerator temperature at 36 to 38 and your freezer at 0 to 5 .
8. When using an oven, minimize door opening while it is in use; it reduces oven temperature by 25 to 30 every time you open the door.
9. Clean the lint filter in your dryer after every load so that it uses less energy.
10. Unplug seldom used appliances.
11. Use a microwave when- ever you can instead of a conventional oven or stove.
12. Wash clothes with warm or cold water instead of hot.
13. Reverse your indoor ceiling fans for summer and winter operations as recommended.
14. Turn off lights, computers and other appliances when not in use.
15. Purchase appliances and office equipment with the Energy Star Label; old refridgerators, for example, use up to 50 more electricity than newer models.
16. Only use electric appliances when you need them.
17. Use compact fluorescent light bulbs to save money and energy.
18. Keep your thermostat at 68 in winter and 78 in summer.
19. Keep your thermostat higher in summer and lower in winter when you are away
20. Insulate your home as best as you can.
21. Install weather stripping around all doors and windows.
22. Shut off electrical equipment in the evening when you leave work.
23. Plant trees to shade your home.
24. Shade outside air conditioning units by trees or other means.
25. Replace old windows with energy efficient ones.
26. Use cold water instead of warm or hot water when possible.
27. Connect your outdoor lights to a timer.
28. Buy green electricity - electricity produced by low - or even zero-pollution facilities (NC Greenpower for North Carolina - www.ncgreenpower.org). In your home-reduce toxicity.

In Your Home – Reduce Toxicity
29. Eliminate mercury from your home by purchasing items without mercury, and dispose of items containing mercury at an appropriate drop-off facility when necessary (e.g. old thermometers).
30. Learn about alternatives to household cleaning items that do not use hazardous chemicals.
31. Buy the right amount of paint for the job.
32. Review labels of household cleaners you use. Consider alternatives like baking soda, scouring pads, water or a little more elbow grease.
33. When no good alternatives exist to a toxic item, find the least amount required for an effective, sanitary result.
34. If you have an older home, have paint in your home tested for lead. If you have lead-based paint, cover it with wall paper or other material instead of sanding it or burning it off.
35. Use traps instead of rat and mouse poisons and insect killers.
36. Have your home tested for radon.
37. Use cedar chips or aromatic herbs instead of mothballs.

In Your Yard
38. Avoid using leaf blowers and other dust-producing equipment.
39. Use an electric lawn- mower instead of a gas-powered one.
40. Leave grass clippings on the yard-they decompose and return nutrients to the soil.
41. Use recycled wood chips as mulch to keep weeds down, retain moisture and prevent erosion.
42. Use only the required amount of fertilizer.
43. Minimize pesticide use.
44. Create a wildlife habitat in your yard.
45. Water grass early in the morning.
46. Rent or borrow items like ladders, chain saws, party decorations and others that are seldom used.
47. Take actions that use non hazardous components (e.g., to ward off pests, plant marigolds in a garden instead of using pesticide).
48. Put leaves in a compost heap instead of burning them or throwing them away. Yard debris too large for your compost bin should be taken to a yard-debris recycler

In Your Office
49. Copy and print on both sides of paper.
50. Reuse items like envelopes, folders and paper clips.
51. Use mailer sheets for interoffice mail instead of an envelope.Use mailer sheets for interoffice mail instead of an envelope.
52. Set up a bulletin board for memos instead of sending a copy to each employee.
53. Use e-mail instead of paper correspondence.
54. Use recycled paper.
55. Use discarded paper for scrap paper.
56. Encourage your school and/or company to print documents with soy-based inks, which are less toxic.
57. Use a ceramic coffee mug instead of a disposable cup.

Ways To Protect Our Air
58. Ask your employer to consider flexible work schedules or telecommuting.
59. Recycle printer cartridges.
60. Shut off electrical equipment in the evening when you leave work.
61. Report smoking vehicles to your local air agency.
62. Don't use your wood stove or fireplace when air quality is poor.
63. Avoid slow-burning, smoldering fires. They produce the largest amount of pollution.
64. Burn seasoned wood - it burns cleaner than green wood.
65. Use solar power for home and water heating.
66. Use low-VOC or water-based paints, stains, finishes and paint strippers.
67. Purchase radial tires and keep them properly inflated for your vehicle.
68. Paint with brushes or rollers instead of using spray paints to minimize harmful emissions.
69. Ignite charcoal barbecues with an electric probe or other alternative to lighter fluid.
70. If you use a wood stove, use one sold after 1990. They are required to meet federal emissions standards and are more efficient and cleaner burning.
71. Walk or ride your bike instead of driving, whenever possible.
72. Join a carpool or vanpool to get to work.

Ways to Use Less Water
73. Check and fix any water leaks.
74. Install water-saving devices on your faucets and toilets.
75. Don't wash dishes with the water running continuously.
76. Wash and dry only full loads of laundry and dishes.
77. Follow your community's water use restrictions or guidelines.
78. Install a low-flow shower head.
79. Replace old toilets with new ones that use a lot less water.
80. Turn off washing machine's water supply to prevent leaks.

Ways to Protect Our Water
81. Revegetate or mulch disturbed soil as soon as possible.
82. Never dump anything down a storm drain.
83. Have your septic tank pumped and system inspected regularly.
84. Check your car for oil or other leaks, and recycle motor oil.
85. Take your car to a car wash instead of washing it in the driveway.
86. Learn about your watershed.

Create Less Trash
87. Buy items in bulk from loose bins when possible to reduce the packaging wasted.
88. Avoid products with several layers of packaging when only one is sufficient. About 33 of what we throw away is packaging.
89. Buy products that you can reuse.
90. Maintain and repair durable products instead of buying new ones.
91. Check reports for products that are easily repaired and have low breakdown rates.
92. Reuse items like bags and containers when possible.
93. Use cloth napkins instead of paper ones.
94. Use reusable plates and utensils instead of disposable ones.
95. Use reusable containers to store food instead of aluminum foil and cling wrap.
96. Shop with a canvas bag instead of using paper and plastic bags.
97. Buy rechargeable batteries for devices used frequently.
98. Reuse packaging cartons and shipping materials. Old newspapers make great packaging material.
99. Compost your vegetable scraps.
100. Buy used furniture - there is a surplus of it, and it is much cheaper than new furniture.

101 Useful Run Commands

2010-02-02T23:12:00.000+05:30

To Access?. Run Command

The shortcut to RUN command is "Windows Key + R"

Accessibility Controls access.cpl

Add Hardware Wizard hdwwiz.cpl

Add/Remove Programs appwiz.cpl

Administrative Tools control.exe admintools

Automatic Updates wuaucpl.cpl

Bluetooth Transfer Wizard fsquirt

Calculator

calc

Certificate Manager certmgr.msc

Character Map

charmap

Check Disk Utility chkdsk

Clipboard Viewer clipbrd

Command Prompt cmd

Component Services dcomcnfg

Computer Management compmgmt.msc

Date and Time Properties timedate.cpl

DDE Shares ddeshare

Device Manager devmgmt.msc

Direct X Control Panel (if installed)* directx.cpl

Direct X Troubleshooter dxdiag

Disk Cleanup Utility cleanmgr

Disk Defragment dfrg.msc

Disk Management diskmgmt.msc

Disk Partition Manager diskpart

Display Properties control.exe desktop

Display Properties desk.cpl

Display Properties (w/Appearance Tab Preselected) control.exe color

Dr. Watson System Troubleshooting Utility drwtsn32

Driver Verifier Utility verifier

Event Viewer eventvwr.msc

File Signature Verification Tool sigverif

Findfast findfast.cpl

Folders Properties control.exe folders

Fonts

control.exe fonts

Fonts Folder fonts

Free Cell Card Game freecell

Game Controllers joy.cpl

Group Policy Editor (XP Prof) gpedit.msc

Hearts Card Game mshearts

Iexpress Wizard iexpress

Indexing Service ciadv.msc

Internet Properties inetcpl.cpl

Java Control Panel (if installed) jpicpl32.cpl

Java Control Panel (if installed) javaws

Keyboard Properties

control.exe keyboard

Local Security Settings secpol.msc

Local Users and Groups lusrmgr.msc

Logs You Out Of Windows logoff

Mcft Chat winchat

Minesweeper Game winmine

Mouse Properties control.exe mouse

Mouse Properties main.cpl

Network Connections control.exe netconnections

Network Connections ncpa.cpl

Network Setup Wizard netsetup.cpl

Nview Desktop Manager (if installed) nvtuicpl.cpl

Object Packager packager

ODBC Data Source Administrator odbccp32.cpl

On Screen Keyboard osk

Opens AC3 Filter (if installed) ac3filter.cpl

Pa**word Properties pa**word.cpl

Performance Monitor perfmon.msc

Performance Monitor perfmon

Phone and Modem Options telephon.cpl

Power Configuration powercfg.cpl

Printers and Faxes

control.exe. printers or control printers

Printers Folder printers

Private Character Editor eudcedit

Quicktime (If Installed)

QuickTime.cpl

Regional Settings intl.cpl

Registry Editor regedit

Registry Editor regedit32

Removable Storage ntmsmgr.msc

Removable Storage Operator Requests ntmsoprq.msc

Resultant Set of Policy rsop.msc

Resultant Set of Policy (XP Prof) rsop.msc

Scanners and Cameras sticpl.cpl

Scheduled Tasks

control.exe schedtasks or control schedtasks

Security Center wscui.cpl

Services

services.msc

Shared Folders fsmgmt.msc

Shuts Down Windows shutdown

Sounds and Audio mmsys.cpl

Spider Solitare Card Game (if installed) spider

SQL Client Configuration cliconfg

System Configuration Editor sysedit

System Configuration Utility msconfig

System File Checker Utility sfc

System Properties sysdm.cpl

Task Manager taskmgr

Telnet Client

telnet

User Account Management nusrmgr.cpl

Utility Manager

utilman

Windows Firewall

firewall.cpl

Windows Magnifier

magnify

Windows Management Infrastructure wmimgmt.msc

Windows System Security Tool syskey

Windows Update Launches wupdmgr

Windows XP Tour Wizard tourstart

Let the world know where we indians stand ....

2010-01-31T23:11:00.002+05:30

Let the world know what we stand for.

There are 3.22 Million Indians in America.
38% of Doctors in America are Indians.
12% of Scientists in America are Indians.
36% of NASA employees are Indians.
34% of MICROSOFT employees are Indians.
28% of IBM employees are Indians.
17% of INTEL employees are Indians.
13% of XEROX employees are Indians.
You may know some of these facts. These
facts were recently published in a German
Magazine, which deals with
WORLD HISTORY FACTS ABOUT INDIA.

India never invaded any country in her last
100000 years of history.
India invented the Number System.
Aryabhatta invented zero.
The World's first university was established in
Takshila in 700BC.More than 10,500 students from
all over the world studied more than 60 subjects. The
University of Nalanda built in the 4th century BC
was one of the greatest achievements of ancient India
in the field of education.
Sanskrit is the mother of all the European
languages. Sanskrit is the most suitable language
for computer software reported in Forbes magazine,
July 1987.

Ayurveda is the earliest school of medicine
known to humans. Charaka, the father of medicine
consolidated Ayurveda 2500 years ago. Today
Ayurveda is fast regaining its rightful place
in our civilization.
Although modern images of India often show
poverty and lack of development, India was the
richest country on earth until the time of
British invasion in the early 17th Century.
The art of Navigation was born in the river
Sindh 6000 years ago.
The very word Navigation is derived from
the Sanskrit word NAVGATIH.
The Word navy is also derived from Sanskrit 'Nou'.

Bhaskaracharya calculated the time taken by the
earth to orbit the sun hundreds of years before the
astronomer Smart.; Time taken by earth to orbit
the sun: (5th century) 365.258756484 days.
Budhayana first calculated the value of pi, and
he explained the concept of what is known as the
Pythagorean Theorem. He discovered this in the
6th century long before the European mathematicians
Algebra, trigonometry and calculus came from
India; Quadratic equations were by Sridharacharya in the
11th century ; The largest numbers the
Greeks and the Romans
used were 10 6(10 to the power of 6) whereas
Hindus Used numbers as big as 1053 (10 to the
power of 53) with specific names as Early as 5000 BCE
during the Vedic period. Even today, the largest
used number is Tera 1012(10 to the power of 12).

According to the Gemological Institute of
America, up until 1896,India was the only source for
diamonds to the world.
USA based IEEE has proved what has been a
century-old suspicion in the world scientifi
community that the pioneer of Wireless
communication was Prof. Jagdeesh Bose and not Marconi.

The earliest reservoir and dam for irrigation was
built in Saurashtra. According to Saka King
rudradaman I of 150 CE a beautiful lake
called 'Sudarshana'
was constructed on the hills of Raivataka during
Chandragupta Maurya's time.

Chess (Shataranja or AshtaPada) was invented in India.

Sushruta is the father of surgery. 2600
years ago he and health scientists of his time
conducted complicated surgeries like cesareans,
cataract, artificial limbs, fractures, urinary
stones and even plastic surgery and brain surgery. Usage
of anesthesia was well known in ancient India.
Over 125 surgical equipment were used. Deep
knowledge of anatomy, etiology, embryology, digestion,
metabolism, genetics and immunity is also found
in many texts.

When many cultures were only nomadic forest
dwellers over 5000 years ago, Indians
established Harappan culture in Sindhu
Valley (Indus Valley Civilization)
The place value system, the decimal system
was developed in India in 100 BC.

QUOTES ABOUT INDIA:

Albert Einstein said: We owe a lot to the
Indians, who taught us how to count, without
which no worthwhile scientific discovery could
have been made.
Mark Twain said: India is the cradle of the
human race, the birthplace of human speech, the mother
of history, the grandmother of legend, and the great
grand mother of tradition. Our most valuable and most
structive materials in the history of man are treasured
up in India only.
French scholar Romain Rolland said: If there is
one place on the face of earth where all
the dreams of living men have found a home from
the very earliest days when man began the dream
of existence, it is India.
Hu Shih, former Ambassador of China to USA said:
India conquered And dominated China culturally
for 20 centuries without ever having to send a single
soldier across her border.
=================================================================
All the above is just the TIP of the iceberg, the
list could be endless. BUT, if we don't see even a
glimpse of that great India in the India That we see
today, it
clearly means that we are not working up to our
Potential and that if we do, we could once
again; be an ever shining and Inspiring country
setting a bright path for rest of the world to follow.
I Hope you enjoyed it and work towards the welfare
of INDIA. PROUD to be an INDIAN.
=================================================================

Let the world know where we indians stand ....

2010-01-31T23:11:00.000+05:30

Tips and TRICKS In GROUP DISCUSSION

2010-01-31T18:03:00.000+05:30

1. Initiation Techniques
2. Body of the group discussion
3. Summarization/ Conclusion
Initiation Techniques
• Initiating a GD is a high profit-high loss strategy.

When you initiate a GD, you not only grab the opportunity to speak, you also grab the attention of the examiner and your fellow candidates.

If you can make a favourable first impression with your content and communication skills after you initiate a GD, it will help you sail through the discussion.

But if you initiate a GD and stammer/ stutter/ quote wrong facts and figures, the damage might be irreparable.

If you initiate a GD impeccably but don't speak much after that, it gives the impression that you started the GD for the sake of starting it or getting those initial kitty of points earmarked for an initiator!

When you start a GD, you are responsible for putting it into the right perspective or framework. So initiate one only if you have in-depth knowledge about the topic at hand.
Body of the group discussion
• Different techniques to initiate a GD and make a good first impression:

i. Quotes
ii. Definition
iii. Question
iv. Shock statement
v. Facts, figures and statistics
vi. Short story
vii. General statement

i. Quotes

Quotes are an effective way of initiating a GD.

If the topic of a GD is: Should the Censor Board be abolished?, you could start with a quote like, 'Hidden apples are always sweet'.

For a GD topic like, Customer is King, you could quote Sam (Wall-mart) Walton's famous saying, 'There is only one boss: the customer. And he can fire everybody in the company -- from the chairman on down, simply by spending his money somewhere else.'

ii. Definition

Start a GD by defining the topic or an important term in the topic.

For example, if the topic of the GD is Advertising is a Diplomatic Way of Telling a Lie, why not start the GD by defining advertising as, 'Any paid form of non-personal presentation and promotion of ideas, goods or services through mass media like newspapers, magazines, television or radio by an identified sponsor'?

For a topic like The Malthusian Economic Prophecy is no longer relevant, you could start by explaining the definition of the Malthusian Economic Prophecy.

iii. Question

Asking a question is an impact way of starting a GD.

It does not signify asking a question to any of the candidates in a GD so as to hamper the flow. It implies asking a question, and answering it yourself.

Any question that might hamper the flow of a GD or insult a participant or play devil's advocate must be discouraged.

Questions that promote a flow of ideas are always appreciated.

For a topic like, Should India go to war with Pakistan, you could start by asking, 'What does war bring to the people of a nation? We have had four clashes with Pakistan. The pertinent question is: what have we achieved?'

iv. Shock statement

Initiating a GD with a shocking statement is the best way to grab immediate attention and put forth your point.

If a GD topic is, The Impact of Population on the Indian Economy, you could start with, 'At the centre of the Indian capital stands a population clock that ticks away relentlessly. It tracks 33 births a minute, 2,000 an hour, 48,000 a day. Which calculates to about 12 million every year. That is roughly the size of Australia. As a current political slogan puts it, 'Nothing's impossible when 1 billion Indians work together'.'

v. Facts, figures and statistics

If you decide to initiate your GD with facts, figure and statistics, make sure to quote them accurately.

Approximation is allowed in macro level figures, but micro level figures need to be correct and accurate.

For example, you can say, approximately 70 per cent of the Indian population stays in rural areas (macro figures, approximation allowed).

But you cannot say 30 states of India instead of 28 (micro figures, no approximations).

Stating wrong facts works to your disadvantage.

For a GD topic like, China, a Rising Tiger, you could start with, 'In 1983, when China was still in its initial stages of reform and opening up, China's real use of Foreign Direct Investment only stood at $636 million. China actually utilized $60 billion of FID in 2004, which is almost 100 times that of its 1983 statistics."

vi. Short story

Use a short story in a GD topic like, Attitude is Everything.

This can be initiated with, 'A child once asked a balloon vendor, who was selling helium gas-filled balloons, whether a blue-colored balloon will go as high in the sky as a green-colored balloon. The balloon vendor told the child, it is not the color of the balloon but what is inside it that makes it go high.'

vii. General statement

Use a general statement to put the GD in proper perspective.

For example, if the topic is, Should Sonia Gandhi be the prime minister of India?, you could start by saying, 'Before jumping to conclusions like, 'Yes, Sonia Gandhi should be', or 'No, Sonia Gandhi should not be', let's first find out the qualities one needs to be a a good prime minister of India. Then we can compare these qualities with those that Mrs. Gandhi possesses. This will help us reach the conclusion in a more objective and effective manner.'
Summarization/ Conclusion
• Most GD do not really have conclusions. A conclusion is where the whole group decides in favor or against the topic.
• But every GD is summarized. You can summaries what the group has discussed in the GD in a nutshell.

Keep the following points in mind while summarizing a discussion:

1. Avoid raising new points.
2. Avoid stating only your viewpoint.
3. Avoid dwelling only on one aspect of the GD.
4. Keep it brief and concise.
5. It must incorporate all the important points that came out during the GD.
6. If the examiner asks you to summaries a GD, it means the GD has come to an end.
7. Do not add anything once the GD has been summarized.

GD Do's & Dont's
• Do's
1. Be as natural as possible. Do not try and be someone you are not. Be yourself.
2. A group discussion is your chance to be more vocal. The evaluator wants to hear you speak.
3. Take time to organize your thoughts. Think of what you are going to say.
4. Seek clarification if you have any doubts regarding the subject.
5. Don't start speaking until you have clearly understood and analyzed the subject.
6. Work out various strategies to help you make an entry: initiate the discussion or agree with someone else's point and then move onto express your views.
7. Opening the discussion is not the only way of gaining attention and recognition. If you do not give valuable insights during the discussion, all your efforts of initiating the discussion will be in vain.
8. Your body language says a lot about you - your gestures and mannerisms are more likely to reflect your attitude than what you say.
9. Language skills are important only to the effect as to how you get your points across clearly and fluently.
10. Be assertive not dominating; try to maintain a balanced tone in your discussion and analysis.
11. Don't lose your cool if anyone says anything you object to. The key is to stay objective: Don't take the discussion personally.
12. Always be polite: Try to avoid using extreme phrases like: `I strongly object' or `I disagree'. Instead try phrases like: `I would like to share my views on…' or `One difference between your point and mine…' or "I beg to differ with you"
13. Brush up on your leadership skills; motivate the other members of the team to speak (this surely does not mean that the only thing that you do in the GD is to say "let us hear what the young lady with the blue scarf has to say," or "Raghu, let us hear your views" - Essentially be subtle), and listen to their views. Be receptive to others' opinions and do not be abrasive or aggressive.
14. If you have a group of like-minded friends, you can have a mock group discussion where you can learn from each other through giving and receiving feedback.
15. Apart from the above points, the panel will also judge team members for their alertness and presence of mind, problem-solving abilities, ability to work as a team without alienating certain members, and creativity.
• Dont's
1. Be as natural as possible. Do not try and be someone you are not. Be yourself.
2. A group discussion is your chance to be more vocal. The evaluator wants to hear you speak.
3. Take time to organize your thoughts. Think of what you are going to say.
4. Seek clarification if you have any doubts regarding the subject.
5. Don't start speaking until you have clearly understood and analyzed the subject.
6. Work out various strategies to help you make an entry: initiate the discussion or agree with someone else's point and then move onto express your views.
7. Opening the discussion is not the only way of gaining attention and recognition. If you do not give valuable insights during the discussion, all your efforts of initiating the discussion will be in vain.
8. Your body language says a lot about you - your gestures and mannerisms are more likely to reflect your attitude than what you say.
9. Language skills are important only to the effect as to how you get your points across clearly and fluently.
10. Be assertive not dominating; try to maintain a balanced tone in your discussion and analysis.
11. Don't lose your cool if anyone says anything you object to. The key is to stay objective: Don't take the discussion personally.
12. Always be polite: Try to avoid using extreme phrases like: `I strongly object' or `I disagree'. Instead try phrases like: `I would like to share my views on…' or `One difference between your point and mine…' or "I beg to differ with you"
13. Brush up on your leadership skills; motivate the other members of the team to speak (this surely does not mean that the only thing that you do in the GD is to say "let us hear what the young lady with the blue scarf has to say," or "Raghu, let us hear your views" - Essentially be subtle), and listen to their views. Be receptive to others' opinions and do not be abrasive or aggressive.
14. If you have a group of like-minded friends, you can have a mock group discussion where you can learn from each other through giving and receiving feedback.
15. Apart from the above points, the panel will also judge team members for their alertness and presence of mind, problem-solving abilities, ability to work as a team without alienating certain members, and creativity.
GD Preparation
While selection tools and techniques like tests, interviews etc. provide good data about an individual, they fall short in providing real life data of how an individual would be performing in a real life situation especially a group situation. Team work being an integral part of the BPO work profile, it is important to ascertain group and inter-personal qualities of an individual. Group discussion is a useful tool to ascertain these qualities and many organizations use GDs as a selection tool along with Personal Interviews, aptitude tests etc. A GD is an activity where
• Groups of 8-10 candidates are formed into a leaderless group, and are given a specific situation to analyse and discuss within a given time limit, which may vary between twenty minutes and forty-five minutes, or
• They may be given a case study and asked to come out with a solution for a problem
• They may be given a topic and are asked to discuss the same
1. Preparing for a Group Discussion: While GD reflects the inherent qualities of an individual, appearing for it unprepared may not augur well for you. These tips would help you prepare for GDs:
Reading: This is the first and the most crucial step in preparation. This is a never ending process and the more you read, the better you are in your thoughts. While you may read anything to everything, you must ensure that you are in good touch with current affairs, the debates and hot topics of discussion and also with the latest in the IT and ITES industry. Chances are the topics would be around these. Read both for the thoughts as well as for data. Also read multiple view points on the same topic and then create your point of view with rationale. Also create answers for counter arguments for your point of view. The electronic media also will be of good use here.
Mocks: Create an informal GD group and meet regularly to discuss and exchange feedback. This is the best way to prepare. This would give you a good idea about your thoughts and how well can you convince. Remember, it is important that you are able to express your thoughts well. The better you perform in these mocks the better would be you chances to perform on the final day. Also try to interact and participate in other GD groups. This will develop in you a skill to discuss with unknown people as well.
2. During the Group Discussion:
What do the panelists assess:Some of the qualities assessed in a GD are:
Leadership Skills - Ability to take leadership roles and be able to lead, inspire and carry the team along to help them achieve the group's objectives.
Communication Skills - Candidates will be assessed in terms of clarity of thought, expression and aptness of language. One key aspect is listening. It indicates a willingness to accommodate others views.
Interpersonal Skills - People skills are an important aspect of any job. They are reflected in the ability to interact with other members of the group in a brief situation. Emotional maturity and balance promotes good interpersonal relationships. The person has to be more people centric and less self-centered.
Persuasive Skills - The ability to analyze and persuade others to see the problem from multiple perspectives.
GD is a test of your ability to think, your analytical capabilities and your ability to make your point in a team-based environment.
These are some of the sub-skills that also get assessed with the skills mentioned above:
• Clarity of thought
• Group working skills (especially during a group task of case study discussion)
• Conflict handling
• Listening and probing skills
• Knowledge about the subject and individual point of view
• Ability to create a consensus
• Openess and flexibility towards new ideas
• Data based approach to decision making
While, it is not possible to reflect all these qualities in a short time, you would do well if you are able to show a couple or more qualities and avoid giving negative evidence on others.
GD Mistakes
Here's a list of the most common mistakes made at group discussions:
Emotional outburst
Rashmi was offended when one of the male participants in a group discussion made a statement on women generally being submissive while explaining his point of view. When Rashmi finally got an opportunity to speak, instead of focussing on the topic, she vented her anger by accusing the other candidate for being a male chauvinist and went on to defend women in general.
What Rashmi essentially did was to
• Deviate from the subject
• Treat the discussion as a forum to air her own views.
• Lose objectivity and make personal attacks.
Her behaviour would have been perceived as immature and demotivating to the rest of the team.
Quality Vs Quantity
Gautam believed that the more he talked, the more likely he was to get through the GD. So, he interrupted other people at every opportunity. He did this so often that the other candidates got together to prevent him from participating in the rest of the discussion.
• Assessment is not only on your communication skills but also on your ability to be a team player.
• Evaluation is based on quality, and not on quantity. Your contribution must be relevant.
• The mantra is "Contributing meaningfully to the team's success." Domination is frowned upon.
Egotism Showing off
Krishna was happy to have got a group discussion topic he had prepared for. So, he took pains to project his vast knowledge of the topic. Every other sentence of his contained statistical data - "20% of companies; 24.27% of parliamentarians felt that; I recently read in a Jupiter Report that..." and so on so forth. Soon, the rest of the team either laughed at him or ignored his attempts to enlighten them as they perceived that he was cooking up the data.
• Exercise restraint in anything. You will end up being frowned upon if you attempt showing-off your knowledge.
• Facts and figures need not validate all your statements.
• Its your analysis and interpretation that are equally important - not just facts and figures.
• You might be appreciated for your in-depth knowledge. But you will fail miserably in your people skills.
Such a behavior indicates how self-centered you are and highlights your inability to work in an atmosphere where different opinions are expressed.
Get noticed - But for the right reasons
Srikumar knew that everyone would compete to initiate the discussion. So as soon as the topic - "Discuss the negative effects of India joining the WTO" - was read out, he began talking. In his anxiety to be the first to start speaking, he did not hear the word "negative" in the topic. He began discussing the ways in which the country had benefited by joining WTO, only to be stopped by the evaluator, who then corrected his mistake.
• False starts are extremely expensive. They cost you your admission. It is very important to listen and understand the topic before you air your opinions.
• Spending a little time analyzing the topic may provide you with insights which others may not have thought about. Use a pen and paper to jot down your ideas.
• Listen! It gives you the time to conceptualize and present the information in a better manner.
Some mistakes are irreparable. Starting off the group discussion with a mistake is one such mistake, unless you have a great sense of humor.
Managing one's insecurities
Sumati was very nervous. She thought that some of the other candidates were exceptionally good. Thanks to her insecurity, she contributed little to the discussion. Even when she was asked to comment on a particular point, she preferred to remain silent.
• Your personality is also being evaluated. Your verbal and non verbal cues are being read.
• Remember, you are the participant in the GD; not the evaluator. So, rather than evaluating others and your performance, participate in the discussion.
• Your confidence level is being evaluated. Decent communication skills with good confidence is a must to crack the GDs.
Focus on your strengths and do not spend too much time thinking about how others are superior or inferior to you. It is easy to pick up these cues from your body language.
Knowledge is strength. A candidate with good reading habits has more chances of success. In other words, sound knowledge on different topics like politics, finance, economy, science and technology is helpful.
Power to convince effectively is another quality that makes you stand out among others.
Clarity in speech and expression is yet another essential quality.
If you are not sure about the topic of discussion, it is better not to initiate. Lack of knowledge or wrong approach creates a bad impression. Instead, you might adopt the wait and watch attitude. Listen attentively to others, may be you would be able to come up with a point or two later.
A GD is a formal occasion where slang is to avoided.
A GD is not a debating stage. Participants should confine themselves to expressing their viewpoints. In the second part of the discussion candidates can exercise their choice in agreeing, disagreeing or remaining neutral.
Language use should be simple, direct and straight forward.
Don't interrupt a speaker when the session is on. Try to score by increasing your size, not by cutting others short.
Maintain rapport with fellow participants. Eye contact plays a major role. Non-verbal gestures, such as listening intently or nodding while appreciating someone's viewpoint speak of you positively.
Communicate with each and every candidate present. While speaking don't keep looking at a single member. Address the entire group in such a way that everyone feels you are speaking to him or her.

windows password ?

2010-01-11T12:52:00.001+05:30

What's the need?1.Forgot passwords!
Many people forgot their password and lost their administrator rights.2.As i told you in post "basics of hacking" ,before attacking we need to prepare our defense.By using given methods , even beginner will get administrative rights and will able to install keyloggers and other tool in computer's at public libraries of school and colleges.
This hack hardly takes 2 minutes. Follow these steps :
1. Open Start and click Run as shown in the pic

2. Type command( in blue color) "control userpasswords2" in run window and press ok.

3. It will pop up User account Windows, in this window click on administrator.

Click on Reset Password Button, here we are! , change password according to you and become administrator.

I hope it helps

windows password ?

2010-01-11T12:52:00.000+05:30

free html tutorials @ www.webreference.com/html/tutorial1/4.html

2010-01-09T11:47:00.001+05:30

what is HTML made of?
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I mentioned that HTML is an application of SGML. This means that every HTML document is also an SGML document. The first thing an SGML document must have is a Document Type Declaration. This means exactly what it sounds like: a Document Type Declaration declares the document to be of a specific type. In our case this type is HTML. I won't go into much depth on Document Type Declarations right now. For the moment, you should use the following declaration:
"http://www.w3.org/TR/REC-html40/strict.dtd">

Do not let the angle brackets confuse you. The above is not an element. If you look carefully, you'll notice that the content of the above construct starts with an exclamation mark; this indicates that this is SGML code. And after looking at it a bit you might be glad you don't have to learn SGML. So just take me on my word for this once, and put this at the top of your document. In a future tutorial, we'll explain what this Document Type Declaration means and show that it's really quite simple.

Now that we have specified that this is an HTML document, we can start adding elements. The first element will always be the HTML element. All HTML documents have an HTML element, which contains all the other elements. Let's put in the start-tag and end-tag for this element and we'll worry about its contents later. Here's what we've got so far:
"http://www.w3.org/TR/REC-html40/strict.dtd">

Every HTML document is split into a head and a body, which are marked by similarily named elements, HEAD and BODY. Every HTML document must have one of each, inside the HTML element. In fact, these two are the only things you can have inside the HTML element. So let's put these in as well and see where we are:
"http://www.w3.org/TR/REC-html40/strict.dtd">

Notice that I've used a slight indent for the HEAD and BODY element tags. This has no special meaning and is only there to make the HTML more legible. You might have noticed that white-space (that is, spaces, tabs and linefeeds) is collapsed in HTML. This means you can add as much of it as you want to, in order to make your HTML easier to read, without any change in the meaning of the document.

The difference between the head and body of a document is that the head contains mostly information about the document, while the body contains the document itself. Before we go on to the body, we'll deal with the one element every document head must contain: a title.

The title of your document is very important. It distinguishes your document and makes it unique, as well as describing it to the reader. In this case, the title "Acme Computer Corp." is unsuitable, because it doesn't describe our document. A more descriptive title would be "About Acme Computer Corp.", but since this is the world of marketing and we can't afford to be bland, we'll give it a title of "Acme Computer Corp.: Who We Are".

The TITLE element is a very simple element. It cannot contain anything but text and that text is the title of the document. So let's insert our title into our document, which is almost complete:
"http://www.w3.org/TR/REC-html40/strict.dtd">

Acme Computer Corp.: Who We Are

We're almost finished! All that remains is to add our document body.

free html tutorials @ http://www.webreference.com/html/tutorial1/4.html

2010-01-09T11:47:00.000+05:30

HTML ---->>Do That SGML Thing You Do

2010-01-09T11:45:00.000+05:30

I mentioned that HTML, as a computer language, has a defined syntax. This syntax was not thought up at random, since the idea of a markup language is hardly new. A language called SGML, the Standard Generalized Markup Language, exists that is used to define markup languages. Make sure you get this right: SGML is a markup language whose sole use is to define other markup languages. And one of those many languages is HTML.

Now wait. Before you click on our sponsor's ad and get out of here, afraid that you'll have to learn another language, let me assure you that no such thing is needed. You do not need to know SGML to learn HTML. It's just useful to know that HTML is an application of SGML, which will explain a few things.

Now let's get down to business. Remember the heading we saw in the last section? If not, here it is again.
Acme Computer Corp.

We know this is a heading, but we need a way of encoding this in the document itself. To do this, we make this heading an element. We do this by writing the following (yes, this is HTML!):

Acme Computer Corp.

This element can be split into three parts. The first part,

is called the start-tag. Then comes the element's content, which in this case is the text "Acme Computer Corp.". Finally,

is the end-tag.

This element is an H1 element, which happens to mean that it is a level 1 heading (we'll get to that later). You need tags to indicate where the element starts and where it ends. Tags always start with a less-than symbol (<) and end with a greater-than symbol (>). A start tag has the element's name in between these symbols, (in this case, H1). An end-tag has a slash (/) followed by the element's name. Here are some more examples of elements:

I'm an element.

So am I!

Hey, me too! And me!

There's another element after me.

The perceptive amongst you may have noticed a few things in the above elements: first, the B element is inside a P element. This is fine; you can have elements inside other elements, as long as you have proper nesting. This means that if an element starts within another element, it must end within that same element, like this:

This is right

This is wrong

The second line above is incorrect because the B element starts inside the P element, but ends outside the P element.

The second curious thing you should have noticed is that the HR element has no content and no end-tag. This is also allowed, for some element types. The HR element in this case is called an empty element. Using only a start-tag is permitted in this case, but only for elements that are empty.

Now that you know what an element is and how to specify it in your document, it's time to learn about some of the elements in HTML and use them to make your first HTML document.

you too can be an HTML author!

2010-01-09T11:41:00.000+05:30

You might have noticed that I've been using the term "HTML document" instead of "HTML file." There's a reason for this; HTML documents aren't necessarily files on a computer. An HTML document is a series of characters that, through its special syntax, defines a document. These characters may be stored in a single disk file, but this is not necessary. They may be created on the fly by a program, or may (as is most often the case) be received over a network.

HTML was designed primarily as a language to be used for creating World Wide Web pages. You're probably learning HTML in order to create a Web page. People have started to use it for other uses, but it might be useful to note (one of my rants, you'll get plenty of these) that it's not a very good language for other uses. There are fine document formats for all kinds of uses, but HTML was created for the Web, and is most suitable for the Web.

So how do you publish a Web page? Well, to do this you need a Web server. A Web server is a program that runs on a computer connected to the Internet, that serves out Web pages. This tutorial will not cover the topic of setting up a Web server or publishing your HTML documents on one. Instead we're going to talk about how to create an HTML document; we'll worry about publishing it later.

Now, how are you going to create your HTML document? The easiest way is to create an HTML file using a text editor. Note that a text editor is not a word processor. A word processor is a program that creates a document ready for printing, and stores it in its own format. Recent times have seen word processors that try to store their documents as HTML, but they usually do a terrible job of doing this. What you need is a program that edits simple text files. An example would be the Windows Notepad, or SimpleText for the Macintosh. It doesn't matter which program you use, as long as it is a simple text editor. In the future, we'll have a look at some of the text editors that you can use, and even some that are specially made for creating HTML documents and can do some of the work for you. Try to avoid these for now - they might confuse you with HTML features we haven't discussed yet. So pick something really, really simple, like the Windows Notepad, SimpleText for the Macintosh, or one of the hundreds of text editors available on Unix systems.

Now create a file, and call it anything you want. On some systems such as Windows, you'll need to give it an extension of .html or .htm to indicate that it's an HTML document. For instance, you might want to call it tutorial.html. As you read through this tutorial, you'll be told to type things into your text file, and by the end of the tutorial you'll have a complete HTML document. In fact, you can stop worrying about typing anything into your file until the end of the tutorial, where we have the complete document listed, so concentrate on reading the tutorial and you can create your HTML file later on if you want to.

In order to view your HTML file, you'll need a program that can do this. The technical term for this is an HTML User Agent. A User Agent is a program that can understand HTML documents and process them is some way. One type of user agent is a Web Browser, or just browser for short. You're probably using one to read this tutorial, so I won't bother with telling you how to get one. After you've created your HTML document, you can open it with your browser and view it. If you can't be bothered to do even that, we've included a link to a copy of the document we're going to create at the end of this tutorial.

Well, here we go. It's time to create our first HTML document

what is this HTML thing?

2010-01-09T11:39:00.000+05:30

In order to be able to write good HTML, you must first understand exactly what HTML is. The most obvious place to look for this information is in its name: HTML stands for HyperText Markup Language. If this doesn't make much sense to you, don't worry, because that's what I'm here to explain. Let's take those initials apart one by one, starting from the right:

HTML is a language, but not the kind you tell your kids to watch. It's a computer language, and as such it has some specific rules that must be followed. In other words, it has a defined syntax, a strict way in which it must be written, and, when the time comes, read. But we'll get to that later.

Now HTML is also a markup language. What this means is that it takes a document and marks specific parts of it, giving them special meaning. Why does it do this? Well, let me give you an example. Take the following piece of text:
Acme Computer Corp.

Acme Computer Corporation is a technology-based
company that seeks to offer its customers the
latest in technological innovation. Our products
are created using the latest breakthroughs in
computers and are designed by a team of top-notch
experts.

We are based in Acmetown, USA, and have offices in
most major cities around the world. Our goal is to
have a global approach to the future of computing.
Have a look at our product catalog for some
examples of our innovative approach.

What you, as a human, see in the above text is a heading ("Acme Computer Corp.") and two paragraphs of text. However, if a computer were to see the above text, all it would see is a bunch of characters, perhaps arranged in a certain way. We have an active interest in letting the computer know that the above is a heading and two paragraphs. I'm sure you can guess some of the reasons. In a larger document, we'd like to be able to have the computer produce an outline of the document containing only the headings; or we might want to display the headings in a different style. However, the computer can't do any of these things unless it can see each part of the document for what it is. Thus, we introduce markup to show the computer what is what.

HTML is also a hypertext markup language. Hypertext is text, in any format, with an added feature: parts of the text is linked to other parts of the text, making it easy to jump from one part of the text to another. For instance, in the Acme example above, the phrase "Have a look at our product catalog" hints that such a catalog exists, but the reader doesn't know where to find it. With hypertext, you can link this phrase with the catalog itself, giving the reader an easy way to get to it if he's interested.

But hypertext links aren't just shortcuts. Just like markup, they mean something. HTML is all about document semantics. A document by itself may be informative, but to be truly useful, you must have a way to get to its meaning. Once you have a way of encoding the document's semantics, you can manipulate it in many interesting ways (don't get any ideas... this isn't that kind of Web site!). By defining the links from a document to a table of contents that lists it, to the next or previous documents if it is part of a series, to a glossary or copyright notice, we give the document itself more meaning, and hence, more value. The primary purpose of any document is to convey information, and by specifying the semantics of a document we supply even more information, which can only be a good thing. That is, if it's done right.

But of course, I can't tell you how to do it right unless I tell you how to do it in the first place, can I?

Hal Paper

2009-12-31T21:48:00.000+05:30

1 :- General Knowledge
1. Lucknow is popular for which fruit: -
a) Mango b) Orange c) Water fruit
2. Governor of utter Pradesh
3. President of India
4. In which district of UP a oil refinery is found
5. Pigeon is the indication of
6. The chemical Formula of salt
7. Republic day in India is celebrated on
8. The “Beating the retreat” hasn’t celebrated on republic day, what is its resion…
9. Lucknow city is located near which river
10. Firozabad is popular for
11. Renukot is popular for
12. The first prime minister of India
13. The first President of India
14. The birth day of Pandit Jawahar Lal Nehru is celebrated as:-
15. The writer of ‘Godan’
16. Word ‘Agfa’ is related to:--

2:- Electronics Questions.

1. When the temperature of the room increased then the energy of semiconductor…

2. Power diode is generally made from: - Silicon/Germanium/Both/None of these

3. Thermionic Emission of electron is due to………….

4. When the both junction of NPN diode is reverse biased, then the diode is in which mode: -
a) Active b) Cutoff c) Saturation d) inverted

5. Which transistor mode gives the inverted output: -
a) Common Emitter b) Common Base c) Common Collector d) None of these

6. Which coupling gives the higher gain in case of amplifier: -
a) Capacitor coupling b) Impedance coupling c) Transformer coupling

7. Which distortion is least objectionable in audio amplification: -
a) Phase b) Frequency c) Harmonic d) Intermediation

8. A narrowband amplifier has a band pass nearly…………of central frequency:-
a) 33.3% b) 10% c) 50%

9. Frequency of wein bridge oscillator

10. Phase shift oscillator consists: -
a) RL b) RC c) RLC

11. Crystal oscillator frequency is very stable due to:-

12. Multivibrater Produces: -
a) Sine wave b) Square wave c) Smooth wave d) sawtooth

13. Convert the 101101 Binary number into octal no: -
a) 65 b) 55 c) 51 d) 45

14. 10 in BCD: -
a) 10100 b) 1100 c) 010111 d) None of these

15. Which PNP device has a terminal for synchronizing purpose:-
a) SCS b) Triac c) Diac d) SUS

16. In 3-Phase full converter, the output during overlap is equal to……………..

17. Addition of indium in semiconductor crystal makes: -
a) PNP b) NPN

18. Free electron exists in which band: -
a) 1 b) 2 c) 3 d) Conduction band

19. Ripple factor of half wave rectifier: -
a) 1.21 b) 0.48 c) 0.5

20. Efficiency of half wave rectifier: -

21. Transistor that can be used in enhancement mode:-
a) NPN b) UJT c) JFET d) MOSFET

21. Following contributes to harmonic distortion in Amplifier: -
a) +Ve feedback b) –Ve feedback c) Defective active device

22. High cutoff frequency: -
a) CB b) CC c) CE

23. Which is used as data selector?
a) Encoder b) Decoder c) modulator d) Demodulator

24. Read write capable memory: -
a) RAM b) ROM c) Both d) None of these

25. When modulation frequency is doubled, modulation index also doubled in case of: -
a) FM b) PM c) AM d) PM & FM

26. Term ’Noise temperature’ is used for

27. Video detector used in TV receiver is a….
a) Diode detector b) Ratio detector c) Phase detector

28. Which isn’t directly operated from mains: -
a) Half wave rectifier b) Centre tap rectifier c) Bridge rectifier d) Voltage doubler

29. In 3-Phase half wave rectifier each diode conducts for duration of: -
a) 30 Degree b) 45 Degree c) 60 Degree d) 180 Degree

30. If number of phase in a multiphase rectifier increased then: -
a) Output increased b) Output decreased c) Output smooth d) No change

31. Which output requires minimum filtration: -
a) Half wave rectifier b) full wave c) SCR half wave rectifier d) Voltage doubler

32. Identify the under circuit
a) Low pass filter b) High pass filter c) Band pass filter

33. Transistor is a following operated device: -
a) Low voltage & low current device
b) Low voltage & high current device
c) Low current & high voltage device
d) High current & high voltage device

ROBERT BOSCH PAPER ON 17th FEBRUARY

2009-12-31T21:47:00.000+05:30

SECTION 1(TECHNICAL)

1. There was a figure of JK flip flop in which ~q is connected to J input and K=1. If clock signal is successively applied 6 times what is output sequence (q=?)
d) 010101

2. Frequency response of a filter is
a) Range of frequencies at which amplification of signal is employed.
b) Output voltage versus frequency (plot)
c) Filter which suppresses particular frequency

3. Gain and bandwidth of an op amp is
a) Independent of each other
b) Gain decreases as bandwidth decreases
c) Gain increases as bandwidth increases till some extent after which stability decreases

4. There was a figure of 4:1 MUX in which A and B are select lines. Inputs S0 and S1 are connected together and labeled as C where as S2 and S3 are connected together and labeled as D. Then which of the following is true?
a) Y= B+C
b) Y= A+C
c) Y= A+B
d) Y= C+D (Where Y is the output)

5. In step up transformer (or Step down… not sure) transformation ratio is 1:5. If the impedance of secondary winding is 16 ohm then what is the impedance of primary winding?
a) 80 b) 3.2

6. There was a circuit consisting of AC voltage source and one inductance. Inductance value=0.2mH (or 0.2uH or 0.2H not sure).AC voltage =150 sin (1000t).what is the current flowing in the circuit?
a) i= 7.5 sin (1000t)
b) i= -7.5 sin (1000t)
c) i= 7.5 cos (1000t)
d) i= -7.5 cos (1000t)

7. Power gain of an amplifier having i/p gain of 20W and output gain of 20mW is
a) 60 b) 25 c) 10 d) 0

8. There was a RC circuit given with AC voltage source. Expression for capacitance was asked for charging condition. Choices were somewhat like this: a) some value multiplied by exp (-t/T)
ans --c i= (Vs/R)exp(-t/ T)

9. 2’s complement of -17
ans -- 01111

10. Instrumentation amplifier is used for--------- --?
a). effective shielding
b). high resective filters
c). high common mode
d). all the above.

11. In “ON CHIP” decoding memory can be decoded to
a) 2^n b)2^n +1 c)2^n -1 d) some other choice

12. Half of address 0Xffffffff is
a) 77777777 b) 80000000 c) 7FFFFFFF d) some other choice

13. Which one of the following is used for high speed power application?
a) BJT b) MOSFET c) IGBT d) TRIAC

14. One question related with SCR rotation angle given ifring angle is 30degree
ans - 150degree

15. SCR is used for
a) To achieve optimum (or maximum ...not sure) dv/dt
b) For high current ratings
c) To achieve high voltage
d) Some other choice

16. State in which o/p collector current of transistor remains constant in spite of increase in base current is
a) Q point b) Saturation c) Cut off

17. A 16 bit monosample is used for digitization of voice. If 8 kHz is the sampling rate then the rate at
which bit is transferred is
a) 128 b) 48 c) d)

18. To use variable as recursive, variable should be used as
a) Static b) Global c) Global static d) Automatic

19. what is the resonant frequency of parrel RLC circuit of R= 4.7 komh L= 2 micro Henry and c=30pf.
a). 20.5 MHz
b). 2.65 KHz
c). 20.5 KHz
d). none

20.for the parallel circuit (one figure is given) Is= 10mA. R1= 2R, R2=3R, R3= 4R. R is artritary
a). 3.076mA
b). 3.76mA

21. main ()
{
int a=0x1234;
a=a>>12;
a=a<<12;
printf (“%x”, a);
}
What is the output?
a)1000 b) 2000 c) d) None of these

22. What does (*fun () []) (int) indicate?
ans...b). an array of pointer to a functions that an int as parameter and return int.

23. #define A 10+10
main ()
{
int a;
a=A*A;
printf (“%d”, a);
}
a) 100 b) 200 c) 120 d) 400

24. One more question related with ADC like voltage is 8 volts. freqency 2 Mega hz. what is the converssion rate

25. Question related with serial in parallel out shift register…What is output sequence?
Ans..... 1010

26. Given one RLC circuit in which values of R, L and C were given. What is the value of frequency f?

27. if (fun ())
{
X++;
}
X gets incremented if and only if
a) fun () returns 0
b) fun () return 1
c) fun () return -1
d) return a value other than 0

28. In dynamic memory
a) Power dissipation is less than that of static memory
b) Clock is needed
c) Refreshing is required
d) All the above

29. Short, int and long integers have how many bytes?
a)2,2,4 b) Machine dependant c)2,4,8 d) Some other choice

30. A (n) is -----------filter combination of
a) Passive b) Active c) AMPLIFIER d) BOOSTER

31.Mobility of electron is
a) Increases as temperature increases
b) Decreases as temp decreases
c) Independent of conductivity
d) Some other choice

32. Structure comparison is done
a). yes
b) no
c) compiler dependent
d)

33. The system in which communication occurs in both ways but not simultaneously in both ways is
a) Half simplex b) Simplex c) Half duplex d) duplex

34. main ()
{
int a=5, b=6;
int i=0;
i=a>b? a:b;
printf (“%d”, i);
}
a) 0 b) 1 c) 6 d)

35. int fun (char c)
{
int i;
static int y ;}
a) c, i are stored in stack and y stored in data segment
b)c stored in stack and i,y are stored in data segment
c) c is stored in text segment, y in data and i in stack

36. main ()
{
int *p;
short int i;
p= (char *) malloc (i*10); (code was showing error here)
p+=10;
printf (“%d”, p);
}
Value of p?

37. main ()
{
int *p,i[2]={1,2, 3};
p=i;
printf (“%d %d %d”, i [0],*p,*p+1) ;
}

38. F = A'B' + C' + D' + E' then
A) F = A+B+C+D+E
B) F= (A+B)CDE
C) F = AB(C+D+E)
D) F= AB+C+D+E

39. how would you insert pre-written code into a current program?
a) #read
b) #get
c) #include
d) #pre

40.structure may contain
a) any other structure
b) any other structure expect themselves
c) any other structure except themselves and pointed to themselves
d) none of the above

41. three boys x,y,z and three girls x,y,z... sit around a round table . but x does not want any girl sitting to him and girl y does not want any boy sitting next. how many ways can they be seated.
a) 2
b) 4
c) 6
d) 8

42. k is brother of n and x. y is the mother of n and z is is the father of k . which of the following statement is not definitely true.
a) k is the son of z.
b) y is the wife of z.
c) k is the son of y.
d) n is the brother of x.

43. find the lateral surface of a prism with a triangular base if the perimeter of the base is 34 cm. and the height is 45 cm.
a) 765 square cm.
b) 3060 square cm.
c) 1530 square cm.
d) none

44.Given a< b< c < d . what is the max ratio of given equation
a) (a+b)/(c+d)
b) (b+c)/ (a+d)
c) (c+d)/ (a+b)
d) (a+c)/ (b+d)

45. A and B starts moving from points X and Y simultaneously at a speed of 5kmph and 7kmph to a destination point which is of 27 km from points X and Y. B reaches Y earlier than A and immediately turns back and met Z. Find the distance XZ.
ans.... 22.5 km

46. Ann is shorter than Jill and Jill is taller than Tom. Which of the following inferences are true?
a) Ann is taller than Tom b) c) d) Data insufficient

47. A and B starts from same point at opposite direction. They will move 6km and take 8km left. How Far is A and B from each other?
ans ...20m

48.6440 soldiers are to be arranged in the shape of square. If 40 soldiers were kept out then the number of soldiers making each straight line is?
Ans --80

49. Sum of squares of two numbers is 404 and sum of two numbers is 22.Then product of two Numbers?
a) 20 b) 40 c) d) (Answer is 40. Two numbers are 20 and 2)

50. In an examination 4 marks are assigned for correct answer and 1 mark is deducted for wrong answer. However one student attempted all 60 questions and scored 130.Number of questions he attempted correct is?
a) 35 b) 38 c) 42 d) 35

51. Each ruby is of 0.3 kg and diamond is of 0.4 kg.Ruby costs 400 crores and diamond costs 500 crores. Ruby and diamonds have to be put into a bag. Bag cannot contain more than 12 kg.Which of the following gives maximum profit (or in terms of wealth) (In crores)
ans ...only ruby 40p

52. the cpu stack is placed in ....
a). cpu resister
b). RAM
c). ROM
d). hard disk

53). (10 | 7) would produce
a). 17
b). 3
c). 11
d). 15

ENGLISH section
fill in the blanks...... .. the answers is
61) a). impounded b). protected c). hounded d). relegated.
62).a). oblinion b). authertiory c). dejection d). deso.......
63).a). subdued b). bountiful c). tentative d). ardem.
64).a). b). esulcant c). emblerratie d). innate
65).manor -- d). n. the landed estate of a land or nobleman.
66).neologism -- c). n. giving a new meaning to an add word.
67).batten -- b). n. a narrow strip of wood.
68).tepid -- d). adj. lacking interest enhusi...., luewarm
69. discerning -- d). adj. distinguishin one thing from another, having good judgment

Wipro Sample Test

2009-12-31T21:46:00.001+05:30

1. An electron moving in an electromagnetic field moves in a

(a) In a straight path
(b) Along the same plane in the direction of its propagation
(c) Opposite to the original direction of propagation
(d) In a sine wave

Ans. (b)

2. The total work done on the particle is equal to the change in its kinetic energy

(a) Always
(b) Only if the forces acting on the body are conservative.
(c) Only if the forces acting on the body are gravitational.
(d) Only if the forces acting on the body are elastic.

Ans. (a)
3. The following unit measure energy:

(a) Kilo-watt hour.
(b) Volt*volt/sec*ohm.
(c) Pascal*foot*foot
(d) (Coulomb*coulomb)*farad

Ans. (a)

4. Astronauts in stable orbits around the earth are in a state of weightlessness because

(a) There is no gravitational force acting on them.
(b) The satellite and the air inside it have an acceleration equal to that of gravitational acceleration there.
(c) The gravitational force of the earth and the sun balance giving null resultant.
(d) There is no atmosphere at the height at which the satellites move.

Ans. (b)
5. An organ pipe, open at both ends and another organ pipe closed at one end,
will resonate with each other, if their lengths are in the ratio of

(a) 1:1
(b) 1:4
(c) 2:1
(d) 1:2

Ans. (c)
6. During an isothermal expansion of an ideal gas

(a) Its internal energy increases.
(b) Its internal energy decreases.
(c) Its internal energy does not change.
(d) The work done by the gas is not equal to the quantity of heat absorbed by it.

Ans. (c)

7. A parallel plate capaciator is charged and the charging battery is then disconnected.
If the plates of the capacitor are moved further apart by means of insulating handles

(a) The charge on the capacitor increases.
(b) The voltage across the plates increases.
(c) The capacitance increases.
(d) The electrostatic energy stored in the capacitor decreases.

Ans. (b)
8. Two equal negative charges q are fixed at point (0,a) and (0,-a) on the y-axis.
A positive charge Q is released from rest at the point (2a,0) on the x-axis. The charge Q will

(a) Execute simple harmonic motion about the origin
(b) Move to the origin and remain at rest
(c) Move to infinity
(d) Execute oscillatory but not simple harmonic motion

Ans. (d)

9. A square conducting loop of length Lon a side carries a current I.
The magnetic field at the centre of the loop is

(a) Independant of L
(b) Proportional to L*L
(c) Inversely proportoinal to L
(d) Directly proportional to L

Ans. (c)
10. The focal length of a convex lens when placed in air and then in water will
(a) Increase in water with respect to air
(b) Increase in air with respect to water
(c) Decrease in water with respect to. air
(d) Remain the same

Ans. (a)

11. The maximum kinectic energy of the photoelectron emitted from the surface is dependant on

(a) The intensity of incident radiation
(b) The potential of the collector electrode
(c) The frequency of incident radiation
(d) The angle of incidence of radiation of the surface

Ans. (c)

12. An electron orbiting in a circular orbit around the nucleus of the atom

(a) Has a magnetic dipole moment
(b) Exerts an electric force on the nucleus equal to that on it by the nucleus
(c) Does not produce a magnetic induction at the nucleus
(d) All of the above

Ans. (d)

13. The X-rays beam coming from an X-ray tube will be:

(a) Monochromatic
(b) Having all wavelengths smaller than a certain minimum wavelength
(c) Having all wavelengths larger than a certain minimum wavelength
(d) Having all wavelengths lying between a minimum and a maximum wavelength

Ans. (c)

14. The mass number of a nucleus is

(a) Always less than its atomic number
(b) Always more than its atomic number
(c) Always equal to its atomic number
(d) Sometimes more and sometimes equal to its atomic number

Ans. (d)

15. Two successive elements belonging to the first transition series have the same number
of electrons partially filling orbitals. They are

(a) V and Cr
(b) Ti and V
(c) Mn and Cr
(d) Fe and Co

Ans. (c)

16. When n+l has the same value for two or more orbitals,the new electron enters the orbital where

(a) n is maximum
(b) n is minimum
(c) l is maximum
(d) l is minimum

Ans. (b)

17. A balloon filled with ethylene is pricked with a sharp pointed needle and quickly placed in a tank
full of hydrogen at the same pressure. After a while the balloon would have

(a) Shrunk
(b) Enlarged
(c) Completely collapsed
(d) Remain unchanged in size

Ans. (b)

18. Which of the following statements is not true?

(a) The ratio of the mean speed to the rms speed is independant of temperature
(b) Tthe square of the mean speed of the molecules is equal to the mean squared speed at a certain temperature
(c) Mean kinetic energy of the gas molecules at any given temperature is independant of the mean speed
(d) None

Ans. (b)

19. Which of the following statements represent Raoult's Law

(a) Mole fraction of solvent = ratio of vapour pressure of the solution to vapour pressure of the solvent
(b) Mole fraction of solute = ratio of vapour pressure of the solution to vapour pressure of the solvent
(c) Mole fraction of solute = lowering of vapour pressure of the solution
(d) Mole fraction of solvent = lowering of vapour pressure of the solution

Ans. (a)

20. Elements having the same atomic number and the same atomic mass are known as

(a) Isotopes
(b) Isotones
(c) Isomers
(d) None of the above

21.Which is the most acidic amongst

(a) Nitrophenol
(b) O-toulene
(c) Phenol
(d) Cresol

22. Pure water does not conduct electricity because it is

(a) Almost not ionised
(b) Low boiling
(c) Neutral
(d) Readily decomposed

Ans. (a)

23. In a salt bridge, KCl is used because

(a) It is an electrolyte
(b) The transference number of K+ and Cl¯ is nearly the same
(c) It is a good conductor of electricity
(d) All of the above

Ans. (d)

24. A depolarizer used in the dry cell batteries is

(a) KCl
(b) MnO2
(c) KOH
(d) None of the above

Ans. (b)

25. The hydrolysis of alkyl halides by aqueous NaOH is best termed as

(a) Electrophylic substitution reaction
(b) Electrophylic addition reaction
(c) Nnucleophylic addition reaction
(d) Nucleophylic substitution reaction

Ans. (d)

26. The hydrocarbon that gives a red precipitate with ammoniacal cuprous chloride is (where 'º' means a triple bond)

(a) CH3-CH2-CH2-CH3
(b) CH3-CºC-CH3
(c) CH2=CH-CH=CH2
(d) CH3-CH2-CºCH

Ans. (d)

27. Which of the following reagents is neither neutral nor basic

(a) Lucas' reagent
(b) Tollen's reagent
(c) Bayer's reagent
(d) Fehling's solution

Ans. (a)

28. The substance which is most easily nitrated

(a) Toluene
(b) Bbenzene
(c) Nitrobenzene
(d) Chlorobenzene

Ans. (a)

29. Carbylamine reaction is a test for

(a) Primary amine
(b) Secondary amine
(c) Tertiary amine
(d) Quarternary ammonium salt

Ans. (a)

30. Which of the following oxides cannot be reduced by carbon to obtain metal

(a) ZnO
(b) Al2O3
(c) Fe2O3
(d) PbO

Ans. (b)

31. Which of the following is not an oxide ore?

(a) Cassiterite
(b) Siderite
(c) Pyrolusite
(d) Bauxite

Ans. (b)

32. Which among the following is called philosopher's wool

(a) Cellulose
(b) Calamine
(c) Stellite
(d) Cerussite

Ans. (c)

33. Out of 10 white, 9 black and 7 red balls, in how many ways can we select one or more balls

(a) 234
(b) 52
(c) 630
(d) 879

Ans. (d)

34. A and B throw a dice. The probabilty that A's throw is not greater than B's is

(a) 5/12
(b) 7/12
(c) 11/12
(d) 5/36

Ans. (b)

35. Given two numbers a and b. Let A denote the single AM between these and S denote the sum of n AMs
between them. Then S/A depends upon

(a) n
(b) n,a
(c) n,b
(d) n,a,b

Ans. (a)

36. If the sum of the roots of the equation ax²+bx+c=0 is equal to the sum of the squares of their reciprocals,
then, a/c, b/a, c/b are in

(a) AP
(b) GP
(c) HP
(d) None of the these

Ans. (c)

In the following questions ~ represents the integral sign-for eg. 1~2[f(x)] means integration of
the function f(x) over the interval 1 to2.

37. Value of -1~2[|2-x²|]dx, ie integration of the function |2-x²| over the interval -1 to 2.

(a) 0
(b) 1
(c) 2
(d) None of the above

Ans. (d)

38. If 0~P[log sinx]dx=k,then the value of 0~P/4[log(1 + tan x)]dx ,where P stands for pi,is

(a) -k/4
(b) k/4
(c) -k/8
(d) k/8

Ans. (c)

39. If a,b,c be in GP and p,q be respectively AM between a,b and b,c then

(a) 2/b=1/p+1/q
(b) 2/b=1/p-1/q
(c) 2=a/p-c/q
(d) None of the above

Ans. (a)

40. A solution of KMnO4 is reduced to MnO2 .The normality of solution is 0.6.The molarity is

(a) 1.8M
(b) 0.6M
(c) 0.1M
(d) 0.2M

Ans. (d)

The questions 41-46 are based on the following pattern.The problems below contain a question
and two statements giving certain data. You have to decide whether the data given in the
statements are sufficient for answering the questions.The correct answer is

(A) If statement (I) alone is sufficient but statement (II) alone is not sufficient.
(B) If statement(II) alone is sufficient but statement(I) alone is not sufficient.
(C) If both statements together are sufficient but neither of statements alone is sufficient.
(D) If both together are not sufficient.

41. What is John's age?

(I) In 15 years John will be twice as old as Dias would be
(II) Dias was born 5 years ago

Ans. (C)

42. What is the distance from city A to city C in kms?

(I) City A is 90 kms from City B
(II) City B is 30 kms from City C

Ans. (D)

43.Is A=C ? A,B,C are real numbers

(I) A-B=B-C
(II) A-2C = C-2B

Ans. (C)

44. What is the 30th term of a given sequence ?

(I) The first two terms of the sequence are 1,1/2
(II) The common difference is -1/2

Ans. (A)

45.Was Avinash early, on time or late for work?

(I) He thought his watch was 10 minutes fast
(II) Actually his watch was 5 minutes slow

Ans. (D)

46. What is the value of A if A is an integer?

(I) A4 = 1
(II) A3 + 1 = 0

Ans. (B)

47. A person travels 12 km in the southward direction and then travels 5km to the right and then travels 15km toward the right and finally travels
5km towards the east, how far is he from his starting place?

(a) 5.5 kms
(b) 3 km
(c) 13 km
(d) 6.4 km

Ans. (b)

48. X's father's wife's father's granddaughter uncle will be related to X as

(a) Son
(b) Nephew
(c) Uncle
(d) Grandfather

Ans. (c)

49. Find the next number in the series 1, 3 ,7 ,13 ,21 ,31

(a) 43
(b) 33
(c) 41
(d) 45

Ans. (a)

50. If in a certain code "RANGE" is coded as 12345 and "RANDOM" is coded as 123678.
Then the code for the word "MANGO" would be

(a) 82357
(b) 89343
(c) 84629
(d) 82347

Ans. (d)

51. If "PROMPT" is coded as QSPLOS ,then "PLAYER" should be

(a) QMBZFS
(b) QWMFDW
(c) QUREXM
(d) URESTI

Ans. (a)

The questions 52-53 are based on the following data

6 people A,B,C,D,E and F sit around a table for dinner.Since A does not like C, he doesn't sit either opposite or beside C.B and F always like to
sit opposite each other.

52. If A is beside F then who is are the two neighbours of B?

(a) D and C
(b) E and C
(c) D and E
(d) Either (a) or (b)

Ans. (c)

53. If D is adjacent to F then who is adjacent to C?

(a) E and B
(b) D and A
(c) D and B
(d) either (a) or (c)

Ans.(d)

54. Complete the sequence A, E ,I ,M ,Q ,U , _ , _

(a) B, F
(b) Y, C
(c) G, I
(d) K, O

Ans.(b)

55. A person travels 6km towards west, then travels 5km towards north ,then finally travels
6km towards west. Where is he with respect to his starting position?

(a) 13km east
(b) 13km northeast
(c) 13km northwest
(d) 13km west

Ans. (c)

56. If A speaks the truth 80% of the times, B speaks the truth 60% of the times.
What is the probability that they tell the truth at the same time

(a) 0.8
(b) 0.48
(c) 0.6
(d) 0.14

Ans.(b)

57. If the time quantum is too large, Round Robin scheduling degenerates to

(a) Shortest Job First Scheduling
(b) Multilevel Queue Scheduling
(c) FCFS
(d) None of the above

Ans. (c)

58. Transponders are used for which of the following purposes

(a) Uplinking
(b) Downlinking
(c) Both (a) and (b)
(d) None of the above

Ans. (c)

59. The format specifier "-%d" is used for which purpose in C

(a) Left justifying a string
(b) Right justifying a string
(c) Removing a string from the console
(d) Used for the scope specification of a char[] variable

Ans. (a)

60. Virtual functions allow you to

(a) Create an array of type pointer-to-base-class that can hold pointers to derived classes
(b) Create functions that have no body
(c) Group objects of different classes so they can all be accessed by the same function code
(d) Use the same function call to execute member functions to objects from different classes

62. A sorting algorithm which can prove to be a best time algorithm in one case
and a worst time algorithm in worst case is

(a) Quick Sort
(b) Heap Sort
(c) Merge Sort
(d) Insert Sort

Ans. (a)

63. What details should never be found in the top level of a top-down design?

(a) Details
(b) Coding
(c) Decisions
(d) None of the above

Ans. (c)

64. In an absolute loading scheme, which loader function is accomplished by assembler

(a) Reallocation
(b) Allocation
(c) Linking
(d) Both (a) and (b)

Ans. (d)

65. Banker's algorithm for resource allocation deals with

(a) Deadlock prevention
(b) Deadlock avoidance
(c) Deadlock recovery
(d) None of these

Ans. (b)

66. Thrashing can be avoided if

(a) The pages, belonging to the working set of the programs, are in main memory
(b) The speed of CPU is increased
(c) The speed of I/O processor are increased
(d) All of the above

Ans. (a)

67. Which of the following communications lines is best suited to interactive processing applications?

(a) Narrowband channels
(b) Simplex channels
(c) Full-duplex channels
(d) Mixedband channels

Ans. (b)

68. A feasibility document should contain all of the following except

(a) Project name
(b) Problem descriptions
(c) Feasible alternative
(d) Data flow diagrams

Ans. (d)

69. What is the main function of a data link content monitor?

(a) To detect problems in protocols
(b) To determine the type of transmission used in a data link
(c) To determine the type of switching used in a data link
(d) To determine the flow of data

Ans. (a)

70. Which of the following is a broadband communications channel?

(a) Coaxial cable
(b) Fiber optic cable
(c) Microwave circuits
(d) All of the above

Ans. (d)

71. Which of the following memories has the shortest access time?

(a) Cache memory
(b) Magnetic bubble memory
(c) Magnetic core memory
(d) RAM

Ans. (a)

72. A shift register can be used for

(a) Parallel to serial conversion
(b) Serial to parallel conversion
(c) Digital delay line
(d) All the above

Ans. (d)

73. In which of the following page replacement policies, Balady's anomaly occurs?

(a) FIFO
(b) LRU
(c) LFU
(d) NRU

Ans. (a)

74. Subschema can be used to

(a) Create very different, personalised views of the same data
(b) Present information in different formats
(c) Hide sensitive information by omitting fields from the sub-schema's description
(d) All of the above

Ans. (d)

RELIANCE PAPER

2009-12-31T21:43:00.000+05:30

Hello freinds, The Reliance Energy Limited came to our college. the process includes
1. written technical -total no of questions is 30 , time 40 minutes no negative marking(elimination round)
2. GD(non elimination round)
3. PI

WRITTEN TECHNICAL (Electronics & Telecommunication)
The questions were mainly from Instrumentation and control system.they require instrumentation engineers.
2 numerical quetions of digital communication 1 from sampling frequency.
1 from optical communication very simple que.
6-7questions from control theory mainly used in power plant related to cascaded controller, PID, second order transfer function, steady error rate in thermometre.
Q The op-amp in non linear mode is used as a) integrator b) schimmt trigger c)active filters d) instrumentation amplifier.
1 quest on IR spectroscopy, stroboscope, pirani guage, 2 questions related to flow measurement, vacuum, Bellow's expansion.
1 numerical on LVDT
1 optical lens magnification numerical
1 question from current transformer(find error rate)
1 question from Geiger tube (quantum efficiency in Angstrom)
Q friction related question of 12th class
Q.The decimal no. -15 can be written in 8-bit signed 2's complement as?
Q The no. of flip- flops used in MOD-10 counter is
Q The deflection sensitivity of CRT is directly/inversely proportional to
QThe common collector configurations characteristics is a) high i/p & low o/p impedance....................................... package 4.5 lakh per annum, and no service bond

NALCO papers on 28th oct. 2007

2009-12-31T21:41:00.000+05:30

Out of 120 questions 60 technical questions, 60 non-tech questions and time duration was 2 and half hours.

1) rms value of AC current is equal to main current at an angle ?

a) 60 b) 45 c) 30 d) 90

2) Corona produce in two transmission line if ?

a) Whn both placed closed b) Whn both placed widely

c) Potential difference is very high d) I did’t remembered

3) Diac is equall to ?

a) Two diode and two resister b) Trice with two gate

c) Two thyristor connect in antiparallel d)Four layer pnpn scr

4) In the Economic transmission point of view Kelvin,s gives ?

A) Voltage drop at feeder b) transmission losses

Two more option I did’t remember.

5) The essential condition for using servomotor for high torque at all?

a) load b) speed

c) voltage d) power

6) If torque of induction motor is 200n-m at slip 12%. Wht is the value of torque at slip 6%.....?

a) 60n-m b) 40

c) 100 n-m d) 150

7) If speed of induction motor at full load is 950rpm then wht is the speed at half load ?

a) 1000 rpm b) 975

Two option I forget……

8) Find the value of ∫│cost +1/2│dt where limit is 0 to ∏/2?

Option I did’t remerm

9) Wht is the slope of –x3+ 6x2+ 2x+1=0?

Option I did’t remerm

10) For the purpose of constant speed we use?

a) Brushless dc motor b) stepper motor

c) shaded pole motor d) sync motor

11) Essential condition for parallel operation of 1phase transformer?

a) Rating b) Polarity

c) Impedance d) Voltage

12) Some of G.K. question

About placed of shriharikota space center.

13) Who did give the name of AL?

14) World literacy day?

15) World environment day?

16) If two heaters of 100W & 110 volt other are 150W & 110 volt connect in series with 220v ac source then what is the total power?

17) What is the GDP rate of India in 2005-06?

18) What is the % rate of production of AL in India?

a) 10 b) 20

c) 7 d) 14

19) One passes which have 4Qs

20) Some fill in blanks

21) One Q about the find the % of fifth harmonics in a current wave equation.

22) 4pole 3phase 50 htz induction motor running at 1460 rpm. Wht is the speed of running slip wrt to rotor?

23) Find the new kva rating of transformer which hav 220 kva, 1watt iron loss and 2watt cu loss?

24) Hay’s bridge used for the measurement of?

25) In R-L-C parallel resonant circuit

a) Impedance is max. b) resistance is min

two more options I did’t remember

26) Two bulb 100w & 50w respectively with constant resistances connect in series with 220v dc source then wht is the value of voltage across first bulb?

27) AC ammeter used for measurement in

a) Pmmc b) induction meter

c) Dynamometer d) moving iron type meter

28) For ac current we use?

Same option

29) we preferred V-V connection over T-T connection why?

30) Suspension type insulator is used whn?

a) high voltage b) low voltage

c) oscillating voltage d) steady voltage

31) One question based on Norton’s theorem?

32) One question based on inverted-V curve of synch motor.

33) one question based on load saturation curve.

34) There were 4 question like if A*B represent A is the son of B, some like that

35) Some questions from BAR-GRAPHS

36) Some questions from probability.

37) Some questions from Data–interpretation

38) Some questions from diodes.

39) One question from winding design for motor (find value of commutator segment)

GD topics for placement

2009-12-31T21:38:00.000+05:30

1. US war on iraq-justified or not.
2. Role of UN in peacekeeping.
3. Position of Women in India compared to other nations.
4. Environment MAnagement.
5. Is China better than India in software.
6. Should SONIA gandhi be made the PM
7. BPOs in INDIA
8. Govt contribution to IT
9. will punch lines rule the Advt
10. premaritial sex
11. is china a threat to indian industry
12. india or west , which is the land of opportunities
13. water resources should be nationalised
14."BALANCE BETWEEN PROFESSIONALISM AND FAMILY"
15. Effect of cinema on Youth
16. Education in India compared to Foreign nations
17. Is it necessary to ban COCOCOLA in India.
18. What is the effect of movies on youth.(is it good or bad)
19. Are studies more benifitial in India or in Abroad.
20."UN's peace activities" and "America's war on Iraq".
21."Environment-Whose Responisibility".
22.Is China a threat to the indian software industry.
23.Role of UN in Peace keeping
24.War on Iraq
25.About Hockey being the primary game in India.
26.Can america occupy iraq
27.Cricket shud be banned or not.
28.IS CHINA A THREAT TO INDIA
29.Present state of Indian Cricket team.
30.Love marriage/Arranged marriage.
31.Advantages of Co-education.
Hot Topics:
-----------
1.How to deal with international terrorism.
2.Should we pursue our policy of dialogue with Pakistan?
3.Is peace and non-violence outdated concepts?

Robert Bosch Joint campus recruitment

2009-12-31T21:37:00.002+05:30

Pattern:
Round 1: The Round consists of 3 parts and total 60 questions
Part 1: Technical (1-35)
Part 2: Numerical(36-47)
Part 3: General English(48-60)
for right answers 2 marks and for wrong answers -0.5 marks
Part 1: Technical
Questions from subjects like Digital Electronics, Microprocessor, Computer Networks, Circuit Analysis, Linear Integrated Circuits were covered in the Question paper. All the questions are basic electronics.....
Give importance to:
Digital: Logic diagrams
, Basic gates, TTL logic levels, K map and memory units
Microprocessor: assembly language programming, instruction sets, architecture
Networks: RIP, BGP, TCP, OSPF, IGMP and so on
Circuit analysis: Impedence measurement, Mesh Analysis, Power, and so on
LIC: working 0f OPAMPs

Part 2: Numerical
Aptitude questions from
Time and distance
Velocity
directions
profit and loss(discount and profit problems)
area, volume(given the sides of the Equilateral triangle and square in terms of x. the perimeter is same calculate x.. like that, then the side of a square is increased by 5 then what is the new area..like that)

Part 3: General English
Data insufficient type questions
Arrange the jumbled sentences
Error detection the given sentence

I have attended Robert Bosch today and hoping for good results.......

All the best friends

ROBERT BOSCH PAPER ON 13th AUGUST 2008

2009-12-31T21:37:00.001+05:30

Hello Friends.....All the best for your career....I'm very happy to post you my experience in Robert Bosch Joint campus recruitment held today (13/08/08).The paper which attended is my core (ECE). Here I have discussed about my experience and I hope it will be help for U....
Pattern:
Round 1: The Round consists of 3 parts and total 60 questions
Part 1: Technical (1-35)
Part 2: Numerical(36-47)
Part 3: General English(48-60)
for right answers 2 marks and for wrong answers -0.5 marks

Part 1: Technical
Questions from subjects like Digital Electronics, Microprocessor, Computer Networks, Circuit Analysis, Linear Integrated Circuits were covered in the Question paper. All the questions are basic electronics.....Give importance to:
Digital: Logic diagrams, Basic gates, TTL logic levels, K map and memory units
Microprocessor: assembly language programming, instruction sets, architecture
Networks: RIP, BGP, TCP, OSPF, IGMP and so on
Circuit analysis: Impedence measurement, Mesh Analysis, Power, and so on
LIC: working 0f OPAMPs

Part 2: Numerical
Aptitude questions from
Time and distance
Velocity
directions
profit and loss (discount and profit problems)
area, volume (given the sides of the Equilateral triangle and square in terms of x. the perimeter is same calculate x.. like that, then the side of a square is increased by 5 then what is the new area.. like that)

Part 3: General English
Data insufficient type questions
Arrange the jumbled sentences
Error detection the given sentence

I have attended Robert Bosch today and hoping for good results.......All the best friends
Alsath M
(13/08/2008)

Robert Bosch interview on 1 sept 2006

2009-12-31T21:36:00.001+05:30

The paper was easy if u have basic knowledge of electronics and C.

Most of the questions were of average difficulty.

45 technical Questions and 35 Aptitude questions

Total Time 1:30 Hours
2 marks for correct answers and -1 for wrong answers.

U have to study the following things (Only Basic)

1. advantage of DRAM over SRAM.
2. about OP-amp Gain and Bandwidth.
3. FET, MOSFET, Transirors.
5. Boolean Algebra.
6. Logic Gates.
7. Digital Arithmetic.
8. If u want to get through C, Study "Let Us C Solutions" By Kanetkar.

and at last for the aptitude test no preparation is required if ur funda's of 8th 9th 10th mathematics r good.

study english carefully should b able to correct sentences grammatically.

5 words were given to find out their meanings.like manor, inept, descerning, etc

Technical - Other Robert Bosch

2009-12-31T21:34:00.002+05:30

30 Technical.
30 Aptitude(quant and verbal).
-->Technical includes question on NETWORK ANALYSIS,ELECTRONIC DEVICE AND CIRCUITS,ANALOG CIRCUITS,C -language,OPERATING SYSTEMS,COMPUTER NETWORKS
,DIGITAL ELECTRONICS etc.
They didn't ask any question on Electromagnetic Theory,Control Systems.
--> Aptitude includes English/Verbal (ANALOGIES,REASONING,SYNONYMS,ANTONYMS,Passage Arrangement,Short Passage and questions on it and Quant.
All the following are the important topics and also some topics covered the some previous Written Tests by Bosch.
TECHNICAL(Topics to be read before attending to the written )
1.Analog Circuits.
--BJT( Biasing,current/voltage series/shunt, Transfer Characteristics Location of Q-point,circuit diagrams of CB,CC,CE Self biasing Circuits).
--ADC/DAC formulas and methods.
--Power amplifiers(class A,AB,B,C,D and Transformer Coupled Amplifier), their efficiency..
--Standard Circuits of IC 741 and 555 timers.
2.Network Analysis/Circuits.
--Basic Circuits Analysis(Nodal Analysis,Mesh Analysis)
--Dependent Souces equivalent circuits.
--Theorems.
3.EDC
--JFET,BJT behaviour with temperature and other parametes.
eg.In n-channel JFET what will be the output if the drain current increase.A) DECREASES, B) INCREASES ,,,,,.......etc.
--Standard effects like Thermal Run away, Avalanche effect etc.
4.Digital and Micro Processor
--Boolean Algebra and code conversions (BCD to Excess -3).
--Semiconductor memories.
--Microprocessor (8085 only).
--TTL,CMOS etc logics voltages
--Address Notations for memories.
5.Computer Networks .
--Network topologies.
--TCP/IP.
--OSI models.
APTITUDE
1.Mathematics or Quantitative
--Boat,Age,Train,Profit and Losses ,Discounts,time and work,Percentage,SI,CI.
2.Verbal
--Antonyms,Synonyms,Passage.. and Logical reasoning like ........All X are Y. Some Y like Z .
--There will be only 7/9 questions on verbal.these are not soo tough but takes time for passage and answering.
NOTE:Its important that you must perfectly use the time. try answering Aptitude/Quant first and then technical.
All those who have poor technical knowledge must improve atleast in the topics like CN,NT,Digital.Also follow the previous papers which I have followed in order to get through this written test.
All the Best!!

ROBERT BOSCH INTERVIEW - 07 JUL 2006

2009-12-31T21:34:00.001+05:30

Hello Frnds,

These are the interview q's i was asked in the Bosch on 07/07/2006 .

First of all he asked 's like tell me abt ur self, ur CET rank, and he also asked my maths score.

Then he started with the tech stuff

1. diff b/w DOS and UNIX?
2. which subject r u proficient in??(me C++)
3. wat is the diff b/w c and c++?
4. wat is features of c++?
5. wat is polymorphism?ex..
6. details abt my project?
7. lot of dbms stuff - wat is normalization?, how many codd rules r
there?, wat r they? , why do nedd them?
8. some queries? - wat does group by do ?
one query was - there is a relation called Salary which has attributes like empid, salary, level of employee. the q was group all the employees according th thier level, and calculate the sum of all salaries in aparticular level??
9. diff types of databases??
10. abt the system software lab project??
11. the some unusual stuff from maths ?? integration, differentiation
12.. Puzzles
a. There r three jars of capacity 8, 5 and 3 ltrs. U have to take out exactly 4 ltrs from the 8 ltr jar which is full initially and the other two jar(5 and 3) are empty.
b. next no. in the sequence 1 2 4 7 12 __

Lastly he asked if i have any q's ?????

Hope all this stuff becomes useful if u have any future interviews with Bosch

Just to give you an idea, these are things aked in my interview, go through more on storage class, volataile, bitwise operation, and project related.

ROBERT BOSCH PAPER ON 29TH JUNE

2009-12-31T21:32:00.002+05:30

We were asked to fill a form…nd den wer given answer sheets…OMR sheets….2HB pencils wer to b used itcms…pencil…AH!!!!!!!none of us had I guess…so dey said pen would do…..hmm…k nd passport photo required too..

Question papers wer distributed….
Test duration:60 minutes…..
No.Of questions:60
2 marks for right answer
0.25 marks minus for each wrong answer……..

I had referred some old papers which were in this site….but none helped me…..it was totally NEW PATTERN>>>NEW PATTERN>…….
U need to b prepared well with the basics…..am fro m ECE stream…so for ECE u need to b well prepared with BASIC ELECTRONICS, lil network analysis,lil Electronic circuits…MICROPROCESSORS, OPERATING SYSTEMS, CCN, APTITUDE AND ENGLISH……..I remember some questions….am not putting dem in order…jus…first fifteen questions wer horrible…rest wer easy kinda…..PREPARE DES QUESTIONS WELL…..SURE U R DONE WITH 32 QUESTIONS….SURE…SURE …so here I go

1.A circuit was given…nd current was asked….applying KCL I guess..

2.A two port network was given..two ports wer arranged serially…so go through 2port networks..

3.Nodal analysis is applicable to
a. KVL & ohms law
b. KVL, KCL &ohms law
c. KVL and KCL.
d.

4.A question on Superposition theorem. simple one…I don remember much…it was somthn like dis…Superposition theorem can be applied to
a. linear networks
b. non linear networks
c.
d.

5.series of resistors are connected across ideal voltages.. If pair of resistors are introduced wot happens to the voltage?
a. halved
b. doubled
c. same
d.

6.Calculation of some OPAMP parameters…I guess calculation of CMRR…don remember….

7.Which of these Class of amplifiers are least efficient?
a. class A
b. class D
c. class B
d. class C

8.In a RC coupled amplifier, coupling capacitor is used
a. changes the frequency response
b. frequency response is unaltered.
c.
d. to reduce DC component but no effect on freq response.

9.Which of these is not a 8-bit processor
a.8085
b.Z-60
c.606666
d.6025

10.How many address lines would u require to access 4K memory?
a.12
b.14
c.16
d.

11.How many times the loop gets executed?
Loop:mov b,#64
NOP
Dec b
Jnz loop
a. infinite
b.64
c.65
d.63

12.Multiprogramming and multithreading….Which of these are better?
a. multiprogramming
b. multithreading
c. depends on the system
d. both are efficient

13.Ethernet has a___________ topology
a. star
b. ring
c. bus
d. linear

14.Which of these are used for high speed and long distance?
a. shielded cables
b. unshielded or twisted cables.. don't remember
c. optical fibres
d. none of the above

15.A Modem converts _______ to _________.
a. analog to serial
b. serial to parallel
c. analog to parallel
d.

16.Which of the following is not a n/w
a. MAN
b. WAN
c. LAN
d. NAN

17.Which of the following is not a type of memory?
a. instuction register
b. instuction stack
c. instuction opcode
d. translation buffer

18.Arrange the TCP/IP OSI layers from bottom to top
a. application, presentation, session, transport, network, datalink, physical
b. network, physical, transport, application
c. physical, datalink, network, transport, session, presentation, application
d. transport, application,……..

19.A question on daemon and background process.
a. both are same
b. daemon is superior dan background
c. daemon created by user?
d. background has a transmitting terminal…..i don remember this properly…

20.When set of interrupts occur
a. one is processed..rest are ignored
b. all are processed according to some scheduling scheme
c. system is aborted/terminated
d. none of the above

21.Which of these schemes introduces a minimum delay?
A, FIFO
b. round robin
c. priority
d.

22.A train problem.

23. A man went 10mts east…and then 20mts north. What is the distance from the start?
--right angled triangle…finding hypotnuese..

24.A car travels at the rate of 50km/hr to reach a places.while comin back it takes 60/hr,what is the avg speed?

25.A’s and B’s income ratios are 2:3.Expenditure ratio is 5:3.What is A’s income?

26.A can finish work in 2 days.B can finish the same work in 6 dayz..Together how much time they take to finish the work…??

27.Another work problem…A and B together take ______days.A takes _____ days.how much B takes alone to finish da work?

28.2 questions on Heights and distance…a question something a pole standing vertical…….sorry don remember

29. A381 is a given number.Which smallest number is ‘A’ such that the number is divisible by 11?
a.5
b.6
c.7
d.9

30.When (9^6+1) is divided by some number(I don remember),wot is the remainder?
a.0
b.1
c.2
d.

Give the SYNONYMS
31.TO MERIT

32.OBSE
I still don know wot it means :)…

Give the ANTONYMS
33.RECANT
34.

Fill in the blanks
35.I don’t know _____ they make money of bill.
a. why
b. how
c, which
d. what

36.The children ran in _________they disappeared.
a. but
b. and
c.
d.

37.Skin: human
a. hide: animal
b. peel: potato
c:
d:

38.Editor:Magazine
a. actor: film
b. captain: ship
c. director: film
d.

ROBERT BOSCH PAPER ON 13TH AUGUST

2009-12-31T21:32:00.001+05:30

Technical Section:
C, C++, Data Structure, Networks and Operating Systems

Aptitude section:
Profit&Loss, Time & Work, Time & Distance, Percentage, A.P, G.P, Average and general maths qs…

English section:
Error Correction, Jumbled Sentences, Inference from the passage…
These are the questions that I remember….
1.main()
{
int const *p = 10;
printf(“%d”, ++(*p));
}
Ans: Compilation Error

2.main()
{
int I;
char *a = “man”;
for(i=0;i<3;i++)
printf(“%c%c%c%c\n”,i[a],a[i],*(a+i),*(i+a));
}
Ans: mmm
aaa
nnn

3.main()
{
extern int i = 10;
printf(“%d”,i);
}
Ans: Runtime error

4.main()
{
float f= 1.1;
double d = 1.1;
if ( f == d)
printf(“I LOVE U”);
else
printf(“I HATE U”);
}
Ans: I HATE U

5.How many queues are required to implement Priority queue
Ans: Two

6.A Binary tree is given and asked to write post order traversal

7.No of Nodes of Perfect Binary Tree :
Ans: 2n-1

8.Which Data Structure is efficient to implement Trees
Ans: Linked List

9. Heterogeneous Linked List is implemented using
Ans: Void pointers

10. #include
void main()
{
int i = 15;
int const & j = i;
cout << i << j << endl;
i = 9;
cout << i << j << endl
}
ANS: 15 15
9 9

11. which of the following function is used to display a character in console
ANS: put

13. To use manipulators we have to include the header file..
ANS: iomanip.h

14. Virtual memory consist of main memory and
ANS: swap files

15. Command used to partition the disk
ANS: fdisk

16. A question from MS-DOS

17. A question from power management

18. A problem in Paging (Given Main and Secondary memory access time, and Hit Ratio)

19. Which of the following is a network layer protocol
a) TELNET b) FTP. Sorry I forget the remaining

20. A question from class addressing

21. A question from routing algorithm

22. Profit & Loss problem

23. Percentage problem

24. General Maths problem

25. Surface area

26. Time & Distance

27. Time & Work

28. A.P

29. G.P

30. Jumbled Sentences – 3 qs

31. Error Correction – 3 qs

32. Inference from paragraph – 2 qs

TIPS: I answered only 48 qs…. I skipped 2 qs in English , 3 qs in Apti, 2 qs in C++, 3 qs in Networks and 2 qs in OS…. Among these 48 qs that I answered, I am sure that all of them are right…. So if u answered more than 45 qs U almost get through… Time is not a problem here…. You can solve technical qs within 20 mins if u r sound in it…
For C: Follow EXPLORING C by Yashwant Kanitkar and also see C – Aptitude qs in FreshersWorld.com
For Aps: Its fully quantitative so Quantitative Aptitude By Agarwal is more than enough
For OS: Concentrate on Paging, Segmentation, Commands in UNIX & MS-DOS… I referred Silberschatz book
For Networks: Concentrate on Duties of Each Layers as well as Protocols used in each layer, Class Addressing, Routing & its algorithms… Forouzan Book will be handy
For Data Structure: Prepare on Trees, Graphs, Stacks, queues, Linked List.. I referred Wicikipedia articles for all the Data Structure topics….
For English: Its all depend on ur English knowledge & some common sense.. For me it’s a cake walk….

So , The Bottom-Line is if u are sound in Technical and having a average problem solving skill and some common sense u can easily get through the Robert Bosch Written Test….All The Best Guys!!!!!!!!!!

ROBERT BOSCH PAPER ON 18TH APRIL

2009-12-31T21:30:00.000+05:30

WRITTEN TEST (89/400)
Questions; 60 Duration: 1 h
CORRECT ANS +1, WRONG no negative
No. of question from diff section are (appx)
1. Basic Electrical-5 q’s
HINTS.
1.Critical circuit (V, I, R) find the voltage between two points.
In one path current source and one R is present. (Easy)
2, KVL,KCL
3. Theory from reciprocity theorem (not sure but one theorem)

2. Basic electronics-5 q’s
i. OPAMP
ii. Different diodes

3. Digital electronics -8 q’s
I. how many min. NAND gate is used to make one EX-OR gate.
Ans 4
ii. Flip flop
iii. Gates

4. Microprocessor and microcontroller -5 q’s
i. which is not a 8 bit mp.
(8085, / / / )
ii. What is the long instruction in mp 8085 instruction set?

6. Computer network and operating system -7 q’s
i. topology
ii. OSI model
iii. Write the 7 layers from top to down,
iv. Some questions from different layers.

7. control system, digital signal processing, etc—5 q’s

8. aptitude-15(very easy) Just remember ur school day. See RS Agawam for 2 h (only for confidences build up)
i. height and distance( 2 no’s) very easy put tan x = ??
Ladder, wall, base
ii. Work, power, energy (no of man, no of women, efficiency, no of work relation)
ii. Train
iii. Boat and stream (u can easily attempt 13 out 15, or 15 /15)

9. English(10 q’s)
i. Synonyms & antonyms ,
ii. Some q’s from prepositions,
iii. Simple grammar.
Total 60 q’s

About 400 candidate were appeared the written test .89 student were got selected.I am one of them (Next day the result was displayed).

GD (89/89)

Don’t worry about GD it’s is formality.No one got rejected.(may be they have time )Topics for all 89 are same.Topics is lets consider u r a CEO of an courier service and u have only 5 laks, how u make it to 5 cr. Within in one year.Don’t worry about this in all times GD is not conducting rarely

Technical Interview (5/89)
This is the main round to crack.There may be two technical.(differ from person to person) For me 2. almost all 2 , few have 1.
Percentage of impotence of diff. subject
1. project- 20 %
i. Describe ur project with block diagram (common for all).This is the best way to make ur impression .They will ask about working of diff parts. (No matter how difficult or easy project u. Have done. They simply check ur knowledge about the project. (20 to 30 min).Depending upon ur project they will go to ur subject.As my project is based on basic electronics, they start from basic electronics.

2. Basic electronics -15 %
i. What is Zener diode, knee current, knee voltage, draw the characteristics, transistor with characteristics, OPAMP, SCR, basics but entire thing. Ideal characteristics of OPAMP, if high i/p impedance and low o/p impedance will not maintain what will happen. can current flow opposite direction of diode , explain with diagram. Relay, 12volt battery supply is given to 6volt bulb, how possible explain with diagram. Multivibator .

3. Digital electronics -20%
h i. about all gates, truth table and diagram, universal gates , why universal, derive all gates from NAND gates,(E-OR from NAND), design a counters, resister, flip flop, conversion of flip flop, magnitude comparator (2 and 3 bit). Design a circuit for square wave; design a circuit for buzzer (buzzer used in quiz competition)

4. Micro processor and micro controller 15%
i. 8085 mp, 8051 mc are very important, difference between microprocessor and microcontroller, interrupt, program counter, how interrupt occur with memory diagram, architecture of 8085 and 8051. (Follow BRAM, and mazide). It is possible to made float variable in 8086 mp and why. Write a program in mp or mc such it will measure room temperature.

5. C (only c) 20%
i. this is the one of the important section. Complete over look is required,What is c, expression, loop and difference between loops, storage class difference between them (very important), structure and union difference between them and memory management of them, some question on files, they must ask to write a program 1. Fibonacci series (for me),2. Sine wave,3. Factorial of a no, 4.prime number, 5. Odd and even, 6. Matrix multiplication, division ,7. Palindrome.8. 2’s compliment .In second technical I got a question that I m not able to answer
if( )
printf(“hello”);
else
printf(“ world”);
do some modification such that out put is hello world. I m not sure , may this will be the answer
if(“hello world”);
printf(“hello”);
else
printf(“ world”);
just check it.

In C language concentrate more on storage classes (Auto, register, static, global).They are given one program on storage classes (better to read balaguruswamy).

6. Computer networks, control system, operating system etc—10%
what is OSI and TCPIP model, difference between them, draw the diagram for OSI model, explain detail about OSI model, some question from topology.
For me no question from Control system.As I am from EEE. At last he ask some question on sensor and signals like what is . tell different sensors , explain thermocouple, what is hall sensor I m not able to answer .
Operating system some basics question only, what is multitasking, scheduling, multithreading,(this is not for me. ECE people) .

Logical questions:
for the manholes why the caps are in circle shapes why not in square or rectangle in other shape.

In tech round if they ask for any questions ask some questions regarding company.In technical round only five people are selected. (Fortunately I am one of them).

HR ROUND (3/5)
In HR they simple check the communication skill.It will take 15 to 20 min.Don’t worry if u done well in technical then no problem. If u r average then u have to strong in communication.
i. tell something about u?
ii. Family background??
iii. Tell continuous 10 min on IT industries
.
iv. Who is the IT minister?
v. future planning?
vi. able to cope with Coimbatore environment?
vii. how much salary u expect from BOSCH.
vii. Do u have any question?

ROBERT BOSCH PAPER ON 24TH MAY

2009-12-31T21:29:00.000+05:30

1) given common mode gain =0.02 and differential gain =40 find CMRR in dB....
ans CMRR=20 log(differntial gain/common mode gain) = 66dB

2) which of the following amplifier has poorest efficiency.
options-A,B,C,D.
ans- A.

3) superposition principle does not apply to :
ans- nonlinear elements.

4) one circuit as shown below….. find V …. The values of resistors and all I don’t remember properly…..

5) one network shown below….. find the voltage V across 15 ohm resisto

ans- -105V hint….. apply KCL and OHMs LAW( this was the only question that I solved at that time for the first time and I was elated…..why not I have done something on my own…..)

6) A network has linear resistors across which a voltage of V is applied if the resistors are doubled the voltage across each resistor gets…….
Ans- no change……. Logic : if by changing resistors if u can change voltage drop across resistors u can create or destroy energy!

7) MVI B,64h
Loop : NOP
DCR B
JNZ Loop
The loop will get executed how may times…..
Ans - 64 times

8) one question on OSI model the layers of the model……

9) Ethernet uses _______ topology.
Ans - BUS topology.

10) how many address lines are used to access 4K of memory spaces…
ans- 12

11) which of the following is not an 8Bit processor…….
Options- 8085,Z-80,8392 etc.

12) one question in which input voltage is given and the gain and was asked to calculate the output voltage in dB

13) 4 questions on OPERATING SYSTEMS ( cant recollect)

14) which of the following logic family has got the lowest noise?
Options - ECL,TTL,DTL etc…

15) how many nand gates are required to realize an XOR gate
ans- 5

16) what is the function of the modem….
Options converts serial to parallel , converts analog signal to digital signals, other options can’t recollect……

17) which of the following gates can be used to realize any Boolean Expression
don’t remember the options…..

18) what is the 2’s complement of (-539) base 10
options were in HEX……

19) which of the following is used for fastest transmission of Data
options - twisted pair , coax cable, optic fiber(ans) etc…

20) which of the following is not a network
options – LAN, MAN, WAN, NAN ( ans)

15 questions on basic maths involving simple trigonometry, work- time , average speed, etc. RS aggarwal will do the job for this……

21) one interesting question in this section is
find the remainder of (8^9 + 1)/7
ans – 2

22) 10 questions on English that includes 2 antonyms, 2 synonyms, 2 analogies, $ one word grammatical questions.

Results of the written test were announced after a week…… I got written and was called for first technical interview in this round I was asked mainly on my project, C ( write code for string concatenation, strcpy etc( using pointers)… also I was asked to write a C code to set a particular bit… the bit number to be taken from the input ( use left shift operator) and about 8051 and basics of control systems… and basics of AD DA converter , a question on Malloc() and Calloc() , one puzzle to test ur problem solving capability……. For me they had asked what is the angle between hour and minute Hand when the time is 3:15 and also they asked me to interchange two variables without using the third variable and it went for more than an hour…..finally they gave me 7 out of 10 which means that u can join the company but requires 1 month of training to pickup. Any ways I cleared the first round

ROBERT BOSCH PATTERN - 16 JULY 2005

2009-12-31T21:28:00.000+05:30

Section 1:
This section had only technical question. They were mainly from Pointers in C, C++, Data Structures and some questions on OS, DBMS. This section also had lots of question on bit conversions (HEX to Binary, decimal) and one question on finding 2's compliment and such others. There were few questions on finding TRUE/FALSE statement which was bit confusing. Question to calculate square root, cube root of a number were also included.

Overall, this section was easy if you are good with your basics. The questions were mainly around the basic concepts only. Each question had four options (A, B, C, D) and we had to find the most appropriate answer. Each correct answer carries 2 marks and there is a negative marking of 1 for each wrong answer. So be careful while attempting these questions.

Section 2:
This had only general English questions. No book will help you to answer these questions; instead you should have good knowledge of grammar and different English works. There was a paragraph with some blanks in between. So we had to find the appropriate words to fill in, from the given options. Other question included comprehension, synonyms, proverbs type question and others.

This section needs some intelligent guessing as you feel that all of the given options are correct and sometime all are wrong. Each correct answer carries 1 mark and there is a negative mark of 0.5. Hope the information given by me here was useful to you.

All the very best, Regards

ROBERT BOSCH PAPER ON 3rd MARCH

2009-12-31T21:26:00.000+05:30

questions are..

1.Big Indian is an----------------(Ans ordering of bytes)
2.76f16+76f16
3.Question on 2s complement
4.inheritance is (forgot choices)
5.if (test())

{

i++

}

i will be incremented only when value of test=1

6.transfering of pages from memory to peripheral devices through buffer is ( spooling)
7.A basic question on dynamic binding
8.many questions on data structures but i dint attend those
9. i[0]=1;i[1]=2

int *pi;

i=pi;

printf("%d %d",*pi+1+(*pi+1)) (repeated)

10.array containingstring is( group of characters ending with /0)
11.Sorry friends i remember ony these but all were repeated just learn basics regarding apti it wAS only 5(avg,time n work,venn dia type)......1 logical reasonong question........1 data interpretation very very easy....english was a bit tough since there is no sectional cut off....no problrm........all the best friends.....

Technical interview will be in ther field of interest only......

BEST OF LUCK

ROBERT BOSCH PAPER - 05 DEC 2004

2009-12-31T21:25:00.000+05:30

Dear all

I have attended the Robert Bosch i applied though the Id That was Given in our group only. I have given the pattern and some questions that i remember.

Pattern:
There were 75 questions.. total Duration 1 hour 20 mins..
1-50 Questions are Technical and in that some 10-15 are Aptitute. it's very easy.
51-75 are Test of English Language..
From 1-50 for Correct answer 2 marks for wrong answer -1 mark..
From 51-75 for correct answer 1 mark for wrong answer -1/4 mark..
Technical questions and some aptitute:
_______________________________________
Totally there are some 35-40 Technical questions and 10-15 questions from Aptitute.. In Technical Questions they r asked about the following topics..
1. 7-10 questions from C (Like What's the output?)
2. DBMS
3. Data Structures some 6-7 questions
4. Networks
5. Operating Systems
6. Some 2-3 questions from Computer architecture
In data Structures they have given a tabular of 4 columns and they have asked like which is NOTCORRECT and which CORRECT.

I have given only the questions that i remember only.
1. The Question was Related to the Computer Architecture. They asked some thing about that..

1.harvard config
2.Van neuman..
2. base 16 is ans. hexadecimal

3. addition of 2 binary nos . and convert it to decimal..

4.structure is 1.can point any other structure except themselves..
2. cannot point any other structure except themselves
3.can point themselves and other structures.
4.none
5.wht is j after execution
int i[0]=1;i[1]=5;*p=I;
j= *p + 1+ (*p++);

6.For 1 parameter to b used as another r function wht to do??
1.declare as global 2.pass parameters 3.either 1 r 2. 4.niether 1 nor 2

7.Wht is a string??
1. array of characters without /0;
2. array of characters ending with /0;

8.DMA 1. data transfer
2.to increase the speed of peripherals
ans.2)

9. In Ethernet every system is called as.
1.client 2.server 3.node 4.system

10.In Ethernet the technique used for transfer data is..
1.CDMA 2.CSMA/CD 3.CSMA 4..

11. The Technique used for data transfer in TCP/IP config is..
1.Socket 2.Stream 3..

12. The Error Control and flow control is used in the following OSI Layer..
1.Transport layer 2.Data link layer 3.physical layer 4..

13.76F base 16 + 76F base 16 (Like this model) what's the Equivalent of this?

14. The technique which is used to model the database into tables..
1.Data model 2.DBMS 3.DML 4..

15. The question related to RDBMS ..answer is RDBMS

16. some binary number's r given like 100101+10023 the they asked about the Hex Equivalent?

17.. 49 base 10 + 49 base 10 then they asked about the Binary equivalent of this?

18. The longest side of the Equilateral triangle is..
1.it's hypotenuse 2.it's opposite side 3..
Then there r few more questions on the basics of mathematics..
Sorry i didn't remember the rest of the questions...please forgive me for that..

TEST OF ENGLISH LANGUAGE..
_____________________________
There were totally 25 questions..
I. Fill in the blanks type question they given a full paragraph..(5 questions)
II. 2 Questions like u have to fill the correct tense
1.Even the roses ________ weathered are .. some thing like that
III. some 5 Pro-verb questions ..
1. don't look __________a gifted horse .
a. Within the mouth b.at c. in the mounth d..

Questions like this was given..
IV.. Meaning of the Given Verb... some 5 questions was there..
1. go wrong----
a. wrong b. fail c. go.. d..
2. go forward---
a. obey the law b. follow the law c.go law d. some thing was given

some 5 questions like that..

Finally there are some questions in which is like fill in the blanks questions..
Like u have to choose the correct option from the given.. the options are like
1.at 2.on 3.within 4.for like this..

This is the pattern and questions that i remember..

BOSCH INTERVIEW ON 25th MAY 2006

2009-12-31T21:24:00.000+05:30

The interview was for people with 2 years of experience, but I with 1 year of experience, took a chance. I was instructed at the beginning of the interview that “Look, though you do not have 2 years of experience, yet I thought I shall spend some time with you”. I said “My pleasure”. Then he went ahead with the interview.

He asked me “What are macros?”
I said Macros are simply replacement texts.

#define clrscr() 500

Now wherever in the program, “clrscr()” appears, it will be replaced by 500.

Then I continued writing a program.

#include
#define clrscr() 500
void main()
{
printf(”%d”, clrscr());
}

I said this will print 500 on screen.

He said “Where are Macros used?”
I said they can be used to replace some magic numbers in the code. We may #define all magic numbers into some sensible macors.

He said “Where else?”
I said instead of simple one line functions. It will omit the overhead attached with the funtions while parallely giving the look and feel of functions.

“OK How much programming have you done?”
I said 30,000 lines of code.

“He said OH MY GOODNESS”
I said that was test code plus the actual code.

“So then what was it all about”
I said it was all about pointers and structures. Pointers is my favourite topic.

“Where do you rate yourself in C on a scale of 10?”
9 out of 10.

“How?”
I browse net specifically for “C riddles C puzzles and C questions” and after finding them, I also share them with my batchmates. So I have been doing that for quite some time. So that way I feel very confident.

“OK What do you understand from static keyword?”
It is used to tell the compiler to preserve the value of a variable between successive calls to a function. You may contrast it with volatile keyword which tells the compiler that its value is going to change and not to optimise it.

“Tell me a situation where pointers come in handy and that situation cannot be handled otherwise?”
No answer.

“Write a function with and without using pointers to swap two numbers?”
Yeah that is the answer to your earlier question..

I worte the two functions.

Then on paper he wrote two statements.
char *p = (char *) malloc(10 * sizeof(char));

char a[10];

He asked,”Tell me the difference between the two?”
I said in first case you are allocating memory dynamically. So you can free it when you are done with it. In second case it is meing allocated statically. So freeing it is not under your control. Both have got their own pros and cons. Using pointers makes your program tough to debug and using static memory allocation means you cannot free memory when you want to.

That was all he asked.
After that was the HR interview. He asked me all about my studies, my future plans to study, my native, Expected salary, and what not.

I am still waiting for offer letter.
Rahul
May 25, 2006

ROBERT BOSCH PAPER ON 17th FEBRUARY

2009-12-31T21:22:00.000+05:30

ROBERT BOSCH PAPER ON 24th JUNE, 2007

2009-12-31T21:21:00.000+05:30

SECTION 1(TECHNICAL)
1. A 16 bit monosample is used for digitization of voice. If 8 kHz is the sampling rate then the rate at which bit is transferred is
a) 128 b) 48 c) d)

1. There was a figure of JK flip flop in which ~q is connected to J input and K=1. If clock signal is successively applied 6 times what is output sequence (q=?)
d) 010101

2. The relationship b/w Gain & Bandwidth?
a) Independent of each other
b) Gain decreases as bandwidth decreases
c) Gain increases as bandwidth increases till some extent after which stability decreases

3. In “ON CHIP” decoding memory can be decoded to
a) 2^n b)2^n +1 c)2^n -1 d) some other choice Ans: a

4. Half of address 0Xffffffff is
a) 77777777 b) 80000000 c) 7FFFFFFF d) some other choice

5. For a 4 bit succe D/A with 16v o/p the resolution is

a)10v b)1v c)1.6v d)16v Ans : b

6. A 20kva transformer at full load conditions have the coreless & cu loss are 250w&300w,To get max efficiency what will be the total loss in tr in w
a) 550 b) 500 c) 600 d) Data insufficient Ans : b

7. There was a circuit consisting of AC voltage source and one inductance. Inductance value=0.2mH AC voltage =150 sin (1000t).what is the current flowing in the circuit?
a) b) c ) d)
8. Power gain of an amplifier having i/p gain of 20W and output gain of 20mW is
a) 60 b) 25 c) 10 d) 0

9. There was a RC circuit given with AC voltage source. Expression for capacitance was asked for charging condition. Choices were somewhat like this: a) some value multiplied by exp (-t/T)
ans --c i= (Vs/R)exp(-t/ T)

10.SCR s are connected in series to get

a) high current rating b) voltage regulation c) high v rating d)some other

Ans: c

11. Which one of the following is used for high speed power application?
a) BJT b) MOSFET c) IGBT d) TRIAC

12. One question related with SCR timing dia
13. SCR is used for
a) To achieve optimum (or maximum ...not sure) dv/dt
b) For high current ratings
c) To achieve high voltage
d) Some other choice

14. State in which o/p collector current of transistor remains constant in spite of increase in base current is
a) Q point b) Saturation c) Cut off

15. what is the resonant frequency of parrel RLC circuit of R= 4.7 komh L= 2 micro Henry and c=30pf.
a). 20.5 MHz
b). 2.65 KHz
c). 20.5 KHz
d). none

16.for the parallel circuit (one figure is given) Is= 10mA. R1= 2R, R2=3R, R3= 4R. R is artritary
a). 3.076mA
b). 3.76mA

17 . Two ques on control sys like the sys will be stable when the roots are
a) real & +ve b) real & -ve c) d)

20 The ques on Niquest plot
21 0ne ques on root locus
22. One more question related with ADC like voltage is 8 volts. freqency 2 Mega hz. what is the converssion rate
23. Question related with serial in parallel out shift register…What is output sequence?
Ans..... 1010
24. Given one RLC circuit in which values of R, L and C were given. What is the value of frequency f?
25. #define A 10+10
main ()
{
int a;
a=A*A;
printf (“%d”, a);
}
a) 100 b) 200 c) 120 d) 400

ans : c

26. main(){
int n=0;
while(n>32767)
pf(“%d”,n);
n++
}

ans ; indefinite loop here ihe loop will continue again & again ie after 32767 the go to -32768

27. if (fun ())
{
X++;
}
X gets incremented if and only if
a) fun () returns 0
b) fun () return 1
c) fun () return -1
d) return a value other than 0 ans ; d
28. In dynamic memory
a) Power dissipation is less than that of static memory
b) Clock is needed
c) Refreshing is required
d) All the above

29. Short, int and long integers have how many bytes?
a)2,2,4 b) Machine dependant c)2,4,8 d) Some other choice ans : b
30. A (n) is -----------filter combination of
a) Passive b) Active c) AMPLIFIER d) BOOSTER

31. according to ohms law the current density is proportional to
a) current b) di/ds c) applied emf d)

32 Two more ques on gravity ie body is projected then find the time of fall and some physics ques
33. Structure comparison is done
a). yes
b) no
c) compiler dependent
d)

34. The system in which communication occurs in both ways but not simultaneously in both ways is
a) Half simplex b) Simplex c) Half duplex d) duplex

35. main ()
{
int a=5, b=6;
int i=0;
i=a>b? a:b;
printf (“%d”, i);
}
a) 0 b) 1 c) 6 d)5

36. int fun (char c)
{
int i;
static int y ;}
a) c, i are stored in stack and y stored in data segment
b)c stored in stack and i,y are stored in data segment
c) c is stored in text segment, y in data and i in stack

37. how would you insert pre-written code into a current program?
a) #read
b) #get
c) #include
d) #pre ans: c

38.structure may contain
a) any other structure
b) any other structure expect themselves
c) any other structure except themselves and pointed to themselves
d) none of the above

39 One ques from pointer declaration

Section II
40 If a>4,b<-1 then which of the following is true
a)2a+b<0 b) 4a<3b c) a>4b d) some oth

41 In a class out of 150 studs participated in various games foot ball ,basket ball, cricket are 120, 130,135 respecly and 5 stud are not playing any one of then then find the least no of studs which are are playing all the games

a)110 b)100 c) 96 Ans a

42. 20 men can do a work in c/2 days & 30 women can do the same work in c/3 days then how many days will take to complete the work when 20 men &30wmen work together
a)5c/6 b)c/6 c) 6c/5

43. find the lateral surface of a cone with diameter of base 12feet . and the latent height is 24feet
a) 1320square f.
b) 260 square f.
c) 96 square f.
d) none ans : c

44. Ann is shorter than Jill and Jill is taller than Tom. Which of the following inferences are true?
a) Ann is taller than Tom b) c) d) Data insufficient ans : d

45. Sum of squares of two numbers is 404 and sum of two numbers is 22.Then product of two Numbers?
a) 20 b) 40 c) d) (Answer is 40. Two numbers are 20 and 2)

46). (10 | 7) would produce
a). 17
b). 3
c). 11
d). 15

ROBERT BOSCH PAPER ON 14th NOVEMBER

2009-12-31T21:20:00.001+05:30

SECTION 1:

1.Two force acting along the same line but in the opposite directions. they r called a) Concurrent b) coplanar c) collinear d) non concurrent (c)

2. Two basic systems of forces (c)

a) Coplanar, concurrent b)non concurrent, concurrent

c) Coplanar, non coplanar d)concurrent , non coplanar

3. Formula for power in HP

4. Forces lying in the same plane but never met (b)

a) concurrent b)parallel c) d)

5. Ger ratio for helical gears one of 30 tooth and the other 60 tooth

a)800:1 b)2:1

6. Four rectangle strips of size 10 X 1 inch r placed on a table as shown in
the figure.what is the area covered on the table. (a)

a)36 b)40 c)1600 d)80

7. If u traveled 45 min at 4 miles per hr and 30 min at 3 miles per hr. what is the

distance covered by u in 1 hr and 15 min. (b)

a) 8 miles b)6 miles c)7miles d)10

8. A ball of weight 10N is dropped freely from rest at a height of 4 mts.what is its

total mechanical energy after it has traveled 1 meter.

9. Two bodies of same mass r traveling at a velocity of 2 m/sec in opposite direction.

After collision one travelled with 1.8 m/sec what abt the other..

a) equal to 1.8 m/sec b)more than 1.8

c) Less then 1.8 d) can not say

10. If X-component of a force is –Ve and Y-component is +Ve, in which plane the

Resultant would be (a) check it

a) First quadrant b) second c) third d) fourth

11. A body is said 2 b in equilibrium (c)

a) if it is in rest b)if it is I uniform velocity

c) if the sum of the forces acting on it is zero d)none

12. X15-X13+ X11-X9+X7-X5+X3-X=7 than value of X16 ?

13. a chain of leanth 1 mt and mass 1 kg is placed on a table.What leanth of the chain

must be hanged freely from the table so that is will drag the remaining chein

freely (d)

a).55 b).66 c) 0.5 d).75

14. Main purpose of governer (d)

a) to live happy life b) to increase engine speed

c) to decrease engine speed d)to run engine smoothly

15.A steel beam of mass m and length l is placed on a surface.a couple u is applied at

the ends to rotate the beam at 90 0 . what is the work done.

16.A & B r two weight lifters .A can lift 100 pounds 10 times a minute, where as b

lifts 100 pounds 10 times a sec.who is doing more power and work. (c)

a)A&B b)A omly c) B only d) none.

17.A ship is 77 Km away from the shore.It had a leak due to ehich water is flowing

into the ship @ 1 ton /min .A pump is used to pump the water out of the ship @

30 ton/ hr The ship sinks if the water Qty reaches 210 ton.what should b the

speed of the ship to reach the shore b4 it sinks. (11)

a)7 mph b)11mph c)15 d)20

18. Very simple ..que..find the hypotenuse …

1) Sides 8, 11, find hyp 2) a similar one..to find side..

19. Which is the largest side in a right angled triangle

a) hypotenuse b)edgescent side c)opposite side

20. The angle of depression from the top a building to another building is 120 .the

distance b/n the two is 50 mts.What is the difference in the height of buildings. (a)

a) 10.6 b)20.3 c)10.4 d)48.9

21. Simplyfy [ 3v--- 27 - v—(27/4)]2

22. Belt drives r mainly used for (d)

a)to reverse the direction of power in belts. b)to transmit it power.

c)to …..d) all

23. BS thread profile has

a)141/2 0 b)450 c)250 d) 90 0

24. The main purpose of bearings (b)

a)to reduse noise. b)to reduce friction c) d)

25.When compared to spur gears bevel gears r more .. (a)

a) noisy b)smooth

26. The angle b/n bore axis of gear and the tangent to the teeth profile is called (b)

a)P.A b) helix c)

27. The circle touching the outer edge of the teeth is called (c)

a) Base circle..b) pitch crcl c) addendum crcl d)….

28. The area of the piston A is 10 cm2 and that of B is 4 cm 2 .If 10n load is applied

on the piston A what is the force on the piston B. (c)

a)100N b)40N c)10N d)none

29. The force exerted by a spring depends on (d)

a)Material of the spring b)Design of the spring c)Length of the spring d)all

30. If a body is isolate from it’s mating parts and all forces acting on it r shown then it

is called. (c)

a)BM diagram b) SF diagram c)free body diagram d)none

31. A coin and a ring of same mass and diameter r allowed to roll smoothly from the

top of a inclined plane. which one reaches the bottom first.

a)coin b) ring c)both reaches same time d) none

32. A ball is thrown horizontally .After hitting the ground it again raises with some

velocity.this is bcoz.. (b)

a)ball is stil having velocity b)it gains momentum from the ground. c) d)

33. a problem on balancing masses..

34. a mass is rotating..with rad /sec..simple problem …on basics..

35 given 0 (c)

a)c+d/ a+b b)c+a/b+d c)a+b/ c+d d)….

36 to 50 sorry..i forgot….

SECTION 2:

51

TO } Passage with blanks..

55

56

to } antother passage with blanks..

60

61 to 65 proverbs..

1. don’t look a gifted horse..

a)at b)in life…. c) in dreams d)

2. more hands……….

a) make more work easy b)more light work c) d)

3. people living in glass houses……

a)may not throw stones b)didn’t throw stones c) cannot throw stones d)should not

throw stones

4.

5.

65 to 75 appropriate meaning

1. Go for …………. (a)

a) try b)get it c)quit d)

2. Go overboard…………….. (a)

a)exaggerated b)exagerate c) exagratred d) none.

3.Go bad ………..

a)….. b)bad c)not good d) none

4. go wrong ……….. (b)

a)fail b) mistake c) d)

5.meaning of batten ….. (a)

a) noun :a long wood of metal strip

b)……….

c)………d)…….

ROBERT BOSCH -Associate Software Engineer

2009-12-31T21:17:00.000+05:30

SECTION 1(TECHNICAL)

1. There was a figure of JK flip flop in which ~q is connected to J input and K=1. If clock signal is successively applied 6 times what is output sequence (q=?)

a) 101010 b) c) and d) some other 6 digit sequences

2. Frequency response of a filter is

a) Range of frequencies at which amplification of signal is employed.

b) Output voltage versus frequency (plot)

c) Filter which suppresses particular frequency

d) Some other choice

3. Gain and bandwidth of an op amp is

a) Independent of each other

b) Gain decreases as bandwidth decreases

c) Gain increases as bandwidth increases till some extent after which stability decreases

d) Some other choice

.

4. There was a figure of 4:1 MUX in which A and B are select lines. Inputs S0 and S1 are connected together and labeled as C where as S2 and S3 are connected together and labeled as D.Then which of the following is true?

a) Y=B+C

b) Y=A+B

c) Y=A+B

d) Y=C+D (Where Y is the output)

5. In step up transformer (or Step down… not sure) transformation ratio is 1:5. If the impedance of secondary winding is 16 ohm then what is the impedance of primary winding?

a) 80 b) 3.2 c) and d) some other choice

6. There was a circuit consisting of AC voltage source and one inductance. Inductance value=0.2mH (or 0.2uH or 0.2H not sure).AC voltage =150 sin (1000t).what is the current flowing in the circuit?

a) i= 7.5 sin (1000t)

b) i= -7.5 sin (1000t)

c) i= 7.5 cos (1000t)

d) i= -7.5 cos (1000t)

7. Power gain of an amplifier having i/p gain of 20W and output gain of 20mW is

a) 60 b) 25 c) 10 d) some other choice

8. There was a RC circuit given with AC voltage source.Expresssion for capacitance was asked for charging condition.

Choices were somewhat like this: a) some value multiplied by exp (-t/T)

c) some value multiplied by 1- exp (-t/T)

9. 2’s complement of -17

10. Instrumentation amplifier is used for--------------------?

11. In “ON CHIP” decoding memory can be decoded to

a) 2^n b)2^n +1 c)2^n -1 d) some other choice

12. Half of address 0Xffffffff is

a) 77777777 b) 80000000 c) 7FFFFFFF d) some other choice

13. Which one of the following is used for high speed power application?

a) BJT b) MOSFET c) IGBT d) TRIAC

14. One question related with SCR rotation angle

15. SCR is used for

a) To achieve optimum (or maximum ...not sure) dv/dt

b) For high current ratings

c) To achieve high voltage

d) Some other choice

16. State in which o/p collector current of transistor remains constant in spite of increase in base current is

a) Q point b) Saturation c) Cut off

17. 16 bit monosample is used for digitization of voice. If 8 kHz is the sampling rate then the rate at

which bit is transferred is

a) 128 b) 48 c) d)

18. To use variable as recursive, variable should be used as

a) Static b) Global c) Global static d) Automatic

19. Question related with successive approximation ADC

20. Question related with resolution of DAC

21. main ()

{

int a=0x1234;

a=a<<12;

a=a>>12;

printf (“%x”, a);

}

What is the output?

a)1000 b) 2000 c) d) None of these

22. What does (*fun () []) (int) indicate?

a) pointer to an array of 4 integer functions

b) c) d)

23. #define A 10+10

main ()

{

int a;

a=A*A;

printf (“%d”, a);

}

a) 100 b) 200 c) 110 d) 400

24. One more question related with ADC

25. Question related with serial in parallel out shift register…What is output sequence?

(This was question appeared in GATE exam…So go through GATE BOOK)

Answer for this might be 1010

26. Given one RLC circuit in which values of R, L and C were given. What is the value of frequency f?

27. if (fun ())

{

X++;

}

X gets incremented if and only if

a) Value of fun () is 1

b) Value of fun () is any value other than 0

c) Value of fun () is greater than 1

d) Some other choice

28. In dynamic memory

a) Power dissipation is less than that of static memory

b) Clock is needed

c) Refreshing is required

d) All the above

29. Short, int and long integers have how many bytes?

a)2,2,4 b)Machine dependant c)2,4,8 d)Some other choice

30. A (n) is -----------filter combination of

a) Active b) Passive c) d)

31. Mobility of electron is

a) Increases as temperature increases

b) Decreases as temp decreases

c) Independent of conductivity

d) Some other choice

32. Structure comparison is done

a) Automatically

b)structcmp is used

c) No automatic function exists

d) Some other choice.

33. The system in which communication occurs in both ways but not simultaneously in both ways is

a) Half simplex b) Simplex c) Half duplex d) duplex

34. main ()

{

int a=5, b=6;

int i=0;

i=a>b? 1:0;

printf (“%d”, i);

}

a) 0 b) 1 c) d)

(Note: The output is 0 in this case. If ‘a’ is greater than ‘b’ (For ex: a=5 and b=4, then o/p is 1)

35. int fun (char c)

{

int i;

static int y ;}

a) c, i are stored in stack and y stored in data segment

b)c stored in stack and i,y are stored in data segment

c) c is stored in text segment, y in data and i in stack

36.main ()

{

int *p;

short int i;

p= (char *) malloc (i*10); (code was showing error here)

p+=10;

printf (“%d”, p);

}

Value of p?

a) b) c) d)

(Note: When I checked this code at home it was giving error as “Cannot convert char * to int *

I have modified coding as p= (int *) malloc (i *10), Then I got o/p “p=1496”)

37. main ()

{

int *p,i[2]={1,2,3};(Note: Some values were given )

p=i;

printf (“%d %d %d”, i [0],*p,*p+1);

}

a) b) c) d)

SECTION 2(APTITUDE)

1. Given 0< P>

a) (a+b)/(c+d) b) (b+c)/ (a+d) c) (c+d)/ (a+b) d) (a+c)/ (b+d)

2. A and B starts moving from points X and Y simultaneously at a speed of 5kmph and 7kmph to a

destination point which is of 22km from points X and Y. B reaches Y earlier than A and

immediately turns back and met Z. Find the distance XZ.

3. Area of lateral surface of a cone given height and radius.

4. Ann is shorter than Jill and Jill is taller than Tom. Which of the following inferences are true?

a) Ann is taller than Tom b) c) d) Data insufficient

5. A and B starts from same point at opposite direction. They will move 6km and take 8km left. How

Far is A and B from each other?

6.6440(Not sure of this number) soldiers are to be arranged in the shape of square. If 40 soldiers were kept out then the number of soldiers making each straight line is? (Question was like this somewhat)

7. If x>5 and y<-1 then which of the following is true?

a) b) c) d) (Choices were some mathematical expression)

8. Sum of squares of two numbers is 404 and sum of two numbers is 22.Then product of two

Numbers?

a) 20 b) 40 c) d) (Answer is 40. Two numbers are 20 and 2)

9. In an examination 4 marks are assigned for correct answer and 1 mark is deducted for wrong

answer. However one student attempted all 60 questions and scored 130.Number of questions he

attempted correct is?

a) 35 b) 42 c) d)

10. Each ruby is of 0.3 kg and diamond is of 0.4 kg.Ruby costs 400 crores and diamond costs 500

crores.Ruby and diamonds have to be put into a bag. Bag cannot contain more than 12 kg.Which of the following gives maximum profit (or in terms of wealth) (In crores)

a) b) c) d)

Finally there were 10 (or 15) questions of English section (Which was included in aptitude section itself).Pattern of English section was like

1. Correcting the sentences

2. Given one passage. We are supposed to answer the questions given on the passage

3. Nearest meaning of highlighted word...

ROBERT BOSCH PAPER ON 24th FEBRUARY 2008

2009-12-31T21:15:00.000+05:30

Questions ; 60

Duration : 1 h

This paper is for EEE,ECE,EIE

Friendas just reffer the previous question papers that is more than enough..... in this written test negative marks is there correct answer carries 2 mrks. and wrong answer carry -1 mraks so be carefully answer.

SECTION 1(TECHNICAL)
1. A 16 bit monosample is used for digitization of voice. If 8 kHz is the sampling rate then the rate at which bit is transferred is
a) 128 b) 48 c) d)

1. There was a figure of JK flip flop in which ~q is connected to J input and K=1. If clock signal is successively applied 6 times what is output sequence (q=?)
d) 010101

2. The relationship b/w Gain & Bandwidth?
a) Independent of each other
b) Gain decreases as bandwidth decreases
c) Gain increases as bandwidth increases till some extent after which stability decreases

3. In “ON CHIP” decoding memory can be decoded to
a) 2^n b)2^n +1 c)2^n -1 d) some other choice Ans: a

4. Half of address 0Xffffffff is
a) 77777777 b) 80000000 c) 7FFFFFFF d) some other choice

5. For a 4 bit succe D/A with 16v o/p the resolution is

a)10v b)1v c)1.6v d)16v Ans : b

6. A 20kva transformer at full load conditions have the coreless & cu loss are 250w&300w,To get max efficiency what will be the total loss in tr in w
a) 550 b) 500 c) 600 d) Data insufficient Ans : b

7. There was a circuit consisting of AC voltage source and one inductance. Inductance value=0.2mH AC voltage =150 sin (1000t).what is the current flowing in the circuit?
a) b) c ) d)

8. Power gain of an amplifier having i/p gain of 20W and output gain of 20mW is
a) 60 b) 25 c) 10 d) 0

9. There was a RC circuit given with AC voltage source. Expression for capacitance was asked for charging condition. Choices were somewhat like this: a) some value multiplied by exp (-t/T)
ans --c i= (Vs/R)exp(-t/ T)

10.SCR s are connected in series to get
a) high current rating b) voltage regulation c) high v rating d)some other

Ans: c

11. Which one of the following is used for high speed power application?
a) BJT b) MOSFET c) IGBT d) TRIAC

12. One question related with SCR timing dia

13. SCR is used for
a) To achieve optimum (or maximum ...not sure) dv/dt
b) For high current ratings
c) To achieve high voltage
d) Some other choice

14. State in which o/p collector current of transistor remains constant in spite of increase in base current is
a) Q point b) Saturation c) Cut off

15. what is the resonant frequency of parrel RLC circuit of R= 4.7 komh L= 2 micro Henry and c=30pf.
a). 20.5 MHz
b). 2.65 KHz
c). 20.5 KHz
d). none

16.for the parallel circuit (one figure is given) Is= 10mA. R1= 2R, R2=3R, R3= 4R. R is artritary
a). 3.076mA
b). 3.76mA

17 . Two ques on control sys like the sys will be stable when the roots are
a) real & +ve b) real & -ve c) d)

20 The ques on Niquest plot

21 0ne ques on root locus

22. One more question related with ADC like voltage is 8 volts. freqency 2 Mega hz. what is the converssion rate

23. Question related with serial in parallel out shift register…What is output sequence?
Ans..... 1010

24. Given one RLC circuit in which values of R, L and C were given. What is the value of frequency f?

25. #define A 10+10
main ()
{
int a;
a=A*A;
printf (“%d”, a);
}
a) 100 b) 200 c) 120 d) 400

ans : c

26. main(){
int n=0;
while(n>32767)
pf(“%d”,n);
n++
}

ans ; indefinite loop here ihe loop will continue again & again ie after 32767 the go to -32768

27. if (fun ())
{
X++;
}
X gets incremented if and only if
a) fun () returns 0
b) fun () return 1
c) fun () return -1
d) return a value other than 0 ans ; d

28. In dynamic memory
a) Power dissipation is less than that of static memory
b) Clock is needed
c) Refreshing is required
d) All the above

29. Short, int and long integers have how many bytes?
a)2,2,4 b) Machine dependant c)2,4,8 d) Some other choice ans : b

30. A (n) is -----------filter combination of
a) Passive b) Active c) AMPLIFIER d) BOOSTER

31. according to ohms law the current density is proportional to
a) current b) di/ds c) applied emf d)

32 Two more ques on gravity ie body is projected then find the time of fall and some physics ques

33. Structure comparison is done
a). yes
b) no
c) compiler dependent
d)

34. The system in which communication occurs in both ways but not simultaneously in both ways is
a) Half simplex b) Simplex c) Half duplex d) duplex

35. main ()
{
int a=5, b=6;
int i=0;
i=a>b? a:b;
printf (“%d”, i);
}
a) 0 b) 1 c) 6 d)5

36. int fun (char c)
{
int i;
static int y ;}
a) c, i are stored in stack and y stored in data segment
b)c stored in stack and i,y are stored in data segment
c) c is stored in text segment, y in data and i in stack

37. how would you insert pre-written code into a current program?
a) #read
b) #get
c) #include
d) #pre ans: c

38.structure may contain
a) any other structure
b) any other structure expect themselves
c) any other structure except themselves and pointed to themselves
d) none of the above

39 One ques from pointer declaration

Section II
40 If a>4,b<-1 then which of the following is true
a)2a+b<0 b) 4a<3b c) a>4b d) some oth

41 In a class out of 150 studs participated in various games foot ball ,basket ball, cricket are 120, 130,135 respecly and 5 stud are not playing any one of then then find the least no of studs which are are playing all the games

a)110 b)100 c) 96 Ans a

42. 20 men can do a work in c/2 days & 30 women can do the same work in c/3 days then how many days will take to complete the work when 20 men &30wmen work together
a)5c/6 b)c/6 c) 6c/5

43. find the lateral surface of a cone with diameter of base 12feet . and the latent height is 24feet
a) 1320square f.
b) 260 square f.
c) 96 square f.
d) none
ans : c

44. Ann is shorter than Jill and Jill is taller than Tom. Which of the following inferences are true
a) Ann is taller than Tom b) c) d) Data insufficient ans : d

45. Sum of squares of two numbers is 404 and sum of two numbers is 22.Then product of two Numbers?
a) 20 b) 40 c) d) (Answer is 40. Two numbers are 20 and 2)

46). (10 | 7) would produce
a). 17
b). 3
c). 11
d). 15

Robert Bosch Off-Campus Recruitment Pattern

2009-12-31T21:13:00.000+05:30

Bosch patter was rarely available in this group, so I thought of sending this mail. This patter below is for computer science (CS) students only.

There was another set of paper for MECH branch students.

Section 1:
This section had only technical question. They were mainly from
Pointers in C, C++, Data Structures and some questions on OS, DBMS.
This section also had lots of question on bit conversions (HEX to
Binary, decimal) and one question on finding 2's compliment and such
others. There were few questions on finding TRUE/FALSE statement which
was bit confusing. Question to calculate square root, cube root of a
number were also included.

Overall, this section was easy if you are good with your basics. The
questions were mainly around the basic concepts only.

Each question had four options (A, B, C, D) and we had to find the
most appropriate answer. Each correct answer carries 2 marks and there
is a negative marking of 1 for each wrong answer. So be careful while
attempting these questions.

Section 2:
This had only general English questions. No book will help you to
answer these questions; instead you should have good knowledge of
grammar and different English works.

There was a paragraph with some blanks in between. So we had to find
the appropriate words to fill in, from the given options. Other
question included comprehension, synonyms, proverbs type question and
others.

This section needs some intelligent guessing as you feel that all of
the given options are correct and sometime all are wrong.

Each correct answer carries 1 mark and there is a negative mark of 0.5.

Hope the information given by me here was useful to you.

Interview Pattern - Robert-Bosch

Here is the pattern
----------------------------
Duration of Test: 80 min
Number of questions: 75
Number of sections: 2
No of questions in section 1: 50 (Technical)
No of questions in section 2: 25 (General English)
* Answers to be marked on the OMR Sheet.

Robert Bosch Company Profile

2009-12-31T21:10:00.000+05:30

The Bosch Group is a leading global manufacturer of automotive and industrial technology, consumer goods, and building technology. It achieved 42 billion euros in sales and has 250,000 associates world wide (as of Jan 2005). It is the world largest automotive supplier and a core AUTOSAR member.

Robert Bosch India Limited is a 100% owned subsidiary of Robert Bosch GmbH, with over 3000 associates. We provide engineering and non-engineering services to the Bosch World such as Electronic Control Unit Development for automotive, industrial, consumer goods and building technology, as well as IT Services, Process Consulting, Mechanical Engineering, IC Design, Shared Services Accounting, and Translation/Documentation. ISO 9001:2000 certified and assessed at CMM Level 4.Three development locations in Bangalore and Coimbatore.

Robert Bosch India plans to set up its own infrastructure on this land to realise the growing opportunities in the IT and ITES space. Construction will start in the last quarter of 2007 and the first phase is targeted to be inaugurated by early 2009. With this, Robert Bosch India has deepened its relationship with Coimbatore which was started in May 2006 when the companies’ second Development Center was inaugurated at Saravanampatti.

While the strength of Robert Bosch India at Coimbatore has grown to more than 480 associates, this new infrastructure will pave way to meet growing demand from 2009 onwards. The new infrastructure will provide employment opportunities
to the people in the region and opportunity to work on the innovative technologies of Bosch in the areas of Software
and Hardware Development, Mechanical Engineering and Business Services-Technologies based on which Bosch applied for more than 3000 patents in 2006.

Financial Highlights
Fiscal Year End: December
Revenue (2007): 68178.40 M
Revenue Growth (1 yr): 18.30%
Employees (2007): 271,000

Employee Growth (1 yr):

5.00%

Headquarters:
Greater Detroit Area
Industry: Automotive
Type: Privately Held
Company Size: 10,001 or more employees

Core businesses

Automotive Technology

Industrial Technology
Consumer goods and power tools

Security Systems

Contact
Robert Bosch Engineering and Business Solutions Limited
123, Industrial Layout
Hosur Road, Koramangala
Bangalore - 560 095
India.

VHDL Not Port

2009-12-03T13:23:00.000+05:30

library ieee;
use ieee.std_logic_1164.all;

entity NotPort is
port(
A: in std_logic;
B: out std_logic
);
end NotPort;

architecture DataFlow of NotPort is

begin

B <= not A;

end DataFlow;

VHDL Or Port

2009-12-03T13:21:00.002+05:30

library IEEE;
use IEEE.STD_LOGIC_1164.all;

entity OrPort is
port( x: in std_logic;
y: in std_logic;
F: out std_logic
);

architecture DataFlow of OrPort is
begin

F<= x or y;

end DataFlow;

architecture DataFlow2 of OrPort is
begin

process(x,y)
begin
if((x='1') or (y='1')) then
F <='1';
else
F<='0';
end if;
end process;

end DataFlow2;

VHDL code for Correct Clock (For instance UART Clock)

2009-12-03T13:21:00.001+05:30

library IEEE;
use IEEE.std_logic_1164.all;

entity UART_CLK is
port(
clk: in std_logic;
rst: in std_logic;
CLK384: out std_logic
);
end UART_CLK;

architecture behav of UART_CLK is
signal clkDiv: std_logic_vector(7 downto 0);
signal CLKt: std_ulogic;
constant baudDivide: std_logic_vector (7 downto 0) :=X"82";

begin

process(clk,rst)
begin
if (rst='1') then
clkDiv <= baudDivide;
CLKt<='0';
elseif (clk='1' and clk'event) then
if (clkDiv=X"00") then
clkDiv<=baudDivide;
CLKt <=not CLKt;
else
clkDiv <=clkDiv -1;
end if;
end if
end process;
CLK384 <=CLKt;
end behav;

VHDL Code for a 2:1 MUX

2009-12-03T13:17:00.000+05:30

library IEEE;
use ieee.std_logic_1164.all;

ENTITY Mux2x1 IS {
PORT (a0, a1, sel: IN BIT; z: OUT BIT); }
END Mux2x1;

ARCHITECTURE conditional OF Mux2x1 IS
BEGIN
z <= a0 WHEN sel = ‘0’ ELSE a1;
END conditional;

VHDL Code for a 4:1 MUX (Using Entity)

2009-12-03T13:15:00.000+05:30

library IEEE;
use ieee.std_logic_1164.all;

ENTITY Mux4x1 IS PORT (
a : IN BIT_VECTOR(3 DOWNTO 0);
sel: IN BIT_VECTOR(0 TO 1) ;
z: OUT BIT); }
END Mux4x1;

ARCHITECTURE mux2x1_based OF Mux4x1 IS
SIGNAL im0, im1 : BIT;
BEGIN
m1: ENTITY WORK.Mux2x1(conditional) PORT MAP (a(0), a(1), sel(0), im0);
m2: ENTITY WORK.Mux2x1(conditional) PORT MAP (a(2), a(3), sel(0), im1);
m3: ENTITY WORK.Mux2x1(selected) PORT MAP (im0, im1, sel(1), z);
END mux2x1_based;

VHDL Code for 3 INPUT AND PORT

2009-12-03T13:11:00.001+05:30

library ieee;
use ieee.std_logic_1164.all;

entity TryOut1 is
port(
X: in std_logic;
Y: in std_logic;
Z: in std_logic;
A: out std_logic);
end TryOut1;

architecture behaviour of TryOut1 is
begin
process(X,Y,Z)
begin
A<= X and Y and Z;
end process;
end behaviour;

VHDL Code for a 1 to 4 Demultiplexer

2009-12-03T13:10:00.000+05:30

library ieee;
use ieee.std_logic_1164.all;

entity TryDeMUX is
port( X: in std_logic;
sel: in std_logic_vector (1 downto 0);
A: out std_logic;
B: out std_logic;
C: out std_logic;
D: out std_logic);
end TryDeMUX;

architecture behaviour of TryDeMUX is
begin
process(sel,X)
begin
case sel is
when "00"=>
A <=X;
B <= '0';
C <= '0';
D <= '0';
when "01" =>
B <=X;
A <='0';
C <='0';
D <='0';
when "10" =>
C <=X;
A <='0';
B <='0';
D <='0';
when others =>
D <=X;
A <='0';
B <= '0';
C <='0';
end case;
end process;
end behaviour;

VHDL CODE - ADDER with carry

2009-12-03T13:09:00.001+05:30

library ieee;
use ieee.std_logic_1164.all;
use ieee.std_logic_arith.all;
use ieee.std_logic_unsigned.all;

entity ADD is
generic(n: natural :=2);
port(
A: in std_logic_vector(n-1 downto 0);
B: in std_logic_vector(n-1 downto 0);
carry: out std_logic;
sum: out std_logic_vector(n-1 downto 0)
);
end ADD;

architecture behav of ADD is

signal res: std_logic_vector(n downto 0);
begin
res <= ('0' & A)+('0' & B);
sum <= res(n-1 downto 0);
carry <= res(n);
end behav;

VIDEO OF LIFE

2009-12-03T09:48:00.001+05:30

MAN WHO R MAKING MISTAKES MUST WATCH

USE youtube downloader software to download this video.........

http://www.youtube.com/watch?v=FT1CKiIVRGg