Wipro Infotech Selection procedure comprises the following sequence of tests:
1. Written Test
2. Tech-Interview
3. HR Round.
1. Written Test
2. Tech-Interview
3. HR Round.
Written Test Contains Three Sections
- Verbal
- Quant
- Technical
Verbal
- It was little bit
tough. - It Contains a passage and 5 Questions Based on that Passage.
- Some synonyms and antonyms
are also there.
Quant
It was easy test out of 20,15 are easily attempting
Some Questions
1. Diff. Between ages of son and father is 30 and their squares is 1560 find their ages.
2. What is probability of choosing a leap year with probability of 7 Fridays.
3. 1 question is on relations
N is brother of X and A . Y is mother of N. Z is wife of Y.
which of the following is not def. true
ans is X is brother of N.
4. 0 is coded as *,
which of the following is not
ans is X is brother of N.
4. 0 is coded as *,
- 1 is coded
as #, numbers more then 1 is coded by using of *
and # like - 2 is coded as #*
- 3 is coded as ##
- 4 is coded as #**
find the lcm of 10 15 30
5. p+q means p is brother of q and p*q means p is mother of q and p/q means p is father of q and p%q means p is sister of q
now make a relation that p is mother in law of q
now make a relation that p
One question on trains,clock, share.
Technical
It was also very easy.
1. One Questions on order of OSI Layers
2. Apache Company Provides____ _____
3. int count =10,sum =0,*temp;
temp=&count;
*temp=20;
temp=∑
*temp=20;
printf("%D%d%d",sum,count,* temp);
*temp=20;
temp=∑
*temp=20;
printf("%D%d%d",sum,count,*
4.
char letter ='a';
if(letter='a')
{
}
else if(letter='b')
{
}
else if(letter='c')
{}
else
{
}
convert it into switch.
char letter ='a';
if(letter='a')
{
}
else if(letter='b')
{
}
else if(letter='c')
{}
else
{
}
convert it into switch.
5. struct a
{
int a;
int b;
}
void main()
{
struct a b={110,10};
print(b);
}
print(struct b)
{
printf("%d %d",b.a,b.b);
}
{
int a;
int b;
}
void main()
{
struct a b={110,10};
print(b);
}
print(struct b)
{
printf("%d %d",b.a,b.b);
}
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