Facts computer software house Interview Questions, Process, and Tips

perimeter of rectangle = 2(l+b)
given l abd b as 18 cm and 26 cm
therefore perimeter of rectangle = 2(44)
perimeter of circle = 2*pi*r
given 2*pi*r=2(l+B)
i.e, 2*(22/7)*r= 2(44)
(22/7)*r=44
r=14 cms
area of the circle = pi*r*r
=(22/7)*14*14
=616 sqcm is the answer

120 (5×4!)

Let the principal be P and rate of interest be R%.
Therefore Required ratio

=P×R×6/100/P×R×9/100
=6PR/9PR
=6/9
=2:3

6 cuts

sample space=36
prob of one of the dice getting face 6
i.e n={(1,6)(2,6)(3,6)(4,6)(5,6)(6,1)(6,2)(6,3)(6,4)
(6,5)(6,6)}=11
p=n/s
p=11/36
ans:11/36

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

here the 30% of 100 is 30
and 10% of the 30 = 3
so 20% = 6%
so ans = 6%

Chambal

Since, there are 10 points on the circle and to draw a chord we need to connect any two points on the circle to make it a straight line, which implies that the number of chords = No of lines connecting any two points out of the 10 points

= 10C2 = 10*9/2 = 45 chords.

boys= x
girls=2x
boys+girls= 60
x+2x=60
x=60/3
x=20
kamal=17th rank
9 girls are ahead of kamal
17-9=8
20-8=12 boys are after kamal’s rank

under pressure a person more effort to his work

72

I belive it is 3:45 hrmins

Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.