Suri & Co Interview Questions, Process, and Tips

5 is the answer….
Remaining all are the multiples of 3 expect 5

Yes answer is 18 days

Let the number be 10Y+X
(10Y+X)/2 = (10X+Y)/3+6
(10Y+X)(1/2-1/3) = 6
10Y+X = 36
Therefore, sum of the digits = 3+6 = 9

Ans: 9

none of these
there will be 20 men

Let the age of the man be x
Then age of his son becomes (x−24)

2 years later from now,

Age of man will be = x+2
and age of his son will be =(x−24+2)=x−22

According to question,

2(x−22)=x+2

i.e., 2x−44=x+2

i.e., x=46

Therefore ,
Present age of man=46 years
And present age of his son =46−24=22 years

North

100

1st gets 15 RS less than average
4x+60=s
x=(s-60)/4 = s/4-15
s/4 is average
1st gets 15 RS less than average

I am Srilaxmi. I born and bought-up in WARANGAL. My father Agriculture cum politician and my Mom homemaker. I blessed with three brothers and one sister. My elder brother was married and working as a TA in MPDO office at Warangal. My first younger brother was also married and working with Sushee Infra Pvt Ltd as a Asset Manager at Hyderabad. Younger brother and sister are into their studies. Coming to me I completed my MBA from ICFAI university in 2008 and worked as a Audit assistant in accounting firm for two years. I had actively participated in bank audits i.e. is in CBI (Concurrent audit) and SBI(Statutory audit) during my services. Later I pursued CS Executive and attempted for five times then I give up to clear my M.Com. At present I was studying postal studies of CS Executive programme. My hobbies are reading books and listening to music.

1261

X = 1/2 * b*h
base incresed by 4 times & height
is devided by 2
X = 1/2 * 4b * h/2
X = h/4 * 4b
X = hb

CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….

71

20&130

0