4 thieves rob a bakery of the breadone after the other.each
thief takes half of what is present ,& half a bread...if
at the end 3 bread remains,what is the no of bread that
was present initially?

Mr.Ramki,
Assume,total no of breads,x=48
first person stole,x/2 = 48/2 = 24.
second person stole,x/4 =(x/2)/2 = 24/12 =12.
third person stole, x/8 = (x/4)/2 = 12/2 =6.
fourth person stole, x/16 = (x/8)/2 = 6/2 =3.
Total breads stolen= 24+12+6+3 =45.
Still at the end there are 3 breads remaining what has
mentioned in the given question.
Hence no of breads initially are,48.

If we assume no of breads initially,x=63.
first person=31
second person=15
third person=7
fourth person=3
Breads stolen=31+15+7+3=56
Remaining breads=63-56=7
But in the question at the end there must be 3 breads
remaining,but in your logic 7 breads remaining,so 63 is
wrong one.

Well i guess 63 is right answer coz @Surya u have calculated
and just made half the breads evertime but u havent
considered the half the totle and also "half bread" !

at the end,total no of breads=3.
Let the no of breads in the bakery before 4th thief robbed
the bakery and after 3rd thief robbed=x.
Then x-(x/2+1/2)=3
=> x=7
similarly,no of breads in the bakery before 3rd thief ,2nd
thief and 1st thief robbed are 15,31 and 63 respectively.
So, total no of breads initially=No of breads before 1st
thief robbed=63.

You have 8 balls. One of them is defective and weighs less
than others. You have a balance to measure balls against
each other. In 2 weighings how do you find the defective
one?

A cow is tethered in the middle of a field with a 14 feet
long rope. If the cow grazes 100 sq feet per day
approximately. What time will be taken by the cow to graze
the grazeable area of the field?
(a) 2 days
(b) 18 days
(c) 24 days
(d) 6 days
(e) None of these

2.) The payment P(n) that a person receives is a function
of the day n. If P(n) = 2 × P(n - 1) and the person
receives Rs.12 on the day two, then what is the total
payment received by a person for 10 days starting from the
day one