ITC
Exam
HR Interview
Interview
Q. There are five thieves, each loot a bakery one after the other such that the first one takes 1/2 of the total no. of the breads plus 1/2 of a bread. Similarly 2nd, 3rd,4th and 5th also did the same. After the fifth one no. of breads remained are 3. Initially how many breads were there?
Read Solution (Total 10)
-
- suppose total no. of breads =x
for 1st thief:x/2+1/2=(x+1)/2
remaining breads=x-(x+1)/2=(x-1)/2
2nd thief:((x-1)/4)+1/2=(x+1)/4
remaining bread=(x-1)/2-(x+1)/4=(x-3)/4
3rd thief:((x-3)/8)+1/2=(x+1)/8
remaining bread=((x-3)/4)-(x+1)/8=(x-7)/8
4th thief:((x-7/16)+1/2)=x+1/16
remainig bread:(x-7/8)-(x+1/16)=(x-15)/16
5th thief:((x-15/32)+1/2)=(x+1)/32
hence x+1(1/2+1/4+1/8+1/16+1/32)+3=x
x=127 - 12 years agoHelpfull: Yes(21) No(2)
- suppose there were x breads 2 thiefs come and perfom the same action.
first = x/2 + 1/2
second=1/2{x-(x/2+1/2)}+1/2
frist + second = x
x/2 + 1/2 + 1/2{x-(x/2 + 1/2)} + 1/2 = x
on solving we will get x = 3
from here we can conclude that for three thiefs = 3 x 2 + 1 = 7
and there were total of if 7 thives steel we will left no breads
for four = 7 x 2 + 1 = 15
for five = 15 x 2 + 1 = 31
for six = 31 x 2 + 1 = 63
for seven = 63 x 2 + 1 = 127
so there was a toatal 127 breads initially - 12 years agoHelpfull: Yes(8) No(12)
- 127
if total no of breads is x then after 1st one the remaing portion is (x/2-1)+1/2
after 2nd one (x-3)/4
after 3rd one (x-7)/8
after 4th one (x-15)/16
after 5th one (x-31)/32
then (x-31)/32=3 - 12 years agoHelpfull: Yes(4) No(3)
- suppose total breads before 1 theft : x
remaining breads after 1 theft : y
so, y = 1/2(x)-(1/2)
=> x = 2(y + 1/2)
now, after the fifth one remaining bread : 3
put 3 in the place of y
x=7,put 7 in the place of y
x=15,put 15 in the place of y
x=31,put 31 in the place of y
x=63,put 63 in the place of y
x=127, initial bread - 10 years agoHelpfull: Yes(2) No(1)
- ans should be 112
before the 5th thief : 2*3+(1/2)=6 and 1/2
" 4th thief : 2*(6 and half)+(1/2)=13 and 1/2
.
.
before the 1st thief: 2*(55 and half)+ 1/2+1/2=112
- 12 years agoHelpfull: Yes(1) No(9)
- suppose total no. of breads =x
for 1st thief:x/2+1/2=(x+1)/2
remaining breads=x-(x+1)/2=(x-1)/2
2nd thief:((x-1)/4)+1/2=(x+1)/4
remaining bread=(x-1)/2-(x+1)/4=(x-3)/4
3rd thief:((x-3)/8)+1/2=(x+1)/8
remaining bread=((x-3)/4)-(x+1)/8=(x-7)/8
4th thief:((x-7/16)+1/2)=x+1/16
remainig bread:(x-7/8)-(x+1/16)=(x-15)/16
5th thief:((x-15/32)+1/2)=(x+1)/32
hence x+1(1/2+1/4+1/8+1/16+1/32)+3=x
x=127 - 10 years agoHelpfull: Yes(1) No(0)
- i think it would be 176!!
- 12 years agoHelpfull: Yes(0) No(20)
- Let the total no. of bread=x
After the 1st thief fled,total no. of bread=x-(x/2+1/2)=(x-1)/2
After the 2nd thief fled,total no. of bread=(x-1)/2-((x-1)/4+1/2)=(x-3)/4
After the 3rd thief fled,total no. of bread=(x-3)/4-((x-3)/8+1/2)=(x-7)/8
After the 4th thief fled,total no. of bread=(x-7)/8-((x-7)/16+1/2)=(x-15)/16
After the 5th thief fled,total no. of bread=(x-15)/16-((x-15)/32+1/2)=(x-31)/32
By the problem,
(x-31)/32=3
or,x=127 (Ans.)
- 10 years agoHelpfull: Yes(0) No(0)
- answer is 127
- 9 years agoHelpfull: Yes(0) No(0)
- GAUTAM could you please explain why in your remaining bread after 1st thief left (x-1)/2 instead of (x+1)/2 ?
- 8 years agoHelpfull: Yes(0) No(0)
ITC Other Question