Latest ITC Aptitude Question SOLUTION: 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 5

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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?

Asked In ITC (11 years ago)

Solved mohd zubair khan Read Solution (10)

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Read Solution (10) :

Gautam  (11 years ago)

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

Is this solution Helpfull? Yes (21) | No (2)  

mohd zubair khan  (11 years ago)

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

Is this solution Helpfull? Yes (8) | No (12)  

RIMI  (11 years ago)

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

Is this solution Helpfull? Yes (4) | No (3)  

Suman Bose  (9 years ago)

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

Is this solution Helpfull? Yes (2) | No (1)  

Shanu Dutta  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No (9)  

Shahrukh  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Rahul singh  (11 years ago)

i think it would be 176!!

Is this solution Helpfull? Yes | No (20)  

Susovon Pal  (10 years ago)

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.)

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Mallikarjuna M  (9 years ago)

answer is 127

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Nitish Kumar  (7 years ago)

GAUTAM could you please explain why in your remaining bread after 1st thief left (x-1)/2 instead of (x+1)/2 ?

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