Total Answers and Comments: 17
Last Update: October 27, 2009
Submitted by : HEMANTHKUMAR VARADHARAJAN Let the no. of breads initially present be x.Then1 st thief takes x/22 nd thief takes x/43 rd thief takes x/84th thief takes x/16.Remaining = 3 breadsEquating x/2 + x/4 + x/8 + x/16 + 3 = x(15x + 48)/16 = x15x + 48 = 16x=> x = 48The number of breads initially present were 48.Above answer was rated as good by the following members: jasu85 , kallam86
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December 20, 2005 05:48:16 #6
sunita
RE: 4 thieves rob a bakery of the breadone afte... ans:
it will be 63
procedure:
let initially der are x breads
1st thief takes x/2 +1/2 so remaining x-(x/2+1/2) x/2-1/2
2nd thief takes 1/2+1/2*(x/2-1/2) 1/2+x/4-1/4 x/4+1/4 so remain x/2-1/2-x/4-1/4 x/4-3/4
3rd thief takes 1/2 +1/2*(x/4-3/4) 1/2+x/8-3/8 x/8+1/8 so remain x/4-3/4-x/8-1/8 x/8-7/8
4th thief takes 1/2+1/2*(x/8-7/8) 1/2+x/16-7/16 x/16+1/16 so remain x/8-7/8-x/16-1/16 x/16-15/16
so x/16-15/16 3
x 48+15 63(ans)
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August 07, 2006 02:04:01 #9
amateur
RE: 4 thieves rob a bakery of the breadone afte... Ans : 48
Let Total no of breads be X.
> 1st Thief took X/2 remaining X/2
> 2nd Thief took 1/2 * X/2 X/4 remaining X/2-X/4 X/4
> 3rd Thief took 1/2 * X/4 X/8 remaining X/4-X/8 X/8
> 4th Thief took 1/2 * X/8 X/16 remaining X/8-X/16 X/16
Now its given that remaining bread 3
Therefore X/16 3
> X 48
Ans 48.
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