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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?
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Answer Posted / resh
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)
(not solved by me but got it from another site)
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