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Childen share nuts. The first child gets one nut plus 1/10

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SVP
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Childen share nuts. The first child gets one nut plus 1/10 [#permalink] New post 09 Aug 2003, 10:09
Childen share nuts. The first child gets one nut plus 1/10 of the remaining; the secong gets two nuts plus 1/10 of the remaining; ans so on. Each child gets equal quantity of the nuts. How many nuts are there if there are nine children?
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Re: NUTS [#permalink] New post 09 Aug 2003, 11:25
stolyar wrote:
Childen share nuts. The first child gets one nut plus 1/10 of the remaining; the secong gets two nuts plus 1/10 of the remaining; ans so on. Each child gets equal quantity of the nuts. How many nuts are there if there are nine children?


Let the total no of nuts be N

Then First child gets 1+ (1/10)*(N-1) nuts
Second gets 2+(1/10)*(N-3-(N-1)/10) nuts

Equating the two and solving for N we get N=81

Sanity check: For 9 kids, each gets 9 nuts
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 [#permalink] New post 09 Aug 2003, 21:59
Well-done

A more direct approach: desribe the nuts of the first child by using both given conditions.

1+ (1/10)*(N-1)=N/9 (the first child)
N=81
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 [#permalink] New post 10 Aug 2003, 20:58
stolyar wrote:
Well-done

A more direct approach: desribe the nuts of the first child by using both given conditions.

1+ (1/10)*(N-1)=N/9 (the first child)
N=81


Definitely a better and faster solution !

:)
  [#permalink] 10 Aug 2003, 20:58
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Childen share nuts. The first child gets one nut plus 1/10

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