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

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SVP
Joined: 03 Feb 2003
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Childen share nuts. The first child gets one nut plus 1/10 [#permalink]  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?
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
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Re: NUTS [#permalink]  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
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 86 [0], given: 0

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
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 1

Kudos [?]: 1 [0], given: 0

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 !

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