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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
Posts: 1603

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

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09 Aug 2003, 11:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Manager
Joined: 24 Jun 2003
Posts: 145

Kudos [?]: 3 [0], given: 0

Location: India

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09 Aug 2003, 12:25
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

Kudos [?]: 3 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 308 [0], given: 0

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09 Aug 2003, 22:59
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

Kudos [?]: 308 [0], given: 0

Manager
Joined: 24 Jun 2003
Posts: 145

Kudos [?]: 3 [0], given: 0

Location: India

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10 Aug 2003, 21:58
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 !

Kudos [?]: 3 [0], given: 0

10 Aug 2003, 21:58
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# Childen share nuts. The first child gets one nut plus 1/10

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