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Joined: 09 Jun 2011
Posts: 103
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m01q14 [#permalink] New post 14 Sep 2011, 04:19
Three piles of 7 beans each are to be made from 10 red, 5 yellow, and 6 green beans. If all of the beans must be used and each stack must contain at least one bean of each color, then what is the maximum number of red beans that can be put in one of the stacks?
Joined: 10 May 2011
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Location: India
Concentration: Strategy, General Management
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Re: m01q14 [#permalink] New post 14 Sep 2011, 18:05
Since each pile must have a bean of each color, max red beans can be 5 in any pile. Rest of the options automatically eliminates themselves because if max red beans is to be 6, 2 more (yellow and green) makes it 8 in each pile...not possible.

Was it that easy or i misinterpreted something?
Senior Manager
Senior Manager
Joined: 11 May 2010
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Location: United Kingdom
Concentration: Entrepreneurship, Technology
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Re: m01q14 [#permalink] New post 20 Sep 2011, 01:33
yep it's that easy - deceptively so... however it wasn't easy to think of the solution under test conditions! I thought it may have been a combinatorics problem initially.
Joined: 04 Aug 2011
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Location: United States
Concentration: Technology, Leadership
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Re: m01q14 [#permalink] New post 14 Jan 2012, 21:08
10 red ball , 1 will go in each stack leaving 7 ,

in any 1 of the stack 4 can go with 1 red, 1 Yellow and 1 Green already there (4 spaces left). So total red makes it to 5 balls.

Re: m01q14   [#permalink] 14 Jan 2012, 21:08
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