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Manager
Joined: 09 Jun 2011
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Question Stats:
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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?
A.5
B.6
C.8
D.9
E.10
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Intern
Joined: 10 May 2011
Posts: 22
Location: India
Concentration: Strategy, General Management
GMAT 1: 690 Q50 V32
Followers: 0
Kudos [?]:
6
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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?
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Senior Manager
Joined: 12 May 2010
Posts: 292
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)
Followers: 4
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26
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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.
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Intern
Joined: 04 Aug 2011
Posts: 48
Location: United States
Concentration: Technology, Leadership
GMAT 1: 570 Q45 V25
GPA: 4
WE: Information Technology (Computer Software)
Followers: 0
Kudos [?]:
10
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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.
IMO - A
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