|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
14 Sep 2011, 05:19
Kudos
Bookmarks
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
A.5
B.6
C.8
D.9
E.10
Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests
for the latest questions.
Still interested? Check out the "Best Topics" block below for better discussion and related questions.
Thank you for understanding, and happy exploring!
hiimmunish
Joined: 10 May 2011
Last visit: 03 Nov 2012
Posts: 14
Location: India
Concentration: Strategy, General Management
GMAT 1: 690 Q50 V32
GMAT 1: 690 Q50 V32
Posts: 14
14 Sep 2011, 19:05
Kudos
Bookmarks
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?
Was it that easy or i misinterpreted something?
n2739178
Joined: 12 May 2010
Last visit: 05 Jul 2013
Posts: 233
Own Kudos:
Given Kudos: 12
Location: United Kingdom
Concentration: Entrepreneurship, Technology
Schools: LBS '14 Oxford '14 Judge Sloan '14 INSEAD '14
GMAT Date: 10-22-2011
GPA: 3
WE:Information Technology (Internet and New Media)
20 Sep 2011, 02:33
Kudos
Bookmarks
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.
