Last visit was: 24 Jul 2024, 00:07 It is currently 24 Jul 2024, 00:07
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Math Expert
Joined: 02 Sep 2009
Posts: 94591
Own Kudos [?]: 643432 [14]
Given Kudos: 86732
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 94591
Own Kudos [?]: 643432 [5]
Given Kudos: 86732
Send PM
General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 94591
Own Kudos [?]: 643432 [1]
Given Kudos: 86732
Send PM
DI Forum Moderator
Joined: 05 May 2019
Status:GMAT Club Team member
Affiliations: GMAT Club
Posts: 997
Own Kudos [?]: 742 [0]
Given Kudos: 1005
Location: India
GMAT Focus 1:
645 Q82 V81 DI82
GMAT 1: 430 Q31 V19
GMAT 2: 570 Q44 V25
GMAT 3: 660 Q48 V33
GPA: 3.26
WE:Engineering (Manufacturing)
Send PM
Re M01-14 [#permalink]
I think this is a high-quality question and I agree with explanation.
Intern
Intern
Joined: 09 Apr 2022
Posts: 13
Own Kudos [?]: 10 [0]
Given Kudos: 43
Send PM
Re: M01-14 [#permalink]
For Red to be maximum in a pile, 3 out of 7 spots of the beans will be filled by each color.
R, G, Y.
With this, we will left with 4 slots out of 7. Since we have total of 10 red beans and 3 piles where each pile must have at least 1 color bean, we need to add another 4 R to fill up the capacity of 7. Hence, 5R + 1Y + 1G = 7 Beans in total.

Answer is A, 5 red beans
GMAT Club Bot
Re: M01-14 [#permalink]
Moderators:
Math Expert
94589 posts
Founder
37875 posts