Last visit was: 12 Sep 2024, 15:23 It is currently 12 Sep 2024, 15:23
Toolkit
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

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.

# M01-14

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 95478
Own Kudos [?]: 658041 [14]
Given Kudos: 87247
Math Expert
Joined: 02 Sep 2009
Posts: 95478
Own Kudos [?]: 658041 [5]
Given Kudos: 87247
General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 95478
Own Kudos [?]: 658041 [1]
Given Kudos: 87247
Retired Moderator
Joined: 05 May 2019
Affiliations: GMAT Club
Posts: 997
Own Kudos [?]: 857 [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)
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 09 Apr 2022
Posts: 12
Own Kudos [?]: 10 [0]
Given Kudos: 43
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
Intern
Joined: 22 Feb 2023
Posts: 13
Own Kudos [?]: 4 [0]
Given Kudos: 63