Last visit was: 31 Dec 2024, 04:24 It is currently 31 Dec 2024, 04:24
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

Ben has 30 pencils in a box. Each of the pencils is one of 5

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30558
Own Kudos [?]: 36907 [8]
Given Kudos: 26113
Send PM
Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink] New post   07 Nov 2019, 10:42
Expert Reply
8
Bookmarks
00:00

Question Stats:

61% (01:05) correct 38% (01:59) wrong based on 42 sessions

Hide Show timer Statistics

Ben has 30 pencils in a box. Each of the pencils is one of 5 different colors, and there are 6 pencils of each color. If Ben selects pencils one at a time from the box without being able to see the pencils, what is the minimum number of pencils that he must select in order to ensure that he selects at least 2 pencils of each color?

A. 24

B. 25

C. 26

D. 27

E. 28
ShowHide Answer
Official Answer
C
Most Helpful Community Reply
avatar
POORVIBANSAL
Manager
Manager
Joined: 09 Oct 2019
Posts: 114
Own Kudos [?]: 60 [8]
Given Kudos: 0
Send PM
Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink] New post   14 Nov 2019, 05:05
7
1
HI,
There's no formula that I used to solve this question.
To select 2 pencils of each colour and keeping max failures in mind, I counted 6+6+6+6+2= 26.
6's denote= all the 6 pencils of each color.
2= pencils of the last colour left.

Please shot a kudos if you like the solution.
Thanks in advance!
General Discussion
avatar
uchihaitachi
Retired Moderator
Joined: 16 Sep 2019
Posts: 187
Own Kudos [?]: 285 [5]
Given Kudos: 0
Send PM
Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink] New post   18 Nov 2019, 21:44
5
Carcass wrote:
Ben has 30 pencils in a box. Each of the pencils is one of 5 different colors, and there are 6 pencils of each color. If Ben selects pencils one at a time from the box without being able to see the pencils, what is the minimum number of pencils that he must select in order to ensure that he selects at least 2 pencils of each color?

A. 24

B. 25

C. 26

D. 27

E. 28


In such type of questions, we take the worst case possible.
Now there are 5 colors and 6 pencils of each color.
The worst case would be that we pick 24 pencils and still we have only 4 colors.
So, to get another 2 pencils such that we have atleast 2 pencils of each color.
the total number of pencils become = 24 + 2 = 26.

OA - C
avatar
S
COolguy101
Senior Manager
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Own Kudos [?]: 172 [0]
Given Kudos: 81
Concentration: , International Business
Send PM
Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink] New post   21 Oct 2021, 05:16
I did not understand any of the solutions here. Hoping someone explain in a more elucidated way.
User avatar
D
rx10
Retired Moderator
Joined: 02 Dec 2020
Posts: 1831
Own Kudos [?]: 2149 [0]
Given Kudos: 140
GRE 1: Q168 V157

GRE 2: Q167 V161
Send PM
Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink] New post   21 Oct 2021, 07:54
This is a very nice sum. You can solve it without lifting a pen.

So Ben has \(30\) pencils with \(5\) different colors so \(6\) pens of each color.

We need the least no of pens so he has at least \(2\) pencils of each color.

Consider this case:

Ben picks all pencils for \(4\) colors which will be \(6 * 4 = 24\) and then for the last color he didn't lift any.

Now when he picks \(2\) pencils of the last color, he will have \(6\) pencils each of \(4\) colors and \(2\) pencils for the last color. So total \(26\)

\(26\) is the min no of pens he can lift so that he can have \(AT\) \(LEAST\) \(2\) pencils of each color.

Answer C

COolguy101 wrote:
I did not understand any of the solutions here. Hoping someone explain in a more elucidated way.
Prep Club for GRE Bot
gmatclubot
Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink]
21 Oct 2021, 07:54
Moderators:
adewale223
GRE Instructor
88 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1126 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact