It is currently 21 Feb 2018, 21:05

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Counting Problem

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
User avatar
Joined: 09 Jan 2009
Posts: 20
Schools: SDSU
Counting Problem [#permalink]

Show Tags

New post 11 Feb 2009, 15:47
1
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

some one help me to solve w/t this pro below, tks for ur help

In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y is

A. 65
B. 55
C. 45
D. 35
E. 25
Director
Director
avatar
Joined: 29 Aug 2005
Posts: 854
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 03:58
75+80+55-m-2n=100
We are after the difference between min and max n.
n is max when m=0 and n is min at n=0
Difference yields 55.
B.
3 KUDOS received
Intern
Intern
avatar
Joined: 11 Feb 2009
Posts: 12
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 04:45
3
This post received
KUDOS
As for me I go for E - 25

Here is my explanation:
The greatest number of households that possibly could possess all three items is 55.
So our task is to find least possible value.
I have attached my poorly drawn graph (dont have any software for this ^_^)

Thus 55-30 = 25
Attachments

graph.jpg
graph.jpg [ 4.71 KiB | Viewed 1658 times ]

Intern
Intern
avatar
Joined: 18 Jan 2009
Posts: 18
Schools: Kellogg, Ross, Darden, Kelley, UNC
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 07:53
SlavaYura: I didn't get how you came to 30 as the minimum value of households that possess all 3 items. Your graph is cool, but can you please explain your logic in further detail?

Baggio: you are bearing the name of one of the most famous italian football players, are you from Italy? And, btw, what's the source for this problem?
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1789
Location: New York
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 08:07
baggio wrote:
some one help me to solve w/t this pro below, tks for ur help

In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y is

A. 65
B. 55
C. 45
D. 35
E. 25




clearly greatest possible nuber = 55


Overlap of DVD AND Cell Phone with least possible numbe r= 75+80 -100= 55

Overlap of (DVD AND Cell Phone with least possible number) and MP3 PLAYER least possible nuber = 55+55-100 = 10



Answer = 55-10=45
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

4 KUDOS received
Intern
Intern
avatar
Joined: 11 Feb 2009
Posts: 12
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 13:30
4
This post received
KUDOS
MBA2012 wrote:
SlavaYura: I didn't get how you came to 30 as the minimum value of households that possess all 3 items. Your graph is cool, but can you please explain your logic in further detail?


I solved it in this way:

1. First we draw 75 and 80.
2. After drawing them we have 20 and 25 from both sides. But the idea is that we find the least possible quantity. That is very tricky because if you begin drawing f55 from the left side you will have 5 items more then if you draw it from the right side(25 and 20) . Since we are asked to find the least possible value, thus drawing from the right side will be most logically (because the intersection(households that have all 3 items)will be less. )
Senior Manager
Senior Manager
avatar
Joined: 08 Jan 2009
Posts: 324
Re: Counting Problem [#permalink]

Show Tags

New post 12 Feb 2009, 17:05
Hi x2suresh

can u please expalin ur method.I didnt really get it.

What wrong with the below one.Why cant the least possible of all the intersection be zero(n=0).

75+80+55-m-2n=100
We are after the difference between min and max n.
n is max when m=0 and n is min at n=0
Difference yields 55.
B.

Thanks a lot.
Manager
Manager
avatar
Joined: 27 May 2008
Posts: 200
Re: Counting Problem [#permalink]

Show Tags

New post 13 Feb 2009, 02:11
pretty straight,

total =100

DVD = 75
cell phone = 80

Max ppl. who can have both - 75
So min ppl. who can have both = 55 (remaining ppl.(20) in village has cellphone)

MP3 - 55
Max ppl. who can have all three = 55
Min = remaining 45 (100-55(both DVD,cell)) can have MP3 so 55-45 = 10

So max - min = 45
Re: Counting Problem   [#permalink] 13 Feb 2009, 02:11
Display posts from previous: Sort by

Counting Problem

  post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderator: chetan2u



cron

GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.