GMAT Prep
In a village of 100 households, 75 at least have 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
OA C
In a village of 100 households, 75 least one DVD player, 80
This topic has expert replies
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Let D = DVD owners, C = cellphone owners, and M = MP3 owners.AAPL wrote:GMAT Prep
In a village of 100 households, 75 at least have 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
T = D + C + M - (DC + DM + CM) - 2(DCM).
The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in D, everyone in C, and everyone in M:
Those in exactly 2 of the groups (DC + DM + CM) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (DCM) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.
In the problem above:
T = 100
D = 75
C = 80
M = 55.
Thus:
100 = 75 + 80 + 55 - (DC + DM + CM) - 2(DCM)
(DC + DM + CM) + 2(DCM) = 110.
MAXIMUM:
To maximize the value of DCM, we must MINIMIZE the value of DC + DM + CM.
If DC + DM + CM = 0, we get:
0 + 2(DCM) = 110
DCM = 55.
MINIMUM:
To MINIMIZE the value of DCM, we must MAXIMIZE the value of DC + DM + CM.
Since D=75, the maximum possible value of CM = 100-75 = 25.
Since C=80, the maximum possible value of DM = 100-80 = 20.
Since M=55, the maximum possible value of DC = 100-55 = 45.
Since the maximum value of DC + DM + CM = 45+20+25 = 90, we get:
90+ 2(DCM) = 110.
DCM = 10.
Thus:
x-y = 55-10 = 45.
The correct answer is C.
For similar problems, check here:
https://www.beatthegmat.com/group-of-stu ... 63753.html
https://www.beatthegmat.com/sets-t148362.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7294
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
The greatest possible number of households that have all three devices is 55 when all the households that have a MP3 player also have a DVD player and a cell phone. So x = 55.AAPL wrote:GMAT Prep
In a village of 100 households, 75 at least have 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
OA C
Since 75 + 55 = 130 and 130 - 100 = 30, then 30 households must have both a DVD player and an MP3 player. Some of these 30 households might have a cell phone and some might not. Since 30 + 80 = 110 and 110 - 100 = 10, then 10 households must have a cell phone in addition to a DVD player and a MP3 player. In other words, a minimum of 10 households must have all three devices. So y = 10.
Therefore, x - y = 55 - 10 = 45.
Answer: C
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews