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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a village of 100 households, 75 have at least one DVD pla [#permalink]

Show Tags

26 Jul 2008, 10:53

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

30% (02:43) correct
70% (01:48) wrong based on 69 sessions

HideShow timer Statistics

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:

Houses with DVD Players only = a Houses with MP3 Players only = b Houses with Cell Phone only = C Houses with DVD Players and MP3 Players only = d Houses with DVD Players and Cell Phones only = e Houses with Cell Phones and MP3 Players only = f Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75 Houses with MP3 Players = b+d+f+g = 55 Houses with Cell Phones = c+e+f+g = 80

Houses with DVD Players only = a Houses with MP3 Players only = b Houses with Cell Phone only = C Houses with DVD Players and MP3 Players only = d Houses with DVD Players and Cell Phones only = e Houses with Cell Phones and MP3 Players only = f Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75 Houses with MP3 Players = b+d+f+g = 55 Houses with Cell Phones = c+e+f+g = 80

Houses with DVD Players only = a Houses with MP3 Players only = b Houses with Cell Phone only = C Houses with DVD Players and MP3 Players only = d Houses with DVD Players and Cell Phones only = e Houses with Cell Phones and MP3 Players only = f Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75 Houses with MP3 Players = b+d+f+g = 55 Houses with Cell Phones = c+e+f+g = 80

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD. The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD. The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

Hence, X-Y = 55 - 30 = 25

OA is 45

will post the OE if we are not able to come up with it. _________________

----------------------------------------------------------- 'It's not the ride, it's the rider'

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD. The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

Hence, X-Y = 55 - 30 = 25

OA is 45

will post the OE if we are not able to come up with it.

Tricky +1.. GMBA.. nice question

(D) 45 makes sense. In the graph I mentioned earlier, we're given CELL=80 DVD=75 MP3=55

Consider two parallel y-axis and these bar graphs between them. X (maximum number of people with all 3) is clearly 55. For Y. Draw a bar graph representing MP3 from left to right. Then draw one for DVD from right to left. Their intersection is 30. The mistake I made earlier is discounting CELL, but you HAVE to account for it. Upon minimizing CELL on this graph, the minimal common area or Y comes down to 10.

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player) Let B be the number of house holds with 2 devices. Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, the maximum number of house holds can be 55 or less. But in order to have the maximum value for X, A should have the highest value. lets try 55, with equation (1) and (2)

we get 2B+ C= 45 B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that. lets try 0, with equation (1) and (2)

we get 2B+C= 210 B+C=100 on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

Consider all the households who have mp3 player also have the other two devices, then X = 55

consider the "worst case", among the 55 owners of mp3 player, 20 of them do not have cell phone. So number of villagers who have both mp3 player and cell phone is 55-20 = 35 Among the 35 villagers who have both mp3 player and cell phone, the worst case is 25 of them do not have dvd player. So number of villagers who have all 3 devices is 35 - 25 = 10 Y = 10

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player) Let B be the number of house holds with 2 devices. Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, [i]the maximum number of house holds can be 55 or less[/i]. But in order to have the maximum value for X, A should have the highest value. lets try 55, with equation (1) and (2)

we get 2B+ C= 45 B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that. lets try 0, with equation (1) and (2)

we get 2B+C= 210 B+C=100 on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

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 the lowest possible number of house holds that have all three devices X- Y=??

A. 25 b 35 c 45 D 55 E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player) Let B be the number of house holds with 2 devices. Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, the maximum number of house holds can be 55 or less. But in order to have the maximum value for X, A should have the highest value. lets try 55, with equation (1) and (2)

we get 2B+ C= 45 B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that. lets try 0, with equation (1) and (2)

we get 2B+C= 210 B+C=100 on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

Trying A=5.

(1) and (2) become:

2B + C= 195 B+C = 95

B= 100, C= -5. Situation cant hold true.

Trying A= 10

(1) and (2) become:

2B+C= 180 B+C=90

B=90, A=10, C=10

Thus situation can hold true.

Thus minimum value for Y = 10

X-Y = 45 answer.

Good explanation. +1 _________________

----------------------------------------------------------- 'It's not the ride, it's the rider'

[quote="chan_nhu78"]How can you have the max is 55? Thanks,

Hey

I took the max as 55 because, as per the given data, its clear that atleast 55 households have a probability of having all the three devices. i.e more than 55 households can not have all the three devices.

Re: In a village of 100 households. 75 have at least one dvd [#permalink]

Show Tags

08 May 2014, 05:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: In a village of 100 households, 75 have at least one DVD pla [#permalink]

Show Tags

08 May 2014, 07:52

Expert's post

1

This post was BOOKMARKED

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

For me the best way to solve this problem is not use Venn diagram or formulas but to draw simple bars (note: each dash is 5):

Min overlap is 10: -------------------- 80 phones; -------------------- 75 DVD's; -------------------- 55 MP3.

Max overlap is 55: -------------------- 80 phones; -------------------- 75 DVD's; -------------------- 55 MP3.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...