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:

100 = No. of households 55 = No. of MP3s for sure (so assuming that min of 55 households have all 3)

This is incorrect.

How can 100 households have all 3 when only 55 households have MP3 players? 55 is the MAXIMUM number of households that can have all 3, not minimum. and 10 is the minimum number of households that must have all 3 (explained in solutions on first page) So 55 - 10 = 45
_________________

Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

Show Tags

11 Dec 2016, 03:52

[quote="VeritasPrepKarishma"]hiredhanak: I am assuming you are looking for a venn diagram solution to this question..

It is pretty simple. First of all maximum number of households: We want to bring the circles to overlap as much as possible. 80 - Cell phone 75 - DVD 55 - MP3 Lets take Cell phone and DVD circles since they will have maximum overlap. They must overlap in 55 households so that total number of households is 100. Now put the MP3 households in a way to maximize all three overlap.

Attachment:

Ques1.jpg

So at most 55 households can have all 3. hi karishma

i do not understand. how did you derive at most 55 households can have all 3 by Van diagram

When discussing the maximum overlap case, none NEEDN'T be 0. It may be, it may not be.

Put the three circles within each other. The 75 circle within the 80 circle and the 55 circle within the 75 circle. The overlap will be 55 in that case and none = 20. The figure only shows one of the possible ways of obtaining the maximum.

In the case of minimum, you would want the circles to lie as far apart as possible. If none is anything other than 0, the circles would need to overlap more. Say none = 10, the circles of 80 and 75 would need to have an overlap of 65. So the 55 circle can occupy 25 but an overlap of 30 will be needed. Hence minimum overlap will increase. To minimize the overlap, we will need None = 0.
_________________

Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

Show Tags

24 May 2017, 05:29

1

This post was BOOKMARKED

I could not load the Venn Diagram, Please try to visualize it. ( The diagram is posted below)

a, b, and c one product each.

x, y, and z two products each.

d all three products.

a + b + c + d + x + y + z = 100 If you assume there is no one who uses two products each, then x = y = z = 0. So, a + b + c + d = 100 ----------------(1) a + d = 80 ------------------------------(2) b + d = 75 ------------------------------(3) c + d = 55 ------------------------------(4)

From 2, 3, and 4, a + b + c + 3d = 210 ----------------(5) a + b + c + d + 2d = 210

From 1 and 5, 100 + 2d = 210 d = 55 --------------Maximum.

If you assume there is no one who uses one product each, then a = b = c = 0. So, d + x + y + z = 100 ----------------(6) x + y + d = 80 ------------------------------(7) x + z + d = 75 ------------------------------(8) y + z + d = 55 ------------------------------(9)

From 7, 8, and 9, 2x + 2y + 2z + 3d = 2(x + y + z + d ) + d = 210 ----------------(10)

From 6 and 10, 2(100) + d = 210 d = 10 --------------Minimum.

Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

Show Tags

16 Aug 2017, 15:30

I often just don't get the complexity of such kind of questions max overlap is 55 (as this is the max number of 3 types of households that can have all three devices) min overlap for 3 devices is 75+55+80 - 100*2 (max number for households that can have 2 devices) = 210 - 200 = 10

Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

Show Tags

25 Oct 2017, 03:54

Here,the maximum households having three will be 55 Now,if we want to get the minimum number of households having three,we need to find at least how many have two of things. So,At least DVD and cell phone=(75+80)-100=55 At least DVD and MP3=(75+55)-100=30 At least cell and MP3 =(80+55)-100=35 now,minimum number=(75+80+55-30-35-55)-100=10 so,the difference is (55-10)=10 Hope it's clear now

In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

Show Tags

23 Nov 2017, 10:27

Bunuel wrote:

Hussain15 wrote:

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.

55-10=45.

Answer: C.

hi Bunuel

maybe it is very obvious, but I am facing bit trouble please guide me through my explanation

at least 55 people own MP3 players, so it is very natural to assume that maximum number of people who can have all three is 55 okay

now, to have the lowest possible overlapping

100 - 80 = 20 people have no phones (but may have DVDs) 100- 75 = 25 people have no DVDs (but may have phones)

so, you have distributed 45 mp3 players to these 45 villagers, causing no overlapping. so far so good

but we are left with 10 more mp3 players to distribute, but one cannot provide these players to the inhabitants who already have the mp3 players because doing so will add up the number of people who own mp3s to 45, a condition that is unwarranted by the question. So, these 10 people must be coming from the groups of DVDs or??? / and ???? phones holders. Here I am stumped, because

these 10 people can have phones earlier and are now getting mp3s OR these 10 people can have DVDs earlier and are now getting mp3s in this case there is no overlapping .....

OR

these people can have phones + DVDs earlier and are now getting mp3s in this case there is overlapping....

hiredhanak: I am assuming you are looking for a venn diagram solution to this question..

It is pretty simple. First of all maximum number of households: We want to bring the circles to overlap as much as possible. 80 - Cell phone 75 - DVD 55 - MP3 Lets take Cell phone and DVD circles since they will have maximum overlap. They must overlap in 55 households so that total number of households is 100. Now put the MP3 households in a way to maximize all three overlap.

Attachment:

Ques1.jpg

So at most 55 households can have all 3.

Now, minimum number of households: We want to take the circles as far apart from each other as possible. Now put the MP3 households in a way to minimize all three overlap. So make the MP3 households occupy the shaded region i.e. region occupied by DVD players alone and cell phone alone. You will be able to adjust 45 MP3s outside the common area but you will need to put 10 of the MP3 households in the common area. So minimum overlap is 10.

Attachment:

Ques2.jpg

x - y = 55 - 10 = 45

Responding to a pm: Explaining the min case.

We want minimum overlap of all three. So we need to spread the 3 around in such a way that all 3 overlap the least. Out of 100 households, 80 have a cell phone. We have 20 households leftover without a cell phone. Now 75 have a DVD so we reduce overlap by putting 20 DVDs in the leftover 20 households. Rest 75 - 20 = 55 will need to be overlapped with DVDs.

So now we have 20 households with just DVDs, 25 with just cell phones and 55 with both. Now we also have to distribute 55 MP3s. Note that overlap of all 3 has to be reduced as much as possible. So we should give minimum to the households that already have both. The rest of the 45 households get MP3s (not these 45 have 2 things each). But we still have 10 MP3s leftover. These will need to be given to the households which have DVDs and cell phones both. So minimum overlap of all 3 is 10.
_________________