In a village of hundred households , 75 have at least one DVD player, 80 have at least one cell phone ,and 55 have at least one MP3 player. Every village has at least one of these three devices.If X and Y are respectively the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A ) 65
B ) 55
C ) 45
D) 35
E) 25
Answer : 45
Any help would be appreciated.
GMAT Forum
- Guest
Re: In a village of 100 households - PS Problem
esseisle Wrote:In a village of hundred households , 75 have at least one DVD player, 80 have at least one cell phone ,and 55 have at least one MP3 player. Every village has at least one of these three devices.If X and Y are respectively the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A ) 65
B ) 55
C ) 45
D) 35
E) 25
Answer : 45
Any help would be appreciated.
This was a really tough one.
I drew three circles and put the respective numbers into each ciricle.
To find the max: I just took the smallest number (55) and put it in the middle of the Venn diagram.
This worked out well:
80 - 55 = 25
75 - 55 = 20
55 - 55 = 0
When we add up all the numbers in the diagram, they're supposed to give us 100
25 + 20 + 0 + 55 = 100
To find the minimum, I thought about the sum of all the numbers (210) and they best way to distribute them to all 6 shared sections without putting something in the middle. Average! 210/6 = 180 and you have 30 leftover. Since you have 30 leftover, you can just put 10 in the middle and each of the sections will get 10 which = 30.
- parvezshah
Re: In a village of 100 households - PS Problem
esseisle Wrote:In a village of hundred households , 75 have at least one DVD player, 80 have at least one cell phone ,and 55 have at least one MP3 player. Every village has at least one of these three devices.If X and Y are respectively the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A ) 65
B ) 55
C ) 45
D) 35
E) 25
Answer : 45
Any help would be appreciated.
The question says about minimum number not maximum amount so let say all the villagers has all the devices .
X= 100 for the minimum say "55 have at least one MP3 player" so lets say that 55 of them had all 3 of them as this assertion also keeps the question true only
100 - 55 = 45
- Guest
This is a toughie - perhaps Stacey or her collegues could weigh in here, it would be helpful.
Here's how I approached this - consider the two limited cases. 1) As much of the "overlap" as possible occurs in three device households and 2) As much of the overlap as possible occurs in exactly two device households, minimizing the number of two device households.
1) This is the easy case - just assume all the overlap occurs in three device households. Let X = number of homes with all three devices. We know that 75-80-55-2X=100. so X=55. You have to use 2x in the equation above because the region where all three overlap is triply counted. So in this case, 20 households have only a DVD player, 25 have only a cell phone, 0 have only a MP3 player and 55 households have all three.
2) Here it's tempting to guess that it's possible to have all the overlap occur in households that have exactly two of the devices, therefore X-Y = 55. Alas, there is no way to distribute the exactly two device households in a way that works without having any 3-device households. After drawing the diagram and interating, you can see that you need at least 10 3-device households, so X-Y= 55-10 = 45. It's important to be careful to keep track if you are dealing with "at least" or "exactly" two device households. If someone has a more intuitive appoach to this case, I'd love to hear it.
cheers
Here's how I approached this - consider the two limited cases. 1) As much of the "overlap" as possible occurs in three device households and 2) As much of the overlap as possible occurs in exactly two device households, minimizing the number of two device households.
1) This is the easy case - just assume all the overlap occurs in three device households. Let X = number of homes with all three devices. We know that 75-80-55-2X=100. so X=55. You have to use 2x in the equation above because the region where all three overlap is triply counted. So in this case, 20 households have only a DVD player, 25 have only a cell phone, 0 have only a MP3 player and 55 households have all three.
2) Here it's tempting to guess that it's possible to have all the overlap occur in households that have exactly two of the devices, therefore X-Y = 55. Alas, there is no way to distribute the exactly two device households in a way that works without having any 3-device households. After drawing the diagram and interating, you can see that you need at least 10 3-device households, so X-Y= 55-10 = 45. It's important to be careful to keep track if you are dealing with "at least" or "exactly" two device households. If someone has a more intuitive appoach to this case, I'd love to hear it.
cheers
- calgmatter
In Fig above,
Step 1: Assign DVD homes from right = 75 homes
Step 2: Assign Cell homes from left = 80 homes
Step 3: We now know 55 homes have D+C (typo in fig)
We have 55 MP3s that we can distribute all to D+C (which gives us max as 55) or distribute 25 to portion IV and 20 to portion V and the rest 10 to portion III and so the minimum would be portion III which is 10 homes.
Answer 45.
- guest
guest
Great way to diagram the problem, thanks!
- Guest
In a village of 100 households - PS Problem
Ron,
Any simple method on how to get the minimum number.. Below is waht I did
100 = 75+80+55- both+all (all have atleast one, so none is 0)
100=210-both+all
110=both-all
Now if I want to minimize "all" then, I have to maximize both, how do I do that ? Regards...
Any simple method on how to get the minimum number.. Below is waht I did
100 = 75+80+55- both+all (all have atleast one, so none is 0)
100=210-both+all
110=both-all
Now if I want to minimize "all" then, I have to maximize both, how do I do that ? Regards...
- rfernandez
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- Dan
That is a tough one! Hopefully more difficult than something I'll see on a test.
This is the approach I found easiest for me to understand.
First we need to figure out what the maximum is. We know it is possible there are 55 with all three, since the 55 MP3 owners is smallest set of the three. (Max = # of MP3)
To calculate the fewest possible, we first need to figure out how many must have all three devices. (max possible - max without all three = minimum with all three).
Of the 80 cell phone owners, 25 cannot have an MP3 player
80 - 55 = 25.
Similarly, 20 of the DVD owners cannot have an MP3 player.
75 - 55 = 20.
If the owners who do not have all three are unique (no overlap), we get the maximum number of owners who do not own an MP3 player, and thus cannot own all three devices.
Max possible - maximum without all three = minimum number of people with all three devices.
55 - 45 = 10
Now we can solve:
55 - 10 = 45.
This is the approach I found easiest for me to understand.
First we need to figure out what the maximum is. We know it is possible there are 55 with all three, since the 55 MP3 owners is smallest set of the three. (Max = # of MP3)
To calculate the fewest possible, we first need to figure out how many must have all three devices. (max possible - max without all three = minimum with all three).
Of the 80 cell phone owners, 25 cannot have an MP3 player
80 - 55 = 25.
Similarly, 20 of the DVD owners cannot have an MP3 player.
75 - 55 = 20.
If the owners who do not have all three are unique (no overlap), we get the maximum number of owners who do not own an MP3 player, and thus cannot own all three devices.
Max possible - maximum without all three = minimum number of people with all three devices.
55 - 45 = 10
Now we can solve:
55 - 10 = 45.
- imanemekouar
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Re: In a village of 100 households - PS Problem
Can somebody please explain it from the beginning. I did not get it .thanks
- RonPurewal
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Re: In a village of 100 households - PS Problem
i'm going to call out this problem as NOT from an official source.
what is the actual source of this problem? if anyone (including the original poster) wants to claim that this problem is from gmat prep, could that poster kindly take a screenshot of the problem, from the prep software, and post it?
if no one claims this problem, and/or posts its actual source, then we will move this thread to the general math folder, or, more likely, delete it altogether.
what is the actual source of this problem? if anyone (including the original poster) wants to claim that this problem is from gmat prep, could that poster kindly take a screenshot of the problem, from the prep software, and post it?
if no one claims this problem, and/or posts its actual source, then we will move this thread to the general math folder, or, more likely, delete it altogether.
- vibhusethi
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Re: * In a village of 100 households - PS Problem
I found this way of thinking easier, does it make sense?
max households with no dvd = 25
no cell = 20
no mp3 = 45
Total max possible that dont have all = 90
In other words, atleast 10 households have all 3.
Finding max was easier, i.e 55.
55-10 is the answer.
Vibhu
max households with no dvd = 25
no cell = 20
no mp3 = 45
Total max possible that dont have all = 90
In other words, atleast 10 households have all 3.
Finding max was easier, i.e 55.
55-10 is the answer.
Vibhu
- mschwrtz
- ManhattanGMAT Staff
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Re: * In a village of 100 households - PS Problem
imanemekouar, did you apply calgmatter's method to he original question? That should get you home.
If the problem is just that you can't easily find Y (the minimum number of households with all three devices), try thinking about it this way: If every family had exactly two devices, then that would account for 200 devices all together. So at least 10 families must have a third device.
To expand: The sum of 55, 75, and 80 is 210. There are only 100 families. The number of "extra devices" is 110. Every family that has exactly two devices accounts for one "extra device" and every family that has three accounts for two "extra devices." To minimize the number of families who have three each, maximize the number who have (at least) two each. If every family had exactly two devices, then that would account for 200 devices all together. At least 10 families must have 10.
If the problem is just that you can't easily find Y (the minimum number of households with all three devices), try thinking about it this way: If every family had exactly two devices, then that would account for 200 devices all together. So at least 10 families must have a third device.
To expand: The sum of 55, 75, and 80 is 210. There are only 100 families. The number of "extra devices" is 110. Every family that has exactly two devices accounts for one "extra device" and every family that has three accounts for two "extra devices." To minimize the number of families who have three each, maximize the number who have (at least) two each. If every family had exactly two devices, then that would account for 200 devices all together. At least 10 families must have 10.