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Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
 05 Feb 2012, 16:50
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Of the 60 families in a certain neighborhood, 38 have a cat. How many of the families in this neighborhood have a dog? (1) 28 of the families in this neighborhood have a cat but not a dog (2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog.
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Hi, there. I'm happy to help.  I solved this with a double-matrix method --- because that can get sloppy in the plaintext of these posting windows, I created a pdf attachment. The double matrix method is a tremendously powerful method for solving these overlapping set problems. At Magoosh, we have a whole series of video lessons going over everything you need to know for GMAT math, including one that explains exactly how to set up the double matrix method of solution. Please let me know if you should have any questions. Mike
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
05 Feb 2012, 21:24
Finally get why it's B.
Total residence = 60.
Those with no pets = X
Those with cats = 38. Those with cats only = 38 -x
With cats n dogs = x.
Dogs = D. Dogs only = D - x
Total residence with pet = Dogs + Cats only.
60 - x = Cats only (38 - x) + D. The reason for doing this is because total amount of pet owners is people with cats + people with dogs plus people with both. If you add total # of dog owners plus total # of cat owners together your adding owners of both pets twice.
Therefore 60 - X + X -38 = D.
D = 22.
B only is significant
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
05 Feb 2012, 23:13
kys123 wrote: Finally get why it's B.
Total residence = 60.
Those with no pets = X
Those with cats = 38. Those with cats only = 38 -x
With cats n dogs = x.
Dogs = D. Dogs only = D - x
Total residence with pet = Dogs + Cats only.
60 - x = Cats only (38 - x) + D. The reason for doing this is because total amount of pet owners is people with cats + people with dogs plus people with both. If you add total # of dog owners plus total # of cat owners together your adding owners of both pets twice.
Therefore 60 - X + X -38 = D.
D = 22.
B only is significant It's true that (# with 1+ pets) = (# with cats only) + (# with dogs only) + (number with both) Using the notation you adopted, 60 - x = (38 - x) + D + x ---> you forgot that last term. 60 - x + x - 38 - x = D 22 - x = D And, thus, we cannot establish the value of D with knowing the value of x, so Statement #2, by itself, is insufficient. As I show in the pdf posted above, the answer is Please let me know if you have any questions. Mike
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
06 Feb 2012, 02:02
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Okay, but answer still true. D who only only dogs. Individuals who only own dogs plus individual who only own cats = Total individual who own dogs. That's what we're trying to find, so therefore B is correct.
22 = Number in the neighbourhood with a dog (D [# of people who own only dogs] +X [# of people who own dogs and cats].
Same equation you stated
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
06 Feb 2012, 02:39
I thought it was C as well... Can someone explain further?
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
06 Feb 2012, 03:08
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Good question. +1 to calreg11. Mike is right that a double-matrix method is probably the easiest way to solve this problem and kys123 is right that the answer to the question is B (+1). Of the 60 families in a certain neighborhood, 38 have a cat. How many of the families in this neighborhood have a dog?Consider matrix below. Numbers in black are given and numbers in red are calculated. Attachment: 
Stem.PNG [ 2.53 KiB | Viewed 1959 times ]
(1) 28 of the families in this neighborhood have a cat but not a dog. Attachment: 
Statement 1.PNG [ 2.68 KiB | Viewed 1958 times ]
So you can see that we can no way get # of the families in this neighborhood who has a dog ( ? in the matrix). Not sufficient. (2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog. Attachment: 
Statement 2.PNG [ 2.94 KiB | Viewed 1960 times ]
You can see that if # of families who have a dog and a cat and # of families who have neither a cat nor a dog is x, then # of families who has cat but not dog is 38-x. Next, total # of families who has no dog is (38-x)+x=38 and # of families who has a dog is 60-38=22. Sufficient. Answer: B. Hope it helps.
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
06 Feb 2012, 03:28
This is what I did. The way that Bunuel solve this problem was a lot more elegant, but for me my way is more intuitive. I know everything inside the matrix should add to 60. Hence my solution.
Attachments
Picture 16.png [ 22.91 KiB | Viewed 1942 times ]
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
06 Feb 2012, 09:46
My apologies to kys123The solution given by Bunuel[/b ]& [b]kys123 is perfectly correct. I realize I was misreading/misinterpreting the question, thinking it was asking for the number of people who owned only a dog, i.e. a dog and no cat, not simply the number of dog owners. A good reminder how crucial careful reading is. If the question were asking for the people who owned only a dog, the answer would be C. As it stands, though, with the question asking for the number of people who own only a dog, the answer is clearly B, as Bunuel and kys123 have shown. Again, my apologies for any confusion. Mike
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
03 Sep 2012, 03:56
very very tricky, Bunuel's response is fantastic(as usual ) well, (i) is insufficient (ii) is sufficient , here is the catch; as we all know that A+B=C so, A=C-B & B=C-A Now check out the file attached
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
03 Sep 2012, 06:35
calreg11 wrote: Of the 60 families in a certain neighborhood, 38 have a cat. How many of the families in this neighborhood have a dog?
(1) 28 of the families in this neighborhood have a cat but not a dog
(2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog. Yes, the question is really good. I like to show a series of diagrams to my students to explain what the statement 'number of families with both = number of families with none' implies. It means the sum of number of families with cat and number of families with dog is constant and is equal to 60. For every one family that has both, there is a family that has none (to keep their numbers equal) Look at the diagrams below. If the number of families that have neither a dog nor a cat is 0, the number of families with a dog is 60 - 38 = 22. Now what happens when you overlap one family? There is one family which has neither a cat nor a dog. The number of families with a cat or a dog or both reduces by 1 and the number of families with neither increases by 1. The sum is kept constant at 60. The following diagrams should make it clear. Attachment: 
Ques3.jpg [ 6.73 KiB | Viewed 1404 times ]
Attachment: 
Ques4.jpg [ 6.91 KiB | Viewed 1402 times ]
Attachment: 
Ques5.jpg [ 7.11 KiB | Viewed 1403 times ]
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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]
23 Jan 2013, 10:32
38+x+Neither-Both=60 Neither=Both 38+x=60 x=22
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Re: Of the 60 families in a certain neighborhood, 38 have a cat.
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23 Jan 2013, 10:32
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