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Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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05 Feb 2012, 15:50

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B

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Question Stats:

40% (02:19) correct
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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.

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.

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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05 Feb 2012, 20:24

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.

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.

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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06 Feb 2012, 01:02

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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].

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 6346 times ]

(1) 28 of the families in this neighborhood have a cat but not a dog.

Attachment:

Statement 1.PNG [ 2.68 KiB | Viewed 6342 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 6348 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.

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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06 Feb 2012, 02:28

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.

The solution given by Bunuel & 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
_________________

Mike McGarry Magoosh Test Prep

Last edited by mikemcgarry on 06 Oct 2013, 12:57, edited 1 time in total.

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.

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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04 Oct 2013, 20:09

mikemcgarry wrote:

My apologies to kys123

The 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

Its strange that a tutor got it wrong and lot of students have got it right

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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25 Sep 2014, 07:41

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Bunuel wrote:

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

(1) 28 of the families in this neighborhood have a cat but not a dog.

Attachment:

Statement 1.PNG

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

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.

If we use Venn Diagram it's a lot faster and space saver. Even though, because of timer, I misread question and thought 'the examiner asked about families with just a dog' and marked wrong answer

Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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08 Oct 2015, 07:04

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Re: Of the 60 families in a certain neighborhood, 38 have a cat. [#permalink]

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12 Nov 2015, 10:20

indeed Venn + formula solve Statement 2 in 15 secs - very straightforward. on the overlapping sets very often matrix is the best approach however this question is the case when Venn is the shortest solution
_________________

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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.

We get a '2by2' table as below:

Attachment:

GCDS calreg11 Of the 60 families in a certain neighborhood (20151113).jpg [ 29.17 KiB | Viewed 1439 times ]

The question asks a+c=? There are 4 variables (a,b,c,d) and 2 equations (a+b=38, a+b+c+d=60). 2 more equations are given by the 2 conditions, so there is high chance (C) will be our answer. Condition 1) a=28 Condition 2) a=d. question is asking for the same thing as whether a+c=d+c? so we get a+b+c+d=60, 38+c+d=60, c+d=22 from the original condition, and the answer becomes (B).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

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.

This is a “2-by-2” question, one of the most common types of questions in GMAT math. From modifying the original condition and the question, we can obtain a table below.

From the table, we can see that the question is a+b=? There are 4 variables (a,b,c, and d) and 2 equations (a+b+c+d=60 and a+c=38). In order to match the number of variables and the number of equations, we need 2 more equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is the answer.

Using both the condition 1) and 2), we get c=28. Also, since a=d, we get a+c=d+c=a+28=d+28=38. Then, a=d=10. So, if we substitute in a+b+c+d=60, we get 10+b+28+10=60. Then, b=12. So, a+b=10+12=22. So, the answer is unique and the condition is sufficient. So we can see how C could be the answer.

However, a question involving hidden integer is one of key questions (integer, statistics, inequality, probability, absolute value) and we have to consider Mistake Type 4(A).

In the case of the condition 2), since a=d, we get a+c=d+c=38. Then, if we substitute it into a+b+c+d=60, we get a+b+38=60. So, a+b=22. The answer is unique and the condition is sufficient. If both C and B are the answers to the question, the final correct answer choice is B.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Attachments

GCDS calreg11 Of the 60 families in a certain (20160103).jpg [ 23.73 KiB | Viewed 1184 times ]

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