Cats and dogs

[DanaJ]

Source: Veritas Prep

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.

Experts: only Veritas Prep experts, please!

[rishab1988]

The answer should be B

Let C represent Cats only,
D represent dogs only,
x represent both dogs and cats,and
N represent neither dogs nor cats

Now from prompt:

C+D+x+N=60 [total families]

C+x=38 [only cats +cats and dogs]

D+N=22

The question asks D+x [be careful.It doesn't say dogs only]

We have 2 equations:

D+N=22
C+x=38

We need N and C to find D+x

1) C=28 [cat but not a dog is same as saying only cat]

x=10

Then what? we have no info about N

N=2 D=20 D+x=30
N=4 D=18 D+x=28

2 values insufficient

2) x [Dog and a cat]=N[neither cat nor dog]

D+x=22 [substituted x for N]

D+x is same as saying Dogs in the neighborhood.

Sufficient

What is the OA?

[tomada]

When you reference 28 families as having only cats, aren't you using Statement 1? Wouldn't that mean the answer is C, since you're using both Statements?

The answer should be B

Let C represent Cats only,
D represent dogs only,
x represent both dogs and cats,and
N represent neither dogs nor cats

Now from prompt:

C+D+x+N=60 [total families]

C+x=38 [only cats +cats and dogs]

D+N=22

The question asks D+x [be careful.It doesn't say dogs only]

We have 2 equations:

D+N=22
C+x=38

We need N and C to find D+x

1) C=28 [cat but not a dog is same as saying only cat]

x=10

Then what? we have no info about N

N=2 D=20 D+x=30
N=4 D=18 D+x=28

2 values insufficient

2) x [Dog and a cat]=N[neither cat nor dog]

D+x=22 [substituted x for N]

D+x is same as saying Dogs in the neighborhood.

Sufficient

What is the OA?

[rishab1988]

When you reference 28 families as having only cats, aren't you using Statement 1? Wouldn't that mean the answer is C, since you're using both Statements?

The answer should be B

Let C represent Cats only,
D represent dogs only,
x represent both dogs and cats,and
N represent neither dogs nor cats

Now from prompt:

C+D+x+N=60 [total families]

C+x=38 [only cats +cats and dogs]

D+N=22

The question asks D+x [be careful.It doesn't say dogs only]

We have 2 equations:

D+N=22
C+x=38

We need N and C to find D+x or Just D+x as expression. [This is the new question]

1) C=28 [cat but not a dog is same as saying only cat]

x=10

Then what? we have no info about N

N=2 D=20 D+x=30
N=4 D=18 D+x=28

2 values insufficient

2) x [Dog and a cat]=N[neither cat nor dog]

D+x=22 [substituted x for N]

D+x is same as saying Dogs in the neighborhood.

Sufficient

What is the OA?

Where did I make a reference to 28 in 2?

The equations that I used in 2 are in bold.From the bold part I know D+N=22.Statement 2 says N=x.I substituted x for N and got D+x=22.Got it.Statement 1 is in no way involved!

[gmatmachoman]

For people who like Boxes!!

St 2 is sufficient!!

[tomada]

I never said you were using 28 in 2 (whatever that means). I just said you were using that information to solve the problem.

Here's a statement in your solution:

1) C=28 [cat but not a dog is same as saying only cat]

x=10



How do you arrive at C=28 without using the first statement given?

When you reference 28 families as having only cats, aren't you using Statement 1? Wouldn't that mean the answer is C, since you're using both Statements?

The answer should be B

Let C represent Cats only,
D represent dogs only,
x represent both dogs and cats,and
N represent neither dogs nor cats

Now from prompt:

C+D+x+N=60 [total families]

C+x=38 [only cats +cats and dogs]

D+N=22

The question asks D+x [be careful.It doesn't say dogs only]

We have 2 equations:

D+N=22
C+x=38

We need N and C to find D+x or Just D+x as expression. [This is the new question]

1) C=28 [cat but not a dog is same as saying only cat]

x=10

Then what? we have no info about N

N=2 D=20 D+x=30
N=4 D=18 D+x=28

2 values insufficient

2) x [Dog and a cat]=N[neither cat nor dog]

D+x=22 [substituted x for N]

D+x is same as saying Dogs in the neighborhood.

Sufficient

What is the OA?

Where did I make a reference to 28 in 2?

The equations that I used in 2 are in bold.From the bold part I know D+N=22.Statement 2 says N=x.I substituted x for N and got D+x=22.Got it.Statement 1 is in no way involved!

[rishab1988]

I never said you were using 28 in 2 (whatever that means). I just said you were using that information to solve the problem.

Here's a statement in your solution:

1) C=28 [cat but not a dog is same as saying only cat]

x=10



How do you arrive at C=28 without using the first statement given?

When you reference 28 families as having only cats, aren't you using Statement 1? Wouldn't that mean the answer is C, since you're using both Statements?

The answer should be B

Let C represent Cats only,
D represent dogs only,
x represent both dogs and cats,and
N represent neither dogs nor cats

Now from prompt:

C+D+x+N=60 [total families]

C+x=38 [only cats +cats and dogs]

D+N=22

The question asks D+x [be careful.It doesn't say dogs only]

We have 2 equations:

D+N=22
C+x=38

We need N and C to find D+x or Just D+x as expression. [This is the new question]

1) C=28 [cat but not a dog is same as saying only cat]

x=10

Then what? we have no info about N

N=2 D=20 D+x=30
N=4 D=18 D+x=28

2 values insufficient

2) x [Dog and a cat]=N[neither cat nor dog]

D+x=22 [substituted x for N]

D+x is same as saying Dogs in the neighborhood.

Sufficient

What is the OA?

Where did I make a reference to 28 in 2?

The equations that I used in 2 are in bold.From the bold part I know D+N=22.Statement 2 says N=x.I substituted x for N and got D+x=22.Got it.Statement 1 is in no way involved!

If you are talking about statement 1.

Yes I used the equation C+x =38 from the prompt and x=10 from statement 1 to arrive at C=28.However,C=28 does not give me the value of D+x because I do not know what D is! Therefore statement 1 is insufficient.

Eliminate answers A and D [A=statement 1 only D= each statement is sufficient alone]

We are left with B,C,and E

When I use statement 2,I will totally forget of statement 1.Since ,after forgetting statement 1,I solved for value of D+x in statement 2.Statement 2 alone is sufficient.This corresponds to answer choice B.

I never said you were using 28 in 2 (whatever that means). I just said you were using that information to solve the problem.

Think carefully.You are contradicting yourself.You said I didn't use this information in statement 2.I said I can solve the problem using statement 2 only,then ,how can you conclude that I used this information to solve the problem??? Got it?

[Night reader]

Source: Veritas Prep

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.

Experts: only Veritas Prep experts, please!

Question: Families with Dogs Only-? Families with Dogs and Cats-?
Answer: C Families with Dogs Only=12; Families with Dogs=22

Solution-
Families=60 Families with Cats=38 Families No Cat=22

Statement (1) Families with Cats Only=28 <> Dogs and Cats=10 (38-28) <> Remaining Families=22 <> How many Dogs with 22 Families-? Decision-Not sufficient;

Statement (2) Families with Dogs and Cats=Families No Dog & Cat <> Families with Cats=38 (-any Dogs ?) Families No Cat=22 (-any Dog ?) Decision-Not sufficient;

Combining Statements (1) & (2)

Statement (2) Families with Dogs and Cats=Families No Dog & Cat <> Families with Cats=38 (-any Dogs ? Yes, 10 Dogs and Cats per Statement (1))

Families with Dogs and Cats=Families No Dog & Cat =10

Families 60 - Cats Only 28 - No Dog & Cat 10 = Remains 22
Remains 22 - Dogs 10 = Still Remains 12
Still Remains 12 = Families with Either Cat or Dog; Families with Cats qualified to 38 (exhausted), therefore Families with Dogs=12

[rishab1988]

Source: Veritas Prep

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.

Experts: only Veritas Prep experts, please!

Question: Families with Dogs Only-? Families with Dogs and Cats-?
Answer: C Families with Dogs Only=12; Families with Dogs=22

Solution-
Families=60 Families with Cats=38 Families No Cat=22

Statement (1) Families with Cats Only=28 <> Dogs and Cats=10 (38-28) <> Remaining Families=22 <> How many Dogs with 22 Families-? Decision-Not sufficient;

Statement (2) Families with Dogs and Cats=Families No Dog & Cat <> Families with Cats=38 (-any Dogs ?) Families No Cat=22 (-any Dog ?) Decision-Not sufficient;

Combining Statements (1) & (2)

Statement (2) Families with Dogs and Cats=Families No Dog & Cat <> Families with Cats=38 (-any Dogs ? Yes, 10 Dogs and Cats per Statement (1))

Families with Dogs and Cats=Families No Dog & Cat =10

Families 60 - Cats Only 28 - No Dog & Cat 10 = Remains 22
Remains 22 - Dogs 10 = Still Remains 12
Still Remains 12 = Families with Either Cat or Dog; Families with Cats qualified to 38 (exhausted), therefore Families with Dogs=12

Does the question in anyway state dogs only?NO

eg D [dogs] ={a,b,c,d} [families that have dogs =4]

If I also told you that

C[cats]={a,e,f} [families with cats =3}

Then the families who own dogs will stop owning dogs??? No

[Night reader]

Rishab with 10 dogs you don't know for sure how many of them left there- Statements (1) and (2) alone are insufficient.

[rishab1988]

your logic is flawed.

First GMAT questions are never ambiguous.

Second,according to you,if the same question were rephrased as only dogs the answer would be same?

do you understand the difference between "how many have dogs?" vs "how many have only dogs?"

Btw the answer is B [checked on other forums]

[Night reader]

your logic is flawed.

First GMAT questions are never ambiguous.

Second,according to you,if the same question were rephrased as only dogs the answer would be same?

do you understand the difference between "how many have dogs?" vs "how many have only dogs?"

Btw the answer is B [checked on other forums]

Rishab: read the Statement (2) again and please think

(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.

How you know how many paired Dogs and Cats are there? the answer is sitting in statement (1); hence you can not switch off 22 and deduce that 22 contains only dogs. 22 may well contain no dog too - you need statement (1).

[rishab1988]

your logic is flawed.

First GMAT questions are never ambiguous.

Second,according to you,if the same question were rephrased as only dogs the answer would be same?

do you understand the difference between "how many have dogs?" vs "how many have only dogs?"

Btw the answer is B [checked on other forums]

Rishab: read the Statement (2) again and please think

(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.

How you know how many paired Dogs and Cats are there? the answer is sitting in statement (1); hence you can not switch off 22 and deduce that 22 contains only dogs. 22 may well contain no dog too - you need statement (1).

Other forums are not better...

Why do I need x when I can solve for the whole expression D+x.I need D+x and not x!

See the prompt itself gave D+N=22 [C+x+D+N=60].I know C+x=38 [because the prompt says cats and NOT only cats]
Statement 2 says N=x.I substitute x for N and get D+x=22 [no of dogs.This is not equal to ONLY dogs].For solving for ONLY dogs I need x.But the prompt asks for DOGS and NOT only DOGS!

Using 1[x=10] ,you are solving for ONLY dogs[Only dogs=12],but did the question ask for solving ONLY dogs? No

Besides I meant to say that the OA posted on other forums is B,when I said the answer on other forums is B!

Here is another D.S question for you;

What is x=?

X^2-y^2=9

1) y=0
2) x-y=9

What would your answer be?

I would say 2 alone is sufficient.You are saying We need y too.This way you can find x!

But I say[from 2] x-y=9.From prompt [x^2-y^2=9] x+y=1; 2x=10 x=5

You are saying "this is not possible'.We use y=0 from 1 and substitute in 1 to get x=9!.Therefore C

[David@VeritasPrep]

OA is B.

The explanations above are correct. For those who need a little slower paced explanation as to why B works, the following gives some more detail.

Correct answer: (B)

Solution: On any Venn diagram problem it is helpful to draw out the two sets and also note the formula Total = Set 1 + Set 2 - Both + Neither. The question stem tells us that the total number of families is 60 and the total number of families with a cat is 38. The question is asking whether we can determine the total number of families who have a dog.

Statement (1) tells us that 28 families have a cat but not a dog. If there are 38 families in total that have a cat, this information tells you that 10 families (38-28) must have both a dog and a cat. It does not, however, give you any information about the number of families who have only a dog or neither a dog nor a cat. It is therefore impossible to determine how many total families have a dog. Note that because the question stem does not say that all the familes own a cat, a dog, or both, some households could own neither.

Statement (2) tells us that the number of families who have both a cat and a dog is the same as the number who have neither. This information does not appear to be sufficient at first glance but a closer look at the Venn diagram formula for two set problems shows that it is. Remember our formula from above: Total Familes = # of Families with a Cat (C) + #of Families with a Dog (D) - Families with Both (B) + Families with Neither (N). Or, with variables, T = C + D - B + N. Plug in the information from the question stem and Statement (2) to find that T = 60, C = 38, and B=N. We can then solve for D using the formula. 60 = 38 + D - B +B. D = 22. Statement (2) alone is sufficient.

So it was actually helpful that the information from the question stem gave the total families with cats and the question was asking about the total number of families with dogs. Because we are working with totals, the Both and the Neither Categories cancel out given the information in statement 2 that these two categories are equal.

Nice work guys!

[Rahul@gurome]

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.

Though the problem had been answered correctly, I received two PMs two look into it. Here is another approach. (This is a more set theoretic approach as an algebraic approach is already there.)

Given: Number of families in a neighborhood = 60.
Say, number of families with cat = C and number of families with dog = D.
Thus number of families with both cat and dog = (C ∩ D)
Number of families with either cat or dog = (C U D) = C + D - (C ∩ D)
Number of families with neither cats nor dogs = 60 - (C U D)

Now, C = 38

Statement 1: 28 of the families in this neighborhood have a cat but not a dog.
Number of families with cat but not dog = C - (C ∩ D) = 28
Thus, (C ∩ D) = C - 28 = 38 - 28 = 10
From this we cannot find D.

Not sufficient.

Statement 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.

Implies (C ∩ D) = 60 - (C U D)
=> (C ∩ D) + (C U D) = 60
=> (C ∩ D) + C + D - (C ∩ D) = 60
=> C + D = 60
=> D = 60 - C = 60 - 38 = 22

Sufficient.

The correct answer is B.