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555-605 Level|   Overlapping Sets|                     
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Bunuel
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P(Total) = 50
P(at least one Dog) = x
P(at least one Cat) = y
P(Both at least one dog and at least one cat) = z
P(Neither) = Households which have neither a dog nor a cat.

P(Total) = P(at least one Dog) + P(at least one Cat) + P(Both at least one dog and at least one cat) + P(Neither)

1) The number of households that have at least one cat and at least one dog is 4 - z = 4
Since we do not have any information about houses with neither, we cannot clearly
tell how many households have at least one cat or at least one dog(but not both). (Insufficient)

2) The number of households that have no cats and no dogs is 14 - P(Neither) = 14
Since we do not have any information about houses with both, we cannot clearly
tell how many households have at least one cat or at least one dog(but not both). (Insufficient)

Combining information from both the statements, 50 = x + y + 4 + 14 | x+y = 32 (Sufficient - Option C)
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As in the attached picture.

Answer should be C.
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IMG_20170703_133141-2.jpg
IMG_20170703_133141-2.jpg [ 52.07 KiB | Viewed 58586 times ]

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Bunuel
Of a group of 50 households, how many have at least one cat or at least one dog, but not both?

(1) The number of households that have at least one cat and at least one dog is 4.
(2) The number of households that have no cats and no dogs is 14.

Total = 50
Households with at least 1 cat OR 1 dog = ?

1) Households with at least 1 cat AND 1 dog = 4
Remaining Households = 46
We cannot determine the number of households with at least 1 cat OR 1 dog.
Insufficient.

2) No Cats and No Dogs = 14
Remaining Households = 36
We cannot determine the number of households with at least 1 cat OR 1 dog.
Insufficient.

1+2)
Total = 50
Households with 1 cat AND 1 dog = 4
Households with NO cats and dogs = 14
=> Households with at least 1 cat OR 1 god = 32

Sufficient. C is the answer.
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pushpitkc
P(Total) = 50
P(at least one Dog) = x
P(at least one Cat) = y
P(Both at least one dog and at least one cat) = z
P(Neither) = Households which have neither a dog nor a cat.

P(Total) = P(at least one Dog) + P(at least one Cat) + P(Both at least one dog and at least one cat) + P(Neither)

1) The number of households that have at least one cat and at least one dog is 4.
z = 4
Since, we do not have any information about houses with neither,
we cannot clearly tell how many households have at at least one cat or at least one dog(but not both). Insufficient.

2) The number of households that have no cats and no dogs is 14.
P(Neither) = 14.
Since, we do not have any information about houses with both,
we cannot clearly tell how many households have at at least one cat or at least one dog(but not both). Insufficient.

Combining information from both the statements,
50 = x +y +4 +14
x+y = 32(Sufficient) (Option C)

I'm not understanding the fact that, isn't the formula for this set problem is total=(a+b-both+neither)? Why are you adding "both" here? Could you please clarify me.
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How to solve this question with a double set matrix ?
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pks02
How to solve this question with a double set matrix ?

waiting for the same.
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Hi sakuac and pks02

Please refer to the attached image I hope it helps.

Posted from my mobile device
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F862B01F-3F62-4163-A022-A0156EB66CB7.jpeg
F862B01F-3F62-4163-A022-A0156EB66CB7.jpeg [ 51.22 KiB | Viewed 51961 times ]

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Salsanousi: How are you getting 32? Because according to your picture -z + z = 0.
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It should be P(Total) = P(at least one Dog) + P(at least one Cat)-P(Both at least one dog and at least one cat) + P(Neither)
Not P(Total) = P(at least one Dog) + P(at least one Cat) + P(Both at least one dog and at least one cat) + P(Neither)

But anyway, you don't need to get an accurate number, Im just saying.
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chetan2u VeritasKarishma
Salsanousi
Hi sakuac and pks02

Please refer to the attached image I hope it helps.

Posted from my mobile device

This is wrong because X + Z + Y + 14 should always be equal to total sum i.e 50

moreover, Z will cancel out and which variable is 32 then ??

I guess the first statement is incorrect, it should rather say

"The number of households that have at least one cat and at least one dog but not both is 4"

because in that scenario

X+Y+Z+14 = 50

we know, X+Y = 4

which gives Z= 32.


please correct me, experts
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Bunuel
Of a group of 50 households, how many have at least one cat or at least one dog, but not both?

(1) The number of households that have at least one cat and at least one dog is 4.
(2) The number of households that have no cats and no dogs is 14.

It is fairly simple.

You have 50 households. No of households with no cats and no dogs = 14.
So 50 - 14 = 36 households have at least one cat or at least one dog or both.

So 36 households lie in the two overlapping circles, some in yellow region (only at least 1 cat), some in blue (only at least 1 dog) and 4 in green (both).

Attachment:
Screenshot 2019-03-31 at 11.50.25.png
Screenshot 2019-03-31 at 11.50.25.png [ 44.06 KiB | Viewed 48488 times ]

Question: "how many have at least one cat or at least one dog, but not both?"
There are 4 households in both region. So number of households in yellow + blue only = 36 - 4 = 32
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Are we not double counting the middle of the diagram? Is it not included when you calculate x or z since those houses are counted? For example, if z = 20 then wouldnt counting the 4 houses that have cats as well as the 4 that have dogs be double counting since we have already included them in the calculation for z? I am reading the equation as this.

P(total) = z+x-y+neither
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Salsanousi
Hi sakuac and pks02

Please refer to the attached image I hope it helps.

Posted from my mobile device

Salsanousi sakuac pks02
This is wrong. The correct one is in the image attached.

The first statement said: The number of households that have at least one cat AND at least one dog is 4.
That means the box where there's both Dog and Cat is 4. Take 50 minus Neither, minus Both, then we have Cats (not dogs) + Dogs (not cats), which is what the question is asking.
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Bunuel
Of a group of 50 households, how many have at least one cat or at least one dog, but not both?

(1) The number of households that have at least one cat and at least one dog is 4.
(2) The number of households that have no cats and no dogs is 14.


This is a easy q.

we need number of houses that have only one animal .
stmnt 1 : says about all houses with both animals. so these are substracted from 50. but it doesnt mention about houses with no animals. so, NS
stmnt 2: mentions houses with no animals. so these are substracted from 50. but doesnt mention about houses with animals.. so, NS

both stmnts, mentions houses with 2 animals, and no animals. so left ones are houses with one animal.

hence, answer is 'C'
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Why is everybody adding both instead of subtracting? Official GMAT also adds both and I'm confused as to why.

I thought the formula was Total = A + B - Both + Neither
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VeritasKarishma Bunuel chetan2u GMATBusters nick1816

ScottTargetTestPrep

Hi Experts!

Hope you all are doing well

Although i am able to understand that we need both "neither" and "both" so as to find what has been asked in this question, I have a doubt..

For point (2) of the question, Isn't "not A and not B" equal to "not(A and B)", which basically means "not(Both A and B)", which definitely isn't the same as "Neither A nor B" ????

Looking forward to hearing from you

Best Regards,
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"not A and not B" means there should not be any A

so only A is also not acceptable

similarly, only B is not acceptable.

Hence "not A and not B" = "not(A or B)"


(when you used "not A and not B" = "not(A and B)", you are accepting only A or only B which is NOT right.

INSEADIESE
VeritasKarishma Bunuel chetan2u GMATBusters nick1816

ScottTargetTestPrep

Hi Experts!

Hope you all are doing well

Although i am able to understand that we need both "neither" and "both" so as to find what has been asked in this question, I have a doubt..

For point (2) of the question, Isn't "not A and not B" equal to "not(A and B)", which basically means "not(Both A and B)", which definitely isn't the same as "Neither A nor B" ????

Looking forward to hearing from you

Best Regards,
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