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# Of the 75 houses in a certain community, 48 have a patio.

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Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
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Kudos [?]: 441 [1] , given: 1

Of the 75 houses in a certain community, 48 have a patio. [#permalink]  30 Mar 2008, 02:58
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Of the 75 houses in a certain community, 48 have a patio. How many of houses in the communit have a swimming pool?

1. 38 of the house in the community have a patio but do not have swimming pool
2. the number of houses in the community that have a patio and a swimming pool is equal to the number of houses in the community that have neither a swimming pool nor a patio.

Friend,
Could you help me to figure out it? I am confused btw the set overlapping using equation: Total = set 1 + set 2... to solve and the set using table.

Many thanks!
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CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [0], given: 359

Re: DS- A Lone Wolf Trap [#permalink]  30 Mar 2008, 05:02
Expert's post
B

+1 for a good question!
When I wanted to pick "C" here I said "wait a minute". Here, we have DS-pattern that is very similar to C-trap: the fist condition is obviously insufficient and help the second condition, and the second seems to be insufficient along.
So, I return to the question that says: How many of houses in the community have a swimming pool? (regardless having a patio).
If we read close the second condition we can reveal that the change of the number of houses in the community that have a patio and a swimming pool actually does not influence on the number of houses with a swimming pool.

Now, the same idea by formulas:

N=75 - the total number of houses
Np=48 - the number of houses with a patio.
Nps -the number of houses with both a patio and a swimming pool.
N(p)(s) -the number of houses without a patio or a swimming pool.
Ns - the number of houses with a swimming pool.

Ns=Nps + (75-48-N(p)(s)) = Nps + 27 - Nps = 27
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SVP
Joined: 28 Dec 2005
Posts: 1583
Followers: 2

Kudos [?]: 88 [0], given: 2

Re: DS- A Lone Wolf Trap [#permalink]  30 Mar 2008, 05:32
i was using a double matrix, fell for the trap and picked C

Now I see why B is good enough. I could have let x represent people with patio and pool, and 27-x represent people with pool but no patio, so my total would have been 27-x+x = 27
SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19

Kudos [?]: 441 [0], given: 1

Re: DS- A Lone Wolf Trap [#permalink]  30 Mar 2008, 19:06
walker wrote:
B

+1 for a good question!
When I wanted to pick "C" here I said "wait a minute". Here, we have DS-pattern that is very similar to C-trap: the fist condition is obviously insufficient and help the second condition, and the second seems to be insufficient along.
So, I return to the question that says: How many of houses in the community have a swimming pool? (regardless having a patio).
If we read close the second condition we can reveal that the change of the number of houses in the community that have a patio and a swimming pool actually does not influence on the number of houses with a swimming pool.

Now, the same idea by formulas:

N=75 - the total number of houses
Np=48 - the number of houses with a patio.
Nps -the number of houses with both a patio and a swimming pool.
N(p)(s) -the number of houses without a patio or a swimming pool.
Ns - the number of houses with a swimming pool.

Ns=Nps + (75-48-N(p)(s)) = Nps + 27 - Nps = 27

Hi Walker and Pmenon,

By reading your post, I found my problem not reading closely and not standing time pressure well. Yesterday, I chose E!

Regards friends,
_________________
Re: DS- A Lone Wolf Trap   [#permalink] 30 Mar 2008, 19:06
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# Of the 75 houses in a certain community, 48 have a patio.

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