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

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Manager
Joined: 17 Dec 2005
Posts: 162
Followers: 1

Kudos [?]: 12 [0], given: 0

Of the 75 houses in a certain community, 48 have a patio. [#permalink]  16 Jun 2006, 21:48
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Of the 75 houses in a certain community, 48 have a patio. How many of the house in the community have a swimming pool?

C1. 38 of the houses in the community have a patio but do not have a swimming pool.

C2. 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 swimminng pool or a patio.
Director
Joined: 06 May 2006
Posts: 780
Followers: 3

Kudos [?]: 17 [0], given: 0

B

#1. We get -

75 = 48 + (swimming pools) - 10 (both patio/swimming pool) + (no patio/swimming pool)
Not sufficient.

#2. We get -

75 = 48 + (swimming pools) - (both patio/swimming pool) + (no patio/swimming pool)
=> 75 = 48 + (swimming pools); since (both patio/swimming pool) = (no patio/swimming pool)
=> swimming pools = 27
Manager
Joined: 17 Dec 2005
Posts: 162
Followers: 1

Kudos [?]: 12 [0], given: 0

B

#1. We get -

75 = 48 + (swimming pools) - 10 (both patio/swimming pool) + (no patio/swimming pool)
Not sufficient.

#2. We get -

75 = 48 + (swimming pools) - (both patio/swimming pool) + (no patio/swimming pool)
=> 75 = 48 + (swimming pools); since (both patio/swimming pool) = (no patio/swimming pool)=> swimming pools = 27

a = only patio
b = both
c = pool
x = neither

Given:
75 = a + c + x - b
a + b = 48

C1. a = 38, b = 10
75 = 48 - 10 + c - x

C2. b = x
75 = a + c
a + b = 48

so we they say that 48 houses have a patio, it can be represented as a = 48 not a + b = 48.
Director
Joined: 06 May 2006
Posts: 780
Followers: 3

Kudos [?]: 17 [0], given: 0

Nope... a = 48 (Set theory; Union formula)

a = all houses that have patio
b = houses that have both
c = all houses that have swimming pool
x = houses that have neither
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