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

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Manager

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

16 Jun 2006, 22:48

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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.

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16 Jun 2006, 22:59

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

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17 Jun 2006, 17:15

paddyboy wrote:

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

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17 Jun 2006, 22:21

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

[#permalink] 17 Jun 2006, 22:21

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

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

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

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