150 students at seward high school. 66 play baseball, 45 : GMAT Problem Solving (PS) - Page 2
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# 150 students at seward high school. 66 play baseball, 45

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CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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Kudos [?]: 3685 [0], given: 360

Re: PS: high school sports [#permalink]

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16 Jun 2008, 20:46
66+45+42-27-6+x=150 ---> x=30

This equation is constructed by following principle: How we can come from (66+45+42) to 150.

1. we have to subtract 27 as we count it twice in (66+45+42)
2. we have to subtract 2*3 as we count it three times in (66+45+42)
3. we have to add x (none) as we do not count it in (66+45+42)
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Re: PS: high school sports [#permalink]

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24 Oct 2009, 22:22
30 too

150 - x = 150 - [66 + 45 + 42 - N2 - 2(3)]

PS: N2 is subtracted only once because any two pairs are counted exactly twice.
N3 is subtracted twice because all three are counted exactly three times.
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Re: 150 students at seward high school. 66 play baseball, 45 [#permalink]

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13 Nov 2011, 10:04
These are the two formulas required in this ques:

1. Three sets A, B, and C, then
P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

2. Number of people in exactly two of the sets =
P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

Using 2,
27 = P(AnB) + P(AnC) + P(BnC) - 3 (3)

Hence, P(AnB) + P(AnC) + P(BnC) = 36

Put this in 1
P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
or
P(AuBuC) = P(A) + P(B) + P(C) – [P(AnB) + P(AnC) + P(BnC)] + P(AnBnC)
=> P(AuBuC) = 66 + 42 + 45 - 36 + 3 = 120

Hence, NONE of sports is
Total - P(AuBuC) = 150-120 = 30

Re: 150 students at seward high school. 66 play baseball, 45   [#permalink] 13 Nov 2011, 10:04

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