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# In a survey, 20% favored candidate A, 30% favored candidate

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Senior Manager
Joined: 30 May 2005
Posts: 373

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

In a survey, 20% favored candidate A, 30% favored candidate [#permalink]

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02 Jul 2005, 07:55
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In a survey, 20% favored candidate A, 30% favored candidate B and 50% favored candidate C. Exactly 5% favored all candidates

What percentage of voters favored atleast two candidates?

S1. 80% of those asked were in favor of atleast one of the candidates A, B and C

S2. 17% of those asked favored both Candidate A and Candidate B

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

Director
Joined: 18 Apr 2005
Posts: 543

Kudos [?]: 37 [0], given: 0

Location: Canuckland

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02 Jul 2005, 12:44
AJ, I see you are getting yourself up to speed with those set problems - good stuff. Knowing how to solve this kind of problems is very important for getting a good math score.

AUBUC + ~(AUBUC) = A + B + C - AnB - AnC -BnC + AnBnC + ~(AUBUC)

we are looking for AnB + AnC + BnC + AnBnC

so S1 gives us ~(AUBUC) = 100 -80 = 20% (didn't favor anyone)
80 = 20 + 30 + 50 - (AnB + AnC + BnC) + 5

so (AnB + AnC + BnC) = 25
and AnB + AnC + BnC + AnBnC = 25 +5 = 30% sufficient

S2 is insuff, we don't know ~(AUBUC)

Kudos [?]: 37 [0], given: 0

Senior Manager
Joined: 30 May 2005
Posts: 373

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

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02 Jul 2005, 14:03
Sparky,

Thanks. Yes, I've sort of learned the concepts and I'm trying to find some tough problems to solve. It'll take me constant practice, I think.

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

Director
Joined: 03 Nov 2004
Posts: 850

Kudos [?]: 57 [0], given: 0

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02 Jul 2005, 17:15
Sparky, Thx for the formula - made me remember something that I learnt decades ago.

Kudos [?]: 57 [0], given: 0

Senior Manager
Joined: 17 Apr 2005
Posts: 372

Kudos [?]: 30 [0], given: 0

Location: India

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03 Jul 2005, 07:20
sparky wrote:

we are looking for AnB + AnC + BnC + AnBnC

sparky, not that It makes a difference to the answer the question, but I think we need to find AnB + AnC + BnC -2(AnBnC).

HMTG.

Kudos [?]: 30 [0], given: 0

Director
Joined: 18 Apr 2005
Posts: 543

Kudos [?]: 37 [0], given: 0

Location: Canuckland

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03 Jul 2005, 11:07
HowManyToGo wrote:
sparky wrote:

we are looking for AnB + AnC + BnC + AnBnC

sparky, not that It makes a difference to the answer the question, but I think we need to find AnB + AnC + BnC -2(AnBnC).

HMTG.

you are right, we are looking for
AnB~C + AnC~B + BnC~A + ABC

Kudos [?]: 37 [0], given: 0

03 Jul 2005, 11:07
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