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If the 4,500 participants in the school election, 1/3 voted for candid

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Math Expert
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If the 4,500 participants in the school election, 1/3 voted for candid  [#permalink]

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New post 10 Jan 2018, 05:46
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

85% (01:33) correct 15% (02:42) wrong based on 43 sessions

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Re: If the 4,500 participants in the school election, 1/3 voted for candid  [#permalink]

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New post 10 Jan 2018, 06:15
4500*1/3=1500

1500*2/5 = 600

Candidate voted for x alone = 1500-600 =900

900/4500 =20%

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Re: If the 4,500 participants in the school election, 1/3 voted for candid  [#permalink]

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New post 10 Jan 2018, 17:25
Total participants = 4,500
1/3 of the total participants voted for candidate X = 4,500*(1/3) = 1,500 participants voted for candidate X
If 2/5 of the participants that voted for candidate X also voted for candidate Y, then 1 - (2/5) = 3/5 of the participants voted for candidate X but did not vote for candidate Y.

(3/5)*1,500 = 900 participants voted for candidate X, but not candidate Y
Hence 900/4500 = 0.2 = 20% (Option A)
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Re: If the 4,500 participants in the school election, 1/3 voted for candid  [#permalink]

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New post 16 Jan 2018, 17:38
Bunuel wrote:
If the 4,500 participants in the school election, 1/3 voted for candidate X and of those that voted for candidate X, 2/5 also voted for candidate Y. What percent of the participants in the election voted for candidate X but not for candidate Y?

A. 20%
B. 18%
C. 15%
D. 9%
E. 5%


We see that 4500 x 1/3 = 1500 voted for candidate X, and 1500 x 2/5 = 600 voted for both candidates.

Thus, 1500 - 600 = 900, or 900/4500 = 20%, voted for only candidate X.

Answer: A
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Re: If the 4,500 participants in the school election, 1/3 voted for candid   [#permalink] 16 Jan 2018, 17:38
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