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GRE Math Challenge #111-When it was found that 150 more

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GRE Math Challenge #111-When it was found that 150 more [#permalink] New post 15 May 2015, 11:54
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Question Stats:

100% (01:52) correct 0% (00:00) wrong based on 6 sessions
When it was found that 150 more tickets for the school play were sold than the seating capacity of the auditorium. It was decided to have two performances. if the total number of tickets sold was equal to the total number who attended and if the auditorium was 2/3 full for each of the two performances, what is the seating capacity of the auditorium?

(A) 100
(B) 200
(C) 225
(D) 300
(E) 450
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #111 [#permalink] New post 27 Aug 2015, 23:18
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x = total capacity of auditorium
(2/3)x + (2/3)x =150 + x
x= 450
Ans. 450

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Re: GRE Math Challenge #111-When it was found that 150 more [#permalink] New post 22 Jul 2018, 17:14
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sandy wrote:
When it was found that 150 more tickets for the school play were sold than the seating capacity of the auditorium. It was decided to have two performances. if the total number of tickets sold was equal to the total number who attended and if the auditorium was 2/3 full for each of the two performances, what is the seating capacity of the auditorium?

(A) 100
(B) 200
(C) 225
(D) 300
(E) 450


We can let x = the seating capacity of the auditorium. So (x + 150) tickets were sold. There were 2 performances, each filling the auditorium at ⅔ capacity, so half of the total number of tickets sold is equal to 2/3 of the capacity of auditorium. Therefore, we have:

1/2(x + 150) = 2/3(x)

x/2 + 75 = 2x/3

Multiplying the equation by 6, we have:

3x + 450 = 4x

450 = x

Answer: E
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Re: GRE Math Challenge #111-When it was found that 150 more   [#permalink] 22 Jul 2018, 17:14
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