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GRE Math Challenge #111-When it was found that 150 more
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15 May 2015, 11:54
15 May 2015, 11:54
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Question Stats:
80% (01:30) correct
19% (01:56) wrong based on 31 sessions
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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
(A) 100
(B) 200
(C) 225
(D) 300
(E) 450
Re: GRE Math Challenge #111
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27 Aug 2015, 23:18
27 Aug 2015, 23:18
1
x = total capacity of auditorium
(2/3)x + (2/3)x =150 + x
x= 450
Ans. 450
Posted from my mobile device
(2/3)x + (2/3)x =150 + x
x= 450
Ans. 450
Posted from my mobile device
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Re: GRE Math Challenge #111-When it was found that 150 more
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22 Jul 2018, 17:14
22 Jul 2018, 17:14
1
Expert Reply
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
(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
