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When it was found that 150 more tickets for the school play were sold

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Bunuel
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When it was found that 150 more tickets for the school play were sold [#permalink] New post   18 Jul 2018, 00:31
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81% (01:43) correct 19% (02:20) wrong based on 59 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
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Re: When it was found that 150 more tickets for the school play were sold [#permalink] New post   18 Jul 2018, 02:15
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Bunuel 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



Let the capacity of the auditorium be x.

Total tickets sold considering more than capacity = x+150

Now they decide to have 2 shows of equal no. of participants. So,

(x+150) /2

Now , we know that 2/3 of the capacity of the auditorium was full. We can form an equation :

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

x +150 = 4x / 3

3x +450 = 4x

x = 450. ( we assume x is the capacity )

The best answer is E.
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Re: When it was found that 150 more tickets for the school play were sold [#permalink] New post   18 Jul 2018, 00:45
+1 for E.

Let,
N = Total Capacity of Auditorium

Therefore,
( 2/3 )*N + ( 2/3 )*N = 150 + N
N = 450

Hence, E.
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Re: When it was found that 150 more tickets for the school play were sold [#permalink] New post   22 Jul 2018, 18:14
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Bunuel 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: When it was found that 150 more tickets for the school play were sold [#permalink] New post   02 Mar 2020, 07:02
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Bunuel 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 s = the number of seats in the auditorium and t = the number of tickets sold. We can create the equations:

s + 150 = t

and

(2/3)s = (1/2)t

Substituting s + 150 into the second equation for t, we have:

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

2s/3 = s/2 + 75

Multiplying the equation by 6, we have:

4s = 3s + 450

s = 450

Answer: E
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Re: When it was found that 150 more tickets for the school play were sold [#permalink]
02 Mar 2020, 07:02
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