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A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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 14 Sep 2015, 23:03
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Joined: 08 Jul 2010
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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14 Sep 2015, 23:59
Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. Orchestra Seats - a Balcony Seats - b a+b = 350 and 12a + 8b = 3320 Solving equations simultaneously (Multiply Equation 1 with 8 and subtract from second equation) 4a = 3320 - 8*350 = 3320 - 2800 = 520 i.e. a = 130 and b = 350-130 = 220 More seats in balcony than orchestra = b-a = 220 - 130 = 90 Answer: option A
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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15 Sep 2015, 00:19
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1 statement gives us x+y=350
1 statement gives us 12x+8Y=3320
we can simplify 2nd statement as 3x+2y=830
mulitply 1st statement by 2 (2x+2x=700)
subtracting the two gives us
x=130
therefore y= 350-130=220
x-y= 90
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Director
Joined: 21 May 2013
Posts: 624
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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15 Sep 2015, 02:25
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Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. Orchestra tickets=x, Balcony tickets=350-x (x*$12)+(350-x)*$8=$3320 12x+2800-8x=3320 4x=520 x=130 350-x=220 Difference in tickets=220-130=90 Answer A
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Posts: 103
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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15 Sep 2015, 02:40
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Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. Solution: 12o + 8(350-o) = 3320 ==> 3o + 2(350-o) = 830 ==> o = 130 and b = 220 b - o = 220 - 130 = 90 Option A
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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15 Sep 2015, 07:20
No. of seats in Orchestra = x & No. of seats in Balcony = y x+y = 350 12x+8y = 3320 Solving for x & y x = 130, y = 220 No. of extra seats sold in the balcony = 220-130 = 90 Answer: A
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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17 Sep 2015, 17:05
Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. Let A be the number of seats in orchestra and B be the number of in balcony. Then, 12A + 8B = 3320 .....(1) A + B = 350 .....(2) Solving equations (1) and (2) We get A= 130 and B= 220 B-A = 90 Answer:- A
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Math Expert
Joined: 02 Sep 2009
Posts: 44657
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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20 Sep 2015, 21:35
Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. KAPLAN OFFICIAL SOLUTION:The first step in this problem is to translate our word problem into math. We can write two equations based on the information in the question stem. If we call balcony seats B and orchestra seats R (we want to avoid using the letter O as a variable because it looks like the number 0), we can write one equation based on the number of seats sold and one equation based on the amount of money made. These equations are: R + B = 350 12R + 8B = 3,320 Next, we need to combine these equations to solve for one of the variables. We can rewrite R + B = 350 as B = 350 – R and substitute (350 – B) in for R in our other equation. This gives us 12R + 8(350 – R) = 3,320. From here, we can solve for R as follows: 12R + 8(350 – R) = 3,320 12R + 2800 – 8R = 3,320 4R = 520 R = 130 Next we plug 130 in for R in our initial equation and solve for B: 130 + B = 350 B = 220 Finally, we need to find out how many more balcony seats than orchestra seats we have by subtracting the two results. This gives us 220 – 130 = 90, which is answer choice (A).
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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18 Sep 2016, 23:44
Bunuel wrote: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 350 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 220
Kudos for a correct solution. Can work through answers too option*8<-(balcony) + 10<-(average 12 and 8) * (350-option) = total cost. 90*8 + 10(350-90) = 3320
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in [#permalink]
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04 Dec 2017, 20:51
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Re: A theater charges $12 for seats in the orchestra and $8 for seats in
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