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A theater charges $12 for seats in the orchestra and $8 for seats in
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14 Sep 2015, 23:03
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Question Stats:
81% (01:45) correct 19% (01:50) wrong based on 108 sessions
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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?
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14 Sep 2015, 23:59
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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 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
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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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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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15 Sep 2015, 02:40
1
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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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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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
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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?
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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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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04 Dec 2017, 20:51
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