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A theater with 600 seats sells tickets at $1.20, $1.80, or $

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A theater with 600 seats sells tickets at $1.20, $1.80, or $ [#permalink] New post 22 Jun 2020, 09:36
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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per seat. On Wednesday evening, 1/3 of the tickets sold were at $1.80 per seat and the total receipts from the sale of 600 tickets was $1,020. How many of the tickets sold were at $2.40 per seat?

A. 150
B. 160
C. 200
D. 250
E. 300
[Reveal] Spoiler: OA

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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $ [#permalink] New post 22 Jun 2020, 09:36
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $ [#permalink] New post 22 Jun 2020, 09:57
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Carcass wrote:
A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per seat. On Wednesday evening, 1/3 of the tickets sold were at $1.80 per seat and the total receipts from the sale of 600 tickets was $1,020. How many of the tickets sold were at $2.40 per seat?

A. 150
B. 160
C. 200
D. 250
E. 300

Given: 600 tickets were sold
One third of those tickets were at $1.80 per seat

1/3 of 600 = 200
So, 200 tickets sold for $1.80 per seat
And the remaining 400 tickets were sold for either $1.20 each of $2.40 each

Let x = the NUMBER of tickets sold for $2.40
So, 400 - x = the NUMBER of tickets sold for $1.20


The total receipts from the sale of 600 tickets was $1,020
In other words: (receipts from the $1.20 tickets) + (receipts from the $1.80 tickets) + (receipts from the $2.40 tickets) = $1,020

(200)($1.80) = $360
So, $360 = the total receipts from the $1.80 tickets

Likewise, (400 - x)($1.20) = the total receipts from the $1.20 tickets
And (x)($2.40) = the total receipts from the $2.40 tickets

Substitute values into our "word equation" to get: (400 - x)($1.20) + $360 + (x)($2.40) = $1020
Expand to get: 480 - 1.20x + 360 + 2.40x = 1020
Simplify to get: 840 + 1.20x = 1020
Subtract 840 from both sides: 1.20x = 180
Solve: x = 180/1.20 = 150

Answer: A

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Brent
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $ [#permalink] New post 26 Jun 2020, 12:25
first thing I do is to multiple all the prices and revenue by 10. it avoids having to work with decimal numbers.

let's be A,B and C the numbers of seats for the tickets costing respectively 12$, 18$ and 24$
we have 12A + 18B + 24C = 10,200
and A + B + C = 600

we know that 1/3 of the tickets sold were at 18$, so 200 seats for a revenue of 200*18$ = 3600
we can replace in our equations

12A + (200*18) + 24C = 10,200
12A + 3,600 + 24C = 10,200
12A + 24C = 10,200 - 3,600
12A + 24C = 6,600

and A + B + C = 600
A + C = 400 therefore A = 400 - C
we can replace in the first equation
12(400 - C) + 24C = 6,600
4,800 - 12C + 24C = 6,600
12C = 1800
C = 150

answer is A
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $   [#permalink] 26 Jun 2020, 12:25
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