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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s

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Bunuel
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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  22 May 2020, 10:21
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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


PS20373
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BrentGMATPrepNow
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  22 May 2020, 10:33
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Bunuel 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


PS20373


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

Cheers,
Brent
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NitishJain
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  24 May 2020, 03:02
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A.

Total Tickets at $1.80 = 1/3 * 600 = 200
Total cost for 200 tickets at $1.80 = $360

Remaining count of tickets sold at $1.20 & $2.40 = 400
Let total count of tickets sold at $1.20= x
total count of tickets sold at $2.40= 400-x

Total cost for tickets sold at $1.20 = 1.20x
Total cost for tickets sold at $2.40 = 2.40(400-x)

As per question: 1.20x + 360 + 2.40 (400-x) = 1020

Solving for x, x = 250

Tickets at $2.40 = 400-250 = 150
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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  Updated on: 23 Jul 2020, 08:41
VeritasKarishma i tried to do this with weights
avg cost per ticket after accounting for 200 tickets sold @ 1.80 is 660/400 = 1.4
1.4-1.2 = 0.2
2.4-1.4= 1
ratio of tickets sold 1.20 and 2.40 = 2/10
tickets sold at 2.40 = 2/12*660 =110

could you help me with where i am faltering?

Originally posted by Kritisood on 04 Jun 2020, 03:19.
Last edited by Kritisood on 23 Jul 2020, 08:41, edited 1 time in total.
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Abhishek009
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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  04 Jun 2020, 09:35
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Bunuel 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


PS20373


\(200*1.80 + 1.20*a + 2.40*b = 1020\)

Or, 1.20a + 2.40b = 660

Or, \(a + 2b = 550\)

Further we know , \(a + b = 400\)

Now substitute values \(400 - b + 2b = 550\)

So, \(b = 150\) , thus \(a = 250\)

Thus, Answer must be (A) 150
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  25 Jul 2020, 14:21
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Bunuel 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


PS20373


Solution:

We see that the revenue generated by the $1.80 seats is 1/3 x 600 x 1.8 = 200 x 1.8 = $360. Therefore, the remaining 400 seats generate a revenue of 1,020 - 360 = $660. Now, if we let m = the number of $1.20 seats sold and n = the number of $2.40 seats sold, we can create the equation:

m + n = 400

and

1.2m + 2.4n = 660

We see that we need to solve for n. Since m = 400 - n, we have:

1.2(400 - n) + 2.4n = 660

480 - 1.2n + 2.4n = 660

1.2n = 180

n = 150

Answer: A
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  13 Oct 2020, 12:49
Let x,y,z be the number of seats costing $1.2,$1.8 and $2.4 respectively.
Given:
=> 1.2x+1.8y+2.4z = 1020
or 2x+3y+4z= 1700-----------1)
Given:
x+y+z = 600-------------------2)
Given:
y = (x+y+z)/3
=> 2y = x+z -------------------3)
Using 2 and 3,
y = 200
using 1,
2(x+y+z) +y+2z = 1700
=>2*600+200+2z = 1700
=> z = 150
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A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  13 Oct 2020, 18:23
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Bunuel 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


PS20373


average price of ticket except the $1.8 ones = (1020-360)/400 = 660/400 = 66/40 = 33/20 = 1.65$
Applying allegation method:
1.20 ||||||||| 2.4
1.65
0.75 |||||||| 0.45
15 |||||||| 9
5 ||||||||||| 3

So, 2.40$ ticket sold were = 400*3/8 = 150

A is the answer.
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  27 Nov 2020, 13:15
A theater sells 600 tickets at $1.20, $1.80, or $2.40 per seat.

1/3 of the tickets were sold at $1.80 = 200 tickets * 1.80 = 360

We have 400 tickets totaling $660 unaccounted for.

Let \(400 - x\) = tickets sold at $1.20
let \(x\) = tickets sold at $2.40

\(1.20(400-x) + 2.40x = 660\)

\(480-1.2x + 2.40x = 660\)

\(480 + 1.2x = 660\)

\(1.2x = 180\)

\(x = 150\)

Answer is A.
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Re: A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per s [#permalink] New post  13 Mar 2021, 18:40
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