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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?
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Bunuel
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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Bunuel
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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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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Bunuel
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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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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Video solution from Quant Reasoning starts at 29:52
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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­Lots of ways to simplify once you have the system of equations up and running. Here's what I did:

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