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13 Jul 2020, 04:13
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B
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Difficulty:
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Question Stats:
73% (03:07) correct
27%
(02:58)
wrong
based on 48
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At a Van Halen concert with $2,275,000 in total ticket sales, the number of tickets sold in Section A was 100 less than three times the number of tickets sold for Section B, and the number of tickets sold in Section C was half the number of tickets sold in Section B. If Section B tickets cost $300, Section A tickets cost $250 and Section C ticket cost $200, how many tickets were sold in Section C?
A. 500
B. 1000
C. 1500
D. 2000
E. 3000
A. 500
B. 1000
C. 1500
D. 2000
E. 3000
13 Jul 2020, 04:21
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Bunuel
Tickets solt in B = 2x
Tickets sold in C = x
Tickets sold in A = 3*(2x)-100
Cost = 250*[3*(2x)-100] + 300*2x + 200*X = 2275000
i.e. 1500x + 600x + 200x - 25000 = 2275000
x = 1000
Answer Option B
13 Jul 2020, 04:28
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Let the number of tickets sold for section B be equal to B. Therefore total revenue earned from section B would be= $300B
Since number of tickets sold in section A is hundred less than three times B, therefore number of tickets for section A= 3B-100. Therefore revenue from section A equals $250(3B-100)
Similarly number of tickets sold in section C equals to B/2. And the revenue should be equal to 200(B/2).
Now the revenue earned from all the three sections is equal to $2,275,000.
Therefore,
250(3B-100) + 300B+ 200(B/2)=2275000
750B-25000+300B+100B=2275000
1150B=2300000
B=2000
We know that C=B/2, therefore C=1000
Answer is option B
Please hit kudos if this helped you
Since number of tickets sold in section A is hundred less than three times B, therefore number of tickets for section A= 3B-100. Therefore revenue from section A equals $250(3B-100)
Similarly number of tickets sold in section C equals to B/2. And the revenue should be equal to 200(B/2).
Now the revenue earned from all the three sections is equal to $2,275,000.
Therefore,
250(3B-100) + 300B+ 200(B/2)=2275000
750B-25000+300B+100B=2275000
1150B=2300000
B=2000
We know that C=B/2, therefore C=1000
Answer is option B
Please hit kudos if this helped you
