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ericavasquez824
14.10.2020 •
Mathematics
A theater owner wants to divide a 3300 seat theater into three sections, with tickets costing $ 60, $70, and $120, depending on the section. He wants to have twice as many $60 tickets as the sum of the other tickets, and he wants to earn $ 224,000 from a full house. Find how many seats he should have in each section.
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Ответ:
Section A = 2,200 seats
Section B = 800 seats
Section C = 300 seats
Step-by-step explanation:
Three sections
Section A cost = $60
Section B cost = $70
Section C cost = $120
Total seat = 3300
Total earnings = $224,000
He wants to have twice as many $60 tickets as the sum of the other tickets
A = 2(B + C)
A + B + C = 3,300
60A + 70B + 120C = 224,000
substitute A = 2(B + C)
2(B + C) + B + C = 3,300
60(2B + 2C) + 70B + 120C = 224,000
2B + 2C + B + C = 3,300
120B + 120C + 70B + 120C = 224,000
3B + 3C = 3,300 (1)
190B + 240C = 224,000 (2)
Multiply (1) by 80
240B + 240C = 264,000 (3)
190B + 240C = 224,000 (2)
Subtract the equations
240B - 190B = 264,000 - 224,000
50B = 40,000
B = 40,000 / 50
= 800
B = 800
Substitute the value of B into (1)
3B + 3C = 3,300 (1)
3(800) + 3C = 3,300
2,400 + 3C = 3,300
3C = 3,300 - 2,400
3C = 900
C = 900 / 3
= 300
C = 300
Substitute values of B and C into
A + B + C = 3,300
A + 800 + 300 = 3,300
A + 1,100 = 3,300
A = 3,300 - 1,100
= 2,200
A = 2,200
Section A = 2,200 seats
Section B = 800 seats
Section C = 300 seats
Ответ:
16+16+16=48 then 48+9=57 then 57+6=66
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