Asked by p00pjoke
A stadium has 53,000 seats. Seats sell for $28 in Section A, $16 in Section B, and $12 in Section C. Th number of seats in Section A equals the total number of seast in Sections B and C. Suppose the stadium takes in $1,118,000 from each sold-out event. How many seats does each section hold?
Section A holds:
Section B holds:
Section C holds:
Section A holds:
Section B holds:
Section C holds:
Answers
Answered by
mathhelper
28a + 16b + 12c = 1,118,000
but a = b+c
28(b+c) + 16b + 12c = 1,118,000
44b + 40c = 1,118,000
11b+10c = 111,800
but we also know that a + b + c = 53000
b+c + b + c = 53000
b+c = 26500
c = 26500 - b
sub in that value of c into 11b+10c = 111,800
and you can find b
after that you can find c
and finally find a in a = b+c
but a = b+c
28(b+c) + 16b + 12c = 1,118,000
44b + 40c = 1,118,000
11b+10c = 111,800
but we also know that a + b + c = 53000
b+c + b + c = 53000
b+c = 26500
c = 26500 - b
sub in that value of c into 11b+10c = 111,800
and you can find b
after that you can find c
and finally find a in a = b+c
Answered by
R_scott
A equals B plus C ... half of the seats are in A ... 53,000 / 2 = ?
value of A seats ... 28 * (53000 / 2) = 742,000
number of B and C seats ... B + C = 53,000 / 2 ... 12B + 12C = 6 * 53,000
value of B and C seats ... 16B + 12C = 1,118,000 - 742,000
subtract the two equations to eliminate C
... 4B = 1,118,000 - 742,000 - (6 * 53,000)
solve for B , then substitute back to find C
value of A seats ... 28 * (53000 / 2) = 742,000
number of B and C seats ... B + C = 53,000 / 2 ... 12B + 12C = 6 * 53,000
value of B and C seats ... 16B + 12C = 1,118,000 - 742,000
subtract the two equations to eliminate C
... 4B = 1,118,000 - 742,000 - (6 * 53,000)
solve for B , then substitute back to find C
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