SOLUTION: At a movie theatre, admission tickets are $5 for children and $8 for adults. The receipts for Friday evening were $2500. The next day, there were three times as many children as th

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Question 177562: At a movie theatre, admission tickets are $5 for children and $8 for adults. The receipts for Friday evening were $2500. The next day, there were three times as many children as the preceding evening and only half the number of adults as the night before, yet the receipts were still $2500. Find the number of children who attended Friday evening.
Found 2 solutions by Mathtut, stanbon:
Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website!
let c and a be the number of children and adult tickets sold, respectively
:
5c+8a=2500.....eq 1
5(3c)+8(a/2)=2500....eq 2--->15c+4a=2500
:
5c+8a=2500.....eq 1
15c+4a=2500....eq 2
:
multiply eq 2 by -2 and add the two equations together thus eliminating the a terms. We are left with -30c+5c=-5000+2500
:
-25c=-2500
:
children attended friday evening

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
At a movie theatre, admission tickets are $5 for children and $8 for adults. The receipts for Friday evening were $2500.
Value Equation: 5c + 8a = 2500
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The next day, there were three times as many children as the preceding evening and only half the number of adults as the night before, yet the receipts were still $2500.
Value Equation: 5(3c) + 8(a/2) = 2500
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Find the number of children who attended Friday evening.
System:
5c + 8a = 2500
15c+ 4a = 2500
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Modify:
5c + 8a = 2500
30c+ 8a = 5000
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Subtract to solve for "c":
25c = 2500
c = 100 (# of child tickets sold)
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Substitute into 5c + 8a = 2500 to solve for "a":
5*100 + 8a = 2500
8a = 2000
a = 250 (# of adult tickets solf)
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Cheers,
Stan H.