SOLUTION: A concert sold 1,068 tickets bringing in total revenue of $15,060. Three categories of tickets were sold. Type "A" sold for $10, type "B" sold for $15 and type "C" sold for $20. Th

Algebra ->  Human-and-algebraic-language -> SOLUTION: A concert sold 1,068 tickets bringing in total revenue of $15,060. Three categories of tickets were sold. Type "A" sold for $10, type "B" sold for $15 and type "C" sold for $20. Th      Log On


   

Question 140792: A concert sold 1,068 tickets bringing in total revenue of $15,060. Three categories of tickets were sold. Type "A" sold for $10, type "B" sold for $15 and type "C" sold for $20. There were 4 times as many "A" tickets sold as there were "B" tickets. How many of each type of ticket were sold and how much money did each type of ticket generate?
Found 2 solutions by scott8148, ankor@dixie-net.com:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
"A concert sold 1,068 tickets" __ A+B+C=1068

"bringing in total revenue of $15,060" __ 10A+15B+20C=15060

"4 times as many "A" tickets sold as there were "B" tickets" __ A=4B

substituting __ (4B)+B+C=1068 __ 5B+C=1068 __ multiplying by 11 __ 55B+11C=11748

substituting __ 10(4B)+15B+20C=15060 __ 55B+20C=15060

subtracting __ [55B+20C=15060]-[55B+11C=11748] __ 9C=3312 __ C=368

substituting __ 5B+(368)=1068 __ B=140

substituting __ A=4(140) __ A=560

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A concert sold 1,068 tickets bringing in total revenue of $15,060. Three categories of tickets were sold. Type "A" sold for $10, type "B" sold for $15 and type "C" sold for $20. There were 4 times as many "A" tickets sold as there were "B" tickets. How many of each type of ticket generate
:
Total tickets equation:
A + B + C = 1068
:
Total Revenue equation
10A + 15B + 20C = 15060
:
"There were 4 times as many "A" tickets sold as there were "B" tickets."
A = 4B
:
Substitute 4B for A in both total equations
2B + B + C = 1068
3B + C = 1068
C = (1068-3B)
:
10(2B) + 15B + 20C = 15060
20B + 15B + 20C = 15060
35B + 20C = 15060
:
Substitute (1068-3B) for C in the above equation:
35B + 20(1068-3B) = 15060
35B + 21360 - 60B = 15060
35B - 60B = 15060 - 21360
-25B = -6300
B = %28-6300%29%2F%28-25%29
B = 252 ea $15 tickets
:
Using equation C = 1068 - 3B)
C = 1068 - 3(252)
C = 1068 - 756
C = 312 ea $20 tickets
:
Find A
A + 252 + 312 = 1068
A = 1068 - 564
A = 504 ea $10 tickets
:
:
You can check our solution:
10(504) + 15(252) + 20(312) =