# 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 ->  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

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 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)   (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(15622)   (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 = 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) =