document.write( "Question 1117544: Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost
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document.write( "$35
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document.write( " and same-day tickets cost
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document.write( "$15
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document.write( ". For one performance, there were
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document.write( "50
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document.write( " tickets sold in all, and the total amount paid for them was
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document.write( "$1350
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document.write( ".How many tickets of each type were sold? \r
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Algebra.Com's Answer #732656 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "By algebra.... \n" ); document.write( "Let x be the number of advance tickets sold. \n" ); document.write( "Then since the total number of tickets was 50, the number of same-day tickets is (50-x). \n" ); document.write( "Then write and solve the equation that says x tickets at $35 each plus (50-x) tickets at $15 each makes a total of $1350: \n" ); document.write( " \n" ); document.write( "I'll let you do the simple algebra from there to find the answer. \n" ); document.write( "By logical reasoning and a bit of mental arithmetic.... \n" ); document.write( "If all 50 tickets had been same-day tickets, the total cost would have been 50*15 = 750; the actual total was 1350-750=600 more than that. \n" ); document.write( "The difference between the cost of an advance ticket and a same-day ticket is 35-15=20. \n" ); document.write( "To make up the \"extra\" 600, the number of advance tickets had to be 600/20 = 30. \n" ); document.write( "So there were 30 advance tickets and 20 same-day tickets. \n" ); document.write( "Check: 30(35)+20(15) = 1050+300 = 1350 \n" ); document.write( " |