document.write( "Question 1189913: Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $55. For one performance, 30 advance tickets and 40 same-day tickets were sold. The total amount paid for the tickets was $1900. What was the price of each kind of ticket? \n" ); document.write( "

Algebra.Com's Answer #821416 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( " It can be solved using arithmetic only, without Algebra.\r
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document.write( "Since the combined cost of one advance ticket and one some-day tickets is $55,\r\n" );
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document.write( "30 pairs of such combined tickets cost  30*55 = 16750 dollars.\r\n" );
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document.write( "It means that remaining 40-30 = 10 same day tickets cost 1900-1650 = 250  dollars.\r\n" );
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document.write( "Hence, one same-day ticket costs 250/10 = 25 dollars.\r\n" );
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document.write( "It means that each advanced ticket costs 55-25 = 30 dollars.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved (mentally).\r
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