document.write( "Question 1122267: The total number of passengers riding a certain city bus during the morning shift is 1100. If the child's fare is $0.50, the adult fare is $1.50, and the total revenue from the fares in the morning shift is $1350, how many children and how many adults rode the bus during the morning shift? \n" ); document.write( "

Algebra.Com's Answer #738346 by ikleyn(52794) $\"\"$ $\"About$

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document.write( "Let assume for a minute that all 1100 passengers in this morning were children.\r\n" );
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document.write( "Then the revenue would be 1100*$050 = $550, making shortage of $1350 - $550 = $800.\r\n" );
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document.write( "The shortage is due to the fact that we counted all 1100 tickets as children tickets.\r\n" );
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document.write( "To compensate the shortage, we should replace some number of children tickets in our counting by adults tickets.\r\n" );
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document.write( "At every such replacement, we diminish the shortage of $800 by $1.00 = $1.50 - $0.50.\r\n" );
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document.write( "So, the number of such replacements is the ratio   = 800.\r\n" );
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document.write( "Hence, the correct number of adult tickets is 800.\r\n" );
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document.write( "Then the number of the children tickets is the rest of 1100 - 800 = 300.\r\n" );
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document.write( "Answer.  800 adults and 300 children.\r\n" );
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document.write( "Check.   800*$150 + 300*$0.50 = $1350.   ! Correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "On solving ticket problems see the lessons\r
\n" ); document.write( "\n" ); document.write( " - Using systems of equations to solve problems on tickets \r
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