document.write( "Question 1197850: Crew A can assemble 2 cars in 5 days, and crew B can assemble 3 cars in 7 days. If both crews
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document.write( "together assemble 100 cars, with crew B working 10 days longer than crew A, how many days
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document.write( "(rounded to the nearest day) must each crew work? \n" );
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Algebra.Com's Answer #831290 by josgarithmetic(39617)![]() ![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "CREW RATE \r\n" ); document.write( "A 2/5 \r\n" ); document.write( "B 3/7 \r\n" ); document.write( "A and B 29/35 \r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Combined Rate A and B \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "B worked 10 hours longer than both crews combined. \n" ); document.write( "The two crews made 100 cars.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Time for B alone to finish, x+10 \n" ); document.write( "Time A and B together, x\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "unfinished -- \n" ); document.write( "x=115.5 days \n" ); document.write( "seems strange \n" ); document.write( " |