document.write( "Question 672642: An oil tanker can be emptied by a main pump in 4hours an auxiliary pump can empty it in 9 hrs if the main pump is started at 9 am when should the auxiliary pump be started so the tanker will be empty by noon? \n" ); document.write( "
Algebra.Com's Answer #418244 by josmiceli(19441) You can put this solution on YOUR website! From 9 AM to noon is 3 hrs \n" ); document.write( "Let \n" ); document.write( "is used pumping by itself \n" ); document.write( " \n" ); document.write( "--------------- \n" ); document.write( "The main pump's rate of pumping is ( 1 tanker / 4 hrs ) \n" ); document.write( "the auxilliary pump's rate is ( 1 tanker / 9 hrs ) \n" ); document.write( "---------------- \n" ); document.write( "( fraction of tanker pumped in x hrs ) + \n" ); document.write( "( fraction pumped by both pumps in 3-x hrs ) = 1 whole tanker pumped \n" ); document.write( " \n" ); document.write( "Multiply both sides by \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The auxiliary pump should be started at 9:45 \n" ); document.write( "check answer: \n" ); document.write( "In 3/4 of an hour the main pump empties \n" ); document.write( " \n" ); document.write( "In 2.25 hrs, both together empty \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "and \n" ); document.write( "OK \n" ); document.write( " \n" ); document.write( " |