document.write( "Question 66869: An oil tanker can be emptied by the main pump in 4 hours and by an auxiliary pump in 9 hours. If the main pump is started at 9 a.m., when should the auxiliary pump be started so that the tanker is emptied by noon? \n" ); document.write( "
Algebra.Com's Answer #47547 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Main pump empties at the rate of 1/4 tanker per hour
\n" ); document.write( "In 3 hours (noon-9am) the main pump empties 3/4 of the tanker and there's 1/4 or 9/36 of the tanker left to pump\r
\n" ); document.write( "\n" ); document.write( "Aux pump empties at the rate of 1/9 or 4/36 tanker per hour\r
\n" ); document.write( "\n" ); document.write( "It will take the aux pump (9/36)/(4/36) hours to finish emptying the tanker:
\n" ); document.write( "(9/36)/(4/36)=(9/36)(36/4)=9/4 hours =2.25 hours\r
\n" ); document.write( "\n" ); document.write( "So the aux pump must come online at noon - 2.25 hours which would be
\n" ); document.write( "9:45am\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
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