document.write( "Question 264824: please help
\n" ); document.write( "an oil tanker can be emptied by the main pump in 4 hours. an auxiliary pump can empty the tanker in 9 hours. if the main pump is started at 9am , when should the auxiliary pump be started so that the tanker is emptied by noon?
\n" ); document.write( "

Algebra.Com's Answer #194908 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
main pump equals 4 hours.
\n" ); document.write( "aux pump equals 9 hours.
\n" ); document.write( "start at 9:00 am and want to end at 12:00 pm.
\n" ); document.write( "total of 3 hours.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "rate * time = units\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the units are 1 tank.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "main pump can empty the tank in 4 hours so x*4 = 1 which means that the main pump can empty 1/4 of the tank in 1 hour.\r
\n" ); document.write( "\n" ); document.write( "the aux pump can empty the tank in 9 hours so x*9 = 1 which means that the aux pump can empty 1/9 of the tank in 1 hour.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the main pump starts at 900 am.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the total time allowed is 3 hours.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the main pump alone would therefore empty 3 * 1/4 = 3/4 of the tank in 3 hours.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "that means that 1/4 of the tank still needs to be emptied.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the aux pump can empty 1/4 of the tank in how many hours?\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the formula is rate * time = units of work.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "we get 1/9 * x = 1/4 where x is the amount of time it will take the aux tank to empty 1/4 of the tank.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "formula is:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "1/9 * x = 1/4\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "multiply both sides of this equation by 9 to get:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "x = 9/4 hours.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "9/4 hours is the same as 2 + 1/4 hours.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "subtract that from 12 pm and you get 9 + 3/4 hours is when the aux pump should be started.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "90 + 3/4 hours is the same as 9:45 am.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "here's what happens.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "at 9:00 am the main pump starts pumping.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "at 9:45 am the aux pump starts pumping.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "by 12:00 pm the main pump has pumped 1/4 * 3 = 3/4 of the tank.
\n" ); document.write( "by 12:00 pm the aux pump has pumped 1/9 * 9/4 = 1/4 of the tank.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "3/4 + 1/4 = 1 which means the tank is empty at 12:00 pm.\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );