document.write( "Question 413815: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 7 pm, when should the auxiliary pump be started so that the tanker is emptied by 10 pm? \n" ); document.write( "
Algebra.Com's Answer #290317 by josmiceli(19441)![]() ![]() You can put this solution on YOUR website! The tanker is to be emptied in 2 stages in 3 hrs. \n" ); document.write( "Let \n" ); document.write( "Amount pumped by main pump = \n" ); document.write( "(1) \n" ); document.write( "That is: (1 tank)/(4 hrs) x (t hrs) \n" ); document.write( "The auxilliary pump + main pump will then pump out: \n" ); document.write( "(2) \n" ); document.write( "If I add (1) and (2), I should get \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( "So, the main pump runs alone for 1 hr 15 min. That means \n" ); document.write( "the auxilliary pump comes on at 8:15 \n" ); document.write( "check answer: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "OK\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |