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)\"\" \"About 
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The tanker is to be emptied in 2 stages in 3 hrs.
\n" ); document.write( "Let \"t\" = hours that main pump alone is working
\n" ); document.write( "Amount pumped by main pump =
\n" ); document.write( "(1) \"%281%2F4%29%2At+=+t%2F4\"
\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)\"%281%2F4%29%2A%283+-+t%29+%2B+%281%2F7%29%2A%283+-+t%29\"
\n" ); document.write( "If I add (1) and (2), I should get \"1\" (1 tank pumped)
\n" ); document.write( "\"t%2F4+%2B+%283+-+t%29%2F4+%2B+%283+-+t%29%2F7+=+1\"
\n" ); document.write( "Multiply both sides by \"4%2A7\"
\n" ); document.write( "\"7t+%2B+7%2A%283+-+t%29+%2B+4%2A%283+-+t%29++=+28\"
\n" ); document.write( "\"7t+%2B+%283+-+t%29%2A11+=+28\"
\n" ); document.write( "\"7t+%2B+33+-+11t+=+28\"
\n" ); document.write( "\"4t+=+5\"
\n" ); document.write( "\"t+=+5%2F4\" hrs
\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( "\"t%2F4+%2B+%283+-+t%29%2F4+%2B+%283+-+t%29%2F7+=+1\"
\n" ); document.write( "\"%285%2F4%29%2F4+%2B+%283+-+%285%2F4%29%29%2F4+%2B+%283+-+%285%2F4%29%29%2F7+=+1\"
\n" ); document.write( "\"5%2F16+%2B+7%2F16+%2B+4%2F16+=+1\"
\n" ); document.write( "\"+1+=+1\"
\n" ); document.write( "OK\r
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