document.write( "Question 766681: A water tank can be emptied by using one pump for 4 hours. A second, smaller pump can empty the tank in 7 hours. If the larger pump is started at 1:00 P.M., how much time should pass until the smaller pump is started so that the tank will be emptied at 4:00 P.M.? \n" ); document.write( "

Algebra.Com's Answer #467156 by stanbon(75887) $\"\"$ $\"About$

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A water tank can be emptied by using one pump for 4 hours. A second, smaller pump can empty the tank in 7 hours. If the larger pump is started at 1:00 P.M., how much time should pass until the smaller pump is started so that the tank will be emptied at 4:00 P.M.?
\n" ); document.write( "--------
\n" ); document.write( "1st pump rate: 1/4 job/hr
\n" ); document.write( "---
\n" ); document.write( "2nd pump rate: 1/7 job/hr
\n" ); document.write( "-----
\n" ); document.write( "Work + work = job done
\n" ); document.write( "(1/4)3 + (1/7)x = 1
\n" ); document.write( "---
\n" ); document.write( "(3/4) + (x/7) = 1
\n" ); document.write( "----
\n" ); document.write( "21 + 4x = 28
\n" ); document.write( "4x = 7
\n" ); document.write( "x = 7/4 hrs
\n" ); document.write( "x = 1 3/4 hrs (amount of time to use the smaller pump)
\n" ); document.write( "----
\n" ); document.write( "Time to start the smaller pump is 4pm - 1 3/4 hrs = 2:15 pm
\n" ); document.write( "================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "

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