Question 940954
Let {{{ R }}} = the rate of the slow pump
{{{ 3R }}} = the rate of the fast pump
{{{ R + 3R = 1/4 }}} 
{{{ 4R = 1/4 }}}
{{{ R = 1/16 }}}
{{{ 3R = 3/16 }}}
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The slow pump's rate is ( 1 tank ) / ( 16 hrs )
The fast pump's rate is {{{ 3/16 }}} = ( 1 tank ) / ( 5 hrs 20 min )
( not that 5 hrs 20 min = {{{ 16/3 = 5.333 }}} hrs