Question 934291
Pipe A's rate of filling is:
( 1 tank filled ) / ( 5 hrs )
------------------------
Let {{{ t }}} = pipe C's time to
empty the tank, so C's rate
of emptying is:
( minus 1 tank filled ) / ( t hrs )
-----------------------------
Pipe B's rate of filling is:
( 1 tank filled ) / ( t - 2 hrs )
-------------------------
With all 3 pipes open, the rate of filling is:
( 1 tank filled ) / ( 3 hrs ), so I can say:
{{{ 1/5 + 1/( t-2 ) - 1/t = 1/3 }}}
Multiply both sides by:
{{{ 15*t*( t-2 ) }}}
{{{ 3t*( t-2 ) + 15t - 15*( t-2 ) = 5t*( t-2 ) }}}
{{{ 5t*( t-2 ) - 3t*( t-2 ) = 15t - 15*( t-2 ) }}}
{{{ 2t*( t-2 ) = 15t - 15t + 30 }}}
{{{ 2t^2 - 4t - 30 = 0 }}}
{{{ t^2 - 2t - 15 = 0 }}}
{{{ ( t - 5 )*( t + 3 ) = 0 }}}
{{{ t = 5 }}} ( can't use negative time )
Pipe C can empty the tank in 5 hrs
check:
{{{ 1/5 + 1/( t-2 ) - 1/t = 1/3 }}}
{{{ 1/5 + 1/( 5-2 ) - 1/5 = 1/3 }}}
{{{ 1/5 + 1/3 - 1/5 = 1/3 }}}
{{{ 1/3 = 1/3 }}}
OK