Question 1129779
Pipe A's rate of filling:
[ 1 tank filled ] / [ 5 hrs ]
Let {{{ t }}} = time in hrs for pipe C to empty tank
pipe C's rate of emptying:
[ 1 tank emptied ] / [ t hrs ]
pipe B's rate of filling:
[ 1 take filled ] / [ t - 2 hrs ]
---------------------------------
Add rates for A % B
Subtract rate for C
{{{ 1/5 + 1/( t-2 ) - 1/t = 1/3 }}}
Multiply both sides by {{{ 3*5*t*( t-2 ) }}}
{{{ 3t*( t-2 ) + 3*5t - 3*5*( t-2 ) = 5t*( t-2 ) }}}
{{{ 3t^2 - 6t + 15t - 15t + 30 = 5t^2 - 10t }}}
{{{ 3t^2 - 6t + 30 = 5t^2 - 10t }}}
{{{ 2t^2 - 4t - 30 = 0 }}}
{{{ t^2 - 2t - 15 = 0 }}}
{{{ ( t - 5 )*( t + 3 ) = 0 }}}
{{{ t = 5 }}} ( choose positive time )
Pipe C empties the tank in 5 hrs
--------------------------------------
check:
{{{ 1/5 + 1/( 5-2) - 1/5 = 1/3 }}}
{{{ 1/5 + 1/3 - 1/5 = 1/3 }}}
{{{ 1/3 = 1/3 }}}
OK