Question 498554
This is a problem of rates
Let {{{c}}} = pipe C's rate of emptying the pool
Pipe A's rate of filling:
{{{ 1/5 }}} ( 1 pool in 5 hrs )
Pipe B's rate of filling:
{{{  1/( c - 2) }}}
Pipe C's rate of emptying pool:
{{{c}}}
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With all 3 pipes open:
{{{ 1/5 + 1/( c - 2 ) - 1/c = 1/3 }}}
Multiply both sides by {{{ 15*c*( c - 2 ) }}}
{{{ 3c*( c - 2 ) + 15c - 15*( c - 2 ) = 5c*( c - 2 ) }}}
{{{ 3c^2 - 6c + 15c - 15c + 30  = 5c^2 - 10c }}}
{{{ 2c^2 - 4c - 30 = 0 }}}
{{{ c^2 - 2c - 15 = 0 }}}
Completing the square:
{{{ c^2 - 2c + (-2/2)^2 = 15 + (-2/2)^2 }}}
{{{ c^2 - 2c + 1 = 16 }}}
{{{ ( c - 1 )^2 = 4^2 }}}
Take the square root of both sides
{{{ c - 1 = 4 }}}
{{{ c = 5 }}}
and, also
{{{ c - 1 = -4 }}}
{{{ c = -3 }}} ( can't use the negative square root )
It will take pipe C alone 5 hours to empty pool
check answer:
{{{ 1/5 + 1/( c - 2 ) - 1/c = 1/3 }}}
{{{ 1/5 + 1/( 5 - 2 ) - 1/5 = 1/3 }}}
{{{ 1/5 + 1/3 - 1/5 = 1/3 }}}
{{{ 1/3 = 1/3 }}}
OK