Question 725146
Add their rates of filling
Let {{{ t }}} = time in hrs for the slowest pipe to fill tank
{{{ t - 5 }}} = time in hrs for the faster pipe to fill tank
given:
( 1 tank filled / t hrs ) + ( 1 tank filled  / t - 5 hrs ) = ( 1 tank filled / 6 hrs )
{{{ 1/t + 1/( t - 5 ) = 1 / 6 }}}
Multiply both sides by {{{ 6t*( t - 5 ) }}}
{{{ 6*( t - 5 ) + 6t = t*( t - 5 ) }}}
{{{ 6t - 30 + 6t = t^2 - 5t }}}
{{{ t^2 - 17t + 30 = 0 }}}
{{{ ( t - 15 )*( t - 2 ) }}} ( just by looking at it )
I get
{{{ t = 15 }}} hrs and
{{{ t = 2 }}} hrs
It can't be {{{ t = 2 }}} because I have to subtract 5 hrs for the slower pipe
and then I get a negative.
So, {{{ t = 15 }}} and
{{{ t - 5 = 10 }}}
The slower pipe fills in 15 hrs
The faster pipe fills in 10 hrs
check:
{{{ 1/t + 1/( t - 5 ) = 1 / 6 }}}
{{{ 1/15 + 1/( 15 - 5 ) = 1 / 6 }}} 
{{{ 1/15 + 1/10 = 1/6 }}}
{{{ 2/30 + 3/30 = 5/30 }}}
{{{ 5/30 = 5/30 }}}
OK