Question 1042119
Let {{{ t }}} = time in hrs  to fill reservoir by 
the slower-filling pipe
{{{ t - 3 }}} = time in hrs to fill reservoir by
the faster-filling pipe
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Rate of filling by slower-filling pipe:
[ 1 reservoir filled ] / [ t hrs ]
Rate of filling by faster-filling pipe:
[ 1 reservoir filled ] / [ t - 3 hrs ]
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Add the rates of filling to get the rate with
both pipes filling together
( first convert minutes to hrs )
{{{ 1/t + 1/( t-3 ) = 1/6.667 }}}
{{{ 1/t + 1/( t-3 ) = 3/20 }}}
Multiply both sides by {{{ t*( t-3 )*20 }}}
{{{ 20*( t-3 ) + 20t = 3t*( t-3 ) }}}
{{{ 20t - 60 + 20t = 3t^2 - 9t }}}
{{{ 3t^2 - 49t + 60 = 0 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 3 }}}
{{{ b = -49 }}}
{{{ c = 60 }}}
{{{ t = ( -(-49) +- sqrt( (-49)^2-4*3*60 ))/(2*3) }}}
{{{ t = ( 49 +- sqrt( 2401 - 720 ))/6 }}}
{{{ t = ( 49 +- sqrt( 1681 ))/6 }}}
{{{ t = ( 49 + 41 ) / 6 }}} 
( note that the negative square root will not work )
{{{ t = 90/6 }}}
{{{ t = 15 }}}
and
{{{ t - 3 = 12 }}}
The pipes can fill the reservoir in 12 hrs and 15 hrs each
working alone
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check:
{{{ 1/t + 1/( t-3 ) = 1/((20/3)) }}}
{{{ 1/15 + 1/12 = 3/20 }}}
{{{ 4/60 + 5/60 = 9/60 }}}
{{{ 9/60 = 9/60 }}}
OK