Question 682048
Let the distance between towns A and B = {{{ d }}}
Mark's total trip:
{{{ d + d - 40 }}}
Nancy's total trip:
{{{ d + 40 }}}
They both traveled these distances in {{{ 3 }}} hrs
So, their rates are
Mark:
{{{ ( 2d - 40 ) / 3 }}} mi/hr
Nancy:
{{{ ( d + 40 ) / 3 }}} mi/hr
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For the 1st meeting, the ratio of their speed is directly 
proportional to the distances they cover, since
{{{ d[1] = r[1]*t }}}
{{{ r[1] = d[1] / t }}}
and
{{{ d[2] = r[2]*t }}}
{{{ r[2] = d[2] / t }}}
and
{{{ r[1] / r[2] = d[1] / d[2] }}}
( Mark's rate ) / ( Nancy's rate ) = {{{ 60 / ( d - 60 ) }}}
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Now I can say:
{{{ 60 / ( d - 60 ) = ( 2d - 40 ) / ( d + 40 ) }}} ( t 's cancel )
{{{ 60*( d + 40 ) = ( 2d - 40 )*( d - 60 ) }}}
{{{ 60d + 2400 = 2d^2 - 40d - 120d + 2400 }}}
{{{ 60d = 2d^2 - 160d }}}
{{{ 2d^2 - 220d = 0 }}}
{{{ d^2 - 110d = 0 }}}
{{{ d*( d - 110 ) = 0 }}}
{{{ d = 110 }}} mi
Mark:
{{{ ( 2d - 40 ) / 3 }}} mi/hr
{{{ ( 2*110 - 40 ) / 3 }}}
{{{ 220 - 40) / 3 }}}
{{{ 180/3 }}}
{{{ 60 }}}
Nancy:
{{{ ( d + 40 ) / 3 }}} mi/hr
{{{ ( 110 + 40 ) / 3 }}}
{{{ 150/3 }}}
{{{ 50 }}}
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Mark's speed is 60 mi/hr
Nancy's speed is 50 mi/hr
check:
( Mark's rate ) / ( Nancy's rate ) = {{{ 60 / ( d - 60 ) }}}
{{{ 60 / ( 110 - 60 ) = 60/50 }}}