Question 1051961
You can think of this as one of them ( either one )
standing still and the other one approching at the 
sum of their speeds
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Let {{{ t }}} = time in hrs until they meet
{{{ 6 = ( 3 + 15 )*t }}}
{{{ 6 = 18t }}}
{{{ t = 1/3 }}} hrs
[ hrs ] x [ min/hr ] = [ min ]
{{{ (1/3)*(60/1) = 20 }}}
They will meet in 20 min
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check answer:
{{{ d[1] = 3*(1/3) }}}
{{{ d[1] = 1 }}} mi
and
{{{ d[2] = 15*(1/3) }}}
{{{ d[2] = 5 }}} mi
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{{{ d[1] + d[2] = 1 + 5 }}}
{{{ d[1] + d[2] = 6 }}} mi
OK