Question 956638
Start at point A with the head of the train at 
point A and the boy at point A
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Make point B where the boy is when the tail end
of the train is also at point B
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Let {{{ d }}} =  the distance in km from A to B 
Let {{{ t }}} = time in hrs for the boy and the tail end
of the train to get to B
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Both going in the same direction:
The equation for the boy is:
(1) {{{ d = 6t }}}
Equation for the train:
(2) {{{ 225/1000 + d = 60t }}}
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Substitute (1) into (2)
(2) {{{ 225/1000 + 6t = 60t }}}
(2) {{{ 54t = .225 }}}
(2) {{{ t = .0041667 }}} hrs
Convert to seconds:
{{{ .0041667*3600 = 15 }}} sec
The train will pass the boy in 15 sec
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In the opposite direction, the boy at A 
turns around and runs toward the tail
end of the train when the head is at A
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You can think of this as the boy standing
still and the train coming toward him at
the sum of their speeds
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{{{ d = ( 60 + 6 )*t }}}
The distance is {{{ 225/1000 }}} km
{{{ .225 = 66t }}}
{{{ t = .0034091 }}} hrs
Convert to seconds
{{{ .0034091*3600 = 12.273 }}} sec
The boy and the tail end of the train
will meet in 12.273 sec
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check:
Boy's equation:
{{{ d[1] = 6*.0034091 }}}
{{{ d[1] = .020455 }}} km
train's equation:
{{{ d[2] = 60*.0034091 }}}
{{{ d[2] = .20455 }}} km
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{{{ d[1] + d[2] = 225 m
{{{ 20.455 + 204.55 = 225 }}}
{{{ 225.005 = 225 }}}
OK