Question 1111066
11 AM to 1:30 PM is {{{ 2.5 }}} hrs
Let {{{ s }}} km/hr = the speed of the southbound train
{{{ s + 20 }}} km/hr is the speed of the nothbound train
Let {{{ d }}} km = distance traveled by southbound train
{{{ 320 - d }}} km = distance traveled by northbound train
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Equation for the southbound train:
(1) {{{ d = s*2.5 }}}
Equation for nothbound train:
(2) {{{ 320 - d = ( s + 20 )*2.5 }}}
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Plug (1) into (2)
(2) {{{ 320 - 2.5s = ( s + 20 )*2.5 }}}
(2) {{{ 320 - 2.5s = 2.5s + 50 }}}
(2) {{{ 5s = 320 - 50 }}}
(2) {{{ 5s = 270 }}}
(2) {{{ s = 54 }}}
The average speed of the southbound train 
is 54 km/hr
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check:
(1) {{{ d = 54*2.5 }}}
(1) {{{ d = 135 }}}
and
(2) {{{ 320 - 2.5*54 = ( 54 + 20 )*2.5 }}}
(2) {{{ 320 - 135 = 74*2.5 }}}
(2) {{{ 185 = 185 }}}
and
{{{ 135 + 185 = 320 }}} km
OK