Question 1064449
Call the slow train T[1]
Call the fast train  T[2]
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Let {{{ d[1] = 50t[1] }}} = the headstart that train T[1] has until
train T[2] leaves city A
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Start a stopwatch when train T[2] leaves city A
Let {{{ t }}} = time on stopwatch in hrs when train T[2] reaches
city C
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Train T[1] travels {{{ 150 - d[1] = 150 - 50t[1] }}} during time {{{ t }}}
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Equation for train T[1]:
(1) {{{ 150 - d[1] = 50*( t - 10/60 ) }}}
Equation for train T[2]:
(2) {{{ 150 = 80t }}}
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Note that {{{ t - 10/60 ) is the actual travel time 
for train T[1] during time {{{ t }}}
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(1) {{{ 150 - d[1] = 50*( t - 1/6 ) }}}
(1) {{{ 150 - d[1] = 50t - 50/6 }}}
(1) {{{ 150 - d[1] = 50t - 25/3 }}}
(1) {{{ 450 - 3d[1] = 150t - 25 }}}
(1) {{{ 150t + 3d[1] = 475 }}}
and
(2) {{{ t = 150/80 }}}
(2) {{{ t = 15/8 }}}
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By substitution into (1):
(1) {{{ 150*(15/8) + 3d[1] = 475 }}}
(1) {{{ 2250 + 24d[1] = 3800 }}}
(1) {{{ 24d[1] = 1550 }}}
(1) {{{ d[1] = 64.5833 }}}
and
{{{ d[1] = 50t[1] }}}
{{{ 50t[1] = 64.5833 }}}
{{{ t[1] = 1.291667 }}} hrs
{{{ .291667*60 = 17.5 }}}
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The express train leaves city A 1 hr 17 min 30 sec after 
the slow train
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check the answer:
(1) {{{ 150 - d[1] = 50*( t - 1/6 ) }}}
(1) {{{ 150 - 64.5833 = 50*( 1.875 - .1667 ) }}}
(1) {{{ 150 - 64.5833 = 93.75 - 8.333 }}}
(1) {{{ 150 - 93.75 = 64.5833 - 8.333 }}}
(1) {{{ 56.25 = 56.25003 }}}
This looks close enough
*Definitely* try to get a 2nd opinion on this
There are a number of ways I could have messed up.