Question 536196
Let {{{t}}} = hours it takes local train to get to Hope
{{{ t - 1 }}} = hours it takes express train to get to Hope
Let {{{s}}} = the speed of the fast train in km/hr
{{{ (2/3)*f }}} = speed of the local train in km/hr
-----------
For local train:
(1) {{{ 180 = (2/3)*s*t }}}
For express train:
(2) {{{ 180 = s*( t - 1 ) }}}
-----------------
(1) {{{ 540 = 2*s*t }}}
(1) {{{ s*t = 270 }}}
and
(2) {{{ 180 = s*t - s }}}
Substitute (1) into (2)
(2) {{{ 180 = 270 - s }}}
(2) {{{ s = 90 }}}
and
(1) {{{ s*t = 270 }}}
{1) {{{ t = 270/90 }}}
(1) {{{ t = 3 }}}
The local train arrives at 3 PM + 3 hrs = 6 PM
check:
(1) {{{ 180 = (2/3)*90*3 }}}
(1) {{{ 180 = 180 }}}
OK