Question 957374
{{{ t = d/r }}}
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Let {{{ r }}} = the train's rate in mi/hr on level terrain
{{{ r - 12 }}} = the train's rate in mi/hr in mountainous terrain
Let {{{ t }}} = the train's time in hrs on mountainous terrain
{{{ 8 - t }}} = the train's time on level terrain
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For mountainous terrain:
(1) {{{ t = 120 / ( r - 12 ) }}}
For level terrain:
(2) {{{ 8 - t = ( 380 - 120 ) / r }}}
(2) {{{ 8 - t = 260/r }}}
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(2) {{{ t = 8 - 260/r }}}
By substitution:
{{{ 120/( r-12 ) = 8 - 260/r }}}
Multiply both sides by {{{ r*( r-12 ) }}}
{{{ 120r = 8*r*( r-12 ) - 260*( r-12 ) }}}
{{{ 120r = 8r^2 - 96r - 260r + 3120 }}}
{{{ 8r^2 - 476r + 3120 = 0 }}}
{{{ 2r^2 - 119r + 780 = 0 }}}
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Use quadratic formula
{{{ r = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 2 }}}
{{{ b = -119 }}}
{{{ c = 780 }}}
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{{{ r = (-(-119) +- sqrt( (-119)^2 - 4*2*780 )) / (2*2) }}} 
{{{ r = ( 119 +- sqrt( 14161 - 6240 )) / 4 }}} 
{{{ r = ( 119 +- sqrt( 7921 )) / 4 }}}
{{{ r = ( 119 - 89 ) / 4 }}}
{{{ r = 30/4 }}}
{{{ r = 7.5 }}} mi/hr
This can't be the answer, since I have to subtract {{{ 12 }}} mi/hr
to get rate on mountainous terrain, so
{{{ r = ( 119 + 89 ) / 4 }}}
{{{ r = 208/4 }}}
{{{ r = 52 }}} mi/hr
{{{ r - 12 = 40 }}} mi/hr
The speed on level terrain is 52 km/hr
check:
(1) {{{ t = 120 / ( r - 12 ) }}}
(1) {{{ t = 120 / ( 52 - 12 ) }}}
(1) {{{ t = 120/40 }}}
(1) {{{ t = 3 }}} hrs
and
(2) {{{ 8 - t = ( 380 - 120 ) / r }}}
(2) {{{ 8 - t = 260 / 52 }}}
(2) {{{ 8 - t = 5 }}}
(2) {{{ t = 3 }}} hrs
OK
Hope I got it!