Question 994696
Let {{{ s }}} = the speed of the train in km/hr
{{{ 4s - 20 }}} = the speed of the plane in km/hr
--------------------
Equation for train:
(1) {{{ 110 = s*t }}}
Equation for plane:
(2) {{{ 385 = ( 4s - 20 )*t }}}
----------------------
(2) {{{ 385 = 4s*t - 20t }}}
and
(1) {{{ s = 110/t }}}
-----------------
By substitution:
(2) {{{ 385 = 4*( 110/t )*t - 20t }}}
(2) {{{ 385 = 440 - 20t }}}
(2) {{{ 20t = 55 }}}
(2) {{{ t = 11/4 }}}
{{{ s = 110/t }}}
{{{ s = 110*(4/11) }}}
{{{ s = 40 }}}
and
{{{ 4s - 20 = 4*(110/t ) - 20 }}}
{{{ 4s - 20 =  4*110*(4/11) - 20 }}}
{{{ 4s - 20 = 160 - 20 }}}
{{{ 4s - 20 = 140 }}}
-----------------
The speed of the train is 40 km/hr
The speed of the train is 140 km/hr
-----------------
check:
(2) {{{ 385 = ( 4s - 20 )*t }}}
(2) {{{ 385 = 140*(11/4) }}}
(2) {{{ 385 = 385 }}}
and
(1) {{{ 110 = s*t }}}
(1) {{{ 110 = 40*(11/4) }}}
(1) {{{ 110 = 110 }}}
OK