Question 698362
Call the increased speed {{{ s + 15 }}}
With speed {{{ s }}}, time is {{{ t + 3 }}}
where {{{ t }}} is on time
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Equation being late by 3 hrs:
(1) {{{ 900 = s*( t + 3 ) }}}
Equation for being on time:
(2) {{{ 900 = ( s + 15 )*t }}}
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(1) {{{ 900 = s*t + 3s }}}
(1) {{{ s*t = 900 - 3s }}}
(1) {{{ t = ( 900 - 3s ) / s }}}
Substitute (1) into (2)
(2) {{{ 900 = ( s + 15 )*( 900 - 3s ) / s }}}
(2) {{{ 900s = ( s + 15 )*( 900 - 3s ) }}}
(2) {{{ 900s = 900s + 13500 - 3s^2 - 45s }}}
(2) {{{ 3s^2 + 45s - 13500 = 0 }}}
(2) {{{ s^2  + 15s - 4500 = 0 }}}
Use quadratic formula
{{{ s = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = 15 }}}
{{{ c = -4500 }}}
{{{ s = ( -15 +- sqrt( 15^2 - 4*1*(-4500) )) / (2*1) }}}
{{{ s = ( -15 +- sqrt( 225 + 18000 )) / 2 }}}
{{{ s = ( -15 +- sqrt( 18225 )) / 2 }}} ( ignore the (-) square root )
{{{ s = ( -15 + 135) / 2 }}}
{{{ s = 120/2 }}}
{{{ s = 60 }}}
{{{ s + 15 = 75 }}}
The increased speed of the train is 75 km/hr
check:
(1) {{{ 900 = 60*( t + 3 ) }}}
(1) {{{ t + 3 = 15 }}}
(1) {{{ t = 12 }}}
and
(2) {{{ 900 = ( 60 + 15 )*12 }}}
(2) {{{ 900 = 75*12 }}}
(2) {{{ 900 = 900 }}}
OK