Question 662490
Let {{{ s }}} = the normal speed in km/hr
Let {{{ t }}} = time to reach town at normal speed in hrs
given:
{{{ s + 6 }}} = faster speed
{{{ t - 1 }}} = time to reach town at faster speed
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Equation for normal speed:
(1) {{{ 224 = s*t }}}
Equation for faster speed:
(2) {{{ 224 = ( s + 6 )*( t - 1 ) }}}
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(1) {{{ s = 224/t }}}
Substitute (1) into (2)
(2) {{{ 224 = ( 224/t + 6 )*( t - 1 ) }}}
(2) {{{ 224 = 224 + 6t - 224/t - 6 }}
(2) {{{ 0 = 6t - 224/t - 6 }}}
(2) {{{ 224/t = 6t - 6 }}}
Multiply both sides by {{{ t }}}
(2) {{{ 224 = 6t^2 - 6t }}}
(2) {{{ 3t^2 - 3t - 112 = 0 }}}
Use quadratic formula to solve
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 3 }}}
{{{ b = -3 }}}
{{{ c = -112 }}}
{{{ t = (-(-3) +- sqrt( (-3)^2 - 4*3*(-112) )) / (2*3) }}} 
{{{ t = ( 3 +- sqrt( 9 + 1344 )) / 6 }}} 
{{{ t = ( 3 +- sqrt( 1353 )) / 6 }}} 
{{{ t = ( 3 + 36.783 ) / 6 }}}
{{{ t = 6.631 }}} 
It normally takes 6.631 hrs to reach town
or 6 hrs 38 min
check:
(1) {{{ 224 = s*t }}}
(1) {{{ 224 = s*6.631 }}}
(1) {{{ s = 33.78 }}}
and
(2) {{{ 224 = ( s + 6 )*( 6.631 - 1 ) }}}
(2) {{{ 224 = ( s + 6 )*5.631 }}}
(2) {{{ s + 6 = 39.78 }}}
(2) {{{ s = 33.78 }}}
OK