Question 591146
{{{ s - c }}} = speed of boat going upstream
{{{ s + c }}} = speed of boat going downstream
Let {{{ t }}} = time going downstream
given:
{{{ c = 3 }}} km/hr
Going downstream:
(1) {{{ 40 = ( s + 3 )*t }}}
Going upstream:
(2) {{{ 40 = ( s - 3 )*( 14 - t ) }}}
-------------------------
(1) {{{ s + 3 = 40/t }}}
(1) {{{ s = 40/t - 3 }}}
and
(2) {{{ 40 = ( 40/t - 3 - 3 )*( 14 - t ) }}}
(2) {{{ 40 = ( 40/t - 6 )*( 14 - t ) }}}
(2) {{{ 40 = 560/t - 84 - 40 + 6t }}}
(2) {{{ 6t + 560/t = 164 }}}
(2) {{{ 6t^2 + 560 = 164t }}}
(2) {{{ 3t^2 - 82t + 280 }}}
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 3 }}}
{{{ b = -82 }}}
{{{ c = 280 }}}
{{{ t = ( -(-82) +- sqrt( (-82)^2 - 4*3*280 )) / (2*3) }}} 
{{{ t = ( 82 +- sqrt( 6724 - 3360 )) / 6 }}} 
{{{ t = ( 82 +- sqrt( 3364 )) / 6 }}} 
{{{ t = ( 82 + 58) / 6 }}} 
{{{ t = 140/6 }}}
{{{ t = 23.333 }}} too big
{{{ t = ( 82 - 58) / 6 }}} 
{{{ t = 24/6 }}}
{{{ t = 4 }}}
and
(1) {{{ s = 40/t - 3 }}}
(1) {{{ s = 40/4 - 3 }}}
(1) {{{ s = 10 - 3 }}}
(10 {{{ s = 7 }}}
The speed of the boat in still water is 7 km/hr
check:
(2) {{{ 40 = ( s - 3 )*( 14 - t ) }}}
(2) {{{ 40 = ( 7 - 3 )*( 14 - 4 ) }}}
(2) {{{ 40 = 4*10 }}}
(2) {{{ 40 = 40 }}}
OK