Question 620028
For now, 
Let {{{ b }}} = the speed of the boat in still water
Let {{{ s }}} = the speed of the stream
{{{ b + s }}} = the speed of the boat going downstream
{{{ b - s }}} =  the speed of the boat going upstream
Let {{{ t }}} = time for the boat to travel upstream
{{{ 2.75 - t }}} = time to travel downstream ( {{{2.75}}} is 2 hrs 45 min )
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Going upstream:
(1) {{{ 12 = ( b - s )*t }}}
Going downstream:
(2) {{{ 12 = ( b + s )*( 2.75 - t ) }}}
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It is given that {{{ b = 11 }}}
(1) {{{ 12 = ( 11 - s )*t }}}
(1) {{{ t = 12/( 11 - s ) }}}
and
(2) {{{ 12 = 2.75b - b*t + 2.75s - s*t }}}
(2) {{{ 12 = 2.75*11 - 11t + 2.75s - s*t }}}
(2) {{{ 12 = 30.25 - 11t + 2.75s - s*t }}}
(2) {{{ t*( 11 + s ) = 30.25 - 12 + 2.75s }}}
From (1),
(2) {{{ (12/( 11 - s ))*( 11 + s ) = 18.25 + 2.75s }}}
(2) {{{ 12*( 11 + s )  = ( 18.25 + 2.75s )*( 11 - s ) }}}
(2) {{{ 132 + 12s = 200.75 + 30.25s - 18.25s - 2.75s^2 }}}
(2) {{{ -2.75s^2 + 30.25s - 18.25s - 12s - 132 + 200.75 = 0 }}}
(2) {{{ -2.75s^2 + 68.75 = 0 }}}
(2) {{{ 2.75s^2 = 68.75 }}}
(2) {{{ s^2 = 68.75/2.75 }}}
(2) {{{ s^2 = 25 }}}
(2) {{{ s = 5 }}}
The speed of the stream is 5 km/hr
check:
(1) {{{ 12 = ( 11 - s )*t }}}
(1) {{{ 12 = ( 11 - 5 )*t }}}
(1) {{{ 12 = 6t }}}
(1) {{{ t = 2 }}} hrs
and
(2) {{{ 12 = ( 11 + s )*( 2.75 - 2 ) }}}
(2) {{{ 12 = .75*( 11 + s ) }}}
(2) {{{ 12 = 8.25 + .75s }}}
(2) {{{ .75s = 3.75 }}}
(2) {{{ s = 5 }}}
OK