Question 696141
I think this should read:
"A fisherman can row a boat 8 km downstream and return in 100 minutes.If the speed of the stream is 2 km per hour.Find the speed of the boat in still water."
Am I right? If so,
Let {{{ s }}} = the speed of the boat in still water in km/hr
Let {{{ t }}} = the time to go downstream
{{{ s + 2 }}} = fisherman's speed going down stream in km/hr
{{{ s - 2 }}} = fisherman's speed going upstream in km/hr
Convert {{{ 100 }}} min to hours
{{{ 100/60 = 5/3 }}} hrs
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Equation going downstream:
(1) {{{ 8 = ( s + 2 )*t }}}
Equation going upstream:
(2) {{{ 8 = ( s - 2 )*( 5/3 - t ) }}}
--------------------------
(1) {{{ t = 8/( s + 2 ) }}}
Substitute (1) into (2)
(2) {{{ 8 = ( s - 2 )*( 5/3 - 8/( s+2 ) ) }}}
(2) {{{ 8 = (5/3)*s - 10/3 - (8s) / ( s+2 ) + 16/( s+2 ) }}}
Multiply both sides by {{{ 3*(s + 2) }}}
(2) {{{ 24*( s+2 ) = 5s*( s+2 ) - 10*( s+2 ) - 24s + 48 }}}
(2) {{{ 24s + 48 = 5s^2 + 10s - 10s - 20 - 24s + 48 }}}
(2) {{{ 5s^2 - 48s - 20 = 0 }}}
Use quadratic formula
{{{ s = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 5 }}}
{{{ b = -48 }}}
{{{ c = -20 }}}
{{{ s = (-(-48) +- sqrt( (-48)^2 - 4*5*(-20) )) / (2*5) }}}  
{{{ s = ( 48 +- sqrt( 2304 + 400 )) / 10 }}}  
{{{ s = ( 48 +- sqrt( 2704 )) / 10 }}} 
{{{ s = ( 48 + 52 ) / 10 }}} ( can't use the negative square root )
{{{ s = 100/10 }}}
{{{ s = 10 }}}
The speed of the boat in still water is 10 km/hr
check answer:
(1) {{{ 8 = ( s + 2 )*t }}}
(1) {{{ 8 = ( 10 + 2 )*t }}}
(1) {{{ 8 = 12t }}}
(1) {{{ t = 2/3 }}}
and
(2) {{{ 8 = ( s - 2 )*( 5/3 - t ) }}}
(2) {{{ 8 = ( 10 - 2 )*( 5/3 - t ) }}}
(2) {{{ 8 = 40/3 - 8t }}}
(2) {{{ 8t = 40/3 - 24/3 }}}
(2) {{{ 8t = 16/3 }}}
(2) {{{ t = 2/3 }}}
OK