Question 851788
It's not easy and probably not a 
test of intelligence
Let {{{ c }}} = the speed of the river ( c for current )
{{{ 15 + c }}} = the speed of the boat going downstream
{{{ 15 - c }}} = the speed of the boat going upstream
Let {{{ t }}} = the time in hours to travel 
from A to B going downstream
{{{ t + 75/60 }}} = the time in hours to travel 
from B to A going upstream
-----------------------------------------
Equation for going downstream:
(1) {{{  45 = ( 15 + c )*t }}}
Equation for going upstream:
(2) {{{ 45 = ( 15 - c )*( t + 75/60 ) }}}
---------------------------------
(2) {{{ 45 = ( 15 - c )*( t + 5/4 ) }}}
(2) {{{ 45 = 15t - c*t + 75/4 - (5/4)*c }}}
(2) {{{ 45 = t*( 15 - c ) + 75/4 - (5/4)*c }}}
and
(1) {{{ t = 45 / ( 15 + c ) }}}
substitute this into (2)
(2) {{{ 45 = ( 45 / ( 15 + c ) )*( 15 - c ) + 75/4 - (5/4)*c }}}
Multiply both sides by {{{ 4*( 15 + c ) }}}
(2) {{{ 180*( 15 + c ) = 180*( 15 - c ) + 75*( 15 + c ) - 5*( 15 + c )*c }}}
(2) {{{ 2700 + 180c = 2700 - 180c + 1125 + 75c - 75c - 5c^2 }}}
(2) {{{ 360c = 1125 - 5c^2 }}}
(2) {{{ 5c^2 + 360c -1125 = 0 }}}
(2) {{{ c^2 + 72c - 225 = 0 }}}
Use quadratic formula
{{{ C = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} ( C and c are different )
{{{ a = 1 }}}
{{{ b = 72 }}}
{{{ c = -225 }}}
{{{ C = ( -72 +- sqrt( 72^2 - 4*1*(-225) )) / (2*1) }}}  
{{{ C = ( -72 +- sqrt( 5184 + 900)) / 2 }}}  
{{{ C = ( -72 +- sqrt( 6084)) / 2 }}}  
{{{ C = ( -72 +- 78) / 2 }}}  
{{{ C = 6/2 }}}
{{{ C = 3 }}} ( can't use the negative square root )
The river flows at 3 km/hr
check:
(1) {{{  45 = ( 15 + 3 )*t }}}
(1) {{{ 45 = 18t }}}
(1) {{{ t = 2.5 }}} hrs
and
(2) {{{ 45 = ( 15 - 3 )*( t + 75/60 ) }}}
(2) {{{ 45 = 12t + 15 }}}
(2) {{{ 12t = 30 }}}
(2) {{{ t = 2.5 }}} hrs
OK
There may be an easier way to solve this- 
I just don't know what it is