Question 161731
Let the rate of the current = {{{c}}}
Let his rate rowing in still water = {{{s}}}
Going upriver is against the current
{{{10/(s - c) = 5}}}
Going back is with the current
{{{10/(s + c) = 2.5}}}
This is 2 equations and 2 unknowns, so I should
be able to solve
Multiply both sides of the 1st by {{{s - c}}}
Multiply both sides of the 2nd by {{{s + c}}}
(1) {{{10 = 5*(s - c)}}}
(2) {{{10 = 2.5*(s + c)}}}
----------------------
(1) {{{10 = 5s - 5c}}}
(2) {{{10 = 2.5s + 2.5c}}}
Multiply (2) by {{{2}}} and add equations
(1) {{{10 = 5s - 5c}}}
(2) {{{20 = 5s + 5c}}}
{{{30 = 10s}}}
{{{s = 3}}} km/hr
And, since
{{{10/(s - c) = 5}}}
{{{10/(3 - c) = 5}}}
{{{10 = 5*(3 - c)}}}
{{{10 = 15 - 5c}}}
{{{5c = 5}}}
{{{c = 1}}} km/hr
His rate in still water is 3 km/hr and the rate of the 
current is 1 km/hr
check:
{{{10/(s - c) = 5}}}
{{{10/(3 - 1) = 5}}}
{{{10/2 = 5}}}
{{{5 = 5}}}
OK