Question 470473
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Let *[tex \Large d] represent the distance for either of the trips which we know to be 60 miles.  Let *[tex \Large r] represent the rate of the boat in still water.  Let *[tex \Large r_c] represent the rate of the current.  Let *[tex \Large t_d] represent the time for the downstream trip which we know to be 2 hours.  And let *[tex \Large t_u] represent the time for the upstream trip which we know to be 6 hours.


When the boat is going downstream, the rate of the boat relative to the stationary bank of the river is the rate of the boat in still water PLUS the rate of the current.  Conversely, going upstream the rate of the boat relative to the bank is the rate in still water MINUS the rate of the current.


We can describe the downstream trip thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \left(r\ +\ r_c)t_d]


and the upstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \left(r\ -\ r_c)t_u]


Now plug in the known values:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 60\ =\ \left(r\ +\ r_c)2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 60\ =\ \left(r\ -\ r_c)6]


A little manipulation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r\ +\ 2r_c\ =\ 60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6r\ -\ 6r_c\ =\ 60]


Solve the first equation for *[tex \Large r] in terms of *[tex \Large r_c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 30\ -\ r_c]


Substitute the RHS expression in place of *[tex \Large r] in the second equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(30\ -\ r_c)\ -\ 6r_c\ =\ 60]


All that is left is to solve for *[tex \Large r_c]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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