Question 118448
Calculating the downstream and upstream rates was a good start, except that you made a slight arithmetic error in calculating the downstream rate -- {{{20/3=6.67}}} approximately, not 6.23.


  Before we do anything else, we need to make a couple of assumptions.  Since this is a sailboat, the boat speed through still water is a function of the wind speed and direction, so let's assume that the wind speed remained constant for the entire trip and the wind direction was at 90 degrees to the direction of travel.  That eliminates any variables due to the wind.  Second we have to assume that the rate of the current was constant for the entire trip.


But let's look at the problem a little differently.  We know that {{{d=rt}}} where d is distance, r is rate, and t is time.  What we need to deal with is two different rates:  One is the rate of the boat through still water, and the other is the rate of the current.  When the boat is travelling downstream, the rate of the current ADDS to the overall speed of the boat, and when traveling upstream the current SUBTRACTS from the overall speed.


Let's say the rate of the boat through still water is {{{r[s]}}} and the rate of the current is {{{r[c]}}}.  The distance traveled in either direction is 20 miles, so we can write:


{{{20 = (r[s]+r[c])3}}} for the downstream trip, and


{{{20 = (r[s]-r[c])4}}} for the upstream trip.


Divide both sides of the first equation by 3 (leaving the left side result in improper fraction form for the time being), and both sides of the second equation by 4.


{{{20/3=r[s]+r[c]}}}


{{{5=r[s]-r[c]}}}


Now, add these two equations term by term:


{{{(20/3)+5=r[s]+r[s]+r[c]-r[c]}}}, and simplify:


{{{35/3=2r[s]}}}
{{{r[s]=35/6}}}, and now we know the speed of the boat through still water, i.e. 5 and 5/6 or roughly 5.83 mph.


Using this value, we can substitute in either of the original equations to solve for the rate of the current.


{{{5=(35/6)-r[c]}}}
{{{(30/6)-(35/6)=-r[c]}}}
{{{r[c]=5/6}}}, and now we know the speed of the current, i.e. 5/6 or roughly 0.83 mph.


Let's check the answer.
If the boat goes {{{(35/6)+(5/6)}}} mph downstream, then using {{{t=d/r}}} or {{{t=20/(40/6)}}}, should give us 3 hours.  I'll let you do the arithmetic.


If the boat goes {{{(35/6)-(5/6)}}} mph upstream, {{{t=20/(30/6)}}} should give us 4 hours.  Again, you get to do the arithmetic.


Hope that helps,
John