SOLUTION: On a typical day in the fall, the scenic riverboat tour travels 11 miles upstream against a 3 mph current. In the spring, the current runs 2 mph faster than in the fall, and the sa

Algebra ->  Test -> SOLUTION: On a typical day in the fall, the scenic riverboat tour travels 11 miles upstream against a 3 mph current. In the spring, the current runs 2 mph faster than in the fall, and the sa      Log On


   

Question 1127534: On a typical day in the fall, the scenic riverboat tour travels 11 miles upstream against a 3 mph current. In the spring, the current runs 2 mph faster than in the fall, and the same upstream trip takes twice as long. What is the speed of the boat in still water?
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = the speed of the boat in still water, in miles per hour.


Then in the fall, the time upstream is  11%2F%28x-3%29 hours.


     in the spring, the time upstream is  11%2F%28x-5%29 hours.


The time equation is


    2%2A%2811%2F%28x-3%29%29 = 11%2F%28x-5%29.


Cancel 11 in both sides


    2%2F%28x-3%29 = 1%2F%28x-5%29.


    2*(x-5) = x - 3


    2x - 10 = x - 3


    x = 10-3 = 7.


Answer.  The speed of the boat in still water is  7 miles per hour.


CHECK.  The spring time is  11%2F%287-5%29 = 11%2F2 = 5.5 hours;


        The fall time is  11%2F%287-3%29 = 11%2F4 = 2.75 hours.


        The spring time is twice the fall time.    ! Correct !