SOLUTION: In Still water, a boat averages 18 miles per hour. It takes the same amount of time to travel 10 miles downstream, with the current. As it does 5 miles Upstream, against the curren

Algebra ->  Rational-functions -> SOLUTION: In Still water, a boat averages 18 miles per hour. It takes the same amount of time to travel 10 miles downstream, with the current. As it does 5 miles Upstream, against the curren      Log On


   

Question 1082577: In Still water, a boat averages 18 miles per hour. It takes the same amount of time to travel 10 miles downstream, with the current. As it does 5 miles Upstream, against the current. What is the rate of the water's current?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +c+ = the rate of the current in mi/hr
+18+%2B+c+ = the rate of the boat going downstream in mi/hr
+18+-+c+ = the rate of the boat going upstream in mi/hr
Let +t+ = time in hrs for both upstream and downstream trips
----------------------------------------------------------
Going downstream:
(1) +10+=+%28+18+%2B+c+%29%2At+
Going upstream:
(2) +5+=+%28+18+-+c+%29%2At+
-------------------------
(1) +t+=+10+%2F+%28+18+%2B+c+%29+
Plug (1) into (2)
(2) +5+=+%28+18+-+c+%29%2A%28+10+%2F+%28+18+%2B+c+%29+%29+
(2) +5%2A%28+18+%2B+c+%29+=+10%2A%28+18+-+c+%29+
(2) +90+%2B+5c+=+180+-+10c+
(2) +15c+=+90+
(2) +c+=+6+
The rate of thr current is 6 mi/hr
--------------------------------
check:
(1) +t+=+10+%2F+%28+18+%2B+c+%29+
(1) +t+=+10+%2F+%28+18+%2B+6+%29+
(1) +t+=+10%2F24+
(1) +t+=+5%2F12+ hrs
and
(2) +t+=+5+%2F+%28+18+-+c+%29+
(2) +t+=+5+%2F+%28+18+-+6+%29+
(2) +t+=+5%2F12+ hrs
---------------------------
Note:
+5%2F12%2A60+=+25+ min
OK