SOLUTION: the rate of the boat in still water is 8 mph. At this rate a 30 mile trip downstream takes the same amount of time as an 18 mile trip upstream. What is the rate of the current?

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Question 518231: the rate of the boat in still water is 8 mph. At this rate a 30 mile trip downstream takes the same amount of time as an 18 mile trip upstream. What is the rate of the current?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = rate in still water
Let +c+ = rate of current
going downstream:
(1) +d%5B1%5D+=+%28+s+%2B+c+%29%2At+
Going upstream:
(2) +d%5B2%5D+=+%28+s+-+c+%29%2At+
-----------------
given:
+d%5B1%5D+=+30+ mi
+d%5B2%5D+=+18+ mi
+s+=+8+ mi/hr
---------------
(1) +30+=+%28+8+%2B+c+%29%2At+
(2) +18+=+%28+8+-+c+%29%2At+
-----------------
(1) +30+=+8t+%2B+c%2At+
(2) +18+=+8t+-+c%2At+
Add the equations
+48+=+16t+
+t+=+3+ hrs
and, since
(1) +30+=+8t+%2B+c%2At+
(1) +30+=+24+%2B+3c+
(1) +3c+=+6+
(1) +c+=+2+
The rate of the current is 2 mi/hr
check:
(1) +30+=+%28+8+%2B+c+%29%2At+
(1) +30+=+%28+8+%2B+2+%29%2A3+
(1) +30+=+10%2A3+
OK