SOLUTION: Rowing with the current of a river, a rowing team can row 25 miles in the same amount of time it takes to row 15 miles against the current. The rate of the rowing team in calm wate

Question 83193This question is from textbook introductory algebra
: Rowing with the current of a river, a rowing team can row 25 miles in the same amount of time it takes to row 15 miles against the current. The rate of the rowing team in calm water is 20 mph. Find the rate of the current. This question is from textbook introductory algebra

Found 2 solutions by checkley75, stanbon:
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
time=distance/rate
seeing as the times are the same we have the following equation:
15/(20-c)=25/(20+c) c=current. now cross multiply.
15(20+c)=25(20-c)
300+15c=500-25c
15c+25c=500-300
40c=200
c=200/40
c=5 rate of the current.
proof
15/(20-5)=25/(20+5)
15/15=25/25
1=1
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Rowing with the current of a river, a rowing team can row 25 miles in the same amount of time it takes to row 15 miles against the current. The rate of the rowing team in calm water is 20 mph. Find the rate of the current
-----------------
With current DATA:
Distance = 25 mi ; Rate = c+20 mph ; Time = 25/(c+20)
----------------------
Against current DATA:
Distance = 15 mi ; Rate = 20-c mph ; Time = 15(20-c)
==================
EQUATION:
time with = time against
25/(c+20) = 15/(20-c)
5/(c+20) = 3/(20-c)
100-5c = 3c+60
8c = 40
c= 5 mph ( rate of current)
================
Cheers,
Stan H.
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