while on vacation Kevin went for a swim. Swimming against the current took 8 min. to go 200 meters coming back took half as long. Find Kevin's average swimming speed and the water current.
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Let x = current speed in meters/sec
Let y = Kevin's speed in same units then we can use the s=rt formula
where s=distance, r = speed, and t = time
Then kevins speed swimming against current is y-x
kevins speed swimming with the current is y+x
The distance was the same both ways 200 meters.
Here are two equations for each trip
200 = 8(y-x) (8 minutes when swimming against current ) eq1
200 = 4(y+x) (4 minutes half the time when swimming with the current) eq2
200=8y-8x eq 1
200=4y+4x eq 2
400=8y+8x eq 2 multiplied by 2
200=8y-8x eq 1
substituting in eq 2
200=4(37.5) + 4x becomes 200=150+4x becomes 4x=50 or x =12.5 meter/min
Change the units in my first post to meters/min as minutes were the units given in the problem.