Question

A river is flowing at the speed of 2 miles per hour. A kayaker moves 528 feet upstream in 20 minutes. At this rate, how fast would the kayaker be paddling in calm water?

Answer 1

D=rt
r=(boatrate+riverrate)=(x+2)mph

d=528ft=528/5280mi=1/10mi
t=1/3hr
remember to keep the same units
so

1/10=(x+2)(1/3)
times both sides by 3
3/10=x+2
minus 2 or 20/10
-17/10=x
-1.7=x
it would be going -1.7mph (means going backwards)