A boat has a speed of 5 mph in still water The boat can trav
A boat has a speed of 5 mph in still water. The boat can travel 21 miles with the current in the same amount of time it takes to travel 9 miles against the current. Find the rate of the current.
Solution
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
(Note: when travelling with the current, we add the speed of the current; when travelling against the current, we subtract the speed of the current)
Let r=rate of the current
Time travelling with the current=21/(5+r)
Time travelling against the current=9/(5-r)
Now we are told that the above times are equal, so:
21/(5+r)=9/(5-r) multiply each side by (5-r)(5+r) or cross multiply
21(5-r)=9(5+r) get rid of parens
105-21r=45+9r subtract 105 and also 9r from each side
105-105-21r-9r=45-105+9r-9r collect like terms
-30r=-60 divide each side by -30
r=2 mph---------------------------speed of the current
CK
21/7=9/3
3=3
