SOLUTION: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?
Will someone please explain this t
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?
Will someone please explain this t
Log On
Question 157512: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?
Will someone please explain this to me? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Sure!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=rate (speed ) of the current
Time to travel upstream=300/(33-r) {note: when we go against the current we must subtract the rate of the current)
Time to travel downstream=360/(33+r) (note: downstream we add the rate of the current)
Now we are told that the above two times are the same, so:
300/(33-r)=360/(33+r) multiply each side by (33-r)(33+r) (or cross-multiply)
300(33+r)=360(33-r) divide each side by 60 (I did this just to reduce the size of the numbers that we have to deal with)
5(33+r)=6(33-r) get rid of parens
165+5r=198-6r subtract 165 from and add 6r to both sides
165-165+5r+6r=198-165-6r+6r collect like terms
11r=33 divide each side by 11
r=3 mph---------------------------------------speed of the current
CK
300/(33-3)=360/(33+3)
300/30=360/36
10=10