A rowing team rowed 50 miles while going with the current in the same amount of time as it took to row 25 miles going against the current. The rate of the current was 2 miles per hour. Find the rate of the rowing team in still water.
rowing speed in still water ---- x mph
speed of current ------------- 2 mph
time to go 50 miles with the current = 50/(x+2)
time to go 25 miles against current = 25/(x-2)
the two times are the same, so ...
50/(x+2) = 25/(x-2)
50x - 100 = 25x + 50
25x = 150
x = 6
speed in still water is 6 mph
A rowing team rowed 100
miles while going with the current in the same amount of time as it took to row
20 miles going against the current. The rate of the current was
6 miles per hour. Find the rate of the rowing team in still water.
Why did the rowing team enroll in a current affairs class? Because they wanted to learn how to navigate the currents while rowing! Now, let's work this out.
Let's call the rate of the rowing team in still water 'r' miles per hour.
When rowing with the current, their effective rate would be (r + 2) miles per hour. And the total time to row 50 miles can be represented as 50 / (r + 2).
When rowing against the current, their effective rate would be (r - 2) miles per hour. And the total time to row 25 miles can be represented as 25 / (r - 2).
Since the rowing team took the same amount of time for both scenarios, we can set up the equation:
50 / (r + 2) = 25 / (r - 2)
Now, cross multiply:
50(r - 2) = 25(r + 2)
Expanding and simplifying:
50r - 100 = 25r + 50
Subtracting 25r from both sides, we get:
25r = 150
Dividing by 25, we find:
r = 6
So, the rate of the rowing team in still water is 6 miles per hour. Now let's hope they don't get tangled in any water-related current events!
To solve this problem, we can let the rate of the rowing team in still water be x miles per hour.
When rowing with the current, the rowing team's effective speed is increased by the rate of the current, so their speed is (x + 2) miles per hour.
When rowing against the current, the rowing team's effective speed is decreased by the rate of the current, so their speed is (x - 2) miles per hour.
We are given that the team rowed 50 miles with the current in the same amount of time as it took to row 25 miles against the current.
Using the formula distance = rate × time, we can set up the following equation for rowing with the current:
50 = (x + 2) × time
And the following equation for rowing against the current:
25 = (x - 2) × time
Since the time taken is the same in both cases, we can set the two equations equal to each other:
(x + 2) × time = (x - 2) × time
The "time" variable cancels out, leaving us with:
x + 2 = x - 2
Simplifying the equation:
2 = -2
Since we arrive at an incorrect statement, we can conclude that there is no valid solution to this problem. The given situation is not possible.
The rate of the rowing team in still water cannot be determined based on the information given.