Question
Noreen can travel 8 miles upstream in the same time it takes into travel 12 miles downstream. Her boat travels 15 mph in Stillwater. What is the rate of the current
Answers
GPT-4o mini
To find the rate of the current, we can start by defining some variables:
- Let \( c \) be the rate of the current in mph.
- The speed of Noreen's boat in still water is 15 mph.
When traveling upstream, the effective speed of the boat is \( 15 - c \) mph, and when traveling downstream, the effective speed is \( 15 + c \) mph.
Given:
- Distance upstream = 8 miles
- Distance downstream = 12 miles
According to the problem, the time taken to travel upstream is equal to the time taken to travel downstream. We can express time as:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}}
\]
Setting up the equation for upstream and downstream travel, we have:
\[
\frac{8}{15 - c} = \frac{12}{15 + c}
\]
Now, we can cross-multiply to solve for \( c \):
\[
8(15 + c) = 12(15 - c)
\]
Expanding both sides gives:
\[
120 + 8c = 180 - 12c
\]
Now, we can combine like terms:
\[
8c + 12c = 180 - 120
\]
\[
20c = 60
\]
Dividing both sides by 20:
\[
c = 3
\]
Thus, the rate of the current is \( \boxed{3} \) mph.
- Let \( c \) be the rate of the current in mph.
- The speed of Noreen's boat in still water is 15 mph.
When traveling upstream, the effective speed of the boat is \( 15 - c \) mph, and when traveling downstream, the effective speed is \( 15 + c \) mph.
Given:
- Distance upstream = 8 miles
- Distance downstream = 12 miles
According to the problem, the time taken to travel upstream is equal to the time taken to travel downstream. We can express time as:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}}
\]
Setting up the equation for upstream and downstream travel, we have:
\[
\frac{8}{15 - c} = \frac{12}{15 + c}
\]
Now, we can cross-multiply to solve for \( c \):
\[
8(15 + c) = 12(15 - c)
\]
Expanding both sides gives:
\[
120 + 8c = 180 - 12c
\]
Now, we can combine like terms:
\[
8c + 12c = 180 - 120
\]
\[
20c = 60
\]
Dividing both sides by 20:
\[
c = 3
\]
Thus, the rate of the current is \( \boxed{3} \) mph.