Walking at 4km/h, Bruce can make the roundtrip between his campsite and Lookout Point in 2.5 h. Rowing on Crooked River, he can row upstream from the campsite to Lookout Point in 1 h and can row back again in 40 min. Find Bruce’s rate of rowing in still water and the speed of the current in Crooked River.........I know how to solve i would really apreciate it if someone can give me the 2 equations.
rate of rowing= 6.25
i don't think my solution is correct though...
Sorry, I had my equations backwards.
Clearly the time has to be less for a faster speed
so 5/(x+y) = 2/3 --->2x+2y = 15
and 5/(x-y) = 1 ----> x-y= 5
You still get x = 6.25, but y is now 1.25
whereas before it would have been negative.
Good for you to doubt the result.
Thank You So MUCH!!!
why did you do 2x and 2y?
To solve this problem, we need to set up two equations. Let's call Bruce's rate of rowing in still water "x" and the speed of the current in Crooked River "y".
Equation 1: Roundtrip by Walking
Since Bruce walks at a constant speed of 4km/h, we can calculate the total distance covered in 2.5 hours.
The distance from the campsite to Lookout Point is the same as the distance from Lookout Point back to the campsite. Let's denote this distance as "d".
So, the equation for the roundtrip by walking is:
2d = 4km/h * 2.5h (speed * time = distance)
Equation 2: Rowing upstream and downstream
When Bruce rows upstream, he has to overcome the current, which reduces his speed.
The distance from the campsite to Lookout Point is still "d". Let's denote the rate of the current as "y".
So, when Bruce rows upstream, his effective speed is (x - y) km/h. The time taken for this trip is 1 hour.
Therefore, the equation for rowing upstream is:
d = (x - y) km/h * 1h
When Bruce rows downstream, the current assists his speed.
So, his effective speed is (x + y) km/h. The time taken for this trip is 40 minutes, which is equal to 40/60 = 2/3 hours.
Therefore, the equation for rowing downstream is:
d = (x + y) km/h * 2/3h
Now, we have two equations:
2d = 4km/h * 2.5h
d = (x - y) km/h * 1h
d = (x + y) km/h * 2/3h
These equations can be used to solve for the values of "x" and "y", which represent Bruce's rate of rowing in still water and the speed of the current, respectively.
The first sentence just tells you that the roundtrip is 4(2.5) or 10 km, so each way is 5 km
let his rate of rowing in still waters be x km/h
let the rate of the current be x km/h
time to go with the current = 5/(x+y) = 1
or x+y = 5
time to go against the current = 5/(x-y) = 2/3
or 2x-2y = 15
There are your two equations.
You said you knew how to solve them, let me know what you got