distance = time * speed
Let
b = boat's speed
c = current's speed
5(b-c) = 100
2(b+c) = 100
b-c = 20
b+c = 50
I guess you can take it from there, yes?
Let
b = boat's speed
c = current's speed
5(b-c) = 100
2(b+c) = 100
b-c = 20
b+c = 50
I guess you can take it from there, yes?
Going upstream:
The speed of the boat relative to the water is reduced by the speed of the current, so the effective speed is (b - c). We know that the boat takes 5 hours to cover 100 miles upstream, so we can set up the equation:
Time = Distance / Speed
5 = 100 / (b - c)
Going downstream:
The speed of the boat relative to the water is increased by the speed of the current, so the effective speed is (b + c). We know that the boat takes 2 hours to cover 100 miles downstream, so we can set up the equation:
Time = Distance / Speed
2 = 100 / (b + c)
Now, we have a system of two equations with two unknowns. We can solve these equations simultaneously to find the values of "b" and "c."
First, let's rearrange the first equation:
5(b - c) = 100
b - c = 20 (Equation 1)
Next, let's rearrange the second equation:
2(b + c) = 100
b + c = 50 (Equation 2)
Now, we have a system of two equations:
b - c = 20 (Equation 1)
b + c = 50 (Equation 2)
We can solve this system of equations by adding Equation 1 and Equation 2:
( b - c) + (b + c) = 20 + 50
2b = 70
b = 35
Substituting the value of "b" in Equation 2, we can find "c":
35 + c = 50
c = 50 - 35
c = 15
Therefore, the rate of the boat in still water is 35 mph and the rate of the current is 15 mph.