rate question?

Question

in still water, a boat averages 7mph. it takes the same amount of time to travel 20 miles downstream with the current, as 8 miles up stream against the current, what is the rate?

how do i set this up to solve it?

Answer 1

rate * time = distance
or
time = distance / rate

x = rate of current

downstream:
rate = 7 + x
time = T
distance = 20 miles

so... T = 20 / (7 + x)

upstream:
rate = 7 - x
time = T
distance = 8 miles

so... T = 8 / (7 - x)

Since the T is the same in each (as we're told in the problem), just set those equal and solve.

20 / (7 + x) = 8 / (7 - x)
20(7 - x) = 8(7 + x)
140 - 20x = 56 + 8x
84 - 20x = 8x
84 = 28x
3 = x

The current travels at 3 mph/

Answer 2

The thing to remember with these types of problems is that the total speed going upstreat is the speed in still water minus the speed of the stream, and going downstream it's the speed in still water plus the stream's sped.

Speed is distance divided by time, so time is distance divided by speed. The distance going downstream is 20 miles, and the speed is 7 + C, where C is the speed of the current. The distance going upstream is 8, and the speed is 7 - C. We're told that the two times are the same, so set one time equal to the other:
20 / (7+C) = 8 / (7-C)

That's how you can set up the problem. Then just cross-multiply and solve for C.

Answer 3

Let speed of current be x mph
Downstream
speed = x + 7 mph
D = 20 miles
t = 20 / (x + 7)

Upstream
speed = 7 - x mph
D = 8 miles
t = 8 / (7 - x)

20 / (x + 7) = 8 / (7 - x)
20 (7 - x) = 8 (x + 7)
140 - 20x = 8x + 56
84 = 28x
x = 3
Current speed = 3 mph

Answer 4

20 = t(7 + c)
8 = t(7 - c), where c = current's speed and t = common time.
20 = 7t + 7c
8 = 7t - 7c
adding, 28 = 14t
t =2
so 20 = 2(7 + c), and c = 3 mph.