A bicyclist rode into the country for 5 h. In returning, herspeed was 5 mi/h faster and the trip took 4 h. Wha

What question do you ask/

Assume the original speed is x

distance = 5x

In returning,
(x + 5) * 4 = distance = 5x

4x + 20 = 5x
x = 20

Wha is the question?!!!

The question is cut off, so I'm not sure what you're asking. However, these problems are all solved using d=rt (distance = rate x time).

Since it's a return trip, the distances are the same. If we let r = her speed going, then r+5 = speed returning. Now we have an equation:

5r = 4(r+5)

both of these are rate x time = distance which is the same.

Solve for r:

5r = 4r + 20
r = 20

Now we know her rate going out to the country was 20 mi/h, returning was 25 mi/h. Also, the distance was 5(20)= 100 miles.

that's it! :)

let the initial speed is x

distance = 5x

(x + 5) * 4 = 5x

4x + 20 = 5x

x = 20 (mi/h)

Set this up as an algebra problem and solve accordingly.