Question

Two towns are 1050 miles apart, a group of hikers start from each town and walk the trail toward each other. They meet after a total hiking time of 200 hours. If one group travels 1 1/2 miles Per hour faster than the other group, find the rate of each group

Answer 1

Rate of the faster group = 3.38 miles per hour

Rate of the slower group = 1.88 miles per hour

Step-by-step explanation:

Let x = rate of the slower group

Therefore the rate of the faster group will be x + 1 1/2 = x + 3/2 = x + 1.5

From the question, we're told that the two groups traveled for a total hiking time of 200 hours.

We know that distance = rate x time

So the distance of the slower group will be = 200x

And the distance of the faster group will be = 200(x + 1.5)

So if the distance between each town is 1050, we can then solve of x as shown below;

`\begin{gathered} 200x+200(x+1.5)=1050 \\ 200x+200x+300=1050 \\ 400x=750 \\ x=(750)/(400) \\ x=1.88\text{ mph} \end{gathered}`

Therefore the rate of the faster group = 1.88 + 1.5 = 3.38 mph.