Question

You drive 4 hours to reach a campground driving home along the same route you increase your speed by 15 miles per hour reducing your driving time by 1 hour how fast were you driving to the camp ground

Answer 1

Answer: 45 mph

Work Shown:

H = home, C = campground

`\begin{array}c \cline{1-4} & \text{distance} & \text{rate} & \text{time}\\\cline{1-4}\text{H to C} & d & r & 4\\\cline{1-4}\text{C to H} & d & r+15 & 3\\\cline{1-4}\end{array}`

Each distance is in miles. Each rate is in miles per hour (mph). Each time value is in hours.

The formula we need is: distance = rate*time

It shortens to d = r*t

Going from H to C, we have the equation d = r*4 or d = 4r

Then from C to H, we have the equation d = 3(r+15)

Both equations involve the same distance d because you traveled the same route both ways.

This means we can equate the right hand sides to solve for variable r.

4r = 3(r+15)

4r = 3r+45

4r-3r = 45

r = 45 mph is the final answer

The portion going from H to C you travel 45 mph. Going for 4 hours means the distance traveled was d = r*t = 45*4 = 180 miles.

Getting back home increases the speed to 45+15 = 60 mph. Then notice how time = distance/rate = 180/60 = 3 hours, which helps us see that you arrive back home 1 hour faster compared to when going from H to C. This helps confirm the answer is correct.