MATH SOLVE

8 months ago

Q:
# Dave walked to his friend's house at a rate of 4 mph and returned back biking at a rate of 10 mph. If it took him 18 minutes longer to walk than to bike, what was the total distance of the round trip?

Accepted Solution

A:

Answer:4 milesStep-by-step explanation:Lets start by trying to get the rates of walk and bike. We know Dave walks at a rate of 4 mph, so, it takes him 60 minutes to walk 4 miles. We can use this proportion to find out how much times takes him to walk 1 mile:60/4 = 15/1 = RwSo, his rate for walking is 1 mile each 15 minutes, or, 15 minutes per mile. Now we do the same for biking, as he bikes at 10 mph. He then bikes 10 miles per 60 minutes:60/10 = 6So, he bikes 1 mile each 6 minutes, or, he needs 6 minutes for per mile.If x is the distance he walked and y the one the biked (in miles), 15x is the time it takes to walk the distance and 6y the time it takes to bike the distance.We know that the time it took him to walk is 18 minutes greater than the time that took him to bike, so:18 + 6y = 15xBut as he traveled between 2 points, the distances are the same, x=y, so:18 + 6x = 15xSubtracting 6x in both sides:18 = 9xDividing by 9 in both:2 = xSo, the distance to his friend's house is 2 miles, and the back distance is also 2 miles. Thus, the trip has a distance of 4 miles.