Question 351962: A family drives at an average rate of 60mi/h on the way to visit relatives and then at an average rate 40mi/h on the way back. The return trip takes 1h longer than the trip there. How many hours did it take to drive there? How many hours did it take to make the return trip? What was the average rate of for the entire trip?
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! When a problem mentions average (or constant) speed we should think of the
distance = rate * time
or
d = rt
equation.
With only the speeds it may seem that we do not have enough information. But if we realize that the distance going to a place is the same as the distance coming back from the same place, we can solve this problem. Since the distances to and from are equal then the "to" rate * time equals the "from" rate * time. So if
x = the time going to the releatives
and if it takes 1 hour longer coming back, then the time coming back would be:
x + 1
With a "to" rate of 60 and a "from" rate of 40,
60 * x = 40 * (x+1)
This equation we can solve. Start by simplifying each side:
60x = 40x + 40
Get the x's on just one side (by subtracting 40x):
20x = 40
Divide both sides by 20:
x = 2
So it took 2 hours to drive there. On the way back it took an hour longer so it took 3 hours.
To find the average rate for the whole trip we will need the total distance and the total time. From above, we know that the total time is 2 + 3 or 5 hours. All we need now is the total distance. The total distance would be twice the distance "to" (or "from"). The distance "to" can be found with the d = rt equation now that we know both the "to" rate and time:
d = 60 * 2
or
d = 120
So the total distance is 2 * 120 or 240 miles. For the average rate for the complete trip we use the d = rt equation again, this time with the total distance and total time:
240 = r * 5
Solving this (by dividing both sides by 5) we get:
48 = r
This makes the average rate 48 mi/h,
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