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Distance/Speed/Weighted Average

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Distance/Speed/Weighted Average

by johndoe88 » Wed Oct 16, 2013 8:15 pm
Susan drove an average speed of 30 miles per hour for the first 50 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip.


Guys, I know how to solve this problem in the traditional way so please I am not looking for the solution as much as the thought process.

We know that for the first 50 miles she drove 30m/h
For the remaining 30 miles she drove at 60 m/h

We are asked for the average speed (m/h). Which is Total Distance/Total Time. Easy to calculate.
But note that in this weighted average problem we are weighting the speeds by time.
(t1*X+t2*Y)/(t1+t2)=Total Distance/Total Time.

But notice the weights t1/(t1+t2)& t1/(t1+t2) are just scalars! It simply a percentage/ratio of t1 to total. IF that is the case why can't I scale the 30m/h and the 60m/h by the distances for each respective portions? So w1=d1/(d1+d2)=50/80 and w2=d2/(d1+d2)=30/80 and multiply the weights by their respective speeds? What does doing so get me?
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by [email protected] » Wed Oct 16, 2013 9:26 pm
Hi johndoe88,

What you're talking about doing is actually a much-more complex calculation than the Average Speed Formula and it uses the exact same information.

It's important to remember the Distance is the end result of the Distance Formula (D = R x T), so you're never going to multiply the distance by anything (even if it's a ratio of part/whole, you're still not going to multiply it by anything).

IF you want to use a ratio of part/whole for the distance, you have to DIVIDE by the speed (just as in the Average Speed Formula. Here's how it would work in this question:

(5/8) / 30 + (3/8) / 60 =
5/240 + 3/480 =
10/480 + 3/480 =
13/480
This fraction is the ratio of Total Time to Total Distance, based on the ratio of each Distance to the Total Distance....

Flip it:

480/13 = Average Speed for the entire trip.

If you think that this is faster than just using the Average Speed Formula, then more power to you.

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by VivianKerr » Wed Oct 23, 2013 4:58 pm
Just for fun, here's two more average speed problems to get more practice!

I got in my car and drove 40 miles to see my cousin and was going 20 mph. It took me 2 hours to get there. Then, I left my cousin's and drove another 30 miles to the store but this time went 10mph. It took me 3 hours to arrive at the store. What was my "Average Speed" for the whole trip?

Average Speed = Total Distance / Total Time. I traveled 40 miles + 30 miles so my Total Distance was 70 miles. I drove for 2 hours + 3 hours so my Total Time was 5 hours. 70/5 = 14. My Average Speed for the whole trip was 14 mph. Think of Average Speed as a weighted average. I spent more time going 10mph than 20mph, so it makes sense that the Average Speed would be closer to 10mph.

Tracey ran to the top of a steep hill at an average pace of 6 miles per hour. She took the exact same trail back down. To her relief, the descent was much faster; her average speed rose to 14 miles per hour. If the entire run took Tracey exactly one hour to complete and she did not make any stops, how many miles is the trail one way?

For the way up the hill, we know that D = 6mph x T. For the way down the hill, we know that D = 14mph x T. Since we went know that the distance up the hill was the same as the distance down the hill, we can pick a number for D. Let's choose "84" since it is a multiple of both 6 and 14. If 84 = 6mph x T, then we know that T = 14 hours. If 84 = 14mph x T, then we know that T = 6 hours. Now we can use another formula, the Average Rate formula, to find the average speed for the WHOLE trip. Average Rate = Total Distance / Total Time

Using our Picked Number of 84, we know that the Total Distance traveled would be 168 miles. The Total Time is 14 hours + 6 hours = 20 hours. So the Average Rate = 168 miles / 20 hours = 8.4 mph. It doesn't matter that Tracey didn't "really" go 168 miles, or that we know she didn't "really" go 20 hours. We Picked a Number just so that we could find the ratio of the Total Distance to the Total Time in order to calculate the Average Rate of the ENTIRE journey.

Now that we have found the Average Rate for the whole trip, we can plug it in to the "DIRT" formula to find the ACTUAL distance for the entire journey. We know that T = 1 hour because the problem told us so. Therefore, the actual distance for the entire trip was 8.4 miles. The problem asks how many miles the trail was one way. 8.4 / 2 = 4.2. The answer to the question is 4.2 miles.
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