<p>A train went 300 miles from City X to City Y at an average rate of 80 mph. At what speed did it travel on the way back if its average for the whole trip was 100 mph?</p>
<p>a 120mph
b 125 mph
c 133 1/3 mph
d 137 1/3 mph
e 150 mph</p>
<p>One way is to use the formula </p>
<p>avg rate = (2ab) / (a+b)</p>
<p>a = speed 1
b = speed 2</p>
<p>In this case avg rate = 100 , a = 80, and we’re trying to find b (the other speed).</p>
<p>100 = (2<em>80</em>b) / (80 + b)
100 = (160b) / (80 + b)
8000 + 100b = 160b
8000 = 60b
133.333… = b</p>
<p>=D</p>
<p>Another way.<br>
We know the trip is 600 miles and total travel time is 6 hours (600/100mph). The first way took 3.75 hours (300/80mph), leaving 2.25 hours to travel 300 miles. Divide the miles by the remaining time, 300/2.25 = 1.33.333.</p>
<p>@Yankee Belle: Why is the total speed is equal to the average speed? The reason I ask this because the time and the distance are the total, but the speed is the average. So it doesn’t seem to match.</p>
<p>Yankee’s logic is fine; that’s the way I solved it.</p>
<p>Never seen that equation, Snorlax. I’ll keep it in mind.</p>
<p>I know that it’s fine. But I need further explanation concerning how it works</p>
<p>total time = time of leg one plus time of leg two = 6 hours</p>
<p>and</p>
<p>time of leg one = 300 miles / 80mph = 3.75 hours, so:</p>
<p>6 = time of leg one + 3.75, therefore time of leg one = 2.25.</p>
<p>speed during leg two = distance of leg two divided by time = 300 miles / 2.25 = 133.3… mph</p>
<p>Why is the total speed is equal to the average speed? The reason I ask this because the time and the distance are the total, but the speed is the average. So it doesn’t seem to match.</p>
<p>Total speed is not equal to average speed.</p>
<p>but didn’t you divide 600 ( total distance) by 100 ( average speed) to get 6 ( total time)?</p>
<p>Well first of all, I don’t know why I said what I said in post #7 because the concept of total speed is not well defined because speed is inherently an average.</p>
<p>But anyways, because the total trip’s distance is 600 miles and the average speed is given to be 100 mph, we know that the time taken is six hours (by d=rt). The individual average speeds for each leg are not relevant in determining this figure.</p>
<p>In order to solve d=rt for r of the second leg, we need to know the time the second leg takes. To figure this out, we subtract the time for leg one from the already determined time for the total trip. In order to determine leg one’s duration, we again use d=rt. </p>
<p>300=80t. t = 3.75, so 6-3.75=2.25; leg two takes 2.25 hours.</p>
<p>Plug in and solve for r. 300=2.25r. r= 133.3… mph.</p>
<p>I hope this explanation helped.</p>
<p>Supposed the answer is 133. Why can’t we plus 133 and leg one’s speed to have 213 mph. Then, what is 100 mph?</p>
<p>100 mph is the total distance divided by the total time. it has nothing to do with the speeds of the legs</p>
<p>100 miles per hour is the AVERAGE speed of the WHOLE trip. One doesn’t simply sum the two legs’ speeds. By that logic, all the choices would be wrong.</p>
<p>The problem is somewhat challenging and counter-intuitive (as are many rate problems) because the leg with the lower speed effectively gets more “weighting” in the calculation of the speed because one spends more time on that leg because he or she is going slower.</p>
<p>Consider the following example to illustrate a point about rates:</p>
<p>Bob is driving a car down a road that is one mile in length. During the first half of the length of his entire trip, Bob averages only 30 miles per hour. However, Bob wants to average 60 mph for the entire trip, finishing the drive in one minute. How fast does he need to go during the second half in order to achieve this average?</p>
<p>The intuitive first response is 90 mph because 30 and 90 average to 60. However, by examining the situation in the average speed equation from post #2, one finds that this is false.</p>
<p>[(2)(30)(90)]/(30+90) = 45 mph because more time was spent on the first half of the drive, thereby lowering the average speed.</p>
<p>Let’s try approaching the problem as I did in post #11. The total time is 1 minute. The first leg takes t hours as determined by d=rt.</p>
<p>.5 miles = (30 mph)(t). t = .01666… hours = one minute</p>
<p>As you can see, the time taken during the first leg is one minute. Look now at d=rt for the second leg, with the time of leg one subtracted from the total time if the average speed is 60 mph.</p>
<p>1 mile = (x mph)(0 hours)</p>
<p>x is undefined because the time it to took travel the first leg is equal to the time that the entire trip must take. Therefore, the answer is that it is impossible to achieve an average speed of 60 mph because all the time (which is a necessary element of speed) has been taken up; no matter how fast Bob goes for the second leg, he can’t finish in a minute. </p>
<p>This is further demonstrated by the fact that no value is great enough to make the equation in post #2 yield an average speed of 60 mph. I hope this helps you understand why the steps I took in solving your original problem are necessary.</p>
<p>Yes, well put silverturtle. Rate problems, I feel, are the hardest in Algebra besides trigonometry because many of those keep sliding under my intuitive.
But again, I still don’t know why people add the time and the distance up, whilst averaging the speed. I wonder if we add the speed up, what will we get.</p>
<p>See if this helps. Say distance 1 is 20 miles, distance 2 is 30 miles, time 1 is 1 hour and time 2 is 2 hours. If I want to add those together, I can. Distance 1 + 2 is 50 miles, and time 1 + 2 is 3 hours.</p>
<p>Speed is different. Speed is distance divided by time. It is defined by the other two variables; you cannot do whatever you want with it. So if I want speed 1 I have to do 20 miles divided by 1 hour, or 20 miles per hour… If I want speed 2 I have to do 30 miles divided by 2 hours, or 15 miles per hour. If I want speed 1 + 2, I need (20 + 30) miles divided by (1 + 2) hours. So you get 16.66 miles per hour, which is somewhere in between speed 1 and speed 2.</p>
<p>Because speed is one quantity divided by another quantity, if you increase both quantities by roughly the same amount, the speed will not change much. That is a simplified explanation but it helps to show why you can’t just add speeds together.</p>