<p>How exactly do I do this problem?
Tracy jogs one lap around a track field at 4 miles per hour. She runs a second lap at 8 miles per hour, What is her average rate of both laps?</p>
<p>Logically I would think it's 6, but I know there is a trick to this.
The Answer is suppose to be 5.33 but I don't know how to get it mathematically.</p>
<p>What were the answer choices for the question?</p>
<p>I don’t remember the choices but the correct answer was 5.33</p>
<p>I can’t think of a certain mathematical process to get to the answer, but I clan ball park it. I know that since Tracy was jogging at 8 miles per hour the second time around the track, it took her less time to run around the track than when she ran at 4 miles per hour. So she ran at 4 miles per hour for a longer period of time than at 8 miles per hour. That means that the average rate has to be less than 6 miles per hour.</p>
<p>yeah, they always use average rate to mess with people. </p>
<p>rate = distance/time
average rate = total distance/total time (just remember this)</p>
<p>so for the problem, just pretend that one lap is 8 miles, it really doesn’t matter. given that assumption, that would mean she ran a total of 16 miles, that lasted a total of 3 hours.</p>
<p>16/3 = 5.33 </p>
<p>(now who really measures their pace in mph? SAT runners.)</p>
<p>^ Much better explanation, haha</p>
<p>Here’s a way to always avoid these problems instantly.</p>
<p>2 * rate 1 * rate 2/ (rate 1 + rate 2)</p>
<p>^Wow that makes a lot of sense, thank you so much 2redpartyhats^</p>
<p>Edit: MasterYster’s strategy works too. I’d also like to know though why MasterYster’s formula works or what is the logic behind it cause I think 2redpartyhats’ formula is more easier to understand.</p>
<p>@MasterYster, can you explain why that works? I just plugged in the numbers using your method and it worked; I just don’t know why.</p>
<p>No idea, just a general formula. It probably arises from manipulating the variables in r=DT and Average rate = Avg D/ Avg T</p>
<p>I’ll jump in for masteryster. it’s called harmonic mean, but I think it was a CC’er who popularized it (forgot who).</p>
<p>the trick is, the two distances you use have to be exactly the same (a lap). it’s derived from that general average rate formula. proof time!</p>
<p>so we have r1 = d/t1 and r2 = d/t2. </p>
<p>so total distance = 2d.</p>
<p>and total time = t1 + t2 = d/r1 + d/r2.</p>
<p>which means that average rate = (2d) / (d/r1 + d/r2).</p>
<p>simplify (common denom), ar = (2d) / [d(r2+r1) / (r1*r2)].</p>
<p>cancel (the d’s), ar = 2(r1*r2) / (r1+r2)</p>
<p>yeah, so just remember that the two distances gotta be the same.</p>
<p>EDIT: it was xiggi who popularized it! whata awesome person.</p>
<p>There are lots of threads on here about this. If you search Xiggi’s formula you’ll probably find lots of information and similar problems. Note that the harmonic mean is just a different kind of average from the arithmetic mean. On the SAT I have only seen it used for averaging two rates.</p>
<p>Be careful. This formula is most useful on a very specific type of problem - finding an average rate when 2 rates are known. It can be used in other situations, but it may not be as straightforward.</p>
<p>Do these type of questions appear on the SAT still?</p>
<p>Yeah. I had one on math in the SAT, one on Math II, and one on Physics. They don’t disappear.</p>