<p>I see. Did you choose 120 because its the LCM of the two rates? So if n = 120 then the first section took 10 hours at 12 pph and the second section took 6 hours at 20 pph. So 240 pages were typed in 16 hours. The rate must have been 15 pph.</p>
<p>Yep - now you got it! The LCM of the two rates is actually 60. This is an even better choice for n. What I meant to say was that 120 is a good choice for the total number of pages. But as you can see 120 works well too. In fact any multiple of the LCM is fine. The LCM just keeps the numbers smaller.</p>
<p>Interestingly, if you mistakenly take the average of the rates (in this case 16 pph), you will <em>always</em> overestimate the correct answer (15 pph), or get it correct if all the rates are equal, as a consequence of the AM-GM-HM inequality.</p>
<p>^^Right. And I remember reading on one of xiggi’s threads, that on rate problems you can usually get the right answer by just guessing the highest number that is less than the average of the rates.</p>
<p>what did i do wrong?
20x12/20+12=7.5</p>
<p>Here’s the correct computation using Xiggi’s formula (harmonic mean):</p>
<p>average rate = 2(rate1)(rate2)/(rate1 + rate2) = 2(12)(20)/(12+20) = 15</p>
<p>Can I pose another, same theme? And easier than it looks…</p>
<p>You drive a distance d miles at an average speed of r miles per hour. Then you drive kd miles at an average of 3r miles per hour. If your average speed for the entire trip is 2r, find k.</p>
<p>How to solve that?</p>
<p>Some hints…</p>
<ol>
<li>You cannot use Xiggi’s formula (harmonic mean) – the two legs of the journey were different lengths.</li>
<li>The longer way: set up an expression for average velocity using total distance over total time. Set it equal to 2r and start doing algebra. The r’s and d’s will all drop out…and you will get an answer that makes you think…</li>
<li>Oh, yeah…notice that in this case, the average of the two separate legs is in fact the arithmetic mean. When does that happen?<br></li>
</ol>
<p>With all due respect (and I mean that sincerely), I think this lesson/concept would be more helpful if it were presented in a more traditional SAT question format</p>
<p>It is easy (and fun) to get caught up and start posting questions way above the SAT level. Didn’t we have a thread going for “level 6” questions? But I am not sure that I have hit level 6 with this one. Could it be a grid-in just the way it is? </p>
<p>@pckeller I disagree - this is definitely a Level 6 problem by my definition. I consider a problem Level 6 if it is just above the level of what would appear on an actual SAT. The problem is the thread you are talking about quickly got “hijacked” by people asking what I consider “challenge questions.” So you are probably thinking of those much harder questions which are not very SAT-like at all (more like competition level). Of course as far as I know these terms are made up by me, so feel free to disagree with them. Just keep in mind that your disagreement will be as irrelevant as my definitions.</p>
<p>Also let me clarify - it is conceivable that a Level 6 problem COULD show up on a future SAT (especially beginning in 2016). But until I see something of that difficulty level on an actual exam I will consider it a Level 6 problem.</p>
<p>OK, I won’t argue. What if I dial it back a level:</p>
<p>You drive 10 miles at an average of 20 mph. How far will you have to drive at 60 mph for your average speed for the entire trip to be 40 mph?</p>
<p>Same concepts, less algebra. Level 5? 4? </p>
<p>^ I’d say around level 4. Let t be the amount of time driving at 60 mph, set up an equation relating total distance and total time, solve for t (multiply by 60 mph at the end).</p>
<p>That will work. But there is another idea I was getting at. Usually, you can’t average the averages. But this time, the overall average, 40 mph was in fact the average of the 20 mph and the 60 mph. That only happens if you spend the same amount of TIME at both speeds. (Same DISTANCE at both speeds is when you use harmonic mean – Xiggi’s Formula). So this time, you spent the same time traveling at 60 mph as you did at 20 mph. That makes things easier… </p>
<p>The answer is 50?</p>
<p>Let’s check: 10 miles at 20 mph takes a half hour. 50 miles at 60 mph takes 5/6 hr. That adds up to a total of 60 miles at a total of 1/2 + 5/6 = 8/6 or 4/3 of an hour. But 60/(4/3) = 45 mph, not 40…
If you were to continue trial and error, you’d want a smaller guess. Try 30 miles… :)</p>
<p>30 is indeed correct, but I wouldn’t advocate using trial and error unless you have the answer choices (in which it is possible to guess at the answers). In either case, the problem can always be solved quickly without guessing.</p>
<p>Level 5 Geometry</p>
<ol>
<li>The lengths of the sides of a triangle are x, 9, and 17, where x is the shortest side. If the triangle is not isosceles, what is a possible value of x?</li>
</ol>