Another math probelms

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1

<p>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?</p>

<p>Erika rode her bike to the grocery store and got stuck in traffic, so her trip was at an average pace of 4 miles per hour. She took the exact same route back home. To her relief, the return home was much faster; her average speed rose to 12 miles per hour. If the entire run took Erika exactly one hour to complete and she did not make any stops, how many miles is her route one way?</p>

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<p>Hey, here are my answers:</p>

<ol>
<li>Assume x = time required to travel the trail at 6 mph. Since (time)(rate)=distance, we can set up an algebraic equation.</li>
</ol>

<p>6x=14(1-x) Note: 6x equals 14(1-x) because the total time for the entire trip was an hour.</p>

<p>Solving that gives us x= 7/10</p>

<p>Plug that answer in our original equation and 6(7/10) = distance we get distance = 42/10 or 4.2 miles.</p>

<ol>
<li>I used the same method for the second problem:</li>
</ol>

<p>Assume x - time required to travel the trail at 4 mph.</p>

<p>My equation:</p>

<p>4x=12-12x</p>

<p>x=3/4</p>

<p>Plug it back in 4(3/4)= 3 miles</p>

<p>Hope that helped.</p>

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<p>Thanks. Now another problem:</p>

<p>Of the numbers between 100 and 999, inclusive, how many have a units digit of 6 and a tens digit of 7?</p>

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<p>

</p>

<p>How about solving the problem without setting up problematic equations?</p>

<p>6<em>14 / 6+14 = answer
6</em>14 / 6+14 = 4.2 </p>

<p>

</p>

<p>Same for the other one</p>

<p>4*12 / 4+12 = 3</p>

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<p>Just to make some trouble…I think there is a flaw in the wording of both of those problems. They are both meant to be of the same variety: you travel one way at a given speed, then return at another speed – what’s your average speed? And I agree that this problem is most easily dealt with using what is now known as Xiggi’s formula, and that this type of question comes up often enough to make the formula worth memorizing…</p>

<p>But here is the flaw: They both say that you go one speed in one direction and then faster the other way so that your average speed increases to some new value. But that makes it sound like they may be referring to your average speed for the whole trip when they were jsut referring to the second half of the trip.</p>

<p>There IS a way to see that they could not have meant it this second way: the new average speed it too high to be referring to the whole trip. Here’s what I mean. Say you go one way at 6 mph. No matter how fast you come back, there is a limit to how high your overall trip average speed can be: even if you could return instantaneously, that would still only double the average speed. And since they gave the new average as 14, which is more than double, they must have meant that 14 was the average for just the second leg of the trip. </p>

<p>But you shouldn’t have to do this level of meta-analysis to determine what they mean. Is this a college-board problem? If so, I’m surprised. And if not, it’s just another reason to avoid non-College-board practice material.</p>

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<p>It is highly unlikely that the questions were written by ETS, especially the third one which was not worth trying to answer.</p>

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<p>Just to make some more trouble…at least among the released QAS tests, the combined rates problem doesn’t occur particularly frequently. I found three instances of this type of problem out of 15 tests previous to May 2010. Only one of these was soluble by direct application of the formula.</p>

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<p>Fignewton, what was needed to solve the other two instances?</p>

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<p>Actually, one of the other two (see the first problem below) could also be done using the formula but this (easy) problem seems easiest to do in another way. The second problem is a related variation of the OP’s problem…</p>

<p>The following are similar to (but different than, of course :)) to these two problems:</p>

<p>1) Bill can eat cabbages at a rate of 5 minutes per cabbage. Aubrey can eat cabbages at a rate of 6 minutes per cabbage. If they both ate cabbages at these rates for one hour, how many cabbages did they eat in total?</p>

<p>2) On a 10 mile race, Aubrey ran 3/4 of the time with a good-luck cabbage, and 1/4 of the time without it. With the cabbage, she could run 6 miles per hour, and without it, she could run 12 miles per hour. How long in hours did it take for Aubrey to complete the race?</p>

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<p>Those are different types of problems, as both require setting up simple additions. </p>

<p>60/5 + 60/6 = 22</p>

<p>(6<em>3/4 x) + (12</em>1/4 x) = 10
x = 4/3 hour </p>

<p>I am not sure why there is a recurrent need to debate the value of knowing the harmonic mean formula. It works very well and elegantly for the … problems it is supposed to address. it will NEVER REPLACE the need to apply the correct reasoning. The SAT is not a fill by the number test, for the math, it is a test of simple mathematical manipulations to support logic and common sense. </p>

<p>As far as how often the formula could come in handy, I believe that the people who took the Jan 08 test and saw a problem appearing on the last question of the grid-in that could be solved in 15 seconds with the simple formula must have been quite happy.</p>

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<p>These questions are not from collegeboard. I practice on an SAT problem website…</p>

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<p>Yes, those problems are a bit different; however, they were the closest matches I could find to the average rate / harmonic mean problem (the Jan 08 problem you mentioned is in fact the one exact match out of the 15 tests).</p>

<p>I see no problem in having strong math people add that formula to their toolkits as long as they are aware of the specificity of it (which they usually are since they are good at math). For the vast majority of test takers, I find the formula not helpful to know, or worse, a disadvantage to know if incorrectly applied.</p>

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<p>

</p>

<p>That makes no sense whatsoever.</p>

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<p>Again, I do not understand why there is a need to discuss this. Based on past history, the precise type of problem that this formula DOES solve shows up, and there are NO faster or better ways to solve. Also, this is a type of problem that continues to trip students and when it shows up on the test, it represents a major … time sink. In fact, ETS does HOPE the student will waste time in establishing equations or … give up. </p>

<p>If you want to establish that students who do not understand this type of problems won’t benefit from learning the formula, you are right. But how would they solve the problem in the first place? Divine intervention? </p>

<p>If you want to establish that the formula won’t help solve UNRELATED problems (such as the two you posted) you are again right. Neither will the formula solve a geometry problem!</p>

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<p>What is the formula?</p>

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</p>

<p>They would typically not attempt a problem at the end of a section, which is where this specific problem would appear. I know people who struggle with applications of d = rt. A person of this type would have difficulty remembering in what situation the formula could be applied. If this means that he or she wastes time on a related but different problem determining whether the formula will work, then the knowledge of the formula had a negative consequence.</p>

<p>Here would be such a problem: Bill drives to work from home in the morning at an average speed of 40 mph. He drives to home from work in the evening along a different route, twice as long as the morning route, at an average speed of 60 mph. What is Bill’s average speed for his entire trip?</p>

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<p>I only wanted to establish that the problem for which the formula is useful appears rarely, using actual data. Since I don’t have access to non-QAS materials, this conclusion could be challenged. However, I could argue that the appearance of this problem on a QAS test could in fact make a future appearance of the exact type of problem even more unlikely. The other problems are not completely unrelated: rate manipulations are involved in all.</p>