About Xiggi's Rate and Distance formula.

<p>The great Xiggi provided us the 2<em>rate1</em>rate2 / ( rate1+rate 2) =average speed per hour.
This formula has no doubt simplified the rate and time formula a bunch loads and turned a 30 second to 1 minute calculations to 10 seconds. Now the problem I have is what if the SAT makers decides to not set the distances traveled equal. Instead of "Martha drives half the distance from A to B at 40 miles per hour and the OTHER HALF at 60 mph." What if it was "Martha drives 40% of the distance from A to B at 40 miles per hour and 60% of the other half at 60mph. what was the average speed?"</p>

<p>Now I know I could do this using the rate x distance but i was wondering if there was another way yet also utilizing xiggi's formula. I know how to do it when it's say 1/4 at x speed and 3/4 at y speed, would simply be 4xy/ (3x+y). 4 serves as the total of the denominator (4) and the 3 serves as the remainder of the other half (3/4). But how do you do it when the other part isn't under a numerator of 1 (ie: 1/4).</p>

<p>Thanks in advance. Sorry if my question is baffling. If it's too confusing just tell me based on:
"Martha drives 40% of the distance from A to B at 40 miles per hour and 60% of the other half at 60mph. what was the average speed?"</p>

<p>“90% of this game is half mental.” – Yogi Berra</p>

<p>So I think you meant to say: “40% of the distance at 40 mph and 60% of the distance at 60 mph…”</p>

<p>Otherwise, I am not sure what you mean by “60% of the other half.”</p>

<p>This is another example where I prefer to make up numbers rather than memorize special-case formulas. Say the trip is 100 miles. 40% is 40 miles. At 40mph, that takes 1 hour. The remaining distance is 60 miles. At 60 mph, that also takes another hour. (You may have made up this problem randomly, but you actually gave it numbers that make it much easier!) So now we know it took 2 hours to travel the 100 miles, so IN THIS CASE, because of THESE PARTICULAR NUMBERS, it happens that the average is 100 miles/2hrs = 50 mph. </p>

<p>But say you made it 80% at 40mph and 20% at 60mph. The same method would work: Call the total distance 100. 80% is 80 miles, at 40 mph is 2 hours. 20% is 20 miles, and at 60 mph, that’s 1/3 hour. So total rate is 100/(2+1/3) mph.</p>

<p>Easiest and simplest way to do rate problems by far:</p>

<p>[SAT</a> Mathematics | Common Problems - Rate - ■■■■■■■■](<a href=“http://■■■■■■■■■■■■/m/common-problems/rate-problems]SAT”>http://■■■■■■■■■■■■/m/common-problems/rate-problems)</p>

<p>I do not understand Mde, I sort of understand pckeller, and I do understand steamedrice.</p>

<p>edit: Unfortunately, as I try to do the problem, I do not undertand anything lol. Not surprising, considering my Math score dropped by more than 150 between the last two practice test I took.</p>

<p>Is the way to solve the typical distance/rate problem on the SAT still debated? </p>

<p>What does the link posted by steamedrice say? </p>

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<p>Do we need six steps to solve this problem? Since when do SIX steps offer the “Easiest and simplest way to do rate problems by far?” That is such a bunch of baloney!</p>

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<p>How many steps are needed to solve that problem? One! How many seconds does one need to solve that problem with a modicum of attention to the proper techniques? Either 4 or 5 or perhaps 20 seconds. What is surely NOT needed is drawing ridiculous boxes and a six step process. </p>

<p>Oh, almost forget to state what the correct technique would entail. All you need to do is verify is that “The whole round trip took 1 hour.” and you can either apply the known formula, or pick the number that is slightly below the straight average. Until the day TCB decides to changes its models, the answer will always remain B. </p>

<p>While there are different methods to solve a problem, in this case it would be hard to find one that is as quick as it is elegant. So, why waste your time?</p>

<p>Hi xiggi,</p>

<p>I’m unsure why you’re attacking the method I posted so vehemently…I wanted to offer a more flexible approach that works for all rate problems, including those that involve solving for the total distance where t=/=1 hour or the time. The formula you describe is fast and easy, but its drawback is that it’s inflexible.</p>

<p>For example, it doesn’t work for the problem posed in the original post. The boxes approach does. </p>

<p>I hope this sheds some light on the issue.</p>

<p>Let me give an example problem straight from College Board: "Jos</p>

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<p>Aren’t we talking about the SAT Reasoning Test? Aren’t we talking about solving problems that actually have appeared on the test? </p>

<p>There is zero inflexibility in correct reasoning. The proper approach to solve the SAT rate problems has never been about applying a formula blindly. It involves understanding why ETS and The College Board present problems in a particular format.</p>

<p>And, fwiw, reasoning might also allow someone to pick the answer the test writer expected, even if the problem was poorly written. Take a look at the problem posted at your “box” site:</p>

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<p>Is the answer to “How many miles was the trip?” one of the five choices?</p>

<p>^ Just wondering, is the answer to that approx. 62.86 km/h? I had no idea how to use the box to this specific problem, so I just used simple averaging…</p>

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[quote]
Jos</p>

<p>Yes, I did it right!!
Xiggi, could you please explain how to use your formula for times other than a total of 1 hr? Or is there no “simple” fix to do that?</p>