The Official SAT QOTD Thread

Test Preparation SAT Preparation
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<p>I'm not really sure what deems this as official, but I'm proclaiming it to be as such, and so it shall be!</p>

<p>I feel that a centralized SAT Question of the Day thread will be of great benefit to many forum users. First, we'll be able to discuss the absolute best methods to solve each question (thus sharpening our collective problem solving skills). Second, this thread could prove to be a great resource that could archive each and every SAT QOTD. Feel free to contribute your personal strategy in solving the daily question and whether or not you got it right.</p>

<p>7/24/2011: Today's question, in my opinion, is most efficiently solved through plugging in numbers. There definitely are alternate solutions, but they just don't seem necessary/as efficient to me.</p>

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<p>You guys should use the techie website where is has all 365 questions throughout the year.</p>

<p><a href=“http://atekkie.com/sat-question-of-the-day-archive-search/[/url]”>http://atekkie.com/sat-question-of-the-day-archive-search/&lt;/a&gt;&lt;/p&gt;

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<p>^That serves as an archive, but it’s not really a viable medium for discussion. A thread will allow forum members to share different strategies for solving the daily SAT questions.</p>

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<p>The past three math QOtD’s have been raw plug-in questions. It’s getting ooooold.</p>

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<p>The July 21st one wasn’t really plug in. You just multiplied 1.2 by 0.8 to find the percent relative to the other rectangle, 0.96 (4 percent less). I don’t see how it could be done any other way, really.</p>

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<p>You just make both sides of the rectangle equal 10, then you get 8x12=96, which is 4% less than the original, which is the correct answer. I prefer to do it that way since you can use logic to make sure the answer makes sense.</p>

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<p>Technically, mine is the same, but with 1 length original sides. You’re right, though; it’s good to make sure things make sense.</p>

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<p>Can someone explain how the explanation for today’s (July 24th) question.</p>

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<p>The College Board’s solution for today’s problem was nebulous and completely unhelpful. Check out this solution from eagles94; it’s very clear/helpful:</p>

<p>Here is how I did it. I think doing it algebraically helps with fundamentals. Meaning you can tackle more problems when plugging in wont work. It also takes a little bit longer than plugging in but worth it.</p>

<p>a/b = 3 & b/c = 7 </p>

<p>now b = 7c. Plug that back in for b in a/b = 3.</p>

<p>a/7c = 3
a = 21c</p>

<p>now we have
a = 21c
b = 7c
and c </p>

<p>just subsitute the values
a+b/b+c </p>

<p>21c + 7c/ 7c + c </p>

<p>= 28c/8c</p>

<p>which = 7/2 cause the c’s cancel out.</p>

<p>simple and effective. takes no more than 20-30 seconds.</p>

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<p>I discovered this way also works for today’s problem:</p>

<p>let a = 21, b= 7 (a/b , 21/7 = 3)
let b=7, c=1 (b/c, 7/1 = 7)</p>

<p>a + b/b + c</p>

<p>21 + 7 / 7 + 1</p>

<p>28/8</p>

<p>7/2</p>

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<p>^That’s exactly how I did it the first time. Plugging in numbers really is the best approach for this problem.</p>