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</a></p>
<p>I spent 30 min on this and still couldn't get it. Is it because nothing's shaded or am I missing something?</p>
<p>I believe the answer is choice B. If |p|>|r|, the line is moving more horizontally than vertically. Since the line is going downwards, the slope will be negative. The only answer choice which satisfies both requirements is B, because since slope is rise/run.</p>
<p>yes, the answer is still B. Same as in the other 2 threads.</p>
<p>Haha, you made 3 threads? Calm down…</p>
<p>Why is it B but not A?</p>
<p>WHOOPS wrong link. wrong question. here is the right one:
<a href=“http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/shade.png[/url]”>http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/shade.png</a></p>
![http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/shade.png[/url]](http://i475.photobucket.com/albums/rr116/watex/shade.png%5B/url%5D){kind=link}
<p>^I think something is wrong as well. But, the answer is either A, C, or E (assuming you can only shade entire boxes). If you want me to explain that, just ask. And further, by my elimination method, the answer is C. E will not be the answer because there is a 5 in it, and the test-maker would not want someone confusing both the numerator/denominator and the y=2x thing to get the problem right. A will not the answer because it is too simple. Therefore, C.</p>
<p>senior- ok can u explain? how did u figure out the answer w/ such a big typo. please just tell me if youre method isnt fullproof because i dont want to use your method and get it wrong. (please reexplain it, i didnt understand wat u were saying from that post ^ sorry -.-</p>
<p>WoW! I got it! sweet! </p>
<p>First, I noticed a pattern with the answers. Each answer has a multiple of 5 or 7 in its denominator. So, I set about trying to see if I could eliminate either of these two denominator options. My explanation is… There are 4 rows with height m. That’s easy enough. I then split the larger boxes in half vertically (just trust me on this for now). So, now you have seven columns, each with a width x (because y=2x, so y divided in half =x). Now, you can pretty much assume that you have to fill in entire boxes, and I’ve never seen a problem where a box is partly shaded, as this would cause the “drawn to scale” and other undesirable (for the writers) problems. So, there are now 7*4=28 boxes in total. You can fill any number of boxes from 0 through 28 even if you didn’t split the y columns (hence why splitting the y columns didn’t screw up the problem). You have two answer choices (B and D) with a denominator that is a multiple of 5 and an integer as a numerator. But this isn’t possible. Because, say you have a fraction a/5, where a is an integer. This, since there are two boxes, would be equivalent to b/28 (where b HAS to be an integer, because you can only fill whole boxes). So, a/5=b/28. Cross multiplying, 28a=5b and a=(5/28)b. However, since B is an integer, the ONLY possible answers from 0 to 28 that would also make a an integer is if b were to equal 28 or 0. Then, a=5 or 0, so the fraction=5/5 or 0/5=the entire board or nothing. Luckily, these are not answer choices. Since 10 is a multiple of 5, the argument still holds (you can do the math if you want. Just switch the first fraction to a/10 and do the same steps). Thus, there can be no possible answer (assuming you fill complete boxes) with a 5 or 10 in the denominator unless the 5/5 is an answer choice, which it is not. So, even before the boxes are even shaded, you realize that two answers can not be correct ever. However, many students may be baited into thinking that a denominator of 5 or 10 is possible, as there are 5 columns. However, these students are not factoring the fact that one column is twice as large as another. </p>
<p>So, you are down to A, C, and E (if you notice, each of these has a denominator with a multiple of 7, which makes complete sense). This is when I started to make educated GUESSES about the design of the problem (because it is flawed, as nothing is filled in). So, I already brought up the fact that some students may be led to think 5 should be in the denominator. There will be dumber students that who get the concepts of denominator and numerator confused, so they will think 5 must go in the numerator. The designer of this problem definitely does not want someone who makes two unrelated mistakes to get this problem correct. So, this is how I “eliminated” E. Also, if only one narrow column were shaded (which would get an answer of 1/7) the problem would be altogether too easy. Smart students would get the problem in an instant, dumber students would realize the answer has to be either 1/5 or 1/7 and might take some time to solve it or guess to get 1/7. So, A is “eliminated”. The only answer choice left is C. Besides being correct, C does not have any tips/hints that would lead a guesser/a dumb person/one who made a mistake, to get the right answer. It is an alltogether unassuming number. </p>
<p>^Remember, the second paragraph I could in no way be sure that any answer was right or wrong. Because of the typo, the answer could have been A, C, or E. But, I was reasoning out the design of the problem. I thought: ‘if i were writing this problem, what answer choices would I put in there to throw the kids off?’</p>
<p>^ Interesting assumption…</p>
<p>But the problem inherently assumes that the line must go through 0,0 and B is an arbitrary point in the 3rd quadrant. Why it doesn’t just tell you, who knows, but it’s the SAT. Only B works, like in the other thread.</p>
<p>^You’re reading the wrong problem. Halfway through the thread the guy admitted he made a mistake, and brought up a different problem with a link. That one has a real typo.</p>
<p>Yep. If you read the new problem correctly, nothing has been shaded in. Senior made an excellent assumption which probably is correct.</p>
<p>Oh hehe, didn’t read past the first post and skipped down to the bottom. Srry.</p>
<p>wow thats very neat. i had something similar concept after i had saw the answer but good job</p>