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3 days isn't a lot of time to improve your Verbal score. What's your reading comprehension strategy?

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What it comes to inference, we must ask "MUST this be true?" So, answer choice D is too strong an inference. Yes, it's POSSIBLE the MDG will not yield intended effects, but that's not 100% guaranteed. Answer choice E, on the other hand, uses weaker language (CAN).

FIVE?!?!?!? Are my eyes deceiving me? I hope this test lasts forever!

Yes it does.

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The answer is definitely A. The main idea here is that, even if y WERE the longest side (which it may or may not be), the value of y would still be smaller than w + x (the sum of the other 2 lengths), which means y is definitely smaller than 2w + x. In other words, we can be certain that y < w + x < 2w + x Does that help?

@@GreenlightTestPrep However, if x is the longest side, then there is not enough information. And it would be too time consuming to do it by this method if I had to plug in each variable as the "3rd side" (i.e., w first, x second, and then y third) to finally find the correct answer of y.

@@TheAncientColossus That's not correct. The triangle rule that appears at 0:11 of the video tells us that, if a triangle has sides with lengths A and B, then |A - B| < 3rd side < A + B So, if we let w and x be sides A and B, and let y be the third side, we can write: |w - x| < y < w + x (Notice that we aren't assuming y is the longest side of the triangle; it's just one of the sides) Now let's focus on this part of the inequality: y < w + y Since w and x must be positive (since side lengths can't be negative), we also know that w + x < w + w + x In other words, w + x < 2w + x Finally, since we already concluded that y < w + y, we can combine the two inequalities to get: y < w + x < 2w + x At this point, it's apparent that y < 2w + x, which means the answer is A.

@@GreenlightTestPrep It's easier when you already know the answer and you can simply, as you say, "let y be the third side." But what if I (me, other test takers that don't have the solution manual) don't know the answer and "let x be the third side." It'll be harder and more lengthy to solve this question, until I come to the conclusion of "let y be the third side." I don't get it. If x is the third side then A can be y and B can be w. Then we won't get your solution.

@@TheAncientColossus First of all, at the beginning of the video I mention that the question concerns the three lengths of a triangle, which means it may be useful to check whether we can apply the 3rd side rule (although there's no guarantee that this will yield the correct answer...yet). In the question, one quantity is y (which is one side of the triangle), and the other quantity concerns the lengths of the other two sides. So, I let y = the third side so that we're comparing y with the other two lengths (just like in the question), and this particular approach yielded the correct answer. That's said, let's examine the other two possibilities. If we let x be the 3rd side, our inequality becomes: |w - y| < x < w + y, however this inequality doesn't help us much because the value of y isn't isolated. Since we are comparing y with some other quantity, the inequality |w - y| < x < w + y doesn't help us much (because y isn't isolated). Similarly, If we let w be the 3rd side, our inequality becomes: |x - y| < w < x + y, however this inequality doesn't help either, because the value of y isn't isolated. So, as we can see, choosing y to be the 3rd length allows us to create an inequality in which y is isolated.

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Thanks for the kind words, Fhella!

Thanks Jatin 🙂

UA-cam adjusts image quality based on the strength of the viewer's internet connection.

@@GreenlightTestPrep exactly. I'm watching it at a good resolution so I was wondering which part had a poor quality

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Thanks!

From the ETS website (www.ets.org/gre/revised_general/test_day/test_center/during): The test administrator will provide you with scratch paper for use during the test only. If you use all of it, you can return it to the test center to receive more. You must return all scratch paper in its entirety to the test center administrator at the end of the testing session. The test center will supply the pen/pencil. I hope that helps.