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The Answer is E.

Test different cases to try to prove each false.

A.
Try this with a negative number. Say x is -2. Then, -0.5 is therefore greater than -2, however -2 is not greater than (-2)^2 which is 4. Hence, false.

B.
Which cases x will be greater than 1/x is when x is a positive integer >1, or x is negative integer between 0 and -1.
Hence, test out both cases. e.g. when x is positive 2, then x is greater than 1/2, and 2(2) is greater than 2. If x is -.5, then x is greater than -1/.5 which is -2, but 2(-0.5) which is -1 is LESS than -0.5. Hence false

C. In this case, x will be greater than 2x if x is a negative number. Test out if x is -0.5, then -1/x is -5 which is LESS than x. Hence false


D.
In this case, x^2 will be greater than x if x is either positive integer greater than 1 or negative integer less than 1. However, if it is a negative integer less than 1, then x^3 is LESS than x^2 as x^3 will be negative. Hence false


E. Correct.
Which cases x will be greater than 1/x is when x is a positive integer >1, or x is negative integer between 0 and -1.
Hence, test out both cases. e.g. when x is positive 2, then x^2 is 4 which is greater than 4. If x is -0.1, then x^2 is -.01 which is greater than -0.1
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This a text book plug-in question if ever there was one. Best way is to quickly test 2 or 3 values and see if the statements always hold true.

A. If \(\frac{1}{x}>x\), is \(x>x^2\)? -

  • \(x=-2\), \(\frac{1}{x}=-\frac{1}{2}\) and \(x^2=4\). Does not Hold

A does not hold in all cases. Eliminate.

B. If \(x>\frac{1}{x}\), is \(2x>x\)? -

  • \(x=-\frac{1}{2}\), \(\frac{1}{x}=-2\) and \(2x=-1\). Does not Hold

B does not hold in all cases. Eliminate.

C. If \(x>2x\), is \(\frac{1}{x}>x\)? -

  • \(x=-1\), \(2x=-2\) and \(\frac{1}{x}=-1\). Does not Hold

C does not hold in all cases. Eliminate.


D. If \(x^2>x\), is \(x^3>x^2\)? -

  • \(x=-1\), \(x^2=1\) and \(x^3=-1\). Does not Hold

D does not hold in all cases. Eliminate.

E. If \(x>\frac{1}{x}\), is \(x^2>\frac{1}{x}\)? -

  • \(x=2\), \(\frac{1}{x}=\frac{1}{2}\) and \(x^2=4\). Holds
  • \(x=-\frac{1}{2}\), \(\frac{1}{x}=-2\) and \(x^2=\frac{1}{2}\). Holds
  • Note that a negative integer cannot be used to test whether E holds because it will not satisfy the initial condition given by the option, e.g. if \(x=-3\) then -3 IS NOT greater than \(\frac{1}{x}=-\frac{1}{3}\)
.
E holds and is thus the correct answer

Now I have a question for the quant experts here chetan2u, VeritasKarishma, Bunuel, gmatbusters, amanvermagmat. Solutions to plug-in questions usually require you to test all options with two or three values, which depending on the complexity can take a pretty long time. Based on your experience, is it a GMAT ploy to hide the answer in the final 2 options to ensure that you take your time going through all of the options before selecting an answer? As a strategy for max/min plug questions, I as a strategy would work on the middle/median option and see where to go from there. In questions such as this, where this is not possible, is it a good idea to start from the bottom rather than the top?
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Which of the following is always true?


A. If \(\frac{1}{x}\) is greater than x, x is greater than \(x^2\).
if X is between -1 and 1, then is x > \(x^2\) ?
e.g. x = 0.5, then x>\(x^2\) - yes
x = -0.5, then x>\(x^2\) - no

B. If x is greater than \(\frac{1}{x}\), 2x is greater than x
If x < -1 or x>1 then is 2x > x
it is true when x>1 but not when x is < -1

C. If x is greater than 2x, \(\frac{1}{x}\) is greater than x.
If X < 0, is \(\frac{1}{x}\) > x.
true if x = -2 but false when x = -0.5

D. If \(x^2\) is greater than x, \(x^3\) is greater than \(x^2\).
if x> 0 or x> 1 , is \(x^3\) > \(x^2\).
true if x = 2, but false when x = 0.5

E. If x is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
if x < -1 or x>1, is \(x^2\) > \(\frac{1}{x}\).
true if x = -2, and also true if x = 2 , this must be true always

E is the answer!
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Which of the following is always true?

We have to test the different cases by plugging in the numbers. I like to start these type of questions from the last option.


A. If \(\frac{1}{x}\) is greater than \(x\), \(x\) is greater than \(x^2\).
B. If \(x\) is greater than \(\frac{1}{x}\), \(2x\) is greater than \(x\).
C. If \(x\) is greater than \(2x\), \(\frac{1}{x}\) is greater than \(x\).
D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).

E. If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
For x to be greater than 1/x the numbers possible for x are 2 to positive infinity.
For any positive integer from 2 onward its square will always be greater than its reciprocal.
So this option is always true. No need to check other options.


Answer E)
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aliakberza
Quote:

Now I have a question for the quant experts here chetan2u, VeritasKarishma, Bunuel, gmatbusters, amanvermagmat. Solutions to plug-in questions usually require you to test all options with two or three values, which depending on the complexity can take a pretty long time. Based on your experience, is it a GMAT ploy to hide the answer in the final 2 options to ensure that you take your time going through all of the options before selecting an answer? As a strategy for max/min plug questions, I as a strategy would work on the middle/median option and see where to go from there. In questions such as this, where this is not possible, is it a good idea to start from the bottom rather than the top?
It is very hard to game the system. When I do make questions, I do find it hard to give the correct answer in the first or second option when number plugging etc could be useful. When the answer is a plain numerical answer for which calculations need to be done, I might place it in the beginning. But the GMAC is aware of all such biases and adjusts for it. Hence, it is hard to find a strategy that would give you any real advantage. So use whatever makes you feel better - until and unless there is an actual logic to taking a particular sequence (such as start from the middle so you know whether to go up or down), it's all the same.
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aliakberza
Based on your experience, is it a GMAT ploy to hide the answer in the final 2 options to ensure that you take your time going through all of the options before selecting an answer?

This is actually a claim I've seen in one or two prep books - these books claim that on "which of the following?" Quant questions, the right answer is normally D or E, so that test takers who work in order from A to E spend longer on the question. That is a myth. I've studied that claim (and any other similar claim I've found) by looking at large pools of official questions. On "which of the following?" questions, D and E are correct about 40% of the time, exactly what you'd expect if each answer choice was equally likely to be right.

The very premise behind the claim should be viewed with suspicion. For one thing, the question designers have no particular interest in making test takers spend more time on a question, so that premise is wrong. I sometimes see experts express an attitude similar to "the test is out to get you", and that just misunderstands the purpose of the test. The GMAT Quant section is trying to assess how good test takers are at quantitative reasoning, nothing more. For another thing, if claims like this one were true, then test takers who read certain prep books (the ones that tell you to start at answer E) would have an advantage over other test takers. The GMAT is not testing you on what prep books you bought - that information isn't helpful to MBA programs when deciding which applicants to admit. So the test designers will purposely foil any gimmicky "strategies" like this one, that only some test takers will be aware of, and only because they read a certain book.
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And it seems like every solution posted so far uses number picking or a lot of algebra on each answer choice. That's a lot of work. When possible, it's better to zero in on the most promising candidate answer first, and only test other answers if that promising candidate turns out to be wrong. When I look at the conclusions were asked to draw, one stands out: answer E says "then x^2 > 1/x". That's the only conclusion where we have an even power of x that is greater than an odd power of x. Naturally that is automatically true for any negative number (since then x^2 > 0 > 1/x no matter what x is), so if we look at E first, we don't need to think about negatives at all, which saves a lot of time and analysis. If x is positive, and x > 1/x, then x > 1, so x^2 > 1/x is true too. So E is certainly correct, and there's no need to consider any of the other options.
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