Which of the following is always true?

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

Which of the following is always true?

We need only 1 value of x for which the option will be false to discard that particular option.

A. If 1/x is greater than x, x is greater than \(x^2\).
if x = -2; 1/-2 > -2 ; BUT -2 is NOT > 4 .... DISCARDED

B. If x is greater than 1/x, 2x is greater than x.
if x = -0.5; -0.5 > -2; BUT -1 is NOT > -0.5..... DISCARDED

C. If x is greater than 2x, 1/x is greater than x.
if x = -0.5; -0.5 > -1; BUT -2 is NOT > -0.5...... DISCARDED

D. If \(x^2\) is greater than x, \(x^3\) is greater than \(x^2\).
if x = -2; 4 > -2; BUT -8 is NOT > 4.... DISCARDED

E. If x is greater than 1/x, \(x^2\) is greater than 1/x.
As all other options are discarded, this has to be the correct answer. Still, below is an alternate explanation:

If x is negative (example -0.5).... the RHS remains negative, but LHS is squared, so LHS will always become positive and will always be greater than RHS (negative)
x cannot be 0 because 1/0 is not defined
If x is positive (example 2) and x > 1/x, then x has to be greater than 1, when LHS is squared, it will always be greater than RHS which has not changed.


ANSWER: E

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?

IMO E

Which of the following is always true?




A. If 1/x is greater than x, x is greater than x^2.
Let x = 1/2, so 1/x = 2 and x^2 = 4
Then, we can see that, 1/x > x, but x is NOT greater than x^2
So A is not always true.



B. If x is greater than 1/x, 2x is greater than x.
Let x = -1/2, so 1/x = -2 and 2x = -1
Then, we can see that, x > 1/x, but 2x is NOT greater than x
So B is not always true.




C. If x is greater than 2x, 1/x is greater than x.

Let x = -1/2, so 2x = -1 and 1/x = -2
Then, we can see that, x > 2x, but 1/x is NOT greater than x
So C is not always true.



D. If x^2 is greater than x, x^3 is greater than x^2.

Let x = -2, so x^2 = 4 and x^3 = -8
Then, we can see that, x^2 > x, but x^3 is NOT greater than x^2
So D is not always true.



E. If x is greater than 1/x, x^2 is greater than 1/x.
Let x = 2, so 1/x = 1/2 and x^2 = 4
Then, we can see that, x > 1/x, and also x^2 greater than 1/x.
Similarly, let x = -1/2, so 1/x = -2 and x^2 = 1/4
Then, we can see that, x > 1/x, and also x^2 greater than 1/x.
For any value of x, if x is greater than 1/x then x^2 will always be greater than 1/x
So E is always true.


IMO-E

(A) If 1/x is greater than x, x is greater than x^2.

1/x > x => (1/x - x) >0
=> (1- x^2)/x>0
Therefore, 0 < x < 1 & x < (-1)
Let x = -2 ; x^2 = 4
So, x < x^2 (Statement false)

(B) If x is greater than 1/x, 2x is greater than x.

x > 1/x => (x- 1/x) >0
=> (x^2-1)/x >0
Therefore, -1 < x < 0 & x > 1
Let x= -0.5 ; 2x = -1
So, 2x < x (Statement false)

(C) If x is greater than 2x, 1/x is greater than x.

x > 2x => x <0
Let x= -0.5 ; 1/x= -2
So, 1/x < x (Statement false)

(D) If x^2 is greater than x, x^3 is greater than x^2

x^2 > x => (x^2 - x) >0 => x(x-1) >0
Therefore, -Infinity < x < 0 & 1 < x < +Infinity
Let x = -2 ; x^3 = (-8) & x^2 = 4
So, x^3 < x^2 (Statement false)

(E) If x is greater than 1/x, x^2 is greater than 1/x
x > 1/x => => (x- 1/x) >0
=> (x^2-1)/x >0
Therefore, -1 < x < 0 & x > 1
Let x = -0.5 ; x^2 = 1 & 1/x = -2 ------- x^2 > 1/x (Statement True)
Let x = 2 ; x^2 = 4 & 1/x = 1/2 ------- x^2 > 1/x (Statement True)

A. when \(\frac{1}{x}>x\),
if \(x>0\) , then \(1 > x^2\) , so \(1>x>0\) , then\(x > x^2\) Yes
if \(x<0\), then \(x^2 > 1\) , so \(0>x>-1\), the \(x < x^2\) No

B. when \(x>\frac{1}{x}\),
if \(x>0\), then \(x^2 > 1\), so \(x>1\) , then \(2x>x\) Yes
if \(x<0\), then \(x^2< 1\), so \(0>x>-1\), then \(2x<x\) No

C. when \(x>2x\), then \(0>x\)
if \(0>x>-1\), then \(x>\frac{1}{x}\) No
if \(x = -1\), then \(x = \frac{1}{x}\) No
if \(-1>x\), then \(\frac{1}{x}>x\) Yes

D. when \(x^2>x\), then \(x(x-1)>0\),
if \(x>0\), then \(x^3 > x^2\) Yes
if \(1>x>0\), then \(x^2>x^3\) No
if \(0>x\), the \(x^2>x^3\) No

E. when \(x>\frac{1}{x}\),
if \(x>0\), then \(x^2 > 1\), so \(x>1\) , then \(x^2>\frac{1}{x}\) Yes
if \(x<0\), then \(x^2< 1\), so \(0>x>-1\), then \(x^2>\frac{1}{x}\) Yes

E

A. If 1/x is greater than x, it must be true that x is greater than x^2.
\(\frac{1}{x} > x\) --> \(\frac{{(1-x)(1+x)}}{x} > 0\) --> \(x<-1\) or \(0<x<1\), e.g. x=\(-2\) --> \(-\frac{1}{2} > -2\) (YES)
\(x > x^2\) --> \(x(1-x)>0\) --> \(0<x<1\), e.g. x=\(-2\) --> \(-2 > (-2)^2 = 4\) (NO)
------> If \(\frac{1}{x}\) is greater than \(x\), it is not necessarily true that \(x\) is greater than \(x^2\).

B. If x is greater than 1/x, it must be true that 2x is greater than x.
\(\frac{1}{x} < x\) --> \(\frac{{(1-x)(1+x)}}{x} < 0\) --> \(-1<x<0\) or \(x>1\), e.g. x=\(-\frac{1}{2}\) --> \(-2 < -\frac{1}{2}\) (YES)
\(2x > x\) --> \(x>0\), e.g. x=\(-\frac{1}{2}\) --> \(2*-\frac{1}{2}= -1 > -\frac{1}{2}\) (NO)
------> If \(x\) is greater than \(1/x\), it is not necessarily true that \(2x\) is greater than \(x\).

C. If x is greater than 2x, it must be true that 1/x is greater than x.
\(x > 2x\) --> \(x<0\), e.g. x=\(-\frac{1}{2}\) --> \(-\frac{1}{2} > 2*-\frac{1}{2}\)= \(-1\) (YES)
As in (A), \(\frac{1}{x} > x\) --> \(x<-1\) or \(0<x<1\), e.g. x=\(-\frac{1}{2}\) --> \(-2 > -\frac{1}{2}\) (NO)
------> If \(x\) is greater than \(2x\), it is not necessarily true that \(\frac{1}{x}\) is greater than \(x\).

D. If x^2 is greater than x, it must be true that x^3 is greater than x^2.
\(\frac{1}{x} <x^2\) --> \(\frac{{1-x^3}}{x} < 0\) --> \(\frac{{(1-x)(x^2+x+1)}}{x} < 0\) --> \(x<0\) or \(x>1\), e.g. x=\(-1\) --> \(-1 <(-1)^2 = 1\) (YES).
\(x^3 > x^2\) --> \(x^2 * (x-1) > 0\) --> \(x > 1\), e.g. x=\(-1\) --> \((-1)^3 =-1 > (-1)^2 =1\) (NO).
------> If \(x^2\) is greater than \(x\), it is not necessarily true that \(x^3\) is greater than \(x^2\).

E. If x is greater than 1/x, it must be true that x^2 is greater than 1/x.
CORRECT ANSWER
As in (B), \(\frac{1}{x} < x\) --> \(-1<x<0\) or \(x>1\)
As in (D), \(\frac{1}{x} <x^2\) --> \(\frac{{(1-x)(x^2+x+1)}}{x} < 0\) --> \(x<0\) or \(x>1\)
------> From the above solution range from each equation, we are certain that if \(x\) is greater than \(\frac{1}{x}\), it must be true that \(x^2\) is greater than \(\frac{1}{x}\).


Answer is (E)

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!

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)

aliakberza

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.

Which of the following is always true?


A. If 1/x is greater than x, x is greater than x^2.
if we take x as any -ve number. then the second equation is incorrect.

B. If x is greater than 1/x, 2x is greater than x.
if we consider x = -1/2 then the first equation is valid
but the second equation is not valid.
Incorrect.

C. If x is greater than 2x, 1/x is greater than x.
If we take x = -1 then equation 1 is valid and the second equation is not valid.
Incorrect.

D. If x^2 is greater than x, x^3 is greater than x^2.
For any -ve value 1st equation is valid but the second equation is not valid.
Incorrect.

E. If x is greater than 1/x, x^2 is greater than 1/x
Any value which satisfies 1st equation satisfies second equation.
Correct

Which of the following is always true?

This is a long question.(unless you start from bottom) . We will focus on (-inf,-1), (-1,0), (0,1) and (1,inf)

A. If 1/x is greater than x, x is greater than x^2...............False.
(if) If 1/x is greater than x: (-inf, -1) & (0,1) satisfy the condition.
(then) x is greater than x^2: not true for (-inf, -1)


B. If x is greater than 1/x, 2x is greater than x.................False.
(if) If x is greater than 1/x: (-1,0) & (1,inf) satisfy the condition.
(then) 2x is greater than x: not true for (-1,0 )


C. If x is greater than 2x, 1/x is greater than x.................False.
(if) If x is greater than 2x: (-inf,-1 ) & (-1,0) satisfy the condition.
(then) 1/x is greater than x: not true for (-1,0 )

D. If x^2 is greater than x, x^3 is greater than x^2..........False.
(if) If x^2 is greater than x: (-inf,-1),(-1,0) & (1,inf) satisfy the condition.
(then) x^3 is greater than x^2: not true for (-inf,-1) & (-1,0)

E. If x is greater than 1/x, x^2 is greater than 1/x............TRUE.
(if) If x is greater than 1/x: (1,inf ) & (-1,0) satisfy the condition.
(then) x^2 is greater than 1/x: true for (-1,0 ) & (1,inf)

We can use the plugin values method for this question.

A. If 1/x is greater than x, x is greater than \(x^2\):
Let x= 1/2, then 1/2 > \((1/2)^2\)
If x= -2, then -2 > \((-2)^2\) This is not true.

B. If x is greater than 1/x, 2x is greater than x.
If x = 2 ,then 2*2 >2
If x = -1/3, then -2/3 > -1/3 This is not true.

C. If x is greater than 2x, 1/x is greater than x.
If x = -2 , then -1/2 > -2
If x = -1/2 , then -2 > -1/2 : This is not true.

D. If \(x^2\) is greater than x, \(x^3\) is greater than \(x^2\).
If x = -2 , then \((-2)^3\) > \((-2)^2\) : This is not true.

E. If x is greater than 1/x, \(x^2\) is greater than 1/x.
This will be always true.
x = 2 , then \(2^2\) > 1/2
x = -1/2 , then 1/4 > -2

IMO the answer is E.

Please hit kudos if you like the solution.

Which of the following is always true?


A. If \(\frac{1}{x}\) is greater than \(x\), \(x\) is greater than \(x^2\).
\(x=-3\)
\(-\frac{1}{3} > -3 \implies -3 >9 \implies\) No

B. If \(x\) is greater than \(\frac{1}{x}\), \(2x\) is greater than \(x\).
\(x=3\)
\(3>\frac{1}{3} \implies 6>3 \implies\) No

C. If \(x\) is greater than \(2x\), \(\frac{1}{x}\) is greater than \(x\).
\(x=-1\)
\(-1>-2 \implies -1>-1 \implies\) No

D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).
\(x=-3\)
\(9>-3 \implies -27>9 \implies\) No

E. If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
\(x=5\)
\(5>\frac{1}{5} \implies 25 > \frac{1}{5} \implies\) Yes

Correct Answer: E

For such questions, I like to test some numbers. To avoid one-sided answers, let's test at least two of the following numbers 5, -5, -1/2, 1/2. Replace x with all four numbers and see which option must be true under all scenario.

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

if x=-5, -\(\frac{1}{5}\)>-5, -5>25 NO, wrong

if x=1/2, 1/1/2 or 2>\(\frac{1}{2}\), \(\frac{1}{2}\)>\(\frac{1}{4}\) YES

Since we have YES and NO, this option cannot always be true.


B. If x is greater than \(\frac{1}{x}\), 2x is greater than x.

if x=5, 5>\(\frac{1}{5}\), 10>5 YES

if x=-\(\frac{1}{2}\), -\(\frac{1}{2}\)>-1/1/2 or -\(\frac{1}{2}\)>-2, -1>\(\frac{-1}{2}\) NO
Here, we got a NO and a YES, which means not always this option is true. Eliminate


C. If x is greater than 2x, \(\frac{1}{x}\) is greater than x.

If x=-\(\frac{1}{2}\), -\(\frac{1}{2}\)>-1, -2>-\(\frac{1}{2}\) NO
If x=-5, -5>-10, -\(\frac{1}{5}\)>-5 Yes
Conflicting information, eliminate


D. If \(x^2\) is greater than x, \(x^3\) is greater than \(x^2\).

if x=5, 25>5, 125>25 YES

if x=-5, 25>-5, -125>25 NO
Eliminate


E. If x is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).

if x=5, 5>\(\frac{1}{5}\) YES, 25>\(\frac{1}{5}\) YES

if x=-\(\frac{1}{2}\)>-2, \(\frac{1}{4}\)>-2 YES.
Option must always be true. E is answer

A. If \(\frac{1}{x}\) is greater than x, x is greater than \(x^2\)
\(\frac{1}{x}\)>x
\(\frac{(1-x^2)}{x}\)>0
x belongs to ( -∞,-1)⋃(0,1)
when x=\(\frac{1}{2}\) then \(x^2\)=\(\frac{1}{4}\)
in this case x>\(x^2\)
but when x= -3 then \(x^2\)=9
then x is not greater than \(x^2\) so this is not always true

B. If x is greater than \(\frac{1}{x}\), 2x is greater than x.
\(\frac{(x^2-1)}{x}\)>0
x belongs to (-1,0)⋃(1,∞)
when x=\(\frac{1}{2}\) 2x=1
then 2x>x
but when x=\(\frac{-1}{2}\) 2x=-1
then 2x<x so this is not always true.

C. If x is greater than 2x, \(\frac{1}{x}\) is greater than x.
x>2x
x<0
when x=-2 then \(\frac{1}{x}\)=\(\frac{-1}{2}\)
in this case \(\frac{1}{x}\)>x
but when x=\(\frac{-1}{2}\) then \(\frac{1}{x}\)= -2
in this case \(\frac{1}{x}\)<x so not always true

D. If \(x^2\) is greater than x, \(x^3\) is greater than \(x^2\).
x(x-1)>0
x belongs to (-∞,0)⋃(1,∞)
when x=3 then \(x^3\)>\(x^2\)
when x=-3 then \(x^3\)<\(x^2\)
not always true

E. If x is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
\(\frac{(x^2-1)}{x}\)>0
x belongs to (-1,0)⋃(1,∞)
for every value of x \(x^2\)>\(\frac{1}{x}\) (correct)

E is the 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.