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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x Updated on: 03 Oct 2022, 05:10
Originally posted by aadikamagic on 05 Oct 2014, 08:12.
Last edited by Bunuel on 03 Oct 2022, 05:10, edited 2 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

48% (02:17) correct 52% (02:08) wrong based on 813 sessions

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Is \(x^5 > x^4\)?


(1) \(x^3 > −x\)

(2) \(\frac{1}{x} < x\)
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C
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 05 Oct 2014, 09:00
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Is x^5 > x^4?

Reduce by x^4 to rephrase the question:

Is \(x > 1\)?

(1) x^3 > −x

\(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x

Multiply by x^2 (it has to be positive, so we can safely do that):

\(x < x^3\);

\(x(x^2 -1) > 0\);

\((x + 1)x(x - 1) > 0\);

\(-1 < x < 0\) or \(x > 1\).

Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 06 Oct 2014, 02:55
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shreyagmat
Bunuel, can you please explain this :"x < x^3 --> x(x^2 -1) > 0 --> (x + 1)x(x - 1) > 0 --> -1 < x < 0 or x > 1. Not sufficient."

"
i understand that " (x + 1)x(x - 1) > 0" and then the three roots are = 1,-1,0. Since the inequality is ">", x is " outside these values. But how did we come to "-1 < x < 0 or x > 1."

I came to :
x(x+1)(x-1)>0
either x>0 or (x+1)(x-1)>0

and then im lost. Please Help
Show more

(x + 1)x(x - 1) > 0 --> the "roots", in ascending order, are -1, 0 and 1, this gives us 4 ranges:

x < -1;
-1 < x < 0;
0 < x < 1;
x > 1.

Next, test some extreme value for x: if x is some large enough number, say 10, then all three multiples will be positive which gives the positive result for the whole expression, so when x>1, the expression is positive. Now the trick: as in the 4th range expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive and finally in the 1st it'll be negative: - + - + . So, the ranges when the expression is positive are: -1 < x < 0 and x > 1.

Theory on Inequalities:
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 05 Oct 2014, 10:07
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Thanks Bunuel. The take away for me here is that we can always divide by squares without bothering about sign changes.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 06 Oct 2014, 02:29
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Bunuel, can you please explain this :"x < x^3 --> x(x^2 -1) > 0 --> (x + 1)x(x - 1) > 0 --> -1 < x < 0 or x > 1. Not sufficient."

"
i understand that " (x + 1)x(x - 1) > 0" and then the three roots are = 1,-1,0. Since the inequality is ">", x is " outside these values. But how did we come to "-1 < x < 0 or x > 1."

I came to :
x(x+1)(x-1)>0
either x>0 or (x+1)(x-1)>0

and then im lost. Please Help
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 07 Oct 2014, 04:54
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aadikamagic
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x
Show more


It can also quickly be solved thinking about ranges.

Is \(

x^5

>

x^4

\)?

If you think about it, the question is implicitly asking whether a value is greater than one. Indeed every other option (numbers smaller than -1; numbers between -1 and 0; numbers between 0 and 1) if plugged into the expression results in a false statement.

1) picking numbers you can figure out that:

I. the inequality is satisfied when our value is greater than zero. ( pick -1/3. (-1/3)^3 is never gonna be greater than [-(-1/3)]. Same reasoning holds for numbers <= than -1)
II. the inequality is satisfied for both proper fractions and values greater than 1.

NS

2) plugging in numbers you can figure out that the inequality is satisfied for values between both -1 and 0 and greater than 1.

NS

1+2) First statement rules out any value smaller than zero. second statement rules out any proper fraction. Our overlapping result is going to be a value grater than one, making us able to claim an answer.

C.

Hope it helps :)
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 07 Oct 2014, 10:31
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
Show more

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0
how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign?
Please explain.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 08 Oct 2014, 01:32
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0
how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign?
Please explain.
Show more

x < x^3
0 < x^3 - x
0 < x(x^2 - 1), which is the same as x(x^2 - 1) > 0 .
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 08 Jan 2016, 11:06
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shasadou
Is x^5 > x^4?

(1) x^3 > −x
(2) \(\frac{1}{x}\) < x
Show more

You are asked is \(x^5 > x^4\) ---> \(x^5-x^4>0\)---> \(x^4(x-1) > 0\) ----> as \(x^4 > 0\) for all x ---> the question basically asks us whether x>1 ?

Per statement 1,\(x^3 > −x\) ---> \(x^3 + x> 0\) ---> x(x^2+1)>0 ---> x>0 (as \(x^2+1 > 0\) for all x) but is x>1 ? no definite answer. Not sufficient.

Per statement 2, 1/x < x ---> 1/x - x<0 ---> (1-x^2)/x < 0 ---> (x^2-1)/x > 0 ---> (x+1)(x-1)/x > 0 ---> -1<x<0 or x>1 , again not sufficient to give you a definite answer.

Combining the 2 statements you get, x>1, thus giving a definite yes to the question asked.

C is thus the correct answer.

Hope this helps.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x Updated on: 28 Nov 2023, 00:07
Originally posted by KarishmaB on 14 Jul 2017, 00:29.
Last edited by KarishmaB on 28 Nov 2023, 00:07, edited 1 time in total.
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aadikamagic
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x
Show more

Responding to a pm:

All the given information can be manipulated to form factors. I have discussed how to deal with multiple factors in this video: https://youtu.be/PWsUOe77__E

Is x^5 > x^4?
Is x^5 - x^4 > 0?
Is x^4(x - 1) > 0?

We ignore the even powers because x^4 will never be negative. But we need to ensure that x should not be 0 for the inequality to hold.

Is (x - 1) > 0?
Is x > 1?

Statement 1: x^3 > −x

x^3 + x > 0
x(x^2 + 1) > 0
x^2 + 1 will always be positive.
So we see that x > 0.
But we need to know whether it is greater than 1. Greater than 0 could be 0.5 or it could be 2 (or any other number). So this is not sufficient to say whether x will eb greater than 1.
Not sufficient.

(2) 1/x < x
When the sign of x is not known, we do not multiply by x. We bring it to the o there side.
(1/x) - x < 0
(1 - x^2)/x < 0
(x^2 - 1)/x > 0
(x + 1)(x - 1)/x > 0
The transition points on the number line are -1, 0, and 1. The expression will be positive when x > 1 or -1 < x < 0. This alone is also not sufficient.

Both together we know that x is positive and x > 1 or -1 < x < 0.
So x > 1 is a must.
Sufficient.

Answer (C)
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 20 Sep 2017, 11:36
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
Show more

Hey Bunuel,

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 20 Sep 2017, 11:46
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hey Bunuel,

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you
Show more

If it were x^5 < x^4, then after dividing by x^4 we'd get x < 1 but x cannot be 0, so it would be x< 0 or 0 < x < 1.

x^5 > x^4 gives x > 1, so it's redundant to mention that x cannot be 0 because we got x > 1.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 27 Dec 2017, 02:40
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
Show more


Can we just reduce the equation by \(x^4\), when we do not the sign of the variables? Will that not be wrong?
If it is right to reduce by \(x^4\) in this question, how do we assess this for other questions? What do we base our judgment on to divide by variables?

Or is you reduced here because \(x^5> x^4\) could only hold if \(x>1\)?
Separate from the question above

if we manipulated the stem this way would it be right?
\(x^5 - x^4>0\)
\(x^4(x - 1)>0\)

Therefore
either \(x^4>0 --> x > 0\)

\(Or x >1\)

Kindly explain.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 27 Dec 2017, 03:15
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.


Can we just reduce the equation by \(x^4\), when we do not the sign of the variables? Will that not be wrong?
If it is right to reduce by \(x^4\) in this question, how do we assess this for other questions? What do we base our judgment on to divide by variables?

Or is you reduced here because \(x^5> x^4\) could only hold if \(x>1\)?
Separate from the question above

if we manipulated the stem this way would it be right?
\(x^5 - x^4>0\)
\(x^4(x - 1)>0\)

Therefore
either \(x^4>0 --> x > 0\)

\(Or x >1\)

Kindly explain.
Show more

When dividing by negative value we should flip the sign of the inequality, when dividing by positive value we should keep the sign of the inequality. x^4 cannot be negative, so we can divide and write x > 1.

If it were x^5 < x^4, then after we divide by x^4, we'd get x < 1 but here we should mention that x ≠ 0 because if x = 0, then x^5 = x^4 = 0.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 16 Jun 2018, 18:04
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
Show more

Hi Bunuel,

From 2 it seems we are able to derive that x cannot equal 0 since 1/x < x and 1< x^2. Why then do we not just do as follows:

1 < x^2
0 < x^2 - 1
(x-1)(x+1) > 0

So, x>1 or x<-1.

Also, keeping in mind that:

1/x < x , then x>0.

Then, from the above x>1.

My tangential questions here would be how did you know to multiply both sides by x^2?

Thanks in advance.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 16 Jun 2018, 23:05
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energy2pe
Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hi Bunuel,

From 2 it seems we are able to derive that x cannot equal 0 since 1/x < x and 1< x^2. Why then do we not just do as follows:

1 < x^2
0 < x^2 - 1
(x-1)(x+1) > 0

So, x>1 or x<-1.

Also, keeping in mind that:

1/x < x , then x>0.

Then, from the above x>1.

My tangential questions here would be how did you know to multiply both sides by x^2?

Thanks in advance.
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Yes, since x is in the denominator, then it cannot be 0 but we still don't know its sign.

If x > 0, then when cross-multiplying 1/x < x we'd get 1 < x^2.
If x < 0, then when cross-multiplying 1/x < x we'd get 1 > x^2 (flip the sign when multiplying by negative value).

Next, when we multiply an inequality by a variable in even power (x^2, x^4, ...), we don't have to worry about its sing because a number in even power cannot be negative.

9. Inequalities




For more check Ultimate GMAT Quantitative Megathread

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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 20 Aug 2019, 15:29
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Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.
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Hey Bunuel

Why were you able to manipulate statement 1 and 2 without knowing the sign?

Is it because the only way for x^5 to be greater than x^4 is for x to be greater than 1? So you tested the statements on this basis?

Secondly,
Why did you multiply statement 2 by x^2? Is it because we don't know the sign but we do know that x^2 > 0?

Third,
The final statements say x>0 x>1 or x>-1 from statement 2, and x^2 +1>0 and x>0 from statement 1, so combined how did we conclude the intersection was >1?

I really found testing values easier with this one, but I'm really unsure how you went about your way.
I don't understand what you mean by second multiple also
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 20 Aug 2019, 21:38
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dcummins
Bunuel
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hey Bunuel

Why were you able to manipulate statement 1 and 2 without knowing the sign?

Is it because the only way for x^5 to be greater than x^4 is for x to be greater than 1? So you tested the statements on this basis?

Secondly,
Why did you multiply statement 2 by x^2? Is it because we don't know the sign but we do know that x^2 > 0?

Third,
The final statements say x>0 x>1 or x>-1 from statement 2, and x^2 +1>0 and x>0 from statement 1, so combined how did we conclude the intersection was >1?

I really found testing values easier with this one, but I'm really unsure how you went about your way.
I don't understand what you mean by second multiple also
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1. We can divide \(x^5 > x^4\) by x^4, because x^4 (x in even power) is non-negative, so we know the sign. This way we get simplified question: is \(x > 1\)?

2. In the firs statement we are not multiplying or dividing x^3 > −x. We are re-arranging it to get x^3 + x > 0, then factoring out x to get x(x^2 + 1) > 0. Here x is the one multiple and x^2 + 1 is another multiple. The product of two multiples to be positive both must have the same sign. Since x^2 + 1 = positive, the x, another multiple, must also be positive.

3. We are multiplying 1/x < x by x^2 to get x < x^3. Again we can do that because x^2 is non-negative.

4.
From (1) -------------------0-----------------------
From (2) --------1---------0---------1--------------------------------
Intersection: -----1---------0---------1--------------------------------
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 03 Nov 2020, 10:28
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Hello guys ,
I have read in a comment on why x can't be 0 , but didn't understand the reasoning, I was wondering if someone could give an alternative explanation? As per my understanding what we have is, Is x^5>x^4 and I don't understand how we can exclude the possibility that x is 0, (and divide with x^4) , without the stem mentioning it.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 24 Dec 2020, 12:45
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UNSTOPPABLE12
Hello guys ,
I have read in a comment on why x can't be 0 , but didn't understand the reasoning, I was wondering if someone could give an alternative explanation? As per my understanding what we have is, Is x^5>x^4 and I don't understand how we can exclude the possibility that x is 0, (and divide with x^4) , without the stem mentioning it.
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let us substitute x=0 in (1)
\(x^3>-x\)
\(0^3>-0\)
\(0>0\) --> this cannot be possible

Similarly, let us substitute x=0 in (2)
\(\frac{1}{x}=\frac{1}{0}\) --> in GMAT we do not worry about undefined quantities, so we can safely assume x is not equal to 0
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