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Mck2023
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 07 Mar 2021, 20:18
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HI Bunuel
Can you please help me understand this:

You reduced by x^4. i.e. x^4(x-1)>0, I guess!!
But I was not confident to do this because I had seen somewhere in 'finding the range of value of question', using wavy line method, where reducing directly like this got me wrong answer!! Can you please contrast between these two scenarios? Thank You!
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 08 Mar 2021, 00:18
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Mck2023
HI Bunuel
Can you please help me understand this:

You reduced by x^4. i.e. x^4(x-1)>0, I guess!!
But I was not confident to do this because I had seen somewhere in 'finding the range of value of question', using wavy line method, where reducing directly like this got me wrong answer!! Can you please contrast between these two scenarios? Thank You!
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We cannot divide an inequality by the variable if we don't know its sign. Here though x^4 cannot be negative, so we can divide and write x > 1.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 28 Apr 2023, 07:09
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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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For option 2, \(-1 < x < 0\) or \(x > 1\).

isn't this range sufficient for the question to answer? IMO x^5 > x^4 for both the ranges. Or am I missing something here?

Bunuel @sayantanck please shed some light
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 28 Apr 2023, 07:42
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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.


For option 2, \(-1 < x < 0\) or \(x > 1\).

isn't this range sufficient for the question to answer? IMO x^5 > x^4 for both the ranges. Or am I missing something here?

Bunuel @sayantanck please shed some light
Show more

The inequality x^5 > x^4 doesn't hold true when -1 < x < 0. For instance, when x = -1/2, x^5 becomes negative while x^4 is positive, making x^5 actually smaller than x^4, not larger.

I hope this helps.
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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x 26 Sep 2025, 04:23
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