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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
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JusTLucK04 wrote:
Bunuel wrote:
Is x > 0?

(1) x^6 > x^7 --> reduce by x^6 (notice that it must be positive, so we can safely do that): 1 > x. Not sufficient.

(2) x^7 > x^8 --> reduce by x^6: x > x^2 --> 0 < x < 1. Sufficient.

Answer: B.


Hey Bunuel..I assumed that u cant divide variables in inequalities(Unless u know whether they are positive or negative)...Just can add or subtract them...
This one is new for me...Can u guide me to some more questions or situations to get it clear as to when can I divide and when not


We cannot divide an inequality by a variable if we don't know its sign.

Now, x^6 > x^7 and x^7 > x^8 imply that x is not zero, so x to the even power (such as x^6) must be positive, which means that we CAN divide.

Theory on Inequalities:
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: inequalities-tips-and-hints-175001.html

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 it helps.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
Hi Bunuel,

Case (ii): x > x^2 --> 0 < x < 1 : I understand the solution from a value point of view (plugging negative & positive fractions). However, I tried to solve it algebraically as shown. Please guide as to where the loophole in my working lies:

x>x^2
x-x^2>0
x(1-x)>0
Hence, the terms could both be either positive or negative. If both are positive, you get the condition: 0 < x < 1
However, incase they are both negative -- x<0, x>1 -- That does not lead to the required condition.

Thanks!
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
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shreyagoenka wrote:
Hi Bunuel,

Case (ii): x > x^2 --> 0 < x < 1 : I understand the solution from a value point of view (plugging negative & positive fractions). However, I tried to solve it algebraically as shown. Please guide as to where the loophole in my working lies:

x>x^2
x-x^2>0
x(1-x)>0
Hence, the terms could both be either positive or negative. If both are positive, you get the condition: 0 < x < 1
However, incase they are both negative -- x<0, x>1 -- That does not lead to the required condition.

Thanks!


You did everything right. There is no solution for x(1-x)>0 if both x and 1-x are negative. In this case (x=negative) < (x^2=positive). So, x > x^2 only when 0 < x < 1.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
JusTLucK04 wrote:
Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8


Solution:
Pre Analysis:
  • We are asked if \(x>0\) or not
  • This is a YES-NO question

Statement 1: \(x^6 > x^7\)
  • We can divide both sides with \(x^6\) without any worry of whether the inequality sign will change or not
  • Because \(x^6\) will always be positive
    \(⇒\frac{x^6}{x^6}>\frac{x^7}{x^6}\)
    \(⇒1>x\)
    \(⇒x<1\)
  • From \(x<1\), we cannot be sure if \(x>0\) or not
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(x^7 > x^8\)
  • We can divide both sides with \(x^6\) without any worry of whether the inequality sign will change or not
  • Because \(x^6\) will always be positive
    \(⇒\frac{x^7}{x^6}>\frac{x^8}{x^6}\)
    \(⇒x>x^2\)
    \(⇒x^2-x<0\)
    \(⇒x(x-1)<0\)
    \(⇒0<x<1\)
  • From this, we can be sure and say YES \(x>0\)


Hence the right answer is Option B
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
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