Is x 0? (1) x^6 x^7 (2) x^7 x^8

Question

Is x > 0?

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

Bunuel · Accepted Answer

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.

JusTLucK04 · Answer

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

Bunuel · Answer

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.

shreyagoenka · Answer

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!

Bunuel · Answer

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.

SatyamOP · Answer

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 ⇒00 Hence the right answer is Option B

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Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8

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