Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 17 Jul 2019, 02:22 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Is x^5 > x^4?

Author Message
TAGS:

### Hide Tags

VP  D
Status: Learning
Joined: 20 Dec 2015
Posts: 1032
Location: India
Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36 GRE 1: Q157 V157 GPA: 3.4
WE: Engineering (Manufacturing)
Re: Is x^5 > x^4?  [#permalink]

### Show Tags

Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

From 1 we can infer that the value of x is greater than 0 but we can not be sure that it is not a fraction .
suppose x=0.2 then 0.008>-0.2 but 0.2^5<0.4 and if we take integer then we will get x^5 > x^4 hence two cases insufficient .

Multiply both the sides of the statement 2 we get x<x^3 or x^3-x>0 or x*(x+1)*(x-1)>0 there are two possible ranges for the equation -1<x<1 or x>1 hence insufficient .
Together they are sufficient as x>1
_________________ S
Joined: 11 Mar 2018
Posts: 86
Location: United States
GMAT 1: 690 Q48 V37 GMAT 2: 710 Q47 V41 GPA: 3.6
WE: Analyst (Energy and Utilities)
Re: Is x^5 > x^4?  [#permalink]

### Show Tags

Bunuel wrote:
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.

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?

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56267
Re: Is x^5 > x^4?  [#permalink]

### Show Tags

energy2pe wrote:
Bunuel wrote:
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.

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?

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

_________________
Manager  S
Joined: 14 Jul 2014
Posts: 90
Location: India
Concentration: Social Entrepreneurship, Strategy
GMAT 1: 620 Q41 V34 WE: Information Technology (Computer Software)
Re: Is x^5 > x^4?  [#permalink]

### Show Tags

Bunuel wrote:
shreyagmat wrote:
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

(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:
Inequality tips: http://gmatclub.com/forum/tips-and-hint ... l#p1379270

http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... 09468.html
http://gmatclub.com/forum/everything-is ... 08884.html
http://gmatclub.com/forum/graphic-appro ... 68037.html

All DS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=184
All PS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=189

700+ Inequalities problems: http://gmatclub.com/forum/inequality-an ... 86939.html

Hope this helps.

Dear Bunuel,

Thank you for the amazing explanation.
But I'm actually confused on this step:
(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.

Kindly explain how does this provide us with the afore mentioned values:
I'm actually stuck here:
(x + 1)x(x - 1) > 0;
So, (x+1)>0
x > -1;
x>0;
(x-1) > 0
x > 1;

Kindly help me with this; Im unable to reach the 4 ranges you mentioned. Re: Is x^5 > x^4?   [#permalink] 09 Aug 2018, 11:37

Go to page   Previous    1   2   [ 24 posts ]

Display posts from previous: Sort by

# Is x^5 > x^4?  