It is currently 17 Dec 2017, 02:16

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is 1/x^5 > y/(y^6+1)?

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 18 Aug 2009
Posts: 413

Kudos [?]: 145 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0

### Show Tags

10 Feb 2011, 13:25
00:00

Difficulty:

45% (medium)

Question Stats:

61% (01:18) correct 39% (01:18) wrong based on 62 sessions

### HideShow timer Statistics

Is 1/x^5 > y/(y^6+1)?

(1) x = y
(2) y > 0
[Reveal] Spoiler: OA

Attachments

Untitled.jpg [ 15.06 KiB | Viewed 2365 times ]

_________________

Never give up,,,

Last edited by Bunuel on 12 Jul 2013, 01:52, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 145 [0], given: 16

Math Expert
Joined: 02 Sep 2009
Posts: 42646

Kudos [?]: 135919 [3], given: 12716

### Show Tags

10 Feb 2011, 13:52
3
KUDOS
Expert's post
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$ --> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

_________________

Kudos [?]: 135919 [3], given: 12716

Senior Manager
Joined: 18 Aug 2009
Posts: 413

Kudos [?]: 145 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0

### Show Tags

10 Feb 2011, 13:58
Thank you, Bunuel!!!
_________________

Never give up,,,

Kudos [?]: 145 [0], given: 16

Manager
Joined: 19 Feb 2009
Posts: 54

Kudos [?]: 139 [0], given: 8

### Show Tags

12 Feb 2011, 08:40
Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$ --> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

HI Bunuel..

in S1: $$x=y$$ , you considered 2 cases , $$y<0$$ and $$y>0$$ but why you did not consider $$x=y=0$$?
_________________

Working without expecting fruit helps in mastering the art of doing fault-free action !

Kudos [?]: 139 [0], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 42646

Kudos [?]: 135919 [1], given: 12716

### Show Tags

12 Feb 2011, 08:47
1
KUDOS
Expert's post
amod243 wrote:
Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$ --> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

HI Bunuel..

in S1: $$x=y$$ , you considered 2 cases , $$y<0$$ and $$y>0$$ but why you did not consider $$x=y=0$$?

x is in denominator it can not be zero.
_________________

Kudos [?]: 135919 [1], given: 12716

Manager
Joined: 05 Jul 2010
Posts: 183

Kudos [?]: 22 [0], given: 18

### Show Tags

12 Feb 2011, 14:06

Kudos [?]: 22 [0], given: 18

Senior Manager
Affiliations: CFA Level 2
Joined: 05 May 2004
Posts: 263

Kudos [?]: 164 [0], given: 0

Location: Hanoi

### Show Tags

14 Feb 2011, 01:18
Hi guys,

Can we do it this way

1/x^5>y/(y^6 + 1) => X^5<(Y^6+1)/Y

S1. Y^5 < Y^5 + 1/Y -> Insufficient
S2 Y>0 -> Insufficient

S1 + S2 -> 1/Y >0 -> Y^5 < Y^5 + 1/Y -> C is the answer
_________________

"Life is like a box of chocolates, you never know what you'r gonna get"

Kudos [?]: 164 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42646

Kudos [?]: 135919 [0], given: 12716

### Show Tags

14 Feb 2011, 01:39
bigtooth81 wrote:
Hi guys,

Can we do it this way

1/x^5>y/(y^6 + 1) => X^5<(Y^6+1)/Y

S1. Y^5 < Y^5 + 1/Y -> Insufficient
S2 Y>0 -> Insufficient

S1 + S2 -> 1/Y >0 -> Y^5 < Y^5 + 1/Y -> C is the answer

You can not cross multiply as you don't know the signs of x^5 and y (y^6+1 is positive). For example if x=1 and y=-1 then: 1/x^5=1>-1/2=y/(y^6+1) but x^5=1>-2=(y^6+1)/y (> not < as you wrote).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.
_________________

Kudos [?]: 135919 [0], given: 12716

Senior Manager
Affiliations: CFA Level 2
Joined: 05 May 2004
Posts: 263

Kudos [?]: 164 [0], given: 0

Location: Hanoi

### Show Tags

14 Feb 2011, 04:21
Bunuel wrote:
bigtooth81 wrote:
Hi guys,

Can we do it this way

1/x^5>y/(y^6 + 1) => X^5<(Y^6+1)/Y

S1. Y^5 < Y^5 + 1/Y -> Insufficient
S2 Y>0 -> Insufficient

S1 + S2 -> 1/Y >0 -> Y^5 < Y^5 + 1/Y -> C is the answer

You can not cross multiply as you don't know the signs of x^5 and y (y^6+1 is positive). For example if x=1 and y=-1 then: 1/x^5=1>-1/2=y/(y^6+1) but x^5=1>-2=(y^6+1)/y (> not < as you wrote).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Got you. Thanks bro
_________________

"Life is like a box of chocolates, you never know what you'r gonna get"

Kudos [?]: 164 [0], given: 0

Senior Manager
Joined: 30 Nov 2010
Posts: 257

Kudos [?]: 121 [0], given: 66

Schools: UC Berkley, UCLA

### Show Tags

15 Feb 2011, 07:42
Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$--> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel!

Mari
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

Kudos [?]: 121 [0], given: 66

Math Expert
Joined: 02 Sep 2009
Posts: 42646

Kudos [?]: 135919 [2], given: 12716

### Show Tags

15 Feb 2011, 07:56
2
KUDOS
Expert's post
mariyea wrote:
Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$--> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel!

Mari

We should determine whether $$\frac{1}{y^5}>\frac{y}{y^6+1}$$ is true. You can do this algebraically or with number plugging (for example test y=1 to get YES and then y=-1 to get NO).

As for algebraic approach: we should somehow simplify $$\frac{1}{y^5}>\frac{y}{y^6+1}$$ (as it looks kind of ugly) to get the answer. For this we consider two cases: $$y<0$$ and $$y>0$$ (y=0 is not possible as y is in denominator and we know that division by zero is undefined). As we proceed, we get NO answer for an assumption $$y<0$$ and we get YES answer for an assumption $$y>0$$.

Hope it's clear.
_________________

Kudos [?]: 135919 [2], given: 12716

Senior Manager
Joined: 30 Nov 2010
Posts: 257

Kudos [?]: 121 [0], given: 66

Schools: UC Berkley, UCLA

### Show Tags

15 Feb 2011, 08:02
Bunuel wrote:
mariyea wrote:
Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$--> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel!

Mari

We should determine whether $$\frac{1}{y^5}>\frac{y}{y^6+1}$$ is true. You can do this algebraically or with number plugging (for example test y=1 to get YES and then y=-1 to get NO).

As for algebraic approach: we should somehow simplify $$\frac{1}{y^5}>\frac{y}{y^6+1}$$ (as it looks kind of ugly) to get the answer. For this we consider two cases: $$y<0$$ and $$y>0$$ (y=0 is not possible as y is in denominator and we know that division by zero is undefined). As we proceed, we get NO answer for an assumption $$y<0$$ and we get YES answer for an assumption $$y>0$$.

Hope it's clear.

Yes! It is very clear now, thanks! I was just wondering whether you considered y -ve... now I understand. Thank you so much for your patience!
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

Kudos [?]: 121 [0], given: 66

Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 866

Kudos [?]: 405 [0], given: 123

### Show Tags

01 Mar 2011, 00:24
Hi Bunuel

If it would have been Is 1/x^5 > 1/(y^6+1) ? Then can I rephrase it as

Is x^5 < y^6+1 ? irrespective of the signs. Am I correct?

Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$ --> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

Kudos [?]: 405 [0], given: 123

Math Expert
Joined: 02 Sep 2009
Posts: 42646

Kudos [?]: 135919 [0], given: 12716

### Show Tags

01 Mar 2011, 01:41
gmat1220 wrote:
Hi Bunuel

If it would have been Is 1/x^5 > 1/(y^6+1) ? Then can I rephrase it as

Is x^5 < y^6+1 ? irrespective of the signs. Am I correct?

Bunuel wrote:
Is 1/x^5 > y/(y^6+1)?

(1) x=y --> is $$\frac{1}{y^5}>\frac{y}{y^6+1}$$? Two cases:

A. $$y<0$$ --> cross multiply and as for negative $$y$$: $$y^5<0$$ and $$y^6+1>0$$ flip the sign (because of negative $$y^5$$). The question becomes: is $$y^6+1<y^6$$? --> is $$1<0$$? In this case the answer would be NO.

B. $$y>0$$ --> cross multiply and as for positive $$y$$ both $$y^5$$ and $$y^6+1$$ are positive remain the sign. The question becomes: is $$y^6+1>y^6$$? --> is $$1>0$$? In this case the answer would be YES.

(2) y>0. Clearly insufficient as no info about $$x$$.

(1)+(2) As from (2) $$y>0$$ then we have case B and the answer is YES. Sufficient.

Check this: inequalities-109031.html#p871109

You can not cross multiply as you don't know the sign of x^5.

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.
_________________

Kudos [?]: 135919 [0], given: 12716

Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 866

Kudos [?]: 405 [0], given: 123

### Show Tags

01 Mar 2011, 02:27
Thanks Bunuel for the brilliant explanation !

Kudos [?]: 405 [0], given: 123

Non-Human User
Joined: 09 Sep 2013
Posts: 14796

Kudos [?]: 288 [0], given: 0

Re: Is 1/x^5 > y/(y^6+1)? [#permalink]

### Show Tags

07 Feb 2014, 09:04
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 288 [0], given: 0

Intern
Joined: 01 Aug 2006
Posts: 34

Kudos [?]: 42 [0], given: 0

Re: Is 1/x^5 > y/(y^6+1)? [#permalink]

### Show Tags

07 Feb 2014, 18:55
Is 1/x^5 > y/(y^6+1)? or is x^5 < (y^6+1)/y? (taking reciprocals changes the sign of inequality)

(1) x = y -> is x^ 5 < (x^6 + 1)/x or is x^5 < x^5 + 1/x? put x = 1, yes. Put x = -1, No.
(2) y > 0 -> no info on x. NOT sufficient

Combining, x = y > 0. For +ve values, x^5 always less than x^5 +1/x.
example: integer values. Put x = 2; Yes. non-integer values: put 1/2. 1/x on the RHS will make it RHS a bigger number.

C

Kudos [?]: 42 [0], given: 0

Re: Is 1/x^5 > y/(y^6+1)?   [#permalink] 07 Feb 2014, 18:55
Display posts from previous: Sort by