GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 11 Dec 2018, 01:11

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.
• ### The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

# If x and y are positive numbers, is x > y?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51096
If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

14 Jun 2016, 01:00
1
4
00:00

Difficulty:

95% (hard)

Question Stats:

45% (02:18) correct 55% (02:08) wrong based on 106 sessions

### HideShow timer Statistics

If x and y are positive numbers, is x > y?

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

_________________
Senior Manager
Joined: 18 Jan 2010
Posts: 252
Re: If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

14 Jun 2016, 02:04
Bunuel wrote:
If x and y are positive numbers, is x > y?

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

Statement 1

$$\frac{x^3}{y} < 1$$

x and y are positive numbers so we can cross multiply.

$$x^3 < y$$

x = 2, y = 9
$$x^3$$ = 8. This is less than y. So statement (1) is satisfied.
Since x=2, and y =9, x is not more than y.

x = (1/2), y = (1/7)
$$x^3$$ = (1/8). This is less than y. So statement (1) is satisfied.
Since x=(1/2), and y =(1/7), x is more than y.

Statement (1) is not sufficient

Statement 2

$$\frac{\sqrt[3]{x}}{y} < 1$$

$${\sqrt[3]{x}} < y$$

x and y are positive numbers so we can raise each of them to cubic powers

so x < $$y^3$$

x = 7, y = 2

$$y^3$$ = 8. x =7. x is less than $$y^3$$. So statement (2) is satisfied.
Since x=7, and y =9, x is more than y.

x = 1, y = 3

$$y^3$$ = 27. x =1. x is less than $$y^3$$. So statement (2) is satisfied.
Since x=1, and y =3, x is not more than y.

Statement (2) is not sufficient

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3 < y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

Current Student
Joined: 13 Aug 2013
Posts: 44
If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

14 Jun 2016, 19:49
Bunuel wrote:
If x and y are positive numbers, is x > y?

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3$$ < $$y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

I am not completely sure about the above. May be I am wrong, but can't we simply multiply the inequalities since we know that they are positive numbers?

$$x^3 < y$$

x < $$y^3$$

we get: $$x^4$$ < $$y^4$$

Hence, x < y and Hence answer should be C and not E

Is there something wrong with this logic?
_________________

Kudos plz if it helped

Manager
Joined: 05 Dec 2015
Posts: 112
If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

26 Jun 2016, 20:33
rockstar23 wrote:
Bunuel wrote:
If x and y are positive numbers, is x > y?

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3$$ < $$y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

I am not completely sure about the above. May be I am wrong, but can't we simply multiply the inequalities since we know that they are positive numbers?

$$x^3 < y$$

x < $$y^3$$

we get: $$x^4$$ < $$y^4$$

Hence, x < y and Hence answer should be C and not E

Is there something wrong with this logic?

Hi - yes, your logic is not correct. If you have an even exponent it masks the sign of the variable, for example:

if x = -1 and y = -2 then x>y
if x = -2 and y = 3 then y>x

(edit b/c I had it backwards ;D)
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India
Re: If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

26 Jun 2016, 21:55
1
2
Bunuel wrote:
If x and y are positive numbers, is x > y?

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

x and y are positive. So we can multiply the inequality by x or y.

(1) $$\frac{x^3}{y} < 1$$
$$x^3 < y$$

How do cubes behave on the right of 0? If x lies between 0 and 1, the cube will be less than x. If x is greater than 1, x^3 will be more than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$
$$\sqrt[3]{x} < y$$

How do cube roots behave on the right of 0? If x lies between 0 and 1, the cube root will be more than x. If x is greater than 1, it will be less than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.

Using both,
x will lie between x^3 and $$\sqrt[3]{x}$$ in either case. Since y is greater than both, y must be greater than x. Sufficient.

_________________

[b]Karishma
Veritas Prep GMAT Instructor

Non-Human User
Joined: 09 Sep 2013
Posts: 9103
Re: If x and y are positive numbers, is x > y?  [#permalink]

### Show Tags

30 Jun 2018, 21:58
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.
_________________
Re: If x and y are positive numbers, is x > y? &nbs [#permalink] 30 Jun 2018, 21:58
Display posts from previous: Sort by

# If x and y are positive numbers, is x > y?

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.