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

 It is currently 18 Nov 2018, 11:00

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat

November 20, 2018

November 20, 2018

09:00 AM PST

10:00 AM PST

The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
• ### The winning strategy for 700+ on the GMAT

November 20, 2018

November 20, 2018

06:00 PM EST

07:00 PM EST

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.

# Is x > y? (1) x^2 > y (2) √x < y

Author Message
TAGS:

### Hide Tags

Manager
Joined: 19 Oct 2011
Posts: 104
Location: India
Is x > y? (1) x^2 > y (2) √x < y  [#permalink]

### Show Tags

Updated on: 19 Feb 2012, 17:39
5
6
00:00

Difficulty:

45% (medium)

Question Stats:

67% (01:50) correct 33% (01:52) wrong based on 229 sessions

### HideShow timer Statistics

Is x > y?

(1) x^2 > y
(2) √x < y

_________________

Encourage me by pressing the KUDOS if you find my post to be helpful.

Help me win "The One Thing You Wish You Knew - GMAT Club Contest"
http://gmatclub.com/forum/the-one-thing-you-wish-you-knew-gmat-club-contest-140358.html#p1130989

Originally posted by dvinoth86 on 19 Feb 2012, 07:00.
Last edited by dvinoth86 on 19 Feb 2012, 17:39, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 50627
Re: Is x > y? (1) x^2 > y (2) √x < y  [#permalink]

### Show Tags

19 Feb 2012, 10:30
2
2
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.

(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that $$x\geq{0}$$.

(1)+(2) $$\sqrt{x}<y<x^2$$ --> both $$x$$ and $$y$$ are between $$\sqrt{x}$$ and $$x^2$$, but we can not say which one is greater. Not sufficient. For example: if $$x=y=4$$ ($$\sqrt{4}<4<4^2$$) then the answer is NO but if $$x=4$$ and $$y=3$$ ($$\sqrt{4}<3<4^2$$) then the answer is YES. Not sufficient.

Hope it's clear.
_________________
Manager
Status: Retaking next month
Affiliations: None
Joined: 05 Mar 2011
Posts: 152
Location: India
Concentration: Marketing, Entrepreneurship
GMAT 1: 570 Q42 V27
GPA: 3.01
WE: Sales (Manufacturing)
Re: Is x > y? (1) x^2 > y (2) √x < y  [#permalink]

### Show Tags

19 Feb 2012, 20:17
Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that $$x\geq{0}$$.

(1)+(2) $$\sqrt{x}<y<x^2$$ --> both $$x$$ and $$y$$ are between $$\sqrt{x}$$ and $$x^2$$, but we can not say which one is greater. Not sufficient. For example: if $$x=y=4$$ ($$\sqrt{4}<4<4^2$$) then the answer is NO but if $$x=4$$ and $$y=3$$ ($$\sqrt{4}<3<4^2$$) then the answer is YES. Not sufficient.

Hope it's clear.

Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.
Math Expert
Joined: 02 Sep 2009
Posts: 50627
Re: Is x > y? (1) x^2 > y (2) √x < y  [#permalink]

### Show Tags

19 Feb 2012, 23:18
1
1
GMATPASSION wrote:
Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that $$x\geq{0}$$.

(1)+(2) $$\sqrt{x}<y<x^2$$ --> both $$x$$ and $$y$$ are between $$\sqrt{x}$$ and $$x^2$$, but we can not say which one is greater. Not sufficient. For example: if $$x=y=4$$ ($$\sqrt{4}<4<4^2$$) then the answer is NO but if $$x=4$$ and $$y=3$$ ($$\sqrt{4}<3<4^2$$) then the answer is YES. Not sufficient.

Hope it's clear.

Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.

First of all: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, number picking strategy can vary for different problems. Generally it's good to test negative/positive/zero as well as integer/fraction to get a YES and a NO answers. If you deal with two variables it's also helpful to test x<y and x>y in addition to the former.

As for this question: you don't really need to test the numbers for it, I just used them to demonstrate that the statements are not sufficient.

From (1)+(2): we have that $$\sqrt{x}<y<x^2$$. Both $$x$$ and $$y$$ are between $$\sqrt{x}$$ and $$x^2$$ ($$x$$ is between them because $$\sqrt{x}<x^2$$, which means that $$x>1$$):
$$\sqrt{x}$$------$$x$$------$$x^2$$, now $$y$$ can be in the green range (answer YES) as well in the red range (answer NO). So, we can not say whether x>y.

Hope it's clear.
_________________
Manager
Joined: 29 Mar 2010
Posts: 121
Location: United States
GMAT 1: 590 Q28 V38
GPA: 2.54
WE: Accounting (Hospitality and Tourism)
Re: Is x > y?  [#permalink]

### Show Tags

18 Aug 2012, 21:43
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y

Can anyone explain a simple method to this could not follovv statement B

\

Bunuel if you can help please

1) Statement 1 only tells us that x is positive and nothing else. So insufficient

2) Statement 2 wants us to go through the process of squaring both sides to make the equation x<y^2, but we do not know anything about the sign so basically it would look like:

x^2 > y

and combined,

x<y^2 or x>y^2

x^2 > y
_________________

4/28 GMATPrep 42Q 36V 640

Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: Is x > y?  [#permalink]

### Show Tags

Updated on: 18 Aug 2012, 22:04
[quote="venmic"]Is x > y?

(1) x^2 > y

(2) √x < y

Can anyone explain a simple method to this could not follovv statement B

(1) For $$x=-2 > y=-3, \, x^2=4>-3.$$ But $$x=2 < y=3$$, although $$x^2=4>y=3.$$
Not sufficient.

(2) From the given inequality it follows that $$x$$ must be non-negative (because of the square root) and since $$y>\sqrt{x}\geq0$$, necessarily $$y$$ is positive.
Therefore, we can square the given inequality and get $$x<y^2.$$

For $$x=4, y=3, \,x=4<y^2=9,$$ and $$x>y.$$
But if $$y=1,$$ we cannot have simultaneously $$x<y^2=1$$ and $$x>y=1.$$
Not sufficient.

(1) and (2) together:
Consider the two cases: $$x=2, y=3$$ and $$x=4,y=3.$$
Again, not sufficient.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Originally posted by EvaJager on 18 Aug 2012, 21:56.
Last edited by EvaJager on 18 Aug 2012, 22:04, edited 1 time in total.
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: Is x > y?  [#permalink]

### Show Tags

18 Aug 2012, 22:03
hfbamafan wrote:
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y

Can anyone explain a simple method to this could not follovv statement B

\

Bunuel if you can help please

1) Statement 1 only tells us that x is positive and nothing else. So insufficient

2) Statement 2 wants us to go through the process of squaring both sides to make the equation x<y^2, but we do not know anything about the sign so basically it would look like:

x^2 > y

and combined,

x<y^2 or x>y^2

x^2 > y

(2) We know about the signs: $$x$$ must be non-negative, otherwise the square root is not defined. Also, because the square root is non-negative, $$y$$ must be positive. Therefore, in this case we can square the given inequality and obtain $$x<y^2.$$
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Joined: 30 Sep 2009
Posts: 93
Re: Is x > y?  [#permalink]

### Show Tags

18 Aug 2012, 22:58
statement 2: in Gmat.... sqrt(x) means x is always positive .... therefore from statement 2 we know x is a positive number.

from statement 1 : x*x=y ..insufficient because we x and y can have any values....

combining 1 and 2:
sqrt(x) < y < x*x .... not possible to determine whether x> y
Non-Human User
Joined: 09 Sep 2013
Posts: 8814
Re: Is x > y? (1) x^2 > y (2) √x < y  [#permalink]

### Show Tags

01 Dec 2017, 00:01
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: Is x > y? (1) x^2 > y (2) √x < y &nbs [#permalink] 01 Dec 2017, 00:01
Display posts from previous: Sort by