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

It is currently 21 May 2019, 11:50

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55230
Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 01:30
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

73% (01:32) correct 28% (01:46) wrong based on 60 sessions

HideShow timer Statistics

Current Student
User avatar
P
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
GMAT ToolKit User
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 01:45
x a positive number?

(1) (x – 2)^2 > 2
Assume x=1 (1-2)^2 > 2 No...
but x=0.1 positive (0.1-2)^2 = (1.9)^2 > 2... True.....
Not sufficient

(2) 2^x > 3^x
Assume x=1 => 2>3.. False
x=0.1 => 2^0.1 > 3^0.1 => False.. All positive powers will give answer "False"
so for 2^x > 3^x ... x is not positive
Sufficient

Answer: B
Director
Director
avatar
P
Joined: 21 May 2013
Posts: 656
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 02:16
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x



Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55230
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 02:28
KS15 wrote:
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x



Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks


Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

\(|x-2|>\sqrt{2}\);

\(x-2>\sqrt{2}\) or \(-(x-2)>\sqrt{2}\)

\(x>2+\sqrt{2}\) or \(x<2-\sqrt{2}\)
_________________
Director
Director
avatar
P
Joined: 21 May 2013
Posts: 656
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 02:34
Bunuel wrote:
KS15 wrote:
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x



Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks


Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

\(|x-2|>\sqrt{2}\);

\(x-2>\sqrt{2}\) or \(-(x-2)>\sqrt{2}\)

\(x>2+\sqrt{2}\) or \(x<2-\sqrt{2}\)


Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks
Current Student
User avatar
P
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
GMAT ToolKit User
Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post Updated on: 05 Sep 2017, 02:52
KS15 wrote:
Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks


KS15

(x – 2)^2 > 2
=> (x – 2)^2 - 2 >0
=>\((x – 2)^2 - \sqrt{2} >0\)
=> \((x-2+\sqrt{2}) (x-2-\sqrt{2}) >0\)

This will term either both term are positive or both terms are negative
both positive :
\(x-2+\sqrt{2} > 0\) and \(x-2-\sqrt{2} > 0\)
\(x>2-\sqrt{2}\) and \(x >2+\sqrt{2}\)

OR

both negative :
\(x-2+\sqrt{2} < 0\) and \(x-2-\sqrt{2} < 0\)
\(x<2-\sqrt{2}\)and \(x <2+\sqrt{2}\)

Combining all 4 equations we can write :

\(x >2+\sqrt{2}\) OR \(x<2-\sqrt{2}\)

Originally posted by Nikkb on 05 Sep 2017, 02:42.
Last edited by Nikkb on 05 Sep 2017, 02:52, edited 3 times in total.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55230
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

Show Tags

New post 05 Sep 2017, 02:45
KS15 wrote:
Bunuel wrote:
KS15 wrote:

Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks


Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

\(|x-2|>\sqrt{2}\);

\(x-2>\sqrt{2}\) or \(-(x-2)>\sqrt{2}\)

\(x>2+\sqrt{2}\) or \(x<2-\sqrt{2}\)


Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks


\((x – 2)^2 > 2\)

\((x – 2)^2 -2> 0\)

\((x – 2-\sqrt{2})(x – 2+\sqrt{2})> 0\).

For the product to be positive both multiples must have the same sign.

When both are positive:
\(x – 2-\sqrt{2}> 0\) and \(x – 2+\sqrt{2}> 0\).
\(x > 2+\sqrt{2}\) and \(x > 2-\sqrt{2}\).

For both above to be true simultaneously, \(x > 2+\sqrt{2}\) should hold (so x must be more than the larger number).

When both are negative:
\(x – 2-\sqrt{2}< 0\) and \(x – 2+\sqrt{2}< 0\).
\(x < 2+\sqrt{2}\) and \(x < 2-\sqrt{2}\).

For both above to be true simultaneously, \(x < 2-\sqrt{2}\) should hold (so x must be less than the smaller number).

So, \(x>2+\sqrt{2}\) or \(x<2-\sqrt{2}\).

Hope it's clear.
_________________
GMAT Club Bot
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x   [#permalink] 05 Sep 2017, 02:45
Display posts from previous: Sort by

Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.