It is currently 19 Mar 2018, 21:08

### 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 a > b^2 ? (1) a>b^4 (2) a> b^(1/2)

Author Message
TAGS:

### Hide Tags

Intern
Joined: 08 Jan 2015
Posts: 28
Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

30 Mar 2015, 11:23
14
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

45% (01:39) correct 55% (01:35) wrong based on 306 sessions

### HideShow timer Statistics

Is $$a > b^2$$ ?

(1) $$a>b^4$$

(2) $$a> \sqrt{b}$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Aug 2017, 04:08, edited 2 times in total.
Edited the question.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7992
Location: Pune, India
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

30 Mar 2015, 22:02
6
KUDOS
Expert's post
6
This post was
BOOKMARKED
Awli wrote:
Is a > b^2 ?

(1) a>b^4

(2) a> \sqrt{b}

We know that exponents behave differently for numbers greater than 1 and for numbers between 0 and 1

(1) $$a>b^4$$
If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 1/4 (which gives b^2 = 1/16) and a = 1/20, then we answer 'No'
Not sufficient

(2) $$a> \sqrt{b}$$
If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 4 (which gives b^2 = 16) and a = 3, then we answer 'No'
Not sufficient

Using both,
if b is between 0 and 1, $$b^2 < b < \sqrt{b}$$. Since a would be greater than $$\sqrt{b}$$, a would be greater than $$b^2$$.
If b is 1 or greater, then, $$b < b^2 < b^4$$. For a to be greater than $$b^4$$, a will be greater than $$b^2$$ too. So in any case, a will be greater than b^2. Note that we don't need to consider negative value for b since $$\sqrt{b}$$ needs to be real.
Sufficient
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 07 Aug 2011 Posts: 577 Concentration: International Business, Technology GMAT 1: 630 Q49 V27 Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink] ### Show Tags 31 Mar 2015, 08:29 1 This post was BOOKMARKED VeritasPrepKarishma wrote: Awli wrote: Is a > b^2 ? (1) a>b^4 (2) a> \sqrt{b} We know that exponents behave differently for numbers greater than 1 and for numbers between 0 and 1 (1) $$a>b^4$$ If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes' If b = 1/4 (which gives b^2 = 1/16) and a = 1/20, then we answer 'No' Not sufficient (2) $$a> \sqrt{b}$$ If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes' If b = 4 (which gives b^2 = 16) and a = 3, then we answer 'No' Not sufficient Using both, if b is between 0 and 1, $$b^2 < b < \sqrt{b}$$. Since a would be greater than $$\sqrt{b}$$, a would be greater than $$b^2$$. If b is 1 or greater, then, $$b < b^2 < b^4$$. For a to be greater than $$b^4$$, a will be greater than $$b^2$$ too. So in any case, a will be greater than b^2. Note that we don't need to consider negative value for b since $$\sqrt{b}$$ needs to be real. Sufficient Answer (C) thanks VeritasPrepKarishma . in this case, is b > 1 ? my approach was as follows : option A : a>b^4 a=1/8 b=1/2 is a>b^2 NO a=1/8 b=1/3 is a>b^2 YES note here , a is clearly a +ive number. option B: a>b^0.5 a=5,b=2 is a>b^2 YES a=5, b=4 is a>b^2 NO note here we know that b is a +ive number. combine : note that we know a and b are +ive. a>b^4 --> minimum value of a is b^4 and a>b^1/2 so , b^4>b^1/2 OR b^8>b .. clearly b is >1 and not a fraction and so if a>b^4 , a must > b^2 too. _________________ Thanks, Lucky _______________________________________________________ Kindly press the to appreciate my post !! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7992 Location: Pune, India Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink] ### Show Tags 31 Mar 2015, 22:40 Expert's post 1 This post was BOOKMARKED Lucky2783 wrote: thanks VeritasPrepKarishma . in this case, is b > 1 ? my approach was as follows : There is no reason why b must be greater than 1. It could be between 0 and 1 too. Say, b = 1/4 1. a > b^4 => a > 1/256 2. a > $$\sqrt{b}$$ => a > 1/2 Since a is greater than 1/2, it must be greater than b^2 which is 1/16. If b is between 0 and 1, $$b^2 < b < \sqrt{b}$$. Since a would be greater than $$\sqrt{b}$$, a would be greater than $$b^2$$. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 10 Mar 2015
Posts: 40
Location: India
GMAT 1: 700 Q49 V37
GPA: 3.9
WE: Web Development (Computer Software)
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

21 Sep 2015, 06:37
My Approach:

a>b^2
my thoughts were - for this to be correct, a should be positive, no other thoughts.

now statement 2: a> \sqrt{b}
\sqrt{b}
This may be positive or negative.. Cannot say which one .
so Statement 2 Not Sufficient.

statement 1: a> b^4
This indicates that b is positive.
But this condition fails when 0<b<1
so Statement 1 Not Sufficient.

combining both, stmnt 1 clearly shows that if a>\sqrt{b} so indirectly implies that b does not lie between 0 and 1 and by statement 2 it implies that it is positive. So C is the answer.
Intern
Joined: 06 Oct 2013
Posts: 46
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

16 Dec 2015, 14:04
Awli wrote:
Is a > b^2 ?

(1) a>b^4

(2) $$a> \sqrt{b}$$

SOLUTION:

I LIKE Bunuel APPROACH FOR THIS KIND OF PROBLEM.

Statement 1) a>b^4.

for $$1<b:---\sqrt{b}--b----b^2---b^4.$$. a is always right side of $$b^2$$. ANS YES

For $$0<=b<=1: 0--b^4---b^2----b-----\sqrt{b}---$$: a Can be on left or right side of $$b^2$$.

Hence Insufficient.

Statement 2)$$a> \sqrt{b}$$

a and b are non negative.

for $$1<b:---\sqrt{b}--b----b^2---b^4.$$ . a Can be on left or right side of $$b^2$$

Hence Insufficient.

Statement 1 and statement 2 together.

for $$1<b:---\sqrt{b}--b----b^2---b^4.$$. from statement 1. a is always right side of $$b^2$$. ANS YES

For $$0<=b<=1: 0--b^4---b^2----b-----\sqrt{b}---$$. from statement 2. a is always right side of $$b^2$$. ANS YES

Hence Sufficient.

ANS C.
Manager
Joined: 24 May 2016
Posts: 170
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

23 Jul 2016, 11:49
VeritasPrepKarishma wrote:
Awli wrote:
Is a > b^2 ?

(1) a>b^4

(2) a> \sqrt{b}

We know that exponents behave differently for numbers greater than 1 and for numbers between 0 and 1

(1) $$a>b^4$$
If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 1/4 (which gives b^2 = 1/16) and a = 1/20, then we answer 'No'
Not sufficient

(2) $$a> \sqrt{b}$$

If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 4 (which gives b^2 = 16) and a = 3, then we answer 'No'
Not sufficient

Using both,
if b is between 0 and 1, $$b^2 < b < \sqrt{b}$$. Since a would be greater than $$\sqrt{b}$$, a would be greater than $$b^2$$.
If b is 1 or greater, then, $$b < b^2 < b^4$$. For a to be greater than $$b^4$$, a will be greater than $$b^2$$ too. So in any case, a will be greater than b^2. Note that we don't need to consider negative value for b since $$\sqrt{b}$$ needs to be real.
Sufficient

We combine statements 1 and 2.

If 0 < b < 1:

Suppose a = 1/4 and that we consider the negative root of $$\sqrt{b}$$.

b = 1/4 ---> -1/2 < 1/ 256 < 1/16 < 1/4 ($$\sqrt{b}$$ < b^4 < b^2 < b)

From this, we cannot conclude that a > b^2.

Hence, the combination of both statements would not be sufficient and the solution would be option E.

Manager
Joined: 24 May 2016
Posts: 170
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

24 Jul 2016, 05:28
Math Expert
Joined: 02 Sep 2009
Posts: 44322
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

24 Jul 2016, 08:09
Expert's post
1
This post was
BOOKMARKED
EBITDA wrote:
VeritasPrepKarishma wrote:
Awli wrote:
Is a > b^2 ?

(1) a>b^4

(2) a> \sqrt{b}

We know that exponents behave differently for numbers greater than 1 and for numbers between 0 and 1

(1) $$a>b^4$$
If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 1/4 (which gives b^2 = 1/16) and a = 1/20, then we answer 'No'
Not sufficient

(2) $$a> \sqrt{b}$$

If b = 4 (which gives b^2 = 16) and a = 1000, then we answer 'Yes'
If b = 4 (which gives b^2 = 16) and a = 3, then we answer 'No'
Not sufficient

Using both,
if b is between 0 and 1, $$b^2 < b < \sqrt{b}$$. Since a would be greater than $$\sqrt{b}$$, a would be greater than $$b^2$$.
If b is 1 or greater, then, $$b < b^2 < b^4$$. For a to be greater than $$b^4$$, a will be greater than $$b^2$$ too. So in any case, a will be greater than b^2. Note that we don't need to consider negative value for b since $$\sqrt{b}$$ needs to be real.
Sufficient

We combine statements 1 and 2.

If 0 < b < 1:

Suppose a = 1/4 and that we consider the negative root of $$\sqrt{b}$$.

b = 1/4 ---> -1/2 < 1/ 256 < 1/16 < 1/4 ($$\sqrt{b}$$ < b^4 < b^2 < b)

From this, we cannot conclude that a > b^2.

Hence, the combination of both statements would not be sufficient and the solution would be option E.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{16}=4$$, NOT +4 or -4. In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4. Even roots have only a positive value on the GMAT.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 6525
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

### Show Tags

08 Aug 2017, 04:06
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 a > b^2 ? (1) a>b^4 (2) a> b^(1/2)   [#permalink] 08 Aug 2017, 04:06
Display posts from previous: Sort by