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

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Intern
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Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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30 Mar 2015, 11:23
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Difficulty:

95% (hard)

Question Stats:

44% (01:39) correct 56% (01:35) wrong based on 302 sessions

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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.

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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30 Mar 2015, 22:02
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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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16925 [6], given: 230 Director Joined: 07 Aug 2011 Posts: 584 Kudos [?]: 523 [0], given: 75 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 !! Kudos [?]: 523 [0], given: 75 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7615 Kudos [?]: 16925 [0], given: 230 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

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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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.

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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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.

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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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.

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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24 Jul 2016, 05:28

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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24 Jul 2016, 08:09
Expert's post
1
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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.
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2)   [#permalink] 08 Aug 2017, 04:06
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