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

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

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New post 30 Mar 2015, 10:23
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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, 03:08, 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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New post 30 Mar 2015, 21:02
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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
Answer (C)
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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New post 31 Mar 2015, 07:29
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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.
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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New post 31 Mar 2015, 21:40
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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 \(b^2\).
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]

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New post 21 Sep 2015, 05:37
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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New post 16 Dec 2015, 13:04
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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New post 23 Jul 2016, 10:49
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
Answer (C)


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.

Please share your views on this observation.

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

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New post 24 Jul 2016, 04:28
Can anyone please answer to my previous post?

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

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New post 24 Jul 2016, 07:09
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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
Answer (C)


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.

Please share your views on this observation.


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, 03:06
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