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

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Awli
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Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink] New post   Updated on: 08 Aug 2017, 04:08
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Question Stats:

44% (02:12) correct 56% (02:04) wrong based on 264 sessions
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Is \(a > b^2\) ?


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

(2) \(a> \sqrt{b}\)
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Official Answer
C

Originally posted by Awli on 30 Mar 2015, 11:23.
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] New post   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
Answer (C)
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink] New post   31 Mar 2015, 08:29
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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] New post   31 Mar 2015, 22:40
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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] New post   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] New post   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] New post   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
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] New post   24 Jul 2016, 05:28
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] New post   24 Jul 2016, 08:09
Expert Reply
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] New post   28 Apr 2020, 11:10
Awli wrote:
Is \(a > b^2\) ?


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

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



Hi Awli

A good Data Sufficiency question I say.

Question stem:
Is \(a > b^2\) ?

No other info is given for us to work with.

Statement 1:
\(a>b^4\)

Let a = 4, b= 1
Then, answer would be yes \(a > b^2\).

In second case, let a = \(\frac{1}{100}\), b= \(\frac{1}{10}\)
Then, although \(a>b^4\), \(a = b^2\); thus, the answer is no.

We can safely conclude that statement 1 is not enough to deduce the answer.

Statement 2:
\(a> \sqrt{b}\)

Let a = 10, b = 9
Then, answer would be no \(a < b^2\).

However, if a =\( \frac{1}{100}\)
\( and b = \frac{1}{100}\)

Then, yes \(a > b^2\).

Therefore, statement 2 is insufficient alone.

Applying both the statements together we get:

\(a>b^4\)
\(a> \sqrt{b}\)

It is clearly seen that in this scenario, \(a > b^2\)

I hope I made it elaborate enough. Is there anything anyone would like me to explain more??
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink] New post   23 Oct 2023, 16:04
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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2) [#permalink]
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