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

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

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Updated on: 08 Aug 2017, 04:08
14
00:00

Difficulty:

95% (hard)

Question Stats:

45% (02:12) correct 55% (02:00) wrong based on 308 sessions

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Is $$a > b^2$$ ?

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

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

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]

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30 Mar 2015, 22:02
5
6
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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Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2)  [#permalink]

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31 Mar 2015, 08:29
1
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

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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31 Mar 2015, 22:40
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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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
Math Expert
Joined: 02 Sep 2009
Posts: 54376
Re: Is a > b^2 ? (1) a>b^4 (2) a> b^(1/2)  [#permalink]

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24 Jul 2016, 08:09
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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