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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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31 Mar 2015, 07:29

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.
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Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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

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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.

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

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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.

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. _________________

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

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08 Aug 2017, 03:06

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