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Re: Given A and B are non negative, is A5 > B2 [#permalink]
G800 wrote:
my approach:
from the question stem: 1. a and b is positive or zero.
we have to find whether a is a/ b are fractions.
Stem 1: not sufficient to decide. as if a is an positive integer then a^5 > B^2 BUT if is a fraction a^5 is not > B^2.
Stem 2: not sufficient for same reason.
Combining these to we can not decide whether a is a fraction or integer. we only can say a^1/3 and a are greater than B^2.
Any comparison between a^2 and a or any power of a would give a hints to solve the question.

E is the pick.

I dont understand how do we know that A and B are less than 0
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Re: Given A and B are non negative, is A5 > B2 [#permalink]
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rahul18 wrote:
G800 wrote:
my approach:
from the question stem: 1. a and b is positive or zero.
we have to find whether a is a/ b are fractions.
Stem 1: not sufficient to decide. as if a is an positive integer then a^5 > B^2 BUT if is a fraction a^5 is not > B^2.
Stem 2: not sufficient for same reason.
Combining these to we can not decide whether a is a fraction or integer. we only can say a^1/3 and a are greater than B^2.
Any comparison between a^2 and a or any power of a would give a hints to solve the question.

E is the pick.

I dont understand how do we know that A and B are less than 0


A and B are not less than zero. The stem says that "A and B are non negative", so each is 0 or positive.
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Re: Given A and B are non negative, is A5 > B2 [#permalink]
Insufficient even by both statements

Ans E. See attached image for calcs.
Attachments

Acube_Bsquare.JPG
Acube_Bsquare.JPG [ 43.63 KiB | Viewed 15695 times ]

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Re: Given A and B are non negative, is A5 > B2 [#permalink]
GMATBaumgartner wrote:
Given \(A\) and \(B\) are non negative, is \(A^5 > B^2\)?

(1) \(A^\frac{1}{3} > B^2\)

(2) \(A > B^2\)



I have solved it verbally , usingg a number line wiith 0 and 1 as breakpoint arears . once greater than 1 and the other etwenn 0 and 1
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Given A and B are non negative, is A5 > B2 [#permalink]
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GMATBaumgartner wrote:
Given \(A\) and \(B\) are non negative, is \(A^5 > B^2\)?

(1) \(A^\frac{1}{3} > B^2\)

(2) \(A > B^2\)


Is A^5 > B^2 ; A > B^(2/5)

The more you increase the power of a number between 0 to 1 , the greater the value of that number will decrease.

Statement 1 says ,
A ^ 1/3 > B^2 ;
A > B^6

B^(2/5) is a smaller number than B^6 when B >1.

But ,when 0 <B <1 , then B ^ (2/5) > B^6. So when 0 < B <1 , then A is greater than a smaller number ( i.e B ^6 ) . A may or may not be greater than B^ (2/5).
So statement 1 is not sufficient.

Statement 2 says ,
A > B^2
In 0 < B <1 , then B^2 < B^(2/5)
A is greater than a smaller number ( i. e B^2). So A may or may not be greater than B^(2/5).
Statement 2 is not sufficient.

Taken two statements together,
the problem does not get resolved. A is greater than two smaller numbers ( i.e B ^ 6 and B^2) .
A may or may not be greater than the bigger number i.e B ^(2/5).

So the answer is option E.

VeritasKarishma GMATNinja Bunuel Please check my solution.

If you find my solution helpful , please give me KUDOs.
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Given A and B are non negative, is A5 > B2 [#permalink]
Taken together:

A = 1000
1000 > B^2
10 > B^2
1000^5 > B^2 ?

Yes, for all values A > 1.


A = 1/1000
1/1000 > B^2
1/10 > B^2
1/1000^5 > B^2 ?

Yes, if B^2 is less than 1/1000^5
No, If B^2 is between 1/1000 and 1/1000^5.

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Re: Given A and B are non negative, is A5 > B2 [#permalink]
For (1), 0.729 > 0.64, so, 0.9 > 0.64.
Main = (0.729)^5 < 0.64, NO

8 > 1, so, 2 > 1.
Main = (8)^5 > (1)^2, Yes

For (2), 10 > (3)^2.
Main = 10^5 > 3^2, YES

0.65 > 0.64.
Main = (0.65)^5 < 0.64, NO

For (1) + (2), a=125, b=2, satisfies both (1) + (2),
Main = (125)^5 > 4, YES

a=0.729, b = 0.64, satisfies both (1) + (2),
Main = (0.729) < 0.64, NO
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Re: Given A and B are non negative, is A5 > B2 [#permalink]
Neither statement alone is sufficient.

Statement 1 tells us \(A > B^6\).

Statement 2 tells us \(A > B^2\).

Combined, there is still a lot of information we don't know.

Is \(B^2 > B^6\)?
Is \(A^5 > A\)?

We don't know whether A and B are integers or fractions. With the given information, we can't conclude \(A^5 > B^2\).

Answer is E.
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Re: Given A and B are non negative, is A5 > B2 [#permalink]
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Re: Given A and B are non negative, is A5 > B2 [#permalink]
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