Given A and B are non negative, is A5 B2

Question

Given  A  and  B  are non negative, is  A^5 > B^2 ?

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

(2)  A > B^2

subhashghosh · Answer

(1) A = 1 and B = 1/2

A^5 > B^2

A = 1/2 and B = 1/2

A^5 < B^2

(1) is insufficient

(B) Again let A = 1/3 B = 1/3

A > B^2

But A^5 < B^2

A = 1 B = 1/2

A > B^2

but A^5 > B^2

(A) and (B) together are insufficient

Answer - E

G800 · Answer

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.

rahul18 · Answer

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

Bunuel · Answer

A and B are not less than zero. The stem says that "A and B are non negative", so each is 0 or positive.

srcc25anu · Answer

Insufficient even by both statements

Ans E. See attached image for calcs.

mohagar · Answer

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

sayan640 · Answer

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^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.

Bambi2021 · Answer

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.

Posted from my mobile device

siddhant12165 · Answer

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

Basshead · Answer

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.

bumpbot · Answer

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Given A and B are non negative, is A5 > B2

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