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

Question

Is  a > b^2  ?

(1)  a>b^4

(2)  a> \sqrt{b}

KarishmaB · Answer

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)

Lucky2783 · Answer

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.

KarishmaB · Answer

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 .

mkarthik1 · Answer

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.

AdlaT · Answer

SOLUTION: I LIKE Bunuel APPROACH FOR THIS KIND OF PROBLEM. Statement 1) a>b^4. for 1 \sqrt{b} a and b are non negative. for 1

EBITDA · Answer

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.

EBITDA · Answer

Can anyone please answer to my previous post?

Bunuel · Answer

When the GMAT provides the square root sign for an even root, such as  \sqrt{x}  or  \sqrt {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.

rsmalan · Answer

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??

bumpbot · Answer

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

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