A:
b4 will always be +ve
so a is positive

b2 will also be positive [Thank God, comples numbers does not exist in gmat land]

a can be greater and less
the problem will come when b is between 0 and 1

Ex:
let b = 1/2
so b^4 will be 1/16
and b^2 will be 1/4

for all numbers (=a) between 1/4 and 1/16
A will be greater than b^4 and less than b^2

b = 1/2 = .5
b^2 = 1/4 = .25
b^4 = 1/16 = .0625

a = 1/8 = .125

so a > b^4 but less than b^2

and for
a = 2 and b =2
a>b^2 and a>b^4

so Insuff



a > sqrt(b)
again
a +ve
b also +ve

again the similar problem

a = 2
b = 1

a > sqrt(b) and a > b^2


Next
a = 3
b = 4

sqrt(b) = 2

a > sqrt(b) and a < b^2
Insuff



Combining both
you can get the ans

so c in IMO

Is question complete..?
Atleast I cannot break it.....

alsergi wrote:Hope you could help me with this:

Thank you all!!!


Is a > b^2 ?

1. a>b^4
c. a >sqrt (b)

The right answer is C. ¿?

@STOP

How you can deduce answer C from both Statements 1 and 2..I was not able to combine both and get final answer.

Thanks in advance

stop@800 wrote:A:
b4 will always be +ve
so a is positive

b2 will also be positive [Thank God, comples numbers does not exist in gmat land]

a can be greater and less
the problem will come when b is between 0 and 1

Ex:
let b = 1/2
so b^4 will be 1/16
and b^2 will be 1/4

for all numbers (=a) between 1/4 and 1/16
A will be greater than b^4 and less than b^2

b = 1/2 = .5
b^2 = 1/4 = .25
b^4 = 1/16 = .0625

a = 1/8 = .125

so a > b^4 but less than b^2

and for
a = 2 and b =2
a>b^2 and a>b^4

so Insuff



a > sqrt(b)
again
a +ve
b also +ve

again the similar problem

a = 2
b = 1

a > sqrt(b) and a > b^2


Next
a = 3
b = 4

sqrt(b) = 2

a > sqrt(b) and a < b^2
Insuff



Combining both
you can get the ans

so c in IMO

In the above scenario ,how can one combine both the question and get the awnser as choice (C).Can anybody put in on this..?

Thanks

alsergi wrote:Hope you could help me with this:

Thank you all!!!


Is a > b^2 ?

1. a>b^4
c. a >sqrt (b)

The right answer is C. ¿?

to make this answer into words

statement one is true only if b is greater than or equal to 1, below one and b^2 is more than b^4

in statement 1

so a>(1/2)^4=A>1/16


a>(1/2)^2=A>1/4

so A in this case could be between 1/4 and 1/16 and would make the original statement false

Statement two is similarbut the oppisite, if B is greater than 1(no negative #'s because it is a sqrt and this is the gmat) there is a similar jump. So lets take an easy number for b, say B=4

In statement 2
a>sqrt(4)= A>2

a>(4)^2= a>16

So if A were between 2-16 in this case, they aren't both true.

And because these are compliments of each other, using both would always assure getting some A>b^2


So answer=C