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. ¿?
Inequities
This topic has expert replies
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
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
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Is question complete..?
Atleast I cannot break it.....
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. ¿?
-
- Senior | Next Rank: 100 Posts
- Posts: 31
- Joined: Mon Sep 15, 2008 6:16 pm
- Thanked: 3 times
@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
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
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- Junior | Next Rank: 30 Posts
- Posts: 16
- Joined: Sun Sep 21, 2008 9:12 am
- Thanked: 1 times
to make this answer into wordsalsergi 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. ¿?
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