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Is x > y? 1) (a^n)*x > (a^n)*y 2) 3n + 1 is odd open

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Is x > y? 1) (a^n)*x > (a^n)*y 2) 3n + 1 is odd open [#permalink] New post 29 May 2003, 04:52
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Is x > y?

1) (a^n)*x > (a^n)*y
2) 3n + 1 is odd

open to debate......i hope i don't screw up on this one like the last one :oops: :oops: :oops:
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 [#permalink] New post 29 May 2003, 12:22
am i missing something here?? seems suspiciously easy...A?
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 [#permalink] New post 29 May 2003, 18:44
Ans : C

1) Insuff, since a and n can be -ve or +ve

2) Insuff, n is even - nothing about x and y

Combined : n is even, and if a is -ve or +ve a^n can be divided out.
=> x>y
Suff
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 [#permalink] New post 29 May 2003, 20:40
In my view, I vote for C
supposed that X and Y are positive 5 and 2 repectively.

Clearly, the statement (I) doesn't tell the values of a^n. If a^n is negative value, X may be less than Y. On the contrary, if a^n is positive, X will be more than Y. Thus, Insufficient.


The statment II only show the value of n which can be negative or positve,the value of a and the value of x and y. Therefore. Insufficient

Combined them together, Thus, the answer is C because a^n always be positive and therefore, X always be more than Y
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 [#permalink] New post 30 May 2003, 01:41
I also had C in mind

to prove that x > y, you have to prove that the common number that is multiplied on both sides is POSITIVE.

From 1, a^n can be negative if n is odd and "a" is negative..

however if n is even than a^n will AlWAYS be positive....

therefore both together are sufficient
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 [#permalink] New post 30 May 2003, 01:50
Hi bhavesh,
I went on the same lines of reasoning as you gave. I put in C. But, another thing crossed my mind.

The question is x > y ?? So it can be YES or NO, correct

So, even if you say if a is negative and n is odd, you are multiplying the same number on both sides. So they get cancelled anyhow and even if the greater than sign reverses to say x < y , you still have an answer.

Am i right??? :roll:
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 [#permalink] New post 30 May 2003, 01:54
if "a" is positive, than x>y...the answer is YES

however if "a" is negative and "n" is odd, than a^n will be negative and in this case x<y.......answer is NO

therefore you cannot say for sure YES or NO.....

depends on sign of "a" and value of "n"
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Re: Hmmmmm... [#permalink] New post 06 Jun 2003, 19:25
Expert's post
neema wrote:
Is it C?


you are still a student and you are getting into MBA?

you feel like you are ready?

or you are a different kind of student ?
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OLD ONE [#permalink] New post 11 Feb 2004, 10:58
This one is a very old one.

Quote:
Is x > y?

1) (a^n)*x > (a^n)*y
2) 3n + 1 is odd

open to debate......i hope i don't screw up on this one like the last one


The agreed upon answer for this question by all the members who responded to this question in the past is C.

I have following questios.

(1) Should the second statement be given as 3n + 1 is positive odd integer ?

If the word positive is not included then we can take

3n + 1 = -1 => n -2/3 = 0.66666..... (Is this odd or even? )


If the word integer is not included,

Can we consider 3n + 1 = 5.5 ? (This is odd number, correct?)
=> n= 4.5/3 => 1.5 (This should be odd)

I think if these conditions are not included, then the answer should be E.

Finally, please let me know (-1)^1.2 = ?

I thought that 1.2 is considered even number so answer should be 1. But the calculator says invalid input function.

(1.2 here is decimal number = 12/10)[/b]
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 [#permalink] New post 11 Feb 2004, 11:05
Quote:
Finally, please let me know (-1)^1.2 = ?

I thought that 1.2 is considered even number so answer should be 1. But the calculator says invalid input function.

(1.2 here is decimal number = 12/10)



Please never mind about this question in my last post. I got it. It will be imaginary number.

But my other questions remain.
  [#permalink] 11 Feb 2004, 11:05
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