Tough Kaplan

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Tough Kaplan

by srisl11 » Sun Dec 07, 2008 10:02 am
Is x odd?

1) 2x-1 is odd
2) x^3 is odd


OA after some discussions please ...
Source :Kaplan

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by cramya » Sun Dec 07, 2008 10:11 am
I am getting B)

Whats the OA?

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by srisl11 » Sun Dec 07, 2008 10:13 am
OA is C

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by niraj_a » Sun Dec 07, 2008 10:14 am
damn...i got B too

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by cramya » Sun Dec 07, 2008 10:17 am
Interesting....


If x is even x^3 can never be odd. x has to be odd. I still feel OA should be B) and not C). If it is for sure C) then I am missing something here!

2X-1 does not help us in any way to determine if x is odd or even. x can be odd or even.

So not sure how combining this with x^3 odd does it...

Sris,
Have they given an official explanation on how its C)? I am curious to know. :roll:

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by srisl11 » Sun Dec 07, 2008 10:27 am
I took B first...
There is explanation in Kaplan (otherwise I would have taken it for granted that there is a "print mistake" and the OA is B. :) )
Its a lengthy explanation for statement 1 ..I will type and post it soon

For B the explanation is
B will fail if x = cubeth root of 7
because in this case x^3 will be odd but x is not odd
Last edited by srisl11 on Sun Dec 07, 2008 10:32 am, edited 1 time in total.

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by cramya » Sun Dec 07, 2008 10:30 am
Ok thanks!

Looking forward to stmt I and II combined sufficiency explanation!

I think we missed it since we considered only integers when no info was given about x. This is where these problems get us...
Last edited by cramya on Sun Dec 07, 2008 10:35 am, edited 1 time in total.

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by Zipper » Sun Dec 07, 2008 10:34 am
Looking at this... what I can think of the top of my head is if x=0 then statement 2 will fail since 0=even.

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by cramya » Sun Dec 07, 2008 10:36 am
Looking at this... what I can think of the top
of my head is if x=0 then statement 2 will fail since 0=even.
Its given x^3 is odd is true so x cant be 0

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by Zipper » Sun Dec 07, 2008 10:39 am
Yes you are right, i rushed with my statement.

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Re: Tough Kaplan

by Stuart@KaplanGMAT » Sun Dec 07, 2008 10:41 am
srisl11 wrote:Is x odd?

1) 2x-1 is odd
2) x^3 is odd


OA after some discussions please ...
Source :Kaplan
Never make assumptions!

To get (B), one has to assume that x is an integer. However, we're never told that's the case.

Is x odd?

(1) 2x - 1 = odd
2x = even

For 2x to be even, x could be an even or odd integer: insufficient.

(2) x^3 = odd

For x^3 to be odd, x could be an odd integer or the cube root of an odd integer: insufficient.

Together: from (1), we know that x is an integer; from (2), we know that if x is an integer, it must be odd. Therefore, we now know that x must be an odd integer: sufficient.

Together sufficient, apart insufficient: choose (C).
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by srisl11 » Sun Dec 07, 2008 10:44 am
Explanation given in Kaplan:

Statement 1 tells you that 2x-1 is odd . That is , 2x must be even since any number that is one more than an odd number must be even. Because 2x is even ,2x is 2 multiplied by an integer N . So 2x=2N. and x=N . THis means that N must be an integer. However , since any integer multiplied by 2 is even , you cannot say whether x is odd or even . So since x can be odd or even this statement is by itself insufficient and you can discard answer choices (1) and (4)

Statement 2 tells you that x^3 is an integer , then since the product of any set of integers is odd if each of those integers id odd , x must be odd. However , x could also be an irrational number like cubeth root 7 .
In this case x^3 is odd but x is not odd ,it's not an integer at all.
So Statement 2 is insuff

Now combine St 1 and St 2 .
From st 1 we know that x is an integer, while from st 2 we know that x is odd.
The product of any set of integers is odd only if each of those integers is odd. So x must be odd and the statements taken together are sufficient.
:shock:

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by Zipper » Sun Dec 07, 2008 10:45 am
Reading the other answers now... as Kaplan explains if x = cubeth root of 7

x^3=odd will fail (it will be odd but x won't be odd, yes it won't be even either but it's not odd)

and since we have no restrictions for x i guess this is really possible.

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Re: Tough Kaplan

by srisl11 » Sun Dec 07, 2008 10:46 am
Stuart Kovinsky wrote:
srisl11 wrote:Is x odd?

1) 2x-1 is odd
2) x^3 is odd


OA after some discussions please ...
Source :Kaplan
Never make assumptions!

To get (B), one has to assume that x is an integer. However, we're never told that's the case.

Is x odd?

(1) 2x - 1 = odd
2x = even

For 2x to be even, x could be an even or odd integer: insufficient.

(2) x^3 = odd

For x^3 to be odd, x could be an odd integer or the cube root of an odd integer: insufficient.

Together: from (1), we know that x is an integer; from (2), we know that if x is an integer, it must be odd. Therefore, we now know that x must be an odd integer: sufficient.

Together sufficient, apart insufficient: choose (C).
THanks Stuart , Your explanation is very clear and easy to understand :)

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by cramya » Sun Dec 07, 2008 10:46 am
Nice problem!

Takeaway for all of us : Like Stuart mentioned dont assume when no info about variables in question are given.