I am getting B)
Whats the OA?
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Tough Kaplan
Source: Beat The GMAT — Data Sufficiency |
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:
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:
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
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.
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 0Looking 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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Never make assumptions!srisl11 wrote:Is x odd?
1) 2x-1 is odd
2) x^3 is odd
OA after some discussions please ...
Source :Kaplan
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).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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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.
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.
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.
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.
THanks Stuart , Your explanation is very clear and easy to understandStuart Kovinsky wrote:Never make assumptions!srisl11 wrote:Is x odd?
1) 2x-1 is odd
2) x^3 is odd
OA after some discussions please ...
Source :Kaplan
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).
Nice problem!
Takeaway for all of us : Like Stuart mentioned dont assume when no info about variables in question are given.
Takeaway for all of us : Like Stuart mentioned dont assume when no info about variables in question are given.
