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Concern with this problem

by knight247 » Wed Jun 22, 2011 5:16 am
Set S = {1, 2, ...,100}, i.e., Set S is constituted of the first 100 natural numbers. Set A is
a subset of set S. If an element is picked up from Set S, what is the probability that it is
also an element of Set A?
(1)The numbers in Set A are not divisible by 3 or 4.
(2)The numbers in Set A are divisible by 2.
(A)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
(B)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
(C)BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
(D)EACH statement ALONE is sufficient to answer the question asked.
(E)Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

I have a concern with the above problem. Lets get everyonez opinions and then i'll mention my concern. Btw OA is D
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by kmittal82 » Wed Jun 22, 2011 5:28 am
(1)
You can easily write down all the numbers in Set A based on the given info
A = {1,2,5,7,10,11,13,14....98}

Point is, you can count the total elements in Set A, and you know the total in Set S, so probability can be found out. Sufficient

(2)
Similar to above, A = {2,4,6,8....100} (all even numbers)
Sufficient

Hence (D)
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by knight247 » Wed Jun 22, 2011 5:38 am
kmittal82 the explanation U gave is the same as in the OA. But my concern with that is as follows:
Lets consider statement 1
(1)The numbers in Set A are not divisible by 3 or 4.
after calculating U get that betn 1-100 there are 50 numbers not divisible by 3 or 4. How does one assume that all of these 50 (numbers not divisible by 3or4) are in set A???? Isn't there a possibility that there are some numbers indivisible by 3and4 which are not in set A??

and the same is my concern with statement 2

Hope u can clarify that
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by Frankenstein » Wed Jun 22, 2011 5:42 am
knight247 wrote:kmittal82 the explanation U gave is the same as in the OA. But my concern with that is as follows:
Lets consider statement 1
(1)The numbers in Set A are not divisible by 3 or 4.
after calculating U get that betn 1-100 there are 50 numbers not divisible by 3 or 4. How does one assume that all of these 50 (numbers not divisible by 3or4) are in set A???? Isn't there a possibility that there are some numbers indivisible by 3and4 which are not in set A??

and the same is my concern with statement 2

Hope u can clarify that
Hi,
If the OA is D, then it is bad source or bad framing of question. Simple
Cheers!

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by knight247 » Wed Jun 22, 2011 5:49 am
Frankenstein can u elaborate a lil more...What would be ur answer to this question?
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by Frankenstein » Wed Jun 22, 2011 5:51 am
knight247 wrote:Frankenstein can u elaborate a lil more...What would be ur answer to this question?
Hi,
It is definitely E. A is not fixed. It can be any of the possible sets.

Another question that is framed on similar lines is:

https://www.beatthegmat.com/refined-s-t84783.html
Cheers!

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by knight247 » Wed Jun 22, 2011 6:13 am
Can we get some expert replies on this??
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by Anurag@Gurome » Wed Jun 22, 2011 8:55 am
knight247 wrote:Can we get some expert replies on this??
In my opinion, the correct answer is E and the explanation is same as yours.
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by Ian Stewart » Wed Jun 22, 2011 1:36 pm
As the question is worded, the answer is definitely E. While we know the elements in the set A are divisible by 2, that does *not* mean that every multiple of 2 is in set A. The set might be {2}, or it might be {2, 10}, or it might be {2, 10, 14}, among many other possibilities. If the question designer wants the answer to be something other than E here, then he or she hasn't worded the question properly. Where is the question from?
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by knight247 » Fri Jun 24, 2011 12:12 am
Hi Ian,
This question is also from the GMAT study material of a coaching institute in Mumbai
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by kmittal82 » Fri Jun 24, 2011 4:43 am
Ah, I see the err of my ways! I just assumed that when set A would contain "all possible" elements from Set S, but ofcourse upon closer inspection, that is not true (unless explicitly stated).

I'm backing (E) now :)
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