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Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
WE: Accounting (Accounting)
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First of all, this is a terrific forum. Good discussions. [#permalink]
14 Sep 2004, 08:50
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
First of all, this is a terrific forum. Good discussions. Thanks.
The mode of a set of integers is x. What is the difference between the median of this set of integers and x ?
(1) The difference between any two integers in the set is less than 3.
(2) The average of the set of integers is x.
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Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
WE: Accounting (Accounting)
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B ??
1) Insufficient
2) when the mode and average are same, i feel the median also the same.
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Director
Joined: 20 Jul 2004
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B.
A challenging problem. Not sure if I could explain clearly.
I. does not give much detail. The numbers can be any where from x-2 to x+2.
II. If the average and mode are the same (x) and when the numbers arrainged ascending, x has to be in the middle (median).
x is the most occuring number (since mode = x) and if this occurs on either side of the middle, the averagewill be tilted to that side.
Hence, median should also be x. And median-mode = x-x = 0
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Manager
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Wow hardworker,
You have helped me a lot today with the medians, average and modes!!
Thanks!!
Regards,
Alex
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Manager
Joined: 31 Aug 2004
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My answer is B.
1) It is not sufficient. Let us assume the following sets
1,2,3,4,5 and 1,3,5,7,9
Both fulfill the requirement that the difference between any 2 integer is less than 3, yet we cannot determine exactly what is difference between the median and mode.
2) Same as hardworker_indian's answer.
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Director
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Alex_NL wrote: Wow hardworker, You have helped me a lot today with the medians, average and modes!! Thanks!! Regards, Alex
My Pleasure!
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Intern
Joined: 02 Sep 2004
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The Answer is C.
Statement 1 tells us that the difference between any two integers in the set is less than 3. This information alone yields a variety of possible sets.
For example, one possible set (in which the difference between any two integers is less than 3) might be:
(x, x, x, x + 1, x + 1, x + 2, x + 2)
Mode = x (as stated in question stem)
Median = x + 1
Difference between median and mode = 1
Alternately, another set (in which the difference between any two integers is less than 3) might look like this:
(x – 1, x, x, x + 1)
Mode = x (as stated in the question stem)
Median = x
Difference between median and mode = 0
We can see that statement (1) is not sufficient to determine the difference between the median and the mode.
Statement (2) tells us that the average of the set of integers is x. This information alone also yields a variety of possible sets.
For example, one possible set (with an average of x) might be:
(x – 10, x, x, x + 1, x + 2, x + 3, x + 4)
Mode = x (as stated in the question stem)
Median = x + 1
Difference between median and mode = 1
Alternately, another set (with an average of x) might look like this:
(x – 90, x, x, x + 15, x + 20, x + 25, x + 30)
Mode = x (as stated in the question stem)
Median = x + 15
Difference between median and mode = 15
We can see that statement (2) is not sufficient to determine the difference between the median and the mode.
Both statements taken together imply that the only possible members of the set are x – 1, x, and x + 1 (from the fact that the difference between any two integers in the set is less than 3) and that every x – 1 will be balanced by an x + 1 (from the fact that the average of the set is x). Thus, x will lie in the middle of any such set and therefore x will be the median of any such set.
If x is the mode and x is also the median, the difference between these two measures will always be 0.
The correct answer is C: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
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Director
Joined: 20 Jul 2004
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Perfect. Yes, it should be C.
I think I made a mistake by overseeing the possibility that the average tilt towards a particular side can be levelled by using a huge number on the other side (like x-10 and x-90 in your examples).
It wouldn't have been clear, if not for your long explanation. Thanks.
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