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Mean, median questions. Median > Mean ?? Etc..

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Mean, median questions. Median > Mean ?? Etc.. [#permalink]

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New post 09 May 2009, 06:10
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K, Fellas need some help here.

DS questions centered on whether median > mean or vice versa are killing me. It looks like there is a concept I am missing or my strategy is wrong. Can some one explain easy ways to solve these DS problems?

Set S is composed of consecutive multiples of 3. Set T is composed of consecutive multiples of 6. Each set contains more than one element. Is the median of set S bigger than the median of set T?

1. The least element in either set is 6
2. Set T contains twice as many elements as set S

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

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Re: Mean, median questions. Median > Mean ?? Etc.. [#permalink]

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New post 09 May 2009, 09:52
Here is my step by step thoughts:

Reading question....

Set S is composed of consecutive multiples of 3.
Great! it is like {3,6,9,12...}. No restriction on the number of elements and we can shift it by 3 as far as we want. In other words, S could be {111, 114, 117,...}.

Set T is composed of consecutive multiples of 6.
The same as about set S. But in set T the distance between the closest elements is twice as large as that in set S.

Each set contains more than one element.
No essential information. Perhaps, we should return to the point latter.

Is the median of set S bigger than the median of set T?
Median is a middle number or average between two for S and T with even number of elements. By the way, S and T are evenly distributed sets for which median = mean.

After reading....
Each of sets is not fixed in the question. In other words, we can have {3, 6} S set and {111111111111, 111111111114}. And median vary in infinite range. So, our conditions should at least fix sets at some point.

Conditions....
1. The least element in either set is 6
Yeah, it fixes the beginning of both sets. But the ranges of sets could vary from 3 (for S) and 6 (for T) to infinity. So, insufficient.

2. Set T contains twice as many elements as set S
It does not fix positions of sets. So, insufficient.

1&2. So, our sets begin at 6. For example, S={6} and T={6,12}. If when S={6,9}, T={6,12,18,24}. We can see that median of T is increasing faster than median of S and it is larger. So, C is sufficient.

It is exactly the way I thought.
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Kudos [?]: 4716 [0], given: 360

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Re: Mean, median questions. Median > Mean ?? Etc.. [#permalink]

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New post 09 May 2009, 10:22
I guess I was not using some thing very simple and that you obviously pointed out

"If sets are not fixed, we cannot determine one way or the other about their Mean and Median"
Correct?

In your example, to the end, you chose one element set. The Q says each set is more than one element. Not that it changes the answer, just wanted to point that.

Kudos [?]: 451 [0], given: 1

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Re: Mean, median questions. Median > Mean ?? Etc.. [#permalink]

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New post 09 May 2009, 12:36
icandy wrote:
I guess I was not using some thing very simple and that you obviously pointed out

"If sets are not fixed, we cannot determine one way or the other about their Mean and Median"
Correct?

In your example, to the end, you chose one element set. The Q says each set is more than one element. Not that it changes the answer, just wanted to point that.

Yeah, I forgot to check this "more than one" statement as it works for all non-empty sets.
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Re: Mean, median questions. Median > Mean ?? Etc..   [#permalink] 09 May 2009, 12:36
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