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T is a set of y integers, where 0 < y < 7. If the average of

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T is a set of y integers, where 0 < y < 7. If the average of  [#permalink]

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New post 28 Mar 2019, 03:36
Bunuel wrote:
as \(y\) is an integer from 1 to 6 then \(\frac{2}{7}y\) is neither an \(integer\) nor \(\frac{integer}{2}\). So \(\frac{2}{7}y\) could not be the median.

Can you please explain the above. "y" is just the number of elements in the set..right? So, how can we conclude the above, just based upon the number of elements?
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T is a set of y integers, where 0 < y < 7. If the average of  [#permalink]

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New post 21 Apr 2019, 22:03
Learnings from this question:
[Assumption: all entities of the dataset are integers]
(1) in case of ONE ENTITY or EVERY ENTITY EQUAL data set, Mean=Median.

(2) Median can an Integer or a fraction in the form of multiple of 0.5.

Median is integer if the dataset has
(a) odd nuodd number of entities
(b) even number of entities with the condition that middle twos are either both odd or both even.

Median is integer if the dataset has
even number of entities with the condition that middle twos are odd and even.


(3) Irrespective of the mean (positive, negative or zero), Median can be anything (positive, negative or zero).
Because mean be affected by extreme values, but Median isn’t affected by extremes. Median is based on POSITION middle value in ascending or descending order).

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Re: T is a set of y integers, where 0 < y < 7. If the average of  [#permalink]

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New post 20 May 2019, 21:19
Zaur wrote:
T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?

A. 0
B. x
C. –x
D. (1/3)y
E. (2/7)y


My method was to find a median in the answer options.

A. -1 0 7
B. 1 1 1
C. -3 -2 11
D. 1 1 1
E. now I couldn't find an integer y so that (2/7)y is an integer and avg. is positive integer x, so E.
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Re: T is a set of y integers, where 0 < y < 7. If the average of   [#permalink] 20 May 2019, 21:19

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