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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]  04 Apr 2009, 03:50
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Apr 2012, 02:39, edited 1 time in total.
Edited the question and added the OA
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Re: ps [#permalink]  04 Apr 2009, 21:48
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

Its E, which is neither an integer nor a multiple of 0.5. Since the set T's elements are integers, median should be either an integer or a multiple of 0.5.
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Re: ps [#permalink]  04 Apr 2009, 23:13
Hi Gmat tiger,
You are right that the median should be either an integer or a multiple of 0.5 ( since the number of integers can be even)
But then, 1/3 is also not an integer and not a multiple of 0.5 !! Couldnt get this

GMAT TIGER wrote:
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

Its E, which is neither an integer nor a multiple of 0.5. Since the set T's elements are integers, median should be either an integer or a multiple of 0.5.
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Re: ps [#permalink]  05 Apr 2009, 05:39
Economist wrote:
Hi Gmat tiger,
You are right that the median should be either an integer or a multiple of 0.5 ( since the number of integers can be even)
But then, 1/3 is also not an integer and not a multiple of 0.5 !! Couldnt get this

GMAT TIGER wrote:
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

Its E, which is neither an integer nor a multiple of 0.5. Since the set T's elements are integers, median should be either an integer or a multiple of 0.5.

y could be 2, 3, 4, 5, or 6. If y = either 3 or 6, y/3 is an integer.
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Re: PS - set of intergers [#permalink]  24 Oct 2009, 20:19
E too

Y/3 is possible if Y = 3 or 6

2Y/7 does not meet the criteria - the remainder must be 0 or 0.5
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Re: PS - set of intergers [#permalink]  30 Apr 2010, 11:40
Can someone break down all the choices as to why they can be median?
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Re: PS - set of intergers [#permalink]  30 Apr 2010, 22:29
ksharma12 wrote:
Can someone break down all the choices as to why they can be median?

Rather than think about whether or not they can be the median, think about whether or not they could be an integer or a multiple of 0.5.

a) 0 integer
b) x could be an integer, all we know is that it's the average
c) –x same as b
d) (1/3)y could be an integer if y=3 or y=6
e) (2/7)y cannot be an integer or multiple of 0.5, as y is 1,2,3,4,5 or 6. Sufficient enough to answer the question
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Re: PS - set of intergers [#permalink]  25 Sep 2010, 08:47
How C) -x could be the median?
Set consists of elements from 1 to 6, the average x is positive, the median is negative which is impossible due to the range from 1 to 6.
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Re: PS - set of intergers [#permalink]  25 Sep 2010, 09:28
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Kronax wrote:
How C) -x could be the median?
Set consists of elements from 1 to 6, the average x is positive, the median is negative which is impossible due to the range from 1 to 6.

Phrase "T is a set of y integers, where 0 < y < 7" doesn't mean that T consist of elements from 1 to 6, it means that number of elements in T is from 1 to 6.

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. $$0$$ --> if $$T=\{0, 0, 3\}$$ then $$mean=x=1$$ and $$median=0$$;

B. $$x$$ --> if $$T=\{3\}$$ then $$mean=x=3$$ and $$median=x=3$$;

C. $$-x$$ --> if $$T=\{-1, -1, 5\}$$ then $$mean=x=1$$ and $$median=-x=-1$$;

D. $$\frac{1}{3}y$$ --> if $$T=\{1, 1, 1\}$$ then $$mean=x=1$$, $$# \ of \ elements=y=3$$ and $$median=\frac{1}{3}y=1$$;

E. $$\frac{2}{7}y$$ --> now, as T is a set of integers then the median is either a middle term, so $$integer$$ OR the average of two middle terms so $$\frac{integer}{2}$$, but 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.

Hope it's clear.
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Re: PS - set of intergers [#permalink]  25 Sep 2010, 22:23
Thank you Bunuel, now its clear!
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Re: T is a set of y integers, where 0 < y < 7. If the average of [#permalink]  17 Oct 2012, 02:04
It's not always the case that E holds true.
If T contains only one element, let's say 3. The mean is 3. The median OUGHT to be 3, never -3. So C is right.
Can anyone help me out?
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Re: T is a set of y integers, where 0 < y < 7. If the average of [#permalink]  17 Oct 2012, 04:02
Ousmane wrote:
It's not always the case that E holds true.
If T contains only one element, let's say 3. The mean is 3. The median OUGHT to be 3, never -3. So C is right.
Can anyone help me out?

The question is "which of the following could NOT be the median of Set T?"
It means we have to find the option for which the given number can NEVER be the median of Set T.

So, $$-3$$ cannot be the median in your example, but there are many other cases when it can be.
Set T contains integers, so negative numbers are not excluded.

$$2/7y$$ can never be an integer when $$0<y<7$$, while the median MUST be an integer, doesn't matter what is $$y$$ and what are the numbers in the set.
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Re: T is a set of y integers, where 0 < y < 7. If the average of [#permalink]  23 Jul 2013, 00:55
Expert's post
From 100 hardest questions.

Bumping for review and further discussion.
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Re: T is a set of y integers, where 0 < y < 7. If the average of [#permalink]  17 Aug 2014, 10:54
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Re: T is a set of y integers, where 0 < y < 7. If the average of   [#permalink] 17 Aug 2014, 10:54
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