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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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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

thanks in advance

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

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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

thanks in advance

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.

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

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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

thanks in advance

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

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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

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

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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.

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.

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

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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?

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

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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]

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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]

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24 Nov 2015, 01:55

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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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10 Dec 2016, 11:27

Median of the set has to be either an integer or an integer when multiplied by 2. Since 0 < y < 7, E doesn't satisfy either condition. Hence, E has to be the answer.
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T is a set of y integers, where 0 < y < 7. If the average of [#permalink]

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23 Jan 2017, 15:19

Bunuel wrote:

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.

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.

Since Set T range from 1 to 6, why the sample of T from part C was T={-1, -1, 5}?

You are asking the same question: Phrase "T is a set of y integers, where 0 < y < 7" does NOT mean that T consist of elements from 1 to 6, it means that the number of elements (the number of terms) in T is from 1 to 6.
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