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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 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 alexsr on 05 Dec 2017, 13:40, edited 4 times in total.
Edited the question and added the OA
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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 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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New post 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.
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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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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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New post 30 Apr 2010, 11:40
Can someone break down all the choices as to why they can be median?
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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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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: T is a set of y integers, where 0 < y < 7. If the average of [#permalink]

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

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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.

Answer: E.

Hope it's clear.
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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 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]

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

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

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New post 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.

Answer: E.

Hope it's clear.


Dear Bunuel,

Since Set T range from 1 to 6, why the sample of T from part C was T={-1, -1, 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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ziyuenlau wrote:
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.

Answer: E.

Hope it's clear.


Dear Bunuel,

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

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New post 24 Jan 2017, 08:05
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 {-1, 0, 4} mean(x)=1; median=0. Possible.
B. x {1, 1, 1} mean(x)=1; median=1. Possible.
C. –x {0, -1, 4} mean(x)=1; median=-1. Possible.
D. (1/3)y Let y be 3. Therefore, median=1 {1, 1, 1} mean(x)=1. Possible
E. (2/7)y The median can either be an integer or multiple of 1/2. Not possible.
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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 08 May 2017, 01:53
Median can be
1. An Integer (if the median is one of the numbers in the set T)
2. A decimal of the form m.5 – where m is any number - (if the median is not one of the number in the set T and hence it is (3rd Term + 4th Term)/2)

So from the above option :

A. 0 possible as obvious
B. x possible as obvious
C. –x possible as obvious
D. (1/3)y possible with y = 3 or 6
E. (2/7)y Not possible as (1/7)y is not an integer since Y cannot be 7 and also 1/7 is NOT equal to a decimal of the form m.5.

So ans is E
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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 16 Jan 2018, 12:11
Hi All,

For this question, you can TEST VALUES to eliminate the 4 answers that COULD be the median of Set T.

We're told that Set T consists of Y integers (where 0 < Y < 7) and that the AVERAGE = X = a POSITIVE INTEGER. We're asked which of the answers could NOT be the median.

Answer A: 0... If Set T is {0,0,3} then the average = 1 and the median = 0. Eliminate A.

Answer B: X... If Set T is {1,1,1} then the average = 1 and the median = 1 = X. Eliminate B.

Answer C: -X... If Set T is {-1,-1,5} then the average = 1 and the median = -1 = -X. Eliminate C.

Answer D: Y/3... If Set T is {1,1,1} then the average = 1 and the median = 1 = Y/3. Eliminate D.

There's only one answer left...

Final Answer:
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E


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Re: T is a set of y integers, where 0 < y < 7. If the average of   [#permalink] 16 Jan 2018, 12:11
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