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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
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Originally posted by Zaur on 04 Apr 2009, 03:50.
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
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25 Sep 2010, 09:28
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
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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
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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.



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Re: T is a set of y integers, where 0 < y < 7. If the average of
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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
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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
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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 integerb) x could be an integer, all we know is that it's the averagec) –x same as bd) (1/3)y could be an integer if y=3 or y=6e) (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
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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.



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



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Re: T is a set of y integers, where 0 < y < 7. If the average of
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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
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Re: T is a set of y integers, where 0 < y < 7. If the average of
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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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Re: T is a set of y integers, where 0 < y < 7. If the average of
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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. 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
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24 Jan 2017, 00:16
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
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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
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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
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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: GMAT assassins aren't born, they're made, Rich
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Re: T is a set of y integers, where 0 < y < 7. If the average of
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21 Dec 2018, 21:15
The set contains 16 elements. A. Consider a set {0, 0 ,6}. Here, mean = 2 and median = 0. B. Consider a set { x }. Here, mean = x and median = x C. Consider a set {1, 1, 5}. Here mean(x) = 1 and median = 1= x D.Consider a set {1, 1, 1}. Here mean = 1, y = 3 and median = 1 = y/3. E. T is a set of y integers where 0 < y < 7. All the elements in the set should be integers. So, the median also needs to be an integer. Now, if 2y/7 become the median it also needs to be an integer which is not possible for any possible value of y i.e. 1,2,3,4,5 or 6. Hence, 2y/7 can’t be the median.



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Re: T is a set of y integers, where 0 < y < 7. If the average of
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21 Dec 2018, 21:35
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 Average of two integers will always be either an integer (if both are odd or both are even) or Integer/2 in lowest form (if one is odd and the other is even). The average can never be integer/7 in lowest terms. Since y is between 0 and 7 exclusive, y is not divisible by 7 and hence, (2/7)y will be (integer/7) in lowest terms. Median of a set of odd number of integers will be one of the integers. Median of a set of even number of integers will be the average of middle two integers. Hence, median cannot be of the form (2/7)y. Answer (E)
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