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T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 Mar 2011, 23:15
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: tisasetofyintegerswhere0y7iftheaverageof77475.html
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Last edited by Bunuel on 14 Mar 2016, 06:54, edited 1 time in total.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 Mar 2011, 23:26
Think it's E. 2*6/7 is not integer
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 Mar 2011, 23:45
2,1,0,3,10  A is out 1,1,3  B is out 3,3,3  C is out 1,1,1  D is out D  2y/7 is the only choice, as y < 7, so y can't be divided by 7
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 Mar 2011, 23:49
Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 Mar 2011, 23:59
To explain further, the answer choices are : A.0 B.x C.x D.y/3 E.2y/7 2,1,0,3,10  A is out Because mean = 10/5 = 2 and media n = 0 (y = 5) 1,1,3  B is out Here mean = 1 and median = 1 (y = 3, x = 1) 3,3,12  C is out Here mean = 2 and median = 3 (y = 3, x = 3, sorry, the extra 3 was a typo) 1,1,1  D is out Here mean = 1 and median = 1 = y/3 (x = 3) D  2y/7 is the only choice, as y < 7, so y can't be divided by 7
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 01:07
gmat1220 wrote: Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E
Posted from my mobile device I know that 2*y/7 is the correct answer. However, I don't understand your logic. Median of 5 integers is not an integer. What about 2 integers, 4 integers and 6 integers. Median can be a noninteger. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5 If you said, median of 2*y/7 would be a nonterminating decimal for 0 Maybe I didn't understand you fully, here. Care to explain? thanks
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 01:18
Hey fluke Consider y=1,3,5 the median of odd number of integers is integer. So you have three cases which can be compared with 2 y/7. Since 2y/7 is never integer for odd values of y I am sure this is the answer. "cannot"be true does not mean "never" be true so you still have cases like y=2,4,6 where the median is non integer.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 01:39
Never be true is the subset of could not be true. Hope that helps.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 02:11
subhashghosh wrote: To explain further, the answer choices are :
A.0 B.x C.x D.y/3 E.2y/7
2,1,0,3,10  A is out
Because mean = 10/5 = 2 and media n = 0 (y = 5)
1,1,3  B is out
Here mean = 1 and median = 1 (y = 3, x = 1)
3,3,12  C is out
Here mean = 2 and median = 3 (y = 3, x = 3, sorry, the extra 3 was a typo)
1,1,1  D is out
Here mean = 1 and median = 1 = y/3 (x = 3)
D  2y/7 is the only choice, as y < 7, so y can't be divided by 7 x=3 or 1? and y is not divided by but how is 2y is not divided by 7.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 07:23
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fluke, This what I found from my notes  Which of the following WOTF (could not be true) has to be done necessarily by elimination. § WOTF cannot be true? Plug in numbers until we find four answers that could be true. Eliminate 4 answers and odd one is the final answer.
You can easily eliminate A,B,C because they are integers. Now lets consider D vs E. Evaluate E  Consider y = 3. The median of odd number of integers MUST be integer and expression 2y/7 is not integer for y=3 Evaluate D  Consider y = 3. y/3 is integer. Hence I have to eliminate D. So in the process I eliminated 4 answer choices A through D to get E. Pls verify the reasoning and let me know if I have missed anything. fluke wrote: gmat1220 wrote: Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E
Posted from my mobile device I know that 2*y/7 is the correct answer. However, I don't understand your logic. Median of 5 integers is not an integer. What about 2 integers, 4 integers and 6 integers. Median can be a noninteger. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5 If you said, median of 2*y/7 would be a nonterminating decimal for 0 Maybe I didn't understand you fully, here. Care to explain? thanks



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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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19 Mar 2011, 10:43
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gmat1220 wrote: fluke, This what I found from my notes  Which of the following WOTF (could not be true) has to be done necessarily by elimination. § WOTF cannot be true? Plug in numbers until we find four answers that could be true. Eliminate 4 answers and odd one is the final answer.
You can easily eliminate A,B,C because they are integers. Now lets consider D vs E. Evaluate E  Consider y = 3. The median of odd number of integers MUST be integer and expression 2y/7 is not integer for y=3 Evaluate D  Consider y = 3. y/3 is integer. Hence I have to eliminate D. So in the process I eliminated 4 answer choices A through D to get E. Pls verify the reasoning and let me know if I have missed anything. fluke wrote: gmat1220 wrote: Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E
Posted from my mobile device I know that 2*y/7 is the correct answer. However, I don't understand your logic. Median of 5 integers is not an integer. What about 2 integers, 4 integers and 6 integers. Median can be a noninteger. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5 If you said, median of 2*y/7 would be a nonterminating decimal for 0 Maybe I didn't understand you fully, here. Care to explain? thanksThanks gmat1220. Yes, I prefer the elimination myself for these type of questions. You have properly eliminated 4 answer choices. Thus, the fifth one got to be the answer.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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it can also be proved very easy. 2y/7 will always be not an integer (y<7) If y is not even  for sure no integer can be = 2y/7 (when y<7) if y is even  it means the median is the sum of the two middle terms so (A+B)/2=2y/7 therefore A+B = 4y/7. if y is an integer between 16, again 4y/7 is always not an integer. and we remember that A and B are integers. the sum of two integers is of course integer as well. Means  A+B can never be equal to 4y/7. Proved.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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20 Mar 2011, 17:30
Nice approach there buddy. 144144 wrote: it can also be proved very easy.
2y/7 will always be not an integer (y<7) If y is not even  for sure no integer can be = 2y/7 (when y<7) if y is even  it means the median is the sum of the two middle terms so (A+B)/2=2y/7 therefore A+B = 4y/7. if y is an integer between 16, again 4y/7 is always not an integer. and we remember that A and B are integers. the sum of two integers is of course integer as well. Means  A+B can never be equal to 4y/7. Proved.



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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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18 May 2011, 23:16
y<7 hence 2y/7 not an integer. Median has been mentioned as an integer. Hence E.
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Re: T is a set of y integers, where 0 < y < 7. If the average of Set T is [#permalink]
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AnkitK 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 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. OPEN DISCUSSION OF THIS QUESTION IS HERE: tisasetofyintegerswhere0y7iftheaverageof77475.html
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