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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 non-integer. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5

If you said, median of 2*y/7 would be a non-terminating decimal for 0<y<6 and median of any set will either be a terminating decimal or an integer, then I would have definitely agreed.

Maybe I didn't understand you fully, here. Care to explain? thanks
_________________

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.

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

1

This post received KUDOS

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 non-integer. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5

If you said, median of 2*y/7 would be a non-terminating decimal for 0<y<6 and median of any set will either be a terminating decimal or an integer, then I would have definitely agreed.

Maybe I didn't understand you fully, here. Care to explain? thanks

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

Show Tags

19 Mar 2011, 10:43

1

This post received KUDOS

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 non-integer. 1,1,2,3,4,6. y=6, Mean=3, Median=2.5

If you said, median of 2*y/7 would be a non-terminating decimal for 0<y<6 and median of any set will either be a terminating decimal or an integer, then I would have definitely agreed.

Maybe I didn't understand you fully, here. Care to explain? thanks

Thanks 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.
_________________

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, 11:42

1

This post received KUDOS

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 1-6, 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.
_________________

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 1-6, 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.

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

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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

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.

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