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T is the 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 the set T? A.0 B.x C.-x D.y/3 E.2y/7 _________________

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 _________________

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.

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

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. _________________

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. _________________

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.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...