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median

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median [#permalink] New post 18 Mar 2011, 22:15
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A
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27% (00:00) correct 73% (01:24) wrong based on 12 sessions
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
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Re: median [#permalink] New post 18 Mar 2011, 22:26
Think it's E. 2*6/7 is not integer

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Re: median [#permalink] New post 18 Mar 2011, 22:45
-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: median [#permalink] New post 18 Mar 2011, 22:49
Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E

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Re: median [#permalink] New post 18 Mar 2011, 22:59
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: median [#permalink] New post 19 Mar 2011, 00:07
gmat1220 wrote:
Median of 5 integers is integer and 2*5/7 is not integer. Hence it is E

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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
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Re: median [#permalink] New post 19 Mar 2011, 00:18
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: median [#permalink] New post 19 Mar 2011, 00:39
Never be true is the subset of could not be true. Hope that helps.

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Re: median [#permalink] New post 19 Mar 2011, 01:11
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: median [#permalink] New post 19 Mar 2011, 06:23
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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

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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
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Re: median [#permalink] New post 19 Mar 2011, 09:43
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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 Image


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.
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Re: median [#permalink] New post 20 Mar 2011, 10:42
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.
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Re: median [#permalink] New post 20 Mar 2011, 16:30
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.
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Re: median [#permalink] New post 20 Mar 2011, 21:34
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Re: median [#permalink] New post 18 May 2011, 22:16
y<7 hence 2y/7 not an integer.

Median has been mentioned as an integer.

Hence E.
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Re: median   [#permalink] 18 May 2011, 22:16
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