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If x is an integer, is the median of the 5 numbers shown gre

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If x is an integer, is the median of the 5 numbers shown gre [#permalink] New post 19 Dec 2009, 07:26
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Question Stats:

60% (02:27) correct 40% (01:54) wrong based on 47 sessions
x,3,1,12,8

If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean ) of the 5 numbers ?

(1) x > 6

(2) x is greater than the median of the 5 numbers.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-is-an-integer-is-the-median-of-the-5-numbers-shown-gre-104134.html#p333039
[Reveal] Spoiler: OA

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medianWithX.GIF
medianWithX.GIF [ 10.75 KiB | Viewed 973 times ]


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Last edited by Bunuel on 15 Apr 2014, 07:02, edited 1 time in total.
Renamed the topic, edited the question and the OA.
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Re: Median with X [#permalink] New post 19 Dec 2009, 10:19
msunny wrote:
How to analyze this question ??

Attachment:
medianWithX.GIF


[Reveal] Spoiler:
E


stmnt 1 - minimum value which x can have is 7. so on arranging the numbers we have 1,3, x,8,12 where x can be b/w 3 and 8 if x=7 or x can be b/w 8 and 12 or after 12

if x = 7, avg of these numbers will be 6.2 and median is 7 so median is > avg . suff
if x= 11, avg of these numbers will be 7 and median is 8 so median is > avg. suff
if x= 100, avg of these numbers will be 24.8 and median is 8 so median is < avg. insuff

so stmnt1 is insuff

stmnt 2 - x is greater than median so x> 8 [ on arranging we get 1,3,8,x,12]

if x= 11, avg of these numbers will be 7 and median is 8 so median is > avg. suff
if x= 100, avg of these numbers will be 24.8 and median is 8 so median is < avg. insuff

so stmnt 2 is insuff

taking together we have x>8 but still insuff. hence E
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Re: Median with X [#permalink] New post 19 Dec 2009, 12:04
Good question I got E as well.
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Re: Median with X [#permalink] New post 15 Apr 2014, 06:55
atish wrote:
Good question I got E as well.


Would you mind elaborating more on your procedure?
Thanks
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Re: Median with X [#permalink] New post 15 Apr 2014, 07:03
Expert's post
jlgdr wrote:
atish wrote:
Good question I got E as well.


Would you mind elaborating more on your procedure?
Thanks
Cheers!
J


x,3,1,12,8

If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean ) of the 5 numbers ?


We have a set: {1, 3, 8, 12, x} Question: is median>mean=\frac{x+1+3+8+12}{5}=\frac{x+24}{5}? Note that as we have odd (5) # of terms in the set then the median will be the middle term when arranged in ascending (or descending) order. So, if x\leq{3}: {1, x, 3, 8, 12} then median=3, if 3<x\leq{8}: {1, 3, x, 8, 12} then median=x and if x\geq{8}: {1, 3, 8, x, 12} then median=8.

(1) x>6. If x=7 then the median will be 7 as well: {1, 3, 7, 8, 12} and mean will be mean=\frac{7+24}{5}=6.2, so median=7>mean=6.2 and the answer is YES BUT if x is very large number then the median will be 8: {1, 3, 8, 12, x=very large number} and mean will be more than median (for example if x=26 then mean=\frac{26+24}{5}=10, so median=8<10=mean) and the answer will be NO. Not sufficient.

(2) x is greater than the median of the 5 numbers --> so median=8: now, if x=11 then mean=\frac{11+24}{5}=7, so median=8>7=mean and the answer is YES. Again it's easy to get answer NO with very large x. Not sufficient.

(1)+(2) Again, x=11 and x=very large number give two diffrent answers to the question. Not sufficeint.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-is-an-integer-is-the-median-of-the-5-numbers-shown-gre-104134.html#p333039
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Re: Median with X   [#permalink] 15 Apr 2014, 07:03
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