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

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Manager
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If x is an integer, is the median of the 5 numbers shown gre [#permalink]

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19 Dec 2009, 08:26
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55% (hard)

Question Stats:

62% (02:45) correct 38% (01:59) wrong based on 73 sessions

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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 [ 10.75 KiB | Viewed 1576 times ]

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Last edited by Bunuel on 15 Apr 2014, 08:02, edited 1 time in total.
Renamed the topic, edited the question and the OA.
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19 Dec 2009, 11:19
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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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19 Dec 2009, 13:04
Good question I got E as well.
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15 Apr 2014, 07:55
atish wrote:
Good question I got E as well.

Would you mind elaborating more on your procedure?
Thanks
Cheers!
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15 Apr 2014, 08:03
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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.

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, 08:03
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