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X,3,1,12,8

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X,3,1,12,8

by amamd » Sat Jul 05, 2008 6:17 am
X,3,1,12,8
If x is an integer, is the median of the 5 numbers shown greater than the average?
1 x>6
2 x is greater than the median of the 5 numbers

The OS is E :?: , Can not figure out. Could some one help? Thanks!
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Re: X,3,1,12,8

by Ian Stewart » Sat Jul 05, 2008 10:35 am
amamd wrote:X,3,1,12,8
If x is an integer, is the median of the 5 numbers shown greater than the average?
1 x>6
2 x is greater than the median of the 5 numbers

The OS is E :?: , Can not figure out. Could some one help? Thanks!
What could the median of the set be? We have the set {1, 3, 8, 12, x}. If x is less than 3, the median will be 3. If x is between 3 and 8, the median will be x. If x is greater than 8, the median will be 8.

What is the average of the set? (24+x)/5.

All right, let's look at the statements:

1) x > 6. This gives two possibilities. Either 6 < x <8, and x is the median, or x >=8, and 8 is the median. Let's only look at the second case, where the median is 8, and find out what must be true for the median to be greater than the average:

8 > (24+x)/5
40 > 24 + x
16 > x

If x <16, then the median will be greater than the average. If x >= 16, the median will not be greater than the average. Insufficient.

2) If x > 8, then x is greater than the median. From the work above, this is insufficient- we really want to know whether x is smaller than 16.

Together, they're still insufficient; we still don't know if x is larger than 16.
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by ildude02 » Sat Jul 05, 2008 4:59 pm
Nice explanation, Ian. I might add, as always :)
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by Thouraya » Tue May 31, 2011 7:42 am
@Ian, if I use the process of plugging numbers; how can I possibly know that once we reach to 16, the statement won't be sufficient anymore? I tried a few numbers, and they all worked, so I was about to assume that A is sufficient.

Also, I did not really understand what they mean by statement 2: Can I assume that it is insufficient, as I don't know what x is and the median depends on the value of x?


Thanks!
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by cans » Tue May 31, 2011 8:30 am
Thouraya wrote:if I use the process of plugging numbers; how can I possibly know that once we reach to 16, the statement won't be sufficient anymore? I tried a few numbers, and they all worked, so I was about to assume that A is sufficient.

Also, I did not really understand what they mean by statement 2: Can I assume that it is insufficient, as I don't know what x is and the median depends on the value of x?
To find if median>mean?
mean = (24+x)/5
Thus to find if median>(24+x)/5
or 5*median>24+x
a)x>6
take x=7, median = 7 and thus 5*7>24+7
Now if we take x>=8, median will always remain 8
and thus to find if 40>24+x
Here while plugging values you must notice that L.H.S is fixed while R.H.S can be as large as possible (no constraint on x other than x>=8)
Thus there will come a point when r.h.s. exceeds l.h.s. and thus not sufficient.
b)x>median => x is not the median and thus median is 8 and x>8
so equation becomes: 40>24+x (again no constraint on x and thus insufficient.)
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by hardikm » Tue May 31, 2011 9:31 am
average = (24+ x)/5
Case I
x = 7 Average =31/5 = 6.2 median = 7 average < median
x = 101 Average= 125/5 = 25 median = 8 average > median

Case II
1,3,8,9,12
x = 9 average = 33/5 = 6.5 median = 8
x = 101 Average= 125/5 = 25 median = 8 average > median

Combining:
still x > 8 so example mentioned for Case II still holds true.
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