Hi...

There are 6 numbers, so total is 6*32=192...
Median=?

Let's see the statements..
1) Six numbers are > or =31..
All numbers are 32..... Median is 32
5 numbers are 31 and 6 is 37.... Median is 31..
Different answers possible..
Insufficient

2) 37 is in set ..
Various combinations possible..
The lowest can be ALL four 0, 5th can be 37 and then 6th will be 192-37=155.... Median=0..
Five of them 31 and 6th 37.... Median 31
Insufficient

Combined..
One is 37...
So TOTAL of remaining 5 is 192-37=155...
Only possibility is 155=31*5..
As nine can be lesser than 31 as per statement I..
Median is 31..
Sufficient

C

==> In the original condition, there are 6 variables and 1 equation. In order to match the number of variables to the number of equations, there must be 5 more equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), you get
31, 31, 31, 31, 31, 37, and so the median=31+310/2=31, hence it is unique and sufficient.

The answer is C.
Answer: C
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You didn't specify that the numbers are integers!!!
With current wording I would answer E.

the stem does not specify these numbers are integers and, in your calculation, you only use integer numbers!

Please check!
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