C Agree with the above poster.
St1] None of the six numbers is greater than 50.
We can have contradictory possibilities here.
The max value any of the 6 nos. is 50. We know that the sum of all 6 = 150
If 1 is 50, the avg. of the remaining 5 is 20. So we could have an AP of 5 nos. with 20 as the median.
In this case if the common difference is 1, then all these 5 nos. have to be less than 25.
Hence our answer is only 1 no. of the 6nos is greater than 25.
If the common difference is 3 then there are 2 nos greater than 25.
Insuff.
St2] Three of the six numbers are 7, 8, and 9, respectively.
This means that the sum of the remaining 3 is 126 and the avg is 42.
Here i could have 1 extremely large no. such as 100 & 2 nos. each smaller than 25 or one as 26 & the other as 1.
Insuff.
Together, We know that the max value of the sum of 2 of the other 3 nos. is 99, hence the 3rd has to be greater than 25.
So we must have 3 nos. greater than 25.
