When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19,

Couple of things to observer before we move into the statements.

1. All the numbers in the set are equally spaced, i.e. the numbers are in arithmetic progression
   In a equally spaced set, the mean is same as the median.
   
2. As the resultant set consists of even number of terms, the median will not be a member of the set as it will be an even number (for this set only).

Let's arrange the set in increasing order

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Statement 1

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

As the median is 10, 9 and 11 (indicated by red) are the middle two terms.

Therefore we must have removed one number whose value is less than 9 and one number whose value is greater than 11.

In other words, we must have removed one number from the set of numbers in green and the other from set of numbers indicated in blue.

However, we do not know which numbers were removed. If the first term i.e. 1 and last term i.e. 10 were removed, the members of the resultant set will still be equally spaced and the mean of the resultant set will be equal to the median of the resultant set.

If any other terms were removed, the members of the set will no longer remain equally spaced, and the mean will not be equal to the median.

The statement is not sufficient.

Statement 2

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Given that the average of the set is 10, hence the total of the resultant set is 8 * 10 = 80

Total of current set = 100

Hence the sum of the two terms that were removed = 100 - 80 = 20

Therefore, the terms that were removed were 1 and 19.

3, 5, 7, 9, 11, 13, 15, 17

The resultant members of the resultant set are equally spaced, hence the mean of the set = median of the set.

Sufficient.

Option B