If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is:

Let us denote the 50 observations by $$x_1,\,x_2,\,x_3,\ldots ,x_{50}$$ and let their arithmetic mean be $$\bar x$$.

We are told that the sum of the deviations of these observations from the number 30 is 50. Writing this information algebraically, we have

$$\displaystyle \sum_{i=1}^{50}(x_i-30)=50.$$

We expand the summation:

$$\sum_{i=1}^{50}(x_i-30)=\sum_{i=1}^{50}x_i-\sum_{i=1}^{50}30.$$

Because 30 is a constant, its sum over 50 terms is simply $$30\times50.$$ Therefore, the right-hand side becomes

$$\sum_{i=1}^{50}x_i-30\times50.$$

So the given equation can be rewritten as

$$\sum_{i=1}^{50}x_i-30\times50=50.$$

We now isolate the unknown sum $$\sum_{i=1}^{50}x_i$$:

$$\sum_{i=1}^{50}x_i=50+30\times50.$$

We calculate the product $$30\times50$$ first:

$$30\times50=1500.$$

Substituting this value back, we obtain

$$\sum_{i=1}^{50}x_i=50+1500.$$

Adding the two numbers gives

$$\sum_{i=1}^{50}x_i=1550.$$

Now, by the definition of the arithmetic mean,

$$\bar x=\frac{\sum_{i=1}^{50}x_i}{50}.$$

Substituting the sum we just found, we have

$$\bar x=\frac{1550}{50}.$$

We perform the division:

$$\bar x=31.$$

Hence, the mean of the 50 observations is $$31$$.

Hence, the correct answer is Option D.