Let $x$ be the number of hard questions and $9x$ be the number of easy questions in the test. Let the total number of questions be $n = x + 9x = 10x$.

Given:
- Easy questions are worth 3 points each.
- Hard questions are worth $D$ points each.

Initial Total Points

The initial total score of the test is given by:
$$T = 3(9x) + Dx = 27x + Dx.$$

Condition 1: All Questions Worth 4 Points

If all questions are worth 4 points, the total score becomes:
$$4 \cdot 10x = 40x.$$

According to the problem, this new total score is 16 points more than the initial score:
$$40x = (27x + Dx) + 16.$$
Simplifying, we have:
$$40x = 27x + Dx + 16 \implies 13x = Dx + 16 \implies Dx = 13x - 16.$$
Thus, we can express $D$ as:
$$D = 13 - \frac{16}{x}.$$

Condition 2: Exchanging Scores

If the easy questions are worth $D$ points and the hard ones 3 points, the new total score is:
$$D(9x) + 3x = 9Dx + 3x.$$

The total score now is $\frac{3}{2}$ times the initial score:
$$9Dx + 3x = \frac{3}{2}(27x + Dx).$$

Equating and simplifying this equation:
$$9Dx + 3x = \frac{3}{2}(27x + Dx) \implies 18Dx + 6x = 3(27x + Dx) \implies 18Dx + 6x = 81x + 3Dx.$$

Subtracting terms appropriately:
$$18Dx + 6x - 3Dx = 81x \implies 15Dx + 6x = 81x \implies 15Dx = 75x \implies Dx = 5x.$$

From the two derived expressions for $Dx$, set them equal:
$$13x - 16 = 5x.$$

**Solve for $x$:**
$$13x - 5x = 16 \implies 8x = 16 \implies x = 2.$$

Conclusion:

The total number of questions is:
$$n = 10x = 10 \times 2 = 20.$$

Thus, the number of questions in the test is:
$$\boxed{20}.$$
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