In a class of 95 students, 38 students are in set P only, 13 students are in set Q only, and the number of students in both P and Q is x. The number of students in P is 2x + 38, and the number of stud

The total number of elements in $P \cup Q$ is the sum of the elements in all three regions of the Venn diagram: the elements only in P, the elements in the intersection of P and Q, and the elements only in Q.

$$n(P \cup Q) = (38+2x) + x + (13+x)$$

We are given that $n(P \cup Q) = 95$.

$$95 = 38+2x+x+13+x$$

Combine the constant terms and the terms with $x$:

$$95 = (38+13) + (2x+x+x)$$ $$95 = 51 + 4x$$

Subtract 51 from both sides:

$$95 - 51 = 4x$$ $$44 = 4x$$

Divide by 4:

$$x = \frac{44}{4}$$ $$x = 11$$

The value of $x$ is $\boxed{\text{11}}$.

$n(P \cap Q)'$ represents the number of elements that are not in the intersection of P and Q. In the context of this problem, where $n(P \cup Q)$ is given as the total, this means the elements within $P \cup Q$ but outside $P \cap Q$.
This is the sum of elements only in P and elements only in Q.

$$n(P \cap Q)' = n(\text{P only}) + n(\text{Q only})$$ $$n(P \cap Q)' = (38+2x) + (13+x)$$

Substitute the value of $x=11$:

$$n(P \cap Q)' = (38+2(11)) + (13+11)$$ $$n(P \cap Q)' = (38+22) + (24)$$ $$n(P \cap Q)' = 60 + 24$$ $$n(P \cap Q)' = 84$$

Alternatively, $n(P \cap Q)' = n(P \cup Q) - n(P \cap Q)$.
Since $n(P \cap Q) = x = 11$:

$$n(P \cap Q)' = 95 - 11$$ $$n(P \cap Q)' = 84$$

The value of $n(P \cap Q)'$ is $\boxed{\text{84}}$.