Let $X$ and $Y$ be the following sums of arithmetic sequences: \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} What is the value of $Y - X$?

$ \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 $
1. **Identify the sequences:**
   - The sequence for \(X\) is \(10, 12, 14, \ldots, 100\).
   - The sequence for \(Y\) is \(12, 14, 16, \ldots, 102\).

2. **Find the number of terms in each sequence:**
   - For \(X\), the first term \(a_1 = 10\) and the common difference \(d = 2\). The last term \(a_n = 100\).
     \[
     a_n = a_1 + (n-1)d \implies 100 = 10 + (n-1) \cdot 2
     \]
     \[
     100 = 10 + 2n - 2 \implies 100 = 8 + 2n \implies 92 = 2n \implies n = 46
     \]
     So, there are 46 terms in \(X\).

   - For \(Y\), the first term \(a_1 = 12\) and the common difference \(d = 2\). The last term \(a_n = 102\).
     \[
     a_n = a_1 + (n-1)d \implies 102 = 12 + (n-1) \cdot 2
     \]
     \[
     102 = 12 + 2n - 2 \implies 102 = 10 + 2n \implies 92 = 2n \implies n = 46
     \]
     So, there are 46 terms in \(Y\).

3. **Calculate the difference between corresponding terms of \(Y\) and \(X\):**
   - Each term in \(Y\) is 2 more than the corresponding term in \(X\):
     \[
     (12 - 10), (14 - 12), (16 - 14), \ldots, (102 - 100) = 2, 2, 2, \ldots, 2
     \]

4. **Sum the differences:**
   - Since there are 46 terms and each difference is 2:
     \[
     \sum_{i=1}^{46} 2 = 46 \cdot 2 = 92
     \]

Conclusion:
\[
Y - X = 92
\]

The final answer is \(\boxed{92}\)