The numbers $ x,\,y,\,z$ are proportional to $ 2,\,3,\,5$. The sum of $ x$, $ y$, and $ z$ is $ 100$. The number $ y$ is given by the equation $ y \equal{} ax \minus{} 10$. Then $ a$ is:

$ \textbf{(A)}\ 2 \qquad 
\textbf{(B)}\ \frac{3}{2}\qquad 
\textbf{(C)}\ 3\qquad 
\textbf{(D)}\ \frac{5}{2}\qquad 
\textbf{(E)}\ 4$
1. Given that the numbers \( x, y, z \) are proportional to \( 2, 3, 5 \), we can write:
   \[
   x = 2k, \quad y = 3k, \quad z = 5k
   \]
   for some constant \( k \).

2. The sum of \( x, y, \) and \( z \) is given as 100:
   \[
   x + y + z = 100
   \]
   Substituting the proportional values, we get:
   \[
   2k + 3k + 5k = 100
   \]
   Simplifying the left-hand side:
   \[
   10k = 100
   \]
   Solving for \( k \):
   \[
   k = \frac{100}{10} = 10
   \]

3. Now, substituting \( k = 10 \) back into the expressions for \( x, y, \) and \( z \):
   \[
   x = 2k = 2 \times 10 = 20
   \]
   \[
   y = 3k = 3 \times 10 = 30
   \]
   \[
   z = 5k = 5 \times 10 = 50
   \]

4. We are given that \( y = ax - 10 \). Substituting the values of \( y \) and \( x \):
   \[
   30 = a \times 20 - 10
   \]
   Solving for \( a \):
   \[
   30 = 20a - 10
   \]
   Adding 10 to both sides:
   \[
   40 = 20a
   \]
   Dividing both sides by 20:
   \[
   a = \frac{40}{20} = 2
   \]

The final answer is \( \boxed{2} \)