Suppose $A = {x|5x-a \le 0}$, $B = {x|6x-b > 0}$, $a,b \in \mathbb{N}$, and $A \cap B \cap \mathbb{N} = {2,3,4}$. The number of such pairs $(a,b)$ is
${ \textbf{(A)}\ 20\qquad\textbf{(B)}\  25\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}} 42\qquad $
1. **Determine the range for \(a\):**
   - Given \(A = \{x \mid 5x - a \leq 0\}\), we can rewrite this as \(5x \leq a\).
   - For \(x \in \{2, 3, 4\}\), we have:
     \[
     5 \cdot 2 \leq a \implies 10 \leq a
     \]
     \[
     5 \cdot 3 \leq a \implies 15 \leq a
     \]
     \[
     5 \cdot 4 \leq a \implies 20 \leq a
     \]
   - Therefore, the most restrictive condition is \(20 \leq a\).
   - Since \(x\) cannot be 5 or more, we also have \(5 \cdot 5 > a \implies 25 > a\).
   - Thus, \(20 \leq a < 25\).
   - The possible values for \(a\) are \(20, 21, 22, 23, 24\).

2. **Determine the range for \(b\):**
   - Given \(B = \{x \mid 6x - b > 0\}\), we can rewrite this as \(6x > b\).
   - For \(x \in \{2, 3, 4\}\), we have:
     \[
     6 \cdot 2 > b \implies 12 > b
     \]
     \[
     6 \cdot 3 > b \implies 18 > b
     \]
     \[
     6 \cdot 4 > b \implies 24 > b
     \]
   - Therefore, the most restrictive condition is \(b < 12\).
   - Since \(x\) cannot be 1 or less, we also have \(6 \cdot 1 \leq b \implies 6 < b\).
   - Thus, \(6 < b < 12\).
   - The possible values for \(b\) are \(7, 8, 9, 10, 11\).

3. **Calculate the number of pairs \((a, b)\):**
   - There are 5 possible values for \(a\) and 5 possible values for \(b\).
   - Therefore, the number of pairs \((a, b)\) is:
     \[
     5 \times 5 = 25
     \]

The final answer is \(\boxed{25}\).