5. Let $S$ denote the set of all positive integers whose prime factors are elements of $\{2,3,5,7,11\}$. (We include 1 in the set $S$.) If

$$
\sum_{q \in S} \frac{\varphi(q)}{q^{2}}
$$

can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, find $a+b$. (Here $\varphi$ denotes Euler's totient function.)
1. **Understanding the Problem:**
   We need to find the sum \(\sum_{q \in S} \frac{\varphi(q)}{q^2}\), where \(S\) is the set of all positive integers whose prime factors are elements of \(\{2, 3, 5, 7, 11\}\), and \(\varphi\) is Euler's totient function. We need to express this sum as a fraction \(\frac{a}{b}\) in simplest form and find \(a + b\).

2. **Using the Multiplicative Property:**
   Since \(\frac{\varphi(q)}{q^2}\) is a multiplicative function, we can write:
   \[
   \sum_{q \in S} \frac{\varphi(q)}{q^2} = \prod_{p \in \{2, 3, 5, 7, 11\}} \left( \sum_{i=0}^{\infty} \frac{\varphi(p^i)}{p^{2i}} \right)
   \]

3. **Simplifying the Inner Sum:**
   For a prime \(p\), \(\varphi(p^i) = p^i - p^{i-1}\) for \(i \geq 1\) and \(\varphi(1) = 1\). Therefore:
   \[
   \sum_{i=0}^{\infty} \frac{\varphi(p^i)}{p^{2i}} = 1 + \sum_{i=1}^{\infty} \frac{p^i - p^{i-1}}{p^{2i}} = 1 + \sum_{i=1}^{\infty} \left( \frac{1}{p^i} - \frac{1}{p^{i+1}} \right)
   \]
   This series is telescoping:
   \[
   1 + \left( \frac{1}{p} - \frac{1}{p^2} \right) + \left( \frac{1}{p^2} - \frac{1}{p^3} \right) + \cdots = 1 + \frac{1}{p}
   \]

4. **Product Over All Primes in the Set:**
   Therefore, we have:
   \[
   \sum_{q \in S} \frac{\varphi(q)}{q^2} = \prod_{p \in \{2, 3, 5, 7, 11\}} \left( 1 + \frac{1}{p} \right)
   \]
   Calculating each term:
   \[
   1 + \frac{1}{2} = \frac{3}{2}, \quad 1 + \frac{1}{3} = \frac{4}{3}, \quad 1 + \frac{1}{5} = \frac{6}{5}, \quad 1 + \frac{1}{7} = \frac{8}{7}, \quad 1 + \frac{1}{11} = \frac{12}{11}
   \]

5. **Multiplying the Fractions:**
   \[
   \prod_{p \in \{2, 3, 5, 7, 11\}} \left( 1 + \frac{1}{p} \right) = \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdot \frac{12}{11}
   \]
   Simplifying the product:
   \[
   \frac{3 \cdot 4 \cdot 6 \cdot 8 \cdot 12}{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11} = \frac{1152}{385}
   \]

6. **Finding \(a + b\):**
   The fraction \(\frac{1152}{385}\) is already in simplest form since 1152 and 385 are relatively prime. Thus, \(a = 1152\) and \(b = 385\).

   Therefore, \(a + b = 1152 + 385 = 1537\).

The final answer is \(\boxed{1537}\).