Gary takes a New York City cab 6 miles to work and must pay `$17.50` for the cab ride. After work, Gary takes another New York City cab 10 miles to visit his family and must pay `$27.50`. During both of these rides, Gary was charged a ''drop fee'' (an initial charge when the cab's meter is activated) of dollars, plus an additional m dollars for each `(1)/(5)` of a mile travelled. What is the value of md?

Answer

To solve the problem, we need to set up two equations based on the information given about Gary's cab rides.

1. **Identify the Variables**:
- Let \( D \) be the drop fee.
- Let \( M \) be the charge per \( \frac{1}{5} \) of a mile.

2. **Set Up the Equations**:
- For the first ride (6 miles to work):
- The total cost is given as $17.50.
- The cost can be expressed as:
\[ D + 6 \times 5M = 17.50 \]
- Simplifying this gives:
\[ D + 30M = 17.50 \quad \text{(Equation 1)} \]

- For the second ride (10 miles to visit family):
- The total cost is given as $27.50.
- The cost can be expressed as:
\[ D + 10 \times 5M = 27.50 \]
- Simplifying this gives:
\[ D + 50M = 27.50 \quad \text{(Equation 2)} \]

3. **Solve the Equations**:
- We have the two equations:
1. \( D + 30M = 17.50 \)
2. \( D + 50M = 27.50 \)

- To eliminate \( D \), subtract Equation 1 from Equation 2:
\[ (D + 50M) - (D + 30M) = 27.50 - 17.50 \]
This simplifies to:
\[ 20M = 10 \]
Thus:
\[ M = \frac{10}{20} = \frac{1}{2} \]

4. **Find the Value of \( D \)**:
- Substitute \( M = \frac{1}{2} \) back into Equation 1:
\[ D + 30\left(\frac{1}{2}\right) = 17.50 \]
This simplifies to:
\[ D + 15 = 17.50 \]
Therefore:
\[ D = 17.50 - 15 = 2.50 \]

5. **Calculate \( MD \)**:
- Now we find \( MD \):
\[ MD = M \times D = \left(\frac{1}{2}\right) \times 2.50 = \frac{2.50}{2} = 1.25 \]
- In fractional form, this can be expressed as:
\[ MD = \frac{5}{4} \]

Thus, the final answer is:
\[ \boxed{1.25} \text{ or } \frac{5}{4} \]