`AtoB` (zero order reaction)
`CtoD` (first order reaction)
In first order reaction gets `75%` completed in 40 min
and zero order reaction gets `75%` completed in 30 min then
calculate the value of Z where
`Z=("Half life period of first order reaction")/("Half life period of zero order reaction")`

Explanation

To solve the problem, we need to calculate the half-life periods of both the first-order and zero-order reactions and then find the ratio \( Z \) as defined in the question.

### Step-by-step Solution:

1. **Understanding the Reactions**:
- We have two reactions:
- \( A \to B \) (Zero Order Reaction)
- \( C \to D \) (First Order Reaction)

2. **Given Information**:
- For the first-order reaction, 75% completion occurs in 40 minutes.
- For the zero-order reaction, 75% completion occurs in 30 minutes.

3. **Calculating Half-life of First Order Reaction**:
- The time taken to reach 75% completion is \( T_{3/4} \).
- For a first-order reaction, the relationship between \( T_{3/4} \) and half-life \( t_{1/2} \) is given by:
\[ T_{3/4} = \frac{t_{1/2} \cdot \ln(4)}{2} \]
- Since \( T_{3/4} \) for the first-order reaction is 40 minutes:
\[ 40 = \frac{t_{1/2} \cdot \ln(4)}{2} \]
- Rearranging gives:
\[ t_{1/2} = \frac{40 \cdot 2}{\ln(4)} = \frac{80}{\ln(4)} \]

4. **Calculating Half-life of Zero Order Reaction**:
- For a zero-order reaction, the relationship is:
\[ T_{3/4} = \frac{3 \cdot t_{1/2}}{2} \]
- Since \( T_{3/4} \) for the zero-order reaction is 30 minutes:
\[ 30 = \frac{3 \cdot t_{1/2}}{2} \]
- Rearranging gives:
\[ t_{1/2} = \frac{30 \cdot 2}{3} = 20 \text{ minutes} \]

5. **Calculating the Value of \( Z \)**:
- Now we can find \( Z \):
\[ Z = \frac{t_{1/2, \text{first order}}}{t_{1/2, \text{zero order}}} = \frac{\frac{80}{\ln(4)}}{20} \]
- Simplifying gives:
\[ Z = \frac{80}{20 \cdot \ln(4)} = \frac{4}{\ln(4)} \]

6. **Final Calculation**:
- Using \( \ln(4) \approx 1.386 \):
\[ Z \approx \frac{4}{1.386} \approx 2.89 \]

### Conclusion:
The value of \( Z \) is approximately \( 2.89 \).