To determine the earliest time all 120 parts can be finished, we must analyze the production rates, setup times, and the constraint that partial parts do not count (meaning machines stop producing once their specific quota is met, or we calculate the time required to produce integer numbers of parts).

### Step 1: Analyze Machine Parameters

**Machine A:**
*   Production Rate ($R_A$): 8 parts/hour.
*   Setup Time ($S_A$): 30 minutes = $0.5$ hours.
*   Effective Start Time: $0.5$ hours.

**Machine B:**
*   Production Rate ($R_B$): 5 parts/hour.
*   Setup Time ($S_B$): 12 minutes = $\frac{12}{60}$ hours = $0.2$ hours.
*   Effective Start Time: $0.2$ hours.

**Total Parts Required:** 120 parts.

### Step 2: Determine Optimal Allocation of Parts

Since the machines run simultaneously, we need to find an integer number of parts $n_A$ produced by Machine A and $n_B$ produced by Machine B such that $n_A + n_B = 120$, which minimizes the total time $T$.

The time taken by Machine A to produce $n_A$ parts is:
$$ T_A = 0.5 + \frac{n_A}{8} $$

The time taken by Machine B to produce $n_B$ parts is:
$$ T_B = 0.2 + \frac{n_B}{5} $$

The total time to finish all parts is determined by the machine that finishes last:
$$ T_{total} = \max(T_A, T_B) $$

To minimize $T_{total}$, we should aim to balance the completion times so that $T_A \approx T_B$. Let's find the approximate equilibrium point where $T_A = T_B$.

$$ 0.5 + \frac{n_A}{8} = 0.2 + \frac{n_B}{5} $$

Substitute $n_B = 120 - n_A$:

$$ 0.5 + \frac{n_A}{8} = 0.2 + \frac{120 - n_A}{5} $$

$$ 0.3 + \frac{n_A}{8} = 24 - \frac{n_A}{5} $$

$$ \frac{n_A}{8} + \frac{n_A}{5} = 23.7 $$

Find a common denominator (40):

$$ \frac{5n_A}{40} + \frac{8n_A}{40} = 23.7 $$

$$ \frac{13n_A}{40} = 23.7 $$

$$ 13n_A = 23.7 \times 40 $$

$$ 13n_A = 948 $$

$$ n_A = \frac{948}{13} \approx 72.92 $$

Since $n_A$ must be an integer, we test the integers closest to 72.92: **73** and **72**.

### Step 3: Calculate Completion Times for Integer Allocations

**Case 1: Machine A produces 73 parts ($n_A = 73$)**
Then Machine B produces $120 - 73 = 47$ parts ($n_B = 47$).

*   **Time for Machine A:**
    $$ T_A = 0.5 + \frac{73}{8} = 0.5 + 9.125 = 9.625 \text{ hours} $$
*   **Time for Machine B:**
    $$ T_B = 0.2 + \frac{47}{5} = 0.2 + 9.4 = 9.6 \text{ hours} $$
*   **Total Time:**
    $$ T_{total} = \max(9.625, 9.6) = 9.625 \text{ hours} $$

**Case 2: Machine A produces 72 parts ($n_A = 72$)**
Then Machine B produces $120 - 72 = 48$ parts ($n_B = 48$).

*   **Time for Machine A:**
    $$ T_A = 0.5 + \frac{72}{8} = 0.5 + 9 = 9.5 \text{ hours} $$
*   **Time for Machine B:**
    $$ T_B = 0.2 + \frac{48}{5} = 0.2 + 9.6 = 9.8 \text{ hours} $$
*   **Total Time:**
    $$ T_{total} = \max(9.5, 9.8) = 9.8 \text{ hours} $$

Comparing the two cases, $9.625$ hours is less than $9.8$ hours. Thus, the optimal allocation is $n_A = 73$ and $n_B = 47$.

### Step 4: Check the Immediately Preceding Completion Time

The prompt asks to "check the immediately preceding completion time." This usually implies checking the time just before the final part is completed or verifying the logic around the integer constraints.

In the optimal scenario ($n_A=73, n_B=47$):
*   Machine A finishes its 73rd part at $T = 9.625$ hours.
*   Machine B finishes its 47th part at $T = 9.6$ hours.

The "completion time" of the batch is determined by the last machine to finish, which is Machine A at **9.625 hours**.

Let's verify the state of production just before this time to ensure no parts are counted incorrectly.
*   At $T = 9.625$ hours, Machine A has just completed the 73rd part.
*   At $T = 9.625$ hours, Machine B has been running for $9.625 - 0.2 = 9.425$ hours.
    *   Parts produced by B: $5 \times 9.425 = 47.125$.
    *   Since partial parts do not count, Machine B has effectively finished its 47th part at $T=9.6$ hours and is idle (or working on a partial 48th part which doesn't count).
    *   Total valid parts = $73 (\text{from A}) + 47 (\text{from B}) = 120$.

If we checked the time *before* Machine A finished its last part (e.g., at $T=9.624$ hours):
*   Machine A has produced $\lfloor 8 \times (9.624 - 0.5) \rfloor = \lfloor 8 \times 9.124 \rfloor = \lfloor 72.992 \rfloor = 72$ parts.
*   Machine B has produced $\lfloor 5 \times (9.624 - 0.2) \rfloor = \lfloor 5 \times 9.424 \rfloor = \lfloor 47.12 \rfloor = 47$ parts.
*   Total parts = $72 + 47 = 119$.
*   Therefore, the 120th part is indeed completed exactly at $T=9.625$ hours.

### Step 5: Convert to Hours and Minutes

Total Time = $9.625$ hours.

$$ 0.625 \text{ hours} = 0.625 \times 60 \text{ minutes} = 37.5 \text{ minutes} $$

So, the time is **9 hours and 37.5 minutes**.

### Final Conclusion

The earliest time all 120 parts can be finished is **9.625 hours** (or 9 hours and 37.5 minutes).

**Summary of Arithmetic:**
1.  **Optimal Allocation:** Machine A produces 73 parts; Machine B produces 47 parts.
2.  **Machine A Time:** $0.5 \text{ h} + \frac{73}{8} \text{ h} = 0.5 + 9.125 = 9.625 \text{ h}$.
3.  **Machine B Time:** $0.2 \text{ h} + \frac{47}{5} \text{ h} = 0.2 + 9.4 = 9.6 \text{ h}$.
4.  **Bottleneck:** Machine A determines the finish time.
5.  **Result:** 9.625 hours.