Five persons A, B,C, D and E (but not necessarily in the same order) have different designations i.e., General Manager (GM), Deputy General Manager (DGM), Assistant Manager (AM), Section Officer (SO), and Clerk in a company. The order of seniority is the same as given above i.e., GM is the senior-most designation and Clerk is the junior-most designation. Each person has different number of pens – 19, 23, 35, 39 and 51 but not necessarily in the same order.

The one who has prime number of pens is not the senior most person but two designations senior to D. C is junior to D. The one who has 39 pens is just senior to E. The difference between the pens of E and C is 4. D has 12 pens more than C. B is junior to the one who has 39 pens.

What is the designation of A?

The correct option is GM.

Step 1: Understand the Given Information

There are five designations arranged from senior-most to junior-most as follows:
1. General Manager (GM) [Senior-most]
2. Deputy General Manager (DGM)
3. Assistant Manager (AM)
4. Section Officer (SO)
5. Clerk [Junior-most]

The available numbers of pens are: 19, 23, 35, 39, and 51.
Note that the prime numbers among these are 19 and 23.

Step 2: Determine the Number of Pens for Each Person

1. We are given that the difference between the pens of E and C is 4:
$|\text{E}-\text{C}|=4$

Looking at our numbers (19, 23, 35, 39, 51), the pair with a difference of 4 is 19 and 23.
So, E and C must have 19 and 23 pens in some order.

2. We are also given that D has 12 pens more than C:
$\text{D}=\text{C}+12$

If C = 19, then D = 19 + 12 = 31 (which is not in our set).
If C = 23, then D = 23 + 12 = 35 (which is in our set).
Therefore, C must have 23 pens, D must have 35 pens, and consequently, E must have 19 pens.

3. The remaining two numbers of pens (39 and 51) belong to A and B.
Since the one who has 39 pens is just senior to E, and B is junior to the one who has 39 pens, the person with 39 pens cannot be B if B is junior to that person, or let's check who holds 39 pens:
The numbers assigned so far: C = 23, D = 35, E = 19.
The remaining pen counts are 39 and 51 for A and B.

Step 3: Determine the Designations

1. "The one who has prime number of pens is not the senior most person but two designations senior to D."
The persons with prime numbers of pens are E (19) and C (23).
Since this person is two designations senior to D, D cannot be in the top two positions.
Also, C is junior to D. If C is junior to D, D must be senior to C.
If D is SO (4th), then the person two designations senior to D is DGM (2nd).
If D is AM (3rd), then the person two designations senior to D is GM (1st). But the clue states that this person is not the senior-most person (GM). Thus, this person cannot be GM.
Therefore, the person two designations senior to D must be DGM (2nd), which means D is SO (4th).

Since C is junior to D, C must be the Clerk (5th).

2. The person two designations senior to D (SO) is DGM. This person has a prime number of pens (19 or 23). Since C (23) is Clerk, the person at DGM must be E (19).

3. Now we know:
- 2nd (DGM): E (19 pens)
- 4th (SO): D (35 pens)
- 5th (Clerk): C (23 pens)

4. We are given: "The one who has 39 pens is just senior to E."
Since E is DGM (2nd), the person just senior to E is GM (1st).
Therefore, the person who is GM has 39 pens.

5. "B is junior to the one who has 39 pens."
The one who has 39 pens is GM (1st). The remaining positions to be filled by persons are GM (1st) and AM (3rd).
Since B is junior to GM, B must be AM (3rd).
Thus, A must be GM (1st), with 51 pens going to B.

Final Arrangement:

1. GM: A (39 pens)
2. DGM: E (19 pens)
3. AM: B (51 pens)
4. SO: D (35 pens)
5. Clerk: C (23 pens)

Therefore, the designation of A is GM.