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 sum of the number of pens does E, C and B has?

The correct option is 93.

Let us solve this puzzle step-by-step to find the number of pens each person has and their respective designations.

1. Understanding the Designations and Seniority Order:
The designations in decreasing order of seniority (senior-most to junior-most) are:
1. General Manager (GM)
2. Deputy General Manager (DGM)
3. Assistant Manager (AM)
4. Section Officer (SO)
5. Clerk

2. Analyzing the Number of Pens:
The available numbers of pens are: 19, 23, 35, 39, and 51.
Among these, the prime numbers of pens are 19 and 23.

3. Analyzing the Clues:

• Clue 1: "D has 12 pens more than C."
Let us check the differences between the given pen counts:
35 - 23 = 12
51 - 39 = 12
So, the pair (C, D) can be either (23, 35) or (39, 51).

• Clue 2: "The difference between the pens of E and C is 4."
Checking the pen counts for a difference of 4:
23 - 19 = 4
39 - 35 = 4
- If C has 23 pens, then E must have 19 pens (since 23 - 19 = 4).
- If C has 39 pens, then E must have 35 pens (since 39 - 35 = 4).

• Clue 3: "The one who has 39 pens is just senior to E."
- If C has 39 pens and E has 35 pens, then the person with 39 pens would be C. But C cannot be "just senior to E" if C is junior to D and other constraints are considered.
Let us test the case where C has 23 pens:
- If C = 23, then D = 23 + 12 = 35.
- Then E = 19 (since |19 - 23| = 4).
- The available numbers left for the remaining two persons are 39 and 51.
- Since the person just senior to E has 39 pens, E cannot have 39 pens. This matches E having 19 pens.

So the distribution of pens is:
• E = 19 pens
• C = 23 pens
• D = 35 pens
The remaining numbers of pens are 39 and 51 for A and B.

• Clue 4: "The one who has prime number of pens is not the senior most person but two designations senior to D."
The prime numbers of pens are 19 (E) and 23 (C).
Since D has a person two designations senior to him, D can be placed at AM, SO, or Clerk:
- If D is AM, then the person 2 designations senior to D is GM (senior-most). But the clue states that the one with prime pens is NOT the senior-most person.
- Therefore, the person two designations senior to D cannot be GM.
- Thus, D must be SO, which makes the person two designations senior to D the DGM.
- DGM has a prime number of pens (either 19 or 23).
- "C is junior to D": Since D is SO, C must be the Clerk.
- Since C is Clerk, C has 23 pens (a prime number). This fits the clue that C (having 23 pens) is junior to D.

Let's verify the positions:
1. GM
2. DGM: Person with prime pens (E, who has 19 pens)
3. AM: The person with 39 pens (just senior to E? No, E is DGM, so GM is just senior to E, meaning GM has 39 pens).
Let's re-verify: "The one who has 39 pens is just senior to E."
Since E is DGM, the person just senior to E is GM. Thus, GM has 39 pens.
Since A and B are the remaining persons:
- GM has 39 pens.
- "B is junior to the one who has 39 pens": B is junior to GM, so B can be AM.
- This leaves A to be GM.
- Therefore, A = 39 pens (GM), E = 19 pens (DGM), B = 51 pens (AM), D = 35 pens (SO), C = 23 pens (Clerk).

4. Summary of Person, Designation, and Pens:
• A: GM – 39 pens
• E: DGM – 19 pens
• B: AM – 51 pens
• D: SO – 35 pens
• C: Clerk – 23 pens

5. Calculating the Required Sum:
We need to find the sum of the number of pens that E, C, and B have:

$\text{Sum}=\text{Pens of E}+\text{Pens of C}+\text{Pens of B}$

$\text{Sum}=19+23+51=93$

Thus, the total sum of pens held by E, C, and B is 93.