Explanation:

Let's analyze all the statements step by step:

- There are 21 people and 6 posts: Chairman > GM > Manager > AM > PO > Clerk.

- At each lower post, there is one more person than in the previous (senior) post.

- So, let's use x as the number of people at Chairman. The posts then have x, x+1, x+2, x+3, x+4, x+5 people.

- Total = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 6x + 15 = 21 ? 6x = 6 ? x = 1.

So, postings:

1. Chairman – 1

2. GM – 2

3. Manager – 3

4. AM – 4

5. PO – 5

6. Clerk – 6

Now let's place names, focusing on S:

- E is Manager, C and G on same post and senior to E, so they're either Chairman or GM. Only two at GM, so C & G are GMs.

- E (Manager), Z is junior to E (so Z is AM), F, Q, R the same post, all junior to Z (so they’re PO or Clerks).

- A just junior to W, just senior to R: (A, W, R in consecutive order). So this likely places them within PO, Clerk, or AM/PO.

- V is junior to Z but senior to J (so V and J must be PO or Clerks, after AM/Z).

- S is senior to D, and D is senior to M. So order: S > D > M. They must be together on three consecutive posts.

- U and B together, K & O together (but not clerks).

- S does not work with U and P. S is not at Clerk or other post having U or P.

Let's find how many are junior to S:

- The only posts S can have are Chairman (1), GM (2), or Manager (3).

- If S is Chairman, all 20 are junior.

- If S is GM, 19-2 = 19 junior, but there are 2 GMs, and S must be senior to D.

- If S is Manager (E is Manager; Z is AM), but S, D, M must be on consecutive posts: S > D > M.

With the rules, S must be Chairman (only one post above D, then M). So S is most senior.

- All others (21 – 1 = 20) are junior to S.

Options:

Option 1: 6

Option 2: 11

Option 3: 15

Option 4: 18

Option 5: 20

Correct Answer: Option 5 (20)

Summary of statements:

- Each lower post increases people by one.

- Assigning the known relationships places S at the top; all others are junior.

- Thus, 20 people are junior to S.