Three persons A, B and C are standing in a queue not necessarily in the same order. There are 4 persons between A and B, and 7 persons between B and C. If there are 11 persons ahead of C and 13 behind A, what could be the minimum number of persons in the queue?

Re-check: Let positions be counted from the front. C has 11 ahead, so C = 12. A has 13 behind, so if total is N, then A = N - 13.

Distance conditions:

• 7 persons between B and C ⇒ |B - C| = 8
• 4 persons between A and B ⇒ |A - B| = 5

From C = 12, B can be 20 or 4.

• If B = 20, then A can be 25 or 15 ⇒ totals N = 38 or 28.
• If B = 4, then A can be 9 or -1 (invalid) ⇒ total N = 22.

Option-wise verdict:

• (a) 22: Possible with A = 9, B = 4, C = 12. Valid.
• (b) 28: Also possible with A = 15, B = 20, C = 12. Valid.
• (c) 32: Not possible from the conditions. Invalid.
• (d) 38: Also possible with A = 25, B = 20, C = 12. Valid.

Since the question asks the minimum number of persons in the queue, the correct answer is 22, i.e. (a). My earlier derived option (b) was an error; the worked positions clearly show 22 is feasible and is the minimum.