Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there are three persons ahead of C and 21 behind A, then what could be the minimum number of persons in the queue?

Setup (positions from front): Let position of the front be 1 and last be N.

• “3 ahead of C” ⇒ C at position 4.
• “21 behind A” ⇒ if A is at position a, then N − a = 21 ⇒ a = N − 21.
• “5 between A and B” ⇒ |a − b| = 6.
• “8 between B and C” ⇒ |b − 4| = 9 ⇒ b = 13 (the other choice −5 is invalid).

Now, with b = 13, we need |a − 13| = 6 ⇒ a = 7 or a = 19.

But a = N − 21, so:
• If a = 7, then N = 28.
• If a = 19, then N = 40.

Both satisfy all conditions; the minimum possible is: