Three persons A, B and C are standing in a queue not necessarily in the same order. There are 4 persons…
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?
22
28
32
38
Solution
With 4 persons between A and B and 7 between B and C, and the constraints of 11 ahead of C and 13 behind A, evaluating the valid cases gives queue lengths of 28, 38 and 22. The minimum is the case C-2-B-4-A with C having 11 ahead and A having 13 behind, giving 11 + 1 + 4 + 1 + 1 + 4 = 22 persons in total.