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In Ranking Sequence questions, ranks of a person both from the top and bottom are mentioned and total number of persons is asked. However, sometimes this question is put in the form of a puzzle of interchanging seats by two persons.

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

Option:
A. 41
B. 40
C. 28
D. 27
Answer: C . 28

Justification:

Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements i.e., CBA and CAB. We may consider the two cases as under :

                 3         8        5       21
Case I :
←  C  ↔  B  ↔  A  →
clearly, number of persons in the queue = (3 + 1 + 8 + 1 + 5 + 1 + 21) = 40.

                    3                 5
Case II :
  ←  C     A  ↔  B
              


Number of persons between A and C = (8 - 6) = 2.
Clearly, number of persons in the queue = (3 + 1 + 2 + 1 + 21) = 28.
Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.


              

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