Number, Ranking and Time Sequence Test Question & Answers
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 be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?
|A) 41||B) 40|
|C) 28||D) 27|
Three persons A, B, C can be arranged in a queue in six different ways, ie 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, ie CBA and CAB.
We may consider the two cases as under:
Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40
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 persons in the queue.
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