QUESTION

CSAT

Hard

Reasoning

Prelims 2022

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?

Select an option to attempt

Explanation

Case 1: A _ _ _ _ B _ _ _ _ _ _ _ C This case has more than 11 persons ahead of C, so it is eliminated.)

Case 2: B _ _ _ _ A _ _ C This case could be valid. Total persons = 11 (ahead of C)+1 (C)+4 (between A and B)+1 (A)+7 (between B and C)+13 (behind A)= $22$

Case 3: C _ _ _ _ _ _ _ B _ _ _ _ A This case could be valid. Total persons = 11 (ahead of C) + 1 (C) + 12 (between C and A) + 1 (A) + 13 (behind A) = $48$

Case 4: C _ _ _ _ A _ _ B This case is invalid because there should be 4 persons between A and B.

Thus, the minimum possible number of people in the queue is $22$ .

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