Question

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

Answer 1

The correct option is C 28

Option (B) is the correct answer.
Explanation:
Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. Butsince there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possiblearrangements 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.
Case II :

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