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Three people M, N and P are standing in a queue. Five people are standing between M and N and eight people are standing between N and P. If there are three people ahead of P and 21 people behind M, what could be the minimum number of people in the queue?


  1. 41

  2. 40

  3. 27

  4. None of these

  5. 28


Solution

The correct option is E

28


Three people M, N, P can be arranged in a queue in six different ways, ie MNP, PNM, NMP, PMN, NPM, MPN. But since there are only 3 people ahead of P, so P should be in front of the queue. Thus, there are only two possible arrangements, ie PNM and PMN.

We may consider the two cases as under:

Case I:

(3 people) P (8 people) N (5 people) M (21 people)

Clearly, number of people in the queue = 3+1+8+1+5+1+21= 40

Case II: 

 (3 people) P M (5 people) N  

Number of people between M and P                  

= (8 - 6) = 2           [Since P (8 people) N, M (21 people)]  

Clearly number of people in the queue = (3+1+2+1+21) = 28

Now, 28 < 40. So, 28 is the minimum number of people in the queue.

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