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Question

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

A
41
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B
40
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C
28
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D
27
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Solution

The correct option is C 28
A, B, C can be arranged in a queue =6!= six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB.
According to question, only 3 persons can be ahead of C,
So, there are only two possible arrangements i.e. CBA and CAB.
Case 1) gives us 40 as number of people in the queue.
Number of persons =3+1+8+1+5+1+21=40.
Case 2) gives us 28 as number of people in the queue.
number of persons =(3+1+2+1+21)=28.
So, minimum number is 28.

524621_109183_ans_0a2117abba774948ad2bca8bcbdf417e.png

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