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
B
40
C
28
D
27

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 $$28.$$Logical Reasoning

Suggest Corrections

0

Similar questions
View More

People also searched for
View More