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GATE2011 MN: GA-63

10 votes
909 views

$L, M$ and $N$ are waiting in a queue meant for children to enter the zoo. There are $5$ children between $L$ and $M$, and $8$ children between $M$ and $N$. If there are $3$ children ahead of $N$ and $21$ children behind $L$, then what is the minimum number of children in the queue?

  1. $28$
  2. $27$
  3. $41$
  4. $40$
in Quantitative Aptitude
edited by 909 views
0
21L5M8N3 For follow the sequence of the queue there are 40 students are there.
1
15---M--5--L--2--N--3    so total 28
0
If in case of maximum children then 40

2 Answers

22 votes
 
Best answer

$L,  M$ and $N$ are waiting in queue that are meant for children, so they are also counted as children.

Correct Answer: $A$


edited by
0
thanks praveen sir
0
18 N 2 L 5 M

18+2+5=35

25+L,M,N=28
2 votes

Answer -> (A) 28

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