in Quantitative Aptitude retagged by
169 views
1 vote
1 vote

Among $100$ students, $x_{1}$ have birthdays in January, $x_{2}$ have birthdays in February, and so on. If $x_{0}= \text{max}\left ( x_{1},x_{2},\dots,x_{12} \right ),$ then the smallest possible value of $x_{0}$ is

  1. $9$
  2. $10$
  3. $8$
  4. $12$
in Quantitative Aptitude retagged by
2.7k points
169 views

1 Answer

1 vote
1 vote

Given that, $x_{0} = \text{max} (x_{1}, x_{2}, \dots, x_{12})$

That means, 

  • $x_{0} \geq x_{1}$
  • $x_{0} \geq x_{2}$
  • $x_{0} \geq x_{3}$
  • $x_{0} \geq x_{4}$
  • $x_{0} \geq x_{5}$
  • $x_{0} \geq x_{6}$
  • $x_{0} \geq x_{7}$
  • $x_{0} \geq x_{8}$
  • $x_{0} \geq x_{9}$
  • $x_{0} \geq x_{10}$
  • $x_{0} \geq x_{11}$
  • $x_{0} \geq x_{12}$

Now, $x_{1} + x_{2} + \dots + x_{12} = 100$

$ \Rightarrow x_{0} + x_{0} + \dots + x_{0} \geq 100$

$ \Rightarrow 12 x_{0} \geq 100 $

$ \Rightarrow x_{0} \geq \frac{100}{12} $

$ \Rightarrow x_{0} = \left \lceil \frac{100}{12} \right \rceil $

$ \Rightarrow x_{0} = \left \lceil 8.33 \right \rceil$

$ \Rightarrow \boxed{x_{0} = 9}$

Correct Answer$: 9$

edited by
10.3k points
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true