retagged by
501 views
1 votes
1 votes

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. $12$
retagged by

1 Answer

1 votes
1 votes

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
Answer:

Related questions

1 votes
1 votes
1 answer
1
soujanyareddy13 asked Sep 16, 2021
527 views
How many $3-$digit numbers are there, for which the product of their digits is more than $2$ but less than $7$?
1 votes
1 votes
1 answer
2
2 votes
2 votes
1 answer
3
soujanyareddy13 asked Sep 16, 2021
450 views
If $\log_4 5=\left ( \log _{4}y \right )\left ( \log _{6}\sqrt{5} \right )$, then $y$ equals
1 votes
1 votes
1 answer
4
soujanyareddy13 asked Sep 16, 2021
741 views
A person spent Rs $50000$ to produce a desktop computer and a laptop computer. He sold the desktop at $20\%$ profit and the laptop at $10\%$ loss. If overall he made a $2...
1 votes
1 votes
1 answer
5
soujanyareddy13 asked Sep 16, 2021
517 views
The area of the region satisfying the inequilities $\left | x \right |-y\leq 1,y\geq 0$ and $y\leq 1$ is