retagged by
698 views

1 Answer

Best answer
3 votes
3 votes
Given that, the average of $30$ integers is $5 \Rightarrow$ Sum of $30$ integers $= 30 \times 5=150$

Among these $30$ integers, there are exactly $20$ which do not exceed $5$.$\Rightarrow 20 \mid \leq 5 \mid$

But $10$ integers might exceed $5$. We need the maximum possible average for $20$ integers, so for the remaining $10$ integers, we can take the minimum values so that the average get balanced.

So, minimum possible sum of remaining $10$ integers $= 10\mid >5\mid  \Rightarrow 10 \times 6 = 60$

So, sum of remaining $20$ integers $\Rightarrow 150-60=90 $

Let the average of $20$ integers be $x.$

$\begin{array}{ccc} \text{Integers} & \text{Average} & \text{Sum of integers} \\ 30 &  5  & 150 \\ 20 & x & 20x \\ 10 & 6 & 60 \end{array}$

$\Rightarrow 20x=90$

$\Rightarrow x= \frac{90}{20}=4.5$

$\therefore$ The highest possible value of the average of these $20$ integers are $4.5.$

Correct Answer: C
edited by
Answer:

Related questions

1 votes
1 votes
1 answer
2
1 votes
1 votes
1 answer
3
go_editor asked Mar 20, 2020
535 views
If x is a real number, then $\sqrt{\log _{e}\frac{4x-x^{2}}{3}}$ is a real number if and only if$1\leq x\leq 2$$-3\leq x\leq 3$$1\leq x\leq 3$$-1\leq x\leq 3$
1 votes
1 votes
1 answer
4
1 votes
1 votes
1 answer
5
go_editor asked Mar 20, 2020
699 views
How many pairs $(m,n)$ of positive integers satisfy the equation $m^{2}+105=n^{2}$ _______