Averages question - Finding Max and Min average

Question

Question:- Among 16 observations, if the least observation is eliminated the average of the remaining observations is 60. If the greatest observation is eliminated the average of the remaining observations is 54. For the 16 observations, find the difference between the maximum possible average and the minimum possible average.

My attempt:-

Let the observations be $a_1<=a_2<=a_3<=.....<=a_{15}<=a_{16}$

Sum of first 15 observations will be $54*15=810$

Sum of last 15 observations will be $60*15=900$

$a_{16}-a_1=90$

How to proceed further? Please guide me.

I am not able to understand the textbook solutions given.

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Textbook solutions:-

Solution 1:-

Min Average

Let last 15 large observations be 60,

$15*54=14*60+a_1$

$a_1=-30$

Average here is = (60*15-30)/16 = 54.375

Max Average

Let first 15 observations be 54

$14*54+a_{16}=15*60$

$a_{16} = 59.625$

Average here will be (144+54*15)/16 = 59.625

Therefore Ans = $59.625-54.375=5.25$

Solution 2:-

The gap between $a_1$ and $a_{16}$ is fixed. Only the middle 14 values can swing by a max of 6 from 54 to 60. This sum in two extreme cases will differ by 14*6 = 84 and thus average of 16 values will differ by 84/16 = 5.25

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How is the author assuming values of terms to be 54 and 60 , and how can the middle terms only range from 54 to 60 ?

Answer 1

You found $\sum\limits_1^{15}a_i =810$ and $\sum\limits_2^{16}a_i =900$.

You also know

• $a_{16} \ge 60$ (the average of the top fifteen, and $a_{16}$ is the biggest of them)
  • so $\sum\limits_1^{16}a_i = a_{16}+\sum\limits_1^{15}a_i \ge 60+810=870$
  • so $\bar a \ge \frac{870}{16}=54.375$
• $a_{1} \le 54$ (the average of the bottom fifteen, and $a_{1}$ is the smallest of them)
  • so $\sum\limits_1^{16}a_i = a_{1}+\sum\limits_2^{16}a_i \le 54+900=954$
  • so $\bar a \le \frac{954}{16}=59.625$

The middle $14$ values are not individually restricted to being between $54$ and $60$, but their average is, so their sum is restricted to being between $756$ and $840$ with a range of $84$.