Question

Consider six distinct natural numbers such that the average of the two smallest numbers is 14 and the average of the two largest numbers is 28. Then,the maximum possible value of the average of these six numbers is

• A. 22.5
• B. 23.5
• C. 24
• D. 23

Answer 1

The Correct Option is A

Let's denote the six distinct natural numbers as a, b, c, d, e, and f. According to the given information:
1. The average of the two smallest numbers (a and b) is 14: (a + b)/2 = 14 
2. The average of the two largest numbers (e and f) is 28: (e + f)/2 = 28 

Solving equation 1 for a + b gives: a + b = 28 
Solving equation 2 for e + f gives: e + f = 56 
To maximize the average of all six numbers, we should maximize the two numbers in between, which are c and d. 

Let's maximize these two numbers by minimizing the other numbers, which are a, b, e, and f. To do this, we can set a and b to be the smallest possible values (1 and 2), and e and f to be the largest possible values (27 and 29). 
So, the arrangement becomes: a = 1, b = 2, c, d, e = 27, f = 29 
Now, we need to find the value of c and d. Since the sum of all six numbers is the sum of these pairs: (a + b) + (c + d) + (e + f) = 28 + c + d + 56 
Simplifying: c + d = 0 
Thus, the maximum possible average is when c = -d, which makes the numbers c and d cancel each other's effects on the average. 
With this arrangement: a, b, 25, 26, 27, 29 
The sum of these numbers is: 1 + 2 + 25 + 26 + 27 + 29 = 110 

And the average is: 110/6 = 22.5 

Therefore, the maximum possible value of the average of these six numbers is 22.5.