Question

Take 6 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 can be how much?

• A. 22.5
• B. 24.5
• C. 21.5
• D. 20
• E. 23.5

Hint:

To maximize the average under constraints, make the middle numbers as large as possible while satisfying the conditions on the extremes.

Answer 1

The Correct Option is B

Step 1:
Let the six numbers in increasing order be \(a < b < c < d < e < f\), all distinct natural numbers.

Step 2:
Average of two smallest = \(\frac{a+b}{2} = 14 \implies a+b = 28\).
Average of two largest = \(\frac{e+f}{2} = 28 \implies e+f = 56\).

Step 3:
To maximize the overall average, maximize the sum \(S = a+b+c+d+e+f\). Since \(a+b\) and \(e+f\) are fixed, we need to maximize \(c+d\).

Step 4:
Since \(a < b\) and \(a+b=28\), choose values that allow \(c\) and \(d\) to be as large as possible.

Step 5:
Take \(a=13, b=15\) (valid distinct pair with sum 28).

Step 6:
For \(e+f=56\), the closest distinct pair is \(e=27, f=29\).

Step 7:
Now maximize \(c\) and \(d\) such that \(b < c < d < e\).
So take \(c=25, d=26\).

Step 8:
Total sum = \(13+15+25+26+27+29 = 135\).
Average = \(\frac{135}{6} = 22.5\).

Step 9:
Thus, the maximum possible average is \(22.5\).

Step 10:
Final Answer: 22.5