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

Question

Accepted Answer

1. **Define the Numbers:**  
   - Let the six numbers be $$a, b, c, d, e, f$$ in ascending order ($$a < b < c < d < e < f$$).  
   - The sum of the two smallest numbers:  
     $$
     a + b = 14 	imes 2 = 28.
     $$  
   - The sum of the two largest numbers:  
     $$
     e + f = 28 	imes 2 = 56.
     $$

2. **Maximizing $$c$$ and $$d$$:**  
   - To maximize the overall average, we need to maximize $$c$$ and $$d$$. This is possible when $$e$$ and $$f$$ are maximized while keeping all six numbers distinct.  
   - The maximum values for $$e$$ and $$f$$ are $$27$$ and $$29$$, respectively (since they must sum to 56 and be distinct).

3. **Assign Values to All Numbers:**  
   - The sequence becomes: $$a, b, c, d, e, f = a, b, 25, 26, 27, 29$$.  
   - Since $$a + b = 28$$, we use this to complete the sequence.

4. **Calculate the Average:**  
   - Total sum of all six numbers:  
     $$
     a + b + c + d + e + f = 28 + 25 + 26 + 27 + 29 = 135.
     $$  
   - Average of these six numbers:  
     $$
     	ext{Average} = \frac{	ext{Total Sum}}{6} = \frac{135}{6} = 22.5.
     $$

Thus, the maximum possible value of the average is **22.5**, and the correct answer is **A**.

Answer

Answer

Answer