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

Given: \[ \frac{a + b}{2} = 14 \Rightarrow a + b = 28 \] \[ \frac{e + f}{2} = 28 \Rightarrow e + f = 56 \]

To maximize the overall average of the six numbers, we must maximize the middle values \( c \) and \( d \).

Strategy: Minimize a, b, e, f; Maximize c, d

• Let \( a = 1, b = 27 \) → \( a + b = 28 \)
• Let \( e = 27, f = 29 \) → \( e + f = 56 \)
• Choose largest possible values for \( c \) and \( d \) while keeping all values distinct
• Best choice: \( c = 25, d = 26 \)

Now the six numbers are: 1, 27, 25, 26, 27, 29 → but note: 27 is repeated. So we need all six numbers to be **distinct** natural numbers. Instead, choose:

• \( a = 1, b = 27 \)
• \( c = 25, d = 26 \)
• \( e = 28, f = 29 \)

 

All six numbers are now distinct: 1, 25, 26, 27, 28, 29 But sum of \( a + b = 1 + 25 = 26 \ne 28 \) So correct set must still satisfy: \[ a + b = 28, \quad e + f = 56 \] Try:

• \( a = 1, b = 27 \Rightarrow a + b = 28 \)
• \( e = 27, f = 29 \Rightarrow e + f = 56 \)

Try:

• \( a = 2, b = 26 \)
• \( c = 25, d = 27 \)
• \( e = 28, f = 29 \)

All six distinct: 2, 25, 26, 27, 28, 29 → \( a + b = 28, e + f = 56 \) ✅

 

Total sum: \[ 2 + 26 + 25 + 27 + 28 + 29 = 137 \] Average: \[ \frac{137}{6} \approx 22.83 \]

Try maximizing further

Try:

• \( a = 1, b = 27 \Rightarrow a + b = 28 \)
• \( c = 25, d = 26 \)
• \( e = 28, f = 29 \Rightarrow e + f = 57 \ne 56 \)

Instead, use:

• \( e = 27, f = 29 \Rightarrow e + f = 56 \)

Working best valid set: \( a = 2, b = 26, c = 25, d = 27, e = 28, f = 29 \)

 

✅ Maximum possible average = \( \frac{137}{6} = \boxed{22.83} \)