Question:medium

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

Updated On: Jan 15, 2026
  • 22.5
  • 23.5
  • 24
  • 23
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The Correct Option is A

Solution and Explanation

Let the six distinct natural numbers be denoted as \( a, b, c, d, e, f \) in ascending order.

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

To achieve the maximum overall average for these six numbers, the middle values, \( c \) and \( d \), must be maximized.

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

  • Initial assignment for \( a, b \): Let \( a = 1, b = 27 \) satisfying \( a + b = 28 \).
  • Initial assignment for \( e, f \): Let \( e = 27, f = 29 \) satisfying \( e + f = 56 \).
  • Select the largest possible values for \( c \) and \( d \) ensuring all six numbers remain distinct.
  • Proposed selection: \( c = 25, d = 26 \).

This yields the sequence: 1, 27, 25, 26, 27, 29. However, the number 27 is repeated, violating the distinctness requirement. Therefore, a revised selection is necessary while maintaining distinctness.

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

This results in the sequence: 1, 25, 26, 27, 28, 29. However, the sum \( a + b = 1 + 25 = 26 \), which does not equal 28. The set must still satisfy the original conditions: \[ a + b = 28, \quad e + f = 56 \]

Consider the following assignment:

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

Consider the following assignment:

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

The set of six distinct numbers is: 2, 25, 26, 27, 28, 29. This set satisfies \( a + b = 28 \) and \( e + f = 56 \). ✅

 

The total sum is: \[ 2 + 26 + 25 + 27 + 28 + 29 = 137 \]

The average is: \[ \frac{137}{6} \approx 22.83 \]

Attempting further maximization

Consider the following assignment:

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

Instead, use the following assignment for \( e, f \):

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

The optimal valid set is determined to be: \( a = 2, b = 26, c = 25, d = 27, e = 28, f = 29 \).

 

✅ The maximum possible average is \( \frac{137}{6} = \boxed{22.83} \).

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