Explanation: Let the 9 numbers in ascending order be \(n_1, n_2, \ldots, n_9\). The middle number is \(n_5\). The problem states that the middle number (\(n_5\)) is the average of the nine numbers. Let \(S\) be the sum of all nine numbers. The average is \(\frac{S}{9}\). So, \(n_5 = \frac{S}{9}\) or \(S = 9n_5\). The five larger numbers are \(n_5, n_6, n_7, n_8, n_9\). Sum of five larger numbers = \(5 \times 68 = 340\). The five smaller numbers are \(n_1, n_2, n_3, n_4, n_5\). Sum of five smaller numbers = \(5 \times 44 = 220\). The sum of all nine numbers, \(S\), is the sum of the five larger numbers plus the sum of the four unique smaller numbers (\(n_1\) to \(n_4\)). The sum of the two sets includes \(n_5\) twice. Sum of 5 larger + Sum of 5 smaller = \(S + n_5\). \(340 + 220 = S + n_5\). \(560 = S + n_5\). We know \(S = 9n_5\). \(560 = 9n_5 + n_5\). \(560 = 10n_5\). \(n_5 = 56\). The sum of all nine numbers, \(S = 9n_5 = 9 \times 56 = 504\). (The solution uses \(9x + x = 340 + 220\) where \(x\) is the average of 9 numbers, which is \(n_5\), correctly leading to \(10x = 560\)).
