9 numbers are written in ascending order, middle number is average of that nine numbers. Average of five larger numbers is 68 and also average of five smaller numbers is 44. What is the sum of all nine numbers?

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 \(10x = 560\)).