The sum of five consecutive even numbers from set X is 440. Determine the total of another group of five consecutive integers if the second smallest number in this group is 121 less than twice the smallest number in set X.

Explanation

Let the five consecutive even numbers be represented as 2(x - 2), 2(x - 1), 2x, 2(x + 1), and 2(x + 2). Their sum is 10x = 440, so x = 44. Therefore, the smallest number in set X is 2(x - 2) = 84. The second smallest number in the other set is 2 * 84 - 121 = 47, which means the smallest number in this new set is 46. The sum of these five consecutive integers is 46 + 47 + 48 + 49 + 50 = 240.