The sum of five consecutive even numbers of set x is 440. Find the sum of a different set of five consecutive integers whose second least number is 121 less than double the least number of set x?

Sure, let's solve this step by step.

Step 1: Find the least number in set x The sum of five consecutive even numbers is 440. Let's denote the least number as 'a'. The next four numbers would be a+2, a+4, a+6, and a+8. So, we have:

a + (a+2) + (a+4) + (a+6) + (a+8) = 440 5a + 20 = 440 5a = 420 a = 84

So, the least number in set x is 84.

Step 2: Find the second least number in the new set The second least number in the new set is 121 less than double the least number of set x. So, we have:

2*84 - 121 = 47

So, the second least number in the new set is 47.

Step 3: Find the sum of the new set The numbers in the new set are 46, 47, 48, 49, and 50. So, the sum is:

46 + 47 + 48 + 49 + 50 = 240

So, the sum of the new set of five consecutive integers is 240.