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?

A. 248
B. 240
C. 228
D. 236
E. None of these

Explanation:

Let the five consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1) and 2(x + 2)
Their sum = 10x = 440
x = 44 => 2(x – 2) = 84
Second least number of the other set = 2(84) – 121 = 47
This set has its least number as 46.
Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50
= 48 – 2 + 48 – 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240