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?
Correct answer: B. 240
- A. 248
- B. 240
- C. 228
- D. 236
- E. None of these
Explanation
Let the least number in the five consecutive even numbers be x. Then 5x + 20 = 440, so x = 84; the second least number of the different set is 2 x 84 - 121 = 47, making its five consecutive integers 46, 47, 48, 49 and 50. Their sum is 46 + 47 + 48 + 49 + 50 = 240. A likely choice is option a, 248, from using the five even numbers rather than constructing the second set of consecutive integers.
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