Problem Solving

theachiever wrote:List S consists of 10 consecutive odd integers

List T consists of 5 consecutive even integers.

If the least integer in S is 7 more than the least integer in T,how much greater is the average(arithmetic mean) of the integers in S than the average of the integers in T?

• A.2
  B.7
  C.8
  D.12
  E.22

Let the list S be 2n + 1, 2n + 3, 2n + 5,.....,2n + 19 where n is an integer.
Let list T be 2k, 2k + 2, 2k + 4,...2k + 8.
Now 2n+1 - 2k = 7. Or 2n - 2k = 6.
Also, average of the integers in S is (4n + 20)/2 = 2n + 10.
Average of integers in T is 2k + 4.
So, difference is (2n + 10) - (2k + 4) = 2n - 2k + 6 = 12

The correct answer is D.