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by theachiever » Thu Dec 13, 2012 11:10 pm
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
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by Anurag@Gurome » Thu Dec 13, 2012 11:34 pm
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.
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by puneetkhurana2000 » Thu Dec 13, 2012 11:49 pm
Take T as (2,4,6,8,10) and S as (9,11,13,15,17,19,21,23,25,27)

Avg of T is (2+10)/2 = 6 and Avg of S is (9+27)/2 = 18

Answer is 18 - 6 = 12.
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