Problem Solving

Need help in getting to the answer on this one!

List S consists of 10 consecutive ODD integers, and 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 integers in T?

a) 2
b) 7
c) 8
d) 12
e) 22

Easiest approach is to try out with real numbers.
Least element of S = Least element of T + 7
Lets say least element of T = 2
Lets say least element of S = 2 + 7 = 9
T = {2,4,6,8,10} and Avg of T = 6
S = {9,11,13,15,17,19,21,23,25,27} and Avg of T = 18
Avg (T) - Avg (S) = 18 - 6 = 12
Ans D

srcc25anu wrote:Easiest approach is to try out with real numbers.
Least element of S = Least element of T + 7
Lets say least element of T = 2
Lets say least element of S = 2 + 7 = 9
T = {2,4,6,8,10} and Avg of T = 6
S = {9,11,13,15,17,19,21,23,25,27} and Avg of T = 18
Avg (T) - Avg (S) = 18 - 6 = 12
Ans D

Thanks alot that is a great methond!

Average of S = (5th term + 6th term)/2
Average of T = 3rd term

Now, assume that smallest integer in S and T are s and t, respectively.
So, s = t + 7

Now, 5th and 6th term of S are (s + 8) and (s + 10)
So, average of S = (2s + 18)/2 = s + 9 = t + 16

And, 3rd term of T = t + 4

Difference in average = (t + 16) - (t + 4) = 12

Answer : D