List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in Sis 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?
Sol: Let List T has the following members : 2,4,6,8 and 10
Then S has : 9,11,13,15,17,19,21,23,25,27
Now If we find the average of List T is 6 and average of List S is (19+17)/2 =18
So Ans is 12.
Suppose if we S also had 5 members and all the other condition remains same then Average of S would have been 13 and diferecne between the 2 would be 7 cause when the same number is added/subtracted from a given set then the average of the new set increases or decreases by the same number
So ans is D.
Average difficulty level of 650 is okay
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