Thanks Bunuel:) ,now i get it.
sunita123 wrote:
Set S contains 100 consecutive integers. If the range of the negative elements of Set S equals 80, what is the average (arithmetic mean) of the positive numbers in the set?
So, we have a set of 100
consecutive integers. Some of them must be negative and some not (if all of them were negative then the range of 100 consecutive negative integers would be 99, not 80). This implies that the greatest negative integer must be -1.
Now, we are told that the range of the
negative elements of Set S equals 80: {range} = {largest} - {smallest} --> 80 = -1 - {smallest} --> {smallest} = -81.
So, the set consists of 81 negative elements (from -81 to -1), 0 and 18 positive integers (from 1 to 18), the total of 81+1+18=100 consecutive integers.
The average of integers from 1 to 18 is (first + last)/2 = (1 + 18)/2 = 9.5.
Hope it's clear.
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