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candygal79
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Sum = (number of integers)(average).candygal79 wrote:The sum of the integers in list S is the same as the sum of the integers in list T. If list S and list T contain ONLY POSITIVE INTEGERS, does list S contain more integers than list T ?
1) The avg of the integers in S is less than the avg of the integers in T
2) The median of the integers in S is greater than the median of the integers in T
Thus:
Number of integers = sum/average.
Let each sum = 100.
Statement 1: The average of the integers in S is less than the average of the integers in T
Test EXTREMES.
Case 1: S's average = 1, T's average = 2
Number of integers in S = sum/average = 100/1 = 100.
Number of integers in T = sum/average = 100/2 = 50.
In this case, S has more integers.
Case 2: S's average = 50, T's average = 100
Number of integers in S = sum/average = 100/50 = 2.
Number of integers in T = sum/average = 100/100 = 1.
In this case, S has more integers.
Implication:
In every case, S will have more integers than T.
SUFFICIENT.
Statement 2: The median of the integers in S is greater than the median of the integers in T
Case 1: S = [20, 20, 20, 20, 20}, T = {1, 1, 1, 1, 96}
Here, S and T have the same number of integers.
Case 2: S = [20, 20, 20, 20, 20}, T = {1, 1, 98}
Here, S has more integers than T.
INSUFFICIENT.
The correct answer is A.
