didieravoaka wrote:The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
I got A. Is this right?
It sure is.
S1: We know the following equation: Average * Number Terms = Sum. Holding the sum constant, the average and the number of terms will have an inverse relationship. As the average gets smaller, the number of terms will grow larger, and vice versa. So if we know that if the average is smaller in S, and the two sums are equal, then there must be more terms in S to compensate for this smaller average. 1 alone is sufficient.
Use simple sets to test statement 2.
Say the sum for both sets is 0.
Case 1
S: {-1,0,1} Median = 0
T:{-1, -1, 2} Median = -1
Here S does not contain more elements than T, so the answer is NO.
Case 2:
S: {-1,0,0,1} Median = 0
T:{-1, -1, 2} Median = -1
Here S does contain more elements than T, so the answer is YES.
If you can get a YES and a NO, statement 2 is not sufficient.
And, as you saw, the answer is
A
