Lists S and T consist of the same number of positive integers. Is the

Original question is:

Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?

Q: is \(Median\{S\}>Mean\{T\}\)? Given: {# of terms in S}={# of terms in T}, let's say \(N\).

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.

From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is \(Mean\{S\}>Mean\{T\}\)? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}.

(2) The sum of the integers in S is greater than the sum of the integers in T.

\(Sum\{S\}>Sum\{T\}\). Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1).

(1)+(2) From (1) question became is \(Mean\{S\}>Mean\{T\}\)? --> As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is \(\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}\) true? --> Is \(Sum\{S\}>Sum\{T\}\)? This is exactly what is said in statement (2) that \(Sum\{S\}>Sum\{T\}\). Hence sufficient.

Answer: C.