We need to compare the mean of $\(t, u \& v\)$, where $\(0<t<u<v\)$ with the median of $\(t, u \& v\)$.


As nothing is said about the values of \($\mathrm{t}, \mathrm{u}\)$ and v , a unique comparison cannot be formed.



For example if we take $\(\mathrm{t}=2, \mathrm{u}=4 \& \mathrm{v}=6\)$, we get mean $\(=\frac{2+4+6}{3}=\frac{12}{3}=4=\)$ median, so column A gets same quantity as column B (When numbers are consecutive/in arithmetic sequence, mean comes same as median)

But if we take $\(t=2, u=12 \& v=14\)$, we get mean $\(=\frac{2+12+13}{3}=\frac{27}{3}=9\)$ (= column A quantity) and \(median =12\)$ (= column B quantity). So, in this case column B gets higher value than column A.

Hence the answer is (D).