We have the frequency of numbers $\(30,40 \& 50\)$ as $\(60,70 \& 80\)$ respectively so we get the arithmetic mean of the numbers as $\(\frac{30 \times 60+40 \times 70+50 \times 80}{60+70+80}=\frac{8600}{210} \approx 41\)$


Next as the median of the set of 210 terms is the average of middle two terms i.e. $\(105^{\text {th \& 106^{\text {th \)$ terms, so we get the median as $\(\frac{40+40}{2}=40$ (The set of numbers is $30,30,30, \ldots 60$ tines...30, 40 , $40,40, \ldots 70\)$ times $\(\ldots, 40\)$ etc, so the $\(105^{\text {th \&\)$ the $\(106^{\text {th \)$ terms are 40 each)

Hence column B (= Mean) has higher quantity when compared with column A (= Median), so the answer is (B).