UPSC NDA I 2022 Mathematics Solved Paper

Concept:The mean $\bar{x}$ of grouped data is $\frac{\sum f_i x_i}{\sum f_i}$.When every frequency $f_i$ is multiplied by the same constant, both numerator and denominator of the mean are multiplied by that constant, leaving the ratio unchanged.Explanation:Let original frequencies be $f_1, f_2, \dots, f_n$ and class marks be $x_1, x_2, \dots, x_n$.Original mean: $\bar{x} = \frac{f_1 x_1 + f_2 x_2 + \dots + f_n x_n}{f_1 + f_2 + \dots + f_n}$.If each frequency is doubled, new frequencies are $2f_1, 2f_2, \dots, 2f_n$.New mean: $\bar{x}_{\text{new}} = \frac{2f_1 x_1 + 2f_2 x_2 + \dots + 2f_n x_n}{2f_1 + 2f_2 + \dots + 2f_n}$.Factor $2$ from numerator and denominator: $\bar{x}_{\text{new}} = \frac{2( f_1 x_1 + f_2 x_2 + \dots + f_n x_n)}{2( f_1 + f_2 + \dots + f_n)} = \frac{f_1 x_1 + f_2 x_2 + \dots + f_n x_n}{f_1 + f_2 + \dots + f_n} = \bar{x}$.Thus, doubling all frequencies does not alter the mean.From the previous part of the problem, the original mean was $50$.Hence, after doubling each class frequency, the mean remains $50$.Answer:$50$ (Option B).