Variance Questions

Suppose a population $A$ has $100$ observations $101,102,\dots ,200$, and another population $B$ has $100$ observations $151,152,\dots ,250$. If $V_A$ and $V_B$ represent the variances of the two populations, respectively, then $\frac{V_A}{V_B}$ is:

Let $x_1,x_2,\dots ,x_n$ be $n$ observations such that $\sum x_i^2=400$ and $\sum x_i=80$. Then a possible value of $n$ among the following is:

In an experiment with 15 observations on $x$, the following results were available: $\sum x^2=2830$, $\sum x=170$. One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is:

The mean of the numbers $a,b,8,5,10$ is $6$ and the variance is $6.80$. Then which one of the following gives possible values of $a$ and $b$?

1: The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$.
2: The sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$ and the sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$.

Let $x_1,x_2,\dots ,x_n$ be $n$ observations, and let $\bar{x}$ be their arithmetic mean and ${\sigma }^2$ be their variance.Statement 1: Variance of $2x_1,2x_2,\dots ,2x_n$ is $4{\sigma }^2$.Statement 2: Arithmetic mean of $2x_1,2x_2,\dots ,2x_n$ is $4\bar{x}$.