Ans: D
Let $\{x_1,~x_2,~\ldots,~x_n\}$ be the set of numbers, where $n$ is an natural number.
Let $\{x_1,~x_2,~\ldots,~x_n\}$ be the set of numbers, where $n$ is an natural number.
The new mean
$\begin{array}{cl}
= & \displaystyle\sum_{i=1}^n \dfrac{3(x_i+5)}{n} \\
= & 3\displaystyle\left(\sum_{i=1}^n \dfrac{x_i}{n} + 5 \right) \\
= & 3 (40+5) \\
= & 135
\end{array}$
The new variance
$\begin{array}{cl}
= & \displaystyle \sum_{i=1}^n \dfrac{[3(x_i+5)-135]^2}{n} \\
= & \displaystyle \sum_{i=1}^n \dfrac{[3(x_i+5)-3(40+5)]^2}{n} \\
= & \displaystyle \sum_{i=1}^n \dfrac{9(x_i-40)^2}{n} \\
= & \displaystyle 9\left(\sum_{i=1}^n \dfrac{(x_i-40)^2}{n}\right) \\
= & 9(9) \\
= & 81
\end{array}$
Let $Q_3$ and $Q_1$ be the upper quartile and the lower quartile of the original set of number.
The new inter-quartile range
$\begin{array}{cl}
= & [3(Q_3+5)-3(Q_1+5)] \\
= & 3(Q_3-Q_1) \\
= & 3(18) \\
= & 54
\end{array}$
