Question

Mark the correct alternative in each of the following:
Given that x, y and b are real numbers and x$<$y, b$>$0, then
(a) $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$b<{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}$}\right.$
(b) $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$b\le {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}$}\right.$
(c) $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$b>{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}$}\right.$
(d) $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$b\ge {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}$}\right.$

Answer 1

Given that x, y and b are real numbers and x$<$y, b$>$0.

Both sides of an inequality can be multiplied or divided by the same positive number.

$\therefore \frac{x}{b}<\frac{y}{b}$

Hence, the correct option is (a).