Question

Which of the following will be the correct ascending order for $-\frac{2}{3},-\frac{3}{5},\frac{3}{5},\frac{2}{3}$?

• A. $\frac{-3}{5}<\frac{-2}{3}<\frac{3}{5}<\frac{2}{3}$
• B. $\frac{-2}{3}<\frac{-3}{5}<\frac{3}{5}<\frac{2}{3}$
  
• C. $\frac{-2}{3}<\frac{-3}{5}<\frac{3}{5}<\frac{2}{3}$
  
• D. $\frac{-2}{3}>\frac{-3}{5}>\frac{3}{5}>\frac{2}{3}$
  

Answer 1

The correct option is B $\frac{-2}{3}<\frac{-3}{5}<\frac{3}{5}<\frac{2}{3}$


To compare two positive rational numbers, the denominators must be the same.
$\text{For}\frac{2}{3}and\frac{3}{5},\text{The LCM of their}$
$\text{denominators 3,5 is 15.}$
$\therefore \frac{2\times 5}{3\times 5}=\frac{10}{15}and\frac{3\times 3}{5\times 3}=\frac{9}{15}$
Now as their denominators are same,
$\frac{10}{15}>\frac{9}{15}\Rightarrow \frac{2}{3}>\frac{3}{5}$

Also, to compare two negative rational numbers, the denominators must be the same.
$\frac{-2}{3}and\frac{-3}{5}becomes\frac{-10}{15}and\frac{-9}{15}.$
We know that -9 > -10.
$\Rightarrow \frac{-9}{15}>\frac{-10}{15}\Rightarrow \frac{-3}{5}>\frac{-2}{3}$

Positive rational numbers are always greater than negative rational numbers.
So, the order will be
$\frac{-2}{3}<\frac{-3}{5}<\frac{3}{5}<\frac{2}{3}.$