Question

Arrange the following rational numbers in ascending order:
$\frac{-7}{10},\frac{-5}{8},\frac{-3}{4}$

• A. $\frac{-7}{10}<\frac{-3}{4}<\frac{-5}{8}$
• B. $\frac{-7}{10}<\frac{-5}{8}<\frac{-3}{4}$
• C. $\frac{-3}{4}<\frac{-7}{10}<\frac{-5}{8}$
• D. $\frac{-5}{8}<\frac{-7}{10}<\frac{-3}{4}$

Answer 1

The correct option is C $\frac{-3}{4}<\frac{-7}{10}<\frac{-5}{8}$

The given rational numbers are, $\frac{-7}{10},\frac{-5}{8},\frac{-3}{4}$
We now convert these numbers to equivalent rational numbers, so that they have common denominator.
For this, we take the LCM of 10, 8 and 4.
The LCM is 40.
Therefore, we have
$\frac{-7}{10}=\frac{-7\times 4}{10\times 4}=\frac{-28}{40}\frac{-5}{8}=\frac{-5\times 5}{8\times 5}=\frac{-25}{40}\frac{-3}{4}=\frac{-3\times 10}{4\times 10}=\frac{-30}{40}$
Comparing the numerators of these numbers, we find that
–30 < –28 < –25.
Therefore, the ascending order is : $\frac{-30}{40}<\frac{-28}{40}<\frac{-25}{40}or\frac{-3}{4}<\frac{-7}{10}<\frac{-5}{8}$