The arrangement of rational numbers $\frac{-7}{10}, \frac{5}{-8}, \frac{2}{-3}$ in ascending order is
(a)$\frac{2}{-3}, \frac{5}{-8}, \frac{-7}{10}$
(b)$\frac{5}{-8}, \frac{-7}{10}, \frac{2}{-3}$
(c)$\frac{-7}{10}, \frac{5}{-8}, \frac{2}{-3}$
(d)$\frac{-7}{10}, \frac{2}{-3}, \frac{5}{-8}$
Answer
Answer (as printed): D
Explanation
$\frac{-7}{10}=-0.7, \frac{5}{-8}=-\frac{5}{8}=-0.625, \frac{2}{-3}=-\frac{2}{3}=-0.66$. Since $-0.7<-0.66<-0.625$, so $\frac{-7}{10}<\frac{2}{-3}<\frac{5}{-8}$.
Explanation as extracted from the printed page; notation may be imperfect.