Question

Simplify: $\left(\begin{array}{c}\frac{-3}{2}\times \frac{4}{5}\end{array}\right)+\left(\begin{array}{c}\frac{9}{5}\times \frac{-10}{3}\end{array}\right)-\left(\begin{array}{c}\frac{1}{2}\times \frac{3}{4}\end{array}\right)$

• A. $\frac{-303}{40}$
• B. $\frac{31}{40}$
• C. $\frac{303}{80}$
• D. $\frac{31}{80}$

Answer 1

The correct option is A $\frac{-303}{40}$

$\left(\begin{array}{c}\frac{-3}{2}\times \frac{4}{5}\end{array}\right)+\left(\begin{array}{c}\frac{9}{5}\times \frac{-10}{3}\end{array}\right)-\left(\begin{array}{c}\frac{1}{2}\times \frac{3}{4}\end{array}\right)$

$=\frac{-3\times 4}{2\times 5}+\frac{9\times (-10)}{5\times 3}-\frac{1\times 3}{2\times 4}$

$=\frac{-3\times 2}{1\times 5}+\frac{3\times (-2)}{1\times 1}-\frac{3}{8}=\frac{-6}{5}+\frac{-6}{1}-\frac{3}{8}$

$=\frac{-6}{5}+\frac{-6}{1}+\frac{-3}{8}$

$=\frac{(-6)\times 8+(-6)\times 40+(-3)\times 5}{40}$

$=\frac{-48+(-240)+(-15)}{40}=\frac{-303}{40}$