Question

The value of $\frac{{\left(0.6\right)}^0-{\left(0.1\right)}^{-1}}{{\left({\displaystyle \frac{3}{2^3}}\right)}^{-1}{\left({\displaystyle \frac{3}{2}}\right)}^3+{\left(-{\displaystyle \frac{1}{3}}\right)}^{-1}}$is

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B

$-\frac{3}{2}$

Simplification

$\frac{{\left(0.6\right)}^0-{\left(0.1\right)}^{-1}}{{\left({\displaystyle \frac{3}{2^3}}\right)}^{-1}{\left({\displaystyle \frac{3}{2}}\right)}^3+{\left(-{\displaystyle \frac{1}{3}}\right)}^{-1}}=\frac{1-{\displaystyle \frac{1}{0.1}}}{{\displaystyle \frac{3^{-1+3}}{2^{-3+3}}}+\left(-3\right)}$ $\left[\because a^0=1,a^{-x}=\frac{1}{a^x},{\left(\frac{a}{b^x}\right)}^y{\left(\frac{a}{b}\right)}^z=\frac{a^{y+z}}{b^{xy+z}}\right]$

$=\frac{1-10}{{\displaystyle \frac{3^2}{2^0}}-3}=\frac{-9}{9-3}=\frac{-9}{{\displaystyle 6}}=-\frac{3}{2}$

The value of $\frac{{\left(0.6\right)}^0-{\left(0.1\right)}^{-1}}{{\left({\displaystyle \frac{3}{2^3}}\right)}^{-1}{\left({\displaystyle \frac{3}{2}}\right)}^3+{\left(-{\displaystyle \frac{1}{3}}\right)}^{-1}}$is $-\frac{3}{2}.$

Hence, option (B) is correct.