0 votes 0 votes If $q^{-a}=\displaystyle{\frac{1}{r}}$ and $r^{-b}=\displaystyle{\frac{1}{s}}$ and $s^{-c}=\displaystyle{\frac{1}{q}}$, the value of $abc$ is $(rqs)^{-1}$ $0$ $1$ $r+q+s$ Quantitative Aptitude gateme-2016-set1 numerical-ability algebra + – Arjun asked Feb 24, 2017 • recategorized Mar 5, 2021 by Lakshman Bhaiya ♦Arjun 28.5k points answer See all 0 reply