Question

# Evaluate: (i) $\frac{-3}{5}×\frac{10}{7}$ (ii) ${\left(\frac{-5}{8}\right)}^{-1}$ (iii) ${\left(-6\right)}^{-1}$

Solution

## $\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\frac{-3}{5}×\frac{10}{7}\phantom{\rule{0ex}{0ex}}=\frac{-3×10}{5×7}\phantom{\rule{0ex}{0ex}}=\frac{-30}{35}\phantom{\rule{0ex}{0ex}}=\frac{-6×5}{7×5}\phantom{\rule{0ex}{0ex}}=\frac{-6}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{-5}{8}{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{-5}{8}\right)}\phantom{\rule{0ex}{0ex}}=1×\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8×-1}{-5×-1}\phantom{\rule{0ex}{0ex}}=\frac{-8}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\left(-6{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{-6}\phantom{\rule{0ex}{0ex}}=\frac{1×-1}{-6×-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{6}\phantom{\rule{0ex}{0ex}}$MathematicsRS Aggarwal (2016)Standard VIII

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