A) \[\frac{-108}{10}\]
B) \[\frac{62}{5}\]
C) \[\frac{51}{5}\]
D) \[\frac{17}{33}\]
Correct Answer: C
Solution :
(c) \[\frac{-17}{18}\times -3\times \frac{108}{\underset{10}{\mathop{\bcancel{30}}}\,}=\frac{-17}{\bcancel{18}}\times (-1)\times \frac{\bcancel{{{108}^{6}}}}{10}\] \[=17\times \frac{\bcancel{{{6}^{3}}}}{\underset{5}{\mathop{\bcancel{10}}}\,}=\frac{51}{5}\]
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