| Column - I | Column - II |
| (P) \[{{\left( {{6}^{-1}}+{{\left( \frac{3}{2} \right)}^{-1}} \right)}^{-1}}\] | (i) \[-\frac{4}{13}\] |
| (Q) \[{{\left\{ {{\left( \frac{4}{3} \right)}^{-1}}{{\left( \frac{1}{4} \right)}^{-1}} \right\}}^{-1}}\] | (ii) \[\frac{9}{32}\] |
| (R) \[\left[ {{\left( \frac{1}{3} \right)}^{-3}}-{{\left( \frac{1}{2} \right)}^{-3}} \right]\div {{\left( \frac{1}{4} \right)}^{-3}}\] | (iii) \[\frac{6}{5}\] |
| (S) \[({{3}^{-1}}\times {{4}^{-1}})\times {{\left( \frac{2}{3} \right)}^{-3}}\] | (iv) \[\frac{19}{64}\] |
A) (P)\[\to \](iii); (Q)\[\to \](i); (R)\[\to \](iv); (S)\[\to \](ii)
B) (P)\[\to \](iv): (Q)\[\to \](i); (R)\[\to \](ii); (S)\[\to \](iii)
C) (P)\[\to \](ii); (Q)\[\to \](iii); (R)\[\to \](iv); (S)\[\to \](i)
D) (P)\[\to \](iii); (Q)\[\to \](i); (R)\[\to \](ii); (S)\[\to \](iv)
Correct Answer: A
Solution :
(P) We have, \[{{\left( {{6}^{-1}}+{{\left( \frac{3}{2} \right)}^{-1}} \right)}^{-1}}={{\left( \frac{1}{6}+\frac{2}{3} \right)}^{-1}}\] \[={{\left( \frac{1+4}{6} \right)}^{-1}}={{\left( \frac{5}{6} \right)}^{-1}}=\frac{6}{5}\] (Q) We have, \[{{\left\{ {{\left( \frac{4}{3} \right)}^{-1}}-{{\left( \frac{1}{4} \right)}^{-1}} \right\}}^{-1}}={{\left( \frac{3}{4}-\frac{4}{1} \right)}^{-1}}\] \[={{\left( \frac{3-16}{4} \right)}^{-1}}={{\left( \frac{-13}{4} \right)}^{-1}}=\frac{-4}{13}\] (R) We have, \[\left[ {{\left( \frac{1}{3} \right)}^{-3}}-{{\left( \frac{1}{2} \right)}^{-3}} \right]\div {{\left( \frac{1}{4} \right)}^{-3}}\] \[=({{3}^{3}}-{{2}^{3}})\div {{4}^{3}}=(27-8)\div 64=\frac{19}{64}\] (S) We have, \[({{3}^{-1}}\times {{4}^{-1}})\times {{\left( \frac{2}{3} \right)}^{-3}}\] \[=\left( \frac{1}{3}\times \frac{1}{4} \right)\times {{\left( \frac{3}{2} \right)}^{3}}=\frac{1}{12}\times \frac{27}{8}=\frac{9}{32}\]
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