| Column - I | Column - II |
| (P) Product of a rational number and its reciprocal is | (i) -1 |
| (Q) If \[\frac{12}{30}\] and \[\frac{x}{5}\] are equivalent, then \[x=\] | (ii) 0 |
| (R) \[\left[ \frac{8}{21}\div \left( \frac{-32}{39}\div \frac{16}{13} \right) \right]\times \frac{7}{4}=\] | (iii) 2 |
| (S) Sum of a rational number and its additive inverse is | (iv) 1 |
A) (P)\[\to \](iv): (Q)\[\to \](iii); (R)\[\to \](i); (S)\[\to \](ii)
B) (P)\[\to \](i); (Q)\[\to \](iii): (R)\[\to \](iv); (S)\[\to \](ii)
C) (P)\[\to \](iv): (Q)\[\to \](iii); (R)\[\to \](ii): (S)\[\to \](i)
D) (P)\[\to \](i); (Q)\[\to \](iv); (R)\[\to \](iii); (S)\[\to \](ii)
Correct Answer: A
Solution :
(P) Product of a rational number and Its reciprocal is 1. \[\text{(Q)}\frac{12}{30}=\frac{x}{5}\times \frac{6}{6}\Rightarrow \frac{12}{30}=\frac{6x}{30}\Rightarrow x=2\] \[\text{(R)}\] We have, \[\left[ \frac{8}{21}\div \left( \frac{-32}{39}\div \frac{16}{13} \right) \right]\times \frac{7}{4}\] \[=\left[ \frac{8}{21}\div \left( \frac{-32}{39}\times \frac{13}{16} \right) \right]\times \frac{7}{4}=\left[ \frac{8}{21}\div \left( \frac{-2}{3} \right) \right]\times \frac{7}{4}\] \[=\left[ \frac{8}{21}\times \frac{3}{-2} \right]\times \frac{7}{4}=\frac{-4}{7}\times \frac{7}{4}=-1\] (S) Sum of a rational number and its additive inverse is 0.
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