

InterviewSolution
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Verify whether the given statement is true or false:(i) \(\frac{13}{5}\div\frac{26}{10}= \frac{26}{10}\div\frac{13}{5}\)(ii) \(-9\div\frac{3}{4}= \frac{3}{4}\div(-9)\)(iii) \(\frac{-8}{9}\div\frac{-4}{3}= \frac{-4}{3}\div\frac{-8}{9}\)(iv) \(\frac{-7}{24}\div\frac{3}{-16}= \frac{3}{-16}\div\frac{-7}{24}\) |
Answer» (i) \(\frac{13}{5} \div\frac{26}{10}= \frac{26}{10}\div\frac{13}{5}\) LHS = \(\frac{13}{5}\div\frac{26}{10}\) = \(\frac{13}{5}\times\frac{10}{26}\) = \(\frac{13\times10}{5\times26}\) = \(\frac{130}{130}= 1\) RHS = \(\frac{26}{10}\div\frac{13}{5}\) = \(\frac{26}{10}\times\frac{5}{13}\) = \(\frac{26\times5}{10\times13}\) = \(\frac{130}{130}=1\) Since, RHS = LHS Therefore, True (ii) \(-9\div\frac{3}{4}=\frac{3}{4}(-9)\) LHS = \(-9\div\frac{4}{3}\) = \(-9\times\frac{4}{3}\) = \(\frac{-9\times4}{3}\) = \(\frac{-36}{3}=-12\) RHS = \(\frac{3}{4}\div(-9)\) = \(\frac{3}{4}\times\frac{1}{-9}\) = \(\frac{3\times1}{4\times-9}\) = \(\frac{3}{-36}=\frac{-1}{12}\) Since, RHS ≠ LHS Therefore, False (iii) \(\frac{-8}{9}\div\frac{-4}{3}= \frac{-4}{3}\div\frac{-8}{9}\) LHS = \(\frac{-8}{9}\div\frac{-4}{3}\) = \(\frac{-8}{9}\times\frac{3}{-4}\) = \(\frac{-8\times3}{9\times-4}\) = \(\frac{-24}{-36} =\frac{2}{3}\) RHS = \(\frac{-4}{3}\div\frac{-8}{9}\) = \(\frac{-4}{9}\times\frac{9}{-8}\) = \(\frac{-4\times9}{3\times-8}\) = \(\frac{-36}{-24}=\frac{3}{2}\) Since, RHS ≠ LHS (iv) \(\frac{-7}{24}\div\frac{3}{-16}=\frac{3}{-16}\div\frac{-7}{24}\) LHS = \(\frac{-7}{24}\div\frac{3}{-16}\) = \(\frac{-7}{24}\times\frac{-16}{3}\) = \(\frac{-7\times-16}{24\times3}\) = \(\frac{112}{72}=\frac{14}{9}\) RHS = \(\frac{3}{-16}\div\frac{-7}{24}\) = \(\frac{3}{-16}\times\frac{24}{-7}\) = \(\frac{3\times24}{-16\times-7}\) = \(\frac{72}{112}=\frac{9}{14}\) Since, RHS ≠ LHS Therefore, False |
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