A) \[1:2\]
B) \[2:1\]
C) \[4:1\]
D) \[1:4\]
Correct Answer: D
Solution :
We have H.M. =\[\frac{2ab}{a+b}\] and G.M. \[=\sqrt{ab}\] So \[\frac{H.M.}{G.M.}=\frac{4}{5}\]\[\Rightarrow \]\[\frac{2ab/(a+b)}{\sqrt{ab}}=\frac{4}{5}\] \[\Rightarrow \] \[\frac{2\sqrt{ab}}{(a+b)}=\frac{4}{5}\]\[\Rightarrow \]\[\frac{a+b}{2\sqrt{ab}}=\frac{5}{4}\] \[\Rightarrow \] \[\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{5+4}{5-4}\]\[\Rightarrow \]\[\frac{{{(\sqrt{a}+\sqrt{b})}^{2}}}{{{(\sqrt{a}-\sqrt{b})}^{2}}}=\frac{9}{1}\] \[\Rightarrow \] \[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{3}{1}\]\[\Rightarrow \]\[\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{3+1}{3-1}\] \[\Rightarrow \] \[\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{4}{2}\]\[\Rightarrow \]\[\left( \frac{a}{b} \right)={{2}^{2}}=4\] \[\Rightarrow \] \[a:b=4:1\] or\[b:a=1:4\]. Aliter: Let the numbers be in the ratio \[\lambda :1\] and let they be \[\lambda a\] and \[a\] Then \[\frac{2(\lambda a)a}{\lambda a+a}.\frac{1}{\sqrt{\lambda a\ .\ a}}=\frac{4}{5}\]\[\Rightarrow \]\[\frac{\sqrt{\lambda }}{\lambda +1}=\frac{2}{5}\] \[\Rightarrow \] \[25\lambda =4({{\lambda }^{2}}+2\lambda +1)\]\[\Rightarrow \]\[(\lambda -4)(4\lambda -1)=0\] \[\Rightarrow \] \[\lambda =4\] or \[\lambda =\frac{1}{4}\].
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