If the ratio of two numbers is 9:1,then the r | vedclass

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(B) Let the two numbers be $a$ and $b$. Given $\frac{a}{b} = \frac{9}{1}$,so $a = 9b$.The geometric mean $G$ is given by $G = \sqrt{ab}$.The harmonic

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$5:3$

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(B) Let the two numbers be $a$ and $b$. Given $\frac{a}{b} = \frac{9}{1}$,so $a = 9b$.The geometric mean $G$ is given by $G = \sqrt{ab}$.The harmonic mean $H$ is given by $H = \frac{2ab}{a+b}$.Substituting $a = 9b$ into the expressions:$G = \sqrt{9b \cdot b} = \sqrt{9b^2} = 3b$.$H = \frac{2(9b)(b)}{9b + b} = \frac{18b^2}{10b} = \frac{9b}{5}$.The ratio of the geometric mean to the harmonic mean is $\frac{G}{H} = \frac{3b}{\frac{9b}{5}} = 3b \cdot \frac{5}{9b} = \frac{15}{9} = \frac{5}{3}$.Thus,the ratio is $5:3$.

If the ratio of two numbers is $9:1$,then the ratio of their geometric mean to their harmonic mean is

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