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If the ratio of H.M. and G.M. of two quantities is \[12:13\], then the ratio of the number is?
D.None of these

Last updated date: 15th Jul 2024
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Hint: Here in this question, given the ratio of harmonic mean and Geometric mean we have to find the ratio of the number. As we know harmonic mean defined as \[\dfrac{{2ab}}{{\left( {a + b} \right)}}\] and geometric mean defined as \[\sqrt {ab} \] by taking the ratio of these two we can find the ratio of a and b by using a basic arithmetic operation.

Complete step-by-step answer:
Harmonic mean (H.M) is defined as the reciprocal of arithmetic mean. Arithmetic mean represents a number that is achieved by dividing the sum of the values of a set by the number of values in the set.
The Geometric Mean for a given number of values containing n observations is the nth root of the product of the values.
Let a and b be the numbers, then
\[ \Rightarrow HM = \dfrac{{2ab}}{{a + b}}\]-------(1)
\[ \Rightarrow \,\,GM = \sqrt {ab} \]
Given that If the ratio of H.M. and G.M. of two quantities is \[12:13\], then
\[ \Rightarrow \,\,\dfrac{{HM}}{{GM}} = \dfrac{{\dfrac{{2ab}}{{a + b}}}}{{\sqrt {ab} }} = \dfrac{{12}}{{13}}\]
On simplification, we have
\[ \Rightarrow \,\,\dfrac{{HM}}{{GM}} = \dfrac{{2ab}}{{\sqrt {ab} \left( {a + b} \right)}} = \dfrac{{12}}{{13}}\]
\[ \Rightarrow \,\,\dfrac{{2ab}}{{\sqrt {ab} \left( {a + b} \right)}} = \dfrac{{12}}{{13}}\]
Where, \[ab\] can be written as \[{\left( {\sqrt {ab} } \right)^2}\], then
\[ \Rightarrow \,\,\dfrac{{2{{\left( {\sqrt {ab} } \right)}^2}}}{{\sqrt {ab} \left( {a + b} \right)}} = \dfrac{{12}}{{13}}\]
On cancelling the like terms i.e., \[\sqrt {ab} \] in both numerator and denominator, then we have
\[ \Rightarrow \,\,\dfrac{{2\sqrt {ab} }}{{\left( {a + b} \right)}} = \dfrac{{12}}{{13}}\]
Divide both side by 2, then
\[ \Rightarrow \,\,\dfrac{{\sqrt {ab} }}{{\left( {a + b} \right)}} = \dfrac{6}{{13}}\]
Taking a reciprocal, we have
\[ \Rightarrow \,\,\dfrac{{a + b}}{{\sqrt {ab} }} = \dfrac{{13}}{6}\]
It can be written as
\[ \Rightarrow \,\,\dfrac{a}{{\sqrt {ab} }} + \dfrac{b}{{\sqrt {ab} }} = \dfrac{{13}}{6}\]
\[ \Rightarrow \,\,\dfrac{{{{\left( {\sqrt a } \right)}^2}}}{{\sqrt a \sqrt b }} + \dfrac{{{{\left( {\sqrt b } \right)}^2}}}{{\sqrt a \sqrt b }} = \dfrac{{13}}{6}\]
On simplification, we have
\[ \Rightarrow \,\,\dfrac{{\sqrt a }}{{\sqrt b }} + \dfrac{{\sqrt b }}{{\sqrt a }} = \dfrac{{13}}{6}\]
It can be written as
\[ \Rightarrow \,\,\sqrt {\dfrac{a}{b}} + \sqrt {\dfrac{b}{a}} = \dfrac{{13}}{6}\]
Let put \[\sqrt {\dfrac{a}{b}} = x\] then \[\sqrt {\dfrac{b}{a}} = \dfrac{1}{x}\] on substituting, we have
\[ \Rightarrow \,\,x + \dfrac{1}{x} = \dfrac{{13}}{6}\]
Taking \[x\] as LCM in LHS, then
\[ \Rightarrow \,\,\dfrac{{{x^2} + 1}}{x} = \dfrac{{13}}{6}\]
On cross multiplication, we have
\[ \Rightarrow \,\,6\left( {{x^2} + 1} \right) = 13x\]
\[ \Rightarrow \,\,6{x^2} + 6 = 13x\]
Subtract \[13x\] on both side, we have
\[ \Rightarrow \,\,6{x^2} - 13x + 6 = 0\]
Now, find a x value by factorization method, then
\[ \Rightarrow \,\,6{x^2} - 13x + 6 = 0\]
\[ \Rightarrow \,\,6{x^2} - 9x - 4x + 6 = 0\]
\[ \Rightarrow \,\,\left( {6{x^2} - 9x} \right) - \left( {4x - 6} \right) = 0\]
Taking out the GCD, then
\[ \Rightarrow \,\,3x\left( {2x - 3} \right) - 2\left( {2x - 3} \right) = 0\]
Take out \[\left( {2x - 3} \right)\] as common, then
\[ \Rightarrow \,\,\left( {2x - 3} \right)\left( {3x - 2} \right) = 0\]
Equate each factor to the zero, then
\[ \Rightarrow \,\,2x - 3 = 0\] or \[3x - 2 = 0\]
\[ \Rightarrow \,\,2x = 3\] or \[3x = 2\]
\[ \Rightarrow \,\,x = \dfrac{3}{2}\] or \[x = \dfrac{2}{3}\]
But \[\sqrt {\dfrac{a}{b}} = x\], then
\[ \Rightarrow \,\,\,\sqrt {\dfrac{a}{b}} = \dfrac{3}{2}\] or \[\dfrac{2}{3}\]
Taking square root on both sides, we have
\[ \Rightarrow \,\,\,{\left( {\sqrt {\dfrac{a}{b}} } \right)^2} = {\left( {\dfrac{3}{2}} \right)^2}\] or \[{\left( {\dfrac{2}{3}} \right)^2}\]
On cancelling the square and root, we get
\[ \Rightarrow \,\,\,\dfrac{a}{b} = \dfrac{9}{4}\] or \[\dfrac{4}{9}\]
Hence, the ratio of the number is \[9:4\] or \[4:9\].
Therefore, option (D) is correct.
So, the correct answer is “Option D”.

Note: To solve these kinds of a problem the student must know about the definition of Arithmetic mean, harmonic mean, geometric means and ratios and simplify by using a factorization, tables of multiplication and simple arithmetic operation like addition, subtraction, multiplication and division.