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Hint: We will compare the ratio between Harmonic mean and Geometric mean by using their general equations and try to solve this question.

Complete step-by-step Solution:

We can use the below formulas to solve the question.

Harmonic mean $=\dfrac{2ab}{a+b}$

Geometric mean $=\sqrt{ab}$

It is given in the question that the harmonic mean between two positive numbers to their geometric mean is 12:13.

We have to find the ratio of the numbers.

Let us assume the two positive numbers as a, b.

Let us consider the case of harmonic mean first. The formula for the harmonic mean for two numbers a and b is given by,

Harmonic mean $=\dfrac{2ab}{a+b}$

Now, let us consider the case of geometric mean. The formula for the geometric mean of two numbers a and b is given by,

Geometric mean $=\sqrt{ab}$

So, from question, we have;

$\dfrac{\text{Harmonic mean}}{\text{Geometric mean}}=\dfrac{12}{13}$

Substituting the known formulas of harmonic mean and geometric mean in the above equation we get,

$\dfrac{\dfrac{2ab}{a+b}}{\sqrt{ab}}=\dfrac{12}{13}$

Multiplying and dividing the LHS with $\sqrt{ab}$, we get,

$\dfrac{2\sqrt{ab}}{a+b}=\dfrac{12}{13}$

Taking 2 to RHS and simplifying further, we get,

$\begin{align}

& \Rightarrow \dfrac{\sqrt{ab}}{a+b}=\dfrac{12}{2\times 13} \\

& \Rightarrow \dfrac{\sqrt{ab}}{a+b}=\dfrac{6}{13} \\

& \Rightarrow \dfrac{a+b}{\sqrt{ab}}=\dfrac{13}{6} \\

& \Rightarrow \dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}=\dfrac{13}{6} \\

\end{align}$

Simplify the terms on LHS using the roots, we get,

$\Rightarrow \sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}=\dfrac{13}{6}.............\left( 1 \right)$

Let us assume $x=\sqrt{\dfrac{a}{b}}$

Substituting $x=\sqrt{\dfrac{a}{b}}$ in equation (1), we get,

$\begin{align}

& \Rightarrow x+\dfrac{1}{x}=\dfrac{13}{6}............\left( 2 \right) \\

& \Rightarrow \dfrac{{{x}^{2}}+1}{x}=\dfrac{13}{6} \\

\end{align}$

On cross multiplying, we get,

$\begin{align}

& 6{{x}^{2}}+6=13x \\

& \Rightarrow 6{{x}^{2}}-13x+6=0 \\

\end{align}$

We have obtained a quadratic equation. We can solve the same using the middle term split method as shown below,

$\begin{align}

& \Rightarrow 6{{x}^{2}}-9x-4x+6=0 \\

& \Rightarrow 6x\left( x-\dfrac{3}{2} \right)-4\left( x-\dfrac{3}{2} \right)=0 \\

\end{align}$

Taking $\left( x-\dfrac{3}{2} \right)$ common from the above equation, we get,

$\Rightarrow \left( 6x-4 \right)\left( x-\dfrac{3}{2} \right)=0$

Hence, we have either, $6x-4=0\ \ or\ \ x-\dfrac{3}{2}=0$

$\begin{align}

& \Rightarrow x=\dfrac{4}{6}\ or\ x=\dfrac{3}{2} \\

& \Rightarrow x=\dfrac{2}{3}\ or\ x=\dfrac{3}{2} \\

\end{align}$

So, $x=\dfrac{3}{2}\ or\ x=\dfrac{2}{3}$

Since we had assumed $x=\sqrt{\dfrac{a}{b}}$. We can now find the ratio of numbers.

Therefore, we have $x=\sqrt{\dfrac{a}{b}}=\dfrac{3}{2}\ \ or\ \ x=\sqrt{\dfrac{a}{b}}=\dfrac{2}{3}\ $

We can take the square and write,

$\begin{align}

& {{x}^{2}}=\dfrac{a}{b}=\dfrac{9}{4} \\

& or \\

& {{x}^{2}}=\dfrac{a}{b}=\dfrac{4}{9} \\

\end{align}$

So, the ratio can be $\dfrac{9}{4}\ \ or\ \ \dfrac{4}{9}$.

Now, in option only $\dfrac{4}{9}$ is given. So, we will consider $\dfrac{4}{9}$ as the correct answer.

Therefore, option (C) 4:9 is the correct option.

Note: There is an alternate method to solve this question. It is shown as below,

Given,

$\begin{align}

& \dfrac{HM}{GM}=\dfrac{12}{13} \\

& \Rightarrow \dfrac{\dfrac{2ab}{a+b}}{\sqrt{a.b}}=\dfrac{12}{13} \\

\end{align}$

Squaring and simplifying both sides, we get,

$\begin{align}

& \Rightarrow \dfrac{ab}{{{\left( a+b \right)}^{2}}}=\dfrac{36}{169} \\

& \Rightarrow \dfrac{ab}{{{a}^{2}}+{{b}^{2}}+2ab}=\dfrac{36}{169} \\

& \Rightarrow 169ab=36{{a}^{2}}+36{{b}^{2}}+72ab \\

& \Rightarrow 36{{a}^{2}}+36{{b}^{2}}-97ab=0 \\

& \Rightarrow 36{{a}^{2}}-16ab-81ab+36{{b}^{2}}=0 \\

& \Rightarrow \left( 9a-4b \right)\left( 4a+9b \right)=0 \\

\end{align}$

Taking positive value,

$\begin{align}

& \Rightarrow \left( 9a-4b \right)=0 \\

& \Rightarrow 9a=4b \\

& \Rightarrow \dfrac{a}{b}=\dfrac{4}{9} \\

& \Rightarrow a:b=4:9 \\

\end{align}$

Complete step-by-step Solution:

We can use the below formulas to solve the question.

Harmonic mean $=\dfrac{2ab}{a+b}$

Geometric mean $=\sqrt{ab}$

It is given in the question that the harmonic mean between two positive numbers to their geometric mean is 12:13.

We have to find the ratio of the numbers.

Let us assume the two positive numbers as a, b.

Let us consider the case of harmonic mean first. The formula for the harmonic mean for two numbers a and b is given by,

Harmonic mean $=\dfrac{2ab}{a+b}$

Now, let us consider the case of geometric mean. The formula for the geometric mean of two numbers a and b is given by,

Geometric mean $=\sqrt{ab}$

So, from question, we have;

$\dfrac{\text{Harmonic mean}}{\text{Geometric mean}}=\dfrac{12}{13}$

Substituting the known formulas of harmonic mean and geometric mean in the above equation we get,

$\dfrac{\dfrac{2ab}{a+b}}{\sqrt{ab}}=\dfrac{12}{13}$

Multiplying and dividing the LHS with $\sqrt{ab}$, we get,

$\dfrac{2\sqrt{ab}}{a+b}=\dfrac{12}{13}$

Taking 2 to RHS and simplifying further, we get,

$\begin{align}

& \Rightarrow \dfrac{\sqrt{ab}}{a+b}=\dfrac{12}{2\times 13} \\

& \Rightarrow \dfrac{\sqrt{ab}}{a+b}=\dfrac{6}{13} \\

& \Rightarrow \dfrac{a+b}{\sqrt{ab}}=\dfrac{13}{6} \\

& \Rightarrow \dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}=\dfrac{13}{6} \\

\end{align}$

Simplify the terms on LHS using the roots, we get,

$\Rightarrow \sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}=\dfrac{13}{6}.............\left( 1 \right)$

Let us assume $x=\sqrt{\dfrac{a}{b}}$

Substituting $x=\sqrt{\dfrac{a}{b}}$ in equation (1), we get,

$\begin{align}

& \Rightarrow x+\dfrac{1}{x}=\dfrac{13}{6}............\left( 2 \right) \\

& \Rightarrow \dfrac{{{x}^{2}}+1}{x}=\dfrac{13}{6} \\

\end{align}$

On cross multiplying, we get,

$\begin{align}

& 6{{x}^{2}}+6=13x \\

& \Rightarrow 6{{x}^{2}}-13x+6=0 \\

\end{align}$

We have obtained a quadratic equation. We can solve the same using the middle term split method as shown below,

$\begin{align}

& \Rightarrow 6{{x}^{2}}-9x-4x+6=0 \\

& \Rightarrow 6x\left( x-\dfrac{3}{2} \right)-4\left( x-\dfrac{3}{2} \right)=0 \\

\end{align}$

Taking $\left( x-\dfrac{3}{2} \right)$ common from the above equation, we get,

$\Rightarrow \left( 6x-4 \right)\left( x-\dfrac{3}{2} \right)=0$

Hence, we have either, $6x-4=0\ \ or\ \ x-\dfrac{3}{2}=0$

$\begin{align}

& \Rightarrow x=\dfrac{4}{6}\ or\ x=\dfrac{3}{2} \\

& \Rightarrow x=\dfrac{2}{3}\ or\ x=\dfrac{3}{2} \\

\end{align}$

So, $x=\dfrac{3}{2}\ or\ x=\dfrac{2}{3}$

Since we had assumed $x=\sqrt{\dfrac{a}{b}}$. We can now find the ratio of numbers.

Therefore, we have $x=\sqrt{\dfrac{a}{b}}=\dfrac{3}{2}\ \ or\ \ x=\sqrt{\dfrac{a}{b}}=\dfrac{2}{3}\ $

We can take the square and write,

$\begin{align}

& {{x}^{2}}=\dfrac{a}{b}=\dfrac{9}{4} \\

& or \\

& {{x}^{2}}=\dfrac{a}{b}=\dfrac{4}{9} \\

\end{align}$

So, the ratio can be $\dfrac{9}{4}\ \ or\ \ \dfrac{4}{9}$.

Now, in option only $\dfrac{4}{9}$ is given. So, we will consider $\dfrac{4}{9}$ as the correct answer.

Therefore, option (C) 4:9 is the correct option.

Note: There is an alternate method to solve this question. It is shown as below,

Given,

$\begin{align}

& \dfrac{HM}{GM}=\dfrac{12}{13} \\

& \Rightarrow \dfrac{\dfrac{2ab}{a+b}}{\sqrt{a.b}}=\dfrac{12}{13} \\

\end{align}$

Squaring and simplifying both sides, we get,

$\begin{align}

& \Rightarrow \dfrac{ab}{{{\left( a+b \right)}^{2}}}=\dfrac{36}{169} \\

& \Rightarrow \dfrac{ab}{{{a}^{2}}+{{b}^{2}}+2ab}=\dfrac{36}{169} \\

& \Rightarrow 169ab=36{{a}^{2}}+36{{b}^{2}}+72ab \\

& \Rightarrow 36{{a}^{2}}+36{{b}^{2}}-97ab=0 \\

& \Rightarrow 36{{a}^{2}}-16ab-81ab+36{{b}^{2}}=0 \\

& \Rightarrow \left( 9a-4b \right)\left( 4a+9b \right)=0 \\

\end{align}$

Taking positive value,

$\begin{align}

& \Rightarrow \left( 9a-4b \right)=0 \\

& \Rightarrow 9a=4b \\

& \Rightarrow \dfrac{a}{b}=\dfrac{4}{9} \\

& \Rightarrow a:b=4:9 \\

\end{align}$

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