# If the ratio of H.M. and G.M. of two quantities is \[12:13\], then the ratio of the number is?

A.\[1:2\]

B.\[2:3\]

C.\[3:4\]

D.None of these

Last updated date: 22nd Mar 2023

•

Total views: 206.1k

•

Views today: 4.84k

Answer

Verified

206.1k+ views

**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)

And

\[ \Rightarrow \,\,GM = \sqrt {ab} \]

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India