Question

If the ratio of H.M to be G.M of two nos. is $6:10$ then the ratio of the nos is

Answer 1

To answer this question, we first define what H.M and G.M are. Then they are represented in equation form: $GM=\sqrt{a\times b}$ and $HM=\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}.$ Here a and b are two numbers. We then take the ratio of these formulae and equate it to the ratio of $6:10$ or $\dfrac{6}{10}.$ We then simplify and solve for the ratio of the numbers $\dfrac{a}{b}.$

Complete step by step solution:
We first define what these terms mean. H.M of two numbers is known as the harmonic mean of two numbers and it is defined as follows: Harmonic mean of two numbers is the ratio of the number of numbers divided by the sum of the reciprocal of the two numbers. This can be represented as:
$\Rightarrow HM=\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}$
Here, a and b are the two numbers whose Harmonic mean is to be calculated. This can also be further simplified and written as,
$\Rightarrow HM=\dfrac{2ab}{a+b}\ldots \ldots \left( 1 \right)$
G.M is another term known as the Geometric mean of two numbers. This is defined as follows: Geometric mean of two numbers is simply the square root of the product of the two numbers which are a part of the geometric series. This can be given as:
$\Rightarrow GM=\sqrt{a\times b}\ldots \ldots \left( 2 \right)$
Here, a and b are the two numbers whose Geometric mean is to be calculated.
Now, it is given in the question that the ratio of this Harmonic mean and Geometric mean are in the ratio $6:10.$ This can be written in equation format by dividing equation $\left( 2 \right)$ and equation $\left( 3 \right),$
$\Rightarrow \dfrac{HM}{GM}=\dfrac{\dfrac{2ab}{a+b}}{\sqrt{a\times b}}=\dfrac{6}{10}$
Dividing ab by $\sqrt{ab}$ gives us $\sqrt{ab}$ we get
$\Rightarrow \dfrac{2\sqrt{ab}}{a+b}=\dfrac{6}{10}$
Cross multiplying the two, we get
$\Rightarrow 10.2\sqrt{ab}=6.\left( a+b \right)$
Taking the product of the terms and simplifying, we get
$\Rightarrow 20\sqrt{ab}=6a+6b$
Dividing the equation by 2 both sides, we get
$\Rightarrow 10\sqrt{ab}=3a+3b$
Taking the $10\sqrt{ab}$ term to the right-hand side, we get
$\Rightarrow 3a-10\sqrt{ab}+3b=0$
We can factorize it as follows,
$\Rightarrow \left( 3\sqrt{a}-\sqrt{b} \right)\left( \sqrt{a}-3\sqrt{b} \right)=0$
We now equate each of these terms to 0 individually,
$\Rightarrow \left( 3\sqrt{a}-\sqrt{b} \right)=0$
Taking the $\sqrt{b}$ term to the RHS, we get
$\Rightarrow 3\sqrt{a}=\sqrt{b}$
Squaring both sides, we get
$\Rightarrow 9a=b$
Dividing both sides of the equation by b and taking 9 to the other side by reciprocating, we get
$\Rightarrow \dfrac{9a}{b}=\dfrac{b}{b}$
$\Rightarrow \dfrac{a}{b}=\dfrac{1}{9}$
Similarly solving the other equation, we have
$\Rightarrow \left( \sqrt{a}-3\sqrt{b} \right)=0$
Taking the $3\sqrt{b}$ term to the RHS, we get
$\Rightarrow \sqrt{a}=3\sqrt{b}$
Squaring both sides, we get
$\Rightarrow a=9b$
Dividing both sides of the equation by b, we get
$\Rightarrow \dfrac{a}{b}=9$
Hence, the ratio of the two numbers is 9 and $\dfrac{1}{9}.$

Note: Students need to know the concept of Geometric mean and Harmonic mean in order to solve this question easily. Care must be taken while factoring and this can be done by referring to the concepts of factorization of a polynomial. We must verify to check once that of the polynomial factored if multiplied together yields back the original equation or not. If it does, the factorization is correct.