Question

The $G.M$ of two numbers is $6$. Then $A.M$. $'A'$ and $H.M$. $H$ satisfy the equation $90A + 5H = 918$then$?$
A.$A = 10,A = 4$
B.$A = \dfrac{1}{5},A = 10$
C.$A = 5,A = 10$
D.$A = \dfrac{1}{5},A = 5$

Answer 1

In this problem given a number of values containing the mean of observation. The arithmetical mean is the number obtained by dividing the sum of the values of the set by the number of values of the set. The Geometric Mean of two numbers and a nadb is $\sqrt {ab}$. The reciprocal mean of the given data values is the harmonic mean.

Formula used:
Arithmetic mean $= \dfrac{{a + b}}{2}$
Geometric mean $= \sqrt {ab}$
Harmonic mean $= \dfrac{{2ab}}{{a + b}} = \dfrac{{{{\left( {G.M} \right)}^2}}}{{A.M}}$
Where $a,b$ is the positive numbers
$G.M$ is the Geometric mean, $A.M$ is the Arithmetic mean and $H.M$ is the Harmonic mean.
Quadratic equation is $a{x^2} + bx + c = 0$
Then, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2a}$
Where, $c$ is the constant

Complete step-by-step answer:
Given $G.M = 6$
$A.M = A$
$90A + 5H = 918$
The given arithmetic mean,
$A = \dfrac{{a + b}}{2}$
On simplifying,
$2A = a + b$
The given geometric mean,
$G.M = \sqrt {ab}$
$G.{M^2} = ab$
The given value is $G.M = 6$
$36 = ab$
On simplifying a harmonic mean,
$\Rightarrow$ $\dfrac{2}{H} = \dfrac{{a + b}}{{ab}}$
Substituting the given value is
$\Rightarrow$ $\dfrac{2}{H} = \dfrac{{2A}}{{36}}$
That left side and right side of the same value is cancel
$36 = AH$…………………$(1)$
Given question $90A + 5H = 918$
Multiplication by $A$ in both sides,
$\Rightarrow$ $A\left( {90A + 5H + 918} \right) = 0$
On simplifying,
$\Rightarrow$ $90{A^2} + 5HA - 918A = 0$
Substituting the $HA$ value in above equation,
We get,
$\Rightarrow$ $90{A^2} - 918A + 5 \times 36 = 0$
Simplifying the above equation,
$\Rightarrow$ $90{A^2} - 918A + 180 = 0$
Divided by $9$,
Then, $\Rightarrow$ $10{A^2} - 102A + 20 = 0$
Again, divided by $2$,
We get,$\Rightarrow$ $5{A^2} - 51A + 10 = 0$
We applying the quadratic equation formula
$\Rightarrow$ $a{x^2} + bx + c = 0$
Then, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2a}$
substituting the given value in above equation
let, $A = \dfrac{{51 \pm \sqrt {{{51}^2} - 4 \times 5 \times 10} }}{10}$
On Simplification
$\Rightarrow$ $A = \dfrac{{51 \pm \sqrt {2601 - 200} }}{10}$
$\Rightarrow$ $A = \dfrac{{51 \pm \sqrt {2401} }}{10}$
The root value of $2401$ is $49$
The given equation is $A = \dfrac{{51 + 49}}{10},A = \dfrac{{51 - 49}}{10}$
We get two separate value,
$A = \dfrac{{100}}{{10}},A = \dfrac{2}{{10}}$
On simplifying,
$A = 10,A = \dfrac{1}{5}$
Thus the $H$ satisfy the equation $90A + 5H = 918$ then $A = \dfrac{1}{5},A = 10$
Hence,option B is Right answer.

Note: One of the methods of average is the harmonic mean and in particular one of the Pythagoreans means.it is ideal for circumstances where the average rates are needed. The three means are always equal to each other if all values in a non-empty dataset are equal. The most important condition for it is that none of the observations should be zero.