Question

There are two numbers $a$ and $b$ whose product is $192$ and quotient of $AM$ by $HM$ of their greatest common divisor and least common multiple $\dfrac{169}{48}$. The smaller out of $a$ and $b$ is?

Answer 1

Firstly, we would be expressing the given notations in mathematical form. And then we have to apply the formula of $AM$ by $HM$ which would be $\dfrac{\dfrac{G+L}{2}\left( G+L \right)}{2GL}$ in which $G$ would be our greatest common divisor and $L$ would be least common multiple. Upon substituting and solving the values, we obtain both the numbers $a$ and $b$. From the obtained $a$ and $b$, we must find the smaller value among them and this would be the required answer.

Complete step by step answer:
Let us have a brief regarding $AM$ by $HM$. $AM$ means arithmetic mean and $HM$ means harmonic mean. Arithmetic means nothing but the value obtained when the sum of a set of values is divided by the number of sets of values. The harmonic mean is simply defined as the value obtained by the reciprocal value of arithmetic mean.
Now let us start finding the numbers and the smaller number among them.
Firstly, let us express the given notations in the mathematical form.
Let product of numbers be denoted as GL$$$$=ab=192
Now let us consider the formula of \dfrac{AM}{GM}=\dfrac{\dfrac{G+L}{2}\left( G+L \right)}{2GL}$$$$=\dfrac{169}{48}
$\dfrac{\dfrac{G+L}{2}\left( G+L \right)}{2GL}=\dfrac{{{\left( G+L \right)}^{2}}}{4GL}$
Now let us substitute the values and solve it. We get,
{{\left( G+L \right)}^{2}}$$$$=$$$${{13}^{2}}\times {{4}^{2}}
\Rightarrow $$$$G+L=52
$\therefore$ $G=4$ and $L=48$.
\therefore $$$$a,b=4,48
From the obtained numbers, $4$ would be a smaller number.

Note: We must solve the problem by considering both the product and the quotient given. the formula of the quotient for the $AM$ by $HM$ should be noted down correctly in order to obtain the accurate numbers.