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**Hint:**We will first assume both the numbers and apply the fact that:

HCF $ \times $ LCM = Product of two numbers. From here, find one number in terms of another. Put this in the equation we can form by a given sum of numbers and thus we will have the answer.

**Complete step-by-step answer:**

Let us first understand the meaning of “Least Common Multiple” and “Highest Common Factor”:

Multiple:

We get a multiple of a number when we multiply it by another number.

Common Multiple:

The common multiples are those that are found lists of multiples of both.

Least Common Multiple:

It is simply the smallest of the common multiples.

Highest Common Factor:

It is simply the largest of the common factors.

Let the numbers be $x$ and $y$ respectively.

\[HCF\; \times LCM{\text{ }} = {\text{ }}x\; \times y = xy\] …… (1)

We have with us that: The HCF and LCM of two numbers are 12 and 72 respectively.

Putting these values in (1), we will have:-

\[12 \times 72 = xy\]

So, we have:- \[xy = 864\]

So, $y = \dfrac{{864}}{x}$ ……..(2)

We also have with us that: sum of the two numbers is 60.

So, we will get:- $x + y = 60$ …..… (3)

Putting (2) in this, we will get:-

$x + \dfrac{{864}}{x} = 60$

Taking LCM to simplify the equation, we will get:-

$\dfrac{{{x^2} + 864}}{x} = 60$

Cross multiplying to get a simpler expression, we get:-

${x^2} + 864 = 60x$

That is ${x^2} - 60x + 864 = 0$

We can rewrite it as:

${x^2} - 36x - 24x + 864 = 0$

$x(x - 36) - 24(x - 36) = 0$

$(x - 24)(x - 36) = 0$

So, either $x = 24$ or $x = 36$.

If $x = 24$, then $y = 36$ and if $x = 36$, then $y = 24$.

**So, the correct answer is “Option B”.**

**Note:**We should always keep in mind that we need as much equations as much unknown variables. Like, here in this question, we had two unknown variables, so we created two equations (1) and (2).

We can solve the quadratic equation either by the method as above that is factorization method or even by using the direct formula which is: If we have $a{x^2} + bx + c$, then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

The students might just find one of the numbers and try finding it among the options which do not even contain it. So, we must find both the numbers so that we choose whichever we see in the option.

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