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A. 2

B. 3

C. 4

D. 5

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Hint: Take two numbers as $Ha$ and $Hb$ or $12a$ and $12b$. Now find their product and equate it to 6336. After that find the values of $a$ and $b$ such that it satisfies the equation. Take care that $a$ and $b$ are coprime to each other.

__Complete step-by-step answer:__

Here, we are given two positive numbers having HCF as 12 and their product as 6336. We have to find the number of possible pairs of the given numbers. Before proceeding with the question, let us find out about HCF and its properties. HCF is the highest common factor of a given number. We can say in other words that HCF is the largest or greatest factor that is common to any two or more given natural numbers. Suppose we have two numbers, say $p,q$, and $H$ be the HCF of the numbers, $p,q$ then we can write $p=Ha,q=Hb$, where $a$ and $b$ are coprime to each other.

Now let us consider our question. Here, we are given two positive numbers whose HCF is 12. So, let the us consider the two positive numbers as $M$ and $N$, so we get,

$\begin{align}

& M=Ha=12a\ldots \ldots \ldots \left( i \right) \\

& N=Hb=12b\ldots \ldots \ldots \left( ii \right) \\

\end{align}$

Where $a$ and $b$ are coprime to each other. Now that we are given that the product of these two numbers is 6336. So, we get, $M\times N=6336$. Substituting the values of $M$ and $N$ in equation (i) and equation (ii), we get,

$\begin{align}

& \left( 12a \right)\times \left( 12b \right)=6336 \\

& \Rightarrow 144ab=6336 \\

\end{align}$

By dividing by 144 on both the the sides of the above equation, we get,

$ab=\dfrac{6336}{144}\Rightarrow ab=44$

We can write 44 as a product of two numbers in these ways:

$\begin{align}

& ab=44=1\times 44 \\

& ab=44=2\times 22 \\

& ab=44=4\times 11 \\

\end{align}$

Here, we select $a=1$ and $b=44$, and $a=4$ and $b=11$. We do not select $ab=2\times 22$ because 2 and 22 are not coprime to each other. So, we get,

For $a=1$ and $b=44$, we get,

$\begin{align}

& M=12a=12\times 1=12 \\

& N=12b=12\times 44=528 \\

\end{align}$

For $a=4$ and $b=11$, we get,

$\begin{align}

& M=12a=12\times 4=48 \\

& N=12b=12\times 11=132 \\

\end{align}$

Hence, we get two pairs of numbers, (12 and 528) and (48 and 132). Therefore, the correct answer is option A.

Note: In this question, students can verify their result by multiplying each pair of numbers and checking if their product is 6336 or not. Also, they can check if the HCF of each pair of numbers is 12 or not. Some students tend to make mistakes by giving $a$ and $b$ as the answer for numbers, while numbers are $M=Ha,N=Hb$ or $12a,12b$. So, this mistake should be taken care of.

Here, we are given two positive numbers having HCF as 12 and their product as 6336. We have to find the number of possible pairs of the given numbers. Before proceeding with the question, let us find out about HCF and its properties. HCF is the highest common factor of a given number. We can say in other words that HCF is the largest or greatest factor that is common to any two or more given natural numbers. Suppose we have two numbers, say $p,q$, and $H$ be the HCF of the numbers, $p,q$ then we can write $p=Ha,q=Hb$, where $a$ and $b$ are coprime to each other.

Now let us consider our question. Here, we are given two positive numbers whose HCF is 12. So, let the us consider the two positive numbers as $M$ and $N$, so we get,

$\begin{align}

& M=Ha=12a\ldots \ldots \ldots \left( i \right) \\

& N=Hb=12b\ldots \ldots \ldots \left( ii \right) \\

\end{align}$

Where $a$ and $b$ are coprime to each other. Now that we are given that the product of these two numbers is 6336. So, we get, $M\times N=6336$. Substituting the values of $M$ and $N$ in equation (i) and equation (ii), we get,

$\begin{align}

& \left( 12a \right)\times \left( 12b \right)=6336 \\

& \Rightarrow 144ab=6336 \\

\end{align}$

By dividing by 144 on both the the sides of the above equation, we get,

$ab=\dfrac{6336}{144}\Rightarrow ab=44$

We can write 44 as a product of two numbers in these ways:

$\begin{align}

& ab=44=1\times 44 \\

& ab=44=2\times 22 \\

& ab=44=4\times 11 \\

\end{align}$

Here, we select $a=1$ and $b=44$, and $a=4$ and $b=11$. We do not select $ab=2\times 22$ because 2 and 22 are not coprime to each other. So, we get,

For $a=1$ and $b=44$, we get,

$\begin{align}

& M=12a=12\times 1=12 \\

& N=12b=12\times 44=528 \\

\end{align}$

For $a=4$ and $b=11$, we get,

$\begin{align}

& M=12a=12\times 4=48 \\

& N=12b=12\times 11=132 \\

\end{align}$

Hence, we get two pairs of numbers, (12 and 528) and (48 and 132). Therefore, the correct answer is option A.

Note: In this question, students can verify their result by multiplying each pair of numbers and checking if their product is 6336 or not. Also, they can check if the HCF of each pair of numbers is 12 or not. Some students tend to make mistakes by giving $a$ and $b$ as the answer for numbers, while numbers are $M=Ha,N=Hb$ or $12a,12b$. So, this mistake should be taken care of.

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