
The L.C.M. of two numbers is $\left( {a + b} \right)$ and their H.C.F. is $k\left( {a - b} \right)$. If one of the numbers is $k$, find the other number
A) $\dfrac{{ka}}{b}$
B) ${a^2} - {b^2}$
C) $\dfrac{{a + b}}{{k\left( {a - b} \right)}}$
D) $\dfrac{{ka + b}}{{ka - b}}$
Answer
505.8k+ views
Hint:
Here, we are required to find one of the two numbers whose LCM and HCF are given. We will use the formula that the product of two numbers is equal to the product of their HCF and LCM. Substituting the given values in the formula, we will get the required value of one of the numbers.
Formula Used:
${\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = m \times n$
Complete step by step solution:
According to the question,
The L.C.M. of two numbers is $\left( {a + b} \right)$
The H.C.F. of two numbers is $k\left( {a - b} \right)$.
Now, If one of the numbers is $k$
Let us assume that the other number is $n$.
Now, as we know, the product of two numbers is equal to the product of their LCM and HCF.
$ \Rightarrow {\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = k \times n$
Since, the L.C.M. of two numbers is $\left( {a + b} \right)$ and their H.C.F. is $k\left( {a - b} \right)$.
$ \Rightarrow \left( {a + b} \right) \times k\left( {a - b} \right) = k \times n$
This can be written as:
$ \Rightarrow k\left( {a + b} \right)\left( {a - b} \right) = k \times n$
Now, by using the formula, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
$ \Rightarrow k\left( {{a^2} - {b^2}} \right) = k \times n$
Cancelling out $k$from both the sides,
$ \Rightarrow \left( {{a^2} - {b^2}} \right) = n$
Therefore the other number, $n = \left( {{a^2} - {b^2}} \right)$
Hence, option B is the correct answer.
Note:
Least Common Multiple or LCM is the smallest possible common multiple of any given natural numbers.
Highest Common Factor or HCF is the largest common factor of two or more given numbers. Now, we have seen the property of LCM and HCF which is:
${\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = m \times n$
Now, this property is applicable for only two numbers.
Also, as a fun fact, HCF of any given number is never greater than those numbers.
And, LCM of any given number is never smaller than those numbers.
Here, we are required to find one of the two numbers whose LCM and HCF are given. We will use the formula that the product of two numbers is equal to the product of their HCF and LCM. Substituting the given values in the formula, we will get the required value of one of the numbers.
Formula Used:
${\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = m \times n$
Complete step by step solution:
According to the question,
The L.C.M. of two numbers is $\left( {a + b} \right)$
The H.C.F. of two numbers is $k\left( {a - b} \right)$.
Now, If one of the numbers is $k$
Let us assume that the other number is $n$.
Now, as we know, the product of two numbers is equal to the product of their LCM and HCF.
$ \Rightarrow {\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = k \times n$
Since, the L.C.M. of two numbers is $\left( {a + b} \right)$ and their H.C.F. is $k\left( {a - b} \right)$.
$ \Rightarrow \left( {a + b} \right) \times k\left( {a - b} \right) = k \times n$
This can be written as:
$ \Rightarrow k\left( {a + b} \right)\left( {a - b} \right) = k \times n$
Now, by using the formula, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
$ \Rightarrow k\left( {{a^2} - {b^2}} \right) = k \times n$
Cancelling out $k$from both the sides,
$ \Rightarrow \left( {{a^2} - {b^2}} \right) = n$
Therefore the other number, $n = \left( {{a^2} - {b^2}} \right)$
Hence, option B is the correct answer.
Note:
Least Common Multiple or LCM is the smallest possible common multiple of any given natural numbers.
Highest Common Factor or HCF is the largest common factor of two or more given numbers. Now, we have seen the property of LCM and HCF which is:
${\text{L}}{\text{.C}}{\text{.M}}{\text{.}} \times {\text{H}}{\text{.C}}{\text{.F}}{\text{.}} = m \times n$
Now, this property is applicable for only two numbers.
Also, as a fun fact, HCF of any given number is never greater than those numbers.
And, LCM of any given number is never smaller than those numbers.
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