
The difference of two numbers is 14. Their LCM and HCF are 441 and 7 respectively. Find the numbers.
Answer
511.5k+ views
Hint: Let the numbers be x and y. Now, the HCF of these numbers is 7, so both these numbers are divisible by 7 and we can write them as 7x and 7y. Now, to find these numbers, we are going to use the relation that the product of these numbers is equal to the product of LCM and HCF. Hence, we will get two equations and solving them we will get the values of x and y.
Complete step-by-step answer:
In this question, we are given the difference of two numbers and their LCM and HCF and we have to find these two numbers.
Let the numbers be x and y.
Now, their difference is 14. So,
$ \Rightarrow x - y = 14 $ - - - - - - - - (1)
And their HCF is 7, so both these numbers will be divisible by 7. So, we can write equation (1) as
$ \Rightarrow 7x - 7y = 14 $
Take out 7 as common, we get
$
\Rightarrow 7\left( {x - y} \right) = 14 \\
\Rightarrow \left( {x - y} \right) = \dfrac{{14}}{2} \;
$
$ \Rightarrow x - y = 2 $ - - - - - - - - - (2)
Now, there is a relation between the numbers and their HCF and LCM, that is
$ \Rightarrow $ Product of two numbers $ = $ HCF $ \times $ LCM
Therefore,
$
\Rightarrow 7x \times 7y = 441 \times 7 \\
\Rightarrow xy = \dfrac{{441 \times 7}}{{7 \times 7}} \\
\Rightarrow xy = 63 \;
$
$ \Rightarrow y = \dfrac{{63}}{x} $ - - - - - - - - (3)
Now, putting this value of y in equation (2), we get
$ \Rightarrow x - \dfrac{{63}}{x} = 2 $
Take LCM, we get
$
\Rightarrow \dfrac{{{x^2} - 63}}{x} = 2 \\
\Rightarrow {x^2} - 63 = 2x \\
\Rightarrow {x^2} - 2x - 63 = 0 \;
$
Solving the above equation, we get
$ \Rightarrow x = 9 $
Therefore, putting this value of x in equation (3), we get
$ \Rightarrow y = \dfrac{{63}}{9} = 7 $
Therefore, our numbers are 7x $ = 7\left( 9 \right) = 63 $ and 7y $ = 7\left( 7 \right) = 49 $ .
Note: The tricky part in this question is that we are using 7x and 7y instead of x and y. As 7 is a factor of these numbers, 7x and 7y will be the multiples of these numbers. So, we can use 7x and 7y instead of x and y. We can also cross verify our answer. We obtained the numbers as 63 and 49. So,
$ \Rightarrow 63 - 49 = 14 $ and the LCM and HCF of 63 and 49 is 441 and 7 respectively. So, our answer is correct.
Complete step-by-step answer:
In this question, we are given the difference of two numbers and their LCM and HCF and we have to find these two numbers.
Let the numbers be x and y.
Now, their difference is 14. So,
$ \Rightarrow x - y = 14 $ - - - - - - - - (1)
And their HCF is 7, so both these numbers will be divisible by 7. So, we can write equation (1) as
$ \Rightarrow 7x - 7y = 14 $
Take out 7 as common, we get
$
\Rightarrow 7\left( {x - y} \right) = 14 \\
\Rightarrow \left( {x - y} \right) = \dfrac{{14}}{2} \;
$
$ \Rightarrow x - y = 2 $ - - - - - - - - - (2)
Now, there is a relation between the numbers and their HCF and LCM, that is
$ \Rightarrow $ Product of two numbers $ = $ HCF $ \times $ LCM
Therefore,
$
\Rightarrow 7x \times 7y = 441 \times 7 \\
\Rightarrow xy = \dfrac{{441 \times 7}}{{7 \times 7}} \\
\Rightarrow xy = 63 \;
$
$ \Rightarrow y = \dfrac{{63}}{x} $ - - - - - - - - (3)
Now, putting this value of y in equation (2), we get
$ \Rightarrow x - \dfrac{{63}}{x} = 2 $
Take LCM, we get
$
\Rightarrow \dfrac{{{x^2} - 63}}{x} = 2 \\
\Rightarrow {x^2} - 63 = 2x \\
\Rightarrow {x^2} - 2x - 63 = 0 \;
$
Solving the above equation, we get
$ \Rightarrow x = 9 $
Therefore, putting this value of x in equation (3), we get
$ \Rightarrow y = \dfrac{{63}}{9} = 7 $
Therefore, our numbers are 7x $ = 7\left( 9 \right) = 63 $ and 7y $ = 7\left( 7 \right) = 49 $ .
Note: The tricky part in this question is that we are using 7x and 7y instead of x and y. As 7 is a factor of these numbers, 7x and 7y will be the multiples of these numbers. So, we can use 7x and 7y instead of x and y. We can also cross verify our answer. We obtained the numbers as 63 and 49. So,
$ \Rightarrow 63 - 49 = 14 $ and the LCM and HCF of 63 and 49 is 441 and 7 respectively. So, our answer is correct.
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