Question

If the difference between two specific numbers is 12, the LCM of the two
numbers is 63 and the HCF of the two numbers is 3, then the sum of the
two numbers is

Answer 1

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• If the difference between two specific numbers is 12, If the LCM of the two numbers is 63 and the HCF of the two numbers is 3, then the sum of the two numbers is ?.

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\sf{Let \: the \: numbers \: be \: x \: and \: y.}$

• $\sf{Difference \: between \: the \: numbers \: = \: 12.}$

• $\therefore{\sf{x \: - \: y \: = \: 12}}$

• $\sf{x \: = \: 12 \: + \: y}$

• $\sf{LCM \: of \: the \: given \: numbers \: = \: 63.}$

• $\sf{HCF \: of \: the \: given \: numbers \: = \: 3.}$

• $\sf{We \: know \: that,}$

• $\boxed{\sf{LCM \: * \: HCF \: = \: a \: * \: b}}$

• $\sf{63 \: * \: 3 \: = \: x \: * \: y}$

• $\sf{189 \: = \: (12 \: + \: y) \: y}$

• $\sf{189 \: = \: 12y \: + \: y^{2}}$

• $\sf{y^{2} \: + \: 12y \: - \: 189 \: = \: 0}$

• $\sf{Finding \: the \: factors \: by \: PSF \: method,}$

• $\sf{P \: = \: - \: 189 \: * \: y^{2} \: = \: - \: 189y^{2}}$

• $\sf{S \: = \: + \: 12y}$

• $\sf{F \: = \: - \: 21y, \: + \: 9y}$

• $\sf{Now,}$

• $\sf{y^{2} \: + \: 9y \: - \: 21y \: - \: 189 \: = \: 0}$

• $\sf{y \: (y \: + \: 9) \: - \: 21 \: (y \: + \: 9) \: = \: 0}$

• $\sf{(y \: - \: 21) \: (y \: + \: 9) \: = \: 0}$

• $\sf{y \: - \: 21 \: = \: 0 \: , \: y \: + \: 9 \: = \: 0}$

• $\sf{y \: = \: 21 \: , \: y \: = \: - \: 9}$

• $\textsf{ Since, the number cannot be negative,}$

• $\boxed{\therefore{\sf{y \: = \: 21}}}$

• $\textsf{We know that,}$

• $\sf{x \: - \: y \: = \: 12}$

• $\sf{x \: - \: 21 \: = \: 12}$

• $\sf{x \: = \: 21 \: + \: 12}$

• $\boxed{\therefore{\sf{x \: = \: 33}}}$

• $\textsf{Therefore, Sum of the 2 numbers =}$

• $\sf{x \: + \: y}$

• $\sf{33 \: + \: 21}$

• $\boxed{\sf{54}}$

$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\boxed{\therefore{\sf{Sum \: of \: the \: numbers \: = \: 54.}}}$

Answer 2

Given :-

Difference between 2 numbers = 12

LCM of the two numbers is 63

HCF of the two numbers is 3

To find :-

Sum of them

Solution :-

Let the number be a and b

a - b = 12

a = 12 + b

Now

LCM * HCF = Product of two numbers

63 * 3 = ab

189 = ab

189 = (12 + b)b

189 = 12b + b^2

$\sf b^{2} + 12b - 189 = 0$

$\sf b^{2} + 9b - 21b-189= 0$

b(b + 9) - 21(b + 9) = 0

(b - 21), (b + 9) = 0

Either

b = 0 + 21

= 21

OR,

b = 0 - 9

= -9

Since number can't be negative.

So

b = 21

a = 21 + 12 = 33