The text{l.c.m.} of two numbers is 14 times t | vedclass

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(A) Let the $	ext{g.c.d.}$ of the two numbers be $x$.Given that the $	ext{l.c.m.}$ is $14$ times the $	ext{g.c.d.}$,so $	ext{l.c.m.} = 14x$.The su

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$80$

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(A) Let the $	ext{g.c.d.}$ of the two numbers be $x$.Given that the $	ext{l.c.m.}$ is $14$ times the $	ext{g.c.d.}$,so $	ext{l.c.m.} = 14x$.The sum of $	ext{l.c.m.}$ and $	ext{g.c.d.}$ is $600$,so $14x + x = 600$.$15x = 600 \implies x = 40$.Thus,$	ext{g.c.d.} = 40$ and $	ext{l.c.m.} = 14 	imes 40 = 560$.We know that for two numbers $a$ and $b$,$	ext{l.c.m.} 	imes 	ext{g.c.d.} = a 	imes b$.Given $a = 280$,we have $560 	imes 40 = 280 	imes b$.$b = \frac{560 	imes 40}{280} = 2 	imes 40 = 80$.Therefore,the other number is $80$.

The $\text{l.c.m.}$ of two numbers is $14$ times their $\text{g.c.d.}$ and the sum of $\text{l.c.m.}$ and $\text{g.c.d.}$ is $600$. If one of the numbers is $280$,then find the other number.

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