The HCF and LCM of two numbers are 11 and 385 | vedclass

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(A) Let the two numbers be $a$ and $b$.We know that the product of two numbers is equal to the product of their $HCF$ and $LCM$,so $a 	imes b = HCF \

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(A) Let the two numbers be $a$ and $b$.We know that the product of two numbers is equal to the product of their $HCF$ and $LCM$,so $a 	imes b = HCF 	imes LCM$.Given $HCF = 11$ and $LCM = 385$.Therefore,$a 	imes b = 11 	imes 385 = 11 	imes (5 	imes 7 	imes 11) = (11 	imes 5) 	imes (11 	imes 7) = 55 	imes 77$.Since both numbers must be multiples of the $HCF$ $(11)$,the possible factors are $11 	imes 1, 11 	imes 5, 11 	imes 7, 11 	imes 35$.The two numbers are $55$ and $77$.We are given that one number lies between $75$ and $125$.Since $77$ is between $75$ and $125$,the required number is $77$.

The $HCF$ and $LCM$ of two numbers are $11$ and $385$ respectively. If one number lies between $75$ and $125,$ then that number is

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