If the sum of two numbers is 55 and the HCF a | vedclass

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(C) Let the two numbers be $x$ and $y$.Given that the sum of the numbers is $x + y = 55$ $....(1)$We are given $HCF = 5$ and $LCM = 120$.We know the p

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$\frac{11}{120}$

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(C) Let the two numbers be $x$ and $y$.Given that the sum of the numbers is $x + y = 55$ $....(1)$We are given $HCF = 5$ and $LCM = 120$.We know the property that for any two numbers,the product of the numbers is equal to the product of their $HCF$ and $LCM$:$x 	imes y = HCF 	imes LCM$$x 	imes y = 5 	imes 120 = 600$ $....(2)$We need to find the sum of the reciprocals of the numbers,which is $\frac{1}{x} + \frac{1}{y}$.$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{x 	imes y}$Substituting the values from equation $(1)$ and $(2)$:$\frac{1}{x} + \frac{1}{y} = \frac{55}{600}$Simplifying the fraction by dividing both numerator and denominator by $5$:$\frac{55 \div 5}{600 \div 5} = \frac{11}{120}$

If the sum of two numbers is $55$ and the $HCF$ and $LCM$ of these numbers are $5$ and $120$ respectively,then the sum of the reciprocals of the numbers is equal to

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