Divide 50 into two parts so that the sum of t | vedclass

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Question

(A) Let the two parts be $x$ and $(50 - x)$.According to the problem,the sum of their reciprocals is $1/12$:$\frac{1}{x} + \frac{1}{50 - x} = \frac{1}

Vedclass · Accepted Answer

$20, 30$

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(A) Let the two parts be $x$ and $(50 - x)$.According to the problem,the sum of their reciprocals is $1/12$:$\frac{1}{x} + \frac{1}{50 - x} = \frac{1}{12}$Taking the common denominator on the left side:$\frac{(50 - x) + x}{x(50 - x)} = \frac{1}{12}$$\frac{50}{50x - x^2} = \frac{1}{12}$Cross-multiplying gives:$50 	imes 12 = 50x - x^2$$600 = 50x - x^2$Rearranging into a standard quadratic equation:$x^2 - 50x + 600 = 0$Factoring the quadratic equation:$(x - 30)(x - 20) = 0$Thus,$x = 30$ or $x = 20$.If $x = 30$,the other part is $50 - 30 = 20$. If $x = 20$,the other part is $50 - 20 = 30$.Therefore,the two parts are $20$ and $30$.

Divide $50$ into two parts so that the sum of their reciprocals is $1/12$.

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