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  • The sum of two numbers is 16 and the sum of their reciprocals is 1/3 .Find the numbers.

The sum of two numbers is 16 and the sum of their reciprocals is 1/3 .Find the numbers.

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\text{Let the two numbers be x}$ and $16-x$ \\ \\ By the given hypothesis, we have $\Rightarrow \frac{1}{x}+\frac{1}{16-x}=\frac{1}{3}$ \\ $\Rightarrow \frac{16-x+x}{x(16-x)}=\frac{1}{3}$ \\ $\Rightarrow 48=16 x-x^{2}$ \\ $\Rightarrow x^{2}-16 x+48=0$ \\ $\Rightarrow x^{2}-12 x-4 x+48=0$ \\ $\Rightarrow x(x-12)-4(x-12)=0$ \\ $\Rightarrow(x-12)(x-4)=0$ \\ $\Rightarrow x =12$ or $x =4$

∴ The two numbers are 4 and 12

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Deependra Verma

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