Question 133

If $$\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}$$= 0, then what is the value of $$\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ $$?

Solution

Given : $$\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$$

=> $$(3-\frac{1}{x})+(5-\frac{1}{y})+(7-\frac{1}{z})=0$$

=> $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3+5+7$$

=> $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=15$$

=> Ans - (C)

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