A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are taken out and transferred to the other. If after the second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z(x+y) = xy$ .
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After the first transfer the first vessel contains $x-z$ wine, and the second contains $z$ wine. After the second transfer the first vessel contains $x-z-\dfrac{x-z}{x}z+\dfrac{z}{y}z$ of wine and the second contains $z-\dfrac{z}{y}z+\dfrac{x-z}{x}z$ of wine. Hence we must have $\dfrac{x-z}{x}z=\dfrac{z}{y}z$ or $xy=(x+y)z$.
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$\begingroup$ how did u get the quantities for the second transfer? @almagest $\endgroup$– FelixMay 3, 2016 at 9:19
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$\begingroup$ @matrikashukla You are assuming the wine is well-mixed after the first transfer. The first vessel contains $x-z$ of wine and $z$ of water after the first transfer. Hence when you remove a total quantity $z$ in the second transfer, the amount of wine removed will be $\frac{x-z}{x}z$. $\endgroup$– almagestMay 3, 2016 at 9:22
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$\begingroup$ why did u add (x/y)z in the first vessel and subtracted (z/y)z in the second vessel?? $\endgroup$– FelixMay 3, 2016 at 10:11
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$\begingroup$ @matrikashukla I didn't add $\frac{x}{y}z$ to the first vessel. I subtracted $\frac{z}{y}z$ from the second vessel because we took out a total of $z$, of which a fraction $\frac{z}{y}$ was wine. $\endgroup$– almagestMay 3, 2016 at 10:18
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$\begingroup$ ok .......... I got it now! thank u so much @almagest $\endgroup$– FelixMay 3, 2016 at 10:27