Question

Nine gallons are drawn from a cask full of wine; it is then filled with water, then nine gallons of the mixture are drawn, and the cask is again filled with water. If the quantity of wine now in the cask be to the quantity of water in it as $16$ to $9$, how much does the cask hold?

Answer 1

Let $x$ be the number of gallons the cask can hold.

After the first drawing and filling,

wine left $=x-9$ gallons

quantity of water $=9$ gallons.

For the second drawing the water and wine will be drawn in the ratio they are present in the cask.

So, Wine drawn =$\left(\frac{x-9}{x}\right)9$

Water drawn=$\left(\frac{9}{x}\right)9$

Wine left in the cask =$x-9-\left(\frac{x-9}{x}\right)9=x-18+\frac{81}{x}$

Water left in the cask=$9-\left(\frac{9}{x}\right)9+9=18-\frac{81}{x}$

Given.

$\frac{x-18+\frac{81}{x}}{18-\frac{81}{x}}=\frac{16}{9}$

$\frac{x^2-18x+81}{18x-81}=\frac{16}{9}$

$9x^2-162x+729=288x-1296$

$9x^2-450x+2025=0$

$x^2-50x+225=0$

$(x-45)(x-5)=0$

$x=5,45$

As the capacity cannot be $5$ according to the question so we have $45$ gallons.