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Bottle $1$ contains a mixture of milk and water in $7: 2$ ratio and Bottle $2$ contains a mixture of milk and water in $9:4$ ratio. In what ratio of volumes should the liquids in Bottle $1$ and Bottle $2$ be combined to obtain a mixture of milk and water in $3:1$ ratio?

1. $27:14$
2. $27:13$
3. $27:16$
4. $27:18$

Let the volume of liquids in Bottle $1$ and Bottle $2$ be $x$ liter and $y$ liter respectively.

Now,  $\dfrac{(\frac{7}{9})x+(\frac{9}{13})y}{(\frac{2}{9})x+(\frac{4}{13})y}= \dfrac{3}{1}$

$\Rightarrow (\frac{7}{9})x+(\frac{9}{13})y=3\left[(\frac{2}{9})x+(\frac{4}{13})y\right]$

$\Rightarrow \frac{7x}{9}+\frac{9y}{13}=\frac{6x}{9}+\frac{12y}{13}$

$\Rightarrow \frac{7x}{9}-\frac{6x}{9}=\frac{12y}{13}-\frac{9y}{13}$

$\Rightarrow \frac{7x-6x}{9}=\frac{12y-9y}{13}$

$\Rightarrow \frac{x}{9}=\frac{3y}{13}$

$\Rightarrow \frac{x}{y}=\frac{27}{13}$

$\therefore$ The Bottle $1$ and Bottle $2$ should be combined in the ratio of $x:y=27:13.$

$\textbf{Short Method:}$

We can solve this question using the allegation method.

Now,

• In Bottle $1$ concentration of milk $=\frac{7}{9}$
• In bottle $2$ concentration of milk $=\frac{9}{13}$
• In mixture concentration of milk $=\frac{3}{4}$

$\therefore$ The required ratio $=\dfrac{\frac{3}{52}}{\frac{1}{36}}=\frac{3}{52}\times{\frac{36}{1}}=\frac{27}{13}.$

$$\textbf{(OR)}$$

• In bottle $1$ concentration of water $=\frac{2}{9}$
• In bottle $2$ concentration of water $=\frac{4}{13}$
• In mixture concentration of water $=\frac{1}{4}$

$\therefore$ The required ratio $=\dfrac{\frac{-3}{52}}{\frac{-1}{36}}=\frac{3}{52}\times{\frac{36}{1}}=\frac{27}{13}.$

Correct Answer $:\text{B}$

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