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A solution, of volume $40$ litres, has dye and water in the proportion $2:3.$ Water is added to the solution to change this proportion to $2:5.$ If one$-$ fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to $2:3$?
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Given that, a solution of volume $40 \; \text{litres}.$

$\begin{array}{ccc} \text{} & \text{Dye} & \text{Water} \\ \text{Ratio:} & 2  & 3 \\ \text{Quantity:} & \frac{2}{5} \times 40 = 16\;\text{litres} & \frac{3}{5} \times 40 = 24\;\text{litres} \end{array}$

Now, water is added to the solution, then new ratio become,

$\begin{array}{ccc} \text{} & \text{Dye} & \text{Water} \\ \text{Ratio:} & 2 & 5 \\ \text{Quantity:} & 16\;\text{litres} & \frac{5}{2} \times 16 = 40\;\text{litres} \end{array}$

When $\frac{1}{4}$ of mixtures is taken out

$\begin{array}{ccc} \text{} & \text{Dye} & \text{Water} \\ \text{Quantity:} &\frac{3}{4} \times 16 = 12 \;\text{litres} & \frac{3}{4} \times 40 = 30\;\text{litres} \end{array}$

Let say $y \; \text{liters}$ of dye will be added, so that ratio becomes $2:3.$

$\Rightarrow \frac{12 + y}{30} = \frac{2}{3}$

$ \Rightarrow 12 + y = 20$

$ \Rightarrow \boxed{y = 8 \; \text{litres}}$

Correct Answer $: 8$
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