The ratio of measurements of water and dettol in $36 \mathrm{litres}$ of water is $5:1$. What volume of dettol in $\mathrm{litres}$ should be added to the mixture so that the ratio of measurements of water and dettol becomes $3:1$?

I prepared two types of beverages in my house. The ratio of measurement of syrup and water in two types of beverage are $2:7$ and $1:5$ respectively. Let's try to calculate what ratio of the measurements of the two beverages needs to be mixed so that the ratio of syrup and water will be$5:21$ 

Answer:

Let's think $x \mathrm{litres}$ of the first beverage mixed with $y \mathrm{litre}$ of the second beverage. 

$\therefore$ The quantity of syrup in the $x \mathrm{litre}$ of the  first beverage$=\frac{2}{9}\times x \mathrm{litres}=\frac{2x}{9} \mathrm{litres}$
And the quantity of water is $=\frac{\boxed{?}}{\boxed{?}}\times \boxed{?} \mathrm{litre}=\frac{7\mathrm{x}}{9} \mathrm{litre}$
$\therefore$ The quantity of syrup in the $y \mathrm{litre}s$ of the second beverage is $\frac{1}{6}\times y \mathrm{litres}$ $=\frac{y}{6} \mathrm{litres}$
And the quantity of water is $\frac{5}{6}\times y \mathrm{litres}=\frac{5y}{6} \mathrm{litres}$
$\therefore$The quantity of syrup in the new beverage is $=\frac{2x}{9} \mathrm{litres}+\frac{y}{6} \mathrm{litres}$

$=\left(\frac{2x}{9}+\frac{y}{6}\right) \mathrm{litre}$
And the quantity of water is $=\frac{7x}{9} \mathrm{litres}+\frac{5y}{6} \mathrm{litres}$

$=\left(\frac{7x}{9}+\frac{5y}{6}\right) \mathrm{litres}$

According to the condition, 

$\frac{{\displaystyle \frac{2x}{9}}+{\displaystyle \frac{y}{6}}}{\frac{7x}{9}+\frac{5y}{6}}=\frac{5}{21}$

$\Rightarrow 21\left(\frac{2x}{9}+\frac{y}{6}\right)=5\left(\frac{7x}{9}+\frac{5y}{6}\right)$

$\Rightarrow 21\left(\frac{4x+3y}{18}\right)=5\left(\frac{14x+15y}{18}\right)$

$\Rightarrow 21\left(4x+3y\right)=5\left(14x+15y\right)$

$\Rightarrow 84x+63y=70x+75y$

$\Rightarrow 84x-70x=75y-63y$

$\Rightarrow 14x=12y$

$\Rightarrow \frac{x}{y}=\frac{12}{14}=\frac{6}{7}$

$\therefore x:y=6:7$