There is a tea-stall of Nilmoni uncle at the crossing of our locality. My brother and I often go to uncle's stall after our study and help him in many ways. Today we have also gone there. We have seen that Nilmoni uncle was mixing some Darjeeling tea with Assam tea in casket No.$2$

In casket No. $2$, the ratio of the quantities of Assam tea and Darjeeling tea is $2:1$ and total quantity of mixture is $21 \mathrm{kg}$.

Let's calculate what quantity of Darjeeling tea needs to be mixed with the blend in casket No. $2$. so that the ratio of Assam tea and Darjeeling tea will be $7:4$ in the new blend.

I prepared two types of beverage 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 calculate the ratio of measurement of syrup and water if $27 \mathrm{litres}$ of first beverage is mixed with the $18 \mathrm{litres}$ of the second.

Answer:
The proportional part of the quantity of first syrup $=\frac{2}{2+7}=\frac{\boxed{?}}{\boxed{?}}$

And the proportional part of the quantity of water is $=\frac{\boxed{?}}{\boxed{?}}$

$\therefore$ The quantity of syrup in the first beverage is $=\frac{2}{9}\times 27 \mathrm{litre}=6 \mathrm{litre}$

And the quantity of water is $\frac{7}{9}\times 27 \mathrm{litre}=21 \mathrm{litre}$

In the same way,

The proportional part of the quantity of syrup of the second beverage is $\frac{\boxed{?}}{\boxed{?}}$
And the proportional pan of the water is $\frac{\boxed{?}}{\boxed{?}}$

The quantity of syrup in the 18 litres of beverage of second type is$=\boxed{?} \mathrm{litres}$

And the quantity of water is$=\boxed{?} \mathrm{litres}$

The quantity of syrup in the $18 \mathrm{litre}$ of beverage is $=6 \mathrm{litres}+3 \mathrm{litres}=\boxed{?} \mathrm{litre}$ 

And the quantity of the water is$21 \mathrm{litres}+15 \mathrm{litres}=\boxed{?} \mathrm{litres}$

In the new beverage the quantity of syrup:quantity of water $9:36=1:4$
 

My brother has prepared a beverage in a jug with syrup and water at a ratio of measurement of  $3:1$. Let's work out what part of the drink will be removed and replaced by water so that the volume of syrup and water becomes equal. 

Let's think $x \mathrm{unit}$ of beverage is there in the jug. $y \mathrm{unit}$ of it is removed and replaced by the same volume of water. 

The quantity of syrup in $x \mathrm{units}$ of beverage is  $=\frac{3}{4}\times x \mathrm{units}$ $=\frac{3x}{4} \mathrm{units}$

And the quantity of water is $=\frac{\boxed{?}}{\boxed{?}} \mathrm{units}$

Again, the quantity of syrup in $y \mathrm{units}$ is $=\frac{3}{4}\times y \mathrm{units}$

$=\frac{3y}{4} \mathrm{units}$ 

And the quantity of water is $\frac{y}{4} \mathrm{units}$

If$y \mathrm{units}$ of syrup is replaced from $x \mathrm{units}$ of syrup,

The remaining quantity of syrup is $\left(\frac{3x}{4}-\frac{3y }{4}\right) \mathrm{units}$
And the quantity of water is $\left(\frac{x}{4}-\frac{y }{4}\right) \mathrm{units}$

Again,$y \mathrm{units}$ of water is added 

Now, the quantity of water is $=\left(\frac{x}{4}-\frac{y}{4}+y\right)$

According to the condition, $\frac{{\displaystyle \frac{x}{4}}-{\displaystyle \frac{y}{4}}+y}{{\displaystyle \frac{3x}{4}}-{\displaystyle \frac{3y}{4}}}=\frac{2}{1}$

$\Rightarrow \frac{x}{4}-\frac{y}{4}+y=\frac{6x}{4}-\frac{6y}{4}$

$\Rightarrow \frac{x}{4}-\frac{6x}{4}=\frac{y}{4}-y-\frac{6y}{4}$

$\Rightarrow \frac{x-6x}{4}=\frac{y-4y-6y}{4}$

$\Rightarrow \frac{-5x}{4}=\frac{-9y}{4}$

$\Rightarrow -5x=-9y$

$\Rightarrow y=\frac{5x}{9}$

$\therefore$Total $\frac{5}{9}$ part of the beverage will be removed and replaced by the water with the same quantity so that the ratio of the quantity of water and syrup will be $2:1$.

The ratio of quantity of syrup and water is $3:1$ in a beverage. Let's work out what part of the beverage will be replaced by the same quantity of water so that the quantities of syrup and water will be same.