A jar of milk contains 40% water. A part of it is replaced by another milk containing 19% water. Now thepercentage of water is 26%. What quantity of milk was replaced?

Initial percentage of water = 40%

Percentage of water in the replaced milk = 19%

Final percentage of water after mixing = 26%

This is a mixture problem where the percentage of water in the original milk and the replaced milk are mixed to give a final percentage of water. 

Solution:  ​

Let amount of solution be 100 

and amount of  replaced solution be x

then,

Amount of water = 40% of 100 = 40 

Amount of milk =60% of 100 = 60 

New solution which  is added has 19% water and 81% milk.

So, according to question,

​$$\begin{aligned}&40 - \frac{40x}{100} + \frac{19x}{100} = 26 \\ &40-26 = \frac{21x}{100} \\ &14 = \frac{21x}{100} \\ &x = \frac{14 \times 100}{21} = \frac{2}{3}\times {100} \end{aligned}$$     

therefore, 

Quantity of milk replaced = $$\frac{ \frac{200}{3}}{100} = \frac{2}{3}$$ 

​