A milkman has $80\%$ of milk in his stock of $800$ litres of adulterated milk. How much $100\%$ milk to be added to give certain purity? Problem: A milkman has $80\%$ of milk in his stock of $800$ litres of adulterated milk. How much $100\%$ milk to be added to it so that the purity of milk is between $90\%$ and $95\%$ 
Let $x$ litres $100\%$ pure  milk need to be added in $800$ litres of milk. 
Please suggest further how to proceed not getting any idea on this. 
 A: Since, it should also be less than $0.95$ , You get $x<1200$ litres. Continuing from the previous answer of Win Vineeth
A: Here's an approach using logic rather than algebra. Consider adding the same volume of milk as the original but at 100% purity. 
800l @ 80 % (original)
800l @ 100%  (adding same volume at 100% purity)... (1)
= = =
1600l@ 90%   (resultant*, by simple average)
1600l@ 100%  (adding same volume at 100% purity)... (2)
= = =
3200l@ 95%   (resultant*, by simple average) 
Hence, from (1) and (1)+(2), you need to add between 800-2400l of 100% milk to give a resultant purity of 90%-95%. 

*NB: you can take the average of the purity if you add an equal volume as the original. 
A: $80\%$ is milk out of $800$ litres. That gives - you have $640$ litres of pure milk.  Now, $640+x\over {800+x}$$>0.9$  That gives $x>800$ litres. Since, it should also be less than $0.95$ , You get $x<1200$ litres.
A: Here you have 800 l total 
640 milk 
160 water. 
Make that 160 10 percent. 
So total 1600. 
Add 1600-640. That is 960l of pure milk 
