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?

Question

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.

Answer 1

$80\%$ 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.

Answer 2

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

Answer 3

Since, it should also be less than $0.95$ , You get $x<1200$ litres. Continuing from the previous answer of Win Vineeth

Answer 4

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%.

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*NB: you can take the average of the purity if you add an equal volume as the original.