Question

A milkman has 80% milk in his stock of 800 L of adulterated milk. How much 100% pure milk should be added to it, so that purity is between 90% and 95% ?

Answer 1

Given adulterated milk contains $80\%$ pure milk.
So, amount of pure milk in $800$ litres of adulterated $=\frac{80}{100}\times 800=640$ litres
Let x litres or 100% pure milk be added to 800 L of adulterated milk then the milk content of mixture form $\frac{640+x}{800+x}$.
Thus, the required purity
$90\%<\frac{640+x}{800+x}<95\%$
Then, $(800+x)\times \frac{90}{100}<(640+x)<(800+x)\times \frac{95}{100}$

$=(800+x)\times 90<(64000+100x)<(800+x)\times 95$

$=72000+90x<64000+100x<76000+95x$

$=72000+90x<64000+100x\text{and}64000+100x<76000+95x$

$=8000<10xand5x<12000$

$=x>800andx<2400$

$=800<x<2400$

For desired purity of milk. the milkman is free to add any quantity from

800L to 2400L of 100% pure milk.