Question 516111
the quantity of antifreeze required to make 20 l to 50% is the amount you need to drain
let that be x l

anifreeze	100	----------------			x	l							
Solution	40	------	20	-	x	l							
Mixture	50.00%	----------------	20										
													
100x	+40(20	-	x	)	=	50	*	20	
													
100x+	800-	40x	=			1000			
100x	-	40x	=	1000	-	800			
60x			=	200								
/	60												
	x=	3.33		l  anifreeze