# SOLUTION: A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?

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 Click here to see ALL problems on Systems-of-equations Question 129672This question is from textbook Algebra structure and meathod : A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?This question is from textbook Algebra structure and meathod Answer by stanbon(57967)   (Show Source): You can put this solution on YOUR website!A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used? ---------------- 20% solution DATA: Amount = x liters; active ingredient = 0.20x L -------------- 60% solution DATA: Amount = "10-x" liters ; active = 0.60(10-x)= 6-0.6x liters ------------------ Mixture DATA; Amt. = 10 L : active = 0.30*10 = 3 liters ---------------------- EQUATION: active + active = active 0.2x + 6 - 0.6x = 3 -0.4x = -3 x = 7.5 liters (amount of 20% solution needed for the mix) 10-x = 2.5 liters (amount of 60% solution needed for the mix) ======================= Cheers, Stan H.