Solving Mixture problems of the GMAT

Let us take a question on mixtures. We will try to solve it using five different methods A 80 liter solution contains 20% alcohol by volume. A part of the solution is removed and replaced with pure alcohol to bring the alcohol concentration to 25%. What is the volume of the solution that was removed? Method 1 - Using basic percentage computations and setting up an equation Volume of alcohol in the 80 liters solution = 20% of 80 = 20 × 80 ÷ 100 = 16 liters Let volume of solution removed = v liters This will also have 20% alcohol ⇒ Volume of alcohol removed = 20 × v ÷ 100 = 0.2v liters Volume of alcohol added = v liters (Note: replace imples adding a quantity equal to what was removed) Thus, total alcohol volume = (16 - 0.2v + v) = (16 + 0.8v) liters Total solution volume = 80 liters Thus, required percent = [(16 + 0.8v) ÷ 80] × 100 Thus, we have: [(16 + 0.8v) ÷ 80] × 100 = 25 ⇒ 16 + 0.8v = 20 ⇒ v = 5 liters - Answer Method 2 - Using the fact that the concentration of alcohol remains u...