Question
A mixture (alcohol + water) contains 60% alcohol and 80 litres water. Find the quantity of mixture that must be removed so that the quantity of alcohol becomes 96 litres in the resultant mixture.
Solution
Initial quantity of mixture = (80/0.4) = 200 litres Initial quantity of alcohol = 0.6 × 200 = 120 litres Let the required quantity of mixture be ‘p’ litres. So, 120 − (6/10) × p = 96 Or, 6p = 24 × 10 Or, p = 40 Therefore, required quantity is 40 litres.
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