Math CentralQuandaries & Queries


Question from Mary, a parent:

How many oz. of solution that is 90% alcohol ( and 10% water) is needed to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?

Hi Mary,

Suppose you use x ounces of the 90% solution then you are adding 0.9x ounces of alcohol. The 5 ounces of 50% solution contains 0.5 × 5 = 2.5 ounces of alcohol. Hence you have put together x + 5 ounces of solution of which 0.9x + 2.5 ounces is alcohol. You want these 0.9x + 2.5 ounces to be 80% of x + 5 ounces. What is x?


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