 |
| ||
| |||
|
| Math Central | Quandaries & 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?
Penny
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.