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Unformatted text preview: 1.5 Þ t = 11.67 hrs So, it looks like it would take John 11.67 hours to clean the complex by himself.
This is the final type of problems that we’ll be looking at in this section. We are going to be
looking at mixing solutions of different percentages to get a new percentage. The solution will
consist of a secondary liquid mixed in with water. The secondary liquid can be alcohol or acid
The standard equation that we’ll use here will be the following. æ Amount of secondary ö æ Percentage of ö æ Volume of ö
è liquid in the water ø è Solution ø è Solution ø
Note as well that the percentage needs to be a decimal. So if we have an 80% solution we will
need to use 0.80. © 2007 Paul Dawkins 78 http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 10 How much of a 50% alcohol solution should we mix with 10 gallons of a 35%
solution to get a 40% solution?
Okay, let x be the amount of 50% solution that we need. This means that there will be...
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- Spring '12