Bored in math class, figured out mixture problems.

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This discussion focuses on solving a mixture problem involving chlorine concentrations. The goal is to create 100 gallons of a 40% chlorine mixture using 30% and 55% chlorine mixtures. The solution involves calculating the ratios of the two mixtures, resulting in a final ratio of 3:2, which translates to 60 gallons of the 30% mixture and 40 gallons of the 55% mixture. The mathematical approach includes setting up equations based on the concentrations and total volume, leading to the determination of the required amounts of each mixture.

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Mixture problem example: You need parts of a 30% chlorine and 55% chlorine mixture to make 100 gallons of a 40% chlorine. mixture.

To figure out what the ratio of 30 to 55 is: 55% - 40% = 15 parts needed of 35% mixture, 40% - 30% = 10 parts of 50% mixture.

15 parts of 30% reduces to 3 parts of 30%
10 parts of 55% reduces to 2 part of 55%

Then you take the ratio of the two (3:2), and multiply the ratio to get 60 gallons of the 30%, and 40 gallons of the 55%. (60:40)

Make sense to anyone? I'm not sure how I do the ratio part for fraction answers, but I have the beginning part out.

If already done, sue me. I figured it out in my own spare 10 minutes of class.
 
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It's easier to understand if you proceed that way:

The mixture z is composed of mixture y and mixture x. We know that z=100 Gallons. That gives x+y=100. You want to find x and y such as 30x/100 + 55y/100=40

So y=100-x

30x + 55(100-x)/100=40
30x + 5500 -55x=4000
-25x=-1500
x=60

so y=40
 

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