Optimizing Antifreeze Mixture: How Much Water to Add?

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To achieve a 30% antifreeze mixture from a 20-gallon solution that is currently 50% antifreeze, one must add approximately 13.3 gallons of water. The original mixture contains 10 gallons of antifreeze, and to maintain this amount at 30%, the total volume must increase to about 33.3 gallons. This calculation shows that the additional water required is 13.3 gallons, not the previously suggested 7.5 or 12 gallons. Understanding the relationship between the antifreeze and water volumes is crucial for accurate mixing. The final answer confirms the need for careful calculation to achieve the desired concentration.
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Problem Solving... Problem

For the love of God I need some help... Here is the question:

How much water should be added to 20 gallons of a solution that is 50% antifreeze in order to get a mixture that is 30% antifreeze?


I came up with two answers but am not sure if either are correct: 7.5 gallons and 12 gallons.

Can anyone help me?
 
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Well think about it, don't be fooled by numbers and percentages :D.

If there's 50% antifreeze in 20 gallons of solution... how many gallons of antifreeze is there? Then, how many gallons overall need to exist for that # of gallons to equal 30%?
 
Pengwuino said:
Well think about it, don't be fooled by numbers and percentages :D.

If there's 50% antifreeze in 20 gallons of solution... how many gallons of antifreeze is there? Then, how many gallons overall need to exist for that # of gallons to equal 30%?


Hmm... still having trouble... :confused:
 
wow here is one where i actually think i can help

Part A: here's a quick tip for figuring it out.

(10gallons of antifreeze)/(20gallons of water-antifreeze solution) = a 50% solution.

so if i wanted to make that 10 gallons of anti freeze only 30% of the solution i would have to add more water to the mixture. So you might have to add X number of gallons to the mixture. Make an equation that shows adding X number of gallons to the mixture and then make that equation equal to X.
 
TheTonik said:
For the love of God I need some help... Here is the question:

How much water should be added to 20 gallons of a solution that is 50% antifreeze in order to get a mixture that is 30% antifreeze?


I came up with two answers but am not sure if either are correct: 7.5 gallons and 12 gallons.

Can anyone help me?
Denote antifr. as AF and water as WR. So you have 10 AF + 10 WR = 20gal. mixture with 50% AF

You want 10 AF + X WR = 33.3 gal mixture with 30% AF (we know this because 10/33.3 = 0.3)

X = 33.3 - 10 = 23.3

Need to add extra WR = 23.3 - 10 = 13.3 gal
 
(10/0.3) - 20

that's the answer!
 
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