Optimizing Antifreeze Mixture: How Much Water to Add?

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Discussion Overview

The discussion revolves around determining how much water should be added to a 20-gallon solution that is 50% antifreeze to achieve a final mixture that is 30% antifreeze. The scope includes problem-solving and mathematical reasoning related to mixtures and percentages.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about their calculations, suggesting two possible amounts of water to add: 7.5 gallons and 12 gallons.
  • Another participant prompts a reevaluation of the problem by asking how many gallons of antifreeze are present in the original solution and how many gallons are needed for that amount to represent 30% of the total mixture.
  • A different participant proposes a method for solving the problem by setting up an equation based on the antifreeze concentration and the total volume after adding water.
  • One participant reformulates the problem using variables for antifreeze and water, leading to a calculation that suggests needing to add 13.3 gallons of water to achieve the desired concentration.
  • Another participant provides a brief mathematical expression as a potential solution without elaboration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct amount of water to add, as multiple calculations and approaches are presented, leading to different proposed solutions.

Contextual Notes

There are unresolved assumptions regarding the calculations and the methods used to derive the amounts of water needed, as well as potential errors in the reasoning presented by different participants.

TheTonik
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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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