Problem Solving.... Problem

[TheTonik]

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?

[Pengwuino]

Well think about it, dont be fooled by numbers and percentages :D.

If theres 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%?

[TheTonik]

Well think about it, dont be fooled by numbers and percentages :D.

If theres 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:

[HappMatt]

wow here is one where i actually think i can help

Part A: heres 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.

[EnumaElish]

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

[plucker_08]

(10/0.3) - 20

that's the answer!