SOLUTION: In solving mixture problems in algebra, is it always true that water is 0% while a pure acid is 100%? (ex: How many grams of a pure acid solution must be added to a 36g of a 49% so

Question 371213: In solving mixture problems in algebra, is it always true that water is 0% while a pure acid is 100%? (ex: How many grams of a pure acid solution must be added to a 36g of a 49% solution to produce a 50% solution?.. In this example, will 'pure acid' be denoted as 100%?) Thank you. =)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
In solving mixture problems in algebra,
"is it always true that water is 0% while a pure acid is 100%?"
:
I would say it is usually true, it is conceivable that you may write an
equation that is based on percent water instead of percent something else
In that case water would be 100% and some solution would be written as percent water.
:
(ex: How many grams of a pure acid solution must be added to a 36g of a 49% solution to produce a 50% solution?.. In this example, will 'pure acid' be denoted as 100%?)
Yes
Write the decimal equiv of the given percentages then pure acid would be 1x or just x.
:
In your example you have 36 liters of the 49% solution
x = amt of pure acid required:
:
1x + .49(36) = .50(x+36)
x + 17.64 = .5x + 18
x - .5x = 18 - 17.64
.5x = .36
x = 2(.36)
x = .72 liters of pure acid required
:
:
You can confirm the solution by substituting .72 for x and check for equality
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