SOLUTION: a graphic artist needs to get a just the right color for a project. How many liters of 50 % dye solution should be add to 10 liters of 80% dye solution to get a 70 % dye solution

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Question 1153358: a graphic artist needs to get a just the right color for a project. How many liters of 50 % dye solution should be add to 10 liters of 80% dye solution to get a 70 % dye solution
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
DYE solutions of 50%? 80%? 70%? Are these from solid dyestuff powders? Is dissolution done just in water? Any use of an alcohol? Are these dispersions and not just "solutions"?

x liters of the 50%
10 liters of the 80%

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Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website!
.

Add " w " liters of the 50% solution. Then you have this equation 0.5*w + 0.8*10 = 0.7*(10+w). Simplify and solve for w 0.5w + 8 = 7 + 0.7w 8 - 7 = 0.7w - 0.5w 1 = 0.2w w = = 5. ANSWER. Add 5 liters of the 50% solution.

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It is a standard and typical mixture problem.

For introductory lessons covering various types of mixture word problems see
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

If an algebraic solution is not required, here is a common sense method that will get you to the answer much faster than algebra.

(1) Use any number of different kinds of calculations to determine that the target 70% is "twice as close to 80% as it is to 50%"

(2) Therefore, the amount of the 80% ingredient must be twice the amount of the 50% ingredient.

(3) Since there are 10 liters of the 80% solution, there must be 5 liters of the 50% solution.

ANSWER: 5 liters