Question 1100733: You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 70% alcohol mixture. How much of each mixture will you need to add to obtain the desired solution?
Found 3 solutions by ikleyn, amalm06, richwmiller:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 70% alcohol mixture.
How much of each mixture will you need to add to obtain the desired solution?
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Let X = "How much of the 30% mixture to mix", in mL.
Then the volume of the 70% mixture to mix is (600-X) mL.
X mL of the 30% mixture contribute 0.3*X mL of the pure alcohol to the new mixture.
(600-X) mL of the 70% mixture contribute 0.7*(600-X) mL of the pure alcohol to the new mixture.
In all, two input mixtures provide 0.3*X + 0.7*(600-X) mL of the pure alcohol in the output mixture.
The output volume is 600 mL, and it contains 0.3*X + 0.7*(600-X) mL of the pure alcohol, as we find out above.
Hence, the concentration of the new mixture is  as the fraction.
According to the condition, it must be 45% or 0.45. It gives you an equation
= 0.45.
It is you major/basic equation, so called "concentration" equation.
As soon you got it (and understand it), the setup part is completed.
To solve the equation, first multiply both sides by 600. You will get
0.3*X + 0.7*(600-X) = 0.45*600.
Simplify and solve it:
0.3X + 0.7*600 - 0.7X = 0.45*600 ====> -0.4X = 0.45*600-0.7*600 = -0.25*600 ====> X =  = 375.
Thus you need 375 mL of the 30% mixture and 600-375 = 225 mL of the 70% mixture.
Check.  = 0.45 ! correct concentration !
Answer. You need 375 mL of the 30% mixture and 225 mL of the 70% mixture.
Solved.
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There is entire bunch of introductory lessons covering various types of mixture problems
- Mixture problems
- More Mixture problems
- Solving typical word problems on mixtures for solutions
- Word problems on mixtures for antifreeze solutions
- Word problems on mixtures for alloys
- Typical word problems on mixtures from the archive
in this site.
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 amalm06(224)  (Show Source):
You can put this solution on YOUR website! There are two ways to solve the problem. One is the long method:
0.30(x)+(0.70)(y)=0.45(600), where x is the required amount of 30% alcohol and y is the required amount of 70% alcohol
x+y=600
By solving this system of equations, you can find the amount of each mixture needed to obtain the desired solution
The other way to solve the problem is with the method of alligation
Let s1 denote the 30% alcohol mixture and s2 denote the 70% alcohol mixture
Then 0.70-0.45=0.25 and 0.45-0.30=0.15
So that s1/s2=0.25/0.15
Thus, for every 0.40 parts of total mixture, there are 0.25 parts of 30% alcohol solution
If there are 600 parts of total mixture, then the amount of 30% alcohol solution can be found as follows:
s1=(0.25/0.40)(600)=375 mL (Answer)
Similarly, the amount of 70% alcohol solution is given by:
s2=600-375=225 mL (Answer)
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! Alligation is the preferred method pharmacists learn for problems like this when the wanted/needed mixture amount and percentage is known.
Algebra is also used but alligation is usually the preferred method.
Alligation can be even be used when three items are mixed to get a fourth.
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