SOLUTION: You need a 65% alcohol solution. On hand, you have a 15 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain t

Question 1144544: You need a 65% alcohol solution. On hand, you have a 15 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain the desired solution?
Answer by ikleyn(52802) (Show Source): You can put this solution on YOUR website!
.


Let x = the amount of the 70%  alcohol needed (in milliliters.

Then the amount of the resulting 65%  alcohol solution will be is (x+15) milliliters.


The amount of the "pure" alcohol in 15 mL of the 25% alcohol solution is 0.25*15 mL.

The amount of the "pure" alcohol in x mL of the 70% alcohol solution is 0.70*x mL.

The total amount of the "pure" alcohol in ingredients is the sum 0.25*15 + 0.70*x mL.


It should be equal to the amount of the "pure" alcohol in the mixture, which is 0.65*(x+15) ml.



So, your "pure alcohol" equation is 

   0.70*x + 0.25*15 = 0.65*(x+15).



Simplify and solve it for x:

    0.70x + 0.25*15 =  0.65x + 0.65*15,

    0.70x - 0.65x = 0.65*15 - 0.25*15,

    0.05x = 6,

    x =  = 120.

Answer. 120 mL of 70% alcohol solution are needed.


Check.  0.07*120 + 0.25*15 = 87.5 mL of the pure alcohol in ingredients, and

        0.65*(120+15) = 87.75 mL of the pure alcohol in the final mixture.   ! Precisely correct !

------------------

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.

RELATED QUESTIONS
You need a 65% alcohol solution. On hand, you have a 25 mL of a 10% alcohol mixture. You... (answered by josgarithmetic)

You need a 45% alcohol solution. On hand, you have a 80 mL of a 40% alcohol mixture. You... (answered by josgarithmetic)

You need a 20% alcohol solution. On hand, you have a 180 mL of a 5% alcohol mixture. You... (answered by jorel555)

You need a 40% alcohol solution. On hand, you have a 375 mL of a 5% alcohol mixture. You... (answered by josgarithmetic)

You need a 30% alcohol solution. On hand, you have a 440 mL of a 5% alcohol mixture. You... (answered by josgarithmetic)

You need a 70% alcohol solution. On hand, you have a 30 mL of a 30% alcohol mixture. You... (answered by ankor@dixie-net.com)

You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You... (answered by ikleyn,amalm06,richwmiller)

You need a 80% alcohol solution. On hand, you have a 25 mL of a 50% alcohol mixture. You... (answered by ikleyn)

You need a 20% alcohol solution. On hand, you have a 400 mL of a 15% alcohol mixture. You (answered by josgarithmetic)