SOLUTION: You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Question 1061149: You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture.
How much of each mixture will you need to obtain the desired solution?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the amount (the volume in mL) of the 25% solution to be mixed. Then the volume of the 60% solution is 420-x. The "alcohol amount" equation is 0.25x + 0.6(420-x) = 0.35*420. Simplify and solve for x. 0.25x + 252 - 0.6x = 147, 0.25x - 0.6x = 147 - 252, 0.35x = 105 ---> x = = 300. Answer. 300 mL of the 25% solution and 420-300 = 120 mL of the 60% solution.

There is entire bunch of 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 the 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 online textbook under the topic "Mixture problems".

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = ml of 25% mixture needed
Let = ml of 60% mixture needed
----------------------------------------
(1)
(2)
----------------------------------------
(2)
(2)
(2)
(2)
----------------------------------
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
-----------------------------

and
(1)
(1)
------------------------------
300 ml of 25% mixture is needed
120 ml of 60% mixture is needed
------------------------------
check:
(2)
(2)
(2)
(2)
(2)
OK

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