SOLUTION: If drink A contains 3% milk and drink B contains 84% milk. How much of each solution should be mixed in order to create 135 liters of a solution that contains 78% milk?

Question 1149432: If drink A contains 3% milk and drink B contains 84% milk. How much of each solution should be mixed in order to create 135 liters of a solution that contains 78% milk?
Found 3 solutions by josmiceli, ikleyn, greenestamps:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = liters of 3% milk needed
Let = liters of 84% milk needed
-----------------------------------------------
(1) liters
(2)
------------------------------------------
(2)
(2)
(2)
Subtract (1) from (2)
(2)
(1)
-----------------------------

and
(1)
(1)
---------------------------
10 liters of 3% milk are needed
125 liters of 84% milk are needed
-----------------------------------------
check:
(2)
(2)
(2)
(2)
OK

Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.

Let x be the volume of drink A and y be the volume of drink B. Drink A contains 0.03*x liters of milk. Drink B contains 0.84*y liters of milk. The mixture has the volume of (x+y) liters. The mixture contains 0.78*(x+y) liters of milk. So you have this system of 2 equations in 2 unknowns x + y = 135 liters (1) 0.03*x + 0.84*y = 0.78*135 (2) Equation (2) says that the volume of the milk in the mixture is the sum of the milk volumes in ingredients. From equation (1), express x = 135 - y and substitute it into equation (1). You will get 0.03*(135-y) + 0.84*y = 0.78*135. From this equation, express y and calculate y = = 125. ANSWER. 125 liters of the 84% drink should be nixed with (135-125) = 10 liters of the 3% drink. CHECK. Calculate concentration of the mixture = 0.78 = 78%. ! Precisely correct !

Solved.

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

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!

Try this as an alternative to the standard algebraic process for solving "mixture" problems, as shown in the other two responses you have received.

If you understand this method (and an algebraic solution is not required) it will get you to the answer much faster and with much less work.

(1) You are mixing 3% milk and 84% milk ingredients to get a mixture that is 78% milk.
(2) 78% is 25/27 of the way from 3% to 84%. (Look at the three numbers on a number line -- 3, 78, and 84. 3 to 78 is 75; 3 to 84 is 81; 75/81 = 25/27.
(3) That means 25/27 of the mixture is the higher percent ingredient.

ANSWER: 25/27 of the 135 liters = 125 liters of the 84% ingredient; 2/27 of 135 liters = 10 liters of the 3% ingredient.

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