SOLUTION: A lab technician needs to make 12 liters of 45% alcohol. How much of 40% alcohol and 60% alcohol must he mix to produce this?

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Question 1175706: A lab technician needs to make 12 liters of 45% alcohol. How much of 40% alcohol and 60% alcohol must he mix to produce this?
Found 5 solutions by josgarithmetic, MathTherapy, ikleyn, ewatrrr, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
40 to 60 can be cut into 4 parts of 5 liters each.
45% is farther away from 60 than it is from 40.
Most of the parts to use must be of the 60% alcohol.

~~Use 3 liters of 40% alcohol and 9 liters of 60% alcohol.~~

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

A lab technician needs to make 12 liters of 45% alcohol. How much of 40% alcohol and 60% alcohol must he mix to produce this?

He's WRONG!!
MORE of the 40% solution than the 60%, needs to be mixed. So, IGNORE these: ~~Most of the parts to use must be of the 60% alcohol.~~
~~Use 3 liters of 40% alcohol and 9 liters of 60% alcohol.~~

Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website!
.

            Both the solution and the answer by @josgarithmetic are INCORRECT.

            I came to bring a correct solution.

You mix x liters of the 40% alcohol solution and (12-x) liters of the 60% alcohol solution. Under this approach, the total volume of 12 liters of the resulting mixture is just provided, and all you need to care about is to provide the requested concentration of 45% for the mixture. So, you write this equation for the concentration = 0.45, (1) or, which is the same 0.4x + 0.6*(12-x) = 0.45*12 (2) From this equation x = = 9 liters. ANSWER. 9 liters of the 40% solution and the rest, (12-9) = 3 liters of the 60% solution. CHECK. = 0.45 = 45% final concentration ! Correct !

Solved.

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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 ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

.60x + .40(12-x) = .45(12L) x = .05(12L)/.20 = 3L of the 60% solution, 9L of the 40%

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

The METHOD used by tutor @josgarithmetic is the fastest and easiest method for solving 2-part mixture problems like this.

However, the conclusion he draws from his calculations is not correct.

The fast and easy solution using his method is this:

45% is 1/4 of the way from 40% to 60%;
therefore 1/4 of the mixture is the 60% alcohol.

ANSWER: 1/4 of the 12 liters, or 3 liters, is the 60% alcohol; the other 3/4 of the mixture, or 9 liters, is the 40% alcohol.