SOLUTION: A 4% solution of local anesthesia is needed. However, 40 fl oz of a 5% solution is the only type of anesthesia in stock. How much neutral solution containing no anesthesia should b

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Question 1106175: A 4% solution of local anesthesia is needed. However, 40 fl oz of a 5% solution is the only type of anesthesia in stock. How much neutral solution containing no anesthesia should be added to the 40 fl oz container of the 5% solution?

Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
4% is 4%2F5 of 5%.
-
40%2A5%2F%2840%2Bv%29=4

50%2F%28v%2B40%29=1

50=v%2B40

v=50-40

highlight%28v=10%29, amount volume of neutral material to add



NOTE: 40%2F%2810%2B40%29=40%2F50=highlight_green%284%2F5%29

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
anesthetic + anesthetic = total anesthetic     (pure anesthetic liquid volume in milliliters)


0.05*40    + 0*w        = 0.04*(40+w)          (0.05 = 5%; 0.04 = 4%;  "w" is the volume of neutral solution to add)


2                       = 1.6 + 0.04w

0.04w = 2 - 1.6  ====>  0.04w = 0.4  ====>  w = 0.4%2F0.04 = 10.


Answer.  10 mL of the neutral solution is needed.


Check.   0.05*40 = 2 mL.   0.04*(40+10) = 0.04*50 = 2 mL.   ! Correct !

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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
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
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) About Me  (Show Source):
You can put this solution on YOUR website!


Tutor ikleyn gave you a solution by the standard algebraic method.

Tutor josgartihmetic hinted at a faster method for solving the problem; but that answer was probably so cryptic that you might well not have understood it.

So let me expand on the solution method the second tutor hinted at.

The key to the fast method of solution for this problem is that no anesthetic (not anesthesia) was added. So the amount of anesthetic is the same after the neutral solution is added.

So all you need to do to solve the problem is answer this question: 5% of 40 fluid ounces is 4% of how many fluid ounces?

Simple mental arithmetic tells you 5% of 40 is the same as 4% of 50.

So the final solution is 50 fluid ounces; since it was originally 40, the amount of neutral solution added was 10 fluid ounces.