SOLUTION: How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 60% copper in order to get an alloy that is 20% copper

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Question 1161874: How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 60% copper in order to get an alloy that is 20% copper

Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
---misread problem---

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Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.

Let x be the amount (ounces) of the 10% copper alloy to mix.


It contains 0.1x ounces of the copper.


The other ingredient is 400 ounces of the 60% copper alloy, which has 0.6*400 ounces of the copper.


The total copper in ingredients is  0.1x + 0.6*400 ounces.


It should be 0.2*(x+400) copper in the resulting alloy:


    0.1x + 0.6*400 = 0.2*(x + 400).


Simplify and solve


    x =  = 1600.


ANSWER.  1600 ounces of the 10% copper alloy should be mixed.


CHECK.   0.1*1600 + 0.6*400 = 400 ounces of the copper in ingredients;

         0.2*(1600 + 400) = 0.2*2000 = 400 ounces of the copper in the resulting mixture.

         ! Precisely correct !

Solved.

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

It is a standard and typical mixture word problem.

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.

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!


Here is a non-algebraic method for solving mixture problems like this.

If you understand this method, it will get you to the answer much faster and with less work than the traditional algebraic method shown by the other tutor.

Think of the problem this way:

You are starting with a 60% copper alloy and you are adding 10% copper alloy, reducing the percentage of copper; you stop when the percentage reaches 20%.

Now model that with numbers on a number line. You are starting at 60 and heading towards 10, but you stop when you get to 20.

What fraction of the distance have you gone? From 60 to 10 is a difference of 50, the distance you went, from 60 to 20, was 40. The fraction is 40/50 = 4/5.

That means 4/5 of the final alloy has to be the 10% alloy that you are adding.

So the 400 ounces you started with is 1/5 of the final alloy; that means the 4/5 of the final alloy that you added is 4*400 = 1600 ounces.

The words of explanation make this sound like a lengthy process; but it is not. Without the words of explanation, here is the complete solution:

(1) 60 to 10 is 50; 60 to 20 is 40; 40/50 = 4/5
(2) 4/5 of the final alloy is the 10% alloy you are adding; that's 4 times the 400 ounces you started with -- so 1600 ounces.


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