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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.
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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.