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Let x be the mass of the 50% silver alloy (in kilograms).
Then the mass of the 25% silver alloy is (5-x) kilograms.
The amount of the pure silver in ingredients is 0.5x + 0.25*(5-x) kilograms.
It should be equal to the amount of the pure silver in the resulting 40% alloy
0.5x + 0.25*(5-x) = 0.4*5 kilograms.
From this equation,
x = = 3 kilograms.
ANSWER. 3 kilograms of the 50% silver alloy and the rest, (5-3) = 2 kilograms of the 25% silver alloy.
CHECK. 0.5*3 + 0.25*2 = 2 kilograms of the pure silver in ingredients, and
0.4*5 = 2 kilograms of the pure silver in the 40% alloy. ! 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
- Advanced mixture problems
- Advanced mixture problem for three alloys
- Unusual word problem on mixtures
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.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
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.