SOLUTION: How many grams of pure silver must a silversmith mix with 100 grams of 55% silver alloy to produce 75% silver alloy? How many grams of 75% silver alloy will result?

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Question 1121352: How many grams of pure silver must a silversmith mix with 100 grams of 55% silver alloy to produce 75% silver alloy? How many grams of 75% silver alloy will result?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = "How many grams of pure silver must a silversmith mix".


Then the total mass is  (100 + x) grams, and the concentration equation is


    %280.55%2A100+%2B+x%29%2F%28100%2Bx%29 = 0.75.      <<<---===  The ratio of the pure silver mass of the alloy to the total mass


Simplify and solve step by step


    0.55*100 + x = 0.75*(100 + x)


    0.55*100 + x = 0.75*100 + 0.75x


    x - 0.75x = 75 - 55


    0.25x = 20  ====>  x = 20%2F0.25 = 80.


Answer.  80 grams of pure silver should be mixed.  The total mass of the resulting alloy then is 180 grams.


Check.   %280.55%2A100%2B80%29%2F%28100%2B80%29 = 0.75     ! Correct !

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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
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(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a completely different way of solving mixture problems like this, where two "ingredients" are being mixed.

I find this method much easier and faster than the traditional algebraic method.

(1) Find where the percentage of the mixture (the final alloy, 75%) lies between the percentage of the original ingredient (55%) and the percentage of the ingredient being added (pure silver, 100%):
100-55 = 45
75-55 = 20
20/45 = 4/9

(2) The percentage of the final alloy is 4/9 of the way from 55% to 100%.

That means 4/9 of the mixture must be the ingredient that is being added.

So let 4x be the amount of the ingredient being added and 9x be the amount of the final alloy; that makes 5x the amount of the original alloy.

Since the amount of the original alloy was 100g, 5x=100g --> x=20g. So the amount of pure silver that needs to be added is 4x = 80g.

And the amount of the final alloy is 5x+4x = 9x = 180g.