The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Silver Alloy

Date: 07/12/2001 at 00:31:55
From: Milad
Subject: Problem Solving

Sterling Silver is 92.5% pure silver. How many grams of pure silver 
and sterling silver must be mixed to obtain 100g of a 94% Silver 

Date: 07/12/2001 at 14:52:38
From: Doctor Greenie
Subject: Re: Problem Solving

Hello, Milad -

Thanks for sending your question to us here at Dr. Math.

I found several pages in the Dr. Math archives where similar problems 
are discussed by doing a search using the keyword  mixture .  You may 
want to perform that search yourself and look at some of the other 
explanations for similar problems provided by other math doctors.   

The traditional method for solving a problem like this (used in all 
the examples I found in the archives) is to write an equation relating 
the amounts of pure silver in the two "input" mixtures and in the 
"output" mixture.

In your problem you have one "input" mixture that is 92.5% silver and 
another that is 100% silver; and your "output" mixture is 94% silver.  
The amounts of the two input mixtures are unknown; the amount of the 
output mixture is 100g.  Let

   x = grams of 92.5% silver alloy

Then, since the total weight is 100g, we have

   (100-x) = grams of 100% silver

We now write an equation relating the amounts of pure silver in the 
two "input" mixtures and in the "output" mixture:

    "x" grams at 92.5% silver + (100-x) grams at 100% = 100g at 94%

    (x)(0.925) + (100-x)(1.0) = (100)(0.94)
             0.925x + 100 - x = 94
                            6 = 0.075x
                         6000 = 75x
                      6000/75 = x
                           80 = x

So to make 100g of an alloy of 94% silver, you need to mix 80g of 
92.5% alloy and 20g of pure silver.

And now here is a completely different approach to the same problem.  
I prefer this method, because I find the calculations are usually 
easier. Understanding why this method works is probably a bit more 
difficult than understanding the traditional method, but it works for 
me, so I use it.  Take a look at this alternative method and see if 
you like it.

We have two "input" mixtures; one of 92.5% silver and the other of 
100% silver. We want to make a mixture of 94% silver. If I think of 
plotting these percentages on a number line, I see that the "distance" 
from 92.5% to 94% is 1.5%, while the "distance" from 94% to 100% is 

And now here is the key to my method: The distances from 92.5% to 94% 
and from 94% to 100% are 1.5% and 6%; these two distances are in the 
ratio 1:4. This means that the two "input" mixtures must be mixed in 
the ratio 1:4 to get the 94% alloy.

If there are to be 100g of the 94% alloy and the two input mixtures 
are in the ratio 1:4, then there must be 20g of one input and 80g of 
the other. Because the resulting alloy is closer to 92.5% than 100%, 
the required amounts of the inputs are 80g of the 92.5% alloy and 20g 
of the 100% silver.

- Doctor Greenie, The Math Forum   
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Ratio and Proportion
Middle School Word Problems

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.