SOLUTION: A jeweller mixes two different alloys of sterling silver, one containing 86% silver with the other containing 76% of silver to make 100g of a new alloy with 82% silver overall. How

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Question 1143954: A jeweller mixes two different alloys of sterling silver, one containing 86% silver with the other containing 76% of silver to make 100g of a new alloy with 82% silver overall. How many grams of each kind of sterling silver alloy did the jeweller use?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
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one containing 86% silver with the other containing 76% of silver to make 100g of a new alloy with 82% silver overall.
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If x is mass of the 86% silver alloy, then
.
Answer by ikleyn(52921)   (Show Source): You can put this solution on YOUR website!
.
A jeweler mixes two different alloys of sterling silver, one containing 86% silver with the other containing 76% of silver
to make 100 g of a new alloy with 82% silver overall. How many grams of each kind of sterling silver alloy did the jeweler use?


Solution

            There are two basic Algebra methods to solve this and similar problems.

            One method is based on "one equation setup".
            The other method is based on "two equations setup".

            I will show both methods below. Which method to use depends on your level; on which of the two methods
            you currently learn and use in the class; or, at other equal conditions, on which of the two methods you prefer.


                        One equation setup. 

Let x be the mass (in grams) of the 86% silver alloy.

Then the mass of the 76% silver alloy is (100-x) grams (since the combined mass should be 100 grams).


x grams of the 86% alloy contains 0.86*x grams of silver.

(100-x) grams of the 76% alloy contains 0.76*(100-x) grams of silver.


Since the resulting alloy of the mass of 100 g should contain 82% of silver,
you have this balance equation for the pure silver amount


    0.86x + 0.76*(100-x) = 0.82*100  grams of the pure silver.    (1)


At this point, the setup for the problem (and for this method) is completed.

Now your task is to solve this equation.


For it, simplify it step by step.


    0.86x + 76 - 0.76x = 82

    0.86x - 0.76x = 82 - 76

    0.1x          = 6

       x          =  = 60.


Thus you just got the ANSWER:   you need to mix  60 grams of the 86% alloy with  100-60 = 40 grams  of the 76% alloy.


CHECK.  To sheck, I will calculate concentration of the resulting alloy, by dividing the mass of the silver

        0.86*60 + 0.76*40 = 82 grams  by the total mass of the alloy of 100 grams

        concentration =  =  = 0.82 = 82%.   ! Correct !

The solution via "one-equation setup" is completed.


                        Two-equations setup. 

Let x be the mass (in grams) of the 86% silver alloy, and

let y be the mass (in grams) of the 76% silver alloy.


Then from the condition, you have these two equations


        x +     y = 100       gram of the alloy total mass                      (2)

    0.86x + 0.76y = 0.82*100  gram of the pure silver in the resulting alloy    (3)


At this point, the setup for the problem (and for this method) is completed.

Now your task is to solve this system of two equations in two unknowns.


For it, from equation (2), express  y = 100-x  and substitute it into equation (2). You will get


        0.86x + 0.76*(100-x) = 0.82*100  grams of the pure silver.    (4)


Thus you get the same single equation for x, as equation (1) in the method above.

Now solve it by the ame way:


    0.86x + 76 - 0.76x = 82

    0.86x - 0.76x = 82 - 76

    0.1x          = 6

       x          =  = 60.


Of cource, you get the same ANSWER:   you need to mix  60 grams of the 86% alloy with  100-60 = 40 grams  of the 76% alloy.


To check it, substitute the found values in each of the original equations (2) and (3).   // Please do it on your own . . .


The solution is completed, and you are familiar now with two basic methods of solution to this problem.

Having this LUXURY explanation, you are now armed to solve any similar mixture word problem.

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

It is a typical mixture problem.

There is entire bunch of 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 dry substances like coffee beans, nuts, cashew and peanuts
    - Word problems on mixtures for dry substances like candies, dried fruits
    - Word problems on mixtures for dry substances like soil and sand
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
    - OVERVIEW of lessons on word problems for mixtures
in this site.

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 the 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 online textbook 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(13216)   (Show Source): You can put this solution on YOUR website!


The standard algebraic solution method for mixture problems like this, shown by the other tutors, is certainly a valid method; and it is a good method for teaching techniques for solving problems using algebra.

But there is a much easier and faster way to solve mixture problems like this involving mixing two ingredients.

Here is this alternate method for solving this example.

(1) The target 82% is six-tenths of the way from the 76% of one alloy to the 86% of the other alloy. (Picture the three percentages -- 76, 82, and 86 -- on a number line; 82 is 6/10 of the way from 76 to 86.)

(2) That means 6/10 = 3/5 of the mixture must be the 86% alloy.

ANSWER: 3/5 of 100g = 60g of the 86% alloy; the remaining 40g of 76% alloy.
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