Question 1203490
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Sornog is an alloy composed of Gold and Tustrite, but contains no Motrium. 
Suppose Yotril is an alloy that is composed of 75% Gold, 13% Tustrite, and 12% Motrium. 
To make a Trophy for a contest, some Sornog is combined with some Yotril to produce a Trophy 
that is 72% Gold, 20% Tustrite, and 8% Motrium. 
What is the percentage of Gold in the Sornog that was used to make the Trophy?
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        From one point of view,  this problem is partly joke and partly entertainment.


        From the other side,  the solution assumes that the reader is a mature person 
        in solving mixture problems and has enough developed common sense or experience 
        to understand the solution below.



<pre>
The mass percentage formulas for participating materials are

    Sornog = (x   g,  y   t,  0   m)

    Yotril = (0.75g,  0.13t,  0.12m)

    Trophy = (0.72g,  0.20t,  0.08m)


         Here letters g, t, and m designate gold, Tustrite and Motrium, as elementary components;
         x and y symbolize the quantities. 


We combine "a" grams of Sornog and "b" grams of Yotril and get a+b grams of Trophy.


Looking for the Motrium component, we see that 

    0*a + 0.12b = 0.08(a+b),

or, simplifying

    0.12b = 0.08a + 0.08b  --->  0.12b - 0.08b = 0.08a  --->  0.04b = 0.08a  --->  b = 2a.

It tells us that "a" grams of the Sornog and 2a grams of the Youtril were melted to get a+2a = 3a grams of the Trophy.


Having this, we can write the mass equation for the gold components

    ax + (2a)*0.75 = (3a)*0.72.


From it, we express "x"

    x = {{{((3a)*0.72-(2a)*0.75)/a}}} = 3*0.72-2*0.75 = 0.66.


It tells us that the percentage of gold in the used Sornog is 66%.    <U>ANSWER</U>
</pre>

Solved.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, &nbsp;this problem is a joke and entertainment problem for specialists.



At the school level, it is accessible only for those advanced students,
who are able (are trained) to solve problems of a Math Olympiad level.