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Question 1203490: 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 trophies?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39630)   (Show Source):
You can put this solution on YOUR website! This is only a start. Not a complete solution.
The percentages of materials in each alloy:
GOLD TUSTRITE MOTRIUM totalpercents
Sornog g t 0 100
Yotril 75 13 12 100
Trophy 72 20 8 100
Look at the values for the Motrium. Notice that 8 is  of 12.
Imagine using 200 parts of Yotril and 100 parts Sornog.
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Answer by ikleyn(52887)  (Show Source):
You can put this solution on YOUR website! .
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.
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 =  = 3*0.72-2*0.75 = 0.66.
It tells us that the percentage of gold in the used Sornog is 66%. ANSWER
Solved.
So, 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.
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
This problem can be worked using a formal algebraic method, as shown in the response from tutor @ikleyn.
It can also be solved quickly and easily by an informal method using the ratios of the two ingredients in the mixture.
The trophy is 8% motrium; sornog contains 0% motrium while yotril contains 12% motrium.
On a number line, 8% is two-thirds of the way from 0% to 12%. That means 2/3 of the mixture for the trophy is yotril; that in turn means there is 2 times as much yotril as sornog in the trophy alloy.
Now use that same idea of percentages on a number line to find the percentage of gold in sornog.
The percentage of gold in sornog is not known; call it x%. The percentage of gold in yotril is 75%; the percentage of gold in the trophy alloy is 72%. Since 2/3 of the trophy alloy is yotril, the calculation we need to do to find x is this:
72% is 2/3 of the way from x% to 75%
Use formal algebra or logical reasoning and simple arithmetic to find that the percentage of gold in sornog is 66%.
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