Question 1203028: By volume, one alloy is 60, copper, 30, zinc, and 10, nickel. A second alloy has percents of 50,
30, and 20, respectively. A third alloy is 30, copper and 70, nickel. How much of each alloy is
needed to make 100 cm3 of a resulting alloy with percents of 40, 15, and 45, respectively?
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39617)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
x = cm^3 of alloy A
y = cm^3 of alloy B
z = cm^3 of alloy C
The copper equation: 60% of x, plus 50% of y, plus 30% of z, is equal to 40% of the total 100 cm^3:
 [1]
The zinc equation: 50% of x, plus 30% of y, and none of z, is equal to 15% of the total 100 cm^3:
 [2]
The nickel equation: 30% of x, plus 20% of y, plus 70% of z, is equal to 45% of the total 100 cm^3:
 [3]
For an algebraic solution, since equation [2] involves only x and y, eliminate z between equations [1] and [3] to get a second equation involving only x and y. Then that pair of equations can be solved for x and y, leading to a solution of the whole problem.
Multiply [1] by 7 and [3] by 3 and find the difference to eliminate z:
 [4]
Now solve [2] and [4].
Multiply [2] by 33 and [4] by 5....
ANSWER: none of alloy A, and 50 cm^3 each of alloys B and C
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Actually, it is very naïve approach to think that volume of an alloy is the sum of volumes of composites.
It is well known fact for all metallurgists just more than several hundreds years
that due to shrinking effect, the volume of an alloy, as a rule, is not equal to the sum
of volumes of composite metals.
I know and remember it from learning Science in my middle school years
(which means that this phenomenon is of the commonly known level of knowledge).
For further information and to develop your knowledge, look, read and learn from these sources
https://www.diva-portal.org/smash/get/diva2:218833/FULLTEXT01.pdf
https://nvlpubs.nist.gov/nistpubs/jres/8/jresv8n1p37_A2b.pdf
(publication of the 1932 year).
Having enough free time, anyone can find hundred of articles, tens textbooks, monographs
and handbooks in metallurgy and material science that teach this phenomenon, which is important
and essential for alloys.
Therefore, a competent Math problems composer will NEVER
propose such a problem, as in this post, to school students.
Masses can be added for alloys, but volumes not.
Adding volumes of alloys to find the volume of the final product is the sign of a general
incompetence in the subject.
If in this problem the numbers / (the quantities) be the masses in consistent units,
it would be a normal / (correct) problem with the solution as in the post by another tutor.
But with the given interpretation, when the given quantities are the VOLUMES,
the problem is like a lame horse and does not suit for proper teaching of students.
The cause WHY the volume of the resulting alloy is not the sum of volumes
of participating components is that crystal structure of the resulting alloy
is DIFFERENT from that of the components.
The phenomenon I am talking about, is close (although is not identical)
to the fact that the volume of ice is greater than the volume of water.
It is why icebergs float on the surface of water . . .
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