Question 1162058: Alcore Inc. has two aluminum alloys on site. One is 52% aluminum and the other is 37% aluminum. How
much of each must they use to make 250 kilograms of an aluminum alloy that is 43% aluminum?
Found 3 solutions by ikleyn, Alan3354, greenestamps:
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
Let x = the amount of the pure aluminium in the 52% alloy, in kilograms.
Then the amount of the pure aluminium in the 37% alloy is (250-x) kilograms.
The amounts of the pure aluminium in the ingredients is the same as in the final alloys.
It gives you this equation
0.52x + 0.37*(250-x) = 0.43*250.
From the equation,
x =  = 100.
ANSWER. 100 kilograms of the 52% alloy and the rest, 250-100 = 150 kilograms of the 37% alloy.
CHECK. 100*0.52 + 150*0.37 = 107.5 kilograms of the aluminium in ingredients, and
250*0.43 = 107.5 kilograms of the aluminium in the final alloy.
! Precisely correct !
Solved.
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It is a standard and typical problem on mixtures. See the lessons
- 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
in this site.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
Read them and become an expert in solution 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 textbook in the section "Word problems" 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 Alan3354(69443)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a quick and easy alternative to the standard algebraic method for solving "mixture" problems like this.
(1) Determine, by whatever simple calculations you choose, that 43% is 6/15 = 2/5 of the way from 37% to 52%.
(2) That means 2/5 of the mixture has to be the 52% aluminum.
ANSWER: 2/5 of 250kg, or 100kg, of the 52% aluminum; the other 150kg of 37% aluminum.
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