SOLUTION: A metallurgist has one alloy containing 22% copper and another containing 53% copper. How many pounds of each alloy must he use to make 51 pounds of a third alloy containing 30% co

Algebra ->  Expressions-with-variables -> SOLUTION: A metallurgist has one alloy containing 22% copper and another containing 53% copper. How many pounds of each alloy must he use to make 51 pounds of a third alloy containing 30% co      Log On


   

Question 1180838: A metallurgist has one alloy containing 22% copper and another containing 53% copper. How many pounds of each alloy must he use to make 51 pounds of a third alloy containing 30% copper? (Round to two decimal places if necessary.)

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
A metallurgist has one alloy containing 22% copper and another containing 53% copper.
How many pounds of each alloy must he use to make 51 pounds of a third alloy containing 30% copper?
(Round to two decimal places if necessary.)
~~~~~~~~~~~~~ 

Let x be the mass of the 53% copper alloy (in pounds).

Then the mass of the 22% copper alloy is (51-x) pounds.


Now you write equation, saying that the sum of amounts of pure copper in ingredients
is the same as that in the final alloy


    0.53x + 0.22*(51-x) = 0.3*51.


From the equation


    x = %280.3%2A51-0.22%2A51%29%2F%280.53+-+0.22%29 = 13.16  pounds (rounded).


ANSWER.  13.16 pounds of the 53% copper alloy and the rest,  (51-13.16) = 37.84 pounds of the 22% copper alloy.


CHECK.   I will check the concentration of the final alloy  %280.53%2A13.16+%2B+0.22%2A37.84%29%2F51 = 0.30, or 30%.

Solved.