A metallurgist has one alloy containing 24% aluminum and another containing 49% aluminum. How many pounds of each alloy

Question

A metallurgist has one alloy containing 24%24% aluminum and another containing 49%49% aluminum. How many pounds of each alloy must he use to make 5959 pounds of a third alloy containing 29%29% aluminum? (Round to two decimal places if necessary.)

Answer 1

Let "a" be the weight of the 24% alloy and let "b" be the weight of the 49% alloy.

Then a + b = 5959 or a = 5959 - b

Now, if you multiply "a" and 0.24 (24% in decimal form) it will give you the amount of pure aluminum in alloy "a". And if you multiply "b" and 0.49 (49% in decimal form) it will give you the amount of pure aluminum in alloy "b". And the sum of those must equal the amount of pure aluminum in the third alloy which will be 0.29 times 5959 so:

0.24a + 0,49b = 0.29(5959)

0.24a + 0,49b =1728.11

Substituting "5959 - b" into this equation in place of "a" we get:

0.24(5959 - b) + 0,49b =1728.11

1430.16 - 0.24b + 0,49b = 1728.11

0.25b = 297.95

b = 1191.8

Then back substituting into a = 5959 - b gives a = 4767.2

So, the metallurgist must use 4767.2 pounds of the 24% alloy and 1191.8 pounds of the 49% alloy