Question

A metallurgist has one alloy containing 21% titanium and another containing 50% titanium. How many pounds of each alloy must he use to make 44 pounds of a third alloy containing 37% titanium? (Round to two decimal places if necessary.)

Answer 1

x = amount of 21% alloy

y = amount of 50% alloy

The metallurgist wants a combination weighing a total of 44 lb, so

x + y = 44

Each pound of either alloy contributes either 0.21 or 0.5 pound of titanium. The final product needs to be comprised of 37% titanium; weighing at 44 lb, this means it should contain 0.37 * 44 = 16.28 lb. So

0.21x + 0.5y = 16.28

From the first equation,

x + y = 44  ==>  y = 44 - x

Substitute this into the second equation and solve for x:

0.21x + 0.5(44 - x) = 16.28

-0.29x = -5.72  ==>  x = 19.72

Substitute this into the first equation to solve for y:

19.72 + y = 44  ==>  y = 24.28