A metallurgist has one alloy containing 40% aluminum and another containing 61% aluminum how many pounds of each alloy must he use to make 57 pounds of a third alloy containing 46% aluminum
Let x be the number of pounds of the alloy containing 40% aluminum.
Then the amount of aluminum in that alloy is 0.4x.
The number of pounds of the alloy containing 61% aluminum is 57-x.
Then the amount of aluminum in that alloy is 0.61(57-x).
The total amount of aluminum in the final alloy is 0.46*57.
So we have the equation:
0.4x+0.61(57-x)=0.46*57
0.4x+34.77-0.61x=26.22
0.21x=8.55
x=40.714285714285715
So he needs to use approximately 40.71 pounds of the alloy containing 40% aluminum, and 16.28 pounds of the alloy containing 61% aluminum.