A metallurgist has one alloy containing 43% aluminum and another containing 63% aluminum. How many pounds of each alloy must he use to make 43 pounds of a third alloy containing 49% aluminum?

Question

A metallurgist has one alloy containing 43%aluminum and another containing 63% aluminum. How many pounds of each alloy must he use to make 43pounds of a third alloy containing 49% aluminum? 

Answer 1

67S + 43F = 43(49)

67S + 43(43-S) = 43(49)

(67-43)S = 43(49)-43(43)

24S = 6(43)

S = 6(43/24 = 43/4 = 10.75 pounds of 67% aluminum

F= 43-S = 43-10.75 = 32.25 pounds of 43% aluminium

check the answers

.67(10.75) +.43(32.25)

= 7.2025 + 13.8675

= 21.07

= .49(43)= 21.07

also just notice 43 is much closer to 49 than 67 is

so much more of 43% will be needed than 67%

49-43= 6

67-49= 18

18/6 =3 times as much 43 as 67%

32.25/10.75 = 3

the 43 showing up twice may cause some arithmetic mistakes

43 lbs and 43% is an annoying coincidence that could throw some people off in the calculations and algebra