a metallurgist has one alloy containing 21% copper and another conatining 60% copper. how many pounds of each alloy must he use to make 47 pounds of a third alloy containg 40% copper

Question

a metallurgist has one alloy containing 21% copper and another conatining 60% copper. how many pounds of each alloy must he use to make 47 pounds of a third alloy containg 40% copper

Answer 1

Approximately 23 1/2 pounds of each

But slightly more of the 21% copper than 60% copper, so a rough guess might be 24 lbs of 21% and 26 lbs of 60%

.6x + .21(47-x) = ..4(47) solve for x = the pounds of 60%

.6x - .21x = .4(47) -.21(47)

.39x = .19(47)

x = .19(47)/.39 = about 22.9 lbs of 60%, or about 24.1 lbs of 21%

you know it should be roughly equal amounts because 40 is about halfway between 21 and 60, but slightly more of the 21 since it is slighter closer to 40 than 60 is

60-40 = 20

40-21 = 19

the ratio of 21 to 60 should be = 20/19

24.1/22.9 = about 1.05

20/19 = about 1.05