SOLUTION: To meet the governments specifications, an alloy must be 65% aluminum. How many pounds each of a 70% aluminum alloy and a 54% aluminum alloy will be needed to produce 640 pounds of

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Question 58109: To meet the governments specifications, an alloy must be 65% aluminum. How many pounds each of a 70% aluminum alloy and a 54% aluminum alloy will be needed to produce 640 pounds of the 65% aluminum alloy?
Any help with this one is appreciated!

Found 2 solutions by hayek, ankor@dixie-net.com:
Answer by hayek(51) About Me  (Show Source):
You can put this solution on YOUR website!
Amount of 70% Al: Al%5B70%5D
Amount of 54% Al: Al%5B54%5D
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The problem tells us to mix the two.
The total amount of pure aluminum in the mix will be: (Eq. 1) Al%5Btot%5D=Al%5B70%5D%2A%28.70%29%2BAl%5B54%5D%2A%28.54%29
The total amount of material in the mix Al%5B70%5D%2BAl%5B54%5D,must equal 640 pounds ==> (Eq. 2) Al%5B70%5D%2BAl%5B54%5D=640
We are looking for a final mix of 640 pounds of 65% Al. Thus, the total aluminum content of the final is:
Al%5Btot%5D=640%2A.65
Al%5Btot%5D=416.
Plugging that back into Eq. 1 gives:
416=Al%5B70%5D%2A%28.70%29%2BAl%5B54%5D%2A%28.54%29
Solving Eq. 2 for Al%5B70%5D==> Al%5B70%5D=640-Al%5B54%5D
Plug that into the previous equation:
416=%28640-Al%5B54%5D%29%2A.7%2BAl%5B54%5D%28.54%29
416=448-.7%2AAl%5B54%5D%2B.54%2AAl%5B54%5D
416=448-.16%2AAl%5B54%5D
416-448=-.16%2AAl%5B54%5D
Al%5B54%5D=%28416-448%29%2F-.16
highlight%28Al%5B54%5D=200%29 pounds.
Plug this into Eq. 2: Al%5B70%5D=640-200 ==> highlight%28Al%5B70%5D=440%29pounds.





Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
To meet the government's specifications, an alloy must be 65% aluminum. How many pounds each of a 70% aluminum alloy and a 54% aluminum alloy will be needed to produce 640 pounds of the 65% aluminum alloy?
:
A typical mixture problem, if you learn this method you will be able to handle most mixture-type problems.
:
Let x = the amt of the 70% alloy; Since the total will be 640 lb,
the 54% amt = (640-x)
:
.70x + .54(640-x) = .65(640)
.70x + 345.6 -.54x = 416
.70x - .54x = 416 - 345.6
.16x = 70.4
x = 70.4/.16
x = 440 lb of 70% aluminum
:
640 - 440 = 200 lb of the 54% aluminum
:
:
Check:
.7(440) + .54(200) = .65(640)
308 + 108 = 416