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

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) (Show Source): You can put this solution on YOUR website!
Amount of 70% Al:
Amount of 54% Al:
-----------
The problem tells us to mix the two.
The total amount of pure aluminum in the mix will be: (Eq. 1)
The total amount of material in the mix ,must equal 640 pounds ==> (Eq. 2)
We are looking for a final mix of 640 pounds of 65% Al. Thus, the total aluminum content of the final is:

.
Plugging that back into Eq. 1 gives:

Solving Eq. 2 for ==>
Plug that into the previous equation:

pounds.
Plug this into Eq. 2: ==> pounds.

Answer by ankor@dixie-net.com(22740) (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

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