Question 954102: A given alloy contains 20% copper and 5% tin. How mmany pounds of copper and of tin must be melted with 100lb. of the given alloy to produce another alloy analyzing 30% copper and 10% tin?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A given alloy contains 20% copper and 5% tin.
How many pounds of copper and of tin must be melted with 100lb of the given alloy to produce another alloy analyzing 30% copper and 10% tin?
:
let c = amt of copper required
let t = amt of tin
then
(100+c+t) = total weight of the resulting alloy
:
Find how many pounds in the original 100 lb
.20(100) = 20 lb of copper
.05(100) = 5 lb of tin
:
Write an equation for each, arrange for elimination
Copper
c + 20 = .30(100 + c + t)
c + 20 = 30 + .3c + .3t
c - .3c = 30 - 20 + .3t
.7c - .3t = 10
Tin
t + 5 = .10(100 + c + t)
t + 5 = 10 + .1c + .1t
t - .1t = 10 - 5 + .1c
.9t = 5 + .1c
-.1c + .9t = 5
:
Multiply the 1st equation by 3, add to the above equation
-.1c + .9t = 5
2.1c + .9t = 30
-----------------Adding eliminates t, find c
2c = 35
c = 35/2
c = 17.5 lb of copper to be added
:
find t using -.1c + .9t = 5
-.1(17.5) + .9t = 5
-1.75 + .9t = 5
.9t = 5 + 1.75
.9t = 6.75
t = 6.75/.9
t = 7.5 lb of tin required
:
:
See of this checks out
Weight of resulting alloy: 100 + 17.5 + 7.5 = 125 lb
Copper: .30(125) = 37.5, (20 oriiginal lbs + 17.5 lbs added)
Tin: .10(125) = 12.5, (5 original lbs + 7.5 lbs added
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