SOLUTION: A chemist wants to create 60 grams of an nickel alloy that is 65% pure. She has one mixture of an nickel alloy that is 45% pure and another mixture that is 75% pure. How much of

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A chemist wants to create 60 grams of an nickel alloy that is 65% pure. She has one mixture of an nickel alloy that is 45% pure and another mixture that is 75% pure. How much of      Log On

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Question 235976: A chemist wants to create 60 grams of an nickel alloy that is 65% pure. She has one mixture of an nickel alloy that is 45% pure and another mixture that is 75% pure. How much of each mixture must she use?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = grams of 45% mixture to be used
Let b = grams of 75% mixture to be used
given:
(1) a+%2B+b+=+60 grams
(2) .45a+%2B+.75b = pure nickel in final mixture
------------------------------
In words:
( grams of pure nickel in final mixture )/(total grams) = 65%
%28.45a+%2B+.75b%29+%2F+60+=+.65
.45a+%2B+.75b+=+39
(3)45a+%2B+75b+=+3900
Multiply both sides of (1) by 45 and subtract
(1) from (3)
(3)45a+%2B+75b+=+3900
(1)-45a+-+45b+=+-2700
30b+=+1200
b+=+40
And, since
a+%2B+b+=+60
a+%2B+40+=+60
a+=+20
20 grams of the 45% mixture and 40 grams of the 75% mixture are needed
check:
%28.45a+%2B+.75b%29+%2F+60+=+.65
%28.45%2A20+%2B+.75%2A40%29+%2F+60+=+.65
%28+9+%2B+30%29+%2F+60+=+.65
39+=+.65%2A60
39+=+39
OK