SOLUTION: Anne and Nancy use a metal alloy that is 17.3% copper to make jewelry. How many ounces of a 17% alloy must be mixed with a 18% alloy to form 80 ounces of the desired alloy?

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Question 1161204: Anne and Nancy use a metal alloy that is 17.3% copper to make jewelry. How many ounces of a 17% alloy must be mixed with a 18% alloy to form 80 ounces of the desired alloy?
Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!


A typical traditional algebraic setup for solving the problem....

x ounces of 17%, plus (80-x) ounces of 18%, equals 80 ounces of 17.3%:

.17%28x%29%2B.18%2880-x%29+=+.173%2880%29

Solve using basic algebra -- although the arithmetic is a bit unpleasant because of the decimals.

Here is a MUCH faster and easier way of solving the problem (and any similar two-ingredient mixture problem), if a formal algebraic solution is not required.

17.3% is 3/10 of the way from 17% to 18%.
Therefore, 3/10 of the mixture should be the 18% alloy.

ANSWER:
3/10 of 80 ounces, or 24 ounces, of 18% alloy; the remaining 56 ounces of 17% alloy.

CHECK:
.17(56)+.18(24) = 9.52+4.32 = 13.84
.173(80) = 13.84