Question 1161204
<br>
A typical traditional algebraic setup for solving the problem....<br>
x ounces of 17%, plus (80-x) ounces of 18%, equals 80 ounces of 17.3%:<br>
{{{.17(x)+.18(80-x) = .173(80)}}}<br>
Solve using basic algebra -- although the arithmetic is a bit unpleasant because of the decimals.<br>
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.<br>
17.3% is 3/10 of the way from 17% to 18%.
Therefore, 3/10 of the mixture should be the 18% alloy.<br>
ANSWER:
3/10 of 80 ounces, or 24 ounces, of 18% alloy; the remaining 56 ounces of 17% alloy.<br>
CHECK:
.17(56)+.18(24) = 9.52+4.32 = 13.84
.173(80) = 13.84<br>