Question 1147512
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A setup for a traditional algebraic solution....<br>
300 pounds at $1.90 per pound, plus x pounds at $0.80 per pound, equals (300+x) pounds at $1.40 per pound:<br>
{{{300(1.90)+x(0.80) = (300+x)(1.40)}}}<br>
Basic algebra; but the decimals will slow down the computations....<br>
Here is a much faster and easier way to solve any mixture problem like this, without algebra.<br>
(1) You are starting with grain that costs $1.90 per pound; you are adding meal that costs $0.80 per pound; you are stopping when the mixture is worth $1.40 per pound.<br>
(2) Picture the three rates on a number line: 1.90, 1.40, and 0.80.  You are starting at the 1.90 and heading towards the 0.80, and you stop when you reach 1.40.  The 1.40 is 5/11 of the way from 1.90 to 0.80 (1.90 to 0.80 is 1.10; 1.90 to 1.40 is 0.50; 0.50/1.10 = 5/11)<br>
(3) That means 5/11 of the mixture must be what you are adding.  So the 300 pounds you started with is 6/11 of the mixture; that means 5/11 of the mixture is 250 pounds.<br>
ANSWER: 250 pounds of the meal need to be added to the 300 pounds of grain.