Single Variable Algebraic Mixture Problem
Date: 05/03/2005 at 21:31:21 From: william Subject: Could you please help? 7th grade teacher here :) Brittany needs a mixture of 76% oats and 24% corn. She currently has 25 pounds of a mixture that is 28% oats. How many pounds of oats should be added to the mixture? How do I identify the x from the y? In most mixture problems you have a % or $ or some coefficent. I think you would need to have 2 equations with 2 variables, like this: x + y = ? Oats x + Corn y = 100% I am sure there is something simple I am missing. I am a 7th grade teacher and asked the students to bring in a problem from the internet to solve. I explained to them I could not solve this but I told them I would try to get help and share with them. I know the answer is 50 pounds. Could you please help? - William
Date: 05/03/2005 at 22:52:26 From: Doctor Peterson Subject: Re: Could you please help? 7th grade teacher here :) Hi, William. First I'll visualize the story and make sure I understand the details. We've got a bag containing 25 pounds of a mixture that is 28% oats, and presumably 72% corn. I could work out just how much of each grain we have; I'm assuming that "percent" here means "by weight", so that 28% of the 25 pounds is oats. We have another bag containing pure oats, and we want to add that in to the mixture to increase the percentage of oats to 76% (with the remaining 24% being corn). Now, the next step is to choose a variable or two. What don't we know? We need to figure out how much (by weight) of the oats to add; I don't see any other obvious unknown, so that one variable may be all we need: Let x = amount of oats to be added, in pounds (I like to define variables precisely, to make sure I always use them the same way, and don't drift into using a different unit, for example.) Now, if I knew how much oats I was adding, how would I figure out the percentage of oats in the final mixture? Well, I'd find the total amount of oats and the total amount of the mixture (including the corn), and then divide. So let's write expressions for each step of that process: Amount of original mixture: 25 lb Amount of oats in original mixture: 0.28*25 = 7 lb Amount of oats in final mixture: 7+x lb (since x lb of oats are added) Total amount of final mixture: 25+x lb (since x lbs of stuff are added in all) (I like to say that the basic idea of algebra is to give a name to anything you don't know, pretend you do know it, and then go through whatever calculations the problem requires, using the variables and writing expressions rather than getting actual numbers. When you're done, you'll have an equation or two.) So the percentage of oats in the new mixture will be (7+x)/(25+x) as a fraction. The goal is for that fraction to be 76%, or 0.76. So we can write an equation (7+x) ------ = 0.76 (25+x) or 7+x = 0.76(25+x) Now I just have to solve this. I'll simplify the right side, and then isolate the variable: 7 + x = 19 + 0.76x 7 + x - 0.76x = 19 7 + 0.24x = 19 0.24x = 19 - 7 0.24x = 12 x = 12/0.24 = 50 And we're done, except for checking to see that the answer actually works in the original problem. If I add 50 lb of oats, I'll have a 75 lb bag containing 57 lb of oats, and the percentage is in fact 0.76. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 05/03/2005 at 23:01:07 From: william Subject: Thank you (Could you please help? 7th grade teacher here :)) Doctor Peterson, Thank you so very much. The children will be thrilled that you answered our problem! - William
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