By DAN MARTENS
Regional Extension Educator
There are usually a couple different approaches to solving the same arithmetic problem. Here are two examples related to corn silage moisture, weight and value. Parents, aunts or uncles might be interested in seeing what math student children can do with this.
1. I’m buying corn silage from a neighbor at a price per ton that is based on corn silage that is 65 percent moisture. We are able to run each load across a scale and to test the moisture. If the moisture tests 70 percent, how many pounds of this corn silage would be the same as one ton of 65 percent corn silage?
First let’s think about what the answer should be like—bigger or smaller. Corn silage at 70 percent moisture will have more water in it than corn silage at 65 percent moisture. If there is more water, it will weigh more and we should get a bigger number—more than 2000 pounds.
For moisture conversion problems, it usually works best to do the arithmetic with dry matter (DM) in mind. First we can ask, how many pound of dry matter (DM) are in 2000 pounds of corn silage at 65 percent moisture?
100 percent minus 65 percent moisture equals 35 percent DM.
Then 35 percent DM X 2000 pounds equals 700 pounds of dry matter.
Second we can ask, if the corn silage is 70 percent moisture, how many pounds does it take to have the same 700 pounds of DM?
100 percent minus 70 percent moisture equals 30 percent DM. Because we are looking for the same pounds of dry matter, it is better to ask, How many pounds of corn silage at 30 percent DM does it take to have the same 700 pounds of DM?
30 percent X?? pounds of corn silage needs to equal 700 pounds DM.
So 700 pounds “divided by” 30 percent equals 2333 pounds.
So 2333 pounds of corn silage at 70 percent moisture would equivalent to 2000 pounds of corn silage at 65 percent moisture—both having 700 pounds DM.
There is a shortcut formula that goes like this:
percent DM in base product “divided by” percent DM in the unknown product “times” the weight of the base product “equals” the weight of the unknown product. 2000 pounds of corn silage at 35 percent DM (65 percent moisture) in this example is the starting base product. I call the 30 percent DM (70 percent moisture) corn silage the unknown product because that’s the weight we are trying to get at. So:
35 percent “divided by” 30 percent X 2000 pounds = 2333 pounds.
It helps to remember that 35/30 X 2000 will give me a bigger number; while 30/35 X 2000 will give me a smaller number. The “/” mark means “divided by.”
The same formula works for converting from a drier product to a wetter product.
2. Sometimes people have a price set for a ton of corn silage at 65 percent moisture (35 percent DM) and they want to know what the price would be for a ton of corn silage at 70 percent (30 percent DM). It is helpful to know that you would pay less for a ton of silage that is wetter than your standard—assuming it is not too wet to make good silage. You would pay more for a ton of silage that is drier than your standard— assuming it is not too dry to make good silage. Here’s the short cut formula:
percent DM unknown silage” divided by percent DM of standard silage X the price of the standard silage “equals” the price of the unknown silage. Let’s use a price of $15/T for corn silage at 35 percent DM. Then:
30 “divided by” 35 X $15 = $12.86 for 1 ton of 70 percent corn silage.
Buyers and sellers need to discuss whether there are other factors that play into the price.
So in problem 1, we are using the same price and converting wet tons in the equivalent number of drier tons. In problem 2, we are using the same weight of 2000 pounds and adjusting the price downward in adjusting for the higher moisture content. These are two different methods to get at the same end value for the harvested crop.
Beyond the arithmetic, do what you can to make sure this adds up to be a SAFE harvest season.

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