When purchasing ground beef at the store, you might see different percentages. Typically, the percentage displayed is the percent of "lean" meat. Different parts of meat from a cow have different percentages of "lean" meat and the rest of the meat is fat. A store employee has to mix the right amount of meat and the right amount of pieces of pure fat in order to create the percentage mixture of ground beef required.
A local butcher needs to create ground beef for the grocery store. The butcher needs to make 80% lean, 20% fat ground beef. Their supplier gives them meat that is 93% meat, 7% fat. They also have pieces of 100% fat on hand, too. How much meat and how much fat should be added to the grinder to achieve this mixture?
Let m be the number of pounds of the 93% lean meat that is used in the mixture. Let f be the number of pounds of 100% fat that is used in the mixture.
There are two variables, m and f. The butcher needs to construct two equations in order to solve for the variables.
The butcher is going to create a 20 pound mixture. Create an equation using m and f to represent the number of pounds needed from each variable in order to equal the 20 pound mixture.
\[\text{number of pounds of meat} + \text{number of pounds of fat} = 20 \text{ pounds}\]
Earlier, we said \(m\) was pounds of the 93% meat and \(f\) the number of pounds of fat. Replace "number of pounds of meat" with \(m\) and "number of pounds of fat" with \(f\).
\[\text{number of pounds of meat} + \text{number of pounds of fat} = 20 \text{ pounds}\]
\[m + f = 20 \text{ pounds}\]
There are two variables, \(m\) and \(f\), and one equation. We need two equations to solve when there are two variables.
We can make a second equation about the percentages in the problem.
\[0.07m + 1f = 0.2 \cdot 20\]
We multiply the amount of meat by \(0.07\), the decimal form of the percentage \(7\%\), because \(7\%\) of the lean meat is fat. We multiply \(f\) by \(1\) because \(1 = 100\%\). We multiply the final amount of pounds by \(0.2\) because we want \(20\%\) of our final product to be lean meat. We could multiply \(0.2\) and \(20\) to make the equation simpler.
\[0.07m + f = 4\]
Put the two equations together.
\[m + f = 20\] \[0.93m + f = 16\]
When we have two equations about the same variables, they are called a system of equations. We can add and subtract the two equations to make one of the variables disappear. We can subtract the top equation by the bottom equation to make the \(f\) variable disappear.
\[m + f = 20\] \[0.07m + f = 4\] \[--------------\] \[0.93m + 0 = 16\]
We write \(0.93m\) because we had one \(m\) in the first equation and we subtracted it by \(0.07m\) in the second equation. We write \(0\) because \(f - f = 0\). We write \(16\) because \(20 - 4 = 16\). Now we can solve for \(m\)!
\[0.93m + 0 = 16\]
\[0.93m = 16\]
Divide both sides by \(0.93m\).
\[0.93m = 16\]
\[\frac{0.93m}{0.93} = \frac{16}{0.93}\]
\[m \approx 17.2\]
We should use approximately \(17.2\) pounds of the meat. That means we should use \(20 - 17.2 = 2.8\) pounds of fat.
1. A local butcher needs to create ground beef for the grocery store. The butcher needs to make 85% lean, 15% fat ground beef. Their supplier gives them meat that is 93% meat, 7% fat. They also have pieces of 100% fat on hand, too. How much meat and how much fat should be used in the grinder to achieve the desired mixture for 20 pounds of meat?
2. A local butcher needs to create ground beef for the grocery store. The butcher needs to make 90% lean, 10% fat ground beef. Their supplier gives them meat that is 95% meat, 5% fat. They also have pieces of 100% fat on hand, too. How much meat and how much fat should be used in the grinder to achieve the desired mixture for 20 pounds of meat?
1. Is it possible to make ground beef that has no fat? Why or why not?
2. If a butcher only had pieces of meat that were 90% meat and 10% fat, and they wanted a 90% meat 10% mixture for ground beef, could they achieve this ground beef mixture?
3. If a butcher only had pieces of meat that were 90% meat and 10% fat, and they wanted a 95% meat 5% mixture for ground beef, could they achieve this ground beef mixture?