1. ABC Feeds has two different types of cattle feed containing crude protein in the amount of 45% and 20% respectively. How many of each type must a feed lot manager mix together to obtain 10 000lbs of a mixture that is 25% protein?
2. Meg has a solution that is 40% alcohol and another solution that is 60% alcohol. If Meg wants 12gal of a solution that is 45% alcohol, how much of the two solutions must he mixed?
10,000*.25=(10,000-x).45 + x*.20
where x is the amount of twenty percent.
2.
12*.45=x*.40+(12-x).60
x is the amount of forty percent
To solve these types of mixture problems, you can use the concept of the weighted average.
1. Let's solve the first problem:
Let's assume the feed lot manager needs to mix 'x' pounds of the cattle feed with 45% protein and 'y' pounds of the cattle feed with 20% protein to obtain 10,000 pounds of a mixture with 25% protein.
To find the solution, we need to equate the total amount of protein in the mixture to the sum of the amount of protein in each type of cattle feed.
So, the equation becomes:
(x * 45% + y * 20%) = 10,000 * 25%
Now let's convert the percentages into decimals:
(0.45x + 0.2y) = 0.25 * 10,000
Simplifying the equation further:
0.45x + 0.2y = 2500
Since we are looking for integer solutions, we will further constrain the problem to find whole number solutions for x and y.
Now we have two equations:
0.45x + 0.2y = 2500 (equation 1)
x + y = 10,000 (equation 2)
To solve these equations, you can use various methods such as substitution, elimination, or matrix methods.
2. Now let's solve the second problem:
Similarly, let's assume Meg needs to mix 'x' gallons of the 40% alcohol solution and 'y' gallons of the 60% alcohol solution to obtain 12 gallons of a mixture with 45% alcohol.
The equation becomes:
(0.40x + 0.60y) = 0.45 * 12
Simplifying the equation further:
0.40x + 0.60y = 5.4
Again, we will further constrain the problem to find whole number solutions for x and y.
So we have two equations:
0.40x + 0.60y = 5.4 (equation 1)
x + y = 12 (equation 2)
You can use the same methods mentioned earlier (substitution, elimination, or matrix) to solve these equations and find the values of x and y.