A dairy farmer wants to mix a
20%
protein supplement and a standard
10%
protein ration to make
1200
pounds of a high-grade
15%
protein ration. How many pounds of each should he use?
by inspection , 15 is the average of 20 and 10
... so, equal amounts of the two
with algebra ... .20 x + .10(1200 - x) = .15 * 1200
To solve this problem, we can use the method of setting up a system of equations:
Let x be the pounds of the 20% protein supplement, and y be the pounds of the 10% protein ration.
Given:
x + y = 1200 (Equation 1) -- Total weight of the mixture is 1200 pounds.
0.2x + 0.1y = 0.15(1200) (Equation 2) -- Total protein in the mixture is 15% of 1200 pounds.
Let's solve this system of equations step by step:
Step 1: Solve Equation 1 for x:
x = 1200 - y
Step 2: Substitute the value of x in Equation 2:
0.2(1200 - y) + 0.1y = 0.15(1200)
Step 3: Simplify the equation:
240 - 0.2y + 0.1y = 180
Step 4: Combine like terms:
-0.1y = -60
Step 5: Divide both sides of the equation by -0.1 to solve for y:
y = -60 / -0.1
y = 600
Step 6: Substitute the value of y in Equation 1 to find x:
x + 600 = 1200
x = 1200 - 600
x = 600
So, the dairy farmer should use 600 pounds of the 20% protein supplement and 600 pounds of the 10% protein ration to make 1200 pounds of a high-grade 15% protein ration.
To solve this problem, we can use a system of equations. Let's call the number of pounds of the 20% protein supplement "x" and the number of pounds of the 10% protein ration "y".
Since the farmer wants to make 1200 pounds of a 15% protein ration, we can set up the following equation:
x + y = 1200 Equation 1
Next, we can calculate the protein content in each ingredient.
The 20% protein supplement contains 20% protein, which means it contains 0.20 * x pounds of protein.
The 10% protein ration contains 10% protein, which means it contains 0.10 * y pounds of protein.
To find the protein content in the final mixture, which is 15%, we multiply the total weight (1200 pounds) by the desired protein percentage (15%):
0.15 * 1200 = 0.20 * x + 0.10 * y Equation 2
Now we have a system of equations:
x + y = 1200
0.20 * x + 0.10 * y = 0.15 * 1200
To solve this system, we can use substitution, elimination, or matrix methods. Let's solve it using the substitution method.
From Equation 1, we can isolate x:
x = 1200 - y
Now substitute this value of x into Equation 2:
0.20 * (1200 - y) + 0.10 * y = 0.15 * 1200
Multiply both sides by 100 to remove the decimals:
20 * (1200 - y) + 10 * y = 15 * 1200
Now distribute and simplify:
24000 - 20y + 10y = 18000
Combine like terms:
-10y = -6000
Divide by -10:
y = 600
So, the farmer should use 600 pounds of the 10% protein ration.
Now substitute this value of y back into Equation 1 to find x:
x + 600 = 1200
x = 600
Therefore, the farmer should use 600 pounds of the 20% protein supplement.
To summarize, the dairy farmer should use 600 pounds of the 10% protein ration and 600 pounds of the 20% protein supplement to make 1200 pounds of a high-grade 15% protein ration.